Zarankiewicz problem explained

The Zarankiewicz problem, an unsolved problem in mathematics, asks for the largest possible number of edges in a bipartite graph that has a given number of vertices and has no complete bipartite subgraphs of a given size.[1] It belongs to the field of extremal graph theory, a branch of combinatorics, and is named after the Polish mathematician Kazimierz Zarankiewicz, who proposed several special cases of the problem in 1951.[2]

Problem statement

G=(U\cupV,E)

consists of two disjoint sets of vertices

U

and

V

, and a set of edges each of which connects a vertex in

U

to a vertex in

V

. No two edges can both connect the same pair of vertices. A complete bipartite graph is a bipartite graph in which every pair of a vertex from

U

and a vertex from

V

is connected to each other. A complete bipartite graph in which

U

has

s

vertices and

V

has

t

vertices is denoted

Ks,t

. If

G=(U\cupV,E)

is a bipartite graph, and there exists a set of

s

vertices of

U

and

t

vertices of

V

that are all connected to each other, then these vertices induce a subgraph of the form

Ks,t

. (In this formulation, the ordering of

s

and

t

is significant: the set of

s

vertices must be from

U

and the set of

t

vertices must be from

V

, not vice versa.)

The Zarankiewicz function

z(m,n;s,t)

denotes the maximum possible number of edges in a bipartite graph

G=(U\cupV,E)

for which

|U|=m

and

|V|=n

, but which does not contain a subgraph of the form

Ks,t

. As a shorthand for an important special case,

z(n;t)

is the same as

z(n,n;t,t)

. The Zarankiewicz problem asks for a formula for the Zarankiewicz function, or (failing that) for tight asymptotic bounds on the growth rate of

z(n;t)

assuming that

t

is a fixed constant, in the limit as

n

goes to infinity.

For

s=t=2

this problem is the same as determining cages with girth six. The Zarankiewicz problem, cages and finite geometry are strongly interrelated.[3]

The same problem can also be formulated in terms of digital geometry. The possible edges of a bipartite graph

G=(U\cupV,E)

can be visualized as the points of a

|U| x |V|

rectangle in the integer lattice, and a complete subgraph is a set of rows and columns in this rectangle in which all points are present. Thus,

z(m,n;s,t)

denotes the maximum number of points that can be placed within an

m x n

grid in such a way that no subset of rows and columns forms a complete

s x t

grid. An alternative and equivalent definition is that

z(m,n;s,t)

is the smallest integer

k

such that every (0,1)-matrix of size

m x n

with

k+1

ones must have a set of

s

rows and

t

columns such that the corresponding

s x t

submatrix is made up only of 1s.

Examples

The number

z(n;2)

asks for the maximum number of edges in a bipartite graph with

n

vertices on each side that has no 4-cycle (its girth is six or more). Thus,

z(2;2)=3

(achieved by a three-edge path), and

z(3;2)=6

(a hexagon).

In his original formulation of the problem, Zarankiewicz asked for the values of

z(n;3)

for

n=4,5,6

. The answers were supplied soon afterwards by Wacław Sierpiński:

z(4;3)=13

,

z(5;3)=20

, and

z(6;3)=26

.[4] The case of

z(4;3)

is relatively simple: a 13-edge bipartite graph with four vertices on each side of the bipartition, and no

K3,3

subgraph, may be obtained by adding one of the long diagonals to the graph of a cube. In the other direction, if a bipartite graph with 14 edges has four vertices on each side, then two vertices on each side must have degree four. Removing these four vertices and their 12 incident edges leaves a nonempty set of edges, any of which together with the four removed vertices forms a

K3,3

subgraph.

Upper bounds

The Kővári–Sós–Turán theorem provides an upper bound on the solution to the Zarankiewicz problem. It was established by Tamás Kővári, Vera T. Sós and Pál Turán shortly after the problem had been posed:

z(m,n;s,t)<(s-1)1/t(n-t+1)m1-1/t+(t-1)m.

Kővári, Sós, and Turán originally proved this inequality for

z(n;t)

.[5] Shortly afterwards, Hyltén-Cavallius observed that essentially the same argument can be used to prove the above inequality.[6] An improvement on the second term of the upper bound on

z(n;t)

was given by Štefan Znám:[7]

z(n;t)<(t-1)1/tn2-1/t+

1
2

(t-1)n+1.

If

s

and

t

are assumed to be constant, then asymptotically, using the big O notation, these formulae can be expressed as

z(m,n;s,t)=O(mn1-1/s+n)

z(m,n;s,t)=O(nm1-1/t+m)

.In the particular case

m=n

, assuming without loss of generality that

s\leqt

, we have the asymptotic upper bound

z(n,n;s,t)=O(n2-1/s).

Lower bounds

One can verify that among the two asymptotic upper bounds of

z(m,n;s,t)

in the previous section, the first bound is better when

m=o(ns/t)

, and the second bound becomes better when

m=\omega(ns/t)

. Therefore, if one can show a lower bound for

z(ns/t,n;s,t)

that matches the upper bound up to a constant, then by a simple sampling argument (on either an

nt/s x t

bipartite graph or an

m x ms/t

bipartite graph that achieves the maximum edge number), we can show that for all

m,n

, one of the above two upper bounds is tight up to a constant. This leads to the following question: is it the case that for any fixed

s\leqt

and

m\leqns/t

, we have

z(m,n;s,t)=\Omega(mn1-1/s)

? [8]

In the special case

m=n

, up to constant factors,

z(n,n;s,t)

has the same order as

ex(n,Ks,t)

, the maximum number of edges in an

n

-vertex (not necessarily bipartite) graph that has no

Ks,t

as a subgraph. In one direction, a bipartite graph with

n

vertices on each side and

z(n,n;s,t)

edges must have a subgraph with

n

vertices and at least

z(n,n;s,t)/4

edges; this can be seen from choosing

n/2

vertices uniformly at random from each side, and taking the expectation. In the other direction, we can transform a graph with

n

vertices and no copy of

Ks,t

into a bipartite graph with

n

vertices on each side of its bipartition, twice as many edges and still no copy of

Ks,t

, by taking its bipartite double cover.[9] Same as above, with the convention that

s\leqt

, it has been conjectured that

z(n,n;s,t)=\Theta(n2-1/s)

for all constant values of

s,t

.[10]

For some specific values of

s,t

(e.g., for

t

sufficiently larger than

s

, or for

s=2

), the above statements have been proved using various algebraic and random algebraic constructions. At the same time, the answer to the general question is still unknown to us.

Incidence graphs in finite geometry

For

s=t=2

, a bipartite graph with

n

vertices on each side,

\Omega(n3/2)

edges, and no

K2,2

may be obtained as the Levi graph, or point-line incidence graph, of a projective plane of order

q

, a system of

q2+q+1

points and

q2+q+1

lines in which each two points determine a unique line, and each two lines intersect at a unique point. We construct a bipartite graph associated to this projective plane that has one vertex part as its points, the other vertex part as its lines, such that a point and a line is connected if and only if they are incident in the projective plane. This leads to a

K2,2

-free graph with

q2+q+1

vertices and

(q2+q+1)(q+1)

edges. Since this lower bound matches the upper bound given by I. Reiman,[11] we have the asymptotic [12]

z(n;2)=(1/2+o(1))n3/2.

For

s=t=3

, bipartite graphs with

n

vertices on each side,

\Omega(n5/3)

edges, and no

K3,3

may again be constructed from finite geometry, by letting the vertices represent points and spheres (of a carefully chosen fixed radius) in a three-dimensional finite affine space, and letting the edges represent point-sphere incidences.[13]

More generally, consider

s=2

and any

t

. Let

Fq

be the

q

-element finite field, and

h

be an element of multiplicative order

t

, in the sense that

H=\{1,h,...,ht-1\}

form a

t

-element subgroup of the multiplicative group
*
F
q
. We say that two nonzero elements

(a,b),(a',b')\inFq x Fq

are equivalent if we have

a'=hda

and

b'=hdb

for some

d

. Consider a graph

G

on the set of all equivalence classes

\langlea,b\rangle

, such that

\langlea,b\rangle

and

\langlex,y\rangle

are connected if and only if

ax+by\inH

. One can verify that

G

is well-defined and free of

K2,t+1

, and every vertex in

G

has degree

q

or

q-1

. Hence we have the upper bound[14]

z(n,n;2,t+1)=(t1/2+o(1))n3/2.

Norm graphs and projective norm graphs

For

t

sufficiently larger than

s

, the above conjecture

z(n,n;s,t)=\Theta(n2-1/s)

was verified by Kollár, Rónyai, and Szabó[15] and Alon, Rónyai, and Szabó[16] using the construction of norm graphs and projective norm graphs over finite fields.

For

t>s!

, consider the norm graph NormGraphp,s with vertex set
F
ps
, such that every two vertices
a,b\inF
ps
are connected if and only if

N(a+b)=1

, where
N\colonF
ps

Fp

is the norm map

N(x)=xxp

p2
x

ps-1
x
(ps-1)/(p-1)
=x

.

It is not hard to verify that the graph has

ps

vertices and at least

p2s-1/2

edges. To see that this graph is

Ks,s!+1

-free, observe that any common neighbor

x

of

s

vertices

y1,\ldots,ys\inF

ps
must satisfy

1=N(x+yi)=(x+yi)

p
(x+y
i)
ps-1
(x+y
i)

=(x+yi)

p)
(x
i
ps-1
(x
ps-1
+y
i

)

for all

i=1,\ldots,s

, which a system of equations that has at most

s!

solutions.

The same result can be proved for all

t>(s-1)!

using the projective norm graph, a construction slightly stronger than the above. The projective norm graph ProjNormGraphp,s is the graph on vertex set
F
ps-1

x

x
F
p
, such that two vertices

(X,x),(Y,y)

are adjacent if and only if

N(X+Y)=xy

, where
N\colonF
ps

Fp

is the norm map defined by
(ps-1)/(p-1)
N(x)=x
. By a similar argument to the above, one can verify that it is a

Ks,t

-free graph with

\Omega(n2-1/s)

edges.

The above norm graph approach also gives tight lower bounds on

z(m,n;s,t)

for certain choices of

m,n

.In particular, for

s\geq2

,

t>s!

, and

n1/t\leqm\leqn1+1/t

, we have

z(m,n;s,t)=\Theta(mn1-1/s).

In the case

m=(1+o(1))n1+1/s

, consider the bipartite graph

G

with bipartition

V=V1\cupV2

, such that

V1=F

pt

x

x
F
p
and

V2=F

pt
. For

A\inV1

and

(B,b)\inV2

, let

A\sim(B,b)

in

G

if and only if

N(A+B)=b

, where

N()

is the norm map defined above. To see that

G

is

Ks,t

-free, consider

s

tuples

(B1,b1),\ldots,(Bs,bs)\inV1

. Observe that if the

s

tuples have a common neighbor, then the

Bi

must be distinct. Using the same upper bound on he number of solutions to the system of equations, we know that these

s

tuples have at most

s!<t

common neighbors.

Clique partitions

Using a related result on clique partition numbers, Alon, Mellinger, Mubayi and Verstraëte [17] proved a tight lower bound on

z(m,n;2,t)

for arbitrary

t

: if

m=(1+o(1))nt/2

, then we have

z(m,n;2,t)=(1+o(1))mn1/2

.For

2\leqt\leqn

, we say that a collection of subsets

A1,...,A\ell\subset[n]

is a clique partition of

H\subset{[n]\chooset}

if
\ell{A
cup
i\choose

t}

form a partition of

H

. Observe that for any

k

, if there exists some

H\subset{[n]\chooset}

of size

(1-o(1)){n\chooset}

and

m=(1+o(1)){n\chooset}/{k\chooset}

, such that there is a partition of

H

into

m

cliques of size

k

, then we have

z(m,n;2,t)=km

. Indeed, supposing

A1,...,Am\subset[n]

is a partition of

H

into

m

cliques of size

k

, we can let

G

be the

m x n

bipartite graph with

V1=\{A1,...,Am\}

and

V2=[n]

, such that

Ai\simv

in

G

if and only if

v\inAi

. Since the

Ai

form a clique partition,

G

cannot contain a copy of

K2,t

.

It remains to show that such a clique partition exists for any

m=(1+o(1))nt/2

. To show this, let

Fq

be the finite field of size

q

and

V=Fq x Fq

. For every polynomial

p()

of degree at most

t-1

over

Fq

, define

Cp=\{(x,p(x)):x\inFq\}\subsetV

. Let

lC

be the collection of all

Cp

, so that

|lC|=qt=nt/2

and every

Cp

has size

q=\sqrt{n}

. Clearly no two members of

lC

can share

t

members. Since the only

t

-sets in

V

that do not belong to

H

are those that have at least two points sharing the same first coordinate, we know that almost all

t

-subsets of

V

are contained in some

Cp

.

Randomized algebraic constructions

Alternative proofs of

ex(n,Ks,t)=\Omega(n2-1/s)

for

t

sufficiently larger than

s

were also given by Blagojević, Bukh and Karasev[18] and by Bukh [19] using the method of random algebraic constructions. The basic idea is to take a random polynomial
s x
f:F
q
s
F
q

Fq

and consider the graph

G

between two copies of
s
F
q
whose edges are all those pairs

(x,y)

such that

f(x,y)=0

.

To start with, let

q

be a prime power and

n=q2

. Let

f\inFq[x1,...,xs,y1,...,ts]

\leqs2
be a random polynomial with degree at most

s2

in

X=(x1,...,xs)

, degree at most

s2

in

Y=(y1,...,ys)

, and furthermore satisfying

f(X,Y)=f(Y,X)

for all

X,Y

. Let

G

be the associated random graph on vertex set
s
F
q
, such that two vertices

x

and

y

are adjacent if and only if

f(x,y)=0

.

To prove the asymptotic lower bound, it suffices to show that the expected number of edges in

G

is

\Omega(q2s-1)

. For every

s

-subset
s
U\subsetF
q
, we let

ZU

denote the vertex subset of
s\setminus
F
q

U

that "vanishes on

f(,U)

":

ZU=\{x\in

s\setminus
F
q

U:f(x,u)=0forallu\inU\}

. Using the Lang-Weil bound for polynomials

f(,u)

in
s
F
q
, we can deduce that one always has

ZU\leqC

or

ZU>q/2

for some large constant

C

, which implies

P(|ZU|>C)=P(|ZU|>q/2)

.Since

f

is chosen randomly over

Fq

, it is not hard to show that the right-hand side probability is small, so the expected number of

s

-subsets

U

with

|ZU|>C

also turned out to be small. If we remove a vertex from every such

U

, then the resulting graph is

Ks,C+1

free, and the expected number of remaining edges is still large. This finishes the proof that

ex(n,Ks,t)=\Omega(n2-1/s)

for all

t

sufficiently large with respect to

s

. More recently, there have been a number of results verifying the conjecture

z(m,n;s,t)=\Omega(n2-1/s)

for different values of

s,t

, using similar ideas but with more tools from algebraic geometry.[20]

Applications

The Kővári–Sós–Turán theorem has been used in discrete geometry to bound the number of incidences between geometric objects of various types. As a simple example, a set of

n

points and

m

lines in the Euclidean plane necessarily has no

K2,2

, so by the Kővári–Sós–Turán it has

O(nm1/2+m)

point-line incidences. This bound is tight when

m

is much larger than

n

, but not when

m

and

n

are nearly equal, in which case the Szemerédi–Trotter theorem provides a tighter

O(n2/3m2/3+n+m)

bound. However, the Szemerédi–Trotter theorem may be proven by dividing the points and lines into subsets for which the Kővári–Sós–Turán bound is tight.[21]

See also

Notes and References

  1. . Reprint of 1978 Academic Press edition, .
  2. . As cited by .
  3. Web site: Archived copy . 2014-09-16 . 2016-03-04 . https://web.archive.org/web/20160304024714/http://www.cs.elte.hu/~hetamas/publ/DHSzFIN.pdf . dead .
  4. .
  5. .
  6. . As cited by .
  7. . As cited by .
  8. .
  9. , Theorem 2.3, p. 310.
  10. , Conjecture 15, p. 312.
  11. .
  12. , Corollary 2.7, p. 313.
  13. .
  14. .
  15. .
  16. .
  17. .
  18. .
  19. .
  20. .
  21. .