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>> Kamal Jain: Hello. My name is Kamal Jain. Today we have a guest, Shayan
Oveis Gharan. And sorry if I mispronounce your name. He came here for an
interview for his fellowship candidate, but he thought he could give a talk since
he had some time. And I think it's a talk prepared on short notice. So people on
the camera, or later, so he proved recently a break-through result in a
long-standing problem which is asymmetric traveling salesman problem. And it
could be considered one of the center problem in approximation algorithms. So
Shayan will explain his new ideas here.
>> Shayan Oveis Gharan: Thank you. Okay. So today I'm going to talk about
the asymmetric traveling salesman problem. And I'll try to tell you some of our
main ideas for this problem. I will tell them in the talk. So that's a start.
So here's an outline of the talk. First I will define the traveling salesman problem,
and I will tell you a brief history about it. And then I will define the concept of thin
trees and its relation to ATSP. And, finally, I'll try to discuss some of our main
ideas for these approximation algorithms for ATSP.
Let's start with the problem definitions. Traveling salesman problem is general
form has a simple definition. Suppose like we have a set of cities. There's some
cost between them. And the traveling salesman who wants to visit each of these
cities exactly once. So, in fact, we want to find the shortest tool such that that
visit vertex is exactly one.
So this problem was initially proved to be implicated by [phonetic] Karp in 1972,
and it was one of the first problems to be proved to be in the complex. Traveling
salesman problem in its general form is very hard to deal with. So we'll relax it a
little bit.
So, in fact, we consider this a little relaxation. So in ATSP, we have a set V
vertices and a nonnegative cost function. We're looking to find the shortest
closest tool that visits each vertex at least once. We have substituted that
exactly once with at least once here.
For example, consider this graph. The cost of each edge is one in this graph.
And as you see, this tool I plotted in red is a tour of cost tool. And it visits some
vertices more than once. In terms that if you have this property you can assume
that the cost functions satisfies the triangle inequality. I.e., this inequality. So
you can do that by easily considering the shortest path distance of going from U
to V as the cost of U.
So, in fact, symmetric traveling salesman problem, which is well-known traveling
salesman problem, we'll call it STSP here, is a special case of ATSP. The main
difference is that in that problem the cost of UV, we call it from the cost of from
VG.
So there are quite a lot of work to solve this ATSP. This is, in fact, one
interesting instance. This tool is found on like over one million cities throughout
the world and the cost of cities only six person of the optimum. In fact, the
heuristics works very well for this problem. But here we'll focus on the
approximation algorithm.
So let me describe, TSP has a lot of applications in various areas. Here is one
application in genome sequencing. In genome sequencing there's a process
called shotgun sequencing which divides DNA into a set of short fragments. And
the question is you want to assemble these short fragments in order to obtain the
original DNA.
So if you think, you'll find out -- sorry, in fact, this problem is the optimization
problem. We want to find the original DNA to be as short as possible, be as
short as possible in length. So if you think you'll find out that this is in fact
minimum super sim problem. Super sim problem, you have a set of streams and
you're looking to find a super string with the smallest length such that given each
string, it's a sub string of the super string and you can convert that problem to the
ATSP problem.
So, in fact, here these five vertices correspond to five different streams. ATG
and C are the DNA basis. And we're looking to find like this super string, each of
these strings is a sub string of it. And the tour is the optimum tour.
One difference is that here what we're looking to is to find the maximum ATSP.
Note here that the cost of going, for example, from this vertex to this vertex is
maximum suffix of this, which is equal to the prefix of the other. So max ATSP
can be solved within constant factor. So this problem can be solved in constant
factor in approximation.
Let me briefly define what do I mean by approximation algorithm. Suppose that
we have a minimization problem. We say that an algorithm is off a factor, is
factor of optimization. If it only produces a solution, which is off a factor of the
optimum.
So the main difficulty is that when we're looking for this approximation algorithm,
it seems to be hard to find actual optimum, otherwise we'll find the optimum. In
fact, if it's hard to find the optimum, how can we compare the cost of our solution
with the optimum. So typically we will use a lower bound on optimum solution.
For that we'll use a linear programming relaxation. This linear program relaxation
always gives you the lower bound for the optimization problem.
So here are now the results for both symmetrics, asymmetric ATSP. You can
see there's a lot of work. I've only mentioned a small portion of them. So this is
the best constant factor approximation algorithm for STSP, which is tree
approximation by Christofides in 1976.
On the other hand, for ATSP, there are a huge number of works. Like four works
here. But as you see, only the factor in front of the log n has improved. So they
all give up an order of magnitude for general approximation.
But for symmetric here, see there are quite a lot of works. It's a special case
when our metrics is defined on a special graph. In particular, polynomial time
approximation is key, having found first for metrics defined on unweighted planar
graphs than for metrics defined on weighted planar graphs. And finally for
weighted bounded genus graphs.
>>: Are they run by STSP, too?
>>: Shayan Oveis Gharan: This is for TSP. There isn't any special results for
ATSP in specially bound [indiscernible] graphs. So here are our main results.
We finally broke the log n barrier. We give the first order of log n over log log n
approximation algorithm for general metric ATSP. This is joint work with
Asadpour and Saberi from Stanford and Mitchell Goemans Alexander Madry
from MIT. Moreover, we give the first constant approximation algorithm for
shortest path metrics defined unweighted bound genus graphs. This is joint work
with [inaudible].
So as you see, similar to the symmetric TSP, where the approximation factors
are much better for the special case, here you have sort of the similar thing for
ATSP where our approximation factor is constant for if it's a special case.
Okay? So before I tell you our approximation algorithm, I need to define the
Held-Karp relaxation because I want to compare our solution with the Held-Karp
relaxation. This is the definition of the Held-Karp relaxation, this Held-Karp
relaxation is proposed by Held and Karp in the 1970s. And it gives a lower
bound for the optimal solution of ASTP. So here's some notation for set S delta
minus of S is a set of edges that are pointing to this set; delta plus of S is those
pointing out of this set.
So, in fact, the fractional solution of these relaxation is the vectors where they
have two properties. The size that -- there are in fact a fractional one flow.
Because the minimum size of the cut is at least one, the size of each cut is
at least one, as seen in this inequality. The inequality says that the amount
of incoming flow to each vertex is equal to the amount of outgoing flow,
equal to one. So it is in fact flow as you see, because the amount of
incoming is equal to outgoing and it is one flow because the amount of
outgoing edges down to flow through these edges is one.
Also remember at L top S is the union of these two sets. It is all the edges
in this cut.
Okay. So in fact if you see, you'll find out that the optimal solution of ATSP
is in fact an integral one flow, not a fractional one flow. So it has both of
these properties, both of these inequalities integrally. This is why this
linear programming is relaxation for ATSP and therefore its optimal value is
a lower bound for the upper bound of ATSP. Moreover, optimum solution
of the relaxation can be computed efficiently using ellipsoid algorithm.
And also integrality gap is defined as the maximum ratio between integral
solution over a fractional solution.
So there are also quite a lot of work to find integrality gap both STSP and
ATSP. As you see, there is this gap here for symmetric TSP. And even
improve it by epsilon factor with greater result. For ATSP core
[indiscernible] conjecture that integrality is equal to four and equal to
symmetric TSP. But then [indiscernible] Goemans [indiscernible] found a
set of counter examples where the integrality gap is at least two where they
show that integrality is at least two. So it's open whether integrality gap is
constant or not. So here's a simple example that shows that integrality gap
is at least one and a half. Consider, this is the same graph I said before.
The cost of minimum tour is 12. But if you consider the half integral
solution in this graph, you'll see that it is in fact a fractional solution of
Held-Karp relaxation, its cut is eight. So the integrality gap instance is one
and a half. And if you play with it, you can find a set of examples where
integrality converges to 2 if N goes to infinity.
Okay. So let's try to work on the approximation algorithm. First I will go
through the thin trees and we'll define it and tell you its relation to ATSP.
So what's the thin tree? A thin tree is that spanning tree that has
fractionally small number of edges in each cut, small number of edges.
So in particular, we say that spanning tree is also comma sigma ten if it has
both of these inequalities. This says that the number of edges of our tree
in any cut is at most alpha times the size of that.
So remember that delta of S is all the edges of this cut. Except L top S is
the sum of the fractions of all the edges of this cut. So I want that my tree
has roughly small number of edges in this cut.
And, moreover, the cost of my tree is within sigma factor cost of X.
So if our tree has both of these properties, I say it's alpha comma sigma.
So is the definition clear? There is this observation which says if you have
an alpha comma sigma thin tree with respect to optimal solution of ATSP,
then it is possible to find an integral invariant tool, variant tool based on
this thin tree where its cost is at most two alpha plus sigma times cost of X.
Remember that X was an optimal solution of Held-Karp relaxation,
therefore its cut of lower bound ATSP, therefore the cost of -- therefore,
this tool is an alpha -- is an alpha plus sigma factor optimal solution of
ATSP. I will not prove this observation for you, but the proof is mainly
maximal mean cut argument and all you need to do is find the minimum cut
invariant augmentation of this tree.
And for that we will use this fractional solution of Held-Karp relaxation.
Okay. So here's an overview of our algorithm. So it has some similarities
with well-known Christofides algorithm for symmetric STP. In fact,
suppose that this is a fractional solution of Held-Karp relaxation and this is
our thin tree, these green edges. So what I need to do, I need to add some
edges to this tree such that the point of graphical formulary. So sometimes
I have to add multiple copies of an edge.
So here are corollaries. First of all, if we had a polynomial time algorithm
that can find constant comma, comma constant thin tree with respect to an
optimal solution of Held-Karp relaxation, it can be a constant factor
approximation algorithm for ATSP.
Moreover, even if we cannot find such a tree, when we can prove its
existence, this implies the constant integrality gap of ATSP. So it's a great
result. So even the existence of this thin tree is a great result. So let me
show you an example what these thin trees are.
Suppose we have this graph. So if you -- in this graph, the cost of each
edge is one. And suppose that I consider half integral solution in this case.
So, in fact, this will be a fractional solution of Held-Karp relaxation.
Suppose we have these two thin trees. It turns out that this has been a
better thinness. And why is that true? Because consider this cost, that's
the previous version from rest of the graph. This tree has two edges in this
cost. This has four edges. But let's see the size of the cost. Because the
fraction of each edge is a half. The size of this cut is two for both of them.
So this tree is 2 over 2-1, 1 comma 4 over 5 thin. But this is 2 comma 4
over 5 thin.
We call this 4 or 5 is related to the cost of our tree versus the cost of the
graph. In fact, this is alpha and this is sigma.
So if you see, this star-shaped tree has fractionally a small number of
edges in many cuts of the graph. So, for example, if you consider this cut
separate this vertex, this tree has two edges, but this one has only one
edge. But you can see there's a very bad cut for this tree, its thinness will
be much worse than the other, than this [indiscernible].
So if you see this example, you'll find that the duration of the edges doesn't
have anything to do with the thin tree. If I drop the direction, I will have the
same thinness, because the definition of the thin tree I've never used -- I
haven't used the direction.
So this is a great result. We can drop the direction with this thinness
property. So if you drop the direction, what you get is a fractionally two
connected graph. Because, remember, that X was a fractional one flow, the
amount of outgoing edges of each set was one. The amount of incoming
edges was one. If you drop the direction, you get fractionally two
connected graph.
Moreover, you can either consider our fractional graph or an integral graph.
So you can convert our fractional graph to integral graph simply by
multiplying the fraction of each edge to some large number of K. For
example, consider this graph, you multiply every fraction by 2, you'll get
this graph.
So, in fact, what we need to do is either to find an alpha comma sigma thin
tree in a fractionally two connected graph, or alpha over K comma sigma
over K in a KH connected graph.
So remember if we multiply everything by K, I get a KH connected graph
but instead I need to prove alpha over K comma sigma over K thin tree,
instead of alpha comma sigma thin tree because we multiply everything by
K.
Also we can do one more simplification. We can get rid of the cost
function. Why is that true? Let me define another distribution for these
thin trees. I say that a thin tree, spanning tree is alpha 10, not alpha
comma sigma. Only alpha 10. If it has only the first inequality. It has
fractions of any one of the edges of the graph. Suppose this is true. I'll
show you that we can get rid of the cost function. How can we do that?
Consider KH connected graph. Suppose it has like C over K thin tree for
some constant C. We can extract this thin tree from our KH connected
graph. And because this tree has small number of edges in any cut, the
graph will still be highly connected.
So we can do this again. We can consider again a thin tree in the new
graph and extract it. So if we do that, you'll find out after some time we'll
get a bunch of thin trees and we get the tree with the smallest cost. And
this will be a good sigma for it, for us. So all I need to do, I need to find
alpha thin tree. I need to have only this inequality for two to the N cuts.
So to recap, I need to do, either to find alpha thin tree in a fraction in a
connected two graph or alpha over K thin tree in a KH connected graph.
Both if I can solve either of these problems I will get alpha approximation
algorithm for ATSP. Okay. So from now I will only focus on how to find
these thin trees.
So let's think about it. For ATSP, first we have some problem in directed
domain. Then we try -- we find this notion of thin tree and we converted the
problem from directed domain to undirected domain. But if you think,
you'll find out that finding these thin trees is not an easy problem. In fact, if
you have a graph and you know the best thin tree in this graph, and you
know its thinness, proving this thinness to someone else is a difficult
problem.
Currently the best approximation algorithm for this we know is log n
approximation for finding the thinness. In fact, this thinness problem does
not seem to be an MP.
So it seems to be a very hard problem. In fact, at least ATSP was an NP.
So it seems to be much harder. But as you will see, in some cases we can
measure the thinness with some tools. And that leads us to approximation
algorithm for ATSP.
So here are our algorithm. First I will show you a deterministic algorithm.
This deterministic works for shortest metrics defined unweighted bound
genus graph. It gives constant over K thin tree with respect to KH
connected bounded genus graph. And for certificate I will use properties
of dual graphs.
Also when we're working on this algorithm, we found out that
mathematician called Louis Goddyn conjectured that any chain connected
graph contains constant of K sentry. If his conjecture is true, then the
integrality gaps of ATSP would be constant.
Then I will present a randomized algorithm. This will work for general
metrics ATSP. And it will find an order of log n over log log n thin tree with
respect to fractionally two connected graphs. I would find constant over K
thin trees that do KH connected graph. But here is fractionally, with
respect to two fractionally connected. Both of them will give us, so this will
give us constant trees -- constant approximation for ATSP will give us -- it
will give us order of log n over log log n approximation. So first I'll discuss
the deterministic algorithm and then randomness.
So okay. We're good. Next I will show you this constant factor
approximation algorithm for finding the thin trees unbounded graph.
So let me define what I mean by bounded genus graph, an orientable
surface with genus gamma is simply in sphere with gamma additional
hand-offs. For example, this surface is the surface for genus tree. And we
say that a graph has genus gamma if it can be embedded on genus gamma,
while edges don't cross each other, don't cross each other. For example, a
genus planar graph is zero, can be embedded on the surface and you'll find
genus of complete graph is an order of n to the 2.
For example, if you have a city with many one-way streets and a few
bridges or underpasses, its genus is bound, or genus is small because you
can easily embed it on a surface, genus embedding by each bridge or pass
by a handoff.
So this is our main theorem. We give a constant factor approximation
algorithm for ATSP metrics defined weighted bounded genus graph or
approximation factor is in order of square root of gamma log of gamma,
and gamma is the genus of the graph. So with that even if gamma is not a
constant but is sufficiently small will get good approximation. So in fact if
gamma is like order of log n we get order of square root of log n
approximation algorithm.
Okay. So truly our algorithm implies constantality gap of Held-Karp
relaxation for bounded genus graphs, prove integrality for thin trees. It will
have more -- it's a bit more general. In fact, even if the inputs instance of
ATSP do not have bounded genus but we can find an optimal solution of
Held-Karp relaxation where it can be embedded on a surface bounded
genus, then again we can use this algorithm. So we do not need the input
graph to have bounded genus, we need to have fractional solution to have
bounded genus.
So here is the overview of the proof. First, I will tell you an unsuccessful
attempt. Here I mainly prove our algorithm for planar graphs. I'll tell you
some of the ideas for extension for non-planar graphs. For planar graphs
I'll tell you unsuccessful attempt and I'll tell you our certificates of
measuring thin tree and I'll tell you our thin tree selection algorithm. For
that I will use the concept of trees defined. So let's use this by
unsuccessful attempt. Suppose we have this graph and we're looking to
find a thin tree in this graph. This is 5-H connected. So what's a
[indiscernible] certificate. So suppose we have this tree in green edges.
So one very simple idea is to chart it at with its parallel edges. And you
can prove that if this edge will -- well, will contain a cut if and only if the
power edges contain that cut too.
So this implies 1 over 5,1 thinness of this tree. Okay. So let's see if we can
generalize this. The answer is no. The answer is yes, but it will not work
for any planar graphs. So suppose we have this graph and consider this
cut and consider four edges in this cut. And because I'm looking to find
the spanning tree I need to select at least one edge from the edges of this
cut.
But none of the edges of this cut have any parallel edges. So I cannot use
the previous idea. So instead what I'm going to do, I'm going to first select
these green edges. The first step I will not have any problem. I can select
these green edges and charge them with the parallel edges. This is the
first. So an idea might be that if we select these edges, what if we contract
and try to solve the problem recursively? That seems to work. Seems to
work. So let's try to do that. So if we do that, we get this graph. Again, we
can select these green edges and charge them with the parallel edges and
then contract it and finally select this edge.
So if you see, you'll find out that these edges will make spanning tree.
Why? Because each time we contract a set of edges we select at least one
of them as an edge of our thin tree. So that planar graph will be spanning.
But let's see -- you might expect the final graph to be 1 over 14 because
each time we charge its edge to parallel edges and have three parallel
edges.
So unfortunately that's not the case. In fact, this is the final graph. And
this is the tree I selected. If you consider this cut, our tree has three edges
of this cut. But the size of the cut is 7. So at best it's 3 over 7, 3, 7, 10. And
why did that happen? The problem is that consider this edge. This is the
edge of selecting the last set of our algorithms. You charge this edge with
these three edges. But in this cut only this edge is contained. None of the
edges we charge is contained. So in fact our charging method will work
only in the first step. When we contract the edges it will not work.
But in the rest of this proof I will show you that the same algorithm will
work with a bit of modification.
Okay. So let's reach our certificate. Let me tell you how we are going to
measure the thinness of our thin tree. So here's an observation. We'll use
a simple observation planar graphs. In the planar graph, dual of any
minimal cut is a cycle. And vice versa. The dual of a cycle, the dual graph
is the minimal original graph. For example, in this graph the dual of this
cut in blue is this red cycle.
So if I want to prove the thinness on my tree, all I need to do is to prove its
thinness in the minimal cuts, just to measure its thinness in the minimal
cuts. For that, all I need to do is to measure its thinness in the cycle of the
dual graph. If my tree has fractionally small number of edges in each cycle
of a dual graph, it will be thin.
So how can I do that? Suppose that I can select a tree where the distance
of its edges in the dual graph is at least P. So this in fact implies 1 over
this thinness. And why is that true? Because our tree can have at most 1
over D fractions of the edges of each cycle in the dual graph. So, for
example, suppose that this is a cycle in the dual graph. This is equal to
tree. So the worst case my tree has at most two edges in the cycle.
>>: [indiscernible].
>> Shayan Oveis Gharan: The distance is the shortest path between the
edges in the tree. So the distance of this dual tree.
>>: Between the edges or?
>> Shayan Oveis Gharan: Between the edges of the dual. So I'm
considering my tree. Considering the edges of the dual graph. If they're
far apart, they're at a distance D with each other. At most my tree would
have 1 over D fractions of edges of n cycle, as you see, for example, on this
example.
So because it has at most one over D fractions in each cycle it will have at
most one over each fraction n cut of the original graph and one over
distance.
So all I need to do is to find a set of edges where there are far apart in the
dual. And also they make connected sub graph in the original graph
course. So, let's see, it's true in this example. Remember I proved this is 1
over 5 thin. So let's see our thinness certificate. If you see -- the dual is
plotted in red here. So these edges are selected. Have distance 5 in the
dual graph.
So this implies perhaps the thinness of our tree is 1 over 5. So next I will
tell you our thin tree selection algorithm. I will tell you how we can find
these far apart edges. So for that I will use tricks. What is a thread? A
thread is a path, you may think of it as a long path, where all of its vertices,
internal vertices has degree 2.
So if you see, you'll find that the dual of the thread is a set of parallel
edges. So suppose that I have a thread like this thread. And this edge, its
middle edge, is the only edge of this thread I've selected as an edge of my
thin tree.
So no matter what the other edges of my thin tree, the distance of this edge
from the rest of the edges will be at least D. At least D over 2. If D is the
length of the thread. Suppose length over thread is 5 here. This is the
distance of this edge from the rest of the edges will be at least 5 over 2.
So what will be an algorithm, if you want to? Each time select a thread dual
graph, pick the edges, the edges of our thin tree and then delete it. And do
it recursively.
And that's exactly our algorithm. These are relatively fine long threads
dual at single edges of the edges of our thin tree and deleted tree. Note
that interestingly the algorithm works with the dual graph. But will show
we'll select the thin tree in the original graph.
So let's go to the proof. First of all, what I need to show is that selected
edges are far apart. This is true by definition of thread, as I said before.
One more thing is that I need to prove that the selected edges would make
a connected spanning sub graph. And why is that true? The reason is so
simple. Consider the thread as the dual of the thread, a set of parallel
edges. So when I'm deleting these parallel edges, sorry, when I'm deleting
the thread edges, in fact I'm contracting these parallel edges in the original
graph.
So if you simulate our algorithm in the original graph, each time you select
a set of parallel edges, one of them as an edge of our thin tree and then
contract it. And we'll do it again. This is an algorithm I said before, but this
time I pick my edge of, pick the edge of my thin tree smartly. And, trivially,
the set of selected edges would make connected spanning sub graph as
before.
So it turns out, if you select it smartly like this, it will give you thin tree.
Also, there is one simpler proof, which says this algorithm will select at
least one edge from any cycle of a dual graph. If you think -- the proof is so
easy, because each time you pick a thread and all the internal methods of a
thread has degree two, the algorithm should select at least one edge of the
edge of dual graph. Now because the dual of any minimal cut is a cycle our
algorithm will select at least one edge from any cut.
Okay. Great. So the only thing I haven't proved yet is the existence of
these threads. And it turns out that they're real easy. So if you can prove
that if we have a planar graph of girth D it contains a thread of length G
over 5.
So this implies that, if you do the calculations, this implies that our tree will
be 1 over G starting. So once G is started you start the gear of the dual.
Remember our algorithm was working with -- we were trying find this
thread dual graph so it's 1 over GS starting thing, order of 1 over GS
starting.
So the only thing I need to show is that GS star is large, and this is the case
in fact. In fact, for planar graph high edge connectivity implies high dual
graph, because the dual of any minimal cut is a cycle. The dual of any
cycle is minimal cut because the minimal cuts have size K, the K of dual is
K. GS star is K and we get order of 1 over K thin tree.
And it is in fact 10 over K for planar graphs. So, yes, this is our thin tree
selection algorithm. So for extending this idea to nonplanar graphs I will
not tell you much about it. The main -- I will tell you the main idea. The
main idea is that for nonplanar graphs the dual can have very small cycles.
It's not the case that if our graph is highly connected the girth of the dual is
so large. In fact, it can contain very small cycle. But fortunately we can get
rid of these small cycles. And the idea is that we can somehow delete their
edges. The minimal algorithm is we consider the smaller cycles, delete
their edges from our original graph. If we do that, because they're small,
they will not disturb the connectivity of our graph and we can convert our
graph to planar graph and everything will be fine.
But there's some tricks. Okay. So this is the ideas of our some of our
concept approximation algorithm. And next I will focus on our order of log
n over log log n approximation algorithm.
Is there any questions? Okay. So let's move on. So again remember that
for in this case we are trying to give a randomized algorithm. It will work
for general metrics ATSP. It considers a fractionally two connected graph
as input and it's looking for an order of log n over log log n thin tree.
So let's start by simple observation. Suppose that this vector of y is the
vector of our fractionally two connected graph. But if you see, you will find
out that this vector Y is in fact not too far from the tree itself. In fact, if we
multiply by 1 minus 1 over n what we'll get is a fractional spanning tree. It's
a spanning tree that's a vector applies to top line vertices of our graph.
And why is that true? In fact, because this vector Y is fractionally two
connected if we multiply by 1 minus 1 over N, still the fractional size of the
edges any cut is bigger than 1, at least 1. Moreover, if you do the
calculations, you'll find out that some of the older fractions in y, some of
the older fractions of edges in y is n.
Because it turns out from the properties of Held-Karp relaxation and you
multiply size 1 minus 1 over n the sum of the fractions edges will be n
minus 1. It will be a fractional spanning tree.
So in fact all we need to do is to consider fractional spanning tree z and
find spanning tree t with respect to z. I mean, that is t with respect to z
across any cut.
Note that previously we were looking to find a tree that is t with respect to
y. But because z is roughly equal to y within some small factors if we find
a tree that's thin with respect to z everything is fine.
So let me start by unsuccessful attempt. Suppose that I round each edge
uniformly at random. Each edge of the -- with its probability. If it is z, if it
has a fraction of ze, I round it with probability z. So if I do that, the
expected number of edges of my tree across any cut, across any cut is
equal to the size of that cut.
So suppose that, for example, this is the cut. This is one of the cuts. So
what I need to have, I need to have a tree that is spanning. I need that this
selection algorithm selects at least one edge across any cut.
But if you want to have that, you need that the size of the cut, expected size
of the cuts to be at least order of log n. Otherwise there will be some cuts
which are not of the edges selected by the algorithm.
So but if you have this, the thinness of our tree cannot be better than order
of log n.
So what we're going to do instead of just uniformly round the edges, you
will select -- we have log n thinness here. So instead we will select a tree T
from some specific discrete distribution that preserves the marginal
probabilities.
So in fact the probability of selecting an edge is equal to z of e, where z is
the vector. So if you have such a distribution, what we'll get is that again
expected number of edges of our tree in any cut is equal to the size of that
cut. But here we will have something more.
Because we are sampling the spanning tree, it will have at least one edge in
any cut. It is connected. So we'll not have any problem for that. But the
only thing we need to prove is that we have concentration for this equality,
only in the right tail. We need to prove that the number of our edges of our
tree is not that large in any cut. So previously we need to -- we had to
prove its concentration in the left side, too, because our tree's spanning it
has at its edge one cut and we need to have concentration in the left tail,
and we need to have concentration on the right and we have 2 to the n
equations.
So this is why we can get better approximation if we want to measure -- if
we want to have concentration on the right tail, because the chain of bound
for the right tail gives us a better bound, comparing the share of bounds for
the left tail.
So let's see how we can get concentration. Previously, because we were
sampling the edges uniformly at random, independently uniformly at
random, we had independence, and we could use shared types of bound.
But here we cannot because we were sampling a tree and the edges are not
independent or are not selected independently. So we need to use
negative correlation.
So if we do not know the definition of the negative correlation, we say that
sampling procedure has negative correlation if the selection of an edge
into our, for example, tree can only decrease the probability of other edges.
In particular, for n is set of f of edges probably that all of these edges are
selected in our tree less than recall their marginal, the multiplication of
their marginals.
So remember that if you had negative correlation, then you would have
equality here. But here -- sorry if you have independence you would have
equality here. But here because you have negative correlation, it gives
inequality. But in fact this is all we need to have the chain of types of
bound. If we have negative correlation, then we can prove the correctness
of any general types of bounds.
And this is the idea. So in fact there are three possible, currently three
possible approaches to sample our random spanning tree while having
negative correlation.
And while having the probability of selecting edges equal to its marginals.
So this is the method maximum distribution used in our paper. It was
previously defined by Amin and Arash [phonetic] in selecting random
matching. There's also a pipeage rounding idea by the work of Calinescu
Checkuri Polin Vondrak. And in fact this -- I mean they proved -- they then
prove exactly this negative correlation, but they proposed a randomized
algorithm and this negative correlation can be found through their proofs
and be applied through their proofs.
Recently, after our paper, Checkuri Vondrak and Zenklusen found some
new approach called randomized swap rounding. This approach is purely
combinatorial. And its running time is better than others. In fact, it's more
general. I won't show you how it will generalize it. But however all of these
algorithms gives us negative correlation concentration bounds.
So after that, suppose that we have negative correlation concentration
bound. Let's see what will be our chain of bound. In fact, the probability
that our tree has more than log n over log log n the number of edges of our
tree is more than log n over log log n of the size of the cut, is less than or
equal to the n minus the constant times the size of that cut.
And this is in fact where the log n over log log n comes from. Because this
chains of types of bound are for the right tail. If you have log n over log log
n, you want to have log n over log log n deviation, the probability will go to
n minus constant.
But, yeah, so by this inequality we'll have concentration for one cut. So all
we need is to have concentration for all the 2 to the n cuts. For that we'll
use Karger formula. And Karger in fact proved that in any graph, through
at most n to the 2 K cuts of size at most K times the minimum cut.
Suppose we have a graph, related graph. The number of cuts of size k
times minimum cut is percent 2 K. We'll use this formula and we'll use it
through a union bound argument.
So in fact in our, as I said before, because we had fraction of 2 connected
graph, its minimum cut is roughly one or two. So the number of size of K,
say between K minus 1 and K is at most 2 to the K.
If you multiplied this n2K by this and N minus constant, it will be K if you
considered the cost of K to the minus K. These two will cancel out. It turns
out you can set this constant C to be large enough just by placing some
factor in front of the log n over log log n. So you can prove easily with high
probability our tree will be thin, will be log n over log log n thin in each of
the cuts of size of size L for L 1 and a half. So if you use union bound for
all the cuts, you'll find out that our tree will be log n over log log n 10 with
high probability.
So let's briefly -- let me briefly tell you where this log n over log log n come
from. I mean, is it -- show you a counter-example that we cannot do better
than log n over log log n. Example is simple. Consider complete graph n
vertices and suppose the fraction of each is 1 over n. So in this graph,
even if you -- if you select each edge uniformly and independently at
random, so suppose that we're not looking for proving anything for the left.
We already proved -- we're only looking to prove something for the right
tail. Suppose only we -- suppose that we select the edges uniformly at
random, but turns out this will be something like [indiscernible]. In fact, for
any edge, because the fraction of each edge is 1 over N, the size of this cut
is 1. So the probability that an edge of our tree will be selected will be
incident to this vertex will be 1 over N. So if you think, we'll select n edges
like n minus 1 over n edges and each time there is 1 over n probability that
this edge will be selected to one specific vertex. So it is like a balls and
bins problem where vertices are bins and edges are balls, and there's 1
over n probability that an edge is incident to a specific vertex.
So it's proved that in a balls and bins, n balls and n bins, there's a bin with
high probability that contains log n over log log n balls.
So in fact if we run this -- if you select our edges uniformly and
independently at random, there will exist a vertex with degree log n over
log log n with high probability. In fact, even if you only consider the cost
that separated the vertex from the rest of the graph you cannot do better
than log n over log log n. In fact, there are divorce costs in this example.
So this is a generalization by Checkuri Chenoff and Zenklusen. They
generalize the idea to any matrix. Not just the spanning tree. So they
proved that suppose that we had a fractional solution of matrix polytope,
and we're looking to sample a base from that matrix also by preserving
marginal probabilities. And they prove that we can do that while getting
negative correlation and therefore we have n type chain of bound for any
linear combination of the variables.
So in fact their algorithm works for a number -- for various problems, and
they give, they drive like log n over log nn approximation for number of
problems.
But always you cannot do better than log n over log log n because of the
right tail of general bounds.
But as I said before, the algorithm is purely combinatorial. You can find the
proof. But their proof is much smaller than comparing to us. You can find
it in Vondrak's home page.
Okay. So any questions? Let's conclude. So here first we consider the
problem in directed domain. Then we try to have some simplification. We
convert it to another problem undirected domain. But although the
probability in undirected seems to be much harder, but as you see we have
found good results for that.
So I mean trying to get rid of this direction would be a very good idea. Also
we've shown some interesting relation between ATSP and graph
embedding problems, dual graphs, planar graphs and these sorts of things.
And, finally, we prove that integrality gap ATSP is constant for shortest
path metric defined unbounded genus graphs.
So the interesting point is that the Checkuri Goemans and Karloff
integrality graph encountered example is planar. So in fact they proved
that integrality gap of ATSP on planar graphs is at least two. Therefore, in
fact at least for planar graphs we know that integrality gap is constant. It's
bounded low by 2, above by 22 as we prove. But for nonplanar graphs, if
there are bounded genus still everything is okay. But if they're not
bounded to genus, they're from general metrics is still open whether their
integrality gap is constant or not.
So let me tell you some often problems for future direction. So, first of all,
is it possible to generalize our ideas to find constant trees for general
metrics. Not only for rated bounded genus graphs. Not rated bounded
genus graphs. And, moreover, is it true, is it possible to get P task for
ATSP on planar graphs, like because we have P task for symmetric genus
planar graphs so it is hard to find P task for problems in directed domain.
There are a lot of difficulties.
But it's a nice problem to work on. Moreover, here we proved that this
guarded conjecture any KH connected graph contains a constant over K
thin tree implies constant integrality graph of the ATSP. Also, there is
some conjecture by Jaeger on number of zero flows. Does not tell you the
details, but this is a well-known conjecture number of zero flow. It says like
any 4-H connected graph contains 2 plus 1 over K number of flow. We can
search for it on the Web. But turns out that this conjecture implies some
version of the Jagger conjecture which has not been proved. This implies
that this conjecture seems to be a very hard problem. But anyway these
are the directions we know, but it's open whether any other direction is
true, whether, for example, we can prove Goemans conjecture from the
constant integrality gap of ATSP. If it's true, at least we know that although
this is a thin tree problem is a hard problem but it is something like ATSP.
And we lose nothing by reducing our problem to thin trees.
Also some directions between these two, these others are very interesting.
>>: [indiscernible] approach at the most you want to ->> Shayan Oveis Gharan: Yeah, there exists six numbers of flows for two
edge connected graphs. But the Jaeger conjecture is that if we increase
the connectivity to some very large number can we get 2 plus epsilon for
very small epsilon or not. 2 plus epsilon over 0 flow.
>>: Epsilon?
>> Shayan Oveis Gharan: 2 plus epsilon.
>>: What does epsilon mean?
>> Shayan Oveis Gharan: Pick any epsilon. Suppose you pick any epsilon.
And I should show that.
>>: No [indiscernible] integral.
>> Shayan Oveis Gharan: You can define for fractional graphs, too. By
setting the -- I mean, allowing the fraction of each edge to be between 1 and
1 plus epsilon. You can -- they are similar. This Jaeger conjecture asks if
you can find -- so, for example, if the graph contains a 2 plus epsilon,
number of zero flow, it is very near Eulerian graph.
>>: They're much bigger things, conjecture. So the types of conjecture s
that if there is a poor connected graph, you have at least three ->> Shayan Oveis Gharan: This top conjecture is special case of Jaeger
conjecture. For K equals 1.
>>: But you don't have to do epsilon. There's still a conjecture -- I led you
to the constant with the [indiscernible], can you show with a thousand
connected graph you can find three normal zero flow.
>> Shayan Oveis Gharan: This is still open.
>>: So you don't even have to talk about epsilon or anything.
>> Shayan Oveis Gharan: But at least for planar graphs you can do that or
for bounded genus graphs. If the connectivity is sufficiently large we can
get 2 plus epsilon.
>>: Planar graph.
>> Shayan Oveis Gharan: Our algorithm implies this. Also, there's some
various works, truly only on Jaeger conjecture that proves that Jaeger
conjecture is true for bounded genus graphs. But our algorithm is a bit
more general because it implies -- Goddyn conjecture, proves the Goddyn
conjecture which is more general than here. Okay. Thank you.
[applause]