Bounds on extremal functions of forbidden patterns
by
Jesse Geneson
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in partial fulfillment of the requirements for the degree of
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MASSACHUSETTS INSTITUTE OF TECHNOLOGY
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@ Massachusetts Institute of Technology 2015. All rights reserved.
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Department of Mathematics
May 10, 2015
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Certified by.................
Peter Shor
Professor
Thesis Supervisor
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Accepted by ..................
Michel Goemans
Chairman, Department Committee on Graduate Theses
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Bounds on extremal functions of forbidden patterns
by
Jesse Geneson
Submitted to the Department of Mathematics
on May 10, 2015, in partial fulfillment of the
requirements for the degree of
Doctor of Philosophy
Abstract
Extremal functions of forbidden sequences and 0 - 1 matrices have applications to
many problems in discrete geometry and enumerative combinatorics. We present
a new computational method for deriving upper bounds on extremal functions of
forbidden sequences. Then we use this method to prove tight bounds on the extremal
functions of sequences of the form (12 ... 1)' for 1 > 2 and t > 1, abc(acb)t for t > 0,
and avav'a, such that a is a letter, v is a nonempty sequence excluding a with no
repeated letters and v' is obtained from v by only moving the first letter of v to another
place in v. We also prove the existence of infinitely many forbidden 0 - 1 matrices P
with non-linear extremal functions for which every strict submatrix of P has a linear
extremal function. Then we show that for every d-dimensional permutation matrix
P with k ones, the maximum number of ones in a d-dimensional matrix of sidelength
n that avoids P is 20(k) dThesis Supervisor: Peter Shor
Title: Professor
3
4
Acknowledgments
Thanks to Peter Shor for advising my thesis research and for improving the bounds on
S2 (m) in Lemma 11. Thanks also to Jacob Fox for research advice and for help proving
lower bounds on extremal functions of multidimensional permutation matrices. I also
thank Henry Cohn for comments to improve the clarity of the 0 - 1 matrix proofs and
for ideas about formation width. Also I thank Peter Shor, Jacob Fox, and Henry Cohn
for being on my thesis committee. In addition, I thank Joe Gallian for introducing
me to extremal functions of 0 - 1 matrices.
Thanks to Peter Tian for collaborating on the multidimensional 0 - 1 matrix
results and on the Python code for computing formation width. Thanks also to Lilly
Shen for collaborating on results about extremal functions of 2-dimensional 0 - 1
matrices, as well as Rohil Prasad and Jonathan Tidor for collaborating on results
about formation width. Moreover, I thank Tanya Khovanova for advice on several
research projects and on my thesis defense, and for collaborating on research about
visibility graphs.
Thanks to PRIMES and RSI for giving me the opportunity to collaborate on
projects and co-author papers as a research mentor. Thanks also to the NSF and
MIT for financial support during graduate school. I also thank Mary Thornton,
David Geneson and Arianna Geneson for their support and advice. In addition, I
thank Katherine Bian for support, advice, and help on my thesis defense.
5
6
Contents
1
2
3
9
Introduction
1.1
1.2
Matrix extremal functions . . . . . . . . . . . . . . . . . . . . . . . .
Sequence extremal functions . . . . . . . . . . . . . . . . . . . . . . .
9
10
1.3
Order of results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
The
2.1
2.2
2.3
maximum number of distinct letters
Upper Bounds . . . . . . . . . . . . . . .
Lower bounds . . . . . . . . . . . . . . .
Bounds on extremal functions of 0 - 1
Schinzel sequences . . . . . . . . . . . .
in ababa-free sequences
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
matrices using Davenport. . . . . . . . . . . . . . . .
13
14
16
The
3.1
3.2
3.3
3.4
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23
25
27
29
31
Computing I . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31
formation width of sequences
An extension of the Erd6s-Szekeres theorem . . . .
Algorithm for computing fw . . . . . . . . . . . . .
Using binary formations to compute fw . . . . . . .
Bounding the formation width of binary formations
3.4.1
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4
Linear extremal functions of forbidden 0 - 1 matrices
4.1 Facts about exs(n, S) and exsk(m, S) . . . . . . . . . . . . . . . . . .
4.2 Bar s-visibility hypergraphs and 0 - 1 matrices . . . . . . . . . . . .
4.3 Infinitely many minimal non-linear 0 - 1 matrices . . . . . . . . . . .
41
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47
5
Extremal functions of forbidden multidimensional matrices
5.1 Upper and lower bounds for d-dimensional permutation matrices
5.2 Sharp bounds on m(n, R . k, d) . . . . . . . . . . . . . . . . . .
5.3 Upper bounds for d-dimensional double permutation matrices .
5.4 Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . .
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57
61
3.5
3.6
3.7
3.4.2 Computing r . . . . . . . . .
Further bounds on extremal functions
Further bounds on fw . . . . . . . . .
Open Problems . . . . . . . . . . . .
7
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using fw
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8
Chapter 1
Introduction
Problems in the area of pattern containment and avoidance focus on determining
whether specific substructures are present in a given structure [1, 5, 15, 27, 29, 34,
36, 43, 441. For example, does the sequence a have a subsequence that is isomorphic
to v? Or, analogously, does the 0 - 1 matrix A have a submatrix that can be turned
into B by possibly changing some ones to zeroes?
We focus mainly on bounding extremal functions of sequences and 0 - 1 matrices
that avoid forbidden patterns. For example, what is the maximum number of ones
in an n x n 0 - 1 matrix that does not contain a forbidden pattern? Or, what is the
minimum length k so that every sequence of length k with n distinct letters and no
adjacent same letters contains an alternation of length 5?
1.1
Matrix extremal functions
An early motivation for bounding matrix extremal functions was to use them for
solving problems in computational and discrete geometry [4, 18, 371. Mitchell wrote
an algorithm to find a shortest rectilinear path that avoids obstacles in the plane [371
and proved that the complexity of this algorithm is bounded from above in terms
of a specific matrix extremal function, which was bounded by Bienstock and Gybri
[4]. Firedi [181 used matrix extremal functions to derive an upper bound on Erd6s
and Moser's [101 problem of maximizing the number of unit distances in a convex
n-gon. Recent interest in the extremal theory of matrices has been spurred by the
resolution of the Stanley-Wilf conjecture using the linearity of the extremal functions
of forbidden permutation matrices [33, 36].
The 0 - 1 matrix A contains a 0 - 1 matrix M if some submatrix of A can be
transformed into M by changing some ones to zeroes. If A does not contain M, then
A avoids M. Let ex(n, M) be the maximum number of ones in an n x n 0 - 1 matrix
that avoids M, and let exk(M, M) be the maximum number of columns in a 0 - 1
matrix with m rows that avoids M and has at least k ones in every column.
Furedi and Hajnal asked for a characterization of all 0 - 1 matrices P such that
ex(n, P) = O(n) [19]. A corresponding problem for the column extremal function
is to characterize all 0 - 1 matrices P with j rows such that for every fixed k > j,
9
exk(m, P) = 0(m). Another related problem is to characterize all 0 - 1 matrices P
for which eXk(m, P) = OF) ).
A method for bounding ex(n, M) by using bounds on the maximum number of
edges in bar visibility graphs was introduced in [171. By using a similar method with
bar visibility hypergraphs, we obtain linear bounds on the extremal functions of other
forbidden 0 - 1 matrices.
Call N a minimal nonlinear 0 - 1 matrix if ex(n, N) = w(n) and ex(n, N') = O(n)
for every N' properly contained in N. Call the k x 2k matrix P a double permutation
matrix if P can be obtained from a k x k permutation matrix by replacing every
column with two copies of itself. We prove that ex(n, P) = 2O(k)n for every k x 2k
double permutation matrix P. Using this result, we show the existence of infinitely
many minimal nonlinear 0 - 1 matrices.
We also study the extremal functions of multidimensional 0 - 1 matrices. A ddimensional ni x ... x nd matrix is denoted by A = (a .... ,id), where 1 < i1 < ne
for e = 1, 2,. .. , d. We may view a d-dimensional 0 - 1 matrix A
(ail ..,id) as a
d-dimensional rectangular box of lattice points with coordinates (i1 , ... , id).
An f-cross section of matrix A is a maximal set of entries ail . ,id with i fixed. A
d-dimensional k x ... x k 0 - 1 matrix is a permutation matrix if each of its i-cross
sections contains a single one for every e = 1,... , d.
A d-dimensional 0 - 1 matrix A avoids another d-dimensional 0 - 1 matrix P if
no submatrix of A can be transformed into P by changing some ones to zeroes. The
maximum number of ones in a d-dimensional n x ... x n matrix that avoids P is
denoted by f(n, P, d).
We exhibit a family of k x
...
x k permutation matrices P for which f(3,d)4
has a
lower bound of 2 Q(kl/d) for n > 2Lkl/dJ/20 Furthermore we improve the upper bound
on A"P") from 2 0(klogk) to 2 0(k) for all k x - x k permutation matrices P, and we
show for every fixed d > 2 that the new upper bound is also true for d-dimensional
double permutation matrices of dimensions 2k x k x ... x k.
1.2
Sequence extremal functions
A sequence s contains a sequence tt if some subsequence of s can be changed into u
by a one-to-one renaming of its letters. If s does not contain u, then s avoids u. The
sequence s is called r-sparse if any r consecutive letters in s are pairwise different.
A Davenport-Schinzel sequence of order s is a 2-sparse sequence that avoids alternations of length s + 2. Upper bounds on the lengths of Davenport-Schinzel sequences
provide bounds on the complexity of lower envelopes of solution sets to linear homogeneous differential equations of limited order [8] and on the complexity of faces in
arrangements of arcs with a limited number of crossings [1.
A generalized Davenport-Schinzel sequence is an r-sparse sequence that avoids a
fixed forbidden sequence with r distinct letters. Fox et al. [161 and Suk et al. [451
used bounds on the lengths of generalized Davenport-Schinzel sequences to prove that
k-quasiplanar graphs on n vertices with no pair of edges intersecting in more than t
10
points have at most (n log n)2a("-)' edges, where a(n) denotes the inverse Ackermann
function and c is a constant that depends only on k and t.
Our main contribution for sequence extremal functions is a new computational
method for proving tight upper bounds using upper bounds that are already known.
We also construct families of sequences avoiding alternations in order to prove lower
bounds on extremal functions of forbidden alternations.
Let As,k(m) be the maximum number of distinct letters in any sequence which
can be partitioned into m contiguous blocks of pairwise distinct letters, has at least
k occurrences of every letter, and avoids alternations of length s. Nivasch [381 proved
that A 5 ,2 d+l(m) = O(mad(m)) for all fixed d > 2. We show that A,+, 8 (m) = ("F1)
for all s > 2, A 5 ,6 (M) =
0(mlog
log m), and A 5 ,2d+ 2 (M) = O(mad(M)) for all fixed
d > 3.
An (r, s)-formation is a concatenation of s permutations of r letters. If u is a
sequence with r distinct letters, then let Ex(u, n) be the maximum length of any rsparse sequence with n distinct letters which avoids u. We introduce a computational
method for deriving tight upper bounds on Ex(u, n): For every sequence u define
fw(u), the formation width of u, to be the minimum s for which there exists r such
that there is a subsequence isomorphic to u in every (r, s)-formation. We use fw(u) to
prove upper bounds on Ex(u, n) for sequences a such that a contains an alternation
with the same formation width as u.
= 2t - 1
We generalize the bounds on Ex((ab)t , n) by showing that fw((12. .)t)
and Ex((12... l)', n) = n2(t2)!a(n)-2O(a(n)t 3 ) for every I > 2 and t > 3, such that
a(n) denotes the inverse Ackermann function. Upper bounds on Ex((12 ... 1)', n) were
used to bound the maximum number of edges in k-quasiplanar graphs on n vertices
with no pair of edges intersecting in more than 0(1) points.
If u is any sequence of the form avav'a such that a is a letter, v is a nonempty
sequence excluding a with no repeated letters and v' is obtained from v by only
moving the first letter of v to another place in v, then we show that fw(u) = 4
and Ex(u, n) = ®(na(n)). Furthermore we prove that fw(abc(acb)t ) = 2t + 1 and
Ex(abc(acb)t , n) = n 2
1.3
I)(n)
O((f)) for every t > 2.
Order of results
The next two chapters are about sequences. In Chapter 2, we prove bounds on
As,k(m), as well as corollaries related to extremal functions of interval chains and
0 - 1 matrices. In Chapter 3, we use formation width to prove tight bounds on
extremal functions of forbidden sequences.
The final two chapters are about 0-1 matrices. In Chapter 4, we show that double
permutation matrices and classes of 0 - 1 matrices corresponding to bar visibility
hypergraphs have linear extremal functions. In Chapter 5, we generalize Fox's results
on permutation matrices, as well as our results on double permutation matrices, to d
dimensions.
The results in these chapters also appear in the papers [20, 21, 22, 23, 24].
11
12
Chapter 2
The maximum number of distinct
letters in ababa-free sequences
The sequence s is called r-sparse if any r consecutive letters in s are pairwise different. Let D.(n) be the maximum length of any 2-sparse sequence with n distinct
letters which avoids alternations of length s. Nivasch [38] and Klazar [32] proved that
lim
D5 (n)
2, such that a(n) denotes the inverse Ackermann function. Agarwal,
Sharir, Shor [21 and Nivasch [38] proved the bounds D8 (n) = n2t!(n)O(o(n)t-1) for
Pettie [42] derived sharp bounds on D (n) for all odd s.
even s > 6 with t == ".
To define the Ackermann hierarchy let A1(n) = 2n and for k > 2, Ak(0) = 1 and
Ak(n) = Akl(Ak(n - 1)) for n > 1. To define the inverse functions let ak(x) =
min {n Ak(n) >
4} for all
k > 1.
We define the Ackermann function A(n) to be A,,(3) as in [38]. The inverse
Ackermann function a(n) is defined to be min {x : A(x) > n}.
Collections of contiguous distinct letters in a sequence are called blocks. Nivasch's
bounds on D8 (n) were derived using an extremal function which maximizes number
of distinct letters instead of length. Let A.,k(m) be the maximum number of distinct
letters in any sequence on m blocks avoiding alternations of length s such that every
letter occurs at least k times. Clearly A,(m) = 0 if m < k and A 8 ,k(M) = o0 if
k < s - 1 and k < m-.
Nivasch proved that A5,2d+1(mn) = O(Mad(m)) for each fixed d > 2 (but noted
that the bounds on A 5,k(m) were not tight for even k). Sundar [46] derived similar
bounds in terms of m on functions related to the Deque conjecture.
In [381, similar bounds were also derived for a different sequence extremal function.
Let an (r, s)-formation be a concatenation of s permutations of r distinct letters. For
example abcddcbaadbcis a (4, 3)-formation. Define Fr,s(n) to be the maximum length
of any r-sparse sequence with n distinct letters which avoids all (r, s)-formations.
Klazar [30] proved that F,2 (n) = O(n) and Fr,3 (n) = O(n) for every r > 0. Nivasch
proved that F,,4 (n) = 0(na(n)) for r > 2. Agarwal, Sharir, Shor [21 and Nivasch [381
showed that F,s(n) n2
*
)'1) for all r > 2 and odd s > 5 with t = Y-2.
Let F,,,,k(m) be the maximum number of distinct letters in any sequence on m
blocks avoiding every (r, s)-formation such that every letter occurs at least k times.
-
13
Clearly F,,,,k(M) = 0 if m < k and F,,,,k(M) = oo if k < s and k K m. Every (r, s)formation contains an alternation of length s + 1 for every r > 2, so AS+1,k(m)
Fr,,k(m) for every r > 2.
Nivasch proved for r > 2 that F,,4,2d+1(M) = O(mad(m)) for each fixed d > 2.
The recursive inequalities for the upper bounds on F,,4,2d+1 (in) in [38] also imply that
F,,4, 6(M) = O(m log log 7n) and F,4,2d+2(mn) = O(Mad(m)).
Similar bounds were also derived on an extremal function related to interval chains.
A k-chain on [1, m] is a sequence of k consecutive, disjoint, nonempty intervals of the
form [ao, a,][ai + 1, a2 ... [ak_1 + 1, ak] for integers 1 < ao K a1 < ... < ak < rM.
An s-tuple is a set of s distinct integers. An s-tuple stabs an interval chain if each
element of the s-tuple is in a different interval of the chain.
Let (,,k(m) denote the minimum size of a collection of s-tuples such that every
k-chain on [1, m] is stabbed by an s-tuple in the collection. Clearly ((,k(m) = 0 if
m < k and (sk(m) is undefined if k < s and k K m.
Alon et al. [3] showed that 3,s(m) = ("
) for s > 1, ( 3 ,4 (m) O(mlogm),
(m log logm), and (3,k(M) = (maLkj (M)) for k > 6.
(3,5(m)
Let rr,s,k(mi) denote the maximum size of a collection X of not necessarily distinct
k-chains on [1, m] so that there do not exist r elements of X all stabbed by the same
s-tuple. Clearly r,,,k(m) = 0 if m < k and 7r,,,k(m) = oo if k < s and k
n.
In Section 2.1 we show that r,,,,,k(M) =F,sl+,k+l(m-i+ 1) for all r > 1 and 1 K s K
k K m. Since (.,k(i)
rq2,,k(m) for all 1 K s K k K m, then AS+ 2 ,k+l(m+1) K (M)
for all 1 K s K k K m. This implies the bounds A,+1 , 8(in) < (" 7il) for all s > 2,
A 5 ,6 (m) = O(m log log m), and A5, 2 d+ 2 (M)= O(mad(in)) for d > 3.
In Section 2.2 we construct alternation-avoiding sequences to prove lower bounds
on As,k(m). We prove that A,+1,8 (m) = ("
) for all s > 2. Furthermore we show
that A 5 ,6 (M) = Q(mloglogim) and A 5 ,2d+ 2 (m) = Q(jMad(M)) for d > 3. Thus the
bounds on A 5 ,a(m) have a multiplicative gap of O(d) for all d.
2.1
Upper Bounds
We show that
F,,s+1,k+l(m + 1) = rlr,s,(M) for all r
1 and 1 K s K k K M using
maps like those between matrices and sequences in [7] and [40].
Lemma 1.
F,,sl+,k+l(M
+ 1)
< Tlr,,k(m)
for all r > 1 and 1 K s K k K m.
Proof. Let P be a sequence with Fr,s +l,k+1(m + 1) distinct letters and m + 1 blocks
1,.. . , m + 1 such that no subsequence is a concatenation of s + 1 permutations of
r different letters and every letter in P occurs at least k + 1 times. Construct a
collection of k-chains on [1, m] by converting each letter in P to a k-chain: if the first
k + 1 occurrences of letter a are in blocks ao, ... , ak, then let a* be the k-chain with
ith interval [ai_ 1 , a, - 1].
Suppose for contradiction that there exist r distinct letters q1,. . ., q, in P such
that q*, ... , q* are stabbed by the same s-tuple 1 < Ji < ... < j 5 K
<m. Let jo = 0
and js+1
m + 1. Then for each 1 K i K s + 1, q,, occurs in some block b,,i such
14
that b
(r, s
+
c,i[j1i-
+ 1, ji] for every 1 < n K r. Hence the letters qi,.. . , q, make an
l
1)-formation in P, a contradiction.
Corollary 2.
AS+ 2 ,k+1(m
+ 1) K Q(m) for all 1 K s < k K m.
The bounds on (,,k(m) in
[3] imply the next corollary.
Corollary 3. A,+i,8 (m) < (1 1 ) for s > 2, A 5 ,5 (m)
for k > 7.
O(m log log m), and A 5 ,k(m) = O(mak21J(m))
'jj
O(mlogrm), A 5 ,6 (m) =
To prove that F,,,+,k+1(M + 1)
ms,k(m) for all r > 1 and 1 K s K k K m,
we convert collections of k-chains into sequences with a letter corresponding to each
k-chain.
Lemma 4. Fr,s+1,k+1(m + 1)
r,s,k(m)
for all r > 1 and 1 K s < k K m.
Proof. Let X be a maximal collection of k-chains on [1, m] so that there do not exist
r elements of X all stabbed by the same s-tuple. To change X into a sequence P
create a letter a for every k-chain a* in X, and put a in every block i such that either
a* has an interval with least element i or a* has an interval with greatest element
i - 1.
Order the letters in blocks starting with the first block and moving to the last.
Let Ai be the letters in block i which also occur in some block j < i and let Bi be
the letters which have first occurrence in block i.
All of the letters in Ai occur before all of the letters in Bi. If a and b are in Aj,
then a appears before b in block i if the last occurrence of a before block i is after
the last occurrence of b before block i. The letters in Bi may appear in any order.
P is a sequence on m + 1 blocks in which each letter occurs k + 1 times. Suppose
for contradiction that there exist r letters q, ..., q,. which form an (r, s + 1)-formation
in P. List all (r, s + 1)-formations on the letters qi, .. . , q, in P lexicographically, so
that formation f appears before formation g if there exists some i > 1 such that the
first i - 1 elements of f and g are the same, but the ith element of f appears before
the ith element of g in P.
Let fo be the first (r, s + 1)-formation on the list and let 7ir (respectively pi) be
the number of the block which contains the last (respectively first) element of the
i < s + 1. Suppose for contradiction that for some
ith permutation in fo for 1
1 < i K s, 7i = Pi+1. Let a be the last letter of the ith permutation and let b be the
first letter of the (i + 1)st permutation.
Then a occurs before b in block ri and the b in 7i is not the first occurrence of b
in P, so the a in wi is not the first occurrence of a in P. Otherwise a would appear
after b in 7ri. Since the a and b in wi are not the first occurrences of a and b in P, then
the last occurrence of a before 7i must be after the last occurrence of b before irr. Let
fi be the subsequence obtained by deleting the a in ir from fo and inserting the last
occurrence of a before 7i. Then fi is an (r, s + 1)-formation and fi occurs before fo
on the list. This contradicts the definition of fo, so for every 1 K i K s, 7ri < Pi+i.
j < r and 1 K i K s + 1, the letter qj appears in some block between
For every 1
w) stabs
71,...,
pi and Tr inclusive. Since 7i < Pj+j for every 1 < i K s, the s-tuple (
15
each of the interval chains q*,.. , q,*, a contradiction. Hence P contains no (r, s + 1)formation.
The idea for the next lemma is similar to a proof about doubled formation free
matrices in [7].
Lemma 5. F,,s,s (m) < (r
-
1)("- IV) for every r > 1 and I < s < m.
Proof. Let P be a sequence with m blocks such that no subsequence of P is a concatenation of s permutations of r distinct letters and every letter of P occurs at least
s times. An occurrence of letter a in P is called even if there are an odd number of
occurrences of a to the left of it. Otherwise the occurrence of a is called odd.
Suppose for contradiction that P has at least 1 + (r - 1) (7ni 1) distinct letters.
+
The number of distinct tuples (i,
.... iLJ) for which a letter could have even
occurrences in blocks i,.. . ,
2 j is equal to the number of positive integer solutions
to the equation (1 I x 1 ) + ... + (1 + xLj) + xI+LJ =
+ 1 if s is even and (1
x1 ) + ... + (1 + X
+H)x1+[IJ = M if s is odd. Then by the pigeonhole principle
there are at least r distinct letters q 1 , ... , q, with even occurrences in the same [LJ
blocks. Then P contains a concatenation of s permutations of the letters q 1 ,... q,
a contradiction.
l
The last lemma is an alternate proof that A,+1, 8(m) < ("r[l) since A8 +,(rm) <
Fr,,,s(mfl) for all s, i > 1 and r > 2.
2.2
Lower bounds
In the last section we showed that A,+1,8 (n)
(
for s > 2. The next lemma
provides a matching lower bound.
("
71) for all s > 2 and m >
s + 1.
Proof. For every s > 1 and m > s + 1 we build a sequence X,(m) with (j"-
)
Lemma 6. As+ 1 ,(n) >
distinct letters. First consider the case of even s > 2. The sequence X,(rn) is the
concatenation of i - 1 fans, so that each fan is a palindrome consisting of two blocks
of equal length.
First assign letters to each fan without ordering them. Create a letter for every
-tuple of non-adjacent fans, and put each letter in every fan in its 1-tuple. Then
order the letters in each fan starting with the first fan and moving to the last. Let Ai
be the letters in fan i which occur in some fan j < i and let B be the letters which
have first occurrence in fan i.
In the first block of fan i all of the letters in Ai occur before all of the letters in
Bi. If a and b are in Ai, then a occurs before b in the first block of fan i if the last
occurrence of a before fan i is after the last occurrence of b before fan i. If a and b are
in Bi, then a occurs before b in the first block of fan i if the first fan which contains
a without b is before the first fan which contains b without a.
16
Consider for any distinct letters x and y the maximum alternation contained in
the subsequence of X,(m) restricted to x and y. Start building the alternation with
a fan that contains only x. Any other fans which contain x without y or y without x
add at most 1 to the alternation length. Any fans which contain both x and y add 2
to the alternation length. If x and y occur together in i fans, then the length of their
alternation is at most (s - i) + (I - i) + 2i = s.
Every pair of adjacent fans have no letters in common, so every pair of adjacent
blocks in different fans can be joined as one block when the rn-1 fans are concatenated
to form Xs(rn). Thus X,(m) has m blocks and ("2_) letters, and each letter occurs
s times.
For odd s > 3, construct X,(m) by adding a block r after Xs-1(m - 1) containing
all of the letters in X,-1(m - 1) such that a occurs before b in r if the last occurrence
of a in X,_(m - 1) is after the last occurrence of b in X,_1(m - 1). Then X,(m)
contains no alternation of length s + 1 since X,_1(m - 1) contains no alternation of
length s. Moreover X,(m) has m blocks and
("
q--+
r
1
)
letters, and each letter occurs
2
s times.
l
The next lemma shows how to extend the lower bounds on As+,,,(m) to F,,,,,(m).
Lemma 7. F,s,k(m)
(r - 1)F2 ,,,k(M) for all r > 1 and 1 < s < k < m.
-
Proof. Let P be a sequence with F 2 ,,,k(m) distinct letters and m blocks such that no
subsequence is a concatenation of s permutations of two distinct letters and every
letter occurs at least k times. P' is the sequence obtained from P by creating r - 1
new letters a 1 , .. , a,-, for each letter a and replacing every occurrence of a with the
sequence a, ... a.
Suppose for contradiction that P' contains an (r, s)-formation on letters q 1 ,... , q,.
Then there exist indices i, j, k, 1 and distinct letters a, b such that qi = a3 and
qk= bl. P' contains a (2, s)-formation on the letters qi and qk, so P contains a (2, s)
formation on the letters a and b, a contradiction. Then P' is a sequence with (r
1)F2,s,k(m) distinct letters and m blocks such that no subsequence is a concatenation
of s permutations of r distinct letters and every letter occurs at least k times.
El
Corollary 8. Fr,,,, (m) = (r - 1)("
) for
all r > 1 and m > s > 1.
The proof of the next lemma is much like the proof that
A 5 ,2 d+1(m)
- Q(Mad(n))
for d > 2 in [381.
Lemma 9. A 5 ,6 (m)
=
Q(m log log m) and A5,2d+ 2 (m) = Q(!mad(m)) for d > 3.
For all d, m > 1, we inductively construct sequences Gd(M) in which each letter
appears 2d + 2 times and no two distinct letters make an alternation of length 5.
This proof uses a different definition of fan: fans will be the concatenation of two
palindromes with no letters in common. Each palindrome consists of two blocks of
equal length.
The sequences G1(m) are the concatenation of m + 1 fans. In each fan the second
block of the first palindrome and the first block of the second palindrome make one
17
block together since they are adjacent and have no letters in common. The first
palindrome in the first fan and the second palindrome in the last fan are empty.
There is a letter for every pair of fans and the letter is in both of those fans. The
letters with last appearance in fan i are in the first palindrome of fan i. They appear
in fan i's first palindrome's first block in reverse order of the fans in which they first
appear. The letters with first appearance in fan i are in the second palindrome of fan
i. They appear in fan i's second palindrome's first block in order of the fans in which
they last appear. By construction G1 (m) contains no alternation of length 5.
For all d > 1 the sequence Gd(1) consists of 2d + 2 copies of the letter 1. The
first and last copies of 1 are both special blocks, and there are empty regular blocks
before the first 1 and after the last 1.
For d, m > 1 the blocks in Gd(m) containing only first and last occurrences of
letters are called special blocks. Let Sd(m) be the number of special blocks in Gd(m).
Every letter has its first and last occurrence in a special block, and each special block
in Gd(m) has m letters.
Blocks that are not special are called regular. No regular block in Gd(m) has
special blocks on both sides, but every special block has regular blocks on both sides.
The sequence Gd(m) for d, m > 2 is constructed inductively from Gd(m - 1) and
Gd_1(Sd(m - 1)). Let f = Sd(m - 1) and g = Sd_1(f). Make g copies X 1 ,..., X
of Gd(m - 1) and one copy Y of Gdl(f), so that no copies of Gd(m - 1) have any
letters in common with Y or each other.
Let Ai be the ith special block of Y. If the 1" element of Ai is the first occurrence
of the letter a, then insert aa right after the Ith special block of Xi. If the 1 t" element
of Ai is the last occurrence of a, then insert aa right before the lPh special block of Xi.
Replace A in Y by the modified Xi for every i. The resulting sequence is Gd(m).
Lemma 10. For all d and m, Gd(m) avoids ababa.
Proof. Given that the alternations in Gi(m) have length at most 4 for all m > 1, then
F-1
the rest of the proof is the same as the proof in [381 that Zd(m) avoids ababa.
Let Ld(m) be the length of Gd(m). Observe that Ld(M) = (d + 1)mSd(m) since
each letter in Gd(m) occurs 2d + 2 times, twice in special blocks, and each special
block has m letters.
Define Nd(m) as the number of distinct letters in Gd(m) and MAd(m) as the number
of blocks in Gd(m). Also let Xd(m) = Ad()
".
We bound Xd(m)
Id(mf)
Sd (M) and Vlj(m) = Ld
and Vd,(m) as in [381.
Lemma 11. For all m, d > 1,
Xd(mn)
< 2d + 2 and V1(m) > M.
Proof. By construction Si(m) = m + 1 for m > 1, Sd(1) = 2 for d> 2, and Sd(7n)
Sd(m -1)Sdl_(Sd(m-1)) for d, m > 2. Furthermore M1 (m) = 3m-+3, MAd(1) = 2d+4
ford> 2, and Aid(m) = A(m-)Sdl(Sd(m--1))+MAd(S(m-1))-Sdl_(Sd(m-
1)) for d, m > 2.
Thus S 2 (m) = S 2 (m - 1)(S 2 (m - 1) + 1)
2(S 2 (m - 1))2 and S 2 (1) = 2. Since
S'(m) = 22-1 satisfies the recurrence S'(1) = 2 and S'(m + 1) = 2(S'(m)) 2 , then
18
22m1 <
3 x
S 2 (m) < 2
For d > 2, Sd(2) = 2Sd_1(2) and S 1 (2) = 3. So Sd(2) =
2 d-1
For d > 2, MAd(2) = (2d + 3)(3 x
(6d + 3 ) 2d-1
2 d-2)
d+2 for all d > 2, and Xd(2) = 2d+ 1 for all
Then X,1(m) = 3 for all m, Xd(1)
d > 2. For d, m > 2, Xd(m)
=
Xd(m
+ Mdl_(2) and M 1(2) = 9. Hence MAd(2) =
-
1) + Xd-1(Sd(m-1))--1
Sd(M-1)
We prove by induction on d that Xd(m)
2d + 2 for all m, d > 1. Observe that
.
the inequality holds for Xi(m), Xd(1), and Xd(2) for all m, d.
Xdj(m - 1) + s2Fix d and suppose Xd1(m) < 2d for all m. Then Xd(m)
1
-~=2 Sd(n)- 1
-2 Sd(n)) = 2d + 1 + (2d - 1)
Hence Xd(m) < Xd(2) + (2d - 1)
2SK(2)-l =
' 2 Sd(n))<
Since Sd(m) > 2Sd(m - 1) for all d, m > 2, then
1
<
1 foral
so
Ld(m)
>
(r)
m1
2d+2. Hence Vd(m) -Xd(m)
all d> ,
x2d-2 - 2d1
_
-(d+1)MSd(rn)
-
The following analysis demonstrates the lower bounds on
(m) for each
A5,2d+2
2, let mi = M2 (i) and ni = N2 (i). Then mi = X 2 (i)S2 (i) < 6S2 (i) <
=
V(i)N 2 (i) > i 2 ()
L 2 (i) 6(221-1) < 22"+2 for i > 1. Then i = Q(log log mi), so ni
6
12
Q(mi log log mi).
d > 2. If d
=
-6
We use interpolation to extend the bound from mi to m. Let i and t satisfy mi <
m < mi+ 1 and t = L[J. Concatenate t copies of G 2 (i) with no letters in common for
a total of at least [- Jni = Q(m log log m) letters. Hence A 5 ,6 (m) = Q(m log log mn).
We prove that S3 (m) < A 3 (2m) following the method of [381. Since S 3 (n)
S3 (m - 1)S 2 (S 3 (m - 1)) < S2 (S 3 (n - 1))2 < 22s3(m-'+-2, then let F(in) = 22m+1_2
and G(m) = 22m. Then 2F(n) = 22m+_1 < 222, = G(2m) for every m > 0. Thus
+
S 3 (m) < F(m-l)(S 3 (1)) < 2F(M-n)(S 3 (1)) 5 G(m- 1)(2S 3 (1)) = A 3 (2mi).
Let mi = A13 (i) and ni = N3 (i). Therefore mi = X 3 (i)S 3 (i) K 8S 3 (i) < A 3 (2i
=_
3()M
i
>3iiMV
(i
.
3
=
,>
2) for i > 1. So i = Q(a3 (Mi)) and ni = N 3 (i) =
Q(mina3 (mi)). Then A 5 ,s(M) Q(ma 3 (m)) by interpolation.
Ad(m + 2) for m > 1 by induction on
For each d > 4 we prove that Sd(m)
d. Since S 4 (in) = S4 (m - 1)S3 (S 4 (m - 1)) < S 3 (S4 (n - 1))2, then let F(m) =
S 3 (im) 2 . Since 4F(m) < A 3 (4m) and A 4 (3) > 4S4 (1), then S 4 (m) < F(rM- (S4 (1)) <
4F(m- 1 ) (S4 (4)) < 4(fl-l)(4S4(1)) < A 4 (m + 2).
Fix d > 4 and suppose that Sd-_(m) < A-1 (m + 2). Define F(m) = Sd-(m) 2
Since 4F(m) < Ad_1(4m) and Ad(3) > 4Sd(1), then Sd(m) < F(m-1 ) (Sd(1)) <
+
4F(m-1 ) (Sd(1)) < A_" 1"(4Sd(1)) < Ad(m + 2).
Fix d > 4. Let mi = Md(i) and ni = Nd(i). Then mi = Xd(i)Sd(i) < (2d
< (2d + 2)Ad(i + 2) < Ad(i + 3). Then i > ad(Mi) - 3, so ni =di
iMA(i) = Q(lMiad(Mi)). By interpolation A,2d+2(m) = Q(1 Mad(M)) for
-
2)Sd(i)
Vd(i)Md(i) >
d > 4.
Corollary 12. For r > 2, F,4 , 6(m) = r.,3,(m)
7Ir,3,2d+1(n) = Q(jm!ed(M)) for d > 3.
19
=
Q(miloglognm) and Fr,4,2d+2(m)=
2.3
Bounds on extremal functions of 0 - 1 matrices
using Davenport-Schinzel sequences
Let IF, (m, n) be the maximum length of a sequence with ?n blocks and n letters which
avoids alternations of length s + 2. Nivasch showed that TV(m, n) < k(HFI'm) + n)
for all k. We modify the matrix-sequence transformations in [401 to show bounds
on exk(m, P) for alternating patterns P. Let P, be the 0 - 1 matrix with s rows
0, ... , s - 1 and 2 columns 0, 1 such that the number in each entry is the sum of its
row and column mod 2.
Lemma 13. T, (m, n) = ex(n,n, P+1) for
for s, k, m > 1.
'm, n,
s > 1 and H(m) = exk(MPs+1)
Proof. The sequence to matrix transformation in [40] starts with a sequence Q with
n letters on m blocks which avoids alternations of length s + 2, and results with a
0 - 1 matrix A with n columns and m rows which avoids the matrix P,+. The letters
of Q are named 0,...,n -I by first occurrence and the entry in column i and row j
of A is a one if and only if the letter i occurs in block j of Q.
Suppose for contradiction that A contains P,+1. Then there is a submatrix of A
with 2 columns co < ci and s + 1 rows ro, ... , r8 such that the entry in column ci
and row rj is one if i + j is odd. The one entries in this submatrix correspond to
an alternation c 1co... of length s + 1 in Q. However, the first occurrence of co is
before the first occurrence of c1 in Q, so Q contains an alternation of length s + 2,
a contradiction. This implies both T, (m, n) < ex(m, n, P,+1 ) for m, n, s > 1 and
Hs(M)
exk(M, P+l) for s, k, m > 1.
)
Pettie used a matrix to sequence transformation to show that ex(m, n, P,+ 1
I(m, n) + n. 0 - 1 matrix A with m rows 0,..., m - I and n columns 0,. .,rn - 1
is converted to a sequence Q with m blocks 0,..., m - 1 and n letters 0,.. .,i - 1.
Letter i occurs in block j of Q if and only if the entry of A in column i and row j is
a one.
Let C be the letters in block j of Q which occur in no block before j and let Dj
be the letters in block j of Q which occur in a block before j. All letters in Di occur
before all letters in Cj in block j, and letters in Dj occur in block j in reverse order of
their last appearance before block j. In [401 the letters in C were ordered arbitrarily,
but here letter x in Ci occurs before letter y in Cj if and only if x < y.
Suppose that Q contains an alternation of length s + 2 on letters x and y such that
x < y. List all alternations of length s+2 on the letters x and y in Q lexicographically,
so that alternation f appears before alternation g if there exists some i > 1 such that
the first i - 1 elements of f and g are the same, but the i''h element of f appears
before the ith element of g in Q.
Let fo be the first alternation on the list and let 7i be the number of the block
which contains the ith element of fo for 1 < i < s + 2. Suppose for the sake of
contradiction that for some 2 < i K s + 1, i = wi+ 1 . Let a be the ith element of fo
and let b be the (i + 1)'t element of fo.
Then a occurs before b in block ri and the b in 7ir is not the first occurrence of b in
Q, so the a in 7i is not the first occurrence of a in Q. Otherwise a would appear after
20
.
b in 1ri. Since the a and b in 7i are not the first occurrences of a and b in Q, then the
last occurrence of a before 7i must be after the last occurrence of b before 7ri. Let fi
be the subsequence obtained by deleting the letter a in 7i from fo and inserting the
last occurrence of a before 7ri. Then fi is an alternation of length s +2 and fi occurs
before fo on the list.
This contradicts the definition of fo, so for every 2 < i < s + 1, 7i < 7rj+ 1 . We
now consider two cases. If the first element of fo is x, then the submatrix of A with
2 columns x, y and s + 1 rows 7r2, -- -,7r,+2 contains P+1 since x < y. If the first
element of fo is y, then suppose for contradiction that 7r, = r2 . The occurrences of x
and y in block 7r, are both first occurrences of x and y in Q. Otherwise, fo would not
be the first alternation on the list. Since x and y are in Cr, then x appears before y
in block 7r1 because x < y, a contradiction. Then the submatrix of A with 2 columns
,r+
7r,... contains Ps+1
x, y and s + 1 rows
Therefore ''(m, n) > ex(m, n, P+1) for m, n, s > 1 and I
>-(m)
exk(M, P+l)
for s, k, m > 1.
for all s > 2, ex6(m, P4 ) = (m loglog m), and
(Mrad(m)) for all fixed d > 3.
Corollary 14. ex,(m,Ps) =("j.)
ex2d+2(m, P4 ) =
6
21
22
Chapter 3
The formation width of sequences
If u is a sequence with r distinct letters, then let Ex(u, n) be the maximum length of
any r-sparse sequence with n. distinct letters that avoids u. If a and b are single letters,
then Ex(a, n) = 0, Ex(ab, n) = 1, Ex(aba, n) = n and Ex(abab, n) = 2n - 1. Nivasch
[381 and Klazar [32] determined that Ex(ababa, n) ~ 2na(n). Agarwal, Sharir, and
Shor [21 proved the lower bound and Nivasch [381 proved the upper bound to show
') for
2O(a(l)t
that if u is an alternation of length 2t + 4, then Ex(u, n) = n9nCdn)
t > 1.
If u is a sequence with r distinct letters and c > r, then let Exc(u, n) be the
maximum length of any c-sparse sequence with n distinct letters which avoids u.
Klazar [30] showed that Exc(u, n) = E(Exd(u, n)) for all fixed c, d > r.
Lemma 15. [30] For every sequence u with r distinct letters, Exd(u, n)
Ex,(u, n)
(1 + Exc(u, d - 1))Exd(u, n) for all n > 1 and d > c > r.
An (r, s)-formation is a concatenation of s permutations of r distinct letters. For
example abcddcbaadbc is a (4, 3)-formation.
Definition 16. F,s (n) is the maximum length of any r-sparse sequence with n distinct letters that avoids every (r, s)-formation.
Klazar [30] proved that Fr2 (n) = O(n) and F,3 (n) = O(n) for every r. Nivasch
[38] proved that F,4 (n) = O(na(n)) for r > 2. Agarwal, Sharir, and Shor [2] proved
the lower bound and Nivasch [38] proved the upper bound to show that F,s()=
n2Aa()+!(a(n)-) for all r > 2 and odd s > 5 with t =
Nivasch [381 proved that Ex(u, n) < Fr,s-.r+i(n) for any sequence u with r distinct
letters and length s by showing that every (r, s - r + 1) formation contains u.
Definition 17. The formation width of u, denoted by fw(u), is the minimum value
of s such that there exists an r for which every (r, s)-formation contains u. The
formation length of u, denoted by fl(u), is the minimum value of r such that every
(r, fw(u))-formation contains u.
By Nivasch's proof, fw(u) < s - r + 1 for every sequence u with r distinct letters
and length s. The next two facts follow from the definition of fw.
23
Lemma 18. If u contains v, then fw(v)
fw(u).
Lemma 19. If u begins with the letter a, then fw(au) = fw(u) + 1.
Lemma 15 implies that fw(u) and fl(u) can be used to obtain upper bounds on
Ex(u, n).
-
Lemma 20. For any sequence u with r distinct letters and fixed integer c with c >
r, Exc(u, it) = 0 (Ffl()
(n))
In this paper we use fw(n) primarily in order to prove tight upper bounds on
Ex (u, n) for several classes of sequences u such that a contains an alternation with
the same formation width as u. We also bound and evaluate fw for various other
families of sequences in order to develop a classification of all sequences in terms of
their formation widths.
If at is an alternation of length t for t > 2, then fw(at) < t - 1 since every
(r, t - 1) formation contains at for r > 2. Any (r, t - 2)-formation in which order
of letters reverses in adjacent permutations avoids at, so fw(at) = t - 1. Pettie
[411 used the fact that every (4,4)-formation contains abcacbc to prove the upper
bound Ex(abcacbc, n) = O(na(n)). Since any (r, 3) formation with order reversing in adjacent permutations would avoid abcacbc, then fw(abcacbc) = 4. Similarly
fw(abcadcbd) = 4.
Definition 21. An (r, s)-formation f is called binary if there exists a permutation p
on r letters such that every permutation in f is either the same as p or the reverse of
p.
Most of the proofs in this paper depend on the fact that if u is a sequence with r
distinct letters, then every binary (r, s)-formation contains u if and only if s > fw(u).
We use the following notation to describe binary formations more concisely.
Definition 22. I is the increasing sequence 1 ... c on c letters and D, is the decreasing sequence c . . 1 on c letters. Given a permutation 7r C Sc, the sequences I, and
D, are 7r(1) ... 7r(c) and 7(c) ... w(1) respectively.
We focus especially on two classes of binary formations in order to derive bounds
on fw(u). The sequence up(l, t) is I, repeated t times, and alt(l, t) is a concatenation
of t permutations, starting with I and alternating between I1 and D1 . For example,
up(3, 3) = 123123123 and alt(3,3) = 123321123.
Definition 23. If u is a sequence with c distinct letters, then l(u) is the smallest k
such that up(c, k) contains a, and r(a) is the smallest k such that alt(c, k) contains
a.
Then fw(u) > l(a) and fw(u) > r(a). We evaluate both 1(u) and r(u) for every
binary formation U.
In Section 3.1 we prove that -y(r, s) = (r-1)2 -1+1 is the minimum value for which
every (-y(r, s), s)-formation contains a binary (r, s)-formation. It follows that if u has
24
r distinct letters, then fw(u) is the minimum s for which every binary (r, s)-formation
contains u.
In Section 3.3 we prove that fw(u) = t - 1 for every sequence u with two distinct
letters and length t. We also determine every sequence u for which fw(u) < 3.
In addition, we show that fw(up(c,t)) = 2t - 1 for all c > 2 and t > 1. This
t3
implies that Ex(up(l, t), n) = n2 -- (n) t 2s(Q(n)
) for all 1 > 2 and t > 3 and that
fw(u) < 21(u) - 1 for every sequence u.
In Section 3.4.1 we compute l(u) and use the result to bound fw(u) up to a factor
of 2 for every binary formation u. In particular we prove the following bounds on
fw(u).
Theorem 24. Fix c > 2 and let u = I," D 2 I
12
...
Cn,
where L is I if n is odd and
D if n is even, and ej > 0 for all i. Define A =
>1 e 2i- 1 and B = E
M = max(A,B) and let m = min(A,B). Then (c- 1)m + M + [ n
e 2i. Let
fw(u)
2(c - 1)m + 2M + 2[nJ - 1.
In Section 3.4.2 we compute r(u) for every binary formation u. Specifically we
prove that if c > 2, then r(Ig1Dc2Ig3 .... Cen) = 2Z" e -nwhere L is I if n is odd
and D if n is even.
In Section 3.5 we use fw(u) to derive tight bounds on Ex(u, n) for other sequences
u besides up(l, t). Let u be any sequence of the form avav'a such that a is a letter,
v is a nonempty sequence excluding a with no repeated letters and v' is obtained
from v by only moving the first letter of v to another place in v. We show that
fw(u) = 4, implying that Ex(u, n) = E(na(n)). We also prove that Ex(abc(acb)t , n) =
for all t > 2.
n 2 (Tt
(n)(a(f)2
In Section 3.6 we compute fw for various classes of binary formations. In particular
we show for c > 2 and k > 1 that fw(IcDcIc) = c + 3, fw(IcDc) = c + 2k - 1,
fw(IcDcIcDc) = 2c+3, fw(alt(c, 2k)) > k(c+2)-1, and fw(alt(c, 2k+1)) > k(c+2)+1.
In Section 3.7 we discuss some unresolved questions.
3.1
An extension of the Erd6s-Szekeres theorem
The following upper bound is obtained by iterating the Erd6s-Szekeres theorem as in
[30].
Lemma 25. Every ((r -
1)2'-
+ 1, s)-formation contains a binary (r, s)-formation.
Proof. We prove by induction on s that every ((r - 1)2'-' + 1, s)-formation contains
a binary (r, s)-formation. Clearly this is true for s = 1. For the inductive hypothesis
fix s and suppose for every r > 1 that each ((r - 1)281 + 1, s)-formation contains a
binary (r, s)-formation.
Consider any ((r - 1)2" +1, s+1)-formation F. Without loss of generality suppose
that the first permutation of F is I(r_1)2-. By inductive hypothesis the first s
permutations of F contain a binary ((r-1)2 +1, s)-formation f. By the Erd6s-Szekeres
theorem, every sequence of (x - 1)2 + 1 distinct integers contains an increasing or
25
decreasing subsequence of length x. Therefore the last permutation of F contains an
increasing or decreasing subsequence of length r on the letters of f. Thus F contains
l
a binary (r, s + 1)-formation.
Corollary 26. If u has r distinct letters, then every binary (r, s)-formation contains
u if and only if s
fw (u).
Proof. If for some s every binary (r, s)-formation contains t, then there exists a
function y(r, s) such that every (?(r, s), s)-formation contains i. Thus fw(u) < s.
If some binary (r, s - 1)-formation f avoids i, then for every z > r the binary
(z, s - 1)-formations which contain f will avoid u. Hence fw(u) > s - 1.
l
Corollary 27. If i is a nonempty sequence and v is obtained from u by inserting a
single occurrence of a letter which has no occurrence in i, then fw(u) =fw(v).
Proof. If 't has r distinct letters, then every binary (2r + 1,fw(u))-formation F with
first permutation I2-+1 has a copy of u using only the even numbers 2,...,2r. Since
there is at least one odd number between every pair of even numbers in F, then the
copy of t in F can be extended to a copy of v using an odd number.
El
1.
Proof. Since every binary (r, fw(u))-formation contains u, then every ((r -1)2mu)-1
1,fw(u))-formation contains u.
+
Corollary 28. If a has r distinct letters, then fl(u) < (r - 1)2ft (u+
l
The next theorem shows that the upper bound in Lemma 25 is tight.
Theorem 29. For every r, s > 1 there exists a ((r
every binary (r, s)-formation.
-
1)2'1,
s)-formation that avoids
Proof. We construct the desired formation Fa(r, s) one permutation at a time. Define
an a-block in Fa(r, s) to be a block of numbers in a permutation from positions
(k-1)(r-1)a+1 to k(r-1) for some k. For k < s-I define a k-swap on a permutation
of length (r - 1)2-1 as follows: For every even i, 1 < i < 2k, a k-swap reverses the
placement of the (i - 1)2'-k-'-blocks in each i2s-k-'-block. For example if (r, s)
(3, 3), then a 1-swap on 1234567890ABCDEF produces CDEF90AB56781234.
-
Let permutation 1 of Fa(r, s) be the identity permutation on the letters 1, . . . , (r
1)2-'.
To form permutation k + 1 of Fa(r, s), perform a k-swap on permutation k.
The next lemma about Fa(r, s) will imply that Fa(r, s) avoids every binary (r, s)formation.
Lemma 30. Consider any set B of distinct numbers occurring in each of the first
k permutations of Fa(r, s) with the same or reverse order in adjacent permutations.
Let i(k) =
ej2k-j~1 where ej = 1 if the elements in B reverse order from
permutation j to permutation j + 1 and ej = 0 otherwise. Then in permutation k the
elements of B are contained in different i(k)2sk-blocks, but the same (i(k) + 1) 2 s-kblock.
26
Proof. We induct on k. When k = 1, i(k) = 0. The entire permutation is a 2~1-block
and 0-blocks are individual elements, so the lemma is true when k = 1.
For the inductive hypothesis, suppose that in permutation k the elements of B
are contained in a single (i(k) + 1)2,-k-block but different i(k)2s-k-blocks. Consider
any set B of distinct numbers occurring in each of the first k + 1 permutations of
Fa(r, s) with the same or reverse order in adjacent permutations.
Now consider the k-swap that sends permutation k of Fa(r, s) to permutation
k + 1. The parts of the swap that reverse the placement of the (j - 1)2s~k-'-blocks
in each j2s-k---block for even j > 2i(k) + 4 do not affect the order of the elements
of B since the elements of B are contained in a single (2i(k) + 2)2s-k-l-block.
The parts of the swap that reverse the placement of the (j - 1)2s-k-'-blocks in
each j2'-k-'-block for even j < 2i(k) also do not affect the order of the elements of B
since the elements of B are contained in different (2i(k))2s-k-'-blocks. Thus the only
part of the swap which is relevant to the order of the elements in B is the reversal of
the placement of the (2i(k) + 1)2s-k-l-blocks inside each (2i(k) + 2)2s-k-l-block.
If the order of elements in B reverses from permutation k to permutation k + 1,
then i(k + 1) = 2i(k) + 1. All the elements of B must be contained in different
(2i(k)+1)2s-k-l-blocks, or else the k-swap would not reverse their order. By inductive
hypothesis the elements of B are contained in the same (i(k + 1) + 1)2s-k-l-block.
If the order of elements in B is the same in permutation k and permutation
k + 1, then i(k + 1) = 2i(k). The elements of B must be contained in the same
(2i(k)+1)2s-k-l-block, or else the k-swap would not preserve their order. By inductive
l
hypothesis the elements of B are contained in different i(k + 1)2s-k-l-blocks.
Given any set B of distinct numbers contained in every permutation of Fa(r,s)
whose order either stays the same or reverses between adjacent permutations, there is
some i such that the elements of B are in different i-blocks, but the same (i + 1)-block
of permutation s. Since there are r - 1 i-blocks in each (i + 1)-block, then r - 1 is
E]
the maximum possible number of elements in B.
3.2
Algorithm for computing fw
The following algorithm for computing fw(u) is an implementation in Python of the
method for computing formation width in Corollary 26. Specifically if u is a nonempty
sequence with r distinct letters, then the algorithm increments s starting from 1 until
it finds that every binary (r, s)-formation contains it. If some binary (r, s)-formation
f contains u, then for every s' > s the algorithm does not check for containment of
u in any binary (r, s')-formations f' for which f' restricted to its first s permutations
is equal to f.
from itertools import permutations
#determines whether one sequence is a subsequence of another:
def issubseq(seq, subseq):
27
if len(subseq) == 0:
return True
else:
if len(seq) == 0:
return False
elif seq[-1] == subseq[-1]:
return issubseq(seq[:-11,subseq[:-1])
elif seq[-1] != subseq[-1]:
return issubseq(seq[:-1],subseq)
#determines the formation width of u:
def fw(u):
1=len(set(u))
v = list(u)
rsformset = set()
rsformsetl = set()
q = tuple(range(l))
qi = q[::-1]
rsformset.add(q)
rsforml=q
while len(rsformset)!=0:
for rsforms in rsformset:
done=False
for perms in permutations(range(l)):
for i in range(len(u)):
v[i] = perms[u[i]]
if issubseq(rsforms, v):
done=True
break
if not done:
rsformsetl.add(rsforms+q)
rsformsetl.add(rsforms+ql)
rsforml=rsforms+q
rsformset.clear()
for rsform in rsformsetl:
rsformset.add(rsform)
rsformsetl.clear()
return len(rsformi)//1
#u must be nonempty tuple with letters 0,1,2,..., e.g.:
print fw((0,1,2,3,4,5,0,2,3,1,4,5,0))
28
3.3
Using binary formations to compute fw
If u has one distinct letter, then fw(u) is the length of it. If u has two distinct letters,
then fw(u) also depends only on the length of u.
Lemma 31. If u has two distinct letters and length t, then fw(u) = t - 1.
Proof. By Lemma 19 it suffices to prove this lemma for sequences with different first
and second letters. The upper bound follows since every (2, t - 1)-formation contains
u. For the lower bound it suffices to construct a (2, t - 1) formation f(u) which only
contains copies of u for which the last letter of the copy of u is the last letter of f(u).
Therefore the (2, t - 2)-formation in the first t - 2 permutations of f(u) avoids u, so
fw(u) > t - 2 by Corollary 26.
Assume without loss of generality that u starts with xy. Construct f(u) by ignoring the leading x and replacing every x in u by ba and every y by ab. Let it' denote
the sequence obtained by deleting the last letter of it. We prove by induction on the
length of i that f(u) contains only copies of u for which the last letter of the copy of
u is the last letter of f(u). The first case to consider is i = xy.
Since f(xy) = ab, then f(xy) contains exactly one copy of the sequence xy and
the last letter of the copy of xy is the last letter of f(xy). Suppose by inductive
hypothesis that f(u') contains only copies of u' for which the last letter of the copy
of u' is the last letter of f(u'). If the last two letters of u are the same, then the first
letter of the last permutation of f(u) is different from the last letter of f(u'), so the
last letter of f(u) will be the last letter of any copy of u in f((u). If the last two letters
of u are different, then the first letter of the last permutation of f(u) is the same as
the last letter of f(,'), so the last letter of f(u) will be the last letter of any copy of
El
it in f(it).
If u has at least three distinct letters, then fw(u) cannot be determined solely from
the length of u and the number of distinct letters in u. For example fw(abcabc) = 3
and fw(abccba) = 4.
The next lemma identifies all sequences i for which fw(u) = 3. As a result of
Corollary 27, deleting any letters which occur just once in u will not change the value
of fw(u) unless a has no other letters. We call a sequence reduced if every distinct
letter in the sequence occurs at least twice.
By Lemma 31, fw(u) = 1 if and only if it is nonempty and each distinct letter in it
occurs once, and fw(u) = 2 if and only if one letter in u occurs twice and every other
distinct letter occurs once.
Lemma 32. If u is reduced and fw(u) = 3, then either there exists some 1 > 2 for
which u is isomorphic to up(l, 2) or u is isomorphic to one of the sequences aaa, aabb,
abba, abcacb, abcbac, abccab, or abcdbadc.
Proof. Since u is reduced, then every distinct letter in u occurs at least twice. If any
letter in u occurs three times, then it is the only letter in it and u is isomorphic to
aaa, or else fw(a) > 4 by Lemma 31. If u is not isomorphic to aaa, then every distinct
letter in it occurs twice.
29
Suppose u is not isomorphic to up(l, 2) for any I > 2. Then there exist two distinct
letters x and y in u for which the subsequence consisting of occurrences of x and y
is isomorphic to aabb or abba. If x and y are the only distinct letters in u, then u is
isomorphic to aabb or abba.
If u has three distinct letters, then t is isomorphic to a sequence obtained by
adding two occurrences of c anywhere in aabb or abba, so we consider 30 cases. If
u had the form xxv or vxx for some letter x and sequence v of length 4 with two
distinct letters not equal to x, then fw(v) = 3 by Lemma 31, so fw(u) = 4 by Lemma
19. This eliminates the cases aabbcc, aabcbc, aacbbc, aabccb, aacbcb, aaccbb, acacbb,
caacbb, accabb, cacabb, ccaabb, abbacc, and ccabba.
The binary (3,3)-formation xyzxyzxyz avoids caabbc, abbcca, accbba, cabbac,
acbbca, and abccba. The binary (3,3)-formation xyzzyxxyz avoids acabcb, abcbca,
and acbeba. The binary (3, 3)-formation xyzzyxzyx avoids acabbc and cacbba. So its
reverse avoids caabcb and abbcac. Thus each of these sequences have formation width
at least 4 by Corollary 26.
If u is one of the remaining sequences abcbac, acbbac, cabbca, or cabcba, then
fw(u) = 3. Thus every reduced sequence ? with three distinct letters for which
fw(u) = 3 is a (3, 2)-formation. Note that acbbac and cabbca are isomorphic to
abccab, and cabcba is isomorphic to abcacb.
If a has four distinct letters, then u is isomorphic to a sequence obtained by adding
two occurrences of d to the sequence abcabc, abcacb, abcbac, or abccab. If u was not a
(4, 2)-formation, then u would contain a reduced sequence v with three distinct letters
which was not a (3, 2)-formation, so fw(7) > 4.
We consider all (4, 2)-formations with first permutation abcd. The binary (4,3)formation xyzwxyzwxyzw avoids abcdadcb, abcdbdea, abcdcbad, abcdcbda, abcddacb,
abcddbac, abcddbca, abcddcab, and abcddcba. The formation xyzwxyzwwzyx avoids
abcdbacd, abcdcabd, and abcdcadb. The formation xyzw'wzyxxyzw avoids abcdacbd,
abcdacdb, abcdbcad, abcdbcda, abcdcdab, and abcdcdba. The formation xyzwwzyxwzyx avoids abcdabdc, abcdadbc, abcdbdac, and abcddabc. Thus each of these (4, 2)formations have formation width at least 4 by Corollary 26.
If u is abcdbadc, then fw(u) = 3. If a had five distinct letters, then a must be a
(5, 2)-formation or else fw(u) > 4. If a was any (5, 2)-formation with first permutation
abcde, then every (4, 2)-formation in u would be isomorphic to abcdbadc or up(4, 2). It
is impossible for a (5, 2)-formation to have both a subsequence isomorphic to abcdbadc
and another subsequence isomorphic to up(4, 2), so every (4,2)-formation in u would
be isomorphic to abcdbadc or else a would be isomorphic to up(5, 2). In particular u
must have both abcdbadc and acdecaed as subsequences, a contradiction.
If a had r distinct letters for some r > 5 and a was not isomorphic to up(r, 2),
then u would contain a subsequence of length 10 with five distinct letters that was
not isomorphic to up(5, 2), so fw(u) > 3.
l
The last lemma can also be verified by using the Python formation width algorithm. The next lemma provides an upper bound on fa (u) for every binary formation
u. It is tight if u = up(1, t) for any 1 > 2 and t > 1.
30
Lemma 33. Let u = Ie1D
23...
that ej > 0 for each i and E',
L-, where L is I if n is odd and D if n is even so
ej = k. Then fw(u) < c(k - em) + 2e, - 1 for all m.
Proof. Let k1 =
ej and k2 = E ml e.. In any binary (c, c(k - em) + 2em - I)formation f, there is a copy of up(c, em) in permutations cki + 1 through ck 1 +2em -I
of f by the pigeonhole principle. This copy of up(c, em) can be extended to make a
copy of u in f by using one letter from each of the remaining cki + ck2 permutations
of f. Thus fw(u)
c(k - em) + 2em - 1 by Corollary 26.
El
Theorem 34.
fw(up(l, t)) = 2t - 1 for every 1 > 2 and t > 1.
Proof. For the lower bound fw(up(l, t)) > fw((ab)t) = 2t - 1 since up(1, t) contains
(ab)t. The upper bound fw(up(l, t)) < 2t - 1 follows from Lemma 33.
l
Therefore fw(u) = 2t - 1 for every sequence a such that a contains (ab)t and there
exists I > 2 for which up(l, t) contains u. As a corollary this implies the upper bounds
in the next result, which gives nearly tight asymptotic bounds on Ex(up(l, t), n). The
lower bounds in the next corollary follow from the lower bounds on Ex((ab)t , n) in [2]
by Lemma 15.
Corollary 35. Ex(up(1, t), n) = n2(
a(n)
+ (a(n)-) for all 1 > 2 and t > 3.
As a result, the constant c improves in the (n log n)2a(n)c upper bound from [451
on the maximum number of edges in k-quasiplanar graphs on n vertices with no pair
of edges intersecting in more than 0(1) points, since their proof used the bounds
Ex(up(l, t), n)
nl21t-3(10l)oa(n)" from [30].
3.4
Bounding the formation width of binary formations
In this section we compute the exact values of l(u) and r(u) for all binary formations
u. This yields upper and lower bounds on fw(u) which differ by at most a factor of
two for each binary formation a.
3.4.1
Computing 1
If 7r E S, and u is a sequence on the letters 1, ... ,c, then let 4,(u) = k if u is
a subsequence of I, but u is not a subsequence of Ig~1. It follows that 1(u) =
min,asc{4,(u)}.
Lemma 36. If 1,(Ic) = a and 1,(Dc)
=
b, then a + b
c +1.
Proof. Represent the permutation K by the set of points (i, r(i)). Connect points
(i, 7r(i)) and (j,7r(j)) if i <j and r(j) = ir(i) + 1. This partitions the points into a
connected sections. In a different representation connect points (i, 7r(i)) and (j, 7r(j))
if i <j and 7r(j) = 7r(i) - 1. This partitions the points into b connected sections.
31
We count the total number of endpoints of connected sections of points in both
representations in two ways so that each connected section of points is considered
to have two endpoints, even when the section consists of a single point. Since every
connected section has two endpoints, then there are a total of 2(a + b) endpoints.
Alternatively every point (i, ir(i)) contributes two endpoints, unless r(i) = 1 or ir(i) =
c, in which case (i, 7r(i)) contributes three endpoints. Thus there are a total of 2c + 2
endpoints, so a + b = c+ 1.
El
Corollary 37. l(ID,) = c + 1 for every c > 1.
Corollary 38. fw(IcDc) = c + 1 for every c > 1.
Proof. The upper bound is trivial. The lower bound follows since I, avoids ID,.
E
If u and v are sequences on the letters 1, . .. , c, then 4,(u) + 4,(v) - 1 < 4r(Uv)
41 (u) + 4,(v). Say that u and v 7r-overlap if 4,(uv) = 1,(u) + 1,(v) - 1. Then u and v
7r-overlap if and only if the last letter of u and the first letter of v 7r-overlap.
For each 7r E Sc, the sequences I, and D, do not 7r-overlap since the last letter of
Ic is the first letter of Dc, and D, and I, do not r-overlap since the last letter of D,
is the first letter of Ic. Furthermore if c > 2, then exactly one of the two sequences I,
or D, T-overlaps itself, depending on the order in which the first and last letters of
I, occur in I,. Moreover for any sequence u, if 1,(u) = 1 then u does not 7r-overlap
itself.
The next theorem implies Theorem 24 since l(u)
fw(u) < 21(u) - 1.
Theorem 39. Fix c > 2 and let u = I,"D 2 13 ... L, where L is I if n is odd and
D if n is even, and ej > 0 for all i. Define A =Li> e 2 i 1 and B =
>1 e 2i. Let
Al = max(A, B) and let rn
min(A, B). Then 1(u) = (c - 1)m + M + Lj.
Proof. Fix an arbitrary
7r
c
S, and let l(Ic) = a and 4,(D,) = b. We show 1,(u) >
L1J by considering two cases depending on whether I, or D, 7r-overlaps
(c- 1)m++A'
itself.
Case 1: Ic ir-overlaps itself.
In this case /,(Iel) = (a - 1)ej + 1 and i,(D'i)
bei. Since I, and D, do not
7r-overlap and D, and Ic do not 7r-overlap, then 41(u)
(a - 1)A + bB+ [+~. Lemma
36 implies that (a - 1) + b = c, while b > 0 and a > 1 since Ic r-overlaps itself. Then
-
,(u) > (c- 1)n + M + [.
Case 2: D, r-overlaps itself.
In this case 41(IL,) = aej and 1,(De) = (b -1)e+1, so 1,(u) = aA+(b- 1)B
Moreover a + (b - 1) = c, a > 0, and b > 1. Then 1,(u) > (c - 1)m + M + [l.
[J.
Thus in either case 1,(u) > (c - 1)m + M + LL J. If A > B, then this value
is attained by letting 7r be the identity permutation. If B > A, then this value is
attained by letting 7r(1) = 1 and ir(i) = c+ 2 - i for 2 < i < c.
l
32
3.4.2
Computing r
For every binary formation u we compute r(u), and we identify when r(u) > l(u).
Theorem 40. If c > 2 and ei > 0 for all i, then r(If1D1I23
where L is I if n is odd and D if n is even.
.. . L-)
=
2
t
ei - n,
Proof. First we show that r(I) = 2x - 1 for every x > 0. The upper bound is trivial.
For the lower bound we also show that alt(c, 2x - 1) has the subsequence Ix only if
r(c) = c.
We proceed by induction on x. Clearly r(Ic) = 1. In addition, , is a subsequence
of Ic only if 7 is the identity permutation, so ir(c) = c. For the inductive hypothesis
assume that r(Ix) = 2x - 1 and that alt(c, 2x - 1) has the subsequence Ix only if
7r(c) = c. We claim that r(I+1 ) = 2x + 1 and that alt(c, 2x + 1) has the subsequence
I ?-+
only if ir(c) = c.
4
Let 7r be an arbitrary permutation. We will first show that Ix+1 is not a subsequence of alt(c, 2x). Suppose for contradiction that Ix+1 is a subsequence of alt(c, 2x).
Then Ix is a subsequence of alt(c, 2x - 1), so 7r(c) = c. Then the last letter in Ix+1
must be the first letter of the last permutation of alt(c, 2x), a contradiction. Thus
r(I+ 1 ) = 2x + 1. We still must show that alt(c, 2x + 1) has the subsequence I+l
only if 'r(c) = c.
Suppose 7r(c) = i for some 1 < i < c, and assume for contradiction that Ix+l is a
subsequence of alt(c, 2x +1). Since Ix is not a subsequence of alt(c, 2x -1), then the
second to last i in Ix+1 must occur in the second to last permutation of alt(c, 2x + 1)
and the last i in I+1 must occur in the last permutation of alt(c, 2x + 1). Since
i < c, then there are at most c - 2 distinct letters between the occurrences of i in the
last two permutations of alt(c, 2x + 1), a contradiction. Thus alt(c, 2x + 1) has the
subsequence Ix+l only if wr(c) = c. This completes the induction.
By symmetry we find that r(Dx) = 2x- 1 for every x > 0. We now prove the claim
that r(Ig1D 2I 3 ... L-) = 2Z
ei - n. The upper bound is trivial since the copy
of Ie, D12I ... L'- can be selected greedily from left to right in alt(c, 2 EZ ei - n).
For the lower bound, suppose for some k and permutation 7r that alt(c, k) has the
subsequence I,, D143 ... e with n sections of the form I4 or Dx. No section I4 or
Dx can occur in fewer than 2x - 1 adjacent permutations of alt(c, k). Furthermore
no different sections have letters occurring in the same permutation. Thus alt(c, k)
contains at least 2ZL 1 ei - n permutations, so k > 2 E= ei - n.
Corollary 41. Fix c > 2 and let u = I D
... fe,
where L is I if n is odd
and D if n is even, and ei > 0 for all i. Define A = E> e 2i- 1 and B = Zi> e 2 i.
Let M = max(A, B) and let m = min(A, B). Then r(u) > l(u) if and only if
M > (c - 3)m + n+ ["].Proof. This follows from setting 2 E=
ei - n > (c - 1)m + M + [nJ since En
m + M.
ei =
l
33
3.5
Further bounds on extremal functions using fw
The lemmas in this section use Corollary 26 to identify sequences u with fw(u) > 3
for which fw(u) provides tight upper bounds on Ex(u, n), starting with an infinite set
of sequences which contain ababa.
Lemma 42. If u is any sequence of the form avav'a such that a is a letter, v is a
nonempty sequence excluding a with no repeated letters and v' is obtained from v by
only moving the first letter of v to another place in v, then fw(u) = 4.
,
Proof. Since t contains an alternation of length 5, then fw(u) > 4. Suppose a has r
distinct letters for r > 2. In order to prove that fw(u) < 4, it suffices by Corollary 26
to show that u is contained in every binary (r, 4)-formation. First note that binary
(r, 4)-formations isomorphic to I, or IrD, contain a copy of i which uses every letter
in the first permutation.
Furthermore if the position in V' of the occurrence of the first letter of v is right
after the occurrence in v' of the ith letter of v, then I DrI, has a subsequence a'
isomorphic to u such that the /th letter of a' is given by (r - i + j - 2 mod r) + 1
for each 1 < j - r. In particular the subsequence u' includes the last i + 1 letters
in the first permutation of I,?DI,, all of the letters except r - i + 1 in the second
permutation, the single letter r - i + 1 in the third permutation, and the first r - i
letters in the last permutation. Thus every binary (r, 4)-formation isomorphic to
I DrI contains a copy of u.
Since every other binary (r, 4)-formation has a subsequence isomorphic to IrD
then it suffices to observe that I,D' contains a copy of a that uses every letter in the
third permutation.
D
Corollary 43. If u is any sequence of the form avav'a such that a is a letter, v is a
nonempty sequence excluding a with no repeated letters and v' is obtained from v by
only moving the first letter of v to another place imuv, then Ex(U, T) = (na(n)).
Proof. The upper bound follows from the last lemma and Lemma 20, while the lower
bound follows by Lemma 15 since a contains ababa.
El
The next corollary is obtained by reversing the sequences considered in the last
lemma.
Corollary 44. If u is any sequence of the form avav'a such that a is a letter, v is a
nonempty sequence excluding a with no repeated letters and v' is obtained from v by
moving a single letter in v to the end of v, then fw(u) = 4 and Ex(u, n) = e(na(n)).
The next lemma implies that if v and v' are nonempty permutations of the same
distinct letters excluding a, then fw(avav'a) = 4 if and only if v' is obtained from v
by only moving the first letter of v to another place in v or by only moving a single
letter in v to the end of v.
Lemma 45. Let u be any sequence of the form avav'a such that a is a letter, v is a
nonempty sequence excluding a with no repeated letters, and v' is a permutation of v
which cannot be obtained from v by only moving the first letter of v to another place
in v or by only moving a single letter in v to the end of v. Then fw(u) > 4.
34
Proof. If x is one of the sequences abcdadbca,abcdadcba,abcdeabdcea, or abcdeacbeda,
then fw(x) > 4 . This can be verified using the formation width algorithm. Suppose
u is a sequence of the form OvOv'0 for which fw(u) = 4, v is the sequence 12.. r, and
v' is the permutation 7r72... Ir, of 12. .. r. Since u avoids abcdadbca and abcdadcba,
then 7ri < i + 1 for each 1 Ki K r.
Consider two cases. In the first case, 7r 1 = 1. If 7r
i for each 1 K i K r
then fw(u) = 4 since fw(up(r + 1,2)) = 3. Otherwise let m be minimal for which
7rm = m + 1. Then 7r
j for each j < m. Since u avoids abcdeabdcea, then r, = m.
Moreover 7r= j + 1 for m K j < r since 7r
i + 1 for each 1 Ki <r. Thus v' can
be obtained from v by only moving a single letter in v to the end of v.
In the second case, 71 = 2. Let m be the index for which lrm = 1. Then 7r = J+1
for 1 < j < m since ri < i + 1 for each 1 < i K r. Since u avoids abcdeacbeda, then
7r1j=j for each j > m. Thus v' can be obtained from v by only moving the first letter
of v to another place in v.
For t < 4 the next lemma exhibits sequences with three distinct letters and t
occurrences of each letter which contain (ab)' and have formation width 2t - 1.
Lemma 46. If t is 2, 3, or 4 and z is any sequence of the form ax1 ax 2 ... axt such
that a is a letter and xi is a sequence equal to either bc or cb for each 1 < i < t, then
fw(z) = 2t - 1.
Proof. The lower bound follows since z contains (ab)t. By Corollary 26, the upper
bound is verified by checking that every binary (3, 2t - 1)-formation contains z. This
can be done with the formation width algorithm.
l
)
Corollary 47. If t is 3 or 4 and z is any sequence of the form ax1 ax 2 ... axt such
that a is a letter and xi is a sequence equal to either bc or cb for each 1 K i K t, then
(n Q
Ex(z, n) = n2 (t 2 )!
Proof. The upper bounds follow from the last lemma and Lemma 20. The lower
bounds follow from the lower bounds on Ex((ab)t , n) in [21 by Lemma 15.
El
There are sequences z of the form axiax 2 ax 3ax 4 ax5 such that a is a letter and xi
is a sequence equal to either bc or cb for each 1 K i < 5 for which fw(z) > 9. For
example fw(abcacbacbabcacb) = 10.
The following lemma presents another infinite class of forbidden sequences with
three distinct letters for which formation width yields tight bounds on extremal functions.
Lemma 48. fw(abc(acb)t ) = 2t + 1 for t > 0.
Proof. The proof is trivial for t = 0, so suppose that t > 0. Since abc(acb)' contains
an alternation of length 2t + 2, then fw(abc(acb)t ) > 2t + 1. In order to prove that
fw(abc(acb)t ) < 2t+ 1, it suffices by Corollary 26 to show that every binary (3, 2t+ 1)formation contains abc(acb)'.
35
.
Consider any binary (3, 2t+1)-formation f with permutations xyz and zyx. Without loss of generality suppose that the last 2t - 1 permutations of f have the subsequence (xyz)'. Then f has the subsequence xzy(xyz) t unless the first six letters of
f are zyxxyz. If the first six letters of f are zyxxyz, then f has the subsequence
zyx(zXy) t
Corollary 49. Ex(abc(acb)t ,n)
-
n2
rO(n)-1
(a(n)") for
t > 2.
Proof. The upper bounds follow from the last lemma and Lemma 20. The lower
bounds follow from the lower bounds on Ex((ab)t , n) in [2] by Lemma 15.
E]
3.6
Further bounds on fw
For c > 2 the bounds on 1(u) imply that (c + 1)k K fw(alt(c, 2k)) < 2(c + 1)k - 1
and (c + 1)k + 1 < fw(alt(c, 2k + 1)) < 2(c + 1)k + 1 for every k. In this section we
derive improved bounds on fw(alt(c, 2k)) and fw(alt(c, 2k + 1)) using Corollary 26.
First we compute fw(alt(c, 3)) for all c > 2. In [411, Pettie proved for all c > 1
that Ex(alt(c, 3), n) = 0(n).
Theorem 50. If c > 2, then fw(IcDcIc) = c + 3.
Proof. First we prove for every permutation 7r c S, that IDI,
is not a subsequence
of the binary (c, c + 2)-formation 1,D2. Assume for contradiction that ID2 has the
subsequence IDI,
for some permutation r E Sc. Since l(IcDc) = c+ 1 by Corollary
37, then the last letter of D, must be in the first Dc in IcD . However, the first
letter of I, is the same as the last letter of D, so the first letter of the last I, in
IDI,
must be in the last D, in IcD2. Then I, = D,, so the last letter of D, is c.
This would imply that Ic has the subsequence I,D,. Since the last letter of I-' is c,
then ID,
would be a subsequence of Ic, a contradiction. Thus IcD2 does not have
1,DI, as a subsequence for any permutation 7 c Sc. Thus fw(IcDcIc) > c + 2 by
Corollary 26.
It remains to show that every binary (c, c + 3)-formation f has a subsequence
I,D,I, for some permutation 7r E Sc. Without loss of generality suppose the first
permutation of f is Ic. If f is I,+3, then f has IcDcIe as a subsequence. If f has
an alternation of Ic and Dc terms of length at least 3, then also f must have IcDcIc
as a subsequence. Otherwise f has the form IcfD' with a + b = c + 3, a > 0 and
b > 0. If a K 2, then f has IcDcIe as a subsequence. If b K 2, then f has DcIcDc as a
subsequence. Otherwise f has the subsequence IDI.,
such that I, is the sequence
Ib2C... (b - 1) consisting of the integers from 1 to b - 2 followed by the integers in
reverse from c to b - 1. In other words I, is obtained by reversing the last a - 1 letters
of Ic. Thus fw (IcDcIc) < c + 3 by Corollary 26.
El
The next two lemmas are used for the lower bounds in the remaining theorems.
Lemma 51. If c > 2 and w E Se, then 1,D, is a subsequence of IcDe if and only if
7r(1) < 7r(2).
36
,
,
Proof. Let 7r E Sc and suppose I,D, is a subsequence of IDc. Then the last letter of
IrD7, namely 7r(1), occurs in the last D, of IDc since l(IcDc) = c + 1 by Corollary
37. If ir(1) is not the only letter of I,D, occurring in the last Dc, then 7r(2)ir(1) is a
subsequence of D,. This is possible only if i(1) < 7(2).
If the final D, contains no letters in ID, besides 7(1), then the last 7r(2) in I, D
occurs in some I,. If ir(1) > 7(2), then the last w(1) in ID 1, can be replaced with the
7Tr(1) in the same permutation as the last 7r(2) in ID,.
This would imply that I, D
is a subsequence of Ic, which is impossible since l(IcDc) = c + 1. Thus ir(1) < 7r(2).
For the other direction suppose that wr(1) < 7r(2). Then I,D, is a subsequence
of I+1 with exactly one letter in the last permutation of Ic+1. Thus I,D, is a
El
subsequence of ItDc.
Define the reverse permutation
7r, E
S, so that
7r,(i) =
c+ 1
-
i for 1 < i < c.
Corollary 52. If c > 2 and w E S,, then I,D, is a subsequence of DcIf if and only
if 7r(2) <wr(1).
Proof. By reflection, ID , is a subsequence of DcIf- if and only if ID, is a subsequence of DcIc. Moreover I,D, is a subsequence of D'Ic if and only if 27,(ID,) is a
subsequence of IDc. By Lemma 51, 7r,(ID,) is a subsequence of IgDc if and only if
7r,(7r(1)) < r(ir(2)). Since 7r,(7(1)) < 7r,(ir(2)) if and only if 7(2) < 7(1), then ID,
is a subsequence of DcIc if and only if r(2) < 7r(1).
0
Using these facts we determine fw(IcDc) and fw(alt(c, 4)). Pettie in [411 showed
bounds of e(na(n)) on the maximum lengths of sequences with n distinct letters
avoiding both ababab and alt(c, 4) for some c. This improved an upper bound by
Ezra, Aronov, and Sharir in [12] on the complexity of the union of n 6-fat triangles.
Theorem 53. If c > 2 and k > 1, then fw(ILkDc) = c + 2k - 1.
Proof. The upper bound follows since fw(IDc) < fw(I,) + c.
For the lower bound let Tk be the (c, c + 2k - 2)-formation obtained by concatenating alt(c, 2k - 2) and I,. We show that Tk avoids 4kDc by induction on k. This
is clearly true for k = 1 since i(IcDc) = c + 1 by Corollary 37. For the inductive
hypothesis assume that Tk avoids IcDc. Suppose for contradiction that Tk+1 has the
subsequence Ik+1D, for some permutation 7r E Sc.
The proof of Theorem 40 showed that r(I)
=2k - 1 and I is a subsequence of
alt(c, 2k - 1) only if ir(c) = c, so the last I,D, of I+ 1D, must be a subsequence of
the rightmost DcIc in Tk+1. Then ir(1) > 7r(2).
Since T avoids IcDe, then the first letter 7(1) of the second I, in I+1D, must
occur in the initial IcDc of Tk 1. Thus 7r(1)7r(2)7r(1) must be a subsequence of IDce.
This contradicts 7(1) > 7r(2), so Tk+1 avoids Ik+lDc. Thus fw(Ik-Dc) > c+2k-2 for
every c > 2 and k > 1 by Corollary 26.
El
Theorem 54. If c > 2, then fw(IcDcIcDc) = 2c + 3.
37
Proof. Since c + fw(IcDcIc) > fw(IcDcIcDc), then 2c + 3 > fw(IcDcIcDc). As for the
lower bound, the (c, 2c + 2)-formation F = IcD'Ic avoids I,D,ID, for all permutations 7r G Sc. To see this assume for contradiction that F contains IDID,
for
some permutation i c Sc. Since Ic does not contain I,D, by Corollary 37, then the
first 1,D, is in the first IDc of F and the second I.D, is in the last DcI; of F. This
is a contradiction by Lemma 51 and Corollary 52. Thus fw(IcDcIcDc) > 2c + 2 by
Corollary 26.
El
We extend the technique used in the last proof to get an improved lower bound
on fw(alt(c, k)) for all c > 2 and k > 5.
Theorem 55. If c > 2 and k > 1, then fw(alt(c, 2k)) > k(c+2)-1 and fw(alt(c, 2k+
1)) > k(c+ 2) + 1.
Proof. Define TI =
, T- = T 2 .D, and T2kTkI
for k> 1. We prove that
avoids alt(c, k) by induction on k. This implies that fw(alt(c, 2k)) > k(c+ 2) - 2
and fw(alt(c, 2k + 1)) > k(c + 2) by Corollary 26. Theorems 50 and 54 proved that
Tk_1
T2 avoids alt(c, 3) and T3 avoids alt(c, 4).
For the inductive hypothesis there are two cases. First assume that T-1 avoids
alt(c, j) for all j < 2k-1, but suppose for contradiction that T2k-1 has the subsequence
(ID, )k for some permutation i E Sc. Let G be the leftmost (I,D,)k-1 in the
subsequence ( 1,D,)k. Since the leftmost T2k-3 in T2k-1 avoids alt(c, 2k - 2), then
the last letter of G must occur somewhere in the rightmost D'Ic in T2k-1. Moreover
the letter directly after G in (I,D,)k is the same as the last letter of G, so these two
letters cannot occur in the same permutation. Thus the last I,D, in (ITD,)k must
be a subsequence of the last DcI; in T2k-1. Then 7r(2) < r(1) by Corollary 52.
Let H be the rightmost (ID,)k-1 in the subsequence (I,D,)k. Since the rightmost
T2k-3 in T2k-1 avoids alt(c, 2k - 2), then the first letter of H must occur somewhere
in the leftmost ID 2 in T2k-1. Moreover the letter directly before H in (I,D,)k
is the same as the first letter of H, so these two letters cannot occur in the same
permutation. Thus the first 1,D,, in (I,D,)k must be a subsequence of the first ID,
in T2k-1. Then 7r(1) < 7(2) by Lemma 51, a contradiction.
For the second case of the inductive hypothesis, assume that T_ 1 avoids alt(c, j)
for all j < 2k, but suppose for contradiction that T2k has the subsequence (ID, )kjI2
for some permutation 7r E Sc. Let G be the leftmost (I,D,)k in the subsequence
(I,D,)kI,. Since the leftmost T2k-1 in T2k avoids G, then the last letter of G must
occur in the last D in T2k. The last letter of G is equal to the first letter of the
last permutation of (I,D, )kI, so the last I, of (I,D,)kI, must be a subsequence of
the final D, in T2k. Therefore 1, = D,, so the last letter of D, is c. This implies
that (I.D )k is a subsequence of T2 k-iC, so (ID,) would be a subsequence of T2k-1,
a contradiction. Thus (I-D,)kI is not a subsequence of T2k for any permutation
7 E Sc.
El
38
3.7
Open Problems
Many questions about formation width are left unresolved by the results in this chapter. We found several classes of sequences u for which u contained an alternation
with the same formation width as u, which implied tight bounds on Ex(u, n). One
problem is to find all sequences u for which u contains an alternation with the same
formation width as u.
We showed that fw(abc(acb)') = 2t + 1 for t > 0. This implied for t > 2 that
Ex(abc(acb)t , n) = n2 t.a(n)t
(o(n)t2. We conjecture the following result, which
would imply nearly tight bounds on Ex(abc(acb)t abc, n).
Conjecture 56. fw(abc(acb)t abc) = 2t + 3 for t > 0.
We identified the set of all sequences u for which fw(u) K 3. These are all the
sequences for which the value of fw(u) implies linear bounds on Ex(u, n). A next step
would be to identify all sequences u for which fw(u) < 4, since these are all of the
sequences for which the value of fw(u) implies O('ntar()) upper bounds on Ex(u, n).
We also determined the values of 1(u) and r(u) for every binary formation u. Since
both of these functions provide lower bounds on fw(u), it would be useful to compute
the values of 1(u) and r(u) for every sequence u.
On a related note, the values of 1(u) implied bounds on fw(u) within a factor of
2 for every binary formation u. What is the exact value of fw(u) for every binary
formation u?
We also obtained bounds on fw(alt(c, k)) for every k > 1. In particular we determined the exact values for k < 4. What is the exact value of fw(alt(c, k)) for each
k > 5?
In addition we proved that fl(u) < (r - 1)
-1+
1 for all sequences u with r
distinct letters. What is the exact value of fl(u) for every sequence u?
39
40
Chapter 4
Linear extremal functions of
forbidden 0 - 1 matrices
For a collection S of 0 - 1 matrices, let the weight extremal function exs(n, S) denote
the maximum number of ones in an n x n 0 - 1 matrix which avoids every matrix in
S, and let the column extremal function exsk(m, S) denote the maximum number of
columns in a 0 - 1 matrix with m rows which avoids every matrix in S and has at
least k ones in every column.
The column extremal function of matrices is the analogue of a sequence extremal
function, the maximum number of letters in a sequence with m contiguous blocks of
distinct letters that avoids a fixed subsequence (or collection of subsequences) and has
at least k occurrences of each letter. This sequence extremal function was bounded for
alternations and collections of sequences called (r, s)-formations in [38]. The column
extremal functions for collections of matrices called doubled (r, s)-formations were
bounded in [7]. Both the weight and the column extremal functions of permutation
matrices were used in [15] to improve upper bounds on the Stanley-Wilf limits of
forbidden permutations.
Since an n x n matrix has n2 entries, then exs(n, S) < n2 for every collection S of
0 - 1 matrices. If S contains only one matrix M, then we write exs(n, S) = ex(n, M)
and exsk(M, S) = exk(m, M). If M has at least two ones, then ex(n, M) ;> n since a
matrix with ones only in a single column or a matrix with ones only in a single row
avoids M.
Fulek [17] found bounds on the extremal functions of the patterns L1 and L 2
(Figure 4-1) using visibility representations constructed by treating rows of a given
n x n matrix as vertices and projections of ones on lower rows as edges. He bounded
the number of ones in the matrix by limiting the multiplicity of edges in the visibility
representation based on the forbidden 0 - 1 matrices.
0 1 1 1 0
1 0 0 0 1
0 0 1 0 0
0 1 1 0
1 0 0 1
0 1 0 0)
Figure 4-1: L1 and L 2
41
In Section 4.1 we prove some general facts about exs(n, S) and exsk(m, S). In
Section 4.2 we define bar visibility hypergraphs, bound their number of edges, and
use these bounds to show that ex(n, L 3) = 0(n) and exk(m, L 3 ) = O(2) (Figure
4-2). We also bound the extremal functions of forbidden collections of 0 - 1 matrices
corresponding to bar visibility hypergraphs.
0 1
1
1 0 0 0
0 0 0 0
0 0 1 0
0
0
1
0
Figure 4-2: L 3
4.1
Facts about exs(n, S) and exsk(m, S)
If k > m, then exsk(m, S) = 0 for any collection S. If K < in and every matrix in L
has at least k rows with ones, then exsj(M, S) = oc for j < k since a matrix with n
rows, any number of columns, and ones in every entry in the first j rows avoids every
matrix in S.
If some matrix in S has k rows and c columns, then exsj(m, S) < (c - 1) (m) for
all j > k. This is because a matrix A with (c - 1) (') + 1 columns, in which every
column contains at least j ones, must contain c columns with ones in the same k rows
by the pigeonhole principle, so A contains the matrix in S with k rows and c columns.
Let ex(m, n, P) (resp. exs(m, n, S)) be the maximum weight of an in x n matrix
which avoids the 0 - 1 matrix P (resp. collection S), so that ex(n, P) = ex(n, n, P).
For every column of P, draw a segment connecting the topmost and bottommost ones.
Call P range-overlapping if, for every pair of columns of P, there exists a horizontal
line passing through the corresponding segments of both columns.
U
U
*
U
*
*
U
U
Figure 4-3: The pattern on the left is range-overlapping. The pattern on the right is
not range-overlapping because its final two columns have disjoint ranges.
Theorem 57. For any range-overlappingP, ex(m, n, P) < k(exk(m, P)
42
+ n).
Proof. Let A be a 0 - 1 matrix with m rows, n columns, and ex(m, n, P) ones which
avoids P. Consider column c in A. Starting from the top of c, divide its ones into
clusters of size k, deleting the at most k - 1 remaining ones in the column. Move
each cluster g after the first cluster horizontally to a new column immediately to the
right of the column containing the cluster which was above g in the original column.
Delete any columns with no ones. Call the newly formed matrix A', and suppose that
A' has n' columns.
Suppose for contradiction that A' contains P, and consider two cases. If the copy
of P in A' contained at least two columns that originated from the same column c,
then P could not be range-overlapping because the construction separated columns
of A into clusters with non-range-overlapping row indices to obtain A'. If the copy
of P in A' only contained columns derived from different original columns, then the
original matrix A also contained P, a contradiction.
Therefore, A' cannot contain P. In our construction, we deleted a maximum of
k(n' + n) <
n(k - 1) ones. Each column of A' contains k ones, so ex(m, n, P)
l
k(exk(m, P) + n).
Lemma 58. If S is a collection of 0 - 1 matrices, c is a constant, and g is a function
) for k > c.
such that exs(m, n, S) < g(m)+cn for all m and n, then exsk(m, S) <
Proof. Fix k > c. Any matrix with m rows, exsk(m, S) columns, and k ones per
column which avoids S has at most exs(m, exsk(m, S), S) < g(m) + exsk(m, S)c
ones. Therefore exsk(m, S)k < g(m) + exsk(m, S)c, so exsk(m, S) < g() for all
El
M.
.
Corollary 59. If S is a collection of 0 - 1 matrices such that exs(n, S) = O(n) and
every matrix in S contains the 2 x 2 identity matrix or two ones in the same row,
then exsk(m, S) = (
Proof. Since exs(m, n, S) < max {exs(m, S), exs(n, S)}, the upper bound follows
from Lemma 58. For the lower bound, let K be the matrix with m rows and [j]
columns obtained from an [J x [IJ identity matrix by replacing every row r in the
identity matrix with the k x [-] submatrix consisting of k adjacent copies of r and
then adding rows full of zeroes at the bottom until K has m rows. Let K' be obtained
- 1 since K' avoids
from K by reflecting over a vertical line. Then exsk(m, S) ;>
El
S.
Marcus and Tardos [36] showed that ex(n, P) = O(n) for any permutation matrix
P, solving the Stanley-Wilf conjecture. Call a 0 - 1 matrix light if it has no pair
of ones in the same column. The Marcus Tardos theorem was generalized in [201 to
show that ex(n, Q)
O(n) for any light 0 - 1 matrix Q such that for every two
columns co and ci in Q with ones in the same row r, all columns between co and ci
also contain ones in row r. Corollary 59 can be applied to the matrices Q to show
that exk(m, Q) = 6(j), as well as to the matrices L 1 and L 2 in [17].
Let Pr,c denote the r x c matrix filled with ones. We find the exact value of
exk(m, Pk,c) using a simple counting argument, and then bound exk(m, P,c) for all
k >r.
43
(c - 1) ()
.
Lemma 60. exk(m, Pk,)
Proof. There exist (7) possible configurations
By the pigeonhole principle, a matrix with (c
configuration of k ones that occurs in at least c
(c - 1) (7). We can obtain the same lower bound
possible configuration c - 1 times; this matrix
of k ones in columns of height m.
- 1) (7) + 1 columns must have a
columns. Therefore, exk(m, P,,) <
by constructing a matrix with every
avoids Ph,c. Hence, exk(m, Pk,c)
l
(c - 1)(7).
Theorem 61. For all fixed k > r, eXk(m, P,,2)
(mr).
,
.
.
Proof. First observe that eXr(M, P,2) = 0(mr), according to the last lemma. Since
exk(m, P) is decreasing in k for every P, exk(M, P,2) = Q(mr) for k > r.
For any 0 - 1 matrix M, let Gm be the graph obtained from M by letting every
column of M be a vertex and adding an edge between two vertices if and only if their
corresponding columns have ones in exactly r - 1 common rows. Note that r - 1 is
the maximum number of rows a pair of columns may share without containing P,,2
We proceed by induction on k to prove that exk(n, Pr,2) = Q
The first case is k = r. The last lemma implies that ex,(m, P,,2 )
("). Observe
that if M is a 0 - 1 matrix with m rows and r ones in each column which avoids Pr,2
then the maximum degree of any vertex of Gm, A(Gm), is at most r(m - r).
Fix k > r. Let M be the matrix obtained in the last inductive step avoiding
Pr,2 with m' < xm rows, (') columns, and k - 1 ones in each column, such that
A(GM) < ym for some constants x and y that depend only on r and k.
Using a greedy algorithm, color the vertices in Gm using A(GA) + 1 colors so
that no two vertices with a common edge have the same color. Let C : VGm -+
{1,... , A(Gm) + 1} be the coloring function. Construct a new matrix A' from M by
adding A(Gm) + 1 new rows to the bottom of Al with a one in row m' + r of column
b if and only if G(b) = r. Since A(GA) < yin, the resulting matrix M' has at most
xm + ym + 1 rows.
Fix a column b of M'. A column c is a neighbor of b in Gm, but not in Gm only
if there are exactly r - 2 rows in M which have ones in both b and c, and b and c
were assigned the same number in the coloring of Gm. Since b has k - 1 rows with
ones in Ak, there are (-1) ways to choose r - 2 rows that contain ones in column b
of Al. For every set of r - 2 rows in which column b has ones in M, there are at most
xm-(k-1)
possible new neighbors c of b in M' with ones in those rows, since every
pair o new neighbors of b in Gm' with ones in those r - 2 rows get the same color
in Gm and thus have no common rows in M containing ones besides the r - 2 rows
each neighbor shares with b. Then A(Gm,) < "--(k-1) k-1) + ym, completing the
induction.
l
Thus for each k > r, there exists a constant ark such that exk(ar,km, P,2) > (M)
To generalize the bounds for all integral values of m, let n satisfy ar,kn < m
ar,k (n+
1) and observe that exk(ar,kn, Pr,2 ) > (n) > (
all fixed k > r.
~k). Hence exk(M, P,2 ) = (ir) for
Corollary 62. For all fixed k > r and c > 2, exk(m, Pc)
44
O(inr).
Proof. For c > 2, exk(rm, Pr,c) > exk(M, P,,2) since P,, contains P,,2. Therfore
exk(m, P,,,) = Q(mr) for all k > r. Since ex,(m, Pr,c)
(c-i)(), then eXk(rM, P,c) =
O(m') for all k > r. This shows that exk(m, P,c) = O(m') for all fixed k > r and
c > 2.
l
Corollary 63. If P is a 0-1 matrix with r rows which contains P,2 , then exk(m, P)
0(n'r) for all fixed k > r.
4.2
Bar s-visibility hypergraphs and 0 - 1 matrices
A bar s-visibility hypergraph's vertices are finite, disjoint, horizontal bars in the plane
and its edges are subsets of vertices of size s + 2 for which a vertical segment exists
which intersects only the vertices in the edge. Using the fact that every bar visibility
graph (i.e. bar 0-visibility hypergraph) is a planar graph, Fulek converted 0 - 1
matrices into bar 0-visibility hypergraphs to prove linear bounds on the extremal
functions of patterns L1 and L 2 [17].
For r, s > 0, define T,,, to be the collection of matrices M which have r + s + 2
rows and r + 2s +2 columns such that Al restricted to the first s +1 columns and rows
2,. . . , s + 2 is a (s 1) x (s + 1) permutation matrix, M restricted to the last s + 1
columns and rows 2, ... , s + 2 is a (s + 1) x (s + 1) permutation matrix, M restricted
to the middle r columns and the last r rows is an r x r permutation matrix, and M
has ones in the middle r columns in row 1. For example T1,0 contains a single 3 x 3
matrix with four ones in a diamond formation. One of the patterns in T4,1 appears
in Figure 4-4, with black squares representing ones and white squares representing
zeroes.
We generalize the technique from [171 to show for all r, s > 0 that exs(n, Tr,,s)
0(n) and exsk(m, T,,,) = 0(1). The first step is to prove a linear bound on the
number of edges in a bar s-visibility hypergraph with n vertices. This proof is similar
to the proof of the maximum number of edges in bar s-visibility graphs with n vertices
in [9]. We assume that all bar endpoints have distinct coordinates since this does not
decrease the maximum number of edges.
Lemma 64. All bar s-visibility hypergraphs with n vertices have at most (2s + 3)n
edges.
Proof. Scan any representation of the given bar s-visibility hypergraph in the plane
from left to right, making a list of distinct edges. Add an edge to the list whenever
the scan for the first time shows part of the representation in which a vertical segment
can be drawn which intersects each of the vertices in the edge. Then edges will only
be added to the list whenever the scan passes the left or right end of some bar.
For each bar B, the maximum possible number of edges added to the list when
the scan passes the left end of B is s + 2 since there are at most s + 2 vertical segments
representing different edges which pass through the left end of B and through s + 1
other bars. The maximum possible number of edges added to the list when the scan
passes the right endpoint of B is s + 1, since at the right endpoint of B there are at
45
Figure 4-4: A matrix in T4,1
most s + 1 vertical segments representing different edges which pass through s + 2 bars
other than B, at least one of which is below B and at least one of which is above B.
Then there are at most (2s + 3)n edges on the list since the representation contains
5
n bars.
In the next lemma, we change 0 - 1 matrices avoiding T,,, into bar s-visibility
hypergraphs, and then show that the resulting hypergraphs have edge multiplicity at
most r - 1.
Theorem 65. For all r > 0 and s > 0, exs(n, Tr,s) = O(n).
Proof. Let M be an n x n matrix which avoids T,,,. Define M' to be the matrix
obtained from M by deleting the first s + 1 and last s + 1 ones in every row, and
the last r ones in every column. Construct a representation of a bar s-visibility
hypergraph H from M' by drawing a bar in each row that contains a one in M' with
left end at the first one of M' in the row and right end at the last one of M' in the
row. For each one in M' with at least s + 1 ones below it in M', draw a vertical line
segment representing an edge starting from the one and extending through s bars
until reaching the (s + 1)" bar below the one.
Suppose for contradiction that H contains some edge e with multiplicity at least
r. Let u, ... , u+ 2 , in order from top to bottom, be the rows of M' which contain the
vertices in the edge e, and let c 1,... , c, be columns of M' which contain r vertical
segments representing the copies of e. Let v 1 ,... , v, be distinct rows of M such that
vi contains one of the bottommost r ones of ci for each i = 1, . . . , r; let di, ... , d,+ 1 be
distinct columns of M such that di contains one of the s+1 leftmost ones of ui for each
i = 2, ... , s+2; and let e, ... , e.+1 be distinct columns of M such that ej contains one
of the s+1 rightmost ones of ui for each i = 2,. . . , s+2. Then the submatrix of M consisting of rows u1 ,...,us+2 , vI,...,v,. and columns ci,...,c,,d, .. ,d s +1,e, ... , e,+
contains a matrix in T,,,, a contradiction.
46
Then every edge of H has multiplicity less than r, so the number of ones in M is
El
at most (3s + 3 + r)n + (r - 1)(2s + 3)(n - r).
Observe that every element of T,,, contains the pattern L 3 for r
which implies the following corollary.
3 and s > 1,
Corollary 66. ex(n, L 3 ) = 0(n).
The next two corollaries follow from Corollary 59.
Corollary 67. exsk(m, T,,,) = 0().
Corollary 68. exk(m, L 3) - 0(M).
4.3
Infinitely many minimal non-linear 0 -1
matrices
In this section we prove that any k x 2k double permutation matrix has a 20(k)n
extremal function. Let the pattern P be any k x 2k double permutation matrix with
k > 1. Fix an integer n that is divisible by k. Choose any n x n matrix A with
ex(n, P) ones that avoids P.
We now partition A into (,)2 submatrices, each with k rows and k columns. Define
block Sij, a square submatrix of A, to be the intersection of rows k(i - 1) + 1 through
ki of A with columns k(j - 1) + 1 through kj of A.
We construct an 1 x n 0 - 1 matrix Q. Define the ones of Q row by row from top
to bottom, in each row from left to right such that qij = 1 if: (1) Sij is the leftmost
block with a one in its row of blocks or (2) Sij has a one and some row of A contained
in the blocks between Sij and Sij inclusive (where j' is the greatest number less than
j such that q, = 1) has at least two ones. Q avoids P, or else A would contain P.
We also define chunks of blocks, Cij, which are sets of blocks in the same blockrow R,. A chunk Cij consists of a block Sij with qij = 1 on the left and all blocks,
including empty ones, between it and Sij, the next block to the right of Sij with
qij = 1. The chunk Cij does not include Sti. If Sij is the rightmost block in its
block-row with qij = 1, then its chunk contains it and all blocks in its block-row to
the right of it. Hence the chunks partition the non-empty blocks of A.
Note that if a chunk has two ones in the same row, then both ones must be in the
same block. Note also that if block Sij has two ones in the same row, then qij = 1
and qij' = 1 for the next non-empty Sij, because the row with two ones is contained
between Sij and Sij inclusive. Hence if a chunk has two ones in the same row, then
there is a single non-empty block in the chunk.
Thus all chunks fall into two categories: (1) chunks with a single non-empty block
or (2) chunks with at least two non-empty blocks and no rows with more than a single
one.
Call chunk Cij wide if it has at least two ones in a single row. Call Cij tall if it
has ones in every row. Chunks that are neither wide nor tall have at most k - 1 ones
total.
47
Lemma 69. The number of wide chunks in A is at most ex(!', P'), where P' is the
k x k permutation matrix obtained from P by removing duplicate columns.
Proof. If A had more than ex(!, P') wide chunks, then the entries in those wide
chunks would form a copy of P. This would be a contradiction.
l
Lemma 70. In each row of chunks Ri = USj for 1 < j
chunks is less than 2k.
1g, the number of tall
Proof. If the number of tall chunks in Ri was at least 2k, then Ri would contain a
copy of P. This would be a contradiction.
El
+
Lemma 71. Suppose that P is the k x 2k double permutation matrix correspondingto
the k x k permutation matrix P. For n divisible by k, ex(n, P) < 2nk+ex(!-, P')k2
(k - 1)ex(n, P).
Proof. We account for all of the ones in A. First, there are at most ex(L, P') wide
chunks in A, so there are at most ex T, P')k2 ones in wide chunks in A. There are
at most 2k2 = 2n tall chunks in A, so there are at most 2nk ones in chunks that are
tall but not wide. The remaining chunks are neither tall nor wide, so they each have
at most k - 1 ones. There are a total of at most ex(, P) chunks since Q avoids P,
so the number of ones in A is at most 2nk + ex(i, P')k2 + (k - 1)ex(11, P).
l
Theorem 72. If P is a k x 2k double permutation matrix, then ex(n, P) = 20 ()n.
)
+
Proof. Let P' be the k x k permutation matrix obtained by removing duplicate
columns from P. Fox proved that ex(n, P') = 2O(k)n [15]. Let c > 1 be a constant such that ex(n, P') < 2ckn for all n. By the last lemma, ex(n, P) < 2nk
2cknk 2 + (k - 1)ex(L, P) < 22 ckn + (k - 1)ex(-, P) for every n divisible by k.
For n < k, it is clear that ex(n, P) 5 2k22 ckn. FLx m, suppose that ex(n, P) <
2k2 2 ckn for all n < m, and consider the case n = m. Let N be the largest multiple of k
less than or equal to n. Then ex(n, P) < ex(N, P)+2kn - 22 ckN +k -e
Ji
2kn < ( 2 2ck + 2(k - 1) 2 2ck + 2k)n < 2k2 2 ckn.
l
Keszegh defined an infinite sequence of 0 - 1 matrices Hk for k > 0 and proved
ex(n, Hk) = Q(n log n) for all k > 0 [28]. Hk is the 3k + 4 x 3k + 4 0 - 1 matrix
with all entries zero except h4 , 1 = h1,2 = hl,3 = h3k+3,3k+4 =h3k+2,3k+4 = 1 and
h3i+4,3i+1= h 3i- 1 ,3 +3 =3i,3i+2 = 1 for each 1 K i K k.
Corollary 73. There exist infinitely many pairwise distinct minimal nonlinear 0 - 1
matrices.
Proof. This result follows because for every k there are at least k +1 ones that can be
deleted from HA to produce a matrix with a linear extremal function. In particular,
for each 0 K i K k, the matrix obtained by deleting the one in Hk at h3 i+4 ,3i+1 has a
linear extremal function. This follows from Theorem 72 and the following fact proven
by Keszegh [28]: if P is any 0 - 1 matrix with a one in its bottom right corner, Q is
any 0 - 1 matrix with a one in its top left corner, and R is obtained by joining P and
Q so that they overlap only in the bottom right corner of P and the top left corner
of
Q, then
ex(n, R) < ex(n, P) + ex(n, Q).
48
l
Chapter 5
Extremal functions of forbidden
multidimensional matrices
A d-dimensional ni x ... x nd matrix is denoted by A = (a.....,id), where 1 < il < ne
for f = 1, 2,... , d. An -cross section of matrix A is a maximal set of entries ail,. ,id
with il fixed. An f-row of matrix A is a maximal set of entries ail . ,,d with ij fixed
for every j
$
1.
A d-dimensional k x ... x k 0 - 1 matrix is a permutation matrix if each of its
i-cross sections contains a single one for every f = 1, ... , d. The Kronecker product of
two d-dimensional 0 - 1 matrices M and N, denoted by M D N, is the d-dimensional
matrix obtained by replacing each 1 in M with a copy of N and each 0 in Al with a
0-matrix the same size as N.
As with two dimensions, we say that a d-dimensional 0 - 1 matrix A contains
another d-dimensional 0 - 1 matrix P if A has a submatrix that can be transformed
into P by changing any number of ones to zeroes. Otherwise, A is said to avoid P.
Let f(n, P, d) be the maximum number of ones in a d-dimensional n x ... x n 0 - 1
matrix that avoids a given d-dimensional 0 - 1 matrix P. It is easy to obtain trivial
lower and upper bounds on f(n, P, d).
Proposition 74. If P is a d-dimensional0- 1 matrix that contains at least two ones,
then nd-i < f (n, P, d) < nd.
Proof. For every P, we can choose a d-dimensional n x
...
x n zero-one matrix A,
with ones in a single 1-cross section for some 1 and zeroes elsewhere, such that A
avoids P. This implies the lower bound, the upper bound is the number of entries in
a d-dimensional n x - - - x n matrix.
L
The upper bound in Proposition 74 is a factor of n higher than the lower bound.
Most work on improving this bound has been for the case d = 2. Pach and Tardos
proved that f(n, P, 2) is super-additive in n [39]. By Fekete's Lemma on superadditive sequences [13], the sequence {"f(nP,2) } is convergent. The limit is known as
the Fiiredi-Hajnal limit [6, 15]. Cibulka [6] showed that this limit is at least 2(k - 1)
when P is a k x k permutation matrix and that the limit is exactly 2(k - 1) when P
is the k x k identity matrix.
49
Marcus and Tardos [36] showed that the Ffiredi-Hajnal limit has an upper bound
of 20(k log k) for every k x k permutation matrix P, and Fox [15] improved this upper
bound to 2 0(k). Klazar and Marcus [341 bounded the extremal function when the
d-dimensional matrix P is a permutation matrix of size k x - x k, generalizing the
d = 2 result [36] by proving that f(n, P, d) = O(nd-1). In particular, they showed
that f(",fPd)
= 2 0(klogk), which generalizes Marcus and Tardos' upper bound on the
ndI
Fiiredi-Hajnal limit [36].
We observe that f(n, P, d) is super-homogeneous in higher dimensions for certain
matrices P, i.e., f(sn, P, d) > Ksd-lf(n, P, d) for some positive constant K. This
super-homogeneity is key to our proof for n> 2 LkI/dJ /20 that f jP has a lower bound
of 2(k1/d) for a family of k x
...
x k permutation matrices, generalizing Fox's result
[14] from d= 2 to d> 2.
For each fixed d > 2 we also show an upper bound of 2 0(k) on f_ 7 Od) for every
k x ...x k permutation matrix P. This improves the previous upper bound of 2 0(k log k)
for d > 2 [341, and it also generalizes Fox's bound 20(k) on the Fiiredi-Hajnal limit
when d = 2 [15]. Moreover for each fixed d > 2 we prove that this upper bound
2 0() is also true for all d-dimensional double permutation matrices P of dimensions
2k x k x .-.x k.
5.1
Upper and lower bounds for d-dimensional permutation matrices
When d > 2, it is still an open problem to prove the convergence of the sequence
{ f(nP,d)
Theorem 75. if P is a d-dimensionalkx -- : x k permutation matrix, then
"aPI
=
20(k)
The proof uses the notions of cross section contraction and interval minor containment [15]. Contracting several consecutive i-cross sections of a d-dimensional matrix
means replacing these i-cross sections by a single i-cross section and placing a one in
an entry of the new cross section if and only if at least one of the corresponding entries
in the original I-cross sections is a one. We say that A contains B as an interval minor
if we can use repeated cross section contraction to transform A into a matrix which
contains B. A avoids B as an interval minor if A does not contain B as an interval
minor.
The containment in previous sections is at least as strong as interval minor containment. Indeed, A contains B implies that A contains B as an interval minor.
However, since a permutation matrix has only one 1-entry in every cross section, containment of a permutation matrix P is equivalent to containment of P as an interval
minor.
Analogous to f(n, P, d), we define m(n, P, d) to be the maximum number of ones
in a d-dimensional n x - - - x n zero-one matrix that avoids P as an interval minor.
50
Let Rkl..'kd be the d-dimensional matrix of dimensions k, x ...
equal to 1.
We observe that
f(n, P, d) m(n, R'..-'k, d)
x kd with every entry
(5.1)
for every k x ... x k permutation matrix P. This follows from the fact that containment
of Rk,.-k as an interval minor implies containment of P.
Lemma 76.
m(tn, R k..,k d) < sdm(n, R'.k---, d) + dntdfk .... k(n, t, s, d),
(5.2)
where fA,,...,kd(nt,s,d) denotes the maximum number of 1-rows that have at least s
ones in a d-dimensional t x n x --- x n matrix that avoids Rkl..-kd as an interval
minor.
Proof. Let A be a d-dimensional In x
...
x In matrix that avoids R .
as an interval
minor and has m(tn, Rk k,.., d) ones. The S-submatrices of A are constructed by
dividing A into nd disjoint submatrices of size t x ... x t and labeling these submatrices
as S(ii, ... , is). The submatrix S(ii, ... , id) is defined to contain the entries a 1,...,d
of A such that t(ie - 1) + I j< < tit for each?= 1,..., d.
The contraction matrix of A is defined to be the d-dimensional n x ... x n matrix
1 if S(ii,i2 ,...,id) is a nonzero matrix and
C = (ci,4 2. d) such that ci, 2 . d
is a zero matrix.
Cil2,...,id = 0 if S(ii, i2 ,... , i)
We do casework based on whether an S-submatrix of A has s nonzero 1-cross
sections for some 1.
We first count the number of ones from the S-submatrices that do not have s
nonzero i-cross sections for any 1. The contraction matrix C has at most m(n, Rk,--) d)
ones, otherwise C would contain Rk---,k as an interval minor, a contradiction. Hence,
A has at most m(n, Rk-.k, d) such S-submatrices, each of which contains at most
(s - 1)d < sd ones. Thus there are at most sdm(n, R k..-'k d) ones from the Ssubmatrices of this type.
We next count the number of ones from the S-submatrices that have s nonzero
i-cross sections for some l. Without loss of generality let 1 = 1. Let A' be the matrix
obtained from A by changing all the ones from the S-submatrices of A that do not
have s nonzero i-cross sections to zeroes. Divide A' into n blocks, each of which is a
t x tn x ... x In submatrix of A'. The Ith block of A' is defined to contain the entries
1,..., n. For each block,
of A' such that t(i - 1) + 1 < j< K ti for each i
a'
contract every t consecutive j-cross sections uniformly for every j # 1 to obtain a
t x n x - x n matrix, which has at most fk,....k(n, t, s, d) 1-rows that contain at least
s ones.
In each block there are at most fk,...,k(n, t, s,d) nonzero S-submatrices. Ranging
over all n blocks as 1 ranges from 1 to d, there are at most dntdfk,..k.,(n, t, s, d) ones
from the S-submatrices of this type.
l
Summing both cases proves the lemma.
Lemma 77. fAk..,kd(n, 2t, 2s, d) < 2 fk 1 1...,kd(n, t, 2s, d) + 2 fk.
51
-,2,...,kd
1
(n, t, s, d).
Proof. Let A be a d-dimensional 2t x n x - x n matrix that avoids Rk, ,kd as an
interval minor and has fki,...,kd(n, 2t, 2s, d) 1-rows, each with at least 2s ones.
The first type of 1-rows have all their ones among their first t or last t entries.
There are clearly at most 2 fk 1 ,. kd(n, t, 2s, d) such 1-rows in A.
The other type of 1-rows must have at least one 1 among both the first t and the
last t entries. Since each 1-row has at least 2s ones, there are at least s ones among
either the first or last t entries. Without loss of generality, we consider those 1-rows
in which the first t entries contain at least s ones.
Let A' be the matrix obtained from A by changing all ones to zeroes in all other
1-rows, and then contracting the last t 1-cross sections. Hence, the last entry in each
nonzero 1-row of A' is a one. The first t 1-cross sections of A' must avoid Rkl-l,k2,...,kd
as an interval minor, otherwise A' contains Rk,..-,kd as an interval minor and so does
A, a contradiction.
Thus, there are at most 2 fk 1 -1,k 2 ,...,kd(n, t, s, d) 1-rows in which both the first t and
last t entries include at least one 1. Summing both cases proves the lemma.
l
Lemma 78. If s, t are powers of 2 and 2'-1 < s K t, then fe,k,...,k(n, t, s, d) <
2f 1t 2 m,(n, Rk,.d 1).
Proof. We induct on f. For e = 1, we prove that fl,k,...,k(n, t, s, d)
m(n, Rk,.
d-1).
Suppose for contradiction that f1,,...,k(n, t, s, d) > m(n, Rk, -. "', d - 1). Then there
is a t x n x ... x n matrix A which avoids Rl',...,k as an interval minor and has more
than m(n, R k,.'k, d - 1) 1-rows with at least s ones. Let B be the 1 x n x ... x n
matrix obtained from A by contracting all the 1-cross sections. Then B, which can
be viewed as a (d - 1)-dimensional matrix, has over it(n, R ak d -1) ones and thus
contains the (d - 1)-dimensional matrix R ,.k as an interval minor. Consequently, A
contains the d-dimensional R,-.-,k as an interval minor, a contradiction. Therefore,
f1,k,..,k(n,
t, s, d) < m(n, R-.', d- 1) <
21 -1t2
d - 1),
m(n, R.',
which proves the base case. We assume for all s and t that are powers of 2 satisfying
2e-2 < s < t that
2t 2
fe_1,,.(n, t, s, d) <
s
r(n,
m Rk'..'k d - 1)
(5.3)
for some f > 2. We need to show that
fe,k,...,k(n, t, s, d) <
t m(n, Rk'.-.k' d- 1)
2
(5.4)
for all s and t that are powers of 2 satisfying 2" < s < t. The case t = s > 2 -1 is
trivial. If fe,k,...,k(n, t, s, d) < 2
m(n, Rk.,k, d - 1) for some t > s > 2 e-1 that is a
52
power of 2, then we prove the same inequality for 2t. By Lemma 77,
ff,k,...,k(n, 2t, s, d)
<
<
2fe,k,...,k (n, t, s, d) + 2fe_1,k,...,k (n, t, s/2, d)
2e-1t 2
2e-2t22
m(n, Rk'--.k
+ 2
,
+
s/2
s
= "
S
, d - 1)
m(n, Rk'--'k d- 1)
where we use both inductive assumptions in the second inequality.
Proof of Theorem 75. We first bound the right hand side of inequality (5.1).
claim for each fixed d > 2 that
n-
20(k)
El
We
(5.5)
Fox proved the base case d = 2 [15]. Assuming that (5.5) is true for d - 1, we
combine Lemmas 76 and 78 to get
m(tn, Rk-.--k, d) < sdm(n, Rk.--k, d) + dtd 2k-1t2 2O(k)nd-1.
S
Choosing t =
and s =
2 dk
2
k-1 yields
m(2dkn, Rk,-.,k d ) < 2(k- 1 )dm(n, Rk.',
In particular, if n is a positive integer power of
d) + d 2 kd(d+2) 2 (k)nd-1.
2 dk,
iterating this inequality yields
Rk.-k, d) + d2 kd(d+2) 2 0(k)( 2 dk)(L-1)(d-1) <
m((2dk)L
kd(d+2) 2 0(k) (1 + 2d(dk2 +1) ( 2 dk)(L-1)(d-1)
2 2(k-1)dm(( 2 dk)L-2, Rk,.-k, d) + d 2
dk)(L-1)(d-1)
+1) + 2 2tdkk+1) +--)
d 2
(1
2 L(k-1)dm(1, Rk'.k, d)+d2kd(d+2)20(k)
k,...,k
-
2
d) <
2 (k-1)d
((2dk)L-1,
0(k)( 2 dk)(L-1)(d-1)*
Suppose that ( 2 kd)L-1 < n < ( 2 kd) L. Then m(n, Rk.k, d) < m(( 2 dk)L k,...,k d)
20(k)nd- 1 . This completes the induction on d, and hence (5.5)
l
is proved.
2 0(k)( 2 dk)(L-1)(d-1)
Next we extend Fox's probabilistic lower bound for the d = 2 case [15]. A key
part of our approach is the observation that there exist d-dimensional permutation
matrices P for which f(n, P, d) is super-homogeneous.
Theorem 79. There exists a family of d-dimensionalk x ... x k permutation matrices
P such that f (nPd)
2 "(k1/d) for n > 2 [kI/dj/20
In two dimensions, the extremal function was shown to be super-additive [39],
i.e., f(m + n, P, 2) > f(m, P, 2) + f (n, P, 2). This implied the convergence of the
sequence { f ,2)} for matrices P whose extremal functions are O(n). The limit is
the well-known Firedi-Hajnal limit.
The super-additivity of f(n, P, 2) also implies its super-homogeneity: f(sn, P, 2) >
sf(n, P, 2) for every positive integer s. Define a corner entry of a ki x . x kd matrix
53
P =(...,id) to be an entry pi2,..., j located at a corner of P, i.e., e1 = 1 or i, = k,
for every 1 < -r d.
Lemma 80. If P is a d-dimensional matrix with a corner 1-entry, then f (sn, P, d) >
S d_1 ,f~ P, d).
-
Proof. Without loss of generality, suppose that pi,...,1 = 1 is a corner 1-entry in
P. Let M be a s x ... x s matrix with ones at the coordinates (ii,... , i ) where
I1 +
+'d i= s +d - 1 and zeroes everywhere else, so that Al has (d-2) >2
ones. Let N be an n x - x n matrix that avoids P and has f(n, P, d) ones. It suffices
to prove that M 0 N avoids P.
Suppose for contradiction that M 0 N contains P. Pick an arbitrary 1-entry p*
in P other than Pi,...,1. Suppose that pi,..1 and p* are represented by el and e 2 in
M 0 N, respectively. In particular, suppose that el and e 2 are in the S-submatrices
S(i 1 ,. ..,i)
and S(ji,.. . ,jd), respectively. Note that i1 -I- -+
.=
+
jd.
Since each coordinate of p* is at least as great as each coordinate of Pi,...,1 in P, then
each coordinate of e 2 must also be at least as great as each coordinate of ei in AM DN,
and hence i4
j, for =1, 2, . . . , d. It then follows from i 1+ - + id= jI + .+ jd
that ir = jr for r = 1, 2,. . . , d, i.e., the two entries ei and e 2 must be in the same
S-submatrix in M 0 N.
Since p* can be any arbitrary 1-entry other than pi,...,1 in P, a single S-submatrix
contains all of P. However, this is a contradiction since each nonzero S-submatrix in
M 0 N is a copy of N, which avoids P. Thus M 0 N avoids P.
El
Lemma 81. If P is a d-dimensional matrix with a corner 1-entry, then for any
positive integers m and n > m,
f (n, P, d)
-1
f(m, P, d)
1
2d-1(d-1)!
rd-1
Proof. Write n as n = sm + r, with 0 < r < m.
f(sm,P,d) >
P) f(nPd) >3_ f(m,) > i f(n,Pd) h Lem
1
(smn+r),d1
-
(d-1)! (sm~r)d
-
(d-1)!
((s+1),n)d-1
-
Then
2dl(d-1)!
mn-,
f(n_,Pd)
-
f(sm+r,Pd)
where
used in the second inequality.
>
ma 80 is
El
The following lemma gives a lower bound on the right hand side of the inequality
in Lemma 81 for a particular m. The proof is based on Fox's methods for the d = 2
case in [14].
( Nd-1/ 2
)
Lemma 82. There exists an N x ...x N matrix A, with N = 2 "(C), that has
ones and avoids R-.. as an interval minor.
+
Proof. We prove the lemma for E that are multiples of 20, the result can then be
easily extended to all f. Let r = -, q = 2
, and N = 2'.
Define a dyadic interval to be a set of consecutive integers of the form {(s - 1) 2 '
1, ... , s2t} for nonnegative integers s and t, and a dyadic hyper-rectangle to be the
Cartesian product of d dyadic intervals. We consider only dyadic intervals that are
subsets of {1,..., N}. Note that there are exactly E_ 2 = 2N - 1 such dyadic
54
intervals and (2N - 1)d such dyadic hyper-rectangles. There are (r + 1)" dyadic
hyper-rectangles containing each entry ai1 . ,,d since each ij is contained in exactly
r + 1 dyadic intervals.
Let R be a random collection of the dyadic hyper-rectangles, each included independently with probability q. Define A to be the N x ... x N d-dimensional 0 - 1
matrix such that aj, . . . 1 if and only if (ii, ... , is) is not contained in any dyadic
hyper-rectangle of R. The expected number of ones in A is
(1
-
q)(r+1)d Nd =
(e-q(-r+1)"d)-
where we use (1 - q)/
=
[(1
-
q)1/q]q(r+1)d N d
2
-(2r/
Nd) =
O(Nd-1/ 2 ),
E(e- 1 ) in the second equality.
Denote by X and Y the events that A contains and avoids Re.
as an interval
minor, respectively. We bound the probability P(X). If B is a set of dyadic intervals,
let X(B) be the number of dyadic intervals on {1,... , N} that contain at least one
interval in B as a subset. Then define h(x) to be the number of sets B containing f
dyadic intervals such that X(B) = x.
If A contains Re.-- as an interval minor, then there are intervals of consecutive
integers, denoted by Wi, .. ., W i ,j, partitioning the set {1, 2,... , N} in the ith dimension such that each submatrix W1,x
j, x ... x Wd,j, of A contains a one. Let Ij be
the unique smallest-length dyadic interval which contains W, 3 . If there are xi dyadic
intervals which contain at least one of the dyadic intervals in Bi = {Ii, .. . , Iq}, then
and there are h(xi) possible sets B. Thus there are h(xi) ... h(Xd) choices
xi >
for B 1 ,..., Bd and x 1 ... Xd dyadic hyper-rectangles which contain at least one of the
d dyadic hyper-rectangles of the form I1,j, x ... X Id,id. Since these X1 ... Xd dyadic
hyper-rectangles contain 1-entries of A, none of them is in R. Hence,
e
P(X) <
Z
(1 - q)x1.".Xdh(xi)
...
h(Xd).
(5.6)
To bound h(x), we must bound the number of sets U = {u1,..., ue} of dyadic
intervals such that X(U) = x. Let ,v1,... , ve be the integers such that vi = X({u1)
andvi = X({u, . . ., uj}) - X({ui,...,
1uj}), for i=
2, . .. , e. Since vi +
+ vf = x,
there are at most (2jey) possible values for vi, . . . , ve. Given v 1 , . . ., ve, there are
2j1-1 choices for ui and at most 2'i+1X({ui, ... , u2 }) choices for uj+1 . Since the order
of the elements in U does not matter, then
h (x) <
hx 1f!<
)2v'-
i - 2i+1X({u1,
. . .
, ui})
(x+f-l)2xxe
e_
Substituting this into (5.6) yields
d
q)(x ..-. )9P
P (X )
d
X1x,-
di
55
(xi+f-12x
i
The summand
r(
,
.
I,
(1 -
)
d xi+ -1
_1
(
q)
) 2Xmi
i=1
is symmetric in its variables. The ratio r(xi + 1, c 2 , . ..
2(1 - q)
4e -qX2
.-.X(1 +
I
1
, xd)/r(Xi, . ..
(I +) 1 )<
2...Xd
X1 + 1
X1
)yj < eqi-
--
2 -2od-I/2&
4e <12,
1/e for q > 0 and (1 + 1/1x)x1 < e for
where we use (1 - q) /
,X) is equal to
1
> 1. We also
note that
,))
rq...
e-fd (2 2e-12ee)d K
2 20d-
le/2( 2 3f-le
(_-)
where we use Stirling's inequality and
r(i,.--
[(Y- fjd
(1- q)d
<
)d
< N-10d,
5 22-. We now use the symmetry of
, Xd) to obtain
00
P(X) <
r(i,...,xd)
d
()
<
< 2 N-
i=O
X1,...,Xd>t
Now we estimate the conditional expectation E( IY), where
E( |Y)P(Y) = E( ) - E(>X)P(X) > O(Nd- 1 / 2 ) - Nd2dN-10d
JAI. Note that
E (Nd-1/ 2 ) So
E( JY) = E(Nd-1/ ). Thus, there exists an N x
x N d-dimensional 0 - 1 matrix
A that avoids Rl-,' as an interval minor and has e(Nd-1/ 2 ) ones.
El
2
Proof of Theorem 79. Let e = [k1/d], and let P be any d-dimensional permutation
matrix of size k x ... x k that contains Rtee as an interval minor and has at least one
corner 1-entry. P contains R.., as an interval minor, so f(N, P, d) > m(N, R....-'', d)
for N = 2 /20, which by Lemma 81 and Lemma 82 implies for n > N that
f(n, P, d)
nd-1
1
2d-1(d -
-
1)!
m(N, R'.,, d) - 2
(k/d)
Nd-1
El
5.2
Sharp bounds on m(n, Rk,.,k) d)
We proved that rn(nR.kld)
20(k) In this section, we prove matching lower bounds
for n> 2(-1)/20
on "("d I'd)
-
Lemma 83. For all s, n> 1, m(sn, R'.k, d) > (d- Fm(n, Rk~.k-, d).
56
...
x s matrix with ones at the coordinates (i, ...
1 and zeroes everywhere else, so that M has
(
, id)
where
d-
d-2)
ones. Let N be an n x - - - x n matrix that avoids Rhi-l,---k-i as an interval minor
d) ones. It suffices to prove that M 0 N avoids Rh'--.k as an
and has m(n, Rhi.1,
interval minor.
Suppose for contradiction that M 0 N contains R k as an interval minor. Let
.
such that r* and r 1 . have no coordinates
trary 1-entry r* in R-..k, other than r,.
and r* are represented by ei and e 2 in MON, respecin common. Suppose that Ti.
tively. In particular, suppose that el and e 2 are in the S-submatrices S(ii, ... , id) and
+jd. Since each coordinate
- id = Ji+
. .. , j), respectively. Note that iI +
S0,
then
of r* is greater than each coordinate of r 1 . in an interval minor copy of Rh.,
each coordinate of e 2 must also be greater than each coordinate of el in M 0 N, and
hence i, < j for T = 1, 2, ... ,d. It then follows fromZ1 + -- + id = ji +
+ id
for T =1, 2,... , d, i.e., the two entries ei and e 2 must be in the same
that iT =
S-submatrix in M 0 N.
Since r* can be any arbitrary 1-entry in Rh'..hk with no coordinates in common with
ri.,
a single S-submatrix contains R-li---k-1 as an interval minor. However, this
is a contradiction since each nonzero S-submatrix in M 0 N is a copy of N, which
avoids Rh-1,..-'k-1 as an interval minor. Thus M 0 N avoids Rhk-.k as an interval
minor.
Lemma 84. For every t > 1 and n > t, "n'f.k"d) >
__
Proof. Suppose that n
sd-1 m(t,R k-l.-k-1,d)
1
(st+r)d(d-i)!
>
-
=
st + r, with 0 < r < t. Then "
__
sd-1 n(t,Rk-l,...,k-1,d)
((s+1)t)d-1
(d-i)!
>
i
-
2d-1(d-1)!
Corollary 85. For n > 2(k-i)/20 mnR.d
Proof. Let N
-
2 (k-)/20.
2
27- '(d-i1)!
-
Then for n > N,
m(tR-l..- k1,
t-
)
arbian
pick
and
minimal,
coordinate
every
with
Rh.-...'
of
1-entry
corner
the
be
Proof. Let M be a s x
= s + d21 + - - +d
> m(stR
',Rkkd)
n
m(t,Rk-1 ,-k-1,d)
td-1
k1
(st+r)d
,d) >
-I
by Lemma 83.
2
mnRk.>d
nI
-
1
2d-(d-)!
m(N,Rk-1--Nd-'
,-1d)
El
P(h)
by Lemma 82 and Lemma 84.
2
Upper bounds for d-dimensional double permutation matrices
5.3
In this section, we bound the extremal functions of double permutation matrices.
Suppose that P is a permutation matrix. A matrix P D Rhi .--. kd is called a j-tuple
permutation matrix generated by P if one of k, ... , kd is equal to j and the rest are
equal to 1.
Let F(n, j, k, d) = maxm f(n, Al, d), such that the maximum ranges over all ddimensional j-tuple permutations matrices M that are generated by d-dimensional
k x
...
x k permutation matrices.
57
Theorem 86. For all fixed d, j > 2, F(n, j, k, d) = 2
We first reduce the problem to the case
j
n(k)rd.
= 2 by applying the following lemma.
Lemma 87. [28] F(n, 2, k, d) < F(n, j, k, d) < (j - 1)F(n, 2, k, d) for j > 2.
By Lemma 87, it suffices to prove an upper bound on f(n, P, d), where P is any
d-dimensional double permutation matrix of size 2k x k x ...x k.
Suppose that A is an arbitrary d-dimensional kn x ... x kn matrix that avoids P.
The S-submatrices of A are constructed by dividing A into nd disjoint submatrices
of size k x ... x k and labeling these submatrices as S(ii,. . . , is). The submatrix
S(ii, . .. , id) is defined to contain the entries aJl,.,d of A such that k(ie - 1) + 1 <
je
kIf for each e =1..
d.
The contraction matrix of A is defined to be the d-dimensional n x ... x n matrix
Co=.(ci2....id) such that cih..,id
1 if S(ii,z 2 ,... ,i) is a nonzero matrix and
C.lh .... i = 0 if S(ii, i2, ... , id) is a zero matrix.
We construct a d-dimensional 'n x - - - x n zero-one matrix Q=(q..d). Each
entry q, . d is defined based on the S-submatrices of A.
.
d,
= 0 if S(ii, ... , id) is a zero matrix.
,..,id = 1 if S(ii, i 2 , . ..., i) is a nonzero matrix and S(1, i2 ,
1, i 2 , ... , id) are all zero matrices.
3. Let x be the largest integer less than i1 for which
, id),
. . .
S(ii
. . .,
-
1. qji
1. Then define
qxi.,'=d
= 1 if the matrix consisting of the submatrices S(x, i 2 , .
. . , Zi),
,...
S(ii, i 2 , . . . , i) contains at least two 1-entries in the same 1-row, and qj, .
otherwise.
q.,
-
Q
avoids P.
Proof. This follows from the definition of Q.
El
For each 1-entry qsii2,.., = 1 of Q, we define the chunk C*(ii, i2 , ...
is a collection of consecutive S-submatrices.
, id),
which
1. If qa1,. . gd = 1 and it is the smallest integer greater than i 1 such that qi,. . '1
1, then the chunk C*(ii,i 2, .. . ,i) consists of the submatrices S(ii, i2 ,. ..,),
17,12,* .. , iid).
2. If qlh,...,d =
*(',
'2, ...
1 and if there is no i" > i1 such that qij
, id) consists of the submatrices S(ii, i 2 ,.
. .. .
. . ,isz), . . . ,
=.
S(r, i 2 ,
1, then
..
.
..,S(i4 -
-
Lemma 88.
=0
We call a chunk wide if the chunk has at least two ones in the same 1-row. Every
wide chunk has a single non-empty S-submatrix.
We call a chunk j-tall, where j = 2, 3,.. ., d, if each of its j-cross sections contains a one. The (d - 1)-dimensional matrix A' = (m 1. . . . . . . . . id) is called the
j-remainder of the d-dimensional matrix M = (mi,. , id) under the condition that
M/=
1 if and only if there exists i1 such that
,
1.
58
Lemma 89. The number of wide chunks of A is at most F(n, 1, k, d).
Proof. If A had more than F(n, 1, k, d) wide chunks, then the entries in those wide
F1
chunks would form a copy of P. This would be a contradiction.
Lemma 90. For eachj 2,3,.. ., d and each m = 1, ... ,n, A has at most F(n, 1+
kd- 2 , k, d - 1) total j-tall chunks C*(ii,. . . , ij-1, m, i+ 1 ,. - ,).
)
)
,
,
+
Proof. Suppose for contradiction that A has r chunks C*,C*,..., C,*, where r >
F(n, 1 + kd-2, k, d - 1), of the form C*(ii, ... , jj 1 , m, ij+1 , - . , id) that have ones in
all of their j-cross sections. Let S1, S2,..., , be the initial S-submatrices of the
, C, respectively. Let A' be the matrix formed by changing all
chunks C*, C2*,ones in A that do not lie in S1,..., S, to zeroes. Let C be the contraction matrix of
the j-remainder of A'. Then C is a (d - 1)-dimensional n x ... x n matrix and it has
r ones, so it contains every (1 + kd- 2 )-tuple (d - 1)-dimensional permutation matrix
of dimensions (1 + kd- 2 )k x k x ... x k.
Since P is a d-dimensional double permutation matrix of size 2k x k x ... x k and
j # 1, the j-remainder of P is a (d - 1)-dimensional double permutation matrix of
size 2k x k x ... x k. Let P' be the (d - 1)-dimensional (1 + kd- 2 )-tuple permutation
matrix of size (1 + kd- 2 )k x k x - -- x k such that P' and the j-remainder of P are
generated from the same (d - 1)-dimensional permutation matrix.
For each pair of ones in a 1-row of P with coordinates (x1, x 2 , ... , Xd) and (x1
1, x2 , . . , Xd), P' has 1 + kd- 2 ones with coordinates (sfi, x 2 , ...
- -,
(ij + k 2 ,1x2
Xj1, +1,.,d),
Xj_1, Xj+1, . , Xe), (-1 + 1, x 2 ,.... ,
. . . , x- _1 , xj+1 , - - - , Xd) in a single 1-row. Since C contains P', each set of (1+ k 2
ones in P' is represented by ones with coordinates (t,(A), t 2 ,..., t._1,
2
tj +,..., td), where A = 1, 2,...,1 + kd- , in the same 1-row of C.
Let S(t 1 (A), t 2 ,..,t_1,M, tj+1, . .. td), for each 1 < A < 1 + kd-2, be the corresponding S-submatrices of A. By the construction, these S-submatrices are the
initial S-submatrices of a subset D of the chunks C1, . .. , C,*. Each of the (1 + kd- 2
chunks in D has ones in every j-cross section; in particular each chunk in D has a
one with the same Jth coordinate (M - 1)k + xj. Among all of the chunks in D, there
are at least 1 + kd- 2 ones with jth coordinate (m - 1)k + xj. Moreover there are kd-2
< k, there
Kj
1-rows in each j-cross section of the chunks in D. Thus for each 1 <
1-row
and
have Jth
are
in
the
same
in
D
that
the
chunks
exists a pair of ones among
coordinates equal to (m - 1)k + xj, by the pigeonhole principle.
Hence, for each pair of ones in the same 1-row of P, there is a corresponding pair
E
of ones in the same 1-row of A. Thus A contains P, a contradiction.
Next we prove a recursive inequality for F(n, 2, k, d) that will imply the desired
upper bounds.
Lemma 91. Let d, s, n be positive integers with d > 2. Then F(kn, 2, k, d) <
(d - 1)nkd-lF(n, 1 + kd-2, k, d - 1) + kdF(n, 1, k, d) + (k - 1)d-lF(n, 2, k, d).
Proof. We count the maximum possible number of 1-entries in A by considering three
cases.
59
Case 1: chunk has two ones in the same 1-row
Such a chunk has only one nonempty S-submatrix, so it has at most kd ones. By
Lemma 89, there are at most F(n, 1, k, d) wide chunks, so wide chunks contain at
most kdF(n, 1, k, d) nonzero entries.
+
Case 2: chunk is j-tall and has no ones in the same 1-row
There are (d - 1) choices for j since j
2, 3,..., d. For each j, the integer m of
Lemma 90 can be 1,.. ., n. A j-tall chunk with no ones in the same row has at most
k-1 ones. For each pair of j and m, there are at most F(n, 1 + kd-2, k, d - 1) j-tall
chunks by Lemma 90. In total, j-tall chunks contain at most (d - 1)nk d-F(n, 1
kd-2, k, d - 1) ones.
Case 3: chunk is not j-tall and has no two ones in the same 1-row
Such a chunk has at most (k - I)d-1 ones. By the definition of a chunk, the
number of chunks is equal to the number of ones in matrix Q. by Lemma 88, Q has
at most F(n, 2, k, d) ones. There are at most (k - 1)diF(n, 2, k, d) ones in chunks of
this type.
Summing the maximum possible number of ones for all cases proves the lemma.
Proof of Theorem 86. We induct on d > 2. The base case of d = 2 is true by Theorem
72. We assume for some d > 2 that F(n,j,k,d 1) = 2O(k)rid 2 for each fixed j
1,
and we prove that F(n, j, k, d) - 2 0(k)nd-1 for each fixed j > 1.
First we simplify the inequality in Lemma 91. The inductive assumption im-
plies that F(n, 1 + kd 2 , k, d - 1) = 20 (k)nr-
2
. We also proved that F(n, 1, k, d)
20(k)nd-1. Hence, there is a sufficiently large constant c such that the sum of the
last two terms on the right hand side of the inequality in Lemma 91 is bounded by
2cknd-1. Therefore, F(kn, 2, k, d) < (k - 1)d F(n, 2, k, d) + 2 "knd-1 for all n.
We use a strong induction on n to prove that F(n, 2, k, d) < k(2ck+dk)n d for all
n. The base case of n < k is trivial. If F(n, 2, k, d) < k(2 ck + dk)nd-1 for all n < m,
then we show that F("1, 2, k, d) < k(2k + dk)md-1.
Let N be the maximum integer that is less than m and divisible by k. A ddimensional m x ... x m matrix has at most md - Nd < Md - (m - k)d < dkmdmore entries than an N x N x ... x N d-dimensional matrix. Thus F(m, 2, k, d) <
F(N, 2, k, d) + dkn 1 . Since F(kn, 2, k, d) < (k
1)d-lF(n, 2, k, d) + 2cknd-1 for all
n, then
F(m, 2, k, d)
<
(k - 1)d- F
(,
2,k,d
+2
ck
k
(k
(k - 1)d-1k( 2ck + dk) (N)k
<
<
(k - 1)( 2ck + dk) Nd- + ( 2
k(2ck + dk)md-1
+2
k
k
-
1
+ dkmd-1
+ dk)md-1
,
<
+ dkmd
-
where we used the strong inductive assumption in the second inequality.
60
L
5.4
Open Problems
Since there has not been much research on d-dimensional 0--1 matrices outside of [34],
bounds on f(n, P, d) are unknown for most d-dimensional 0 - 1 matrices P. Besides
finding bounds on f(n, P, d) for specific d-dimensional 0 - 1 matrices P, there are a
few more general questions about f(n, P, d) that are left unresolved by our work.
Question 92. Does the sequence
matrix P?
{f ("i d} converge
Question 93. Does the sequence {,,Pd)
matrix P?
for every d-dimensional 0 - 1
converge for every d-dimensional 0 - 1
Question 94. What is the maximum possible value of
k x .. x k permutation matrices P?
61
f (?d)
n
for all d-dimensional
62
Bibliography
[1] P.K. Agarwal and M. Sharir. Davenport-Schinzel sequences and their geometric
applications. Cambridge University Press, Cambridge, 1995.
[2] P.K. Agarwal, M. Sharir, and P. Shor. Sharp upper and lower bounds for the
length of general Davenport-Schinzel sequences. Journal of CombinatorialThe-
ory Series A, 52:228-274, 1989.
131
N. Alon, H. Kaplan, G. Nivasch, M. Sharir, and S. Smorodinsky. Weak epsilon-
nets and interval chains. J. ACM, 55 (2008).
[4] D. Bienstock and E. Gy6ri, An extremal problem on sparse 0-1 matrices, SIAM
J. Discrete Math. 4 (1991), 17-27.
[5] M. B6na, Combinatorics of Permutations, second edition, CRC Press-Chapman
Hall, 2012.
[6] J. Cibulka, On constants in the Ffiredi-Hajnal and the Stanley-Wilf conjecture,
J. Combin. Theory Ser. A, 116 (2) (2009), 290-302.
[71 J. Cibulka and J. Kyncl. Tight bounds on the maximum size of a set of permutations with bounded vc-dimension. Journal of Combinatorial Theory Series A,
119 (2012) 1461-1478.
[8] H. Davenport and A. Schinzel. A combinatorial problem connected with differential equations. American Journal of Mathematics, 87:684-694, 1965.
[91 A. M. Dean, W. Evans, E. Gethner, J. D. Laison, M. A. Safari, W. T. Trotter, Bar
k-visibility graphs. J. Graph Algorithms and Applications, 11(1): 45-59 (2007).
[10] P. Erd6s and L. Moser, Problem 11, Canadian Math. Bull. 2 (1959), 43.
[111 P. Erdos and J. Spencer, Probabilistic Methods in Combinatorics, Academic
Press, New York, 1974.
[12] E. Ezra, B. Aronov, and M. Sharir. Improved bound for the union of fat triangles. Proceedings of the twenty-second ACM-SIAM Symposium on Discrete
Algorithms, 1778-1785, 2011.
63
[13] M. Fekete, Uber die Verteilung der Wurzeln bei gewissen algebraischen Gleichungen mit. ganzzahligen Koeffizienten, Mathematische Zeitschrift 17 (1) (1923),
228-249.
[14] J.
Fox,
Combinatorics
of
permutation,
http://www.math.uni-
frankfurt.de/dm2014/Fox.pdf, 2014
[15] J. Fox, Stanley-Wilf limits are typically exponential. CoRR abs/1310.8378 (2013)
[16] J. Fox, J. Pach, and A. Suk. The number of edges in k-quasiplanar graphs. SIAM
Journal of Discrete Mathematics, 27:550-561, 2013.
[171 R. Fulek, Linear bound on extremal functions of some forbidden patterns in 0-1
matrices. Discrete Mathematics 309(6): 1736-1739 (2009).
[18] Z. Furedi, The maximum number of unit distances in a convex n-gon, Journal of
Combinatorial Theory Ser. A 55 (2), 316-320 (1990).
[19] Z. Furedi, P. Hajnal, Davenport-Schinzel theory of matrices, Discrete Mathemat-
ics, v.103 n.3, p. 2 3 3 - 2 5 1 (1992)
[201 J.T. Geneson, Extremal functions of forbidden double permutation matrices,
Journal of Combinatorial Theory Series A, v.116 Issue 7, 1235-1244 (2009)
[211 J. Geneson, Improved bounds on maximum sets of letters in sequences with
forbidden alternations. CoRR abs/1401.0063 (2014)
[22] J. Geneson, R. Prasad, and J. Tidor. Bounding Sequence Extremal Functions
with Formations. Electr. J. Comb. 21(3): P3.24 (2014)
[23] J. Geneson and L. Shen, Linear bounds on matrix extremal functions using
visibility hypergraphs. CoRR abs/1410.3147 (2014)
[24] J. Geneson and P. Tian, Improved bounds on extremal functions of forbidden
multidimensional 0 - 1 matrices. Preprint. (2014)
[25] A. Hesterberg,
preprint, 2009.
Extremal functions of excluded block permutation matrices,
[26] A. Hesterberg, Extremal functions of excluded tensor products of permutation
matrices, Discrete Math., 312 (10) (2012), 1646-1649.
[27] S. Heubach and T. Mansour, Combinatorics of Compositions and Words, CRC
Press, 2009.
[28] B. Keszegh, On linear forbidden submatrices, Journal of Combinatorial Theory
Series A, v.116 n.1, p. 2 3 2 - 2 4 1 (2009)
[29] S. Kitaev, Patterns in Permutations and Words, Springer-Verlag, 2011.
64
[30j M. Klazar. A general upper bound in the extremal theory of sequences. Commentationes Mathematicae Universitatis Carolinae, 33:737-746, 1992.
[31J M. Klazar, Extremal problems for ordered (hyper) graphs: applications of
Davenport-Schinzel sequences, European J. Combin. 25 (2004) 125-140.
[321 M. Klazar. On the maximum lengths of Davenport-Schinzel sequences. Contemporary trends in discrete mathematics (DIMACS series in discrete mathematics
and theoretical computer science), 49:169-178, 1999.
[331 M. Klazar, The Fiiredi-Hajnal conjecture implies the Stanley-Wilf conjecture,
in: D. Krob, A.A Mikhalev, A.V. Mikhalev (Eds), Formal Power Series and
Algebraic Combinatorics, Springer, Berlin, 2000, 250-255.
[34] M. Klazar, A. Marcus, Extensions of the linear bound in the Furedi-Hajnal con-
jecture, Adv. in Appl. Math. 38, no. 2, 258-266 (2006)
[351 T. K6vdri, V. T. S6s and P. Turin, On a problem of K. Zarankiewicz. Colloquium
Math. 3 (1954), 50-57.
[361 A. Marcus, G. Tardos, Excluded permutation matrices and the Stanley-Wilf conjecture, Journal of Combinatorial Theory Series A, v.107 n.1, p. 1 5 3 - 1 6 0 (2004)
[371 J. Mitchell: Shortest rectilinear paths among obstacles, Department of Operations Research and Industrial Engineering Technical Report No. 739, Cornell
University, Ithaca, New York (1987)
[381 G. Nivasch. Improved bounds and new techniques for Davenport-Schinzel sequences and their generalizations. Journal of the A CM, 57: article 17, 2010.
[391 J. Pach, G. Tardos, Forbidden paths and cycles in ordered graphs and matrices,
Israel J. Math. 155, 309-334 (2006)
[40J S. Pettie. Degrees of nonlinearity in forbidden 0-1 matrix problems. Discrete
Mathematics 311 (2011) 2396-2410.
[411 S. Pettie. On the structure and composition of forbidden sequences, with geometric applications. Proceedings of the twenty-seventh annual symposium on
computational geometry, 370-379, 2011.
[421 S. Pettie. Sharp bounds on Davenport-Schinzel sequences of every order. Symposium on Computational Geometry (2013): 319-328.
[431 R. P. Stanley, Increasing and decreasing subsequences and their variants, in:
Proceedings of the International Congress of Mathematicians, Plenary Lectures,
vol. I, Madrid, Spain, 2006, 545-579.
[441 E. Steingrimsson, Some open problems on permutation patterns, in: Surveys in
combinatorics 2013, London Math. Soc. Lecture Note Ser. Cambridge University
Press, 2013, 239-263.
65
[45] A. Suk and B. Walczak. New bounds on the maximum number of edges in kquasi-planar graphs. Twenty-first InternationalSymposium on Graph Drawing,
95-106, 2013.
[46] R. Sundar. On the Deque conjecture for the splay algorithm. Combinatorica
12(1): 95-124 (1992).
66