Analyzing Monotone Space Complexity Via the
ARCHIVES
Switching Network Model
MASSACHUrTS INSTITUTE
OF TECH' OLOLGY
by
JUN 3 0 2015
Aaron H. Potechin
Submitted to the Department of Mathematics
in partial fulfillment of the requirements for the degree of
LIBRARIES
Doctor of Philosophy in Mathematics
at the
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
June 2015
Massachusetts Institute of Technology 2015. All rights reserved.
A uthor ........
Signature redacted
.................
Department of Mathematics
May 8, 2015
Certified by..........
Signature redactedJonathan A. Kelner
Associate Professor
Thesis Supervisor
Accepted by ...................
Signature redacted
Wichel X. Goemans
Chairman, Department Committee on Graduate Theses
2
Analyzing Monotone Space Complexity Via the Switching
Network Model
by
Aaron H. Potechin
Submitted to the Department of Mathematics
on May 8, 2015, in partial fulfillment of the
requirements for the degree of
Doctor of Philosophy in Mathematics
Abstract
Space complexity is the study of how much space/memory it takes to solve problems. Unfortunately, proving general lower bounds on space complexity is notoriously hard. Thus, we instead consider the restricted case of monotone algorithms,
which only make deductions based on what is in the input and not what is missing. In this thesis, we develop techniques for analyzing monotone space complexity
via a model called the monotone switching network model. Using these techniques,
we prove tight bounds on the minimal size of monotone switching networks solving
the directed connectivity, generation, and k-clique problems. These results separate
monotone analgoues of L and NL and provide an alternative proof of the separation
of the monotone NC hierarchy first proved by Raz and McKenzie. We then further
develop these techniques for the directed connectivity problem in order to analyze
the monotone space complexity of solving directed connectivity on particular input
graphs.
Thesis Supervisor: Jonathan A. Kelner
Title: Associate Professor
3
4
Acknowledgments
This thesis is based on work supported by a National Science Foundation Graduate
Research Fellowship under grant No. 0645960.
The author would like to thank his advisors Jonathan Kelner and Boaz Barak and
his collaborator Siu Man Chan for their guidance on this project. The author would
also like to thank his family and friends for their support during his PhD studies.
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6
Contents
1
2
Introduction
13
1.1
Known Results on Space Complexity . . . . . . . . . . . . . . . . . .
13
1.2
Approaches to Space Lower Bounds . . . . . . . . . . . . . . . . . . .
14
1.3
Known Results on Monotone Complexity . . . . . . . . . . . . . . . .
15
1.4
Overview of Our Results . . . . . . . . . . . . . . . . . . . . . . . . .
17
Background Material
19
2.1
Switching Networks: From Practice to Theory . . . . . . . . . . . . .
20
2.2
Turing machines and space complexity
. . . . . . . . . . . . . . . . .
21
2.3
Switching-and-rectifier networks and nondeterministic space complexity 23
2.4
Branching programs, switching networks, and deterministic space complexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
3 Bounds on monotone switching networks for directed connectivity
29
3.1
D efinitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30
3.2
Certain Knowledge Switching Networks . . . . . . . . . . . . . . . . .
34
3.2.1
The certain knowledge game for directed connectivity . . . . .
35
3.2.2
Adapting the certain knowledge game for monotone switching
netw orks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36
3.2.3
A partial order on knowledge sets . . . . . . . . . . . . . . . .
37
3.2.4
Certain knowledge switching networks and reachability from s
39
3.2.5
An upper bound on certain-knowledge switching networks
41
3.2.6
A lower size bound on certain-knowledge switching networks
7
.
44
3.4
Elementary results on monotone switching networks . . . . . . . . .
48
3.3.1
The knowledge game for directed connectivity . . . . . . . .
49
3.3.2
A partial order on states of knowledge
50
3.3.3
Knowledge description of monotone switching networks
3.3.4
Reduction to reachability from s . . . . . . . . . . . . . . . .
55
3.3.5
Reduction to certain knowledge switching networks . . . . .
57
.
47
.
.
.
.
.
60
3.4.1
Function descriptions of sound monotone switching networks
.
61
3.4.2
Fourier analysis . . . . . . . . . . . . . . . . . . . . . . . . .
64
3.4.3
Invariants and a quadratic lower bound . . . . . . . . . . . .
65
3.4.4
General lower bounds . . . . . . . . . . . . . . . .
68
3.4.5
Conditions for a good set of functions . . . . . . .
72
A Superpolynomial Lower Bound . . . . . . . . . . . . .
75
.
.
.
.
.
. . . . . . . . . . . . . . . . . . . . . . . . .
3.5.1
From certain knowledge decriptions and knowledge d escriptions
. . . . . .
76
3.5.2
A criterion for E-invariance
. . . . . . . . . . . .
. . . . . .
78
3.5.3
Choosing Fourier coefficients via dot products . .
. . . . . .
81
3.5.4
Proof of Theorem 3.5.1 . . . . . . . . . . . . . . .
. . . . . .
82
.
.
.
.
to function descriptions . . . . . . . . . . . . . . .
An 001g) lower size bound
. . . . . . . . . . . . . . . .
. . . . . .
84
3.7
Monotone analogues of L and NL . . . . . . . . . . . . .
. . . . . .
89
3.8
Finding paths with non-monotone switching networks . .
. . . . . .
90
.
.
.
3.6
Lower bounds on monotone switching networks for gen eration and
95
4.1
Circuit Depth and Switching Network Size . . . . . . . .
95
4.2
Our Results and Comparison with Previous Work . . . .
97
4.3
Lower bounds on monotone switching networks for Generation
99
4.4
Lower bounds on monotone switching networks for k-clique . .
.
k-clique
.
.
.
4
54
Fourier Analysis and Invariants on Monotone Switiching Networks For
Directed Connectivity
3.5
. . . . . . . . . . . .
.
3.3
Simplified bounds on certain-knowledge switching networks .
.
3.2.7
8
104
5
Upper Bounds: Dealing with Lollipops via Parity
5.1
Motivation for considering particular input graphs for directed connectiv ity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
109
5.2
Lollipops and certain knowledge switching networks . . . . . . . . . . 110
5.3
A general upper bound . . . . . . . . . . . . . . . . . . . . . . . . . .111
Improved Lower Bound Techniques for Monotone Switching Net123
works for Directed Connectivity
7
109
6.1
Permutations on input graphs, cuts, functions, and Fourier coefficients 124
6.2
Calculating Permutation Averages . . . . . . . . . . . . . . . . . . . .
125
6.3
Lower bounds from permutation averages . . . . . . . . . . . . . . . .
132
6.4
Equations on sum vectors
6.5
Freedom in choosing a base function
6.6
Building up a base function
. . . . . . . . . . . . . . . . . . . . . . . . 134
. . . . . . . . . . . . . . . . . .
139
. . . . . . . . . . . . . . . . . . . . . . .
145
6.7
Constructing a base function . . . . . . . . . . . . . . . . . . . . . . .
155
6.8
The cost of extending a base function . . . . . . . . . . . . . . . . . . 159
6.9
Putting everything together: A lower bound . . . . . . . . . . . . . .
165
167
Conclusion and Future Work
169
A Proof of Lemma 3.2.32
9
10
List of Figures
3-1
A monotone swiitching network solving directed connectivity . . . . .
31
3-2
A sound but not complete switching network . . . . . . . . . . . . . .
32
3-3
A complete but not sound switching network . . . . . . . . . . . . . .
33
3-4
A non-monotone switching network solving directed connectivity . . .
34
3-5
A certain knowledge switching network G' solving directed connectivity
on V(G)
3-6
=
{s, t, a, b} together with a certain knowledge description of it 38
A certain knowledge switching network G' solving directed connectivity
on V(G) ={s, t, a, b, c} together with a certain knowledge description
ofit .........
3-7
=
{s, t, a, b, c} together with a certain knowledge description
ofit .........
....................................
55
An illustation of the partial reduction of monotone switching networks
to certain knowledge switching networks
3-9
40
A monotone switching network G' solving directed connectivity on
V(G)
3-8
....................................
. . . . . . . . . . . . . . . .
60
Representing cuts of V(G) . . . . . . . . . . . . . . . . . . . . . . . .
63
3-10 A certain knowledge switching network and the reachability function
description for it. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
64
3-11 A monotone switching network together with the reachability function
description for it and the Fourier decomposition of each function . . .
65
3-12 An illustration of the linear independence argument . . . . . . . . . .
68
3-13 An illustartions of the ideas used to prove Lemma 3.5.12 . . . . . . .
80
11
12
Chapter 1
Introduction
The central topic of this thesis is space complexity, i.e. how much space/memory
it takes to solve various problems. We are particularly interested in the space complexity of the directed connectivity problem, which asks if there is a path from from
some vertex s to another vertex t in a directed input graph. Determining the space
complexity of the directed connectivity problem would resolve the fundamental open
question of whether the complexity class L (Logarithmic Space) is equal to the complexity class NL (nondeterministic logarithmic space). Unfortunately, proving general
lower bounds on space complexity is notoriously difficult and is currently beyond our
reach. Nevertheless, if we place restrictions on what our algorithms can do we can obtain bounds on the space complexity of these algorithms, which gives insight into the
general problem. In this thesis, we use a model called the switching network model to
analyze the space complexity of monotone algorithms, which make deductions based
on what is present in the input but not what is absent from the input.
1.1
Known Results on Space Complexity
Although no non-trivial lower space bounds are known for directed connectivity and
most other problems, we do have considerable knowledge about deterministic and
nondeterministic space complexity. It is folklore that the directed connectivity problem is NL-complete. This is the basis for the statement above that to resolve the
13
L versus NL problem, which asks whether non-determinism helps in space-bounded
computation, it is sufficient to understand the space complexity of the directed connectivity problem. In 1970, Savitch [23] gave an upper bound with an O(log 2 n)-space
deterministic algorithm for directed connectivity. This proved that any problem that
can be solved nondeterministically using g(n) space for some function g(n) can be
solved deterministically in O(g(n)2 ) space. In 1988, Immerman [11] and Szelepcsenyi
[26] independently gave an O(log n)-space nondeterministic algorithm for directed
non-connectivity. This proved the surprising result that for any problem which can
be solved nondeterministically in g(n) space, it's complement can also be solved nondeterministically in g(n) space. In particular, NL=co-NL.
For the problem of undirected connectivity (i.e. where the input graph G is undirected), a probabilistic logarithmic space algorithm was shown in 1979 using random
walks by Aleliunas, Karp, Lipton, Lovisz, and Rackoff [1]. In 2004, Reingold [22] gave
a deterministic O(log n)-space algorithm for undirected connectivity, showing that
undirected connectivity is in L. Trifonov [27] independently gave an 0 (lg n lg lg n) algorithm for undirected connectivity. As discussed in Chapter 2, Reingold's result was
an inspiration for our choice of the switching network model (which is an undirected
graph whose edges depend on the input) for analyzing space complexity.
1.2
Approaches to Space Lower Bounds
For the problem which we are considering, proving space lower bounds, there have
been two general approaches so far. One approach is to use the branching program
model, which is an intuitive way to represent the different branches an algorithm
can take depending on the input. Bounds are then proved on restricted branching
programs/algorithms.
For example, one common restriction is that the branching
program/algorithm must look at the bits of the input in order and may only pass
through the input k times for some constant k. For a survey of some of the results
on branching programs, see Razborov [20].
Another approach which is specialized to analyzing the directed connectivity prob14
lem is to use the JAG (Jumping Automata on graphs) model or variations thereof.
The JAG (Jumping Automata on Graphs) model was introduced by Cook and Rackoff [7] as a simple model for which we can prove good lower time and space bounds but
which is still powerful enough to simulate most known algorithms for the directed connectivity problem. This implies that if there is an algorithm for directed connectivity
breaking these bounds, it must use some new techniques which cannot be captured by
the JAG model. Later work in this area has focused on extending this framework to
additional algorithms and more powerful variants of the JAG model. Two relatively
recent results in this area are the result of Edmonds, Poon, and Achlioptas [8] which
shows tight lower bounds for the more powerful NNJAG (Node-Named Jumping Automata on Graphs) model and the result of Lu, Zhang, Poon, and Cai [16] showing
that Reingold's algorithm for undirected connectivity and a few other algorithms for
undirected connectivity can all be simulated by the RAM-NNJAG model, a uniform
variant of the NNJAG model.
In this thesis, we take a third approach, using the switching network model to
analyze space complexity. There are two main reasons for this choice. First, unlike
the branching program model and the JAG model we may restrict switching networks
to be monotone, only using what is present in the input and not what is absent from
the input. This means that the switching network model may be used to analyze
monotone space complexity. Second, switching networks have the important property
that every step which is made on them is reversible. This property plays a crucial
role in our lower bound techniques.
1.3
Known Results on Monotone Complexity
Since we are analyzing monotone algorithms and switching networks, our work is
part of monotone complexity theory. We now describe some of the known results in
monotone complexity theory and how our work fits in.
Looking at time complexity rather then space complexity, in 1985 Razborov [21]
used an approximation method to show that any monotone circuit solving the k-clique
15
problem on n vertices (determining whether or not there is a set of k pairwise adjacent
vertices in a graph on n vertices) when k
proving that mP
#
=
[19--] must have size at least N'), thus
mNP. This method was later improved by Alon and Boppana
[2] and by Haken [10].
Looking at monotone circuit depth instead of monotone circuit size, Karchmer and
Wigderson [12] used a communication complexity approach to show that any monotone circuit solving undirected connectivity has depth at least Q((lg n) 2 ). Recalling
that monotone-NC is the class of problems which can be solved by a monotone circuit
of depth O((lg n)'), this proved that undirected connectivity is not in monotone- NC
and separated monotone-NC and monotone-NC 2 . Raz and McKenzie [19] later sep-
#
arated the entire monotone NC hierarchy, proving that for any i, monotone-NC'
monotone-NCi+l.
Our results are closely related to the results of Karchmer, Wigderson, Raz, and
McKenzie, as there is a close connection between monotone circuit depth and the
size of monotone switching networks. As discussed in Chapter 4, any problem which
can be solved by a monotone circuit of depth O(g(n)) can be solved by a monotone
switching network of size 2 0(g(n)), so our lower size bounds on monotone switching
networks prove corresponding lower bounds on monotone circuit depth. Conversely,
any problem which can be solved by a monotone switching network of size
be solved by a monotone circuit of depth
Q(g(n)
2
2 0(g(n))
can
), so the previous lower bounds on
monotone circuit depth imply corresponding lower bounds on the size of monotone
switching networks.
That said, while our results are closely related to previous work on monotone
circuit depth, our approach is very different and thus gives further insight into the
problem. Also, as discussed in Chapter 4, comparing our bounds with previous work,
our size lower bounds on monotone switching networks are stronger and our corresponding depth lower bounds on monotone circuits are equally strong. Thus, our
work strengthens the previous results of Karchmer, Wigderson, Raz, and McKenzie.
16
1.4
Overview of Our Results
We now give an overview of the remainder of this thesis. In Chapter 2 we present
background material on switching networks and their relationship to space complexity.
In chapter 3 we prove an no(lgn) lower size bound on monotone switching networks
solving the directed connectivity problem, separating monotone analogues of L and
NL. In chapter 4 we apply the techniques developed in chapter 3 to the generation
problem and the k-clqiue problem. For the generation problem, we show that any
monotone switching network solving the generation problem on pyramid graphs of
height h < n' where e <
must have size
nQ(h).
This gives an alternative proof of the
separation of the monotone-NC hierarchy. For the k-clique problem, we show that any
monotone switching network solving the k-clique problem for k < nE where 6 <
must
have size n.(k). In chapters 5 and 6 we return to the directed connectivity problem,
focusing on the difficulty of using monotone switching networks to solve directed
connectivity on particular input graphs.
In chapter 5 we prove an upper bound,
showing that monotone switching networks can easily solve directed connectivity on
input graphs in which the vast majority of vertices are either easily reachable from
s or can easily reach t. Finally, in chapter 6 we further develop the lower bound
techniques in chapter 3 to show that monotone switcing networks are no more efficient
than Savitch's algorithm for input graphs where the vertices are not too connected
to each other.
17
18
Chapter 2
Background Material
In this chapter, we present background material on switching networks and their connection to space complexity. Along the way, we present background material on two
closely related models, switching-and-rectifier networks and branching programs, and
their connection to nondeterministic space complexity and space complexity respectively. The definitions in this chapter for switching networks, switching-and-rectifier
networks, and branching programs are slightly modified from Razborov [201.
Intuitively, a switching network is an undirected graph with distinguished vertices
s' and t' where each edge is switched on or off depending on one bit of the input.
Given an input, a switching network outputs 1 if there is a path from s' to t' for that
input and 0 otherwise. The precise definition is as follows.
Definition 2.0.1. A switching network is a tuple ZG', s', t', P') where G' is an undirected multi-graph with distinguished vertices s', t' and /t' is a labelingfunction that associates with each edge e' C E(G') a label p(e') of the form xi = 1 or xi = 0 for some i
between 1 and n. We say that this network computes the function f : {0, 1} -+ {0, 1},
where f(x)
=
1 if and only if there is a path from s' to t' such that each edge of this
path has a label that is consistent with x.
We take the size of a switching network to be
{0, 1}
-+
|V(G')|,
and for a function f
{0, 1}, we define s(f )(n) to be size of the smallest switching network com-
puting f.
19
Definition 2.0.2. We say that a switching network is monotone if it has no edge
labels of form xi = 0.
The main result of this chapter is the following theorem, which says that any
program using g(n) space can be simulated by a switching network of size
2 0(g(n)).
This means that if we can show a lower size bound on switching networks solving a
given problem, this gives us a corresponding space lower bound for that problem.
Theorem 2.0.3. If f E DSPACE(g(n)) where g(n) is at least logarithmic in n,
then s(f)(n) is at most 2 0(g(n))
In the remainder of this chapter, we first give a brief history of the switching
network model. We then present proofs of this theorem and related results.
2.1
Switching Networks: From Practice to Theory
Although the switching network model has fallen into relative obscurity, it has a rich
history. Switching networks were widely used in practice in the early 20th century.
In the words of Shannon [24],
"In the control and protective circuits of complex electrical systems it is frequently
necessary to make intricate interconnections of relay contacts and switches. Examples of these circuits occur in automatic telephone exchanges, industrial motor control
equipment and in almost any circuits designed to perform complex operations automatically."
Since switching networks were widely used in practice, the early research on switching networks was devoted to finding ways to construct efficient switching networks.
A key observation of Shannon [24], [25] is that if we can write a function f as a
short boolean algebra equation such as
f=
(x 1 V x 2 ) A (-,(x 1 A
x3 )
V X 3 ) then it can
be computed with a corresponding series-parallel switching network. An alternative
technique for constructing switching networks was later found by Lee [13]. Lee [13]
observed that we can construct switching networks from branching programs.
20
However, with the development of the boolean circuit model, which is vastly more
powerful, switching networks became obsolete. As switching networks became obsolete, branching programs were no longer needed for constructing switching networks.
Nevertheless, branching programs found a theoretical use when Masek [17] discovered
that branching programs could be used to analyze space complexity.
Just as the branching program model is now used theoretically, the switching network model can also be used theoretically. As noted later in this chapter, combining
Masek's observation that branching programs can be used to analyze space complexity with Lee's construction of switching networks via branching programs shows that
switching networks can also be used to analyze space complexity. As noted in Chapter
4, Shannon's series-parallel construction of switching networks allows us to analyze
circuit depth with switching networks as well.
2.2
Turing machines and space complexity
Since we are analyzing space complexity, we use Turing machines which have two
tapes, a read-only input tape and a work tape. A formal definition is given below.
Definition 2.2.1. A deterministic Turing machine T = (Q, F, {, , -1}, }, , qo F)
is a 7-tuple where
1.
Q is the
set of states.
2. qo E Q is the initial state.
3. F C
4.
Q is
the set of final states.
F is the set of alphabet symbols.
5. 0 E F is the zero symbol, HE F marks a left boundary, and -iE F marks a right
boundary.
6. E C F \ {H, -1} is the set of input symbols.
21
7. 6 : (Q \ F) x F2
-
Qx
F x { L, S, R} 2 is the transitionfunction which takes the
Turing machine's current state and the symbols currently under the input tape
head and the work tape head and determines which state the Turing macine goes
to next, what symbol is written on the work tape, and whether the input tape
head and the work tape head shift left, stay stationary, or shift right.
Additionally, we assume the following:
1. A Turing machine head cannot shift left when it is on a left boundary and cannot
shift right when it is on a right boundary. Left and right boundaries cannot be
overwritten with anything else.
2. The input tape initially has H at position 0, the input symbols at positions 1
through n, and H at position n + 1. The work tape initially has F- at position 0
and 0 everywhere to the right of it.
3. Both the input tape head and the work tape head start at position 0.
nondeterministic Turing machines are defined in the same way as deterministic Turing machines except that 6(q, 71, '72) is now a subset of
a single element of
Q
Q x F x { L, S, R} 2
rather than
x F x {L, S, R} 2 . This represents the fact that at any given
point there can be several possibilitiesfor what the Turing machine does next.
Definition 2.2.2. We say that a deterministic Turing machine T computes the function
f
: {0, 1}
-+ {0, 1} in space g(n) if
1. E = {0, 1} and F = {ACCEPT, REJECT}.
2. For any input, T halts.
3. Whenever f(x) = 1, the Turing machine ends at ACCEPT on input x.
4.
Whenever f(x) = 0, the Turing machine ends at REJECT on input x.
5. For any input x E {0, 1}, the work tape never moves to the right of position
g(n).
22
-
We say that a nondeterministic Turing machine T computes the function f: {0, 1}"
{0, 1} in space g(n) if
1. E
{O, 1} and F = {ACCEPT, REJECT}.
2. Whenever f(x) = 1, the Turing machine has a nonzero probability of ending at
ACCEPT on input x.
3. For any input x
c {0,
1}", the work tape never moves to the right of position
g(n).
2.3
Switching-and-rectifier networks and nondeterministic space complexity
Before analyzing switching networks, we first consider a closely related model, switchingand-rectifier networks, and show how they capture nondeterministic space complexity.
Definition 2.3.1. A switching-and-rectifier network is a tuple (G, s,t, p) where G
is a directed multi-graph with distinguished vertices s, t and A is a labeling function
that associates with some edges e E E(G) a label p(e) of the form xi = 1 or xi = 0
for some i between 1 and n.
{0, In
_,
We say that this network computes the function f :
{0,1}, where f(x) = 1 if and only if there is a path from s to t such that
each edge of this path either has no label or has a label that is consistent with x.
We take the size of a switching-and-rectifiernetwork to be
f : {0,1}"
|V(G)
, and for a function
{0, 1}, we define rs(f)(n) to be size of the smallest switching-and-
rectifier network computing f.
Remark 2.3.2. Note that switching networks are the same as switching-and-rectifier
networks except that all edges are undirected and we cannot have edges with no label.
However, allowing edges with no label does not increase the power of switching networks, as we can immediately contract all such edges to obtain a smaller switching
network computing the same function where each edge is labeled.
23
The following theorem is folklore.
Theorem 2.3.3. If f c NSPACE(g(n)) where g(n) is at least logarithmic in n,
then rs(f)(n) is at most 2 0(g(n))
Proof. Let T be a nondeterministic Turing machine computing
f
using g(n) space.
To create the corresponding switching-and-rectifier network, first create a vertex vj
for each possible configuration cj of T, where a configuration includes the state of
the Turing machine, where the Turing machine head is on the input tape, where the
Turing machine head is on the read/write tape, and all of the bits on the read/write
tape. Now add edges in the following way:
For each configuration cj1 of T, the head on the input tape is on some bit xi.
All information except for the input is included in the configuration and the Turing
machine T only sees the bit xi, so which configurations T can go to next depends
only on xi. For each other configuration cj, of T, add an edge from v, to vj, with
no label if the T can go from cj, to c12 regardless of the value of xi and add an edge
from vj, to v32 with label xi = b if T can go from cj, to cj, if and only if xi = b, where
b E {0, 1}.
After we have added all of these edges, we obtain our switching-and-rectifier network G by merging all vertices corresponding to an accepting configuaration into one
vertex t. Now if there is a path from s to t in G whose edge labels are all consistent
with the input x, there must have been a nonzero chance that T accepts the input
x. Conversely, if there is a nonzero chance that T accepts the input x then there
must be a path from s to t in G whose edge labels are all consistent with the input
x. Thus, G computes
f.
Now we just need to bound the size of G. There are a constant number of states
of T, there are 29(n) possibilities for what is written on the work tape, there are n + 2
possibilities for where the head is on the input tape and there are g(n) + 1 possibilities
for where the head is on the read/write tape. Thus, there are at most
2 0(9(n))
possible
configurations of T which implies that JV(G)J is at most 20(9(n)), as needed.
l
24
Proposition 2.3.4. Directed connectivity can be solved nondeterministicallyin logarithmic space.
Proof. It is easily verified that the following algorithm (which just takes a random
walk from s) solves directed connectivity nondeterministically in logarithmic space:
1. Start at s
2. Keep track of the current location v, the destination t, and the number of steps
taken so far. At each point, choose randomly where to go next. If we ever arrive
at t, accept. If we get stuck or take at least JV(G)J steps without reaching t, then
D
reject.
Corollary 2.3.5. Directed connectivity is NL-complete with respect to AC0 reductions.
Proof. Looking at the construction in the proof of Theorem 2.3.3, determining whether
a given edge in the switching-and-rectifier network is there or not and what its label
is can be computed by unbounded fan-in circuits of constant depth (where the depth
needed depends on the Turing machine T). This implies that the directed connectivity problem is NL-hard with respect to AC 0 reductions.
By Proposition 2.3.4,
directed connectivity is itself in NL so directed connectivity is NL-complete.
E
Remark 2.3.6. Essentially, the space needed to conpute a function f nondeterministically is log(rs(f)). The only reason this is not quite true is that Turing machines
are a uniform model while switching-and-rectifiernetworks are a nonuniform model.
2.4
Branching programs, switching networks, and
deterministic space complexity
We now define branching programs and show how branching programs and switching
networks capture deterministic space complexity.
Definition 2.4.1. A branchingprogram is a switching-and-rectifiernetwork (G, s, t, it)
which satisfies the following conditions:
25
1. G is acyclic
2. Every vertex either has outdegree 0 or 2. For each vertex with outdegree 2, one
edge has label xi = 0 and the other edge has label xi = 1 for some i.
3. t has outdegree 0.
We take the size of a branchingprogram to be
|V(G)|,
and for a function f:
{0, 1}
-+
{0, 1}, we define b(f)(n) to be size of the smallest branching program computing f.
Theorem 2.4.2. If f C DSPACE(g(n)) where g(n) is at least logarithmic in n,
then b(f)(n) is 20
Proof. By definition, there is a deterministic Turing machine T computing
f in
space
O(g(n)). First take the switching-and-rectifier network G given by the construction
in the proof of Theorem 2.3.3. Since T is deterministic, this switching-and-rectifier
network will satisfy conditions 2 and 3 of Definition 2.4.1 but may not be acyclic.
To fix this, let m = JV(G)j and take m copies of G. Now for each i, take the
edges in the ith copy of G and do the following. Have them start at the same vertex
but end at the corresponding vertex in the (i + 1)th copy of G. This preserves the
structure of G while making it acyclic. Finally, merge all copies of t into one vertex
t and take s to be the first copy of s.
For a given input x, if G has no path from s to t then our modified switching-and-
rectifier network won't either. If G has a path from s to t then it has length at most
m - 1 so we can follow this same path in our modified switching-and-rectifier network
to go from s to t. Thus, our modified switching-rectifier network computes f. It still
satisfies properties 2 and 3 and is now acyclic as well, so it is a branching program.
It has size at most JV(G)
2
and JV(G)j is 20(9(n)), so our branching program has size
20(9(n), as needed.
E
We now give a proof of Lee's [131 observation that we can construct switching
networks from branching programs.
Lemma 2.4.3. If we take a branching program and make all of its edges undirected,
we will obtain a switching network computing the same function.
26
Proof. Let G be a branching program. For each input x, consider the directed graph
GX where V(Gx)
=
V(G) and E(Gx) is the set of edges of G whose labels are consistent
with x. The vertex t corresponding to the accepting state has outdegree 0 and since
G is a branching program, each other vertex of Gx has outdegree at most 1.
Now note that we make the edges of G, undirected, it does not affect whether
or not there is a path from s to t in G,. To see this, note that if there was a path
from s to t in Gx, there will still be a path in Gx from s to t after we make all of the
edges of Gx undirected. For the other direction, if there is a path P from s to t in
Gx after we make all of the edges of Gx undirected, look at whether the edges of P
are going forwards or backwards (with respect to the original direction of the edges
in Gx). Each vertex of Gx has outdegree at most 1 so if we had a backwards edge
followed by a forwards edge this would just get us back to where we were. This is
a contradiction as P cannot have repeated vertices. This implies that P must be a
sequence of forward edges followed by a sequence of backwards edges. But P cannot
end with a backwards edge as t has outdegree 0. Thus P consists entirely of forward
edges and thus P was a path from s to t in G to begin with.
This implies that if G' is the switching network constructed by making all of the
edges in G undirected, then for all inputs x there is a path from s' to t' in G' consistent
with x if and only if there is a path from s to t in G consistent with x. Thus G' and
E
G compute the same function, as needed.
Corollary 2.4.4. For any function
f
: {0, 1} -+ {0, 1}, s(f)(n) < b(f)(n).
Theorem 2.0.3 now follows immediately.
Theorem 2.0.3. If f c DSPACE(g(n)) where g(n) is at least logarithmic in n,
then s(f)(n) is 20(9
Proof. By Theorem 2.4.2, b(f)(n) is
2 0(g(n)).
By Corollary 2.4.4, s(f)(n) < b(f)(n)
E
so s(f)(n) is 20(g(n))
Remark 2.4.5. Looking at the constructions in the proof of Theorem 2.0.3, determining whether a given edge in the switching network is there or not and what its label
27
is can be computed by unbounded fan-in circuits of constant depth (where the depth
needed depends on the Turing machine T). This implies that the directed connectivity
problem is L-hard with respect to AC0 reductions. Reingold [22] proved that undirected connectivity is itself in L so this gives a proof that the undirected connectivity
problem is L-complete with respect to AC0 reductions.
Remark 2.4.6. Essentially, the space needed to conpute a function f deterministically is O(log(s(f)(n))). The only reason this is not quite true is that Turing machines
are a uniform model while switching networks are a non-uniform model.
28
Chapter 3
Bounds on monotone switching
networks for directed connectivity
In this chapter, we present our original work [18] on monotone switching networks
for the directed connectivity problem. The techniques developed here form the foundation for all of our later results. The main result of this chapter is the following
theorem.
problem has size n
)
Theorem 3.0.7. Any monotone switching network solving the directed connectivity
We now give an overview of the remainder of this chapter. In Section 3.1 we
develop definitions specifically for analyzing the directed connectivity problem. In
Section 3.2 we define and prove bounds on a restricted case of monotone switching
networks called certain knowledge switching networks. In Section 3.3 we partially generalize the techniques and results of Section 3.2 to all monotone switching networks.
In Section 3.4 we introduce a very different way of analyzing monotone switching networks, Fourier analysis and invariants. In Section 3.5 we show how all of the previous
material can be synthesized to give a superpolynomial lower size bound on monotone
switching networks solving directed connectivity. Finally, in Section 3.6 we carry out
our analysis more carefully to prove Theorem 3.0.7.
After proving Theorem 3.0.7, we discuss two limitations of this result. As discussed
29
in Section 3.7, there is some ambiguity in how to define monotone-L, so instead of
saying that we have separated monotone-L from monotone-NL, we say that we have
separated monotone analogues of L and NL. In Section 3.8 we note that non-monotone
switching networks can easily solve input graphs consisting of just a path from s to
t and no other edges. Thus, our monotone bounds for solving directed connectivity
on these input graphs cannot be extended to non-monotone switching networks and
bounds for other input graphs are needed to have a chance at proving non-monotone
lower bounds. We return to this question in Chapters 5 and 6.
3.1
Definitions
Since we will be working extensively with the directed connectivity problem, it is
convenient to develop definitions specifically for directed connectivity.
Definition 3.1.1. A switching network for directed connectivity on a set of vertices
V(G) with distinguished vertices s, t is a tuple < G', s', t', pa' > where G' is an undirected multi-graph with distinguished vertices s',t' and p' is a labeling function giving
each edge e' c E(G') a label of the form v1
G
v 2 or
-(V
1
-+
v 2 ) for some vertices
V(G) with v1 -A v 2
.
v 1 ,v 2
-+
Remark 3.1.2. For the remainder of this chapter, all switching networks G' will be
assumed to be switching networks for directed connectivity on a set of vertices V(G)
with distinguised vertices s, t, so we will just write switching network for brevity.
Definition 3.1.3. Whenever we consider a switching network, we assume that our
input graphs G all have vertex set V(G) (though they may have different edges).
We take Vred(G) = V(G)
\
{s,t} and we take n = |Vred(G)|.
We exclude s,t when
considering the size of V(G) because it makes our calculations easier.
Definition 3.1.4.
1. We say that a switching network G' accepts an input graph G if there is a path
P' in G' from s' to t' such that for each edge e' E E(P'), p/(e') is consistent
30
with the input graph G (i.e. of the form e for some edge e C E(G) or
-e
for
some e 0 E(G)).
2. We say that G' is sound if it does not accept any input graphs G which do not
have a path from s to t.
3. We say that G' is complete if it accepts all input graphs G which have a path
from s to t.
4. We say that G' solves directed connectivity if G' is both complete and sound.
5. We take the size of G' to be n' =IV(G')|.
6. We say that G' is monotone if it has no labels of the form -(vi
-
v 2 ).
A monotone switching network solving directed connectivity on V(G) = {s, t, a, b}
s -+ t
Figure 3-1: In this figure, we have a monotone switching network G' solving directed
connectivity, i.e. there is a path from s' to t' in G' whose labels are consistent with the
input graph G if and only if there is a path from s to t in G (where V(G) = {s, t, a, b}).
For example, if we have the edges s -* a, a -- b, and b -+ t in G, so there is a path
from s to t in G, then in G', starting from s', we can take the edge labeled s
-+
a,
then the edge labeled a -+ b, then the edge labeled s -+ a, and finally the edge labeled
b -+ t, and we will reach t'. On the other hand, if G has the edges s -+ a, a -+ b,
b -+ a, and s -+ b and no other edges, so there is no path from s to t, then in G' there
is no edge that we can take to t', so there is no path from s' to t'.
31
A monotone switching network for directed connectivity on
V(G) {s, a, b, t} which is sound but not complete
0,
G1
Figure 3-2: In this figure, we have another monotone switching network G'. This G'
accepts the input graph G if and only G contains either the edge s -+ t, the edges
s
-4
a and a
-
t, or the edges s
-4
b and b
-
t. Thus, this G' is sound but not
complete.
Remark 3.1.5. Figures 3-2 and 3-3 illustrate why solving directed connectivity on a
set of vertices V(G) requires a larger switching network. Roughly speaking, the reason
is that we are trying to simulate a directed graph with an undirected graph.
Remark 3.1.6. It is natural to ask where the examples of Figures 3-1 and 3-4 come
from. As we will see, there are many different ways to construct switching networks
solving directed connectivity on a set of vertices and we will give the particular constructions leading to the switching networks in Figures 3-1 and 3-4 later on. For now,
the readershould just make sure that he/she understands Definition 3.1.1. That said,
it is a good exercise to verify that these switching networks have the claimed properties
and to try and figure out what they are doing.
In this thesis we analyze monotone switching networks. However, rather than
looking at all possible input graphs, we focus on particular sets of input graphs. To
do this, instead of assuming that the switching networks we analyze solve directed
connectivity, we only assume that these switching networks solve the promise problem
where the input graph G is guaranteed to either be in some set I of input graphs
which contain a path from s to t or to not contain a path from s to t.
Definition 3.1.7. Given a set I of input graphs which all contain a path from s to t,
let m(I) be the size of the smallest sound monotone switching network which accepts
32
A monotone switching network for directed connectivity on
V(G) = {s, a, b, t} which is complete but not sound
I -- + b b -- +a
8
a
G
Figure 3-3: In this figure, we have another monotone switching network G'. This G'
accepts the input graph G whenever G contains a path from s -+ t (where V(G) =
{ s, t, a, b}), so it is complete. However, this G' is not sound. To see this, consider the
input graph G with E(G) = {s -+ a, b -+ a, b -+ t}. In G', we can start at s', take
the edge labeled s - a, then the edge labeled b -+ a, then the edge labeled b -+ t,
and we will reach t'. Thus, this G' accepts the given input graph G, but G does not
contain a path from s to t because the edge from b to a goes the wrong way.
all of the input graphs in I.
In this chapter, we focus on input graphs which contain a path from s to t and no
other edges, as they are the minimal YES instances and are thus the hardest input
graphs for a monotone switching network to accept. We have the following definitions.
Definition 3.1.8. If V(G) is a set of vertices with distinguished vertices s, t,
1. Define PV(G),l to be the set of input graphs G such that E(G) = {vo V2, * - - ,v 1 _1
-+
v 1} where v0 = s, vi = t, and vo,
v..
,vi
v 1 , vi
are distinct vertices of
V(G).
2. Define PV(G),<l = U j=1PV(G),j
3. Define PV(G)
=
1
-
PV(G),j
Proposition 3.1.9. A monotone switching network solves directed connectivity on
V(G) if and only if it is sound and accepts every input graph in PV(G)Corollary 3.1.10. The size of the smallest monotone switching network solving directed connectivity (on V(G)) is m(PV(G))-
33
A non-monotone switching network solving
directed connectivity on V(G) = {s, a, b, t}
b__'-( a t
t
G'
Figure 3-4: In this figure, we have a non-monotone switching network G' solving
directed connectivity. Note that the edge with label -,(b - t) and the edge with label
-i(a -+ t) are necessary for G' to be complete. To see this, consider the input graph
G with E(G) = {s -+ a, a -+ b, b -+ t}. To get from s' to t' in G' we must first take
the edge labeled s -- a, then take the edge labeled a -- b, then take the edge labeled
-(a -+ t), and finally take the edge labeled b -÷ t.
In chapters 5 and 6 we will analyze the difficulty of directed connectivity for a
given input graph G. To do this, we consider the promise problem where the input
graph is guaranteed to either be isomormphic to G or to have no path from s to t.
Definition 3.1.11. Given an input graph G, let m(G) = m(I) where I = {G 2
G 2 is obtained from G by permuting the vertices of Vred(G)}
We give this definition here because it will make it easier for us to state some of
the results of this chapter.
3.2
Certain Knowledge Switching Networks
In this section, we define certain knowledge switching networks, a subclass of sound
monotone switching networks for directed connectivity which can be described by
a simple reversible game for solving directed connectivity.
We then prove upper
and lower bounds on the size of certain knowledge switching networks, showing that
certain-knowledge switching networks can match the performance of Savitch's algorithm and this is tight. The main result of this section is the following theorem.
34
Definition 3.2.1. Given a set I of input graphs all of which contain a path from s
to t, let c(I) be the size of the smallest certain-knowledge switching network which
accepts all of the input graphs in I.
Theorem 3.2.2. Let V(G) be a set of vertices with distinguished vertices s, t. Taking
n
Vre(G)J, if I > 2 and n > 2(l - 1)
1
n
then
1)_u
) F* 1 <- C(PV(G),1) <- C( V(G),<1) < nF11
2.
3.2.1
2
lgn+
c(PV(G))
-2
The certain knowledge game for directed connectivity
We can use the following simple reversible game to determine whether there is a path
from s to t in an input graph G.
Definition 3.2.3 (Certain knowledge game). We define a knowledge set K to be a
set of edges between vertices of V(G). An edge u
that there is a path from u to v in G.
-4
v in K represents the knowledge
We do not allow knowledge sets to contain
loops.
In the certain knowledge game, we start with the empty knowledge set K
=
{}
and
use the following types of moves:
v 2 e E(G), we may add or remove v 1
v 2 from K.
1. If we directly see that v1
-
2. If edges v 3
v 5 are both in K and v 3 -4 v 5 , we may add or remove
V3 -+ v 5
-+
v 4 , v4
-+
-
from K.
We win the certain knowledge game if we obtain a knowledge set K containing a path
from s to t.
Proposition 3.2.4. The certain knowledge game is winnable for an input graph G if
and only if there is a path from s to t in G.
35
3.2.2
Adapting the certain knowledge game for monotone switching networks
Intuitively, certain knowledge switching networks are switching networks G' where
each vertex v' E V(G') corresponds to a knowledge set Kv, and the edges between
the vertices of G' correspond to moves from one knowledge set to another. However,
there are two issues that need to be addressed. First, if G' is a switching network,
I',
v', w' c V(G'), and there are edges with label e between u' and v' and between v'
and w', then we may as well add an edge with label e between u' and w'. This edge
now represents not just one move in the game but rather several moves. Thus an
edge in G' with label e should correspond to a sequence of moves from one knowledge
set to another, each of which can be done with just the knowledge that e E E(G).
The second issue is that the basic certain knowledge game has many winning
states but a switching network G' has only one accepting vertex t'. To address this,
we need to merge all of the winning states of the game into one state. To do this,
we add a move to the game allowing us to go from any winning state to any other
winning state.
Definition 3.2.5 (Modified certain knowledge game). In the modified certain knowledge game, we start with the empty knowledge set K
=
{}
and use the following types
of moves:
1. If we directly see that v1
-+*v 2
2. If edges v 3
v 5 are both in K and v 3 f v 5 , we may add or remove
v3
-+
3. If s
-+
v4 , v 4
-+
E E(G), we may add or remove v1
-
v 2 from K.
v 5 from K.
-4
t E K then we can add or remove any other edge from K.
We win the modified certain knowledge game if we obtain a knowledge set K containing
a path from s to t.
Remark 3.2.6. In the modified certain knowledge game, an edge v 1
.-
+
v 2 in K now
represents knowing that either there is a path from v1 to v 2 in G or there is a path
from s to t in G.
36
Proposition 3.2.7. The modified certain knowledge game is winnable for an input
graph G if and only if there is a path from s to t in G.
Definition 3.2.8. We say a monotone switching network G' is a certain knowledge
switching network if we can assign a knowledge set Kv, to each vertex v' G V(G') so
that the following conditions hold:
1. Ks' = {}
2. KtI
contains a path from s to t (this may or may not be just the edge s
3. If there is an edge with label e = v1
-
-
t)
v 2 between vertices v' and w' in G', then
we can go from Kv, to K&' with a sequence of moves in the modified certain
knowledge game, all of which can be done using only the knowledge that v1
v 2 c E(G).
We call such an assignment of knowledge sets to vertices of G' a certain knowledge
description of G'.
Proposition 3.2.9. Every certain knowledge switching network is sound.
Proof. If there is a path from s' to t' in G' which is consistent with an input graph G
then it corresponds to a sequence of moves in the modified certain knowledge game
from K,, =
{}
to a K,, containing a path from s to t where each move can be done
with an edge in G. This implies that the modified certain knowledge game can be
won for the input graph G, so there is a path from s to t in G.
3.2.3
D
A partial order on knowledge sets
There is a partial order on knowledge sets as follows. This allows us to restate the
definition of certain knowledge switching networks more cleanly and will be important
in Section 3.3
Definition 3.2.10. Define the transitive closure R of a knowledge set K to be
37
a}
K={s
s
'
K =}
a
K=ese
a
b
an
a, s - b})
b
b
t
a
K =Is -+
b
K
b}
-+fs
=
tK
t
Figure 3-5: In this figure, we have a certain knowledge switching network G' solving directed connectivity (on V (G) = {s, t, a, b}) together with a certain knowledge
description for it.
k.
=
{V1
-4 V2 : V1,V2 c
are all in K} if s--4 t
2. K = {vji
V2 : V1,iV2 E
V (G), v1
V2,
there is a path from v1 to V2 whose edges
K
V(G), vi
V2}
if s -+ t E K
1. We say that K 1 < K 2 if K 1 C K
.
Definition 3.2.11.
2. We say that K 1 = K 2 if K1 = K
2
.
2
Remark 3.2.12. Intuitively, K describes all knowledge which can be deduced from
K. The statement that K1 < K2 means that all of the information that K1 represents
about G is included in the information that K 2 represents about G.
Proposition 3.2.13. If K 1 , K 2 , K3 are knowledge sets for V(G), then
1. K1 < K1 (reflexivity)
2. If K1 < K 2 and K2 < K1 then K1
K2
(antisymmetry)
3. If K1 < K 2 and K 2 < K3 then K1 < K3 (transitivity)
Proposition 3.2.14.
38
1. For knowledge sets K1 , K 2 , there is a sequence of moves from K1 to K2 in the
modified certain knowledge game which does not require any information about
the input graph G if and only if K1
K2.
2. For knowledge sets K1, K2 and a possible edge e of G, there is a sequence of
moves from K1 to K 2 in the modified certain knowledge game which only requires
the information that e E E(G) if and only if K1 U {e}
1K2U {e}.
Corollary 3.2.15. We can restate the definition of certain knowledge switching networks as follows. A monotone switching network G' is a certain knowledge switching
network if we can assign a knowledge set Kv, to each vertex v' e V(G') so that the
following conditions hold:
1. KSI
2. Kt,
=
{}
{s -+ t}
3. If there is an edge with label e = v1
-
v 2 between vertices v' and w' in G', then
Kv, U {e} = Kw U {e}
3.2.4
Certain knowledge switching networks and reachability
from s
While certain knowledge switching networks can consider all paths in G, most of
our examples only consider reachability from s. For these examples, the following
definition and lemma are extremely useful.
Definition 3.2.16. For each subset V C Vred(G), define Kv = {s
Lemma 3.2.17. For any v 1 ,v 2 E Vred(G) and V
C
-
Vred(G) with v1
v : v E V}
c V there is a
sequence of moves in the modified certain knowledge game from Kv to Kvu{v 2} which
only requires the knowledge that v 1
Proof. Starting from K
1. Add v,
-*
=
-+
v2
G
E(G)
Kv, take the following sequence of moves:
v 2 to K.
39
2. Add s
-+
v 2 to K (we already have s -- v, and vi
3. Remove v,
--
v 2 in K)
v 2 from K.
-
El
We are now at K = KvuV{ 2}
Remark 3.2.18. Note that there is no sequence of moves in the modified certain
knowledge game from Kvl to KIV 2} which only requires the knowledge that v1 -+
V2
E E(G). The reason is that this is not reversible. We are forgetting the fact that
there is a path from s to v1 in G or a path from s to t in G and forgetting information
is irreversible. Also, while we can deduce that there is a path from s to v 2 in G from
the fact that there is a path from s to v1 in G and v 1
-+
v 2 e E(G), we cannot deduce
that there is a path from s to v1 in G from the fact that there is a path from s to v 2
in G and v 1
v2
C
E(G).
s-t
s
b
at
Kf a,bj
Kja)
i
s +b
Kjbj
a
t
C
t
tl
Kfa,c)
Ct
K{c)j
K(b,c)
Figure 3-6: In this figure, we have a certain knowledge switching network G' solving
directed connectivity (on V(G) = {s, t, a, b, c}) together with a certain-knowledge
description for it. By default, we take K., = {} and Kt, = {s -+ t}.
Remark 3.2.19. Figure 3-6 shows how removing old information can help in finding
a path from s to t with the modified certain knowledge game. If G is the input graph
40
with V(G) = { s, t, a, b, c} and E(G) = {s
-+
a, a
-+
s' to t' in G', we must first take the edge labeled s
b, b -+ c, c -+ t}, then to get from
-
a to reach K = K{a}, then take
the edge labeled a -+ b to reach K = K{a,b}, then go "backwards" along the edge labeled
s
a to reach K =
-
K{b},
then take the edge labeled b -s c to reach K = K{b,,} and
finally take the edge labeled c -+ t to reach t'.
3.2.5
An upper bound on certain-knowledge switching networks
While the certain knowledge condition is restrictive, certain knowledge switching
networks nevertheless have considerable power. In particular, the following certain
knowledge switching networks match the power of Savitch's algorithm.
Definition 3.2.20. Given a set of vertices V(G) with distinguished vertices s, t, let
G'(V(G), r) be the certain knowledge switching network with vertices t' U {v' : V C
< r} and all labeled edges allowed by condition 3 of Definition 3.2.8,
each v'v has knowledge set Kv, s' = v',, and K, = {s - t}.
Vred(G), V|
where
Definition 3.2.21. Define G'(V(G)) = G'(V(G), n)
Example 3.2.22. The certain-knowledge switching network shown in Figure 3-5 is
G'({s, t, a, b}) = G'({ s, t, a, b}, 2). The certain-knowledge switching network shown
in Figure 3-6 is G'({s, t, a, b, c}, 2).
Theorem 3.2.23. For all 1 > 1, c(Pv(G),<l)
() + 2
(f
Proof. Consider the switching network G' = G'(V(G), [lg 11). G' is a certain-knowledge
switching network with IV(G') =
(n) +2,
so to prove the theorem it is sufficient
to show that G' accepts all of the input graphs in PV(G),<l.
Consider an input graph G - Pv(G),<l. E(G) = {vi -÷ vj
vo
=
s', vj
=
t', and
j
: 0 < i < J - 1} where
< 1. We will show that G' accepts G by showing that we can
win at the following reversible pebble game without placing more than [lg 1] pebbles
on the vertices vi,. -,
vj_ 1 . This pebble game was introduced by Bennet
time/space tradeoffs in computation.
41
[3
to study
Definition 3.2.24. In the reversible pebble game for directed connectivity, we start
with a pebble on s.
We win the game if we can place a pebble on t.
We are only
allowed to use the following type of move:
1. If there is a pebble on a vertex v1 C V(G) and an edge v1
-+
v 2 C E(G) where
v 1 # v 2 , we may add or remove a pebble from the vertex v 2
Remark 3.2.25. Note that after putting a pebble on a vertex we do NOT remove the
pebble on the previous vertex. This is a key difference between this game and the JAG
model.
Definition 3.2.26. Let G be the input graph with vertices {vo, v 1 , v 2 , -
{Vi -+
v 1 : i
>
} and edges
0}. Assume that we start with a pebble on s = vo.
1. Let f (i) be the minimal number m such that we can place a pebble on vi without
ever having more than m pebbles on the vertices v 1 , - - - , vi1.
2. Let g(i) be the minimal number m such that we can reach the game state where
there is a pebble on v 0 and vi and no other pebbles without ever having more
than m pebbles on the vertices v 1 ,- -- ,vi.
We need to show that f(l) < [log 1] for all 1 > 1. This follows immediately from
the following lemma. This result was first proved by Bennet [31 and we present a
proof here for convenience.
Lemma 3.2.27. For all integers i > 1,
1.- f (i <- [log ( I
2. g(i) < [log (i)] + 1
Proof.
Proposition 3.2.28. For all i > 1, g(i) < f(i) + 1
Proof. We have a sequence of moves allowing us to place a pebble on vi while placing
at most f(i) pebbles on the vertices v, ... , vi_ 1 . After completing this sequence, run
42
the sequence in reverse except that we do not remove the pebble on vi. When we are
done, we only have pebbles on vo and vi and at all times we have at most f(i) + 1
pebbles on the vertices vi, -.. , vi, as needed.
Proposition 3.2.29. For all i, j > 1, f (i + j) < max {g(i), f (j) + 1}
Proof. We first take a sequence of moves allowing us to reach the state where there
are pebbles on vo and vi and no other pebbles without ever having more than g(i)
pebbles on vi,. -, vi. We then take the sequence of moves allowing us to put a pebble
on vj without ever having more than f(j) pebbles on v 1 , -- - , vj_1 except that we shift
the sequence of moves to the right by i. This sequence now allows us to start from
the state where there are pebbles on vo and vi and no other pebbles and put a pebble
on vi+j without ever having a pebble on v,, - -- , vi_ 1 or having more than f(j) + 1
pebbles on vi, - -- , vi+j-1. Composing these two sequences of moves gives a sequence
of moves putting a pebble on vi+j while never having more than max {g(i), f(j) + 1}
pebbles on vi, -- - , vi+-1
Applying Proposition 3.2.28 to Proposition 3.2.29 we obtain that for all i, j, f(i+
j)
< max {f(i), f(j)} + 1. Clearly f(1) = 0 so we can easily show by induction that
for all i > 1, f(i) 5 [log (i)]. Applying Proposition 3.2.28 again we obtain that for
E
all i > 1, g(i) < [log (i)] + 1, as needed.
By Lemma 3.2.27 there is a sequence of moves for the reversible pebble game
on G allowing us to win without placing more than [ig 1] pebbles on the vertices
vi, - -- , vj_ 1 . We now translate this winning sequence of moves into a sequence of
moves in G'. If we have pebbles on a set of vertices V C {vO,
...
, vj_ 1} with vi E V
and our move is to place a pebble on vi+ 1 , this corresponds to moving from Ky to
Kvulm,
} in G' along an edge labeled vi
-
vj+1 (we know such an edge exists because
of Lemma 3.2.17). Similarly, if we have pebbles on a set of vertices V C {vo, ** , vi _1 }
with vi, vj+1 E V and our move is to remove a pebble from vi+1, this corresponds to
moving from Kvyf,2 1 } to Kv in G' along an edge labeled vi -* vi+1 . Finally, if we
have pebbles on a set of vertices V C {vo, *** , vj1} with vj_ 1 E V and our move is
43
to place a pebble on vj = t, this corresponds to moving from Ky to Kt, in G' along
an edge labeled vy_1 -+ t. The entire sequence of moves corresponds to a walk from s'
to t' in G' whose edge labels are all consistent with G so G' accepts the input graph
G, as needed.
3.2.6
A lower size bound on certain-knowledge switching networks
In this subsection, we prove lower bounds on c(PV(G),<l).
Definition 3.2.30. For a knowledge set K such that K 0 Kt,, define
V(K) ={v : v C Vre(G),-]w C V(G) : w # v,v -+w E E(G) or w -+v c E(G)}
)
Proposition 3.2.31. If K1 = K2 * Kt, then V(K1 ) = V(K 2
The heart of our lower bound argument is the following lemma, which says that if
we have an input graph G containing a path P from s to t and no other edges, then
for any walk W' from s' to t' in G' whose edge labels are all in E(P), there is one
vertex v' on W' such that V(Kv') contains many vertices of P and no vertices not in
P, which gives a lot of information about P.
Lemma 3.2.32. Let G' be a certain knowledge switching network. For any certain
knowledge description of G' and any path P = s -+ v 1
+ -..
--+ v1 _ 1
-
t, if G is the
labels are all in G from a vertex v'tart with K,
- Ks, to a vertex v' n with K,
Kt, then W' passes through a vertex v' such that Kv' 0 Kt,, V(Kv,)
C
{v 1,
, 1
-
input graph with vertex set V(G) and E(G) = E(P), if W' is a walk in G' whose edge
-
and |V(Kv')| > lg(l)|.
However, the proof of this lemma is long, so we relegate it to the appendix.
Instead, we show here that this lemma holds if all Kv' are of the form Kv where
V
C
Vred(G) or are equivalent to KtI. This is equivalent to proving the following
result about the reversible pebbling game.
44
Lemma 3.2.33. f(l) > Flg 1]
This result was proved by Li and Vitanyi [15]. We give a short alternative proof
of this result here which emphasizes the role of reversiblity.
Proof. We have that f(1)
=
0, f(2)
=
1, and f(i) is an nondecreasing function of i,
so this follows immediately from the following claim.
Lemma 3.2.34. For all i> 2, f(2i - 1) > f(i) + 1
Proof. Note that if we merge all of the vertices vi,--. , vi_
1
playing the reversible pebbling game with the vertices s, vi, ...
with s
,
=
vo, we are
v2 -1
1 and trying to
pebble v 2i-1. By definition, this requires placing at least f(i) pebbles on the vertices
vi ...
,
v 2i- 2 at some point.
If at this point, we have any pebble on the vertices v 1 , - -,
vi_ 1 , then we have
used f(i) + 1 pebbles. Thus, we may assume that we have no pebbles on the vertices
v 1 , ** , vi_ 1 . Now let vj be the leftmost vertex that is pebbled and run the sequence
of moves we used to reach this state in reverse. At some point, we must remove a
pebble on vj so that we can reach the initial state of no pebbles anywhere. However,
to do this, we must have first had f(j)
f(i) pebbles on the vertices vi, -- - , vji.
Moreover, at this point we still had a pebble on vj so we had a total of at least f(i) + 1
D
pebbles placed, as needed.
This completes the proof of Lemma 3.2.32 when all Kv' are of the form Ky where
D
V Q; Vred(G) or are equivalent to Kt,.
The other part of our lower bound proof is finding a large collection of paths such
that each pair of paths has very few vertices in common.
Lemma 3.2.35. If m, k1 , k2 are non-negative integers and there is a prime p such
that k 2 < k1 < p and m > pk 1 , then there is a collection of pk2+1 subsets of [0, m - 1]
such that each pair of subsets has at most k 2 elements in common.
Proof. To obtain our collection, first take the set of all polynomials in Fp[x] of degree
at most k 2 where Fp is the integers modulo p. There are pk
45
2
+1 such
polynomials. For
each such polynomial f(x), let Sf
polynomials fi and
f2
=
{(x, f(x)) : 0 < x < ki}. For any two distinct
of degree at most k 2 , S OSf2 = {(x,
-f 2 (x) = 0}.
is a non-zero polynomial in Fp[x] of degree at most k 2 , so there are at most
-
k2
x E Fp such that fi(x) - f 2 (x) = 0. Thus,
Isf, n sf2I
k2
-
fi
f2
fi(x) : fx(x)
We now translate these sets into subsets of [0, pki - 1] by using the map
#:[0,pkl
O(x)
- 1] --+ [0, ki - 1] x Fp
([x], (x mod p))
Taking our subsets to be
#F1 (Si)
for all f E lF[x] of degree at most k 2 , each such
subset of [0, pki -- 1] ; [0, m-1] has ki elements. Since
any distinct fl, f2 c Fp[x] of degree at most k 2 ,
# is injective
J4O-(Sf 1 )n#- I(S
2
and surjective, for
)1 = I|f1OSf2|I
k2
D
and this completes the proof.
then there is a collection of at least
( n )k 2 +1
'
Corollary 3.2.36. If n > 2 and k 1 , k 2 are non-negative integers with k 2 < k1
paths from s to t of length k 1 + 1 on
the vertices V(G) such that each pair of paths has at most k2 vertices in common
(excluding s and t).
Proof. We prove this result using Lemma 3.2.35 and a suitable prime number p chosen
using Bertrand's postulate.
Theorem 3.2.37 (Bertrand's postulate). For any integer m > 3 there is a prime p
such that m <p < 2m - 2
Corollary 3.2.38. For any real number m > 1 there is a prime p such that m < p <
2m.
Proof. By Bertrand's postulate, for any real number m > 3 there is a prime p such
that m < rm] < p < 2[rm] - 2 < 2m. If rn c [1, 2] then m < 2 < 2m. If m E [2, 3]
then m < 3 < 2m.
By Corollary 3.2.38 we can take a prime p such that
that n > pk1 and since k, <
we have that k 2
The result now follows from Lemma 3.2.35.
<
<p <
. We now have
which implies that k, <
-
< p.
1
We now prove the following lower bound on certain knowledge switching networks:
46
Theorem 3.2.39. If 1 > 2 and n > 2(1 - 1)2, then c(P,1 ) > (2(1-
Proof. Taking ki = I - 1 and k2
collection of at least
(2 n)k 2 +1
=
Flg 1] - 1 and using Corollary 3.2.36, we have a
( 2 (l 1 ) F11 paths of length 1 from s to t on the set of
vertices V(G) such that each pair of paths P, P has at most k 2 vertices in common
(excluding s and t). However, by Lemma 3.2.32, for any certain knowledge switching
network G' which accepts all of the input graphs in P,,, we can associate a vertex
v' in G' to each path Pi in our collection such that IV(K,,)l > k 2 and V(K,,) is a
subset of the vertices of P. This implies that we cannot have v' = v for any i = j
as otherwise we would have that IV(Kv')l = IV(Kv')l > k 2 and IV(K,')I is a subset
of the vertices of both P and P, which is impossible as any two distinct paths in
our collection have at most k 2 vertices in common. Thus, IV(G')
> (2(1n-1)) [g
E
needed.
3.2.7
1, as
Simplified bounds on certain-knowledge switching networks
We now use Theorems 3.2.23 and 3.2.39 to prove Theorem 3.2.2.
Theorem 3.2.2. Let V(G) be a set of vertices with distinguished vertices s, t. Taking
n = |Vred(G)|, if 1 > 2 and n > 2(l - 1)2 then
1. (12(2.
1 ))
lgl]
n41L"-12
C(PV(G),l)
c(PV(G)) <
2
+- ~11911
c(PV(G),<1)
gn+
2
Proof. For the first statement, the lower bound is just Theorem 3.2.39. To prove the
upper bound, note that by Theorem 3.2.23 we have that c(P",
If [lgl] =1 then 1 = 2 so
n > 2(l
1)2 >
zY41
()
-
n
2[lg 1]. This implies that E
g1 1]
1
1
) < E 1
(n)
+ 2.
= n. If [lgl] > 1 then 1 > 2 so
(n)
< [lgl1 ([Inl])
< W1911, as needed.
For the second statment, the upper bound follows immediately from the upper
bound of the first statement.
For the lower bound, taking 1 = []
47
by Theorem
3.2.39 we have that
C(PV(G))
ng-4
c(PV(G),i)
0
S gn-
n
2(1 - 1)
(-
DO
>
(
jy~ ig-
2,1
g
-1gn
Remark 3.2.40. As noted in Remark 2.4.6, a size bound of E(s(n)) on switching
networks solving a problem roughly corresponds to a space bound of 9(lg (s(n))) on
algorithms solving that problem. Thus, the size bounds of Theorem 3.2.2 correspond
to a space bound of E( lg 1] Ig n) for finding all paths of length at most 1 and a space
bound of E((lg n) 2 ) for finding all paths, which is exactly the performance of Savitch's
algorithm.
3.3
Elementary results on monotone switching networks
In this section, we see which of our techniques and results about certain knowledge
switching networks can be generalized to all monotone switching networks.
The most important thing to note is that for general sound monotone switching
networks, at any given vertex v' we may not be certain of which paths are in G.
Instead, we will have several possibilities for which paths are in G and will only
know that at least one of them holds.
We begin by defining a knowledge game
for directed conectivity which takes this difficulty into account. We show that any
sound monotone switching network can be described using this game. We then show
a nice simplification of sound monotone switching networks which can be done by
increasing their size by a factor of at most n. Finally, we show a partial reduction
of sound monotone switching networks to certain knowledge switching networks. By
itself, this reduction is not strong enough to prove good lower size bounds. However,
it will play an important role in Section 3.5.
48
3.3.1
The knowledge game for directed connectivity
Definition 3.3.1. A state of knowledge J is a multi-set {K1 , - - - , K,} of knowledge
sets (we can have duplicates in J). In the knowledge game for directed connectivity, J
represents knowing that for at least one i E [1, m] the knowledge about G represented
by Ki is true.
Example 3.3.2. If J = { K{a}, K{b}, K{} then J represents knowing that either
there is a path from s to a in G, a path from s to b in G, a path from s to c in G, or
a path from s to t in G.
Definition 3.3.3. In the knowledge game for directed connectivity, we start at J.,
{{}}
and we win if we can get to J, = {{s
--
t}}. We are allowed to use the following
types of moves. If J = { K1 , - - - , Kn} then
1. If we directly see that an edge v1 -+ v 2 is in G we may add or remove v1 -+ v 2
from any K.
2. If v 3
3. If s
v4 ,v 4
-+
-
-+
v5
e Ki andV3 : v 5 we may add or remove v 3 --+ v 5 from K.
t G Ki we may add or remove any other edge from Ki.
4. If i, j C [1,m], i
#
j, and Ki C K then we may remove K from J.
5. If K is a knowledge set such that Ki
C
K for some i E [1, m] then we may add
K to J.
Remark 3.3.4.
The knowledge game for directed connectivity is a generalization of
the certain knowledge game for directed connectivity. The moves which are new are
the moves of types 4 and 5. Moves of type 4 make sense because if we know that
Kj implies Ki and have the statement that Ki OR K is true then this statement is
equivalent to the statement that Ki is true. Moves of type 5 are the inverse of moves
of type
4 so we still have reversibility.
Proposition 3.3.5. It is possible to win the knowledge game for directed connectivity
for an input graph G if and only if there is a path from s to t in G.
49
3.3.2
A partial order on states of knowledge
We have a partial order on states of knowledge as follows.
Definition 3.3.6.
1. We say thatJi = {K 11 ,-
, Kimi} < J2 ={K2 1 ,-
,Kim 2 } if for all j E [1, m 2 ]
there is an i E [1,m1 ] such that K1 i < K 2
We say that J
= J2 if J1
<
J2 and J2
<
J1
.
2.
Remark 3.3.7. Intuitively, the statement that J1 < J2 means that all of the information that J1 represnents about G is included in the information that J2 represents
about G.
Proposition 3.3.8. If Ji, J2 , J3 are states of knowledge then
1. J1 < J1 (reflexivity)
2. If J1 < J2 and J2 < J1 then J1 = J2 (antisymmetry)
3. If J1 < J2 and J2 < J3 then J1 < J3 (transitivity)
We now show the connection between this partial order and the knowledge game
for directed connectivity.
Proposition 3.3.9. For states of knowledge J1 , J2 , if we can go from J1 to J2 in the
knowledge game for directed connectivity with no information about the input graph
.
G then J1 = J2
Proof. By transitivity, to show that if we can get from J1 to J2 in the knowledge
game for directed connectivity with no information about the input graph G then
Ji - J 2 it is sufficient to show that if we can get from J1 to J2 in the knowledge
game for directed connectivity with a single move then J - J2 . J can be written as
Ji = {Ki,
...
,
Kim} and we have the following cases:
1. If we use a move of type 2 or 3 altering some knowledge set Kj to reach J2 then
J2 = {K 2 1 ,- .. , K 2m} where K2 j = Kli for all i
Kii =-K2j so J 1 _ J2
50
# j and Kjj = K 2j. For all i,
2. If we use a move of type 4 to delete some knowledge set Kj from Ji then
J2 =
{K 2 1 , ... , K 2(-1), K 2(j+1), -- - , K2m} where for all i # j, K 2i
there exists a
J1
<
12
#j
J2 . For all i #
=
K1l and
such that K 2 i 2 = K1 j2 < K1 j. For all i z j, Ki < K 2i so
j,
K 2 i < K1i and
K2i2
< Ki so J2 < J1 . Thus, J - J2 as
needed.
3. If we use a move of type 5 to add some knowledge set K to J then
J2 =
{K2,
, K2m,
K2(m+1)} where for all i E
[1, m], K 2i
=
Kli and there is
a j E [1, m] such that Ki < K2 (m+) = K. For all i E [1, m], K 2 < Kii so
J2 < Ji. Kij < K 2 (m+l) and for all i E [1, m], Kii < K2 so J,
<
J2 . Thus,
J, - J2 as needed.
To show the converse to Proposition 3.3.9, we use the following lemma.
Lemma 3.3.10. If J = {K 11 ,... ,Kimi} = J2 = {K2 1 ,.--- ,K2m 2 } then there is a
set I C [1, mi], a set
a function
f2
12 C
: [1, M 2 ] \
[1,
12 -+
n 2]
of equal size to 1, a function fi : [1, m1 ] \ I1
> 11,
I2 such that
I2, and a perfect matching < : I1 -
1. For all i c [1, in
1 ] \ 1, Klf1i() < Ki
2. For all j
G [1, iM 2] \ I,
K 2f 2 (,)
<
K 23
3. For all i E 1, K1l = K20(i).
Proof. Consider the graph formed as follows.
The vertices of this graph will be
the knowledge sets {Kii, i E [1, m1]} U {K 2j, j E [1, m 2]}. Since J1 < J2 , for each
j
E [1, M 2] there is an i E [1, m1 ] such that K1l < K 2j. Draw a directed edge from
each K 23 to the corresponding Kii (if there are more then one possible i, just choose
one of them). Since J2 < J 1, for each i C [1, m1 ] there is a
j
E [1, M 2 ] such that
K 23 < K1 i. Draw a directed edge from each Ki to the corresponding K 23 (if there
are more then one possible
j, just choose
one of them).
After adding all of these edge, we have a bipartite graph where each vertex has
outdegree 1. This graph must have the structure of a set of cycles along with paths
51
leading into the cycles. Choose 11 and I2 such that for each cycle C there is exactly
one ic E
AI
and exactly one Jc C 12 such that Klic is in C and K 2jc is in C. Then
#(ic)
for all cycles C set
= jc. We know we can do this because there cannot be a
cycle consisting entirely of vertices of the form Kli or a cycle consisting entirely of
vertices of the form K2,. We then choose the functions fi and
f2
as follows.
1. For all i E [1, m1 ] \ I, there is a cycle C such that there is a path from Kli to
We take fi(i) = ic.
C and thus a path from Kii to Ki.
2. For all j C [1, M 2] \ 12 there is a cycle C such that there is a path from K 2 j to
C and thus a path from K2j to K 2 jc. We take
f2(j) =
jc.
Now note that an edge from a knowledge set K, to a knowledge set K 2 implies that
K2
< K 1 . By transitivity, a path from a knowledge K1 to a knowledge set K 2 also
implies that K 2 < K1 . This implies that for any cycle all knowledge sets in the cycle
are equivalent. The result now follows immediately because
1. For all cycles C, Klic and K 24(ic) = K2jc are in the same cycle.
2. For all i E [1, mi]
\
3. For all j E [1, iM2]
\
1 there is a path from K1 2 to Klf 1 (i)
there is a path from K 2 to K 2f 2 (j)
2
Corollary 3.3.11. For states of knowledge Ji, J2 , we can go from J1 to J2 in the
knowledge game for directed connectivity with no information about the input graph
.
G if and only if J1 = J2
Proof. The only if part is just Proposition 3.3.9. For the if part, assume that J1
{K 1 1 , ...
, KimI}
- J2 = {K 2 1 , ...
,
K2m2 }.
By Lemma 3.3.10 there is a set I, C
[1, m1 ], a set I2 C [1, m 2] of equal size to I1, a function fi : [1, ml] \ I, -
function
f2 :
[1, M 2 ]
\ 12
-
12,
and a perfect matching 0 : 1 -+ 12 such that
1. For all i c [1, mi 1 ] \11, KIf 1(i) < Ki
'2,
K 2f 2 (j) < K 2
,
2. For all j E [1, M 2 ] \
52
=
1, a
3. For all i E I,, Ki
=-
K2+()
We will go from J= J to J2 in the knowledge game for directed connectivity using
the following steps.
1. Use moves of type 2 and 3 to replace each Kli with K1 i.
2. For all i E [1, m 1 ] \ I,, Kif,(j) < Kli which implies that Kif(j) C K12 . We can
thus use moves of type 4 to delete K 1i for all i E [1, Mi] \ I,.
i E 1}. For all i E A1, Kii - K24(j) so
3. We are now at J = {Ki2
Thus, J
=
{k
: j E
2
12}.
For all
j
,\
E [1,iM 2
12, K 2f2(j)
=
kii
k24(j).
K 2j which
implies that R 2f 2 (j) C K 2j. We can thus use moves of type 5 to add K 2j for all
j E [1, m 2 ] \ 124. We finish by using type 2 and 3 to replace each K 21 with K 2j and obtain J = J2
We have similar results when we directly see that an edge e is in the input graph
G.
Definition 3.3.12. For a state of knowledge J = {K 1 , - -- , K}
and an edge e, define
J U {e} = {K1 U {e}, ... , Km U {e}}
Lemma 3.3.13. For states of knowledge J1 , J2 , we can go from J1 to J2 in the
knowledge game for directed connectivity with the information that v 1
if and only if J1 U {v 1
Proof. If J1 U {v 1
J1 U {v 1
-
-*
-
v2 }
J1 U {v 1
v2}
v 2} to J2 U {v 1
J2 U {v 1
-
-*
-*
-
v 2 C E(G)
v2}
v 2 } then by Lemma 3.3.10 we can go from
v2 } in the knowledge game for directed connectivity with
no information about the input graph G. With the information that v 1
we can go from J1 to J1 U {v 1
-+
v 2 } and from J 2 U {vi
-÷
-
v 2 E E(G)
v 2 } to J 2 in the knowledge
game for directed connectivity using moves of type 1. Thus, we can go from J1 to J2
in the knowledge game for directed connectivity, as needed.
53
For the converse, note that for any states of knowledge J1 , J2 , for any sequence of
moves in the knowledge game for directed connectivity to go from J = J to J= J2,
if we replace J with J U {v l
v 2} at each step we will still have a correct sequence
of moves. Moreover, all moves of type 1 now correspond to doing nothing. This
implies that we can get from J U {vi
-a
v 2} to J2 U {vi
-
v 2 } in the knowledge game
for directed connectivity without knowing anything about the input graph G so by
Proposition 3.3.9 we have that J1 U {vI
v2}
-+
J2 U {vi
-
v 2 }, as needed.
D
Knowledge description of monotone switching networks
3.3.3
In this subsection, we show that all sound monotone switching networks can be described in terms of the knowledge game.
Definition 3.3.14. If G' is a monotone switching network, we call an assignment of
states of knowledge J,, to vertices v' of G' a knowledge description if the following
conditions hold:
1. J',{{}}
2. Jt/
{{s
-+
t}} or Jt, =
{}
3. If there is an edge with label e = v 1
-
v 2 between vertices v' and w' in G' then
Jv' U {e} - Jw, U {e}.
Remark 3.3.15. It is impossible to reach the state of knowledge J
{{}}
in the knowledge game for directed connectivity. If J,, =
{}
=
{}
from J,=
this says that the
vertex v' is impossible to reach from s' regardless of the input graph G.
Proposition 3.3.16. A monotone switching network G' has a knowledge description
if and only if it is sound.
Proof. If G' has a knowledge description then it is sound because we can only win
the knowledge game for directed connectivity if the input graph G has a path from s
to t. Conversely, given a sound monotone switching network G', set J
is a walk from s' to v' in G' whose edge labels are all in E}.
54
=
{E : there
If there is an edge with label e between vertices v' and w' in G' then for every
K E J', K U {e} E Jw/.
Obviously, K U {e} < K U {e} so this implies that
Jw, U {e} < Jv, U {e}. By a symmetrical argument, we also have that Jv,
<
Jw, U {e}.
Thus, Jv, U {e} = J, U {e}, as needed.
{{}} and Jt,
Now we just need to check that J,,
{}
{{s
-+
t}} or J
{{}}. Since G'
J,, and can be used to delete everything else in J,,. Thus J,,
is sound, for every K E Jt', K contains a path from s to t so K -{s
moves of type 2 and 3 we can transform every K in Jt' into K
can use moves of type 4 to delete all but one copy of {s
had Jt =
{}
or we are left with {{s
-
-+
t}}. Thus Jtl
{s
-
= {}.
-+
--+ t}. Using
t} and then we
t}, so either we originally
{{s
-
t}} or J,=
{}, as
D
needed.
s
t
Kj.1
K(a,b)
K~b,,)
Kja,c)
Kg,1
K, c,a)
Figure 3-7: A monotone switching network that solves directed connectivity on
V(G) = {s, a, b, c, t} together with a knowledge description of it. The label inside
each vertex gives the J for that vertex, with each line corresponding to one of its K.
By default we take J,, {{}} and J, = {{s - t}}.
3.3.4
Reduction to reachability from s
In this subsection, we prove the following theorem which shows that there is little
loss in only considering monotone switching networks G' which only make deductions
based on reachability from s.
55
Theorem 3.3.17. If (G', s', t', pi') is a sound monotone switching network, then there
is a sound monotone switching network (G', s', t', /p') such that G' accepts exactly the
same inputs as G', |V(G')| :
(n + 1)IV(G')j, and G' has a knowledge description
where for any vertex v' of G', for any K in Jv,, K
{Kv : V
C
Vred(G)} U { Kt}
Proof. We construct G' by taking n +1 copies of G' and making the s' of each copy
equal to the t' of the previous copy. We take s' for G' to be the s' of the first copy
of G' and t' for G' to be the t' of the last copy of G'. Clearly, G' accepts exactly the
same inputs as G' and we have that IV(G') < (n + 1)1V(G')|.
Now for a vertex v' we construct J, as follows. For each walk W' from s' to v' in
G'2, create a K for that walk as follows:
1. Start with the set Xo = {s} of vertices in G.
2. Let ei = vi -+ wi be the edge in G which is the label of the ith edge in W'. Take
Xi = Xi_
1
if vi 0 Xi
1
and take Xi = Xi_ 1 U {wi} if vi E Xi- 1 . Let X be the
set obtained after taking the final edge in W'.
3. Set K = Uvex\{sf{s --4 V}.
Now take Jo, to be the set of all such K.
Following similar logic as was used to prove Proposition 3.3.16, it can be verified
that this assignment of states of knowledge to vertices of G' satisfies condition 3 of
Definition 3.3.14 and that J,
Jt, =
{{}}. We just need to show Jt,
{{s --+ t}} or
{}.
To show that Jt,
{{s
-+
t}} or Jt, = {}, consider a given walk W' from s' to
t' in G' and look at how the set {Xi} changes as we go along W'. Let Y be the set
of vertices we have when we first reach the vertex t which was the t' of the ith copy
of G'. Yo
=
{s}. Since G' is sound, if t 0 Y then the portion of W' from t to t+
must have at least one edge which crosses the cut between Y and V(G) \ Y. If ek is
the first edge on this portion of W' crossing this cut, then Y 9 Xk_1
C
X
C
Yj+l-
Thus either t E Y C Y+1 or Y C Y+1. There are only n vertices except for s and t
so this implies that t E Y,+1 . Thus for all K c Jt', s -+ t c K. Using the same logic
as before, Jt,
{{s -÷ t}} or Jt,= {}, as needed.
56
Reduction to certain knowledge switching networks
3.3.5
Finally, we prove a theorem that shows that in some sense, monotone switching
networks can be reduced to certain-knowledge switching networks.
Although this
theorem is not strong enough to prove any lower size bounds, the reduction used in
this theorem is deep and will play a crucial role in Section 3.5.
nectivity together with a knowledge description and a path P' = {s' -÷ v',v'
,...
,v
, 2
v,
-+
1 ,v, 1
-+
t'}
-
Definition 3.3.18. Given a sound monotone switching network G' for directed con-
from s' to t' in G', define the certain-knowledge
switching network H'(G', P') as follows:
{{}}
First, if we do not already have that J8, =
{{}}
and Jt,
k E [0, l'], Jv, = { K, 1 ,.--
sets K,,
,
K
.
{{s
=
, Kv,}
t}}. Now let v= s' and let v',
=
J,,
=
: k c [0,l'], S
{{}}
t'. For each
for some positive integer mk and some knowledge
For each non-empty subset S of [1, Mk] let Kvs
We take V(H'(G'P')) = {
knowledge set Ks.
-+
and J,=
C
{{s
=
UjEsK,
.
take J,=
and Jt, = {{s -+ t}}, then
[1,mk], S f 0} where each w',Ss has
-+ t}} so we take s' = w,
and
in H'(G', P'). We take all possible edges which are allowed by condition 3
t'= w',/
of Definition 3.2.8.
Theorem 3.3.19. If G' is a sound monotone switching network for directed connectivity with a given knowledge description and P'
V/,_),V/,_1
{s' -+ v',v'
-+
v,--- ,v-
2
-+
_ t'} is a path from s' to t' in G', then it is possible to take a subset of
the edges of H'(G', P') and assign a direction to each edge to obtain a directed graph
Hried(G', F') for which the following is true:
1. He(G', P') consists of a directed path from s' to t' and directed cycles.
2. Every vertex in H'ed(G', P') is on a path or cycle.
3. For all vertices w',
(a) If w/,/s
74
where |SI is odd,
s' then the incoming edge for w'IS has the same label as the
edge from v_ 1 to v' in P' and its other endpoint is either of the form
57
where |T| =Sj or the form w's where S2 is obtained by adding
W',/
or deleting one element from S.
#
(b) If w',,
t' then the outgoing edge for w's has the same label as the
edge from v' to v'k
in P' and its other endpoint is either of the form
1
where |T| = |S| or the form w's where S2 is obtained by adding
w'it
or deleting one element from S.
4.
For all vertices w's where
|S|
is even,
(a) If wis 4 t' then the incoming edge for w't s has the same label as the
edge from v' to v'k
1
in P' and its other endpoint is either of the form
where S2 is obtained by adding
where |T\ = \S| or the form wits2
kS
w'Vk+lT
or deleting one element from S.
(b) If w'i' s f s' then the outgoing edge for w', s has the same label as the
edge from v-_
to v' in P' and its other endpoint is either of the form
where S 2 is obtained by adding
where |T| = |S| or the form w',
w'/
or deleting one element from S.
Proof. For all k, letting ek be the label of the edge from v' to v'+1 we apply Lemma
and Jv,k+1 U {ek}. This gives us a set Iki
3.3.10 to the states of knowledge Jv' U {ek}
k
[1, Mk],
a set Ik2
C
[1, mk+1] of equal size to Iki, a function fkl
function fk2 : [1, mk+1] \ 1 k2
k2,
-4
[1, imk] \ Ik1 -4 Iki,
and a perfect matching #k :ki
Ik2
1
C
a
such that
1. For all i C [1, nk] \ 41, Kv'f,,(i) U {ek} < Kv,'i U {ek}
2. For all j C [1, mnk+1] \
3. For all i C
'ki,
12, Kv'+lfk 2 (j) U
Kvj U {ek} - K
{ek}
10k(i)
K
U {ek}
U {ek}.
Proposition 3.3.20.
1. For all S C
Ik1,
Kv, s U {ek} = K
2. For all S C [1, m 1 ] and i
{ek} and if fkl(i)
c
4
(s) U {ek}
S\Ik1, if fkl(i)
C S then Kv's U
{ek}
58
S then Kv SU{ek}
Kv'(s\{fk1(i)}) U {ek}
KV (Su{fk1(i)})U
3. For all T C [1,m 2 ] and j e T \ Ik2, if fk2(j)
Kv/+i(TUffk 2 (j)}) Ufek}
and if fk2(j) E T then Kvi
T then K,,
TU
{ek}
T U {ek}
Kv'(T\{fk
2 (j)})U
{ek}
We now choose the edges of H ed(G', P') and assign directions to them as follows. For
each vertex w', s,
then take the edge with label ek between w', and w'V
to~ ~ ~ ~ ~
~I W" IfIIi vnte aetis
then have this
S|
is
even
to
w'
(.If
edge
go
from
w',
odd then have this
1. If S C
Ikl
to /s
tos)
edge go from w,
2. If S Z Iki then take the first i E S \
between w',s and w'/ s
and SA{fkl(i)}
=
(i)})
'kl
and take the edge with label
where SA{fkl(i)} = S U {fi(i)} if
S \ {fki(i)} if fkl(i)
fkl(i) 0
c S. Have this edge go from
S
w', S to
W/(sa{fkl(i)}) if ISI is odd and have this edge go from w'(sa{fkl(i)}) to w' S
ISI
ek
if
is even.
For each vertex w',Vk+1 T'
1. If T C Ik2 then take the edge with label ek between w',iT and w', _q. If ITI
to w'Vk~lT*. If ITI is even then have
is odd then have this edge go from w', _q(T
Vko-'(T)
this edge go from w',
+1T
to WI,
_
VkO-
2. If T Z Ik2 then take the first
WT
TA{fk
to w'"
j2
(T)'
E T\Ik2 and take the edge with label ek between
and we',(TA{fk 2 (j)}) where TA{fk 2 (j)}
2
(j)} = T\{fk 2 (j)} if fk2(j)
T U {fk2(j)} if fk2(j)
c T. Have this edge go from
T and
w',i(TA{fk 2 (j)})
odd and have this edge go from w', (TAIfA (j)}1) to w v'
if ITI is
Vk+ITvk+
2
1
IT
if
ITI is even.
Conditions 3 and 4 of Theorem 3.3.19 are now satisfied by the edges we have chosen.
All vertices have indegree one except for s' and all vertices have outdegree one except
for t'. This implies that Hred(G', P') consists of a path from s' to t' and directed
cycles and that every vertex is on a path or cycle, as needed.
59
D
c'
b/
:S)a
Kj}
K(b
Kfb)
K{c}
d
K{
K
t')
cd
K{,I
K1
K{f}'
Kja}
K{!}>
S
s-a
a
Kt')
Kjc,d}:
K{,}
b'
a-+b
b-*c
c'
c-*d
dl
d-*t
t'
Figure 3-8: This is an illustration of the ideas used in the proof of Theorem 3.3.19.
Above, we have the path P' from s' to t' in G', where the J for each vertex is
given below that vertex with each line corresponding to one of its K. Below, we
have the arrows between all of the knowledge sets from the argument used to prove
correspond to going along a bidirectional
Lemma 3.3.10. Here the functions {0}
edge. The functions {fkl} and {fk2} correspond to going along a unidirectional
edge and then going the opposite direction along a bidirectional edge. To get from
s' to t' in H ed(G', P') we have the following sequence (not shown): K.1 11} = {},
Ka,{i} = {s -+ a}, Ka'{1,2} = {s -+ a, s - b}, Ka'{2} = {s - b}, Kb'{1} = {s -+ b},
Kb/{1,2} = {s
=
{s
b, s
{s
-+
-+ c}, Ka'{2,3} = {s
a, s -+ c}, Ka'{3}
-+ c, s -+ d}, Kt'{1} = {s
C}, Ka'{1,3} =
Kd{1}i
4
b, s -4 c}, Ka'{1,2,3} = {s -
-
c}, Kb'{2}
= {s-
= {S -+
a, s
-+
b, s
-+
-+
c},
c}, Kc'{ 1} = {s
t}.
Corollary 3.3.21. If G' is a sound monotone switching network for directed connectivity with a given knowledge description and P = {s
-4
v1 , v1
-+
v 2 ,... , v1
1
-+
t}
is a path from s to t in G, then any path P' in G' from s' to t' using only the
edges of P must pass through at least one vertex a' such that Ja' /
Jt' and if
Ja' = { Ka'i, - - - , Kaim} then V = UT V(Ka'i) contains at least [lg 11 of v 1 , , 1 _Proof. This follows immediately from Theorem 3.3.19 and Lemma 3.2.32.
3.4
El
Fourier Analysis and Invariants on Monotone Switiching Networks For Directed Connectivity
Unfortunately, Theorem 3.3.19 is not sufficient to prove a superpolynomial lower size
bound on monotone switching networks solving directed connectivity.
60
To prove a
good lower size bound, more sophisticated techniques are needed. In this section,
we introduce a very different way of analyzing the problem: Fourier analysis and
invariants.
While these techniques are less direct, they allow us to give lower size bounds on
all monotone switching network for directed connectivity, not just certain knowledge
switching networks. In this section, we use Fourier analysis and invariants to prove
a quadratic lower size bound and show how more general lower size bounds can be
obtained.
3.4.1
Function descriptions of sound monotone switching networks
The following tautology is trivial yet illuminating:
For any yes/no question, the
answer is yes if and only if it is not no.
Before, we analyzed each vertex of the switching network in terms of how much
progress has been made towards showing directly that the answer to the question is
yes. Here, we will analyze each vertex of the switching network in terms of which
NO instances have been eliminated. For monotone switching networks, we only need
to consider maximal NO instances, as once these have been eliminated all other NO
instances must have been eliminated as well. For directed connectivity, the maximal
NO instances correspond to cuts. Thus, we will analyze each vertex of the switching
network in terms of which cuts have been crossed. We make this rigorous below.
Definition 3.4.1. We define an s-t cut (below we use cut for short) of V(G) to be
a partition of V(G) into subsets L(C), R(C) such that s e L(C) and t E R(C). We
say an edge v1 -s v 2 crosses C if vi C L(C) and v 2 c R(C). Let C denote the set of
all cuts C of V(G).
Definition 3.4.2. We define a function description of a monotone switching network
to be an assignment of a function hv to each vertex v' E V(G') such that
1. Each h', is a function from C to {0, 1}.
61
2. VC c C,s'(C) =1 and t'(C)
=
0.
3. If there is an edge e' e G' with label e between vertices v' and w' in G', for all
C E C such that e does not cross C, v'(C) = w'(C).
For convenience we identify each vertex v' with its associatedfunction h', i.e. we take
v'(C) = h',(C) for all C E C
Proposition 3.4.3. Any monotone switching network which has a function description is sound.
Proof. Assume that G' has a function description yet accepts some input graph G
which does not have a path from s to t. Then there is some path P' in G' from s' to
t' whose labels are all in E(G). Now let C be the cut such that L(C) = {v E V(G) :
there is a path from s to v in G}. Note that E(G) cannot have any edge crossing C
as otherwise there would be a path in G from s to some vertex in R(C). This implies
that for any two adjacent vertices v' and w' in P', v'(C) = w'(C). But then we must
have that s'(C) = t'(C), contradicting the fact that s'(C) = 1 and t'(C) = 0.
E
We now show the converse to this proposition, that every sound monotone switching network has a function description.
Definition 3.4.4. For a cut C, define the input graph G(C) to be the graph with
vertex set V(G) and edge set
E(G(C)) ={e : e does not cross C}
Definition 3.4.5. Define the reachabilityfunction descriptionfor a sound monotone
switching network G' to be the assignment of the function hv, to each vertex v' C V(G')
where
hV, : C -+ {0, 1}
hv,(C) = 1 if there is a path from s' to v' in G' whose edge labels are all in E(G(C))
and 0 otherwise.
Proposition 3.4.6. For any sound monotone switching network G', the reachability
function description is a function description of G'.
62
Proof. Consider the reachability function description for G'. For all C E C, s'(C)
=
1.
Assume that t'(C) = 1 for some C E C. If so, there must be a path P' in G' from s'
to t' such that no edge label in P' crosses C. If so, then since G(C) contains all edges
which do not cross C, all edge labels in P' are contained in E(G(C)) so G' accepts
G(C) and is thus not sound. Contradiction. Thus, t'(C) = 0 for all C.
To see that the third condition for a knowledge description holds, assume that it
does not hold. Then there is an edge e' in G' with endpoints v' and w' and a cut
C such that the label e of e' does not cross C but v'(C)
#
w'(C). Without loss of
generality, v'(C) = 1 and w'(C) = 0. But then there is a path P' from s' to v' such
that none of the labels of its edges cross C. w' is not on P' as otherwise we would
have that w'(C) = 1. But then taking P2 to be the path P' with the edge e' added
at the end, P2 is a path from s' to w' such that none of the labels of the edges of W2
cross C, so we should have w'(C)
=
D
1. Contradiction.
Remark 3.4.7. Reversibility is crucial here. If we defined a similar reachability
function description for a switching-and-rectifier network H and had an edge with
label e from vertex v to vertex w in H where e does not cross some cut C, we could
have v(C) = 0 but w(C) = 1 if there is a no path from s to v in H whose edges are
all in G(C) but there is a path from s to w in H whose edges are all in G(C).
a
s t
b
C1
t C3 C4
C2
C3
C4
Figure 3-9: Here we show how to represent all of the cuts of V(G) simultaneously
when V(G) = {s, a, b, t}. The column determines whether a is with s or t and the
row determines whether b is with s or t.
63
S-*t
K1.1
s
8 at
s 1
'1
a
b
a
-t
0
0
KK.,bb
b
b8b
t
1
1
8 1 0
t 0 0
Kjbj
Sa
tt
1t
bS 0 0
t 0 0
t
Figure 3-10: This is the certain-knowledge switching network G'({s, a, b, t}, 2) together with its certain-knowledge description and the reachability function description
for it.
Fourier analysis
3.4.2
The space of functions from C to R is a vector space which we will use Fourier analysis
to analyze. We have a dot product, a Fourier basis, and Fourier coefficients as follows:
Definition 3.4.8. Given two functions f, g : C -4 R, f . g = 2-n
cec f (C)g(C)
Proposition 3.4.9. If G' is a sound monotone switching network for directed connectivity with a given function description, then for any vertex v' E G', ||v'|| = Vv' -v' <
1.
Definition 3.4.10. Given a set of vertices V C Vre(G), define ev : C -+ R by
ev(C) = (-l)|VnL(C)I
Proposition 3.4.11. The set {ev, V
C
Vred(G)} is an orthonormal basis for the
vector space of functions from C to R.
Definition 3.4.12. Given a function f : C -+ R and a set of vertices V C Vred(G),
define
fv
= f - ev.
Proposition 3.4.13 (Fourier Decomposition and Parseval's Theorem). For any function f : C
-+
R, f = XVCVred(G) fvev and fV f =
64
VVred(G)fV
t
S -4
s
a
a
b
sat
t
s
t
s
s
-e{)
e{}
-
(efa + etb} + e{,b})
a-*t
ba
sb
t
a
sa+tb--
at
1(e(
-
etb})
t
(e(
-
t
e(.1 - e{b} + eaa
)
0
Figure 3-11: In this figure, we have the monotone switching network solving directed
connectivity on V(G) = {s, a, b, t} shown in Figure 3-1 together with the reachability
function description for it and the Fourier decomposition of each function.
Remark 3.4.14. The switching network shown in Figures 3-1 and 3-11 determines
whether or not there is a path from s to t by checking all of the cuts one by one
to determine if there is an edge in G crossing that cut. This can be generalized to
give a monotone switching network of size 2' solving directed connectivity. While the
size of such a switching network is enormous, this extreme construction is interesting
because the switching network is analyzing all of the possible paths from s to t at the
same time.
3.4.3
Invariants and a quadratic lower bound
We now introduce functions and invariants which we will use to analyze sound monotone switching networks.
Definition 3.4.15. Given a switching network G', a directed path P' from s' to t' in
G', and a set of edges E, recall that pu' is the labeling function for edges in G' and
define
AE(P)
(nd
(FI
e)
-
vtart(P', e))
e'EE(P'):p'(e')E
where v'tart(P', e') and v'nd(P', e') are the vertices in G' such that e' goes from V'tat(P', e')
to V'end(P', e') in P'.
Lemma 3.4.16. Let G' be a sound monotone switching network for directed connectivity on V(G), let G be an input graph on the set of vertices V(G), let (E1 , E2 ) be a
65
partition of E(G), and let P' be a directed path from s' to t' in G' whose edge labels
are all in E(G).
-
1. If C E C is a cut which cannot be crossed by any edge in E1 , then (A E1 (P')
AE 2 (P))(C)
-
1.
-
2. If C E C is a cut which cannot be crossed by any edge in E2, then (ZAE 1 (P')
AE 2 (P))(C)
=
-1.
Proof. For any cut C,
(ZAE 1 (P) + AE 2 (P')) (C)
-1.
To see this, note that since
all edge labels of P are in E1 U E2,
(AE 1 (P') + AE2
(P'))(C)
e')
vend (P',
=
- V'tart(P', e'))(C) =
(t
- s')(C) =
-1
e'EE(P')
Now if C E C is a cut which cannot be crossed by any edge in El, then AE1 (P') (C)
0 so
(AE 1 (P')
A (P', E2 ))(C) = -(AEI(P') + AE 2
-
(P'))(C)
1
Similarly, if C E C is a cut which cannot be crossed by any edge in E 2 , then
AE 2 (P')(C)
=
0 So
(A E1 (P)
-
AE 2
(P))(C)
(AEI(P')
AE 2 (P))(C)
We now use the fourier analysis and invariant approach to prove a quadratic lower
bound using a linear independence argument.
Theorem 3.4.17. Let V(G) be a set of vertices with distinguishedvertices s, t. Taking
n
= Vred(G)|, m(PV(G),<2) =
n + 2 and m(PV(G), 3)
Proof. The upper bounds on m(PV(G),
2)
=
(n) +
and m(PV(G),
3)
n + 2.
follow immediately from
Theorem 3.2.23. We prove the lower bounds using the following proposition:
Proposition 3.4.18. IV(G')l > dim(span{V(G')}) + 1
66
Proof. t' = 0 so IV(G')l = IV(G') \ {t'} + 1 > dim(span{V(G') \ {t'}}) + 1 =
E
dim(span{V(G')}) + 1
If P is a path of length 2 in G from s to t, then P has the form P = s
Take E= {s
-
v -- t.
v} and E2 = {v -+ t}. For any cut C,
-
1. If v C L(C) then C cannot be crossed by any edge in E1 so by Lemma 3.4.16,
(AE 1 (P')
- EE
2 (P))
1
(C)
2. If v E R(C) then C cannot be crossed by any edge in
(AE, (P')
This implies that AE 1
AE 2 (P')) (C)
-
AE 2 (P)
(P') -
=-
2
so by Lemma 3.4.16,
1
Note that AE 1 (P/)
-v}.
-
AE 2 (P') is
a linear combination of vertices of G' so if G' accepts all inputs in PV(G),2 then
e{v} E span{V(G')} for all v C Vred(G).
e{} = s' E span{V(G')} as well so by
Proposition 3.4.18, IV(G') \ {s', t'}I > n + 1 + 1
=
n + 2.
If P is a path of length 3 in G from s to t, then P has the form P = s v2
t. Take Ej={s -v
1
,v 2
v,
-- t} and E2 = {vi --+ v 2}. For any cut C,
1. If v1 ,v2 E L(C) then C cannot be crossed by any edge in E2 so by Lemma
3.4.16,
(AE, (P)
-AE
-1
2 (PT))(C)
2. If vi E L(C), v 2 c R(C) then C cannot be crossed by any edge in El so by
Lemma 3.4.16,
(ZAE 1 (P) -
AE 2 (PT))(C)
=
1
3. If v 2 E L(C), v1 E R(C) then C cannot be crossed by any edge in E2 so by
Lemma 3.4.16,
(AE 1 (P)
-
IAE 2 (PT))(C)
=
1
67
4. If v 1 , v 2 E R(C) then C cannot be crossed by any edge in E2 so by Lemma
3.4.16,
(L\E 1 (P')
-
AE 2 (PT(C)
=-
By direct computation, this implies that (AE 1 (P
AE 2 (e
V1}+e{V2
e{ 1,V 2 }). Thus, we have that if G' accepts all inputs in PV(G),<; 3 , since we already have
that ell = t' E span{V(G')} and e{v} E span{V(G')} for all v E Vred(G), we also
have that for all
E Vred(G), e{V 1 ,V 2} E span{V(G')}.
Vi, V 2
By Proposition 3.4.18,
IV(G')I > (n) + n + 1 + 1 = (n) + n + 2, as needed
(2 a -+ t
s at
11
s at
e{)
-
s at
00
0
E(P) = {s -+ a, a -+t}
El = {s -+ a}, E2 = {a-+t}
s at
-
P/s
A(P', E1 ) -
b -+t
Ssat
bs
b
b
e)
A(P', E2) = -eta}
a-+ b
s-*a
P/
E]
(e{} -
e{.}) 1(e{
-
sats
0bS
e{a) -
E(P) = {s -+ a, a -+ b, b -+ t}
E= {s -+ a, b-+ t},E2 ={a -+b}
at
0
efb} + e{a,b})
bs
s at
1 1
A(P', E1 ) - A(P', E2) = !(-e}
- e{a) + e{b) - e(.,bl)
Figure 3-12: This figure demonstrates the reasoning used to prove Theorem 3.4.17.
Edges whose label is in E1 are colored blue and edges whose label is in E2 are colored
red.
3.4.4
General lower bounds
Unfortunately, the linear independence argument breaks down for longer paths. The
problem is that for paths P of length greater than 3, we can no longer find a non-trivial
partition (E1 , E2 ) of the edges of P such that AE1 (P') and AE2 (P') are invariant over
all sound monotone switching networks G' and paths P' from s' to t' whose edge
labels are all in P. Thus, for longer paths we need a more sophisticated approach.
68
For this approach, we partition the edges of E(G) into several sets {E} and look
at the dot product of vertices v' E V(G') with a carefully chosen set of functions
These functions are chosen so that for all i,
gG,Ei
-s' = 1 and whenever there
is an edge between vertices v' and w' in G' with label e E Ej,
V'
gG,Ei
= WI
9G,E
-
{gG,Ei}.
We now imagine the following game. There are several players, one for each set of
vertices E. At each vertex v', the ith player has value gG,Ei -v'.The players are trying
to go from all having value 1 at s' to all having value 0 at t' in a sound monotone
switching network G' while only taking edges in G' whose labels are in E(G). While
doing this, they are trying to keep their values as close to each other as possible.
However, since every edge the players take has label in Ej for some i, for any
given move there will be some player whose value remains fixed. This means that
their values cannot all change at the same time so there will be some point where
there is a significant discrepency between their values. This corresponds to a vertex
v' and i, j such that v' - (gG,Ej
9G,Ei)
-
is non-negligible, which we can use to prove
our lower bounds. We make this intuition rigorous below.
Definition 3.4.19.
1. We say a function g : C -- R is e-invariant for some pos-
sible edge e if g(C) = 0 for any cut C which can be crossed by e.
C --+ R is E-invariant for a set of edges E if g is e-
2. We say a function g
invariantfor every e
c
E.
Proposition 3.4.20. If P' is a path from s' to t' in G' and E is a set of edges, then
if g is an E-invariantfunction, zAE(P')
g
=
0
Theorem 3.4.21. Let I = {Gj} be a set of input graphs on V(G) all of which
contain a path from s to t and let M be a constant. If for each j we have a partition (E1 , ... , Ekj) of the edges of Gj, functions gGj,E 1 '' ,gGj,Eq,
constants {zj} and { Mj} such that
1. For all j, YgGj,Ei 9 is Eij -invariantfor i C {1, qj}
2. For all j and all i E [1, q]
,
gGj,Eij - e{} = Zj
69
> 0
C -
R, and
#
(gGjE 1i3
4. For all i, j, ||gGj,Eij
j2 and all i 1 ,i 2
,
3. For all J1, j2 where j1
gGjEj ) * (YGj 2 ,Ei 2 j 2
-
|
- YGj,E1j
-
gG 32 ,E1
0
2
<; M
Then
m(I)2
j2
Proof.
Lemma 3.4.22. Let G be an input graph containing a path from s to t. If we have
a partition (E1 ,--- , Eq) of the edges of G and functions gG,Ei such that gG,Ei
is
Ei-
invariant for all i then for any sound monotone switching network G', for any path
P' in G' from s' to t' whose edge labels are all in E(G),
q
AE(G)\E (P')
(G,Ei
-
(q
9G,E 1 )
-
2)(gG,E 1 -e})
9G,Ei
-
i=2
e{}
i=2
Proof. Let P' be a walk from s' to t' in G' whose edge labels are all in E(G). Since
9G,Ei
is Ei-invariant,
Vi, AE(G)\Ei(P)
9G,Ei
=
AE(G)(P)
gG,Ei =
9G,Ei (t'
-
s') = -gG,Ei
e{}
Since gG,E 1 is E1 -invariant,
k
q
S
AE(G)\EjP)
gG,E1
= ((q - 2) 5AE,(Y) + (q
i=2
=
((q
=
(q
-
'AE1
' gG,E 1
i=2
- 2)
-
AE(P))
2)(gG,E 1
(t'
-
g G,E 1
s')) =
-(q
- 2 )(gG,E 1 '
e{})
Putting all of these equations together gives the needed equality.
70
Dl
Corollary 3.4.23. Let G be an input graph containing a path from s to t. If we
have a partition (E1 ,--- , Eq) of the edges of G and functions gG,Ej such that gG,Ei
Zs
Ei-invariantfor all i and YG,Ei e{} is the same for all i, then for any sound monotone
switching network G' which accepts G, taking z = gG,E1 - e{},
E
I|V' - (gG,Ei - gG,E 1 ) >Z
i=2 V'eV(P')
In particular, there must be some i E [2, q] such that
E
v'EV(G')
|v'
(gG,Ei -g G,E1 )
>
q-i
Proof. This follows immediately from Lemma 3.4.22 and the fact that for all i,
|v' - (gG,E,
-
9G,E 1) I
AE(G)\Ei(F') - (G,Ei
-
9G,E1
)
E
v'EV(P')
because AE(G)\Ej (P') is a linear combination of the vertices in P' where each vertex
has coefficient -1,
0, or 1..
We now prove Theorem 3.4.21 using Corollary 3.4.23, an orthogonality argument,
and the Cauchy-Schwarz inequality.
Proposition 3.4.24. If {gj} is a collection of nonzero orthogonal functions from C
to R, then for any function h : C -+ R where
Ih|
=
h h < 1,
Z
(g
1 h) 2
< 1
Proof. If {gj} is a collection of nonzero orthogonal functions, we can extend it to
an orthogonal basis {gj} U {f } for the vector space of functions from C to R. Now
h=E
gf+f , so
1 > h - h = J7jjl9 2 +
i.h
Z(fh)
as needed.
>,,
Now let G' be a sound monotone switching network which accepts all of the inputs
in I = {C3 }. By Corollary 3.4.23, Vj, ]i :Zv'V(G') (gG,EZ
Using the Cauchy Schwarz inequality (EZ, f(v')g(v'))
71
2
< E,
- gGEi 3 )
' > J
f(v')
2
E,
g(v')
2
with
f
(v') = 1 and g(v')
=
(gG,,Ejj
gGj,Elj) - v',
-
((gGj,Eij -
Vj,
we have that
gGj,E 1 j)
')
((qj-1Z,)2
2
v'cV(G')
This implies that
z((gG,Eij
-
i
v'cV(G')
H9GJEi
2
YGj,E 1 ) v')
-gG 3 ,E
(z
q(G-1
>
--
2
2V
)2
(G')|
However, by Proposition 3.4.24,
zz((gqGj,Ej
G,E 1
I G,Eij- - gGj,Elj)
j
v'CV(G')
1 r2 2
v)
HGJEiJ~-YGj,E1
v'<V(G')
(
Putting these inequalities together, IV(G')I >
Z3
)2
2
,U1(G)
so
IV(G') I>
M3 i
as needed.
3.4.5
0
Conditions for a good set of functions
The simplest way to use Theorem 3.4.21 is to take one input graph G, find a set of
functions {gG,E} and then obtain the other input graphs and sets of functions by
symmetry. We now give conditions which are sufficient to ensure that we can do this
and deduce that if such sets of functions exist for paths P of arbitrary length then any
monotone switching network solving directed connectivity must have superpolynomial
size.
Theorem 3.4.25. Let V(G) = {s, t, V 1i, - , Vm}.
If there is a partition E1 , - - - , Eq
of the edges of G, functions {gG,Ej}, a value z > 0, a value M, and a value r K m
such that:
1. gG,Ei is E-rinvariantfor i E [1, q]
2. For all i G [1, q] and all V C Vred(G) with |V| < r, (gG,Ei
3.
gG,E1
e{} = Z
72
-
gG,E 1 )
- eV
=
0
4. For all i e [1, q], |gG,E - 9G,E
1
< M
then for all n > 2m 2 , if W is a set of vertices such that V (G) 9 W and |W\{s, t}I = n
then letting H be the input graph with V(H) = W and E(H) = E(G), m(H) >
2
(q-1l)M
Proof. We first show that we can add additional isolated vertices to the input graph G
while still keeping the same functions (expressed in terms of their Fourier coefficients).
Proposition 3.4.26. For any U, V C
symmetric difference function, i.e. VAU
Vred(G),
=
euev = evAu where A is the set-
(U U V) \ (U n V)
Proposition 3.4.27. For all v, w E Ved(G), for all C E C,
1. (e{} + e{ W)(C)
=
2 if w c R(C) and 0 if w E L(C).
2. (e{} - e{v)(C) = 2 if v E L(C) and 0 if v E R(C).
3. ((e{} - e{fv)(e{} + e{w))(C) = 4 if v G L(C) and w G R(C) and 0 otherwise.
Corollary 3.4.28.
+
1. If e = s -+ w for some w E Vred(G) then g is e-invariant if and only if (e{}
efw})g = 0. Equivalently, g is e-invariant if and only if .vu{w} = -- $v whenever
w c V.
t for some v E Vred(G) then g is e-invariant if and only if (e{}
-
-
2. If e = v
e{v})g = 0. Equivalently, g is e-invariant if and only if .vu{v}
V.
3. If e = v -+ w for some v, w
E Vred(G)
then g is e-invariant if and only if (e{}
-
v
- =v whenever
e{v})(e{} + e{w})g = 0. Equivalently, g is e-invariant if and only if jvu{v,w}
-jvu{v}
+
Yvu{w}
We now write
+ jv whenever v, w $ V.
ZVCVed(G)
9G,E
graph H and take gH,Ej
=
=
VCVred(G)
civev. By Corollary 3.4.28 if we have the input
CiveV
then all conditions of Theorem 3.4.25 are
still satisfied by {gH,Ei}. Moreover, for all i and all V Z Vred(G), gH,Ei ' ev = 0.
We now take our set of input graphs I = {Hj} so that
73
1. Each H is obtained from H by applying some permutation oa to the vertices
W \ {s, t}.
2. For all distinct ji and j2,
Qj1
(Vred(G))
n ogi (Vred(G)) < r
By Corollary 3.2.36, we can take at least (,)r such graphs.
Proposition 3.4.29. If we take Eij = oj(Ei ) and gHj,Eij
Vrd(G)
=
CiVeoj(v) then
1. For all j, gHj,Eij is Ejj-invariantfor i c [1, q]
2. For all j and all i G [1,q]
3. For all i, j, (gGj,Eij
4.
For all i, j, I|gGj,Eij
-
,
e{} = z
gGj,Ei
gG,,E 1 )
0 whenever VI <
' ev
- gGj,E1IH
r or V
Z Jj(Vred(G))
M
Proof. This follows immediately from the properties of the functions {gH,Ej} and the
1:
fact that for all i, j and all V, gHj,E, - eo. (y) = 9H,E, - ey
Now note that since (gHj,E- gHj,Eij)-eV= 0 whenever IVI < r or V
and for all distinct ji and
il, i2, J, j2
where j
1
j2,
i(Vred(G))
n Uj(Vrecj(G))
Z
O (Vred(G))
< r, we have thatfor all
J2,
(gHj1,Eijj
--
YHjl ,Ei3 )
9H 2
(g2,Ei2j2
,E132 )
0
Applying Corollary 3.4.21,
z
)
m(H) > m(I) >
( Z 1)2
ZM
(q
2
-
1)M 2m
D
as needed.
...
-+ vi_ 1
-+
1,
t} and let P be the path s -+ v1
-
Corollary 3.4.30. Take V(P) = {s,v1, - - - , vi
t. If n > 2(1- 1)2 and we can find a partition {E 1 , -- - , Eq} of the edges
of P, functions {gPEj}, values z, M, and a value r < 1 such that:
1. gP,E is Ei-invariantfor i G [1, q]
74
2. (gP,
-
gPE1 )
3. gPE1 e{}
-ev = 0 for all i C [1,q] and all V C Vred(G) with |VI < r
=Z>
4. For all i, |YgP,E
0
M
~- 9P,E1
'
then m(PV(G),l) > (q-i~)M 2(1-1)
El
Proof. This follows immediately from Theorem 3.4.25.
Example 3.4.31. For 1 = 2 and r
=
1 we can take P = s
--
v 1 }, E2
-
e{ 1}), and
(e{
9P,E2 -
=
{v 1
-+
t}, gP,E1
(e{}
=
{s
-+
+ e{v 1 }) This gives
gPE1 = -ev 1 . Using Proposition3.4.28 it can be verified directly that YP,E
E-invariantfor i E {1, 2}. 1lgp
for all n > 2, m(P,1) >
{s
=
1 and z = gP,E1 - e{} =
v 1 , v 2 -+ t}, E 2
=
3 and r
{vI
-+
=
so by Theorem 3.4.25,
2 we can take P = s
v 2 }, gPE2
=
4(e{}
-
v 1 --
V2
t,
-+
e{V 1} I e{V 2 } -F 3e{ v,V 2 }), and
-
e{- 1 } - e{v 2 } ~ e{i,V 2 }) . This gives 9P,E2
,-
Zs
j
Example 3.4.32. For 1
E=
=
YP,E 2
v, -+ t, El =
-
9P,
=
-e{V
1
,V2}.
Using
Proposition 3.4.28 it can be verified directly that gp,E is El-invariantfor i c {1, 2}.
gp
=
1 and z
9PE1 - e{} =
4 so by
Theorem 3.4.25, for all n > 8, mPf,2) >
8
Remark 3.4.33. These bounds are around the square root of the bounds obtained
from the linear independence argument. This square root comes from the Cauchy-
Schwarz inequality and so far we have not found a way to avoid having this square
root. Nevertheless, getting a lower bound for m(PV(G)) which is around m(PV(G))c
for some c > 0 is good enough for our purposes and unlike the linear independence
argument, we can use these techniques for longer paths.
3.5
A Superpolynomial Lower Bound
While the Fourier analysis and invariant approach of Section 3.4 is powerful, we need
to actually find suitable functions {gpEI}. There are several possibilities for how we
could do this. One possibility is to look directly at the values gPE (C) for all C E C.
75
However, this gives us very little control over the Fourier coefficients of each gPEi. A
second possibility is to work directly with the Fourier coefficients of each gp,E. This
approach is viable, but it would involve analyzing how to satisfy many equations for
Ei-invariance simultaneously. Here we take a third approach. We look at the dot
products of each 9PE, with the vertices of the certain knowledge switching network
It turns out that all of the conditions of Corollary 3.4.30 correspond to
G'(V(G)).
simple conditions on these dot products. Furthermore, we have complete freedom in
choosing these dot products, which enables us to construct suitable {gp,E}, and thus
prove the following theorem.
Theorem 3.5.1. For all 1 > 2, if we have V(G) = {s,t,v1,--- ,v 1 1} and let P be
v1 ---
...- 4 v1_1 -+t
path
l| ands taking the, partition
then takingthe
r = [lg
Ej =
{v-_
vi} of the edges of E (where vo = s and v, = t), we can find functions
{gP,EjI such that:
1. 9P,E, is Ei-invariantfor all i
2. (gP,E
-
9PE1 )
ev for all i and all V
C
Vred(G) with
|VI < r
3. YP,E l e{} > 0
Combined with Corollary 3.4.30, this immediately proves the following theorem,
which implies a superpolynomial lower bound on monotone switching networks solving directed connectivity on n vertices.
Theorem 3.5.2. For any integer 1 > 2, there is a constant ci such that if V(G) is a set
of vertices with distinguishedvertices s, t and n =|Vred(G)
3.5.1
then m(Pv(G),l) > cin
FIg 11
2
From certain knowledge decriptions and knowledge descriptions to function descriptions
For the proof of Theorem 3.5.1, we adapt results from Sections 3.2 and 3.3 to the
Fourier analysis and invariants approach.
We begin by taking certain knowledge
descriptions and knowledge descriptions and giving a corresponding function description.
76
Definition 3.5.3. For a given knowledge set K, define the function
K :C
{O, 1}
K(C) = 1 if there is no edge in K which crosses C and 0 otherwise.
Proposition 3.5.4. If we can get from K1 to K 2 in the certain knowledge game using
only the knowledge that some edge e is in G and e does not cross some cut C then
K2(C) = K1(C).
Proof. This follows immediately from the fact that if e does not cross C, then for any
knowledge set K, no individual move on K in the certain knowledge game which can
be done with only the knowledge that e is in G changes the value of K(C).
F]
Corollary 3.5.5. If a monotone switching network G' has a certain-knowledge description {K,,} where each vertex v' is assigned the knowledge set Kv, then if we
assign each vertex v' the function Kv,, this is a function description of G'.
Definition 3.5.6. For a given state of knowledge J, define the function
J: C
{0, 1}
J(C)
0 if there is no K in J such that K(C) = 1 and 1 otherwise.
Proposition 3.5.7. If we can get from J1 to J2 in the knowledge game for directed
connectivity using only the knowledge that some edge e is in G and e does not cross
some cut C then J2 (C) = J1 (C).
Proof. This follows immediately from the fact that if e does not cross C, then for any
state of knowledge J, no individual move on J in the knowledge game for directed
connectivity which can be done with only the knowledge that e is in G changes the
l
value of J(C).
Corollary 3.5.8. If a monotone switching network G' has a knowledge description
where each vertex v' is assigned the state of knowledge Jv, then if we assign each v'
the function Jr', we have a function description of G'.
Remark 3.5.9. If we take the knowledge description given in the proof of Proposition
3.3.16 and take the correspondingfunction description we will obtain the reachability
function description.
77
3.5.2
A criterion for E-invariance
Now that we have translated certain knolwedge descriptions and knowledge descriptions into function descriptions, we give and prove the following criterion for Einvariance.
Theorem 3.5.10. If g is a function from C to R and E is a set of edges between
vertices in V(G) then g is E-invariant if and only if g - v' = g - v' whenever v', v'
are vertices of G'(V(G)) such there is an edge between v' and v' in G'(V(G)) whose
edge label is in E.
Before proving Theorem 3.5.10, we need a few preliminary results.
Lemma 3.5.11. If g is a function from C to R and E is a set of edges between
vertices in V(G) then g is E-invariant if and only if g -J1 = g -J2 whenever J1 , J2 are
states of knowledge such that J1 U {e} = J2 U {e} for some e E E and all knowledge
sets in J1 and J2 are either equal to Kt, or have the form Kv where V C Vred(G).
Proof. The only if direction follows immediately from Proposition 3.5.7. To prove the
if direction, assume that g - J1 = g - J2 whenever J1 , J2 are states of knowledge for
V(G) such that all knowledge sets in J1 and J2 are either equal to Kt, or have the
form Ky where V
C
Vred(G) and there is an e e E for which it is possible to go from
J1 to J2 in the knowledge game for directed connectivity using only the knowledge
that e is in G. Given a cut C which can be crossed by an edge e c E, if R(C) # 0,
take J, = {KL(c)} and J 2 = UvR(C){KL(c)u{v}}. If R(C) = 0, take J=
J2
= Jt'. We have that J2 (C)
=
{KL(c)},
0, Ji(C) = 1, all knowledge sets in J1 and J2 are
either equal to Kt, or have the form Ky where V C Vre(G), and J U {e} = J2 U {e}.
Now consider any cut C2 C C.
If L(C) n R(C2 ) is nonempty then Ji(C2 )
J2 (C2 ) = 0. If R(C2 ) C R(C) then J1 (C2 )
Ji(C2 )
=
J2 (C2 ).
J 2 (C2 )
together, 0 = g - J2- g -
-
=
J2 (C 2 )
-
.
By our assumption, g - J, = g - J2
= 1. Thus, if C2 # C then
J1 (C2 ) # 0 if and only if C2 = C. Putting everything
=
24(J2 (C) - Ji(C))g(C) so g(C) = 0. Thus, g(C) = 0
for any C which can be crossed by an edge e c E, as needed.
78
E
Lemma 3.5.12. Let J1 = {K 11 ),.
{e}
-
J2 U
, Kim1 }
and let J2
=
{K2 1 ,- -
,
K2m 2 }- If J1 U
{e} for some possible edge e then we may write J2 - J1 as a sum of terms of
the form K 2 -K
1
where K1U{e} = K 2 U{e} and both K1 and K2 are either of the form
U3-sK1j where S C [1,mi], S # 0 or the form
UkGETK2k
where T C [1,m 2 ],T
# 0.
Proof. We first give a proposition which allows us to express J2 - J in terms of these
knowledge sets.
Proposition 3.5.13. If J = {K 1 , K 2 , J(C) =
E
,
Km} where m # 0, then for any C E C,
(-1)Is l+((UicsKi)(C))
Proof. This is just the inclusion-exclusion principle. Note that J(C) = 0 if Ki(C) = 0
for every i and 1 otherwise. If Ki(C) = 1 for some i, then we can add or remove i
from S without affecting (UiesKi)(C).
But then all terms in the sum on the right
cancel except Ki(C), which is 1.
If Ki(C) = 0 for all i, then for all non-empty subsets S of [1, m], (UiesKi)(C) = 0.
Choosing an arbitrary i, we can add or remove i from S without affecting (UicsKi)(C),
so we again have that everything cancels except Ki(C), which is 0.
E
Lemma 3.5.12 now follows directly from Theorem 3.3.19. We can easily create a
sound monotone switching G' which has a path P' from s' to t' such that there are
vertices v', V+
vo+
1
1
oni P' with
J,ij+1= J1 and J,
i+1
= J2 and there is an edge e' from v' to
with label e. By Proposition 3.5.13 we have that
J2 - J1=
(-1)Il+1((UjeTK 2j)(C))
TC[1,m 21,
(-1)sl+1((uicsKii)(C))
SC[1,m1],
S/4J
TO0
By Theorem 3.3.19, if we let Ee, be the set of directed edges corresponding to e' in
Hr' (G', P'),7
A(e')
ke /EE.
=
E
(--1)'S+1((UiSsKi)(C
(-1)+l(UJETK 2 )(C)) SC[,m1],
TC[1,M2]
Soo
T=AO
79
where if e' goes from w' to w' in Hrd(G', P') then A(e')
Thus, J2
-
J=
w-
w.
ZCE., A(ek) and the result follows.
e2
1e
,
I
V3
e'
/q
(G
' S'
s a K eb, 0
t2
8Kb~
I
et
-*
/
e
e, }
V/I
V/
Hj
K{} e2
V1
P1
Figure 3-13: In this figure, we illustrate the ideas used in the proof of Lemma 3.5.12.
It can be verified that A(e'1) = A(e'4) + A(e' 0 ), A(e) = A(e's) + A(e'7) + A(e'g), and
e(ee)
A(e') =
+ Z(e').
With these results, we can now prove Theorem 3.5.10.
Theorem 3.5.10. If g is a function from C to
and E is a set of edges between
vertices in V(G) then g is F-invariant if and only if g -
=
g -v; whenever vi, v;
are vertices of G'e(V(G)) such there is an edge between vi and v6 in G'e(V (G)) whose
edge label is in F.
Proof. The only if direction follows immediately from Proposition 3.5.4. To prove
the if direction, assume that g(C)
0 for some cut C which can be crossed by an
edge e E E. By Proposition 3.5.11, there exist states of knowledge J1 , J2 for V(G)
such that J U {e}
2
U 1J
{e} and all knowledge sets in J1 and J2 are either equal
to K or have the form K where V e Vre(G), and g- J1
Lemma 3.5.12, we may write J2
K1 U {e}
g- J2 . But then by
as
fJi a sum of terms of the form K 2
-
K 1 where
K2 U {e} and both K 1 and K 2 are either Ke' or of the form Ky where
V G Vred(G). Since g -(J2
that g- (K2
-
#
-
K1)
-
Ji) # 0, there must be at least one such pair K1 , K 2 such
0. But then taking vi and v to be the corresponding vertices
in G'e(V(G)), there is an edge with label e between v and v and g v($g - v, as
needed. T
80
3.5.3
Choosing Fourier coefficients via dot products
In this subsection, we prove the following theorem relating the Fouerier coefficients
{v} of a function g to the dot products of g with the functions {Kv}
Theorem 3.5.14. For any set of values {av : V C
function g : C -+ R such that for all V C
r, v = 0 for all V such that
|V| <
Vred(G),
Vred(G)},
g -Kv
there is a unique
av. Furthermore, for any
r if and only if av =g - Kv = 0 for all V such
that |V| < r.
Proof.
Proposition 3.5.15. For any V
Vred(G), Ky =
2v
EZ
v (-1)Iuleu
Proof. Note that Kv(C) = 1 if V C L(C) and 0 otherwise. Now for all cuts C E C,
E(-1)Ue (C) = I E (-1)IUI(-1)UnL(C)I
I
S
Ucv
Ucv
If V Z L(C) then all terms will cancel so
2v
EUv (-1)UWeu(C) = 0. If V C L(C)
then
(-1)IUI(-1)j nL(C)1
Ucy
This implies that Kv
1
Ucy
EUCV (-1)UIeu, as needed.
C
Corollary 3.5.16.
1. For all V C
Vred(G),
ev - Kv
# 0
2. For all subsets U, V of Vred(G), if U g V then eu - Kv = 0.
To see the first part of Theorem 3.5.14, pick an ordering {V} of the subsets
V C Vred(G) such that if i <
j
then V/
(t V4. We now pick the Fourier coefficients
.v, in increasing order of i. By statement 2 of Corollary 3.5.16, for all subsets U, V
of Vred(G), if U t V then eu Kv = 0. This means that for each i, once we pick
this determines the value of g Kv
=
av, as for any
j
> i, V9
v,,
Vi so ev, -KV, = 0. By
statement 1 of Corollary 3.5.16, for all i, ev, -Kv, $ 0. This means that we always have
a unique choice for each coefficient
v, which gives g - Kv, = av. Putting everything
81
together, there is a unique function g : C --+ R such that for all V C Vred(G),
g - Ky = av, as needed.
Now we just need to show that if av
v
#
=
0 for all V such that IVI < r then
0 for all V such that IVI < r. To see this, assume it is false. Take a minimal
=
subset V of Vred(G) such that v
v
g -Kv
#
0 and au = g - Ku = 0 for all U C V. Then
0, U = 0 for all U C V, and g - KU = 0 for all U C V. However, by Corollary
3.5.16, if v f 0 and 'u = 0 for all U C V then g- Kv
3.5.4
#
0. Contradiction.
D
Proof of Theorem 3.5.1
We are now ready to construct the functions
{gP,Ei}
and prove Theorem 3.5.1, which
we recall below for convenience.
Theorem 3.5.1. For all 1 > 2, if we have V(G) = {s, t, v 1 ,-- ,vi
1
} and let P be
the path s -
v - --
Ei = {vi_
vi} of the edges of E (where vo = s and v, = t), we can find functions
1
-
->
vi_1
-+ t then taking r =lg 1] and taking the partition
{gP,Ei} such that:
1. gPEi is Ei-invariantfor all i
2. (gPEi - 9P,E1 ) ey for all i and all V C Vred(G) with
|VI
< r
3. gPE1 e{} > 0
Proof. Note that by Theorem 3.5.10 and Theorem 3.5.14, the three conditions of
Theorem 3.5.1 are equivalent to the following three conditions:
1. For all i, gP,E 'U
9PEi vVfor
any vertices u', v' of G' (V(G)) which have an
edge between them whose label is ei.
2. gPEi KV
=
gPE1 - Kv for all i and V C Vred(G) with IVI < r = [lg l
3. gP,E1 ' KfI > 0
By Theorem 3.5.14, we can choose the values {gp,Ei - Ky : V C Vred(G)} freely, so it
is sufficient to give a function b' : V(G'(V(G))) x {[1, l]} -4 R such that
82
1. If there is an edge between vertices u' and v' whose label is in E then b'(u', i)=
b'(v', i)
2. b'(v', i) = b'(v', 1) for all i whenever K E {KV : V C
Vred(G),
V| < r} U {Kt'}
3. b'(s', 1) = 1 and b'(t', 1) = 0
We choose the values {b'(v', i)} by looking at connected components of certain graphs
which are closely related to V(G'(V(G)). Let H' be the graph with
1. V(H')
{v' E V(G'(V(G))) : v' has knowledge set K c {K
: V C Ved(G),
IVI < r} U {K}}
2. E(H') = {(U', v') : u', v' E V(H'), there is an edge between u' and v' whose
label is in E(P)}
For each i let Hj be the graph with
1. V(Hj) = V(G'(V(G)))
2. E(Hl) = E(H) U {e' E E (G' (V(G))) :up'(e')
E}
Proposition 3.5.17. If u',v' E V(H') and u' and v' are in the same connected
component of Hj for some i then u' and v' are in the same connected component of
H'.
Proof. Assume that we have u' and v' which are in different components of H' but
are in the same component of Hj for some i. If so, choose u' and v' to minimize
the length of the shortest path in Hj from u' to v'. Note that there cannot be any
W' E V(H') on this path, as otherwise w' is either in a different component of H'
than u' in which case we could have taken the shorter path from u' to w' instead or
w' is in a different component of H' than v' in which case we could have taken the
shorter path from w' to v' instead. Thus all edges of the path between u' and v' in
Hj are not edges of H' and thus must have label ej. But then since G'(V(G)) has all
allowable edges, there must be an edge between u' and v' with label ej so u' and v'
are actually in the same connected component of H'. Contradiction.
83
El
This proposition implies that we may first choose any set of values {b'(v')} such that
b'(u') = b'(v') whenever u' and v' are in the same connected component of H' and then
choose any set of values {b'(v', i)} such that if v' E V(H') then b'(v', i)
all i and b'(u', i)
=
b'(v') for
b'(v', i) whenever u' and v' are in the same connected component
of Hj. One way to do this is to take b'(v', i)
=
b'(u') whenever v' is in the same
connected component of Hi as u' for some u' E V(H') and take b'(v', i)
v' is not in the same connected component as any u'
E V(H').
=
0 whenever
This is guaranteed to
satisfy the first two conditions of Theorem 3.5.1.
For the third condition, we need to check that s' and t' are in different connected
components of H' so that we may choose b'(s') = 1 and b'(t') = 0. To check this,
assume that s' and t' are in the same connected component of H'. Then there is a path
from s' to t' in H'. However, this is impossible by Lemma 3.2.32. Contradiction.
D
Remark 3.5.18. In choosing the values b'(v') for v' E H', we are essentially picking
the Fourier coefficients v : V < r for a "base function" g which we then extend to
an Ei-invariantfunction gi for every i. The crucial idea is that if we only look at the
Fourier coefficients (gi)v for |VI < r, all of the gi look identical to g and thus look
identical to each other.
3.6
An ONQ(") lower size bound
In this section, we prove an nn(lgn) lower size bound on monotone switching networks solving directed connectivity by explicitly finding the functions {gPEj
given
by Theorem 3.5.1 and then modifying them by "cutting off" high Fourier coefficients.
Theorem 3.6.1.
m(Pv(G),l) >
n
1)2)
64(l -_1
1 1 n_3
-
m(PV(G)) >
_n16
FlT
2
4
Proof. The first step in proving this lower bound is to gain a better understanding of
the functions given by Theorem 3.5.1.
84
Definition 3.6.2. For all V C
\ {s})
1. gv(C) = 0 if (L(C)
2. gv(C)
Vred(G),
define the function gv : C -> R so that
V
if (L(C) \ {s}) C V
= 2 n(--)IV\L(C)l
The most important property of these functions is as follows.
Lemma 3.6.3. If V1 , V2 C Vred(G), Kv1 - gV2
=
1 if V
=
V2 and 0 otherwise.
Proof. We have that Ky(C) = 1 if V C L(C) and Kv(C) = 0 otherwise. Now note
that
Ky, - gV2
=
(-1)IV2\L(C)l
CEC:V C(L(C)\{s})CV2
Ths implies that K, -gv2 = 1 if V = V2 and 0 otherwise, as needed.
C
We can now construct the functions {gpE} in terms of the functions {gv} and
analyze their Fourier coefficients.
Lemma 3.6.4. If {gp,I are the functions given by Theorem 3.5.1 then we have that
1.
gPPE
2.
YP,Ei
=VCVred(G) b(V,
-ev =
EVc
i)gv
b(U, i)(gu . ev) = Euv b(U, i)(-2)UJ
Proof. The first statement follows from the definition of b(V, i) and Lemma 3.6.3. For
the second statement, we use the following proposition.
Proposition 3.6.5. For all U, V C Vred(G), gu - ev = (- 2 )uI if U C V and 0
otherwise
Proof.
gu * ey
(-1)U\L(C) (_,IlvnL(C)l
=
CEC:(L(C)\{s})CV
If there is some i E U \ V, then shifting i from L(C) to R(C) or vice versa changes
(-i)U\L(C)(_))VnL(C)
by a factor of -1.
Thus, everything cancels and we have
gU * ey =
Es (-1)U\L(C)(-1)IvnL(C)
CeC:(L(C)\{s})9U
85
-
gu ev = 0. If U C V then
E
CEC:(L(C)\{s})CU
(_1)IUI
= (- 2 )IUI
F
The completes the proof of Lemma 3.6.4
El
Corollary 3.6.6. If we take each b(V, i) to be 0 or 1 when choosing the functions
} then for all V C
|P,Ei - ev
Vred(G),
< 2 2v
If we take the functions {gP,Ei} given by Theorem 3.5.1 directly, then
YP,E 1
YgPE
-
{gP,Ei
may be very large. The key observation is that as shown below using Corollary
3.4.28, we can cut off all of the Fourier coefficients ev where IVI > r
Lemma 3.6.7. Taking r = [lg
11,
[lg 1].
there exist functions gi such that
1. For all i, gi is Ei-invariant.
2. For all i and all V such that |V| < r, (gi - gi) -ev =0
3. g 1 - ell = 1
||gi - gjiH < (1 - I)F191--1 2 41g 1+2
4. For all i,
Proof. We repeat Corollary 3.4.28 here for convenience.
Corollary 3.4.28.
=
s
-*
w for some w E Vred(G) then g is e-invariant if and only if (e{}
+
1. If e
elw})g = 0. Equivalently, g is e-invariant if and only if
.vu{w}
= -jv whenever
w V V.
-
2. If e = v -+ t for some v E Vred(G) then g is e-invariant if and only if (e{}
efv})g = 0. Equivalently, g is e-invariant if and only if
.vu{v}
= .v whenever
v V V.
3. If e = v -+ w for some v, W G Vred(G) then g is e-invariant if and only if
(e{} - e{ })(e{} + e{w})g
jvu{v,w}
= -gvu{v}
+
=
0. Equivalently, g is e-invariant if and only if
Yvu{w}
+ .v whenever v, w $ V.
Definition 3.6.8. Define the functions {gi} so that
86
1. gi eV = gPE - eVifV < r
2. gi - ev = 0 if |V| > r
3. If i = 1 (so that E= {s
-+
v1}), and |VI =r then
(a) If vi E V then g- ev = -gi - ev\{v1 } if v1 E V
(b) If v1 V V then g -ev = 0
4. If i
= 1 (so that E = {v1 _ 1
(a) If vi_
1
C V then gi
(b) If vi_1
(a) If vi_ 1 , vi
(c) If vi
c
t}) and |V| = r then
ev = gi - ev\{fvl}
V then gi ev = 0
5. If i V {1, l} (so Eji
(b) If vi
-
{vi_1 -+ vi}) and |V|
e V then gi - ev =
=
r then
ev\{lv,,vi} - gi
V and vi_ 1 $ V then gi ev = -gi
ev\{Vp} + gi -v\{Vj
1}
ev\{v,}
V then gi - ev = 0
Proposition 3.6.9. gi is Ej invariantfor all i.
Proof. We can show that gi is Ej invariant using Corollary 3.4.28. When vi_ 1 , vi
and IV U {vi_ 1, vi} \ {s, t}
VV
;> r we can check directly that the associated equation
in Corollary 3.4.28 holds. When vi- 1 , vi V V and IV U {vi_ 1 , vi} \ {s, t}I < r we use
the fact that the associated equation in Corollary 3.4.28 must hold for gp,E and thus
holds for gi as well.
E
Corollary 3.6.10. The functions {gi} have the following properties:
1. For all i, gi is Es-invariant.
2. For all i and all V C Vred(G) where
|V|
< r, gi - ev = 91 - ev
3. For all i, gi ev
=
0 whenever V C Vred(G) and |V| > r
4. For all i, gi - e{}
=
1
87
5. For all i, gi , ev
6. |gi - ev| < 3-
#
0 for at most
2 2 [g 1--2
<
22[ 1g 1
V with |VI = r
for all V with |V| = r
Proof. The first statment is just Proposition 3.6.9. The second, third, and fourth
statements all follow from the definition of the functions {gj} and the properties of
the functions {gP,}Ei. For the fifth statement, note that every time we fix a nonzero
Fourier coefficient gi - ev where IVI = [ig l1 we use Fourier coefficients of the form
gi ew where IWI < r to determine its value. Moreover, we never use the same Fourier
coefficient twice, so the number of nonzero Fourier coefficients gi -ev where IVI = r is
at most (-_).
Finally, the sixth statement follows from the definition of gi and our
bounds on the Fourier coefficients of the functions {9P,Ei}-
Note that when we look at gi2 - g 1 , all Fourier coefficients with V < r cancel.
From this, it follows that for any i, IIg-gi
2
< 21gi 12 +21g11
2
(1_ I)r-124r+2.
D
We now prove Theorem 3.6.1. By Corollary 3.4.30, taking M = (1 - 1)r2122r+1
-
1)2,
1n
'M(PV(G),1)>)2>51+(
- M(l -
n2
1) 2(l - 1)
2 5 2+ 1 (
-
-
+.
)
for all n > 2(1
Recalling that r = [lg 11 and using the fact that 2r > l, we may reexpress this bound
as
I 1l1
[lgl]
m(Pv(G),l) >
>
' 26g%+1(/_iii
-
1)
64(1 -1)2
- 2
Note that we may ignore the condition that n > 2(1
because the bound is trivial
-1)2
if n < 2(l - 1)2. Taking 1 = [inl], we have that
M (PV(G)
1
>
1Ign-3
2
--
2
as needed.
1
-n
2
ign3
6
1
4
E
88
3.7
Monotone analogues of L and NL
We would love to say directly that monotone L does not equal monotone NL. Unfortunately, as discussed in Gringi and Sipser [9], while it is relatively straightforward
to define monotone NL, it is not at all clear how to define monotone L. In this
section, we note that whether or not we have separated monotone L from monotone
NL depends on how we define monotone L.
We define the following uniform and non-uniform monotone analogues of NL,
which are the same as the ones given by Gringi and Sipser
[91.
Definition 3.7.1. A nondeterministic Turing machine T is monotone if whenever T
is examining an input bit xi and can make a given transition when xi = 0, it can also
make that transition when xi = 1.
Definition 3.7.2. Define mNSPACE(g(n)) to be the class of all problems solvable
by a monotone nondeterministic Turing machine in O(g(n)) space.
Theorem 3.7.3. If f
E
mNSPACE(g(n)) where g(n) is at least logarithmic in
n, then there are monotone switching-and-rectifiernetworks of size at most
2 0(9(n))
computing f.
Proof. This can be proved by applying the exact same construction we used to prove
Theorem 2.3.3 and noting that if the initial nondeterministic Turing machine T is
monotone then so is the resulting switching-and-rectifier network.
E
By Theorem 3.7.3 and the fact that the directed connectivity problem is in mNL, we
can define a non-uniform monotone analogue of NL as follows:
Definition 3.7.4. Define mNLswitching-and-rectifiernetworks to be
{f
: {0, 1}*
-
{0, 1} : f can be computed by monotone switching-and-rectifiernet-
works of polynomial size}
Unfortunately, we do not know how to define a nice uniform monotone analogue
of L. However, we can mimic Definition 3.7.4 and define a non-uniform monotone
analogue of L as follows:
89
Definition 3.7.5. Define mLswitching networks to be
{ f : {0,
1}*
--
{0, 1} : f can be computed by monotone switching networks of polyno-
mial size}
Theorem 3.7.6. mLswitching networks
C mNLswitching-and-rectifier networks
Proof. This follows immediately from Theorem 3.6.1 and the fact that the directed
connectivity problem is in mNL.
D
However, Gringi and Sipser [9] define a different non-uniform monotone analogue
of L. They define a monotone analogue of L as follows:
Definition 3.7.7. A boolean circuit is monotone if it has no NOT gates.
Definition 3.7.8. A boolean circuit is leveled if we can partition the gates into levels
such that the outputs of the gates at each level depend only on the inputs and the
outputs of the gates at the previous level. Define the width of a leveled boolean circuit
to be the maximum number of gates in a level.
Definition 3.7.9. Define mLcircuits to be
{f : {0, 1}* -+ {0, 1} : f can be computed by a monotone leveled circuit of polynomial
size and logarithmic width}
As far as we know, our analysis does not imply any lower bounds on the minimum
width and size of monotone leveled circuits solving the directed connectivity problem,
so we are currently unable to prove that mLcircuits $ mNLswitching-and-rectifier
networks-
Thus, although we have separated monotone analgoues of L and NL, we cannot say
directly that monotone L is not equal to monotone NL.
3.8
Finding paths with non-monotone switching networks
Unfortunately, proving lower size bounds on all switching networks solving the directed connectivity problem is much harder than proving lower size bounds on monotone switching networks solving the directed connectivity problem. Non-monotone
90
switching networks can use the information that edges are not there in the input
graph, which can be very powerful.
In this section, we note that there are small
sound non-monotone switching networks for directed connectivity on n vertices which
accept all of the inputs in PV(G), so the bound of Theorem 3.6.1 does not hold for
non-monotone switching networks.
Definition 3.8.1. Given a set I of input graphs on a set V(G) of vertices with
distinguished vertices s, t where each graph in I contains a path from s to t, let s(I)
be the size of the smallest sound switching network for directed connectivity on V(G)
which accepts all of the input graphs in I.
Theorem 3.8.2. For all n, if V(G) is a set of vertices with distinguished vertices s, t
and |Vred(G)J = n then s(PV(G)) < n _+2.
Proof. The intuitive idea is as follows. If the input graph consists of just a path from
s to t, it is easy to find this path; we just have to follow it. If we are at some vertex
v, and see that there is an edge from v, to v 2 and no other edges going out from vi,
then we can move to v 2 and we can forget about v, because the only place to go from
v 1 is v 2. We can thus easily follow a path as long there are no branches in the path.
While doing this, we only need to remember around 3 ig n bits of information. We
need to remember what v, and v 2 are and we need to remember how many other
possible edges going out from v, we have confirmed are not in G. We now give a
rigorous proof:
Definition 3.8.3. Define Gpathfinder(V(G)) to be the non-monotone switching net-
work for directed connectivity on V(G) constructed as follows:
1. Start with G'(V(G), 2)
2. For each pair of distinct ordered vertices v 1 , V 2
n - 1 between v',1, and v
V/
and vj'
1 .
G Vred(G),
add a path of length
in parallel to the edge labeled v1
Give the edges in this path the labels {,(v 2
{ s, vi, v 2 }}
91
-
--
U)
v 2 between
: u C V(G)\
n3 + 2
Proof. There are n vertices of the form v, where v c Vred(G).
there are n - 1 added paths which start at v'f
vertices v'
adds n -2 vertices. Thus, IV(Gathfinde,(V(G)))
V(G' (V(G), 2))l
n+2=n 3
= n(n-1 +n+2
+n+2<n
_n2
3
-
For each of these
and each of these paths
V(G' (V(G), 2))I = n(n-1)(n-2).
so IV(Gpathfinder(V(G)))= n(n -1)(n--2)+n(n21)
+2asneeded.
+
Proposition 3.8.4. 1V(Gathfinde,(V(G)))
E
Proposition 3.8.5. G'athfinder(V(G)) accepts all input graphs in PV(G).
Proof. If G is an input graph with vertex set V(G) and edges E(G)
i
=
{vi
-*
vi+1
E [0, 1 - 1]} where vo = s and v, = t then we have a path from s' to t' whose edges
are all consistent with G as follows.
2. If we are at v
vi
-
for any i E [1, 1- 2] then go to o
-+
v1
along the edge labeled
vi+1.
3. If we are at v
vi+1
along the edge labeled s
.
1. If we are at s' then go to v'
-e
U
for any i C [1, 1 - 2], then for all u c V(G)
\
{s, vi, vi+}
E(G). Thus we can go to vjov 1 } along the path between vvg and
Vt
{vi,vi+1}.
4. If we are at v'
then go to t' along the edge labeled v 1 _1
-
t
Lemma 3.8.6. G'pthfinder(V(G)) is sound.
Proof.
Definition 3.8.7. Given an input graph G, create an input graph Ga as follows. Let
Ea
=
{v
-*
Take V(Ga)
w : V,W E Vred(G),V
=
w,Vu c V(G)\{s,v,w},w -÷ u $ E(G)}.
V(G), E(Ga) = E(G) u Ea-
Proposition 3.8.8. If Grathfinder(V(G)) accepts an input graph G then G'(V(G), 2)
accepts the corresponding input graph Ga.
92
Using Proposition 3.8.8, to prove Lemma 3.8.6 it is sufficient to show that for any
input graph G there is a path from s to t in Ga only if there is a path from s to t
in G. To show this, assume that there is no path from s to t for some input graph
G. Then let V be the set of all vertices v such that there is a path from v to t in G.
Since there is no path from s to t in G, s V V. Let C be the cut with R(C) = V.
Note that there cannot be any edge in G that crosses C.
Assume there is an edge in Ga which crosses C. Then it must be an edge v
-
w in
Ea and we must have v E L(C), w E R(C). This implies that there is a path from w
to t. However, by the definition of Ea, w y t and Vu E V(G)\{s, v, w}, (w, u) V E(G).
Thus, any path from w to t must go through v, so there must be a path from v to
t and we should have that v c R(C). Contradiction. There is no edge in Ga which
crosses C, so there is no path from s to t in Ga, as needed.
D
Theorem 3.8.2 follows immediately from Proposition3.8.4, Proposition 3.8.5, and
D
Lemma 3.8.6.
Example 3.8.9. The switching network in Figure 3-4 is G'athfinder({st,a,b}) with
some edges removed.
93
94
Chapter 4
Lower bounds on monotone switching
networks for generation and k-clique
The material in this section comes from joint work between Siu Man Chan and the
author [5] on applying the techniques developed in Chapter 3 to other problems,
in particular the generation problem and the k-clique problem. This work gives an
alternate proof of the monotone circuit depth lower bounds and separation of the
monotone NC-hierarchy proved by Raz and McKenzie.
4.1
Circuit Depth and Switching Network Size
We begin by showing the connection between circuit depth and switching network
size mentioned in the introduction and Chapter 2.
Theorem 4.1.1. If a function f can be computed by a (monotone) circuit of depth
d, then it can be computed by a (monotone) switching network of size at most
2d
+ 1.
Proof. This is proved using Shannon's [24], [25] construction of a series parallel switching network from a boolean formula/circuit. This construction is recursive and works
as follows.
1. In the base case where the circuit is a single input bit b or -b, take the switching
network G' to have the vertices s' and t' and a single edge with label b = 1 or
95
b = 0 respectively between s' and t'.
2. If the final gate of the circuit is an OR gate, take the two switching networks
G' and G' correspnding to the two inputs of this gate. Now merge the s' of
G' and G' together and let this be the new s'. Also merge the t' of G' and
G' together, letting this be the new t'. This effectively places G' and G' in
parallel.
3. If the final gate of the circuit is an AND gate, take the two switching networks
G'1 and G'2 correspnding to the two inputs of this gate. Take s' to be the s' of
G1, merge the t' of G' with the s' of G', and take t' to be the t' of G'. This
effectively places G' and G'2 in series.
4. If the final gate of the circuit is a NOT gate, if there are two NOT gates in
a row then just delete both of them. Otherwise, use De Morgan's laws that
-,(A V B) = -,A A -B and -,(A A B) = -A V -B to reduce to one of the two
cases above.
Now we just need to check the size of the resulting switching network. We check this
using induction. For the base case, IV(G')I = 2 = 20 + 1. For the other cases, using
the inductive hypothesis, IV(G')j < IV(G')l + IV(G') - 1 <
2d
+ 1, as needed.
E
Theorem 4.1.2. If a function f can be computed by a (monotone) switching network
of size n', then it can be computed by a (monotone) circuit of depth O((lgn') 2 ).
Proof. This is proved using Savitch's algorithm. Savitch's algorithm checks whether
there is a path of length at most 1 from s to t as follows. If 1 = 1 then just check
whether there is an edge from s to t in G. Otherwise, for each possible midpoint m,
recursively check whether there is a path of length at most
path of length at most
[j]
[]
from s to m and a
from m to t. If the answer is yes for any midpoint m,
accept. Otherwise, reject.
Savitch's algorithm was used implicitly in the switching networks G'(V(G), [lg 1])
from Section 3.2. Savitch's algorithm can be implemented with circuits as follows.
For each m, recursively create the circuit checking for a path of length at most []
96
from s to m and the circuit checking for a path of length at most [12] from m to t. Put
an AND gate between these two outputs. Now take the OR of all of the ooutputs of
these circuits. This can be done with [lg n] additional levels, where n is the number
of vertices of the graph.
This means that each recursive level adds depth at most [ign] + 1 to the circuit.
There are at most [ig n] levels of recursion in Savitch's algorithm, so the total circuit
depth is at most ([lg n'])([lg n'] + 1).
Now given any problem which can be solved by a (monotone) switching network
of size n', take the circuit corresponding to Savitch's algorithm on this switching
network and this will be a (monotone) circuit of depth O((lg n') 2 ).
4.2
E
Our Results and Comparison with Previous Work
Raz and McKenzie [19] proved the following lower depth bounds on monotone circuits
(the first bound was first proved by Karchmer and Wigderson [12]).
Theorem 4.2.1.
1. Any monotone circuit solving undirected connectivity has depth Q((lgn) 2 ).
2. For some e > 0, any monotone circuit solving the generation problem on pyrmamids of height h < nE has depth Q(hlgn).
3. For some E > 0, any monotone circuit solving the k-clique problem where k < nE
has depth Q(k lg n)
Using Theorem 4.1.2, this implies the following lower size bounds on monotone
switching networks.
Corollary 4.2.2.
1. Any monotone switching network solving undirected connectiivty has size n"N'.
2. For some e > 0, any monotone circuit solving the generation problem on pyrmamids of height h < nE has size n"
97
1n.
3. For some e > 0, any monotone circuit solving the k-clique problem where k < n'
has size n" Vg.
In this chapter, we prove the following bounds.
Theorem 4.2.3.
1. Any monotone switching network solving directed connectiivty has size n"(n).
2. For some e > 0, any monotone circuit solving the generation problem on pyrmamids of height h < nE has size nQ(h).
3. For some c > 0, any monotone circuit solving the k-clique problem where k
<
nE
has size nQ(k).
Using Theorem 4.1.1, this implies the following depth lower bounds on monotone
circuits.
Corollary 4.2.4.
1. Any monotone circuit solving directed connectivity has depth Q((lgn) 2 ).
2. For some c > 0, any monotone circuit solving the generation problem on pyrmamids of height h < nE has depth Q(h lg n).
3. For some e > 0, any monotone circuit solving the k-clique problem where k < n
has depth Q(klgn)
Thus, our bounds for directed connectivity, generation, and k-clique are a strengthening of the results of Raz and McKenzie. For the undirected connectivity problem,
the bound of Karchmer, Wigderson, Raz, and McKenzie is tight for both monotone
circuit depth and monotone switching network size as undirected connectivity can
be solved by a monotone circuit of depth O((lg n)2 ) and trivially by a monotone
switching network of size O(n). This example shows that Theorem 4.1.2 cannot be
strengthened and that bounds on monotone circuit depth cannot always be strengthened to a corresponding bound on monotone switching network size. This in turn
98
implies that considerably different ideas are used in the communication complexity
approach for proving lower bounds on monotone circuit depth and our approach for
proving lower size bounds on monotone switching networks.
4.3
Lower bounds on monotone switching networks
for Generation
The generation problem is a generalization of the directed connectivity problem.
Definition 4.3.1. In the generation problem on a set of vertices V(G) with distinguished vertices s,t, we have an axiom s and a set of implications of the form
V =-
w where V C V(G)
\ {t}
and w E V(G)
\ (V U {s}).
We have a YES instance
if we can deduce t from the axiom s and we have a NO instance otherwise.
The YES instances we will be considering for this problem are pyramid instances.
Definition 4.3.2. A pyramid P of height h consists of vertices {vij : i
C [0,
h],j E
[0, h-i]} where vij has children vi_ 1,j and vi-1,j+1. We call VhO the root of the pyramid.
Definition 4.3.3. Given a pyramid P where V(P) C Vred(G), we define the YES
instance G(P) so that
1. V(G(P)) =V(G)
2. E(G(P)) = {s =
[0, h - i]} U{vh0
vo0
j E [0, h]} U {{vi-1j, vi-, +}
-- > vij :
[1, h],j E
4 t}
All of the techniques we used in analyzing directed connectivity apply to this
problem with only a few small changes. First, the certain knowledge game becomes a
logical proof system except that we are not allowed to delete information arbitrarily.
The reversible pebbling game changes slightly as follows.
Definition 4.3.4 (Reversible Pebbling Game for Generation). In the reversible pebbling game for generation, we start with a pebble on s and can make the following
types of moves
99
1. If we have an implication V
(V U
-- > w where V C V(G) \ {t} and w E V(G) \
{s}) and we have pebbles on all vertices in V, we can add or remove a
pebble from w.
2. If we have a pebble on t we can add or remove any other pebble.
We win the reversible pebbling game if we have a pebble on t.
Just as the heart of our lower bound for directed connectivity was a lower bound
on the number of pebbles needed to win the reversible pebbling game for a path
graph, the heart of our lower bound here is a lower bound on the number of pebbles
needed to win the pebbling game on a pyramid of hieght h. This bound was proved
by Cook [61 and we present the proof here for convenience.
Lemma 4.3.5. At least h + 1 pebbles are required to win the reversible pebbling game
on a pyramid of height h.
Proof. In order to win the reversible pebbling game, we must reach a state where
every path from the top of the pyramid to the bottom of the pyramid contains a
pebble. Consider the first time we reach such a state. Note that moves where we
pebble or unpebble a vertex when its children are pebbled cannot affect whether there
is a pebble on these paths, so must have just done a move where we pebbled a vertex
at the bottom of the pyramid. We now have that
1. For every path P from the top of the pyramid to the bottom of the pyrmaid,
there is a pebble on one of its vertices.
2. There exists a path P from the top of the pyramid to the bottom of the pyrmaid
which only has a pebble on its bottom vertex
We now show that these conditions require having h + 1 pebbles on the pyrmamid,
which immediately implies the result.
Lemma 4.3.6. Assume we have a pyramid and pebbles on the pyramid such that
1. For every path P from the top of the pyramid to the bottom of the pyrmaid,
there is a pebble on one of its vertices.
100
2. There exists a path P from the top of the pyramid to the bottom of the pyrmaid
which only has a pebble on its bottom vertex.
then we must have at least h + 1 pebbles on the pyramid.
Proof. We prove this lemma by induction. The base case h = 0 is trivial. If h > 0
then let P be a path from the top of the pyramid to the bottom of the pyrmaid
which only has a pebble on its bottom vertex. Without loss of generality we may
assume that P starts at VhO and goes to V(h_1)O. Now note that if we consider the
subpyramid rooted at V'(h1)o, this pyramid has height h - 1 and must satisfy both
conditions given in the lemma, so by the inductive hypothesis there must be at least
h pebbles on this pyramid. Furthermore, if we consider the path P, which goes along
the vertices Vi(h-i), this path must have at least one pebble on it, giving a total of at
1I
least h + 1 pebbles, as needed.
l
This completes the proof of Lemma 4.3.5.
Now if we say that an implication V
=>
w crosses a cut C if V C L(C)
and w c R(C), all of our techniques can be applied in exactly the same way! The
only remaining difference is that the analysis is slightly different when we explicitly
compute the Ei-invariant functions.
Lemma 4.3.7. Taking r = h + 1 and E= {e=} where e 1 ,
, eIE(G(P))l are
the edges
of G(P), there exist functions {gi} such that
1. For all i, gi is Ei-invariant.
2. For all i and all V C Vrd(G) such that |V| < r = h+ 1, (gi - g) -ev = 0
3. gi -ell = 1
4.
For all i, gi
-
gill < h2h24h+8
Proof. Taking the functions {gP,Ej given by the analogue of Theorem 3.5.1 for the
generation problem, when we take all of the b'(V) to be 0 or 1,
1. gP,E is Ei-invariant
101
2. gP,-
3.
e{} -= 1
(gPE --
,E
1
4. |gP,E - ev| <
) -
ev = 0 whenever IVI < r = h + 1
2 21vI
Corollary 3.4.28 must be adjusted as follows.
Proposition 4.3.8.
1. If e = s
>
w for some w G Vred(G) then g is e-
invariant if and only if (e{} + e{f})g = 0. Equivalently, g is e-invariant if and
only if .vu{w}=
2. If e
(e{}
=
-
v
-
v whenever w
$
V.
t for some v C Vred(G) then g is e-invariant if and only if
=>
eqv})g = 0.
Equivalently, g is e-invariant if and only if
vu{v} =
v
whenever v $ V.
3. If e = {u, v}
w for some u, v, w
==>
Vred(G) then g is e-invariant if and
only if
(e{} - e{u})(e} - e{t})(e} + e{w})g = 0
Equivalently, g is e-invariant if and only if
9Vu{uVw}
= -VU{u,v}
whenever u, v, w
+
YVU{u,w}
+ Vu{v,w} +
jvu{u}
V.
Definition 4.3.9. Define the functions {gi} so that
1. gi - ev = gP(G),i'ev if IVI <h + 1
2. gi - ev = 0 if |V| > h + 1
3. If ei is of the form s ==> v and |V| = h+1 then
(a) If v E V then gi - ev = -gi - eV\{V}
(b) If v V V then gi ev = 0
4. If ei is of the form v
=->
t and |V| = h + I then
102
+
9VU{v} - jvu{w} - jV
(a) If v E V then gi ev = gi - ev\{v1_ 1}
(b) If v V V then gi ev = 0
5. If ej is of the form u,v -- >w and |VI = h + 1 then
(a) If U),v,w G V then
(gi)v
=
+ (gi jyfV) + (gi j yx, + (gi y ,
-(giV\{w}
+ (Yi)V\{UUW - (9 )V\u,vl - (g )yg{s,,,,
(b) Ifv,wE V anduVV then
(Yiyv
-(gi)VU{U}\V,W}
(c) If w G V and u, v
(d) If w
-
(9i)V\{W} + (gi)V\{v
+(
V then (gi)v = -- (92)V\{w}
V oruC V butv
V then (g)
=O
Proposition 4.3.10. gi is Ei-invariantfor all i.
Proof. We can show that gi is ei-invariant using Proposition 4.3.8. When u, v, w $ V
and IV U {u, v, w} \ {s, t}| ;> r we can check directly that the associated equation in
Proposition 4.3.8 holds. When u, v, w
V and |V U {u, v, w} \ {s, t}j < r we use
the fact that the associated equation in Corollary 3.4.28 must hold for gP,E, and thus
U
holds for gi as well.
Corollary 4.3.11. The functions {gi} have the following properties:
1. For all i, gi is Ei-invariant.
2. For all i and all V C Vred(G) where V| < h + 1, (Ei)v
3. For all i, (9i)V =0 whenever V
C Vred(G)
and |V| > r=h+1
4. For all i, (9i)f} = 1
5. For all i, (gi)v # 0 for at most ('2)
103
(Y1)y
V with |V| = h +1
6. |(gi)v| < 7.
2
(h+1)-2
<
2 2h+3
for all V with |V| = h + 1
Proof. The first statment is just Proposition 4.3.10. The second, third, and fourth
statements all follow from the definition of the functions {ge} and the properties of
the functions {gp,,}.
For the fifth statement, note that every time we fix a nonzero
Fourier coefficient (gi)v where IVI = h + 1 we use Fourier coefficient(s) of the form
gi - ew where IWI = h to determine its value. Moreover, we never use the same
Fourier coefficient twice, so the number of nonzero Fourier coefficients gi - ev where
VI = h +1 is at most
(l).
Finally, the sixth statement follows from the definition
of gi and our bounds on the Fourier coefficients of the functions {gPE I
Note that when we look at gi-g, all Fourier coefficients with V < r = h+1 cancel.
From this, it follows that for any i, I i - g,112 < 21|g11 2 + 21|g,1i
2
((2 )) 2 4h+8 <
h2h24h+8.
Applying Theorem 3.4.25 with q
M = h 2 2h+4, whenever r ; 2(+2
1
2
Choosing h
4.4
(h+1) + 1, r = h + 1, z = 1, m = (h+1), and
we get a lower bound of
n
1
h
n
(h+1)hh2 2h+4 (h+1)
-1 we get a lower bound of
2
h+1
1)
- 4(h + 1) 8(h
2
Lower bounds on monotone switching networks
for k-clique
We can also use this framework to prove lower bounds on monotone switching networks for the k-clique problem. Although we do not have a certain knowledge or
pebbling game that we can use as a guide, the symmetry of the problem allows us to
give and analyze the sets of functions directly.
For the k-clique problem, we analyze yes instances containing a clique
Q of size
k. Our NO instances are k - 1 colorings C of the vertices Vred(G). Letting C be the
set of all such colorings, we have an inner product and Fourier coefficients as follows:
104
Definition 4.4.1. Given f, g : C -+ C, define (f, g)
Ecc f(C)g(C).
||fil =
(f, f)
Definition 4.4.2. Given a function U :
WZvEVred(G)
C(v)U(V)
where w
=
Vred(G)
[0, k - 2], define eu(C) =
-
ek-i
Proposition 4.4.3. The functions {eu} form an orthonormal basis for the space of
functions from C to C.
Definition 4.4.4. Given a function f : C -+ C and a function U : Vred(G)
+
[0, k-2],
define fu = (f, eu).
Proposition 4.4.5 (Fourier Decomposition and Parseval's Theorem). For any function f : C
C, f
-+
=
E
U:Vred(G)-+[O,k-2]
fueu and (f, f)
=
U 12
ZU:Vre d(G)-+[O,k-2]
Lemma 4.4.6. For any vo G Q there exist functions {g, : v C Q} such that
1. For all v C Q, gv is Ev-invariant where Evv is the set of edges of
Q
incident
with v.
2. For all e and all U:
k -2
Vred(G) -*
[0, k - 1] such that U(v) is nonzero for at most
v E Q, (gv-g,eu) =0
3. For all v c Q, (gv, 1) = 1
4. For all v c Q, ||gv - gvn|| < 4k
Proof.
Definition 4.4.7. Define q(k)
=
(k-
_k-2.
Definition 4.4.8. For every vertex v in a k-clique Q, we define gv(C) to be #(k)(-1)t(t2)! if C restricted to
Q contains a monopoly of size t not
Lemma 4.4.9. (1, gv) = 1
105
containing v and 0 otherwise.
Proof. There are (k - 1) (k- ) -
choices for how to color the vertices of
Q so that
there is a monopoly of size t not containing v. Thus,
1
k - 1) (k -2)!
k-i
1)k
(k
- (k)(--1)'(t - 2)!
(t - 2)!
-
(1, gV) = (k
t=2
1)
0(k) (k - I)! k(1) t ( k
(k - 1
t
t=2
Note that
k-1
k-1
(lI)t (k
(-1)t (k t 1)
t=2
Thus, (1 g)
(2)!(k-2)
=
1 + k - 1 = (1 + - 1 )k-1 + k - 2 = k - 2
t
t=o
1, as needed.
D:
k
Lemma 4.4.10. jjvj < k-2
Proof. Recall that 0(k)
Again,
there are (k
Agiteeae(
(k 1)k(k-2)!(k-2)*
how to color the vertices of
Q so that
-
1)
choices for
_-2
(k-1)
there is a monopoly of size t not containing v
Thus,
=
(k - 1)k (k -k)
(k
-
(k -
1 )1
2) 2
k-
k -1)
t
- 1
(k
t
(t -
-
2)'
)!
-
(ge,Ig9)
(t - 2)!
(k - 2)!
Note that (k-1) (t-2)! < k - I, so this is at most (k - 1)k < kk.
Thus,
|gV1
=
(gv, g) < k , as needed.
Lemma 4.4.11. For all distinct v, w E Q and any function D :
" ECCC:VuCQ\{w},C(u)=D(u)9V(C)
Q \ {w}
-
[0, k - 2],
is 0$(k) if D maps every vertex in Q\{w} to a different
color and 0 otherwise.
Proof. Consider the positions of the vertices in
Q \ {w}.
If some color has a monopoly
of size m > 2 on these vertices, then if w has this color as well, then for this cut C,
ge(C)
= 0(k)(-1)m+l(m - 1)!.
Otherwise, there are m - 1 choices for the color
106
w which give w its own color. For these C, g,(C) = 0(k)(-1)'(m - 2)!. For the
remaining choices for the color of w, at least two colors will have at least two vertices,
so g(C) = 0. Adding everything up, we get 0 as claimed.
If no color has a monopoly on these vertices, all choices for the color of w pair it
with one other vertex, so g,(C) =
E
(k).
Corollary 4.4.12. Whenever U(v)
#
0 for at most k - 2 v E
Q,
for all v E
Q,
(gv)U = (gvo)u.
Proof. Choose some v 2 such that U(v 2 )
such that w
(9V)
U
=
z
v, w / v 2 , and U(w)
#
0. Now for any v E
Q,
there must be a w
0. From the above lemma, this implies that
(gV 2 )u. This holds for any v E Q so (gv)u = (gv2)u = (gv) u
D
Putting everything together, Lemma 4.4.6 follows.
Applying Theorem 3.4.25 with q = k, r = k - 1, z = 1, m = k, and M= 4k,
whenever n > 2k 2 we get a lower bound of
()
k 21)
8(k-1)k-2
Choosing k=
L2
we get a lower bound of
1--)2 >
8(k - 1)
n 2k2
107
-2
4n
4
108
Chapter 5
Upper Bounds: Dealing with
Lollipops via Parity
5.1
Motivation for considering particular input graphs
for directed connectivity
We have two reasons for analyzing the difficulty of particular input graphs for directed
connectivity. First, the problem of what makes an input graph hard for monotone
switching networks/algorithms to solve directed connectivity on is interesting in its
own right. Second, obtaining such bounds is a necessary first step for obtaining nonmonotone bounds. As discussed in Section 3.8, non-monotone switching networks can
easily solve input graphs which consist of an isolated path from s to t and no other
edges, so we cannot use such input graphs to prove non-monotone lower bounds.
Thus, we must consider other input graphs and a necessary first step is to prove
monotone lower bounds for these graphs.
In chapter 6 we will develop the techniques in chapter 3 further to prove such lower
bounds. Here we first prove an upper bound which shows that the minimal length of
a path from s to t in an input graph G is not the only factor which makes it easy or
hard for monotone switching networks/algorithms to solve directed connectivity on
G.
109
Definition 5.1.1.
We call a vertex v E Vred(G) a lollipop if there is an edge from s
to v in G or an edge from v to t in G but not both.
In this chapter, we show that adding lollipops to an input graph G does not significantly increase the difficulty of solving directed connectivity on G using monotone
switching networks.
5.2
Lollipops and certain knowledge switching networks
As a warmup, we show that lollipops can help even certain knowledge switching
networks solve directed connectivity.
Theorem 5.2.1. If G is an input graph such that
1. There is a path P in G from s to t of length 1
2. For all v E V(G)
\ V(P),
s -+ v
c
E(G)
then c(G) < l!nlg n + 2
Proof.
Definition 5.2.2. Given an ordering v 1 ,---
,vn
of the vertices of Vred(G),
define
P'(v1,--- ,vn) to be the certain knowledge switching network with vertices s',v',--and all allowed edges where K,= {s -+ t}, K,, = Ku
, and Kt,
{s
-+
,v',
t}.
Lemma 5.2.3. If G 2 is an input graph such that
2. For all v C V(G)
\
V(P), s
- - --+v
then P'(v1, - - - , v,) accepts G 2 if u 1 ,
u 1 _1 -+ t in G from s to t of length l
E E(G 2
)
1. There is a path P = s - u 1
, ul-1 appear in order in v 1 ,-
,Vn
Proof. Take uo= vo = s, U 1 = v =t, v = s', and v'+1 = t'- If ul,-
, ul-1 appear
j
. To see this,
in order in vi,-
,
vn then for all
E [0, n] we can go from K , to K,
note that we have the following cases
110
t'
1. If vj+1 V U'_ 1 {ui} then s
2. If v3 +1 =
v1 ,-
Uk+1
-
vj+1 E E(G 2) so we can go from K,' to K,'
for some k c [0, 1 - 1] then since u1, - -, u, appear in order in
-,Vnwe know that
Uk
E U
U
0
{vi}. Thus, since Uk
-
Uk+1 E
E(G2 ) we
can go from K,, to K,,+i, as needed.
This implies that P'(vi,
) accepts G 2 , as needed.
,
E
We now construct G' by taking q random orderings of the vertices Vred(G), taking
the corresponding P', merging all s' together, and merging all t' together. IV(G')=
nq + 2.
By Lemma 5.2.3, for a given permutation G 2 of G, when we choose each P' the
probability that it accepts G 2 is at least (1'.
Thus, the probability that no P'
accepts a given permutation G 2 of G is at most (1-
)q.
There are at most ni-1
different permutations of G, so by the union bound the probability that G' does not
accept all permutations of G is at most n1-1(1
-
1
)
)q.
1 )] then n'-'(1 - (
If we take q = [(l - 1)! ig (n-'
1 1 ))
< 1 so there is a nonzero
chance G' accepts all permutations of G. Thus, since [(l - 1)! lg (ni- 1)] < 1! lg n,
c(G) < l!n lg n + 2.
5.3
A general upper bound
We now show a general upper bound on input graphs G in which the vast majority
of vertices are lollipops.
Theorem 5.3.1. For any input graph Go containing a path from s to t, if we can
win the reversible pebbling game on Go using at most k pebbles then letting m
V(G0 )
\ {s, t},
=
for any input graph G such that G includes Go and all other vertices
of G are lollipops, m(G) is nO(1)mO(k). More precisely,
m(G) <
2
3
2
k+2 m k+l
n(n+ 2 2k+1mk+1)(gn)
Proof.
111
2
Definition 5.3.2. We call a vertex v E Vred(G) even if there is a path from s to v in
G. and odd if there is a path from v to t in G.
Proposition 5.3.3. If any vertex v C Vred(G) is both even and odd then there is a
path from s to t in G.
Definition 5.3.4. We define a parity knowledge set K to be a set of pairs (V, b) where
V C Vred(G) and b E {0, 1}.
(V, 0) represents the knowledge that either there is a
path from s to t in G or the number of odd vertices in V is even. (V, 1) represents the
knowledge that either there is a path from s to t in G or the number of odd vertices
in V is odd.
Definition 5.3.5. In the parity knowledge game, we can make the following types of
moves on a parity knowledge set K
1. If we directly see the edge e = s -+ v in G then we may add or remove ({v}, 0)
from K
2. If we directly see the edge e =v -+ t in G then we may add or remove ({v}, 1)
from K
3. If ({v}, 0) E K and we directly see the edge e = v -+ w in G then we may add
or remove ({w},0) from K
4. If ({w}, 1) E K and we directly see the edge e = v -+ w in G then we may add
or remove ({v}, 1) from K
5. If (V, b 1), (W, b 2 ) E K then we may add or remove (VAW, b1 + b 2 mod 2) from
K
6. If ({s}, 1) C K or ({t}, 0) C K then we may add or remove anything else from
K.
We win the parity knowledge game if we reach a parity knowledge set containing
({s}, 1) or ({t}, 0)
112
Proposition 5.3.6. We can win the parity knowledge game for an input graph G if
and only if there is a path from s to t in G.
Just as we did for certain knowledge switching networks, we can use this game to
construct switching networks where each vertex v' corresponds to a parity knowledge
set Kv, and each edge with label e between vertices v' and w' in G' corresponds to a
sequence of moves in the parity knowledge game between Kv, and K"1, all of which
can be done using only the knowledge that e E E(G).
We first note that given a set of vertices V
C
V(G) and a distinguished vertex v,
if all vertices in V \ {v} are lollipops then it is relatively easy to recover the parity of
v from the parity of V.
Definition 5.3.7. If all vertices U C Vred(G) are lollipops then we take b(U) = 0 if
there are an even number of u C U such that there is an edge from u
-
t and we take
b(U) = 1 otherwise.
Lemma 5.3.8. For any V
C
Vred(G), we can construct a switching network G' such
that
1. G' has a vertex with parity knowledge set {(V,0)} and a vertex with parity
knowledge set {(V, 1)}
2. For all v c V, G' has a vertex with parity knowledge set {({v}, 0)} and a vertex
with parity knowledge set {({v}, 1)}
3. If G is an input graph such that for some v E V, all vertices u E V \ {v} are
lollipops, then for each b G {0, 1} there is a path P' in G' from {(V, b+b(V\{v})
mod 2)} to {(v, b)} whose edge labels are all in E(G) and of the form s -+ u or
u -+ t where u e V \ {v}
4.
The number of vertices in G' excluding the ones mentioned above is at most
Proof. We prove this lemma by induction. Let f(x) be the number of additional
vertices needed if IVI = x. The base case IVI = 1 is trivial. For larger IVI, split
113
V into two halves, VL and VR. If v
C
then taking an ordering of wi, -, wIVRI of
VL
the vertices in VR, take one vertex for each of the additional parity knowledge sets
K
=
(VL U {wi : i < j}, b) where j E [0, IVRI
-
1] and b E {0, 1}.
If we are at the parity knowledge set (VL U {wi : i < j}, b) and j > 1, we may
use the following sequence of moves to reach the parity knowledge set (VL U {w i<
j
-
1},b+b({wj})
mod 2)
1. Add ({w}, bw,) to K
2. Since (VL U Iwi : i < j}, b) C K and ({wj}, bw,) c K, add (VL U {wi :
j-
1}, b + bj,
<
mod 2) to K. Then remove (VL U{wi : i < J}, b) C K from K.
3. Remove ({w 3 }, be,) from K
+
Applying this process repeatedly, if we start at the parity knowledge set (V, b
b(V \ {v}) mod 2) we will reach the parity knowledge set (VL, b + b(V \ {v}) + b(VR)
mod 2) = (VL, b + b(VL \ {v}) mod 2). We can now use the inductive hypothesis on
VL.
Following similar logic, if v E VR we can taking an ordering w, ... , wiye of
the vertices in VL, take one vertex for each of the additional parity knowledge sets
K = (VR U {wi : i
j}, b)
where
j
c [0, VL
-
1] and b C {0, 1}, and then use the
inductive hypothesis on VR. Putting everything together,
f (|V|) < 21VLI
f(IVL
)+
21VR + f (|VR|) < 2|VL|(1 + [Ig |VLfl) + 21VRI(1 + [lg VRf)
< 2|VLI [Ig VJ+ 21VRI[lgJVJ]
< 2|V||Ig|IV|]
as needed.
El
We also note that if we have subsets V, W and all vertices in VAW are lollipops,
it is easy to obtain the parity of W from the parity of V. We will use this as follows.
If we have a distinguised vertex v and subsets V, W such that v E V n W and all
114
other vertices of V and W are lollipops, then we can switch from storing infromation
about v in the parity of V to storing information about v in the parity of W.
Lemma 5.3.9. For any V, W C Vred(G) such that V
nW
#
0, we can construct a
switching network G' such that
1. G' has a vertex with parity knowledge set {(V, b)} for b E {0, 1}
2. G' has a vertex with parity knowledge set {(W, b + b(VAW)
mod 2)} for b e
{o, 1}.
3. If G is an input graph such that all vertices in VAW are lollipops, then for
each b E {0, 1} there is a path P' in G' from the vertex with parity knowledge
set {(V, b)} to the vertex with parity knowledge set {(W, b + b(VAW)
4.
mod 2)}
The number of vertices in G' is at most 2(V| + |W|).
Proof. Letting vi, - -- , v,, be the vertices of Vred(G), take one vertex for each distinct
parity knowledge set of the form ((U -j+ 1vi
n V)
all allowable edges. Note that starting from j
=
U (Uj~vvi n W), b) and then add
1, each new
j
gives distinct parity
knowledge sets if and only if vj C VAW. Thus, we have at most 2(fVAW| + 1) <
2(V + WI) vertices. Also note that whenever vj E VAW, since vj is a lollipop we
can go from {((Ug7igvj n V) U (U _ Vi n W), b + b ((VAW) n (Uj_-Ivi))
{((U-j+1 vi nV) U (uivi nW), b+ b ((VAW) n (uj v.))
from
j
= 0 to
j
mod 2)} to
mod 2)} Using this to go
D
= n gives the result.
Corollary 5.3.10. For any subsets V1 ,all i G [1, r], Vi n Wi
#
1. For all values b 1 ,-
, Vr,
W1 , ...
, Wr C
Vred(G) such that for
0, we can construct a switching network G' such that
br, G' has a vertex with parity knowledge set {(V, bi) : i e
[1, r]}
+
2. For all values b 1,-- br, G' has a vertex with parity knowledge set {(Wi, bi
b(VXAWj)
mod 2): i E [1, r]}
115
3. If G is an input graph such that for all i, all vertices in V1AW are lollipops,
then for each set of values b 1 , - , b, there is a path P' in G' from the vertex with
parity knowledge set {(Vi, bi)
{(W, bi + b(VIAW)
i C [1, r]} to the vertex with parity knowledge set
mod 2): i E [1, rl}
4. The number of vertices in G' is at most 2' Z, ( ViI + |Wi) < 2r+ln.
Proof. We prove this corollary by induction on r. The base case r = 1 is Lemma
5.3.9. If r > 1, for each possible set of values b 2 , - , b, create a copy of the G' given
by Lemma 5.3.9 between the vertices with parity knowledge sets {(Vi, bi) : i E [1, r]}
and new vertices with parity knowledge sets {W 1, bi + b(VAW 1 ) mod 2} U {(V, bi)
i E [2, r]}. The total number of vertices in these copies is at most 2(1I+IW1|). Now
for b1 = 0 and b, = 1, we use the inductive hypothesis on the vertices with parity
knowledge sets {W 1, bi + b(V1 /AW1 )} U {(Vi, bi) : i C [2, r]} and the vertices with
parity knowledge sets {(W, bi + b(VAWj) mod 2) : i E [1, r]}. The total number of
vertices required is 2'
2r
E-2
( Vi +
IWi 1).
Summing these up gives an upper bound of
E (1Vi + IWi), as needed.
We now define switching networks which are effective at solving inputs with lots
of lollipops.
Definition 5.3.11. Given disjoint sets of vertices {Vi : i C [1, q], j E [1, m]} where
for all i, Vi1,..-
Vi
is a partition of Vred(G), we define the switching network
G,,k({Vij}) so that
1. s' has parity knowledge set {} and t' has parity knowledge set {(t, 0)}
2. For every i, every I C [1, m] such that |I| < k, and every set of values {bj : j C
I}, there is a vertex in G' with parity knowledge set UjEI{(Vij, bj)}.
3. For every i, every I C [1,m] such that
Va C
Via, and every set of values
{bj : j
\I\
< k, every a E I, every vertex
C I}, there is a vertex in G' with parity
knowledge set UjcI\{a}{(Vij, bj)} U {({Va, ba})}.
116
4. For every i, every I
|I
C [1,m] such that
< k, every a E I, and every set
of values {bj : j E I \ {a}}, there is a copy of the switching network given
by Lemma 5.3.8 between the vertices with parity knowledge set of the form
UjcI\{a}{(Vij, bi)} U
{({ Via, b})} where b e {0, 1} and the vertices with parity
knowledge set of the form UjIe\{a}{(Vi, b)}} U {({Va, b})}
where Va e Via and
bE {0, 1}
5. For every i, every I C [1, m] such that I|
k, every pair of distinct a, c e I,
every vertex Va G Via, every vertex v, E Vc, and every set of values {b:
j C I}, there is a vertex in G' with parity knowledge set UjzI\{a,c}{(Vi, bj)} U
{({Va, ba})} U {({ve, bc})}.
6. For every i, every I C [1, m] such that II < k, every distinct a, c E I, every
vertex Va E Va, and every set of values {bj : j e I \ {c}}, there is a copy of the
switching network given by Lemma 5.3.8 between the vertices with parity knowledge set of the form UjE1\{a,c}{(Vij, bj)} U {({Va, ba})} U
{({Vi,
{0, 1} and the vertices with parity knowledge set of the form
{({Va, ba})}
b})} where b c
UiEI\{a,c}{(Vi
,
b )}U
U {({ve, b})} where v, E Vi, and b E {O, 1}
k and for every i1 ,i 2 E [1,q] such that
7. For every I C [1,m] such that I|
i1 < i2 and for all j G I, V(i,)j n V( 1 )i
$
0, we have a copy of the switching
network given by Corollary 5.3.10 between the vertices with parity knowledge
sets Ujys{(V(i1 )i bj)} and the vertices with parity knowledge sets
UjEI(Vi2 ),
bi +
b(V(i)jAV(i 2 )j)
mod 2)}
Proposition 5.3.12. For any disjoint sets of vertices V1 ,.--
,Vm
C Vred(G), the
switching network GWk({Vij}) is sound.
Proof. Assume that there is a path P' from s' to t' in G,k(V1, V2 , -
, Vm)
whose edge
labels are all in E(G) for some input graph G. Note that any such path corresponds
to a sequence of moves in the parity knowledge game from K = {} to K = {({t}, 0)}
where each move can be done using an edge in E(G). Since we can only win the
117
parity knowledge game when there is a path from s to t in G, there must be a path
from s to t in G. Thus, G,,(V1, V 2 , -
, Vm)
is sound.
Lemma 5.3.13. Let Go be an input graph with distinguished vertices s, t such that
IV(Go)\{s, t} = m and it is possible to win the reversible pebbling game on Go without
placing more than k pebbles. Letting v 1,-.
,vm
be the vertices of V(Go)
\{s, t},
if G
is an input graph such that
1. E(Go) C E(G)
2. All vertices in V(G)
\
V(Go) are lollipops
and {Vij : i E [1,q], j C [1,m]} are sets of vertices such that
1. For all i, Vi1,. -vim
I is a partitionof Vred(G)
2. For all I C [1,m] such that I| < k, there is an i e [1,m], such that v E Vi for
all j C I and Va
Vij
whenever a V I and j E I.
then Gk({Vi }) accepts the input graph G.
Proof. If we have pebbles on vertices vj: j E I in Go then this corresponds to being
at a vertex with parity knowledge set {(Vi, b(Vij
v3 C Vi for all j E I and Va V Vi whenever a V
\ {vj}))
I and j
: j E I} for an i such that
c I. We now match each
reversible pebbling move on Go with a sequence of moves in G,,
fVi
1.
Proposition 5.3.14. If vj E V(1), n V(,)j and all other vertices of V(ij 3 and V(,)j
are lollipops then
b(V(l)j \ {vj}) + b(V( 1 )jAV(i 2 )j) mod 2 = b(V( 2 )j \ {v 3 })
To add a pebble on
Vb
for some b E [1, m], we first use the copy of the switching
network given by Corollary 5.3.10 to shift to an i such that vj E Vi for all j E I U {b}
and Va 0 Vi2 whenever a V I U {b} and
j
c I U {b}. We now have the following cases
118
1. If we have an edge s -
Vb
G,,({Vij}) we first add
in E(GO) and are using it to add a pebble on
({Vb},
Lemma 5.3.8 to transform
({Vb,
in
0) to K. We then use a copy of the G' given by
0}) into
(Vib, b(Vi
2. If we have a pebble on va and an edge va
add a pebble on vb, in
Vb,
G,,k({Vi})
-* Vb
\ {Vb})).
in E(GO) and are using it to
we first use a copy of the G' given by Lemma
5.3.8 to transform (Via, b(Via \ {va})) into ({Va}, 0) in K. We then use the edge
Va --
Vb
to add ({vb}, 0) to K. Finally, we use a copy of the G' given by Lemma
5.3.8 to transform ({Va}, 0) into (Via, b(Via
\
{va})) and then use a copy of the
G' given by Lemma 5.3.8 to transform ({vb}, 0) into (Vib, b(Vi \ {Vb})).
To remove a pebble on vb, we run these steps in reverse except that we do not need
to shift to a different i.
If we have a pebble on va and an edge va -+ t in E(G) which we use to win, in
G,,({Vij}) we first use a copy of the G' given by Lemma 5.3.8 to transform (Via, b(Via\
{Va}))
into ({Va}, 0) in K. We then use the edge vi -
t to add ({t}, 0) to K and
win.
D
We now analyze the size of G
Lemma 5.3.15.
IV(G,,k(IVij}))
k({Vij}).
2qn2 g n(2m) k + 2q 2 n(2m)k
Proof. We count how many of each type of vertex G({Vi})
1. There are at most q E1k
has.
2r (') possible parity knowledge sets of the form
Uj I{(Vi, bj)}.
2. There are at most qrt n
k
UjCI\{a}{(Vi, bj)} U{({Va,
{bj} and
2r (m) possible parity knowledge sets of the form
ba})} as we have the same choices for I and the values
are now choosing one vertex Va-
3. For each fixed a, there are at most q E_- 2'(7) copies of the switching network given by Lemma 5.3.8 between vertices with parity knowledge set of the
form UjCI\{a}{(Vij, b 3 )} U
{({ Via, b})}
and vertices with parity knowledge set
of the form UjGI\{a}{(Vij, bj)} U {({Va, b})}.
119
Each of these copies has at most
2 Va lg n additional vertices so the total number of additional vertices is at most
21Val ig n Ek- 2r (m). Summing this up over all a gives 2qn lg n
4. There are at most qn 2
UicI\{ac}{(Vij bj)} U
2 ' (in)
k_ 2r(7) possible parity knowledge sets of the form
{({Va,
ba )U
I{({vc b})}
as we have the same choices for
I and the values {bj} and are now choosing two vertices Va and
5. For each fixed c, there are at most qn j _- 2
(7)
vc.
copies of the switching network
given by Lemma 5.3.8 between vertices with parity knowledge set of the form
UjEI\{a}{(Vij,
bj)}U{({Va, ba})}U{({Vc, b})}
set of the form UjEI\{a}(
j
bj)} U
{({a,b
and vertices with parity knowledge
Each of these
a})} U {({vc, b})}.
copies has at most 2 Vc Ig n additional vertices so the total number of additional
vertices is at most 2qnIVc lg n j$k- ' 2r (7). Summing this up over all c gives
2qn2 lg n E _- 2r(r)
6. There are at most
(q)
_1=Z
(7) copies of the switching network given by Corol-
lary 5.3.10. Each such switching network has at most 2r+ln vertices giving us
a total of at most q2
k
2'(7) vertices
Summing up all of these numbers and adding s' and t' gives
k
2
q(1+n+n +qn)
k-i
2
+ 2q(n2+n)lgnE2
+ 2
r=O
r=1
If n > m > 4, this is at most 2qn 2 lg n(2m)k + 2q2 n(2m)k, as needed.
D
Finally, we put an upper bound on how large q has to be in order to have a nonzero
chance that for all possible v, - --vm, for all I C [1, m] such that III < k, there is an
i E [1, m], such that vj E Vi for all j E I and va
Vi whenever a
I and j E I.
To do this, we choose each Vil, -.- , Vin by taking each v E V(G) and randomly
selecting which Vpj it will be part of. For a given vi, -
, Vm
and I of size k, the
probability of success is (y)k (ink)m-k. Now
mn
-)k
k
(m
M- k)
k\ m\
>-(
> m-k
1 -- --
120
2k
> mn-k 2--2k __(Tn -k
Thus, for a given vi, - - , vm and I of size k, the probability that we fail for all i is at
most (1 - ( 4 m) k) . There are at most mknm possibilities for vi,- - , vm and I, so
the expected number of v,- - , vm and I which fail is at most
mknm (1 - (4m)-k)q
If q > (4m)k ig
(ik
nm) then this is less than 1 so there must be some chance of success.
-
Thus we may choose q so that q < 2m(4m)k ig n. Plugging this into 2qn 2 ig n(2m)k
2q2 rn(2m)k we get an upper bound of
2
23 k+ 2 m 2 k+ rn2 (lg n)
+
2 5k+ 3 m 3k+2 n(lg n) 2
as needed
23 k+ 2 m 2 k+lrn(n
+
2
2
k+lmk+l)
(lg n) 2
D
121
122
Chapter 6
Improved Lower Bound Techniques
for Monotone Switching Networks for
Directed Connectivity
In this chapter, we prove the following theorem, which shows that if G is a directed
acyclic input graph, we get a lower bound on m(G) based on the length of the shortest
path from s to t as long as the vertices of G aren't too connected to each other.
Theorem 6.0.16. If r and m are constants such that m < 2000r4 and G is a directed
acyclic input graph such that
1. There is no path of length at most 2'-1 from s to t.
2. For any vertex v G V(G), there are at most m vertices w E V(G) such that
either there is a path of length at most
2r-2
from v to w in G or there is a path
of length at most 2 r-2 from w to v in G
then
m (G) >
(9mn)
4
201 E(G) V/2(r+ 1)! 9m
The key idea in proving this theorem is as follows. In chapter 3 (see Theorem
3.4.25), we found functions gG,E, corresponding to each input graph G such that
whenever G' accepts G it must contain a vertex v' such that gG,Ei
123
-
9G,E1
V'
is large.
We then obtained a lower bound based on an orthogonality argument. Unfortunately,
the orthogonality argument depended on the input graphs consisting almost entirely
of isolated vertices.
To prove Theorem 6.0.16 we instead argue that if we take a
random permutation o- of Vred(G),
9G,E1 ) - v']
E[(gG,Ei -
is small for any possible v'
and this is sufficient to prove lower bounds.
6.1
Permutations on input graphs, cuts, functions,
and Fourier coefficients
We first define what we mean by permutations of functions, cuts, and input graphs
and show how permutations affect the Fourier coefficients.
Definition 6.1.1. We take
SVred(G)
to be the set of permutations c- of V(G) such that
u(s) = s and -(t) = t.
Definition 6.1.2. For each permutation - E
1. Given an edge e = v
-
SVred(G),
w, define o-(e) = o-(v)
-
o-(w).
2. Given an input graph G with vertices V(G), define -(G) to be the graph with
vertices V(G) and edges {ou(e), e
e
E(G)}.
3. Given a cut C E C, define -(C) to be the cut such that L(o-(C)) = o-(L(C)) and
R(u(C)) = o-(R(C)).
4. Given a function g : C -+ R, define the function -(g) : C
o-(g)(a-(C))
=
g(C), i.e. a-(g)(C)
=
-+
R so that
g(a--1 (C)).
Proposition 6.1.3. For all - C SVed(G),
1. An edge e crosses a cut C if and only if o-(e) crosses o-(C).
2. g : C -÷ R is e-invariant for some edge e if and only if o-(g) is o-(e)-invariant.
3. For allf,g :C->R, -(f)-o(g) =f g
124
4. For all V
C
Vred(G), o(ev) = eo(v)
5. For all g :C -+ R, for all V c Vred(G), OY),(A) =A
Proof. Statements 1 and 3 are clear, statement 2 follows from statement 1, and statement 5 follows from statements 3 and 4. Thus we only need to show statement 4. To
see statment 4, note that
-(ev)(C) = ev(o- 1 (C)) =
(-1)-,(oa(C)nv1
(_
1
)7-1 (L(C))nVi
-
(_i)L(C)no(V)
_
D
eo(V) (C)
Finally, we recall the definition of m(G).
Definition 6.1.4. For an input graph G containing a path from s to t, m(G) =
m({-(G) : - E SVde,(G)})
Remark 6.1.5. The function m(G) is a complexity measure on G which measures
how hard it is for sound monotone switching networks to solve directed connectivity
on G.
6.2
Calculating Permutation Averages
For our lower bounds, we will need bounds on expressions of the form EGSVred(G)
-(g)] 2 where
f, g
[f
are functions from C to R and c-(g) is a permutation of g. Here
we obtain bounds on EacsVred(G)[f . -(g)] 2 in terms of the norm I fII of f and certain
sums associated to the function g. This will be extremely useful because the function
f
will correspond to a vertex in an arbitrary sound monotone switching network G'
so we will have no control over anything except for I f 1. However, we will have a
large degree of control over g and will thus be able to adjust the values of many of
the sums associated with g to give us the bounds we need. We now state our bound
on
EEsV
d(G)[(f
Definition 6.2.1. Given functions gi,92 : C -+ R, define
Sk,u1,u 2 (Y1, 92 ) =
E
A,B,CCVred(G):AI=kIBI=ul,IC=u2,
AnB=AnC=BlC =
125
g1AUBg2AUC
Theorem 6.2.2. For any function g : C < - 1, for any function f : C ->
IVR,
R such that v is only nonzero when
;
EEsVe(G)[(f
o(g)) 2 ] <
+ U2)!
-u)!(k
-V'(k +
4
n
k,ui,u2
.
Sk,ul,u2 (gg)
k+u1+u2
2
lfHl
Proof.
Lemma 6.2.3. For any functions f, g : C - R,
k!ui!u 2!(n - k - ui
2)!
-
-
EovsVred(G)[(f . o(g)) 2 ]
k,ui
U2
Proof.
EaESV,-d(G)
2]=
.
F
(Z
I
f
or (g) V
)
EO'ESVre(G)[(f
Vi CVred(G)
E
fV 2 a (g) V2
V2 CVred(G)
Taking A =v nV2 , B = V1 \ V2 , C = V2 \ V, this is equal to
jAUB
ETSsVred(G)
k,u2,Uz
.
AUC
9 gAUB
S
AUC
A,B,CCVred(G):
LA|=k,\B|=u1,BC =u2,
AnB=AnC=BnC=O
which is equal to
Y,
k,ui,u 2
EAUBACEEESVred(G) U(Y)AUB
(g) AuC]
(6.1)
A,B,CCVred(G):
|A|=k,|B|=u1,|C|=U2,
AnB= AnC=BnC=O
Now by statement 5 of Proposition 6.1.3, for any A, B, C C Vred(G) such that JAI
|B|= u1 , |C|= u2 , and AnB=AnC= BnC=0,
EassVreS(G) U(g)AUB
()AUC]=
EaCsVreS(G) [-
1(AUB)Y-1(AUC)]
k!ui!u2 !(n - k - ui- U 2 )!
--
126
I
sk,ui,U2
(g, g)
=
k,
Plugging this into (6.1) we deduce that
Eo-csV, ed(G)
[(f
k!ui!u 2 !(n -
=
k - u1
-
2 )!
k,ui,u2
which completes the proof of Lemma 6.2.3.
E
We now wish to bound how large sk,ul,U2
(f,
f)
can be in terms of I|f 1l.
Definition 6.2.4. Given a function g : C --+ R and a subset A C Vred(G), define
SA,u 1 (9) =AUB
BCVred(G):
IB|=u1,AnB=O
Definition 6.2.5. Given functions 9 1 ,92 : C -+ R and a subset A C V,d(G), define
SA,u1,u2
(91, 92)
=
E
g1AUB2AUC
B,CCVred(G):
IBju1,ICI=u2,
AnB=AnC=BnC=0
Proposition 6.2.6. For all A,
8A,ui (91)SA,u 2 (g 2
SB,ui-U,u 2 -u(91,
92)
u>O ACBCVred(G),
|BI=AI+u
,
Corollary 6.2.7. For all k, u1 , and u 2
Z:
A:IAI=k
SA,ui (91)SA,u 2 (92)
=
5 (kI- )Sk+u,ui
-U,U2 -U
(91, 92)
u>O
Proof. This follows immediately when we sum Proposition 6.2.6 over all A of size
D
k.
Surprisingly, Corollary 6.2.7 has an almost identical inverse formula.
127
,
Lemma 6.2.8. For all k, u1 , and U2
Sk,u,u2
(- 1)U (k
(91, 92)
SA,ui -u (91) SA,u 2 -u (92)
U
u>O
A:|AI=k+u
Proof. This lemma corresponds to the fact that the inverse of a matrix like
(k+1)
(k+2)
(k+3)
o
1
(k+2)
(k+3)
0
0
1
(k+3)
0
0
0
1
is
(k+1)
1
k+2)
2
(k+3)
(k+3)
(k+2)
0
1
0
0
1
0
0
0
(k+3)
1
Calculating directly, every entry in the product of these matrices has the form
(-1)u
U=O
where
j
+ u)
U
(j+2)
m-U
is k plus the row number and m is the column number minus the row number.
Now if m > 1,
m
(-1)"
(Ij +
+ Mn
U)
j!
U=0
m)'
_(j+
-(I
f!m!
U=0
(-1 )"
+ (-1))"
1
=0
If m = 0 then
m
1
E
-1
UO-
E
To bound how large ISk,u,U 2 (f, f) can be in terms of If 11, we just need bounds
128
for how large expressions of the formZAjAjk SA,uj(g)sA,u 2 (g) can be.
,
Proposition 6.2.9. For all k, ul, u 2
>
sA,ui(9)sA,u 2 (9)
A:IAI=k
1:
(SA,ui (_9))2
A:IAI=k
(A,U2
(g) 2
A:IAI=k
Proof. This follows immediately from the Cauchy-Schwarz inequality.
D
Proposition 6.2.10. For all k,u1 and all A with A| = k,
(SA,ui(g))
k)
2
I:
)2
( AUB
BCVred(G):IBI~ul,
AnB=O
Proof. This follows from the Cauchy-Schwarz inequality
(BCVed(G):IBI=
2
I
<
f(B)h(B)
)
E3
BCVred(G) IBI=ui,
AnB=O
AnB=0
E3
(f(B)) 2
(h(B)) 2
BCVred(G):IBI=ui,
AnB=0
El
with f (B) = 1 and h(B)= .AUB.
Corollary 6.2.11. For all k, ul,
(SA,u 1 (g)) 2
I:
A:IAI=k
k)
k+u1)
U1
(
B:IBI=k+ui
<
k +1
n- k
2
(yB1)
Proof. This follows immediately when we sum Proposition 6.2.10 over all A of size
El
k.
,
Corollary 6.2.12. For all kU 1 , U2
A:IA|=k
SA,ui (9)SA,u
k) n -k)
(n
2
(9
U1
U2
k
(k +U1
U1
)
(
I:
+U2
U2)
gI
Proof. This follows immediately from Proposition 6.2.9 and Corollary 6.2.11
129
E
f
is a function f : C -
S
(k
U>0
f)
k
-
U)
U
("U1i-
+
Sk,u,U 2 (,
,
R, we have that for all k, u 1 , u 2
-
u
U
)
Corollary 6.2.13. If
n - k- u
k+u1
k+U2
U2-U )(U -U||U2-|
1l
Proof. This follows immediately from Corollary 6.2.12 and Lemma 6.2.8 with k replaced by k + u, u, replaced with u, - u, and u 2 replaced with a2
-
D
U.
Corollary 6.2.14. If n > k + max{ui, u 2} + 2(max{ui, u 2}) 2 then
(f, f)l
< 2
(k+u)
Ui (n)
U2
(n)
U1
k +U2
U2 lfil
)
ISk,ulu2
Proof. The idea is to show that each term in the sum in 6.2.13 is at most half of the
previous term. This follows from the equations
1
k+u+1
-_____
+1
)k+u
'
U
2U-1
u2-u-1
(ul-u)(U2-u)
2
(n-k-u)
n-k-
UjUk+u 1 1 U2
4(ul-u)(u2-u)
n2 k-u_
k+u 2
(ui -U) (U2 -u)
(k+u+1)2
-
32.
1
-
FD
Theorem 6.2.2 now follows easily. By Lemma 6.2.3,
EgcsVred(G)[(f . o(g)) 2 ]
k!ui!U2 !(n - k
=
ni!
k,u1 ,U2
130
-
ui-
U2)!
Sk,u2 (f i
)S
,u2 (9
g)9
Assuming that n > 2k + u -+u 2 , plugging Corollary 6.2.14 into this gives
Eas
~Sd
(G)
[(f
g)2
.
(k-+ i- l!kU2!(n
!k+9)
<2<2S
E -(+
(n - u n -U2)!
kulu2
V(k +-2-k+21+u2
-u)!(k
+ U2)!
<2
u2
<2
-
U2)!
k
V
4<4
ISk,ui,U2 (g,
k+U1 +U2
n
ku,2(gg)IIfII
sku1 ,u2 (g, g)U
'f
+
2---2n
k,u,,U2
as needed.
2
(ui+u 2 +k)(k+u1+2)
k,un
(k +U2)!
k
vl-f(k + u 1)! (k +
ki 2 u
9
g)I-II
2
To check that the bounds on n hold when needed, note that by our
assumption that .v is only nonzero when IV I
k + max{ui,
2
}>
-1.
- -- 1, we may ignore all terms where
Thus, for all of our terms, n
2(k+max{ui,u 2}
+ 1)2
This is bigger than both k + max{ti, U2} + 2(max{ui, u 2}) 2 and 2k + ui + U2, as
needed.
E
For the functions g we will be looking at, it is difficult to bound |sk,ul,u2 (9, g)1
directly. We would like a bound in terms of the sums SA,u(g). To obtain such a
bound, we apply Lemma 6.2.8 and Proposition 6.2.9 to Theorem 6.2.2.
Corollary 6.2.15. For any function g : C -+ R such that v is only nonzero when
IV I
r for some r < -
EUESVred(G)[(f
- 1, for any function
O(g))
2]
f
: C -+ R,
< 4(r + 1)f 11E 2'
k,u
(SAu(g)) 2
(k + u)!
E
k+u
A:IAI=k
Proof. By Lemma 6.2.8 and Proposition 6.2.9,
(k +-i)!(k +
n
ku1,U2
<SE
ku ,U 2
u:!min {U 1 ,u 2 }
k+u1+u2
2
(
k~
k--\
2)!
sk,ui,u 2 (9 9)
k k+ul)!(k
ul~u2 +
n
2
(5A,ul -u(9)) 2
U22)!
A:IAI=k+u
,
2(k +ui)!
-
f II
ISku172(1911
Isk,ui,u 2 (9, g)| '|f|
-
ku1 ,u 2 (n
u1 - U2 - k)! sk,ul,U2 (9, 9)1 -1
131
>1?
E
A:IAIu~k+u
(AU
(
g-())
2
Replacing k + u with k, u1 - u with u 1 , and u 2 - u with u 2 on the right hand side,
z (k +
,U
uI:
n
k,ui,U2
ui)!(k+ U 2 )!
k+
Sk,u 1 ,u 2 kyg)
2
1U
2
+ui)! (k + U2)!(g)
k)(k
A:|AI=k
(sA,u 1 (9)) 2
S
(sA,u 2 (g)) 2
A:|AI=k
(
k,u1,2
_
n k+u_ 2 u2
2
n
u1
k
2
A:IAI=k
2
u)!g))2
1)+
(r+
sA,ul
+
k
= ((
u
n
i
+1)! 2S~u (g)
AI (A|=k
2
2
k,u1
A:A|I=k
Plugging this in to Theorem 6.2.2 gives the desired result.
6.3
D
Lower bounds from permutation averages
In this section we show how we can obtain lower size bounds on monotone switching
networks using Corollary 6.2.15.
Definition 6.3.1. For an input graph G with a path from s to t, we say that FG =
{g : e e E(G)}
is a set of invariant functions for G if
1. For all e c E(G), ge is e-invariant.
2. For all e E E(G), g, - e
= 1
Theorem 6.3.2. Let G be an input graph containing a path from s to t and let
FG= {g, : e E E(G)} be a set of invariant functions for G. If for all e E E(G) we
have that 4 ev is only nonzero when
|V| <
r for some r <
_-
1, then for any edge
e o c E(G),
m(G) >
1
1max
|E(G)j -I1
e(EE(G)\{e,,}
(k +
+ u)!
4(r + 1)5 k 2k(k
k
132
n k+u
AI)
A:A
(sA,u(ge - geo))2
k
Proof.
Lemma 6.3.3. Let G be an input graph containing a path from s to t and let FG
=
{g, : e c E(G)} be a set of invariant functions for G. For any edge eo E E(G), for
any sound monotone switching network G', for any path P' in G' from s' to t' whose
edge labels are all in E(G),
1.
Zv',Gp') ZeE E(G)\{eo}
2.
Ev'CV(G),ecE(G)\{eo} 1V'
3. EvCV(G'),eEE(G)\{eo}[IV'
K'
(ge
(g -
-
geo)f > 1
g)
>
(ge - Yeo) 2 -
1)V(G)
(lE(G)I-1) 2 1V(G')
2
Proof. The first statement follows from Corollary 3.4.23. The second statement follows immediately from the first statement. The third statement follows immediately
D
from the second statement and the fact that for any X, E[X 2 ] > (E[X]) 2 .
Now if G' is a monotone switching network accepting all of the input graphs
{o-(G) : - G
SVred(G)}
EorSVred(G)
then by statement 3 of Lemma 6.3.3,
[Ev'EV(G')\{s',t},eE(G)\{eo}[V - U(ge
(E(G)
- ge 0 ) 2]]
1
-
1)2V(G')2
Applying Corollary 6.2.15 to g, - ge0 and v' for each e E E(G) \ {eo} and v' E V(G'),
since IIv'I < 1 for all v' E V(G'),
|VG'|
EeCEG\eO [4(r + 1)
> E(G)I1V(G')l
- I EnEG\e}ATk+u
S 2k(k + u)!
k,u
()2
(sA,u fge -
eo
2
A:IA|=k
0
Theorem 6.3.2 now follows immediately.
Theorem 6.3.2 says that for our lower bound, we want to find a set of invariant
functions FG
the sums ZA:A(
{eg : e E E(G)} such that for some eo C E(G), for all e E E(G) \ {eo}
k (sA,u(ge - geo)) 2 are as small as possible.
133
6.4
Equations on sum vectors
Rather than choosing the functions {ge : e E E(G)} directly, we choose the sums
{SA,u(ge)
A C Vred(G),u > 0,e E E(G)} and have these sums determine the
functions {ge : e
c
E(G)}. However, our choices are not entirely unrestricted. In this
section, we show which equations on the sums {SA,u(ge) : A C Vred(G), u
0, e E
E(G)} we need to obey. These equations are most easily expressed in terms of vectors
with sums as their coordinates.
Definition 6.4.1. For a function g and k, u > 0, define Sk u,g to be the vector with
one coordinate for each A C Vred(G) of size k such that
(Sk,u,g)A = sA,u(g)
Proposition 6.4.2. For any function g : C -+ RI, ISk,u,g|1
Vred(G), AI = k}
k + 1}. Take
Pk
-A:IAI=k
(sA,u(g))
2
be the matrix with rows correspondingto the subsets { A C
and columns corresponding to the subsets {B C Vred(G), B|
(Pk)AB =
-
Definition 6.4.3. Let
2
1 if A C B and 0 otherwise.
Proposition 6.4.4. For any function g, sk,u,g
Sk+1,u-1,g
Proof.
(Sk,u,g)A = SA,u()
Z
C
C:ACC,
IC|=IA|+u
( PkSk+1,u-1,g)A
Z
=
= 1
sB,u-l(g
:
B,C:ICI=IAI+u,
B:|B|=|A+1,
IB|=IAI+1,
ACB
C
C
C:ACC,
|C|=A|+u
ACBCC
Definition 6.4.5. For any collection of vectors {Sku,g : k, u > 0}, for all k and all
U>
1, define the error vector
1
ek,ug = Sk,u,g
-
-Pk Sk+1,u-1,g
As we will show, to guarantee that we have an actual function g it is sufficient
to have all of the error vectors be equal to 0. However, we also need to ensure that
134
we get an e-invariant function for each e c E(G). e-invariance gives us another set
;> 0}. We repeat Corollary 3.4.28 here for
of equations on the vectors {ku,9 : k,
convenience.
Corollary 3.4.28.
e{w})g
w
-
w for some w E Vred(G) then g is e-invariant if and only if (e{}
+
1. If e = s
0. Equivalently, g is e-invariant if and only if
=
.vu{w}
= -V
whenever
V.
-
2. If e = v -+ t for some v E Vred(G) then g is e-invariant if and only if (e{}
efv})g = 0. Equivalently, g is e-invariant if and only if .VU{v}
v
= .v
whenever
V.
3. If e = v -
w for some v, w
C
Vred(G) then g is e-invariant if and only if
(e{} - e{v})(e{} + e{fw)g = 0. Equivalently, g is e-invariant if and only if
gvu{v,w}
+ vu{w} + v whenever v, w
= -vu{v}
V.
Lemma 6.4.6. If g is an e-invariantfunction for some e E E(G), A ; Vred(G), and
|=
k then
1. If e= s -
w for some w C Vred(G) and w C A then for all u,
(Skug)A -
2. If e =v
(s'k,u-1,g)A
t forsomev G Vrd(G) and v E A then for all u,
(Sk,u,g)A =
3. If e = v
-
-(k_1,u,g)A\{w}
-
(8
_.1,,,g)A5(vj -
(Sk,u_1,g)A
w for some v,w E Vred(G) and v,w G A then for all u,
(Sk-1,u,g)A\{v}
(Sk,4u,g)A
--
-
(-k-1,u,
1,u_1,g)Axv1 -
135
9
)A\fwj +
(Sk-1,u_1,g)A\{w
(Wk-2,u,g)A\{v,w}
+
(IS,u-2,g)A
Proof. To show statement 1, note that
E
=
E
B
+
(k-1,u,g)A\{w}
B: ACB,
|B|=|AI+u-1
YB
B: A\{w}CB,
wVB,|B|=|AI+u-1
E
-
= (Sk,u-1,g)A
YBU{w}
B:A\{w}CB,
w O B , B| =j A|+u -1
5YB
-
= (Sk,u-1,g)A
B:ACB,
IBI=IAI+u
= ('k,u-1,g)A
('k,u,g)A
-
Rearranging now gives the desired statement. Similarly, to show statement 2, note
that
S
B
+
5
-:
BBAC
B:AQB,
|BI=|A|+u-1
vIB,|B|=|AI+u-1
E
+
(Sk,u-1,g)A
v
YBU{v}
B:A\{v}CB,
B,|B|=|A|+u-1
(Sk,u-1,g)A
59B
+
=
YB
B:A\{v}CB,
B:ACB,
IBI=IAI+u
=
(Sk,u-1,g)A
+
(Sk,u,g)A
Rearranging now gives the desired statement. The proof for statement 3 is more
complicated but uses similar ideas. In particular, note that
S
(*k-2,u,g)A\{v,w}
B -+-
IBI=IAI+u-2
E
YB+
B:ACB,
B:A\{v}CB,
B+
B:A\{w}CB,
viZB,|B|=|A\+u-2
w-B,IBI=IAI+u-2
YB
I:
B:A\{v,w}CB,
v,wOB,IBI=IAI+u-2
Now let's consider these terms one by one.
1-
5
B = (8k,u-2,g)A
B:ACB,
IBI=IAI+u-2
2.
E
B:A\{v}CB,
v VB,IB I= IAI+u-2
YB
=
BB:A\{v}CB,
IBI=IAI+u-2
E
B:A\{v}CB,
vEB,|B|= AI+u-2
136
YB =
k-,u-1,g)A\{v}-
k,u-2,g)A
3.
I
9B =
B:A\{w}QB,
w i B,|B| IA +u-2
YB
=
(jBU{v,w}
+
YB-
k-,u-,g)A\{w}-Sk,u--2,g)A
B:A\{w}QB,
B:A\{w}CB,
wEB,IB|=|A|+u-2
IB=IAI+u-2
4. By statement 3 of Corollary 3.4.28,
YB =
.BU{v}
-
YBU{w})
B:A\{v,w}CB,
v,wOB, B|=|IA|+u-2
v,wOB,|BI=|A +u-2
5B
E
E
-+
B:ACB,
YB
B
-
B:A\{v,w}CB,
B:A\{v}CB,
B:A\{w}gB,
wVB,IBI=IAI+u
\BI=\Ai-+u
gvB,IBI=\AI+u-1
-1
We now have that
1.
YB
(kug)A
B:ACB,
IBI=iA|+u
2.
Y
YB
=
5
(8k-1,u,g)A\{w}-
iB|=|
=
(SBk-1,u,g)A\{w}-- (k,u-1,g)A
wEB,IBI=IAI+u-1
YB
=
(8k-1,u,g)A\{v}
E
-
3.
YB
B:A\{w}CB,
B:A\{w}CB,
A|+u -1
w$B ,
YB
=
(Sk-1,u,g)A\{v} -
('k,u-1,g)A
B:A\{v}CB,
B:A\{v}CB,
vOB,|B|=|AI+u-1
vEB,IBI=IAI+u-1
Putting these three statements together,
I:
YB
=
(Sk,u,g)A
+
( k-1,u,g)A\{w} -
(k--1,u,g)A\{v}
B:A\{v,w}CB,
v,wOB,|B =|AI+u-2
Putting everything together,
(Sk,u,g)A
(8k-2,Ug)A\{v,w}
-
+
(Sk-1,u,g)A\{w}
(Sk-1,u-1,g)A\{v}
+
Rearranging now gives the desired result.
-
(8k-1,u,g)A\{v}
(k-1,u-1,g)A\{w}
-
('k,u-2,g)A
El
For each possible edge v --> w, we define difference vectors which show far a
collection of vectors {Sk,u,g : k
> 0} is from representing a (v -+ w)-invariant
function.
137
;> 0}, define the vectors
Definition 6.4.7. Given a collection of vectors {kug : k,
{Akug,v-_w : k,u > 0, v, u)
1. If w - A then
Otherwise,
# wv # tw =
V(G),v
(Ak,,,gs*w )A = (S-k,u,g)A -
s,v
s
w
-4
((s'k,u-1,g)A -
-4
t}
as follows
(sk-1,u,g)A\{w})
(Ak,u,g,s-+w)A = 0
2. If v c A then
(Ak,u,g,v-+t)A = (Sk,u,g)A -
Otherwise,
= 0
(Ak,u,g,v-+t)A
((8k-1,u,g)A\{v}
-
(Wk,u-1,g)A)
3. Ifv,wEA then
(k,u,g,v-÷w)A
=
--
Otherwise,
( k,u,g)A
--
(Sk-1,u-1,g)A\{v}
(1Ak,u,g,v-*w)A
-
(Sk-1,u-1,g)A\{w}
+
+
(k-2,u,g)A\{v,w}
(sk,u-2,g)A)
= 0
Lemma 6.4.8. For any edge e = v
and v -+ w z
(k-1,u,g)A\{w}
(Qk-1,ug)A\{v} -
-
w where v C V(G) \ {t}, w c V(G) \ {s, w},
k U > 0} corresponds to an
s -+ t, the collection of vectors {SI, ,:
e-invariantfunction g if and only if the error vectors ek,u,g and the difference vectors
Ak,u,g,e are all 0.
Proof. The only if statement follows from Proposition 6.4.4 and Lemma 6.4.6. For
the if statement, we define g by taking
A =
(sAIo,g)A.
The fact that EjAIOge
=
0
now corresponds precisely to the criteria for e-invariance in Corollary 3.4.28 so g is
e-invariant. Using the fact that 'k,u,g = 0 for all k and u, we show by induction on
u that for all k, u and all A with JA = k, ('k,U, 9 )A
case u
=
= ZB:ACBB|=AI+U5B-
The base
0 is trivial. For u > 0, assuming the result holds for u - 1 and following the
same logic as before,
I(Pkk+1u-1,g)A -
SB,u-(g)
B:|B|=IA+1,
ACB
5
=C
B,C:IC|=|A|+u,
IB=|AI+1,
C
C:ACC,
ICI=IAI+u
ACBCC
Since
(ek,u,g)A = 0, (Sk,u,g)A =
(Pk4k+1,u-1,g)A
138
EC:ACC,IC|=|A|+uC,
as needed.
D
6.5
Freedom in choosing a base function
In this section we analyze what freedom we have in choosing a base function g which
we will use to choose our functions {ge}. In turn, this will determine what freedom
we have in choosing the coordinates of our vectors {sk,,}.
Recall the graph H' used in the proof of Theorem 3.5.1. Adapting the definition
of H' to our current probelm, let H' be the graph such that
1. V(H')
{v' E V(G'(V(G)))
: v' has knowledge set K E {Kv : V
C Vred(G),
IVI < r}U {Kt}}
{(u', v') : u', v' c V(H'), there is an edge in G'(V(G)) between u' and
2. E(H')
V' with label in E(G)}
By the reasoning used in the proof of Theorem 3.5.1 we may choose the values of g-Kv
on each connected component of H' except for the connected component containing
t' where we must have that g - Kt,
=
0. We now determine what these connected
components are by finding a representative for each connected component of H'.
In doing so, our reasoning is very close to the reasoning about reachable pebbling
configurations in Li and Vitanyi [151 and in Li, Tromp, and Vitanyi [141.
Definition 6.5.1. Given a fixed r and a set of vertices V C Vred(G), we say an
w is relevant for V if it is from a vertex v E V U {s} to a vertex
edge e = v
-4
w C V U {t}
\ {v}
and there is a path in G from v to w of length at most 2 r-IVI-1
Definition 6.5.2. For a fixed r, we say that a set of vertices V C V(G) \ {s} is
irreducible if V = {t'} or V
C
Vred(G) and there are no relevant edges for V.
Theorem 6.5.3. Each connected component of H' has exactly one vertex v' such that
if V is the set of vertices such that K, = Ky (where we take KtI = Kt}) then V is
irreducible.
Proof. If V
C
Vred(G)
and there is a relevant edge e = v
w for V where w
-
$
t
then we may go from Ky to KV\{w} along edges in H'. This effectively reduces V to
V \ {w}. If V
C
Vred(G)
and there is a relevent edge e = v
139
-
t for V then we may
go from Kv to Kt, along edges in H'. This effectively reduces V to {t}. We will now
show that the process of reducing a set of vertices V
C
Vred(G) until we obtain an
irreducible set of vertices U always yields the same U.
Lemma 6.5.4. If there is a relevant edge e for V of the form v -s t then when we
reduce V to an irreducible set of vertices U, we always obtain U = {t}.
Proof. If we ever use an edge v
-
t which is relevent, we will reach U = {t}. Thus,
it is sufficient to show that no matter what other reductions we do, we always have a
relevant edge v
-+
t for V. To show this, note that if we have a relevant edge v
-
t
for V, the only way to make this edge irrelevant is to remove v from V. We can only
do this by reducing V with a relevant edge of the form u
-
v where u E V U {s} \ {v}.
However, once we do this, since we reduce the size of V by 1 the edge u
->
t must
now be relevant so we still have a relevant edge u -+ t for V.
E
Lemma 6.5.5. In the process of reducing V by relevant edges, if we can reach an
irreducible set of vertices U by starting with one relevant edge, then we can reach U
regardless of which relevant edge we start with.
Proof. We prove this result by induction on JVJ. The base case IVI
Assume that there is a relevant edge v 1
-+
=
0 is trivial.
v 2 such that if we start with v, -+ v 2 we
may reach U. We show that for any other relevant edge v 3
-+
v 4 , we may reach U if
we start with it. To show this, we first show that if we start by reducing V to V
using vi
->
\{v2 }
v 2 , we can then reduce V \ {v 2} to V \ {v 2 , v 4} by using another relevant
edge. Since we can reach U from V \ {v 2}, by the inductive hypothesis we can reach
U regardless of which edge we use next from V \ {v 2 }, so we must be able to reach
U from V \ {v 2 , v4 }. We then show that we can also reach V \ {v 2, v4 } by starting
with the relevant edge v 3
-
v 4, so we can reach U by starting with the relevant edge
v 3 -+ v 4 , as needed.
The proofs for these two parts are below. Note that we do not have to consider
the case where v 2 or v 4 equals t because in this case by Lemma 6.5.4 we always end
up at {t}.
140
1. If we start by using the edge v,
the edge V3
-*
v 2 to reduce V to V \ {v 2 }, we can then use
-+
v 4 to reduce V \ {V2} to V \ {v 2 , V4} unless v 3 = V2. However, in
this case, the edge v, to V4 is relevant for V
\ {V2}
so we can use it to reduce
V \ {v 2} to V \ {v 2 , v 4 }. Thus, we can always reach V \ {v 2 , V4} by starting with
v1
--
v 2 . Note that we cannot have vI
= 04
because G is acyclic.
2. Following similar logic, if we start by using the edge v 3
V \ {V4}, we can then use the edge v
->
V2
-
V4
to reduce V \
{V4}
to reduce V to
to V \ {V2,
unless vl= V4. However, in this case, the edge v 3 to V2 is relevant for V \
so we can use it to reduce V \
V\
{v 2 , V4} by starting with V3
to V \ {V2,
{V4}
-+
V4.
V4}.
V4}
{V4}
Thus, we can always reach
Note that we cannot have v 2 = v 3 because
G is acyclic.
Corollary 6.5.6. Regardless of how we choose to reduce V, we always end up at the
same irreducible set of vertices U.
Proof. Lemma 6.5.5 implies that the possible sets of vertices we end with is unchanged
by our first reduction and thus by any reductions that we do. However, once we reduce
to some irreducible set of vertices U, U is the only possible set of vertices to end with.
Thus, we must have always had that the only possible end set of vertices was U, as
needed.
E
We now prove Theorem 6.5.3 using Corollary 6.5.6.
Corollary 6.5.6 says that
we may define a mapping R which takes each set of vertices V and maps it to the
irreducible set of vertices that it reduces to.
Proposition 6.5.7. For all V, R(V) is in the same component of H as V
Proof. This follows from the fact that every reduction step can be carried out using
a path in H'.
Corollary 6.5.8. Every connected component of H has at least one irreducible set of
vertices.
141
Proposition 6.5.9. If V and W are adjacent in H then R(V) = R(W)
Proof. If V is adjacent to W in H then either we can obtain V from a reduction step
on W or vice versa. Without loss of generality we may assume that we can obtain
V from a reduction step on W. This immediately implies that R(W) = R(V), as
D
needed.
This means that R is invariant on all connected components of H, so there cannot
be a connected component of H with two irreducible sets of vertices U1 and U2 as
otherwise we would have R(U1 ) = U1 and R(U2 )
=
U2 which would mean that R
would not be invariant on this connected component.
E
By Theorem 6.5.3 and the reasoning used in the proof of Theorem 3.5.1, we
may freely choose the Fourier coefficients (P)v where V is irreducible and this will
determine the other Fourier coefficients
v where IV| < r. We now analyze what
freedom this gives us in choosing the coordinates {Sk,,g,}.
Definition 6.5.10.
We say that e = v -4 w is relevant for ('k,u, g )v if v G V U {s},
w E V U {t}, and there is a path of length at most 2 r-IVI-u-1 from v to w in G.
Definition 6.5.11.
1. We call a coordinate ('ko,g)v free if IVI = k < r and V is irreducible. Otherwise,
we say that (Sko,g)v is fixed.
2. We call a coordinate ('k,u, g )v free if there are no relevant edges for ('k,u,g)v.
Otherwise, we say that (Sku,g)v is fixed.
We now show that we may choose the free coordinates ('k,o,g)v and then determine
all of the other coordinates (Sku,g)v accordingly.
Theorem 6.5.12. If we choose the coordinates {(Sku,g)A} in increasing order of k+u
up to k + u = r - 1 where coordinates with equal k + u are chosen in increasing order
of u then if:
142
1. We choose the fixed coordinates {(4,u,)A} by choosing any relevant path v
for
(Sk,,,,g)A
and setting
(Ak,u,g,v-w)A
-+
w
= 0.
2. We choose the free coordinates { (,O,g)A}
arbitrarily.
3. We choose the free coordinates {(Sk'u,g)A} where u > 0 so that ( K,u,)A
=
0
then
1. For all edges e e E(G) there exists a function g, with these coordinates which
is e-invariant.
2. For any function g with these coordinates, for all V such that |Vt < r, g -Ky
=
g * KRMv)
Proof. We prove this theorem by induction on k + u. The base case k + u = 0 is
trivial. Assume that we have already chosen the values for k + u <
V such that IVt <
j, g -Kv
j
and that for all
= g -KR(v). Now consider the coordinates with k +u =
First assume that we choose the coordinates {(Sk,o,g)V
: |VI
j.
= j} so that for
any g with these coordinates, g - Ky = g - KR(v) (this choice is arbitrary for the
free coordinates). For any e = v
-+
w such that there is a path of length at most
2r-3-1 from v to w in G, whenever IVI < j -
1 and v E V U {s} we have that
R(V U {w}) = R(V) so g - Kvu{w} = g - Ky = g - KR(v). Using the same reasoning
used to prove Theorem 3.5.1, there is a function g with these coordinates which is
e-invariant.
For each such e, choose such a function g,. Now note that since g, is e-invariant,
(Ak,u,ge,e)V =
0 for all k, U, V. In particular,
implies that if we choose these coordinates,
(Ak,u,g,,e)V
=
(Ak,u,g,e)v=
0 whenever k+u < j. This
0 whenever e is relevant for
(Sk,u,g,e)V.
Now note that when we choose a fixed coordinate ( k,o,g)v, we set (Eko,,,-_w)v
where v
--
0
w is relevant for (sKo,g)v. There is a unique (Sk,o,g)v for which this is true.
From the above, we know that choosing (Sk,o,g)v so that g - Ky = g - KR(v) makes
(Ak,o,g,v_w)y =
all IV I
j, for
0. Thus, this choice must be our actual choice. This implies that for
all g with these coordinates, g -Kv
143
=
g - KR(v).
At this point, we just need to show that the choices for ('k,u,g)v where u > 0
correspond to our functions {gej, which we show with the following lemma.
Lemma 6.5.13. For all of the functions g, described above, if we choose the coordinates {(S,,g)v} as described above then ('k,u,g)v
= (s,u,g,)v whenever k + u < j.
Proof. We prove this by induction on u. Note that all of the functions {g,} have the
same coordinates for k + u < j so we just have to prove equality for a single e. The
base case u
=
0 is the fact that all of these g, have the coordinates {(Sko,g)v}. For
the free coordinates
that (,)
u,g)v
with u > 1, since we are choosing these coordinates so
(S4ku,g)v
0v =(0P(k+1,u-1,g)v = (Pk
k+1,u_1,g,)v = (su,g,)v. For
the fixed coordinates (k,g)v with u > 1, we chose (sk,u,g)v so that (A,,,,e)v = 0
for some e which is relevant for (sk,u,g)v. Now note that
(Ak,u,ge,e)v
the inductive hypothesis, ('k,,g)v = (sku,g)v whenever IVI +u <
that
(aku,g)v
= (SkU,g)V,
= 0 as well. By
j, so
we must have
El
as needed.
l
This completes the proof of Theorem 6.5.12.
Theorem 6.5.12 says that we may choose the coordinates (sk,u,g)v in increasing
lexicographic order of k + u and u and may choose the free coordinates
arbitrarily. However, this does not allow us to control the vectors Skug for
To control the vectors
,
(Sko,g)v
u
> 1.
for u > 1 we now show that we may instead choose the
coordinates ('ku,g)v in increasing lexicographic order of k + u and k and choose all
of the free coordinates (,u,g)v
as long as we make sure that the error vectors are 0
on the free coordinates (they will automatically be 0 for the fixed coordinates).
Corollary 6.5.14. If we choose the coordinates {(k,,g)v} in increasing order of
k + u up to k + u
=
r - 1 where coordinates with equal k + u are chosen in decreasing
order of u then if:
1. We choose the fixed coordinates { Sug)v} by choosing any relevant path v
for
(Sk,u,)A
and setting
(Akug,'vW)A
=
-4
w
0.
2. We choose the free coordinates {(4,,g)v} so that ('k-1,u+1,g)v = 0 whenever
(Wk-1,u+1, 9 )V
is free.
144
then
1. For all edges e G E(G) there exists a function g, with these coordinates which
is e-invariant.
2. For any function g with these coordinates, for all V such that |V| < r, g -Kv
g - KR(v)
Proof. This is exactly the same as Theorem 6.5.12 except that we are choosing the
coordinates in decreasing order of u rather than increasing order of u. This has no
effect on our choices for the fixed coordinates as all of our choices only depend on
coordinates ('ku,g)v where k +u < j. For the free coordinates, note that in Theorem
6.5.12 the free coordinates (sko,g)v are defined by the coordinates (8k,o,g)v (which
are arbitrary) and the requirement that (sk-1,u+1,g)v
0 whenever (k-lU+l,
9
)V
is
free. As long as these properties are preserved, the order in which we choose the
free coordinates (Sku,g)v does not matter. Thus, we may choose these coordinates in
decreasing order of u as long as we ensure that (Sk-1,u+1,g)v
=
0 whenever (sk-l,u+l, g )v
E
is free.
6.6
Building up a base function
Corollary 6.5.14 allows us to construct a base function g while having considerable
control over all of the sum vectors {Sk,g,}. This is extremely useful because we will
try to make II,,ug I small for every k and u. However, in order to use Theorem 6.5.12
we need a way to choose the free coorindates of the sum vectors {
kug}
to make these
vectors small while also making the error vectors 0. In this section, we show how to
do this. For this section, we are always working with collection of sum vectors {k,u,}
where we have already decided which coordinates of these vectors are fixed and which
coordinates of these vectors are free.
Definition 6.6.1. For each k, U,
1. Define
l7k,ufixed
to be the projection which projects any vector in the same vector
space as Sk,,,, onto the fixed coordinates of Sk,u,g.
145
2. Define rk,u,free to be the projection which projects any vector in the same vector
space as Sk,u,g onto the free coordinates of sk,,,.
3. Define Sk,u,g,fixed
=
4. Define Sk,u,,free
=Fk,u,free(sk,u,g)
Proposition 6.6.2.
7k,u,fixed(Sk,u,g)
S'k,u,g
=
Sk,u,g,fixed
+
Sk,u,,,free
Proposition 6.6.3. The free coordinates of the error vectors are all 0 if and only if
Sk,u,g,free =
Wk,u,free
(Pksk+1,u-1,g)
U
whenever u > 0
Definition 6.6.4. Let Pk,u,fr,,ee be the matrix Pk except that we make all of the rows
corresponding to a set of vertices A such that
(Sk,u,g)A
is fixed 0 and we make all of
the columns corresponding to a set of vertices B such that (Sk+1,u-1,g)B is fixed 0.
Definition 6.6.5. When u > 0 define the adjustment vector
ak,u,g
Proposition 6.6.6.
7
~-k,u,free(Pk Sk+1,u-1,g,fixed)
+
k,u,free(Pks'k+1,u-1,g) = -k,u,g
Pk,u,freeSk+1,u1,gfree
Corollary 6.6.7. The free coordinates of the error vectors are all 0 if and only if
Pk,u,freeSk+1,u-1,g,free = ak,u,g
+
Usk,u,g,free
whenever u > 0
Proof. This follows immediately from Proposition 6.6.6 and Proposition 6.6.3.
If we let P be the matrix
Pk,u,free
E
after we delete all the rows and columns
corresponding to fixed coordinates (which were all 0 rows and columns by definition), let X' be the vector dkn,g + USk,u,g,free after we delete all fixed coordinates
146
of -k,ug (which were also all 0) and let ' be the vector sk+1,u-1,g,free after deleting all fixed coordinates of Sk+l,u-1,g (which were also all 0), then we now have the
matrix equation PF? =
If P does not have full row rank then it may be im-
?.
possible to find a ' such that Py' =
. We will take care to avoid this case by
showing that PPT has no zero eigenvalues. If P does have full row rank then we
want to find the Y with smallest norm such that P/ = ? and then find a bound on
Sk+1,u-1,g,free1| =
we can get is
MY
2
1Y 2
in terms of
- USk,u,g,free 2 -
Hakd,,
The best bound
2.
{
; maxsyj{minv
We have now reduced our problem to a problem of the following form. For a given
real matrix P with full row rank, obtain bounds on maxyoo{ming.p#g
{
}}
Lemma 6.6.8. If P is a real matrix with full row rank then letting { A} be the
eigenvalues of PTP,
2
2
.r
max
m mmg_x+O 1 y:Py=x
max {-}
}}= i:,\i=4o
1:#0k Ai
Proof. Note that PT has full column rank, so PI = X if and only if pTp.
-
PTE
Thus, for a given i we are looking for the ' with smallest norm such that PTPYI
PT-X.
Since pTp is a positive semidefinite real symmetric matrix, if
j
is the number
of columns of P then there is an orthonormal basis of Ri consisting of eigenvectors
el, ..
- -ej
of pTp with eigenvalues Ai,
Proposition 6.6.9. If PT
-
=
, Aj. We now write pT
cie' then ci
=
=
_ cii.
0 for every zero eigenvector ei of
pTp
Proof. Note that if pTP
= 0 then -pTpi = Pei - Pei = 0 so Pei = 0. But then
eWPT=c = X-T P- =0
E
Writing
=
Zi=
iei and solving the equation PTP= pTj
pTp -pTp(d.
=
i=1
Zidis
i=1
147
=
Zcis
i=1
we obtain that
This says that di =-
whenever Ai # 0 and di can be anything when Ai =0 (as ci = 0).
To minimize the norm of ', we take y= ZI
2 = SpTpO
=
Ei:AiO
$ ge'. Now
2
2 and
(C,)2
Comparing term by term, we have that
ZA.O
(A)2
(maxA=o
{&})
Z
1
and this inequality is an equality when ci is nonzero for this particular i and 0 for all
other i, so maxj:o{minVppy
}
=
maxo
{y}
Thus, we are interested in the smallest nonzero eigenvalue of pTp. Since P is real
with full row rank, the eigenvalues of ppT are precisely the nonzero eigenvalues of
pTp, so we can find the smallest nonzero eigenvalue of pTP by finding the smallest
eigenvalue of ppT.
We give an exact answer when P =
Pk
and give bounds for many other P.
Theorem 6.6.10. If k < 1 then PkPi has eigenvalues (n - k - i)(k + 1 - i) with
multiplicity (n)
-
(i_2) for i E [0, k]
Proof. Given V C Vred(G) where
lVi
< k, let s'v be the vector such that for any
A C V if B C Vred(G), |B| = k, and BnV = A then ('v)B =
(-)|k,
5
where BnV = A have total weight
In other words, we make the coordinates ( V)B
(-l)IAI
where the weight is spread evenly among these coordinates.
Lemma 6.6.11. If V
C Vred(G)
and |V|
=
m < k then 5 v is an eigenvector of PkPk'
with eigenvalue (n - k - m)(k + 1 - m)
Proof. Consider where the weight is in PkPkx'v. Letting m = JVJ, the coordinates
where IBn VI = j have total weight (-i)i().
Multiplying by PkP[ multiplies
and shifts this weight as follows. For a given B with IB n Vl =
m+
j)j ways to
j
there are (n - k
-
(Xv)B
add a vertex to B then remove a vertex and obtain a set of vertices
C such that IC n VI = j - 1. There are (m - j)(k - j) ways to add a vertex to B then
remove a vertex and obtain a set of vertices C such that IC n VI = j + 1. Finally,
since there are (n - k)(k + 1) total ways to add a vertex to B and then remove a
vertex from B there must be (n - k) (k +1) - (n - k - m+ j)j - (m - j)(k - j) ways
to add a vertex to B then remove a vertex and obtain a set of vertices C such that
148
C n V=
j.
Thus, from the original total weight of (-I)i(') on the coordinates
( V)B where B n Vl = j, we get
1. A total weight of (-1)i(7)(--k -m+j)j on the coordinates (PkPkj 'V)c where
|CnV= j-l
)(m - j)(k -j) on the coordinates (PkP[?v)c where
2. A total weight of (-1)j
|CfnV =j+1
((n - k)(k +1) - (n - k - m +j)j - (m - j)(k - j))
3. A total weight of (-1)()
on the coordinates (PkP,[yv)c where |C n V = j
Turning this around, in PkPTzv we have the following contributions to the total
weight of the coordinates (PkPkTxj)c where IC n VI = j:
1. A contribution of (-1)j+'
(TV)B
where B n VI
(iN) (n -
j+
k - m+ j+ 1)(j + 1) from the coordinates
1
2. A contribution of (-1)--' (g)(m-- J+1)(k - j+1) from the coordinates ('v)B
where IBnVl=j-1
3. A contribution of (-1)j(')((n - k)(k + 1) - (n - k - m+j)j - (m - j)(k - j))
from the coordinates ('V)B
where B
nV
=
j
Summing these contributions together, we get the following total weight for the co149
ordinates (PkPk,7jv)c where |C n V| = j
m
+ (-1)
(((n-k)(k+1)-(n
+
\\j)
-
j+1)(j+1)-
+ I(n -k-rn
:+1m-+
(m
-
j+
1)(k - j + 1)
-k-m+j)J-(m-j)(k-j))
(_)
(-)
rn
(n - k - m + j + 1)(j + 1) + (1)3-1)(
Ir+ 1)(kr--j
+(m-j+1)(k--j+1)
+
(-i)i(
+(n--k)(k+ 1)- (n-k-m+J)j-(m-j)(k-j))
=
+ (n-k)(k+1)
=(-1)()(
j+
j)(n - k - m +j + 1) - (n - k - m + j)j - j(k -
(-1)-
1)
(m-j)(k- j))
-m(n--k-m+j
+1)+j-jk+J2 -j+nk+n-k
2
-k-mk
+jk+jm-j2)
=(-1)j (7)(rmn+M2
=
(-1)
-rm+nk+n-k 2 -k)
( (n-k-m)(k+1 -im)
By symmetry the weight on the coordinates (PkPkiv)c where IC n VI =
j
will be
+
spread evenly. Thus, 'v is an eigenvector of PkP[ with eigenvalue (n - k - M) (k
D
1-rm) where m = IVI, as needed.
Now we need to check that the vectors {v : IVI = m} span the eigenspace of
PkP[ with eigenvalue (n - k - m)(k +1 - m) and that these are the only eigenvalues.
We also need to find the dimension of these eigenspaces.
Definition 6.6.12. For each V C Vred(G) where IV| < k, take /v to be the vector
on subsets B C Vred(G) of size k where (fv)B
=
1 if V C B and 0 otherwise.
Lemma 6.6.13. For all j < k, the vectors {Yv
V|
=
are a basis for the vector
space spanned by the vectors {V : V| - j}
Proof. We first check that each 'y with V
vectors { v: IV <j}.
150
=
j is
in the vector space spanned by the
Lemma 6.6.14. For any V C Vred(G), the vectors {A:ACVIAj
are a basis for the vector space of vectors & such that ()A
C
XA
only depends on
[0, V|]}
|A
V|.
Proof. Note that all of these vectors are nonzero and live in different eigenspaces
of PkP[, so they are all linearly independent.
(EA:ACV,IAI=j
XA)B
only depends on |V
By symmetry, for all j c [0, VI],
n B| so all of these
vectors are in this vector
space. This vector space has dimension IV + 1 so these vectors must be a basis for
this vector space, as needed.
D
Corollary 6.6.15. For all V C Vred(G) of size at most k, ,v E span{ZA.ACVIAj
j
aE[0, Vl]}
We now check that the vectors {
the vectors {y : VI <
VI =
j} span
the vector space spanned by
j}.
Lemma 6.6.16. The span of the vectors {v : V C Vred(G),IV I= j} contains all of
the vectors {sy
j}
V C Vred(G), V|
Proof. For all V C Vred(G) where VI
i c [0, V
1}.
j, consider the sums {ZA:ACVIAnvI=i
YA
By symmetry, the values of these sums on a coordinate B will depend
only on IB n VI. Moreover, this sum is 0 on a coordinate B if IB n VI < i and
nonzero if IB n VI = i. Thus, if we take a linear combination of these sums by
choosing the coefficients in increasing order of i, each choice determines the value on
the coordinates B where IB n VI = i and does not affect the value on the coordinates
B where B n VI < i, so we may arbitrarily determine all of these values. This implies
that s'v is a linear combination of these sums, as needed.
Finally, we check that the vectors {V
independent. First note since k <
the vectors {
,
pT k has
: V
C
Vred(G),VI
D
=
}
are linearly
full column rank. To see this, note that
:V = k} are a basis for the vector space which PkPkj acts on. From
the above, this vector space is spanned by the vectors {V : VI < k} so the only
eigenvalues of PkP are {(n - k - m)(k + 1 - m) : m E [0, k]}. In particular, PkPkT
has no zero eigenvalues so
PTk
has full column rank. Now assume that there are
coefficients {cV : V C Vred(G), Vl
=
j} such that ZV:VVre(G),IVI=j cvv = 0. If we
151
let ' be the vector with value cv in every coordinate V,
since k < a2 all of the matrices Pk-1'
k-1
-
PC
0. However,
PT have full column rank and thus '= 0.
I'
All cv are 0 so the vectors {f/v : V C Vred(G), V
-
j}
are linearly independent, as
needed.
E
Using Lemmas 6.6.11 and 6.6.13, for every
j, the
sum of the dimensions of the
eigenspaces of PkP[ with eigenvalues (n-k-m)(k+ 1-rm) where m < j is equal to the
number of vectors {yV : V C Vred(G), IVI = j} which is ('). This immediately implies
that the dimension of the eigenspace of PkPkj with eigenvalue (n - k - j)(k + 1 - j)
is
(n)
-
E
(j).
For our results we are not dealing with the matrix Pk itself but rather matrices
which are close to Pk. We now bound the smallest eigenvlue of such matrices.
Theorem 6.6.17. Let P be a matrix obtained from Pk as follows: Take a collection
of bad vertices V and bad pairs of vertices E. Delete the rows and columns containing
bad vertices and pairs of vertices from
Pk.
If
1. |V| < n
2. There is an m <
vertices w
2QgJok
such that for every vertex v
V, there are at most m
V with {v,w} c E
then the smallest eigenvalue of ppT is at least n2
Proof. The proof idea is as follows. The smallest eigenvalue of ppT is miny {I
12}
This quantity can only be reduced when we subtract a symmetric positive semidefinite
matrix M from PpT. Thus, we will choose M so that M is a symmetric positive
semidefinite matrix and ppT
-
M has a large minimum eigenvalue, and this will give
the claimed bound. We construct M as follows:
Definition 6.6.18. Let U, W be subsets of Vred(G)
\V
such that |UI = k - 1 and
U n W = 0. Define Mu,w so that MU,w has the same rows and columns as ppT and
1. If no pair of vertices in U U W is bad then (MU,W)A 1 A2 = 1 if U C A 1 C U U W
and U C A 2 C U U W and 0 otherwise.
152
2. If a pair of vertices in U U W is bad then Mu,w = 0
Lemma 6.6.19. If we take M =
EU,W:U,WCVred(G)\V,
|U=k-1,|W|=j,UnW=0
MUw and have |A 11 = |A 2 | = k
then
1. If A 1 U A 2 has no bad vertices or pairs of vertices and A 1 n A 21 = k - 1 then
n -|JV|
2mj( j + k)
- k - 1
<_~A
-|V|
_n
j-
j -2
n
- k - 1
2
2. If A has no bad vertices or pairs of vertices then
+ k))k n -- V - k
(-2mJ(j
MAA
k n -V
I-
- I
n
- k
I
3. If A 1 or A 2 has any bad vertices or pairs of vertices, MA 1 A2 is undefined. Otherwise, if A 1 U A 2 has any bad pairs of vertices or JA1 n A 2 | < k - 1 then
MA 1 A 2
=
0
Proof. Statement 3 follows immediately from the definitions. The upper bounds in
statements 1 and 2 follow by noting that these are the maximal possible number of
Muw which have
(MUW)A
1
A2
or
(MUW)AA
equal to 1. For the lower bounds, for any
A 1 , A 2 such that A 1 U A 2 has no "bad" vertices or pairs of vertices and IAinA 21 > k-1,
take random U, W with U, W
C
Vred(G) \ V, |UI = k - 1, JW| = j, U C A 1 C UU W,
UnW = 0 and U C A 2 C UU W. To do this, first randomly choose U. After choosing
U, start with Wo = A 1 U A 2 \ U and add vertices to W one at a time. We add at
most j vertices and the probability that each new vertex adds a bad pair of vertices
is at most
m(j+k)
n-IVj-(k+j)
Thus by the union bound the probability that (MU,W)A
1 A 2 is
(U)l2i
mj&j+k)
0 rather than 1 is at most n-IVI-(k+j)
<
2mj(j+k)
n4
-
as IV| < i and kj << n.
El
Corollary 6.6.20. If we instead take
M
_n-_V-k-) ZU,W:U,WC Ve(G)\VIUI=k-1,w|=j,Unw=0 MUw
=
1. If A 1 U A 2 has no bad vertices or pairs of vertices and
0
< (ppT
-
M)AM)AjA2
1 A 2 <-
2mj(j+k)
n
153
then
|A1 n A 21 =
k - 1 then
V|
-
2. If A has no bad vertices or pairs of vertices then (ppT - M)AA > n - k km - k(n-k-IVI)
1
j-
3. If A 1 U A 2 has any bad vertices, (ppT
M)A 1 A2 is undefined. Otherwise, if
-
A 1 UA 2 has any bad pairs of vertices or |A 1 nA 2 | < k-1 then (PPT -M)AA
2
= 0
-
Proof. This follows immediately from Lemma 6.6.19, the fact that (PPT)AA > n
k - IVI - km whenever A has no bad vertices or bad pairs of vertices, and the fact
that
VI
n-
-1
j
(n-IVI-k-1)
-
-
k
1
The final step is as follows. For each column, we subtract all of the non-diagonal
elements from the diagonal element and then set all non-diagonal elements to 0. This
will not increase the minimum eigenvalue as
(1
0
is a positive semidefinite matrix. Note that each column has at most k(n - k - IVI)
nonzero non-diagonal elements, so by Corollary 6.6.20, all of the diagonal elements
are still at least
n- k
-
V|
-
km
-
k(n-k-IVI)
1
J-
_
2mj(j+k)k(n-k-IVI)
n
k-VI)_ 2mj(j+k)k(n-k-IVI)
n then n-k-|V|-kmTaking j = 8k, if m <
- 2000k0 then n
Thus, the minimal eigenvalue of ppT is at least
n,
as claimed.
>
2
E
Corollary 6.6.21. If G is an input graph such that
1. There are at most 14 vertices v such that there is a path from s to v or a path
from v to t in G of length at most 2'-1
2. For any vertex v E Vred(G) there are at most 2O3
W G Vred(G)
such that there
is a path from v to w or a path from w to v in G of length at most 2 r2
154
then for all k, u such that k + u < r, given any Sk,u,g,free and aku,
Sk+,u-1,g,free
we can choose
so that
1-
PkSk+1,u-1,g,free
2.
Sk+1,u-1,gfree
I I J ,12 <4U2
11 12
+411- 2
USk,u,g,free+ ak,u,g
12
2
USk,u,g,free +
ak,u,g
Proof. Recall from before that the equation
-
n
-
Sk,ugfree
Pkk+1,u-1,g,free
k,,2
Usk,u,g,free
+ akug
is
equivalent to the equation Pj = i where P is the matrix Pk,u,free after we delete all the
rows and columns corresponding to fixed coordinates, - is the vector
after we delete all fixed coordinates of
and gyk,u,g,
is the vector
dk,u,g+tsk,u,gfree
Sk+1,u-1,g,free
after
deleting all fixed coordinates of Sk+1,u-1,gNow for a given k and u, let V be the set of vertices V such that s
is relevant for (Sk,u,g)A when v
w
-
E
6.7
v or v -+ t
A. Let E be the set of (v, w) such that v -+ w or
v is relevant for (Sk,u, 9 )A when v, w
6.6.17 we may take ' so that
-
c
A. Applying Lemma 6.6.8 and Theorem
yfl I I -I|I and the result follows.
D
Constructing a base function
We now use the tools we have developed to construct our base function g.
Theorem 6.7.1. If r and m are constants such that m
<
2000r4
and G is a directed
acyclic input graph such that
1. There is no path of length at most 2'-1 from s to t.
2. For any vertex v
e V(G), there are at most m vertices w E V(G) such that
either there is a path of length at most 2r-2 from v to w in G or there is a path
of length at most 2 r-2 from w to v in G
then there is a function g such that
1. yf
2.
= 1
'v = 0 whenever
|V|
> r
155
3. For every e E E(G) there is a function g, such that
(a) ge is e-invariant.
(b)
4.
4 ev
= Yv whenever
(9mn)
Isk,u,gI2
|V| <
r
2
Proof. We construct this function g by using Corollary 6.5.14 and Corollary 6.6.21 to
choose the vectors sk,u,g in increasing lexicographic order in (k + u, k). For k + u = 0
we take
=
1I0,Og
and for k + u
{}
j
=
> 0 we start with s0 ,j,g = 0. Corollary
6.5.14 guarantees that when we do this, every statement except the last one will hold,
so we just need to show that skU,g2 5 (9mn)
2
. This follows immediately from the
following lemma.
Lemma 6.7.2. If k + u > 1 then
<
1. ISku,g,fixedI' K(9mn)
2.
Sku,g,free2 <
!(9mm)
2
k
Proof. We prove this by induction. We prove bounds on each term in terms of previous
terms and then verify that if the bound holds for the previous terms it holds for the
current term as well.
Lemma 6.7.3. For any k > 1 and any u,
Sk,u,g,fixed112
< 2s
_,u_1,g 12
+ 61 [
k2g
Proof. By our definitions, for every term
(4m + 6km) IIu',
9
+ 6km|Sekl,u_1,g
2+
2 +
(Sk,u,g)A
,12
3nm|k- 2,u,g| 2
which is fixed one of the following
is true
1. w E A and there is a relevant path from s to w. In this case,
-
(sk-1,u,g)A\{w1
2(('k1,u,g)A\{w})
This implies that
2
156
((Sk,ug)A)
2
< 2((s',_1,g)A) 2
+
(Sk,u-1,g)A
=
(S'k,u,g)A
-
(Wk_1,ug)A\{v
(_k,_u1,g)A
implies that
This
In this case,
((4k, ,,)A)
2
=
('k,u,g)A
2(Sk_1, ,,)A\{V1)
2
+
2. v E A and there is a relevant path from v to t.
2(Sku--,g)A)2
3. v, w E A and there is a relevant path from v to w. In this case,
= (Sk_1,u,g)A\f{v
(Sk,u,g)A
-
-
(8_k-1,u,g)A\{w
-
(Sk-1,u_1,g)A\{v
-
(Sk-1,u_1,g)A\{w
(Sk-2,u,g)A\{v,w}
+ (k,u-2,g)A
This implies that
6((k-1,u,g)A\{w})2
+ 6((k-2,u,g)A\{v,w}) 2(8k,u,g)A
2
- 6((k-1,u,g)A\{v})
+
k
6(('k-1,u_1,g)A\jv1 ) 2
+
6((k_1,_1,g)A\fw
)2
+ 6(('k,u-2,g)A)
To bound I|Sk,u,gfixed 12 we sum the above inequalities over every A such that
2
(Sk,,,g)A
is fixed. Consider how many times each term can appear in this sum. Terms of the
form
((1k,
1,g)V)
2
and
(('k,.-2,g)V)
2
will only appear once (with the appropriate
constant in front) as they only occur when A = V. Terms of the form ((Sk_1,,,
g
)v)2
will appear up to 2m times from cases 1 and 2 (these terms can occur when A = V Uv
where v is (r - k - u - 1)-linked to s or t) and up to km times from case 3 (these
terms can occur when A = V U v where v is (r - k - u - 1)-linked to a vertex in A).
Terms of the form
((Sk-1,u-1,g)A)
2
cannot occur from cases 1 and 2 but will appear
up to km times from case 3 (these terms can occur when A = V U v where v is
(r - k - u - 1)-linked to a vertex in A). Finally, terms of the form
appear up to
n- 2
((8k-2,u,g)A)
2
will
times (these terms can occur when A = V U {v, w} where v and w
are (r - k - u - 1)-linked to each other). Putting everything together,
Sk,u,g,fixed 112
< 21 Sku_1,g 12 + 611
k,1-2,g|2
+ (4m + 6km)|IIk _1,u,g 12
+ 6km|Skl,u_1,g
2
+ 3mSk-2,u,g
F157
Corollary 6.7.4. For a given k > 1 and any u, if the bounds hold for earlierterms
then ||Sk,u,g,fixeds|2 < I(9mn)
Proof. If k = 0 then Sk,u,gfixed = 0. By Lemma 6.7.3 and the bounds of Lemma 6.7.2,
2(9mn)
k+u-1
2
6km(9n))
2
<
+ 6(9mn)
_ __
22
+
-v/9mn
k+u-2
2
+ (4m + 6km)(9mn)
k+u-1
2
) k+ u - 2
2
+ 3nm(9mn)
6
100k 2 m 2
9mn
9mn
+
2
Sk,u,g,fixed 2
Since n > 2000mr4 this is less than - (9mn)
+
6km
1
+9mn3
k+u
(9mn
2
2-u
Lemma 6.7.5. For any k > 1 and any u,
k-
1,u+1,g 2
< m(k + 1)k Ik,u,g,fixed
2
Proof. Recall that ak-1,u+1,g is the projection of -Pisk,u,g,fixe
of Sk-1,u+1,g. Now
For the A such that
is free, there are at most (k - 1 + 2)m = (k + 1)m B such that A C B,
(Sk-1,u+1,g)A
= k, and
=BI
(Pk-1Sk,u,g,fixed)A = ZACBB~k ('ku,g,ixed)B.
onto the free terms
('k,u,g)B
is fixed. Thus, for all such A, the sum ZACB,IBI=k
(Sk,u,g,fixed)B
has at most (k + 1)m terms so using the Cauchy-Schwarz inequality,
((P-18k,u,g,fixed)A)
2
A
(k + 1)m
((Sku,g,fixed)B)
2
ACBjBI=k
Summing over all such A, each term ((sk,u,g,fixed)B)
I*k-1,u+1,g|
< m(k +
2
appears at most k times so
1)k||I ,u,g,fixed I2
E
Corollary 6.7.6. For any k > 1 and any u,
Sk
I 2 <4u
ug,f
ree
1
2
sk-1,u+1,g,free
n Vn
158
2
+ 4m(k + 1)k
I k,u,g,fixed
2
Proof. By Corollary 6.6.21,
2
2k<u4U2free
2+4
+
2
II -nk-1,u+1,g,free
2
+
skUgjree
44
k-1,u+1,g 2
Pluggin in Lemma 6.7.5,
r-
112 <4Uu2
sk,u,g,free
g,free skugfixed
2
4m(k + 1)k
112
2
E
as needed.
Corollary 6.7.7. For any k > 1 and any u if previous bounds hold then
ISk,u,g,free
2
<
1(9mn)
.
Proof. This follows immediately from Corollary 6.7.6 and the fact that n > 2000mr4
This completes the proof of Lemma 6.7.2 and thus Theorem 6.7.1
6.8
The cost of extending a base function
We now have our base function g which we know can be extended to an e-invariant
functions {g.} for every e E E(G) (we assume that s -+ t
E(G) and E(G) has no
edges of the form v -+ s or of the form t -+ v). However, while we were careful to
make the sum vectors Sk,u,g as small as possible for k + u < r, we have little control
over the sum vectors
,
for k + u = r. In this section, we show that when we
extend g to ge, the sum vectors sk,u,g, are not too large.
For all of these lemmas, let r be a constant and let g be a function such that
v = 0 whenever IV ;> r.
Lemma 6.8.1. If e is of the form s
-+
w and .vu{w}
V| < r - 1, if we take g, so that
159
= -v
whenever w
$
V and
1. g'ev = v whenever
2. gev = -PV\{w}
|V|
r and
if w E V, V| = r and 0 if w V V,|V| = r
then g, is e-invariant. Moreover,
1. If w $ A and |A| + u
=
r,
2. If w E A and A| + u = r,
SA,ut(ge) = -SA,u-1(g)
SA,u(ge) = -SA\{w},u(g)
+ SAu{w},u-2(9)
+
SA,u_1(g)
Proof. The fact that g, is e-invariant follows from Corollary 3.4.28. For the moreover
statement, note that if w $ A and JAI + u = r,
SA,u (e)
=
4eBU{w}
E
B:ACB,|B|=r-1,w B
5
-
YB
S
B:ACB, BI=r-1
YB
B:ACB,|B|=r-1,weB
:
+
-SA,u-1
5:
B
+
-
YB
B :AUjw}CB,IBjz~r-1
=
-SA,u-1(g)
+ SAU{w},u-2(9)
If w C A and IAI + u = r then by Lemma 6.4.6,
SA,u(ge) = -SA\{w},u(ge)
+
SA,u-1(ge) = -SA\{w},u(g)
Lemma 6.8.2. If e is of the form v -+ t and .vu{v}
=
#
r and
2. gev = gv\{v} if v G V,|V| = r and 0 if v
then g, is e-invariant. Moreover,
160
SA,u-1(9)
v whenever v 0 V, |V| < r -1,
if we take g, so that
1. gev = gv whenever |V|
+
V,|V| = r
1. If v 0 A and A| +u
=
r,
2. If v G A and AI +u = r,
SA,u(ge) = SA,u-1(g) -
SA,u(ge)
SAu{v},u-2(g)
SA\{v},u(g) -
SA,u-1(Y)
Proof. The fact that ge is e-invariant follows from Corollary 3.4.28. For the moreover
statement, note that if v
A and JAI + u = r,
SA,u(e)
B
YeBU{v}
S
YB
B:ACB,JBr-l,v B
=
B:ACB,|B|=r-1,vgB
=
B
-
YB
YB
B:ACB,|B|=r-1
B:ACB,IBI=r-l,vCzB
SA,u-1()
S
--
YB
B?:Aufv}CB,jBj=r-1
= SA,u-1(9) -
SAu{v},u-2(9)
If v E A and IAI + u = r then by Lemma 6.4.6,
SA,u(ge) = SA\{v},u(ge)
-
SA,u-1(9e) =
SA\{v},u(9)
Lemma 6.8.3. If e is of the form v -+ w and .vu{v,w}
4
ev =
Yv whenever
2. 4ev = -.
v\{w}
3. ev = -v\{w}
4.
ev=OifwVIV|
+
|VI
.V\{v}
# r
+
Yv\{v,w}
if v, w G V,
if v V V, w c V|IV| = r
=r
then ge is e-invariant. Moreover,
161
|VI
SA,u-1(Y)
= -vu{v}
whenever v,w $ V,| V < r - 2, if we take ge So that
1.
-
= r
+
Yvu{w}
+ Yv
1. If v,w
A and |A +u
SA,u(ge)
r,
-SA,u-1(9)
+
2SAu{w},u-2(9)
SAU{v},u-3(9) -
-
SAU{w},u-3(g)
-
SAU{v,w},u-3 (9)
+
SAU{v,w,u-4(9)
+
SA,u-2(9)
2. If w E A,v $ A and JA| +u = r then
SA,u(ge)
3. If w
+
=
-SA\{w},u(g)
-
SA\{w}U{v},u-2(9)
2SA,u-1(9) -
SA,u-2(g)
SAU{v},u-2(9) + SA\{w},u-1(9)
+ SAu{v},u-3(9)
A,v ( A and AI+u-=r,
SA,u(Ye)
=
SAU{w}\{v},u-1(9) - SA,u-1(9)
+ SA\{v},u-1(9)
-
SAU{w}\{v},u-2(9) -
+ SAu{w},u-3(Y)
SA,u-2(g)
4. Ifv,w c A and IA +u==r,
SA,u(Ye)
SA\{v},u(9) -
SA\{w},u(9)
+
SA\{v,w},u(g)
SA\{,u_1(g) - SA\{w,u_1(9)
+ SA,u-2(g)
Proof. The proof of this lemma is similar but more complicated. Again, the fact that
ge
is e-invariant follows from Corollary 3.4.28. For the moreover statement, note that
ifv,w
A and |A|+u=r,
SA,u(9e)
z
4eBU{w}
BACB,
|B|=r-1,v,wOB
+
z
B:ACB,
IB|==r-2,v,w B
162
9eBU{v,w}
Let's consider each term separately. For the first term,
YB
gB
E
BACB,
B:ACB,
IB|=r-1,veB
|BI=r-1
-SA,u-1(9)
-
YB
B
I:
-
BACB,
|B|=r-1,V,w
+
E eBU{w}
B:ACB,
IB|=r-1,v,
w B
B:ACB,
Bj =r-1,v,weB
B:ACB,
IB|=r-1,wEB
+ SAU{v},u-2(9)
+ SAu{w},u-2(9)
-
SAU{v,w},u-3(9)
For the second term,
4eBU{v,w}
E
B:ACB,
lBI=r-2,v,wVB
(.BUfw} -
YBU{v}
+
YB)
B:ACB,
|BI=r-2,vw B
Consider each of these terms separately. By the same logic as above,
YB
SA,u-2(g)
-
SAU{v},u-3(9) -
SAU{w},u-3(9) + SAU{v,w},u-4(9)
B:ACB,
|BI=r-2,v,w B
For the first term,
E
E
YBU{w}
B:ACB,
YB
B:AU{w}CB,
|BI=r-2,
IBI=r-1,
v,w B
vVB
~
--
B
B:AU{w}CB,
B|=r-1
=
sAU{w},u-2(g)
B
B:AU{w}CB,
vEB,
IB =r-1
-
SAU{v,w},u-3(9)
Following the same logic,
5
YBU{v}
SAU{v},u-2(9)
BACB,
B\=r-2,v,wVB
163
-
SAU{v,w},u-3(g)
B
Putting everything together,
SA,u(ge)
=
+ SAU{v},u-2(9) + SAU{w},u-2(Y)
(-SA,u-1(9)
+ (SA,u-2(g)
SAU{v},u-3(9)
-
+ (SAU{w},u-2(g)
-
SAU{v},u-3(g)
2SAU{w},u-2(g)
-
SAU{w},u-3(g)
SAU{v,w},u-3(Y))
-
+
-sA,u-1(9)
-
SAU{w},u-3(9)
-
-
-
SAU{v,w},u-3(9))
+ SAU{v,w},u-4(g))
(sAU{v},u-2(9)
-
SAU{v,w},u-3(g))
SAU{v,w},u-3(9) + SA,u-2(g)
+ SAU{v,w},u-4(Y)
Now that we have done this calculation, note that if w E A, v
A and JAl + u = r
then
SA,u(Ye) = SA\{w},u+1(ge)
-SA\{w},u(Y)
-
SA,u-2(g) -
+
2
sA,u-1(9) - SAU{v},u-2(g)
+ SA\{w},u-1(9)
SA\{w}U{v},u-2(g) + SAU{v},u-3(9)
For the fourth part, by Lemma 6.4.6, if v, w E A and IA + u = r then
SA,u(se) = SA\{v},u(g)-SA\{w},u(g)+SA\{v,w},u(Y)-SA\{v},u-1(9)-SA\{w},u-1(9)-+SA,u-2
Finally, note that if w
SA,u(ge)
=
(g)
A, v E A and IAl + u = r then
SAU{w},u-1(9e)
+ SA\{v},u-1(g)
=
SAU{w}\{v},u-1(Y)
-
SA,u-1(9)
-
SAU{w}\{v},u-2(Y)
-
SA,u-2(g) + SAU{w},u-3(g)
Corollary 6.8.4. If g, is defined according to Lemma 6.8.1, Lemma 6.8.2, or 6.8.3
then for all k,
Z:
A:|A|=k
(SA,r-k(ge))
2
{
< 200 max max
(SB,r-k 2 -1(g))
B:IBI=k2
164
2
(SB,r-k 2 -2 (g)) 2
,
B:IBI=k2
Proof. Consider the equation in the first moreover statement of Lemma 6.8.3:
SA,u(9e)
-SA,u-1(9)
2sAU{W},u-2(g)
SAU{v},u-3(Y)
-
SAU{w},u-3(g)
-
-
SAU{v,w},u-3(Y)
+ SA,u-2(g)
+ SAU{v,w},u-4(g)
Applying a Cauchy-Scwarz inequality to this,
(SA,u(ge))
10((SA,u-1(g)) 2
2
-
(SAU{v},u-3 (g) 2
(AU{w},u-2(g) )
-
2
(g) )2
(SAU{W},u-3
+ (SAU{v,w},u-3(g))
2
+ (sA,u-2(g))
2
(sAU{v,W},u-4 (g)) 2)
Similarly, we can apply a Cauchy-Scwarz inequality to all of the other equations.
In this way, each sum
of the form
each term
(SB,u(ge) ) 2
A:IAI k (SA,rk(ge))
(SB,u(ge)) 2
2
can always be bounded by a sum of terms
where IBI+-u
= r - 1 or r - 2 and I(IBI - k)|I
must come from an A with JA|
=
2. Moreover,
k, AAB C {v, w}. Thus, for
each B there are at most two A which will give a term with that B. This implies
that the coefficient for each term
(SB,u(9e) ) 2
has magnitude at most 2, so for all k we
have the inequality
S
(SA,r-k(ge) )
A:IAI=k
2
20
k 2 :|k-k 2 <2
(SB,r-k
B:IBI=k
2 -1
(g))
2
2
(SB,r-k 2 -2(9))2
+
B:IBI=k 2
The result now follows immediately as there are 5 possible k 2 with Ik - k2<
< 2. E
6.9
Putting everything together: A lower bound
We now put everything together and prove our lower bound on m(G)
Theorem 6.0.16. If r and m are constants such that m < 2000r4 and G is a directed
acyclic input graph such that
1. There is no path of length at most 2'-1 from s to t.
2. For any vertex v G V(G), there are at most m vertices w
165
c
V(G) such that
either there is a path of length at most 2r-2 from v to w in G or there is a path
of length at most
2 r-2
from w to v in G
then
(
(9mn)
(
nr
)
m(G) ;>
201 E(G) IVr (r + 1)! 9m
Proof. We obtain the base function g from Theorem 6.7.1 and obtain the e-invariant
functions {ge} using Lemmas 6.8.1, Lemma 6.8.2, and 6.8.3. Now by Corollary 6.8.4,
for all e E E(G), if k + u = r then IIsk,u,geI1
all el, e 2 E E(G), IIsku,9ge2-ge 1
12
2
< 200(9mn)
9
. This implies that for
< 800(9mn)2 1 if k + u = r and is 0 otherwise.
By Theorem 6.3.2,
m(G) >
E
-
JEGj- 1
{4(r
2k (k + u)I
(
1
ma
(\eGE(G)\{eo}
k,u
nk+u
2
sk,u,ge -geo 12
With some algebra, noting that we only need to consider terms where k + u = r, this
implies that
m (G) >
(9mn)
(n7
)4
(30mE(G) I2r(r + 1)! 9m
as needed.
EI
166
Chapter 7
Conclusion and Future Work
In this thesis, we developed a powerful new framework for analyzing monotone space
complexity via the switching network model. Using this framework, we separated
monotone analogues of L and NL and gave an alternative proof of the separation of
the monotone-NC hierarchy. We also investigated the monotone space complexity
of directed connectivity more deeply, proving upper and lower bounds on m(G) for
particular types of input graphs G. To the best of our knowledge, this focus on
particular types of input graphs is novel.
However, this work leaves many open questions. Most importantly, can this framework be extended to prove non-monotone lower bounds? Such an extension would
have to consider all NO instances, not just maximal NO instances. This will almost
certainly be extremely difficult and require many new techinques, but if successful, it
would be a major breakthrough.
Even for monotone space complexity, there are several open problems remaining.
One open problem is to determine the relationship between mLcircuits and
mLswitchingnetworks.
In other words, what is the relationship between the power of
monotone switching networks of polynomial size and the power of leveled monotone
circuits with polynomial size and logarithmic width? Answering this question would
help resolve the question of how monotone space complexity should be defined. Another open problem is to prove tight bounds on the size of monotone switching networks solving other problems such as the k-matching problem which asks whether or
167
not a graph has a matching of size k. Finally, for the directed connectivity problem,
can we prove better bounds on m(G)? In particular, can we prove tight or almost
tight bounds on m(G) when G is an acyclic directed graph? Brakensiek and the
author [4] have shown partial results in this area by proving almost dight bounds on
m(G) when G is an acyclic directed tree, but problem is still wide open for general
acyclic G.
Thus, it is our hope that this thesis represents not only an introduction of a new
framework for analyzing monotone space complexity and a summary of what has
been achieved so far, but also a foundation for further research on the fundamental
problem of space complexity.
168
Appendix A
Proof of Lemma 3.2.32
Lemma 3.2.32. Let G' be a certain knowledge switching network. For any certain
knowledge description of G' and any path P = s -+ vi -+
.
--
vi_ 1
-
t, if G is the
input graph with vertex set V(G) and E(G) = E(P), if W' is a walk in G' whose edge
labels are all in G
from
a vertex v'tar with Kv'
K
to a vertex v'nd with K,
Kt, then W' passes through a vertex v' such that K,, i Kt1, V(Kv') C {v 1 ,---
=
,-1
and IV(Kv)| ;> [lg(l)].
Proof. In this proof, we will split the path P in two and use induction on each half.
This will require projecting onto each half of P in two different ways.
Definition A.0.1.
1. Call the vertices L = {V1,
2. Call the vertices R = {v
, vi 2 1 } the left half of P.
I.
,- -1]1
, vi1} the right half of P.
Definition A.0.2.
1. We say an edge e = u
-+
v is a left edge if u,v G L U {s}
2. We say an edge e = u
-+
v is a right edge if u,v G R U {t}
3. We say an edge e = u
-
v is a left-jumping edge if u = s and v E R. Note that
t c R.
169
4. We say an edge e = u
that s
-*
v is a right-jumping edge if u =G L and v = t. Note
L.
Out first projections focus on the progress we have made towards showing that
there is a path from s to R U {t} and L U {s} to t, respectively.
Definition A.0.3. Given a vertex v E V(G),
1. Define pi(v) =v
if v $ R and pi(v) = t if v ER.
2. Define pr(v) =v if v
L and pi(v) = s if v
L.
Definition A.0.4. Given an edge e =u -+ v where u, v E V(G),
1. Define pi(e)
pi(u)
2. Define pr(e)
Pr(U)
-+
pi(v).
Pr(V).
Definition A.0.5. Given a knowledge set K,
1. Define pi(K) = {pi(e) : e G K,pi(e)
# t -*
2. Define pr(K) = {pr(e) : e E K,p,(e)
#s-
t}.
s}.
Definition A..6. Given a certain knowledge switching network G' together with
a knowledge description of G', define pi(G') to be the certain knowledge switching
network formed from G' with the following steps:
1. Replace all edge labels e with pl(e)
2. Replace all knowledge sets Kr, in the certain knowledge description with p,(Kvi).
3. Contract all edges in the switching network which now have label t
-
t. When
contracting an edge e' with endpoints v' and w', we may choose either Kv, or
Kw, to be the knowledge set for the resulting vertex.
Similarly, given a certain knowledge switching network G' together with a knowledge
description of G', define pr(G') to be the certain knowledge switching network formed
from G' with the following steps:
170
1. Replace all edge labels e with p,(e)
2. Replace all knowledge sets K,, in the certain knowledge description with p,(Kv').
3. Contract all edges in the switching network which now have label s -+ s. When
contracting an edge e' with endpoints v' and w', we may choose either Kv' or
K., to be the knowledge set for the resulting vertex.
Proposition A.0.7. Given a certain knowledge switching network G' for directed
connectivity on V(G),
1. pi(G') is a certain knowledge switching network for directed connectivity on
V(G)
\ R.
Furthermore, for any vertex w' E V(pi(G')), for all of the vertices
V' E V(G') which were contracted into w', K,' = Pi(Kv,).
2. p,(G') is a certain knowledge switching network for directed connectivity on
V(G)
v'
c
\ L.
Furthermore, for any vertex w' E V(p,(G')), for all of the vertices
V(G') which were contracted into w', Kw,
pr(Kv,).
Proof. We prove the first claim, the proof for the second claim is similar. To prove
the first claim, it is sufficient to show the following.
1. p, (K,,) = Ks,
2. p, (Kt) = Kt,
3. For any knowledge sets Ku,, K,, and any possible edge e which is not a right
edge, if Ku/ U {e} = K,, U {e} then pi(Kn') U {p1 (e)} - pi(Kg) U {p1 (e)}.
4. For any knowledge sets Ku,, Kv, if K,, = K,, or Ku, U {e} - K,, U {e} for some
right edge e then p,(Ku,)
p(Kvi)
The first two statements are trivial. For the third and fourth statements, we consider
the effect of pi on each type of move in the modified certain knowledge game.
1. If we originally added or removed an edge e from K after directly seeing e, if e
was not a right edge then we now add or remove p,(e) from p,(K) after directly
seeing pl(e). If e was a right edge then we now do nothing.
171
2. If we originally added or removed an edge v 3
v3
-4
v4 , v 4
-+
or remove pi(v
v5 from K after noting that
-+
v5 c K, then if p1(v3 ),p 1 (v4 ),p 1 (v5 ) are all distinct we now add
3
-+
v 5 ) from p1 (K).
If p(v 3 ), p(v 4 ), p(v 5 ) are not all distinct
two of them must be equal to t. In all of these cases we now do nothing.
If p(v) = pi(v 4 ) = t then p1(v3 --+ v5 ) = pi(v 4 -+ v5 ). This means that
pi(K U {v 3
-
V5 })
= pi(K) = pi(K \ {v 3
-+
Similar logic applies if
v 5 }).
pi(v 4 ) = p(v) = t. Finally, if p(v 3) =p(v 5 ) = t then p1(v 3
we again have that pl(K U {v 3
-
V5 }) =
pi(K)
=
p1 (K \ {v
v5 ) = t
-+
3 -+V5)
3. If we originally added or removed an edge e f s -+ t after noting that s
K, if e was not a right edge we now add or remove an pl(e)
t so
-
#
s
-
-
tE
t after noting
that s -+ t C pl(K). If e was a right edge then we now do nothing.
Using Proposition 3.2.14, statements 3 and 4 follow directly from these observations.
El
We now define a slightly different projection to each half. These projections will
help us look at the progress towards removing obsolete information after obtaining a
left-jumping or right-jumping edge.
Definition A.0.8. Given a knowledge set K,
1. Define p*(K) = {pi(e) e E K, pi(e)
t
2. Define p*(K) = {p,(e)
s
e E K, p,(e)
-
t, pi(e)
#
s
-
t}.
s, p,(e) 7 s -+ t}.
Definition A.0.9. Given a certain knowledge switching network G' together with
a knowledge description of G', define p*(G') to be the certain knowledge switching
network formed from G' with the following steps:
1. Delete t' and all other vertices v' such that Kv, = Kt from G'
2. Delete all edges e' such that e' has an endpoint v' and label e and Kv U {e} = Kt
3. Replace all edge labels e with pi(e)
4.
Replace all knowledge sets Kv, in the certain knowledge descriptionwith p*(K,).
172
5. Contract all edges in the switching network which now have label t
-
t. When
contracting an edge e' with endpoints v' and w', we may choose either Kv' or
K&' to be the knowledge set for the resulting vertex.
6. Add the vertex t' to G', assign it the knowledge set Kt= {s
-
t}, and add all
labeled edges with endpoint t' to G' which are allowed by condition 3 of Definition
3.2.8.
Similarly, given a certain knowledge switching network G' together with a knowledge
description of G', define p*(G') to be the certain knowledge switching network formed
from G' with the following steps:
1. Delete t' and all other vertices v' such that Kv, - Kt, from G'
2. Delete all edges e' such that e' has an endpoint v' and label e and K,'U{e} = Kt,
3. Replace all edge labels e with p,(e)
4.
Replace all knowledge sets Kv, in the certain knowledge description with p*(Kv).
5. Contract all edges in the switching network which now have label s -+ s. When
contracting an edge e' with endpoints v' and w', we may choose either K1 or
K&' to be the knowledge set for the resulting vertex.
6. Add the vertex t' to G', assign it the knowledge set Kt, = {s -+ t}, and add all
labeled edges with endpoint t' to G' which are allowed by condition 3 of Definition
3.2.8.
Proposition A.0.10. Given a certain knowledge switching network G' for directed
connectivity on V(G),
1. p*(G') is a certain knowledge switching network for directed connectivity on
V(G)
\
R. Furthermore, for any vertex w' G V(p*(G')), for all of the vertices
V' E V(G') which were contracted into w', K& -= p*(Kv).
173
2. p*(G') is a certain knowledge switching network for directed connectivity on
V(G)
\
L. Furthermore, for any vertex w' e V(p*(G')), for all of the vertices
V' E V(G') which were contracted into w', Kw, - p*(K,,).
Proof. We prove the first claim, the proof for the second claim is similar. To prove
the first claim, it is sufficient to show the following.
1. p*(Ks,) = Ksl
2. For any knowledge sets Ku,, K,, and any possible edge e which is not a right
edge, if K' U {e}
Kv U {e}
K then p*(K') U {p1 (e)}I
3. For any knowledge sets Ku', Kvr, if KI
Kt, for some right edge e then p*(K,,)
p*(Kv,) U {p 1 (e)}.
Kv, 0 K, or K,' U {e} = K, U {e}
p*(K,)
The first statement is trivial. For the second and third statements, we consider the
effect of p* on each type of move in the modified certain knowledge game.
1. If we originally added or removed an edge e from K after directly seeing e, if e
was not a right edge then we now add or remove p,(e) from pi(K) after directly
seeing pl(e). If e was a right edge then we now do nothing.
2. If we originally added or removed an edge v3
v3
-
V4 V 4
-+
V5
K, then if p1(v3),p 1(v4 ),pj(v 5 ) are all distinct we now add
V5
or remove pi(v 3
v5 from K after noting that
-4
-
v 5 ) from pi(K). Note that we cannot have v3 = s and
= t because we are assuming that we never have a knowledge set K such
that K = Kti. If p(v 3 ), p(v 4 ), p(v 5 ) are not all distinct two of them must be
equal to t. Following the same logic as before, in all of these cases we now do
nothing.
3. We do not have to consider moves where s
-4
t
C
K because we are assuming
that we never have a knowledge set K such that K = Kt,.
Using Proposition 3.2.14, statements 2 and 3 follow directly from these observations.
E174
Now that we have defined these projections, we give two more useful definitions
and then prove Lemma 3.2.32.
Definition A.0.11.
1. We say a vertex v' on a walk W' satisfies the lemma for the left half if K,, # Ktl,
V(Kv') C {v 1 , --- ,vi_ 1}, and |V(Kv') n L| > [lg(l)] - 1.
%
2. We say a vertex v' on a walk W' satisfies the lemma for the right half if K,,
Ktl, V(Kv ) C {v 1 ,-
,vi_}, and |V(Kvi) n R| > [lg(l)] - 1.
Definition A.0.12.
1. We say a knowledge set K is left-free if K 0 Kt, and V(K) n L = 0.
2. We say a knowledge set K is right-free if K # Kt, and V(K) n R = 0.
We now prove Lemma 3.2.32 by induction. The base case 1 = 2 is trivial. If 1 > 2
then given a walk W' from v'tr to V',n whose edge labels are all in E(P), first modify
W' and G' slightly as follows. Let u' be the first vertex on W' such that if e is the
label of the edge after u' then Ku, U {e} - Kt,. If u' is not the vertex immediately
before v'fd then add an edge from u' to v',d in G' with label e and replace the portion
of the path from u' to v' n with this single edge. Note that if the lemma is still
satisfied now then it was satisfied originally. This modification ensures that we do
not have to worry about moves in the modified certain knowledge game where we
have s -÷ t E K.
We now show that W' must have at least one vertex v' which satisfies the lemma
a
K81
for the left half. To see this, apply the projection p, to G' and W'. pl(K
and pl(Kvi)l
)
Kt,, so by the inductive hypothesis there must be some vertex w'
on p1 (W') such that V(K,,) C L and IV(Kw,) I
[Ilg l - 1]. Choose a v' which was
contracted into w' by pi. V(Kv') C {v 1 , - -- , vi-1 } and IV(K,) n L|I
[lg l - 1], so
v' satisfies the lemma for the left half, as needed. Following similar logic, W' must
also contain a vertex satisfying the lemma for the right half.
Now take b' to be the last vertex on W' which either satisfies the lemma for the
left half or satisfies the lemma for the right half. Without loss of generality, we may
175
assume that b' satisfies the lemma for right half. We may also assume that Kb, is
left-free, as otherwise b' satisfies Lemma 3.2.32. There are now two cases to consider.
Either Kb, contains a left-jumping edge, or it does not.
If Kb, does not contain a left-jumping edge, then apply pi to the portion of W'
between b' and t'. p(Kb,) =
{}
and p,(Kt,) = Kt, so using the inductive hypothessis
there must be a vertex a' on this portion of W' which satisfies the lemma for the left
half. However, this constradicts the definition of b'.
If Kb, does contain a left-jumping edge then choose a sequence of moves in the
certain knowledge game for going along W'. Let K be the first knowledge set we
obtain such that k conatins a left-jumping edge and for every K 2 after K but before
Kb', K 2 contains a left-jumping edge. K occurs when we are transitioning between
some vertices v' and w' in W' along an edge e' with label e.
Note that p*(K) = Kt,. This implies that p*(Kv') Up,(e) = p*(K.,) Upi(e) = Kt,.
Now take the portion of p*(W') from p*(v') to p*(b') and replace p*(v') with t'. Since
Kb, is left-free, p*(Kb,) = K,,. Using the inductive hypothesis, there must be a vertex
a' between v' and b' on W' which satisfies the lemma for the left half. Ka, occurs
beetween K and Kb', so Ka, also contains a left-jumping edge. Thus, a' satisfies
Lemma 3.2.32, as needed, and this completes the proof.
176
El
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