Table Of ContentOn the Biclique cover of the complete graph
Farokhlagha Moazami and Nasrin Soltankhah
Department of Mathematics
Alzahra University
Vanak Square 19834 Tehran, I.R. Iran
[email protected]
3
[email protected]
1
0
2 Abstract
n
Let K be a set of k positive integers. A biclique cover of type K of a
a
J graphGisacollectionofcompletebipartitesubgraphsofGsuchthatforevery
1 edge e of G, the number of bicliques need to cover e is a member of K. If
2 K = {1,2,...,k} then the maximum number of the vertices of a complete
graph that admits a biclique cover of type K with d bicliques, n(k,d), is the
]
maximum possible cardinality of a k-neighborly family of standard boxes in
O
Rd. In this paper, we obtain an upper bound for n(k,d). Also, we show that
C
the upper bound can be improved in some special cases. Moreover, we show
.
h that the existence of the biclique cover of type K of the complete bipartite
t graphwith a perfect matching removedis equivalentto the existence ofa cross
a
m K-intersection family.
[ Key words: biclique cover- multilinear polynomial- cross K-intersection fami-
lies.
1
v Subject classification: 05C70.
0
9
7 1 Introduction
4
.
1
Throughout the paper, we consider only the simple graph. Let V(G) and E(G)
0
3 denote the vertex set and edge set of the graph G. As usual we will use the symbol
1 [n] to denote the set {1,2,...,n}. By a biclique we mean the complete bipartite
:
v graph. WewilldenotebyK− thecompletebipartitegraphwithaperfectmatching
m,m
Xi removed. A biclique cover of a graph G is a collection of bicliques of G such that
each edge of G belongs to at least one of the bicliques. In the literature, there are
r
a
several ways to define a biclique cover problem for different purposes, see [2, 3, 4,
5, 7, 8, 9, 10, 14, 13]. In this paper we consider the biclique cover of order k and in
general of type K that was introduced by Alon [2].
Definition 1. Let K be a set of k positive integers. We say that a biclique cover of
the graph G is of type K if for every edge e of the graph G, the number of bicliques
that cover e is an element of the set K. ♠
Thenumberof bicliques in a biclique cover is the size of the cover. We writen(K,d)
for the maximum number of vertices of a complete graph that admits a biclique
cover of type K and size d. In the aforementioned definition if we take the set
K = {1,2,...,k} then this biclique cover is called of order k. Alon [2] used the
concept of the biclique cover and provided a relation between the biclique cover of
1This research was in part supported bya grant from research institute for ICT.
1
order k and size d of the complete graph and k-neighborly family of standard boxes
in Rd, (see [2] for more details.) The following result is due to N. Alon [2].
Theorem A. [2] Let d be a positive integer and 1≤ k ≤ d then
1. d+1= n(1,d) ≤ n(2,d) ≤ ··· ≤ n(d−1,d) ≤ n(d,d) = 2d.
2. (d)k ≤ k−1(⌊d+i⌋+1) ≤ n(K,d) ≤ k 2i d < 2(2ed)k.
k i=0 k i=0 i k
Q P (cid:0) (cid:1)
To prove the upper bound in the second relation he constructed multilinear poly-
nomials that are linearly independent. The proof can be easily extended to the
biclique cover of type K. In this paper, we add some other linearly independent
multilinear polynomials and obtain a slightly improvement boundfor thementioned
upper bound.
Theorem 1. For 1≤ k ≤ d, assume that K is a set of k positive integers. Then
k−1
d d−1
n(K,d) ≤ 2k + 2i .
(cid:18)k(cid:19) (cid:18)i−1(cid:19)
Xi=1
Also, it was shown by Alon [2] that for K = {2,4,...,2i} there exists a biclique
cover H ,H ,...,H of type K for the complete graph on (1+ d + d +···+ d )
1 2 d 2 4 2i
vertices which improves the lower boundin TheoremA.Tosee t(cid:0)h(cid:1)is, a(cid:0)ss(cid:1)umethat(cid:0)th(cid:1)e
vertices of the complete graph is denoted by all subsets of cardinality 0,2,4,...,2i
of [d]. Construct the biclique cover H ,H ,...,H which H has (X ,Y ) as vertex
1 2 d i i i
classes such that X is the set of all subsets that contain i and Y is the set of all
i i
subsets that do not contain i. These results are far from being optimal, but the
above construction motivated us to define the following definition.
Definition 2. Suppose that {G ,...,G } is a biclique cover of the graph G where
1 d
G has (X ,Y ) as its vertex set. If for every 1 ≤ i ≤ d, X ∪Y = V(G), then this
i i i i i
biclique cover is called a regular biclique cover. ♠
We will denote by n (K,d) the maximum possible cardinality of the vertices of a
r
complete graph such that there exists a regular biclique cover of type K and size d.
We prove the following upper bound of n (K,d).
r
Theorem 2. If K is a set of k positive integers, then
d d−1
n (K,d) ≤ 2k−1 +2k−1 +2k −1.
r
(cid:18)k(cid:19) (cid:18) k (cid:19)
It is interesting to consider the biclique cover of type K of other graphs besides the
complete graph. Assume that K is a set of k positive integers and X is an arbitrary
set of d points. Suppose that A = {A ,A ,...,A } and B = {B ,B ,...,B } are
1 2 m 1 2 m
two collections of subsets of X such that |A ∩B | ∈ K for i 6= j and |A ∩B | = 0
i j i i
for every i. The pair (A,B) is called a cross K-intersection families. The following
theorem shows that a cross K-intersection family can be formulated in terms of
biclique cover of type K of the graph K− .
m,m
2
Theorem 3. Let K = {l ,l ,...,l }, there exists a cross K-intersection families
1 2 k
with m blocks on a set of d points if and only if there exists a biclique cover of type
K and size d of K− .
m,m
In [12] Snevily made the following conjecture.
Conjecture 1. [12] Let A = {A ,A ,...,A } and B = {B ,B ,...,B } be two
1 2 m 1 2 m
collections of subsets of an d-element set. Let K = {l ,l ,...,l } be a collection of k
1 2 k
positive integers. Assume that for i6= j we have |A ∩B |∈ K and that |A ∩B | = 0,
i j i i
then
d
m ≤ .
(cid:18)k(cid:19)
By Theorem 3, we can state the above conjecture in terms of the biclique cover as
follows. The maximum number of the vertices in each part of a complete bipartite
graph with a perfect matching removed that admits a biclique cover of size d is at
most d . Note that this bound is sharp by taking all k-element subsets of [d] as A
k
and a(cid:0)ll (cid:1)(d−k)-element subsets of [d] as B. In [6] William Y.C. Chen and Jiuqiang
Liu proved the following theorem.
Theorem B. [6]Letpbe aprime numberand K = {l ,l ,...,l } ⊆ {1,2,...,p−1}.
1 2 k
Assume that A = {A ,A ,...,A } and B = {B ,B ,...,B } are two collections
1 2 m 1 2 m
of subsets of X such that |A ∩B |(modp) ∈ K for i 6= j and |A ∩B | =0 for every
i j i i
i. If maxl < min{|A |(modp)|1 ≤ i≤ m}, then
j i
d−1 d−1 d−1
m ≤ + +···+ ,
(cid:18) k (cid:19) (cid:18)k−1(cid:19) (cid:18)k−2r+1(cid:19)
where r is the number of different set sizes in A.
Clearly, for a prime number p greater than d, and r = 1 the above theorem is true
and the following corollary is straightforward.
Corollary 1. [6] Let K = {l ,l ,...,l } be a set of k positive integers and max l <
1 2 k j
s. Suppose that A= {A ,A ,...,A } andB = {B ,B ,...,B } are two collections
1 2 m 1 2 m
of subsets of [d] such that |A ∩B | ∈ K for i 6= j and |A ∩B | = 0 for every i. If
i j i i
either A is s-uniform or B is s-uniform, then
d
m ≤ .
(cid:18)k(cid:19)
This corollary shows that if we have a biclique cover of type K = {l ,...,l } of a
1 k
complete bipartite graph with a perfect matching removed such that every vertex of
this graph lies in exactly s bicliques and max l < s. Then the maximum possible
j
d
cardinality of the vertices of this graph is at most . The structure of the rest
k
of this paper is to prove Theorem 1, 2, and 3. The p(cid:0)r(cid:1)oofs based on the concept of
applying linear algebra method that is used in [1, 6, 11, 12].
3
2 Proof of Theorem 1
A polynomial in n variable is called multilinear if every variable has degree 0 or
1. Observe that when each variable in a polynomial attains values 0 or 1, if each
p
variablex (p >1)isreplacedbyx ,wecanconsiderthispolynomialasamultilinear
i i
polynomial. For a subsetA of [n], thecharacteristic vector of A is the vector v =
i i Ai
(v ,...,v ), where v = 1 if j ∈ A and v = 0 otherwise. Let {H ,H ,...,H } be a
1 n j i j 1 2 d
biclique cover of type K for the graph K such that H has X and Y as its vertex
n i i i
classes. For every 1 ≤ i≤ n define
A := {j | i ∈X } & B := {j | i ∈ Y }.
i j i j
Now, with each pair (A ,B ) we associate a polynomial P (x,y) defined by:
i i i
P (x,y) = (v .x+v .y−l ).
i Ai Bi j
ljY∈K
Where v (resp. v ) is the characteristic vector of the set A (resp. B ), x =
Ai Bi i i
(x ,x ,...,x ) and y = (y ,y ,...,y ). The key property of these polynomials is
1 2 d 1 2 d
P (v ,v ) 6= 0 and P (v ,v ) = 0 for all i6= j, (1)
i Bi Ai i Bj Aj
which follows immediately from this fact that {H ,...,H } is a biclique cover of
1 d
type K. Let
A= {(M,N) | M,N ⊆[d], M ∩N = ∅, d∈ N, |M ∪N| ≤ k}.
It is easy to see that the cardinality of A is equal to k−12i d−1 . For every
i=0 i
(M,N) ∈ A, the polynomial Q(M,N)(x,y) is defined by P (cid:0) (cid:1)
Q (x,y) = x y .
(M,N) i i
iY∈M iY∈N
Throughout the paper we set x = y = 1 when A is an empty set. We
i∈A i i∈A i
will now show that the polynoQmials in theQset
P = {P (x,y) | 1 ≤ i≤ n}∪{Q (x,y) | (M,N) ∈A}
i (M,N)
, as the polynomials from {0,1}2d to R, are linearly independent. For this purpose,
we set
n
α P (x,y)+ β Q (x,y) = 0.
i i (M,N) (M,N)
Xi=1 (MX,N)∈A
We can rewrite the above equality as follows:
α P (x,y)+ α P (x,y)+ β Q (x,y) = 0. (2)
i i i i (M,N) (M,N)
dX∈Ai dX6∈Ai (MX,N)∈A
The proof will be divided into 3 steps.
4
Step 1. We begin by proving that for every i which d 6∈ A , it holds that
i
α = 0. In the contrary assume that i is a subscript such that α 6= 0 and
i 0 i0
d 6∈A . Substituting (v ,v ) in the equation 2, according to the relation 1 and
i0 Bi0 Ai0
since d ∈ N, all terms in the relation 2 but α P (v ,v ) vanish. In this way,
i0 i0 Bi0 Ai0
α P (v ,v ) = 0. Finally, as P (v ,v ) 6= 0, we have α = 0 which is a
i0 i0 Bi0 Ai0 i0 Bi0 Ai0 i0
contradiction.
Step 2. We will show that for every i which d ∈ A , it holds that α = 0.
i i
According to the step 1 we have
α P (x,y)+ β Q (x,y) = 0. (3)
i i (M,N) (M,N)
dX∈Ai (MX,N)∈A
Assume that i is a subscript such that α 6= 0. Let v′ = v −(0,...,0,1) and
0 i0 Ai0 Ai0
evaluate the equation 3 in (v ,v′ ). As A ∩B = ∅ we have that d 6∈ B so for
Bi0 Ai0 i i i
every iin theequation 3we have P (v ,v′ ) = P (v ,v ). From this andsince
i Bi0 Ai0 i Bi0 Ai0
d ∈N we conclude that α P (v ,v ) = 0, hence α = 0.
i0 i0 Bi0 Ai0 i0
Step 3. Note that Q(M′,N′)(vM′,vN′) = 1 and Q(M,N)(vM′,vN′) = 0 for any
(M,N) ∈ A with M∪N 6= M′∪N′ and |M∪N|≥ |M′∪N′|. Soall thepolynomials
oftheset{Q (x,y)|(M,N) ∈ A}arelinearlyindependent. Therefore,forevery
(M,N)
(M,N) ∈ A we have β = 0.
(M,N)
We have thus proved the polynomials of the set P as the polynomials with
domain {0,1}2d are linearly independent. So we can consider these polynomials as
multilinear polynomials. On the other hand every polynomial in the set P can be
written as a linear combination of the multilinear monomials of degree at most k.
Furthermore, they do not contain any monomials that contain both x and y for
i i
the same i. The number of such monomials are k 2i d and hence,
i=0 i
P (cid:0) (cid:1)
k−1 k
d−1 d
n+ 2i ≤ 2i .
(cid:18) i (cid:19) (cid:18)i(cid:19)
Xi=0 Xi=0
Now, by a straightforward calculation the formula of Theorem 1 will be achieved.
3 Proof of Theorem 2
Before embarking on the proof of Theorem 2, we will establish the following lemma.
Lemma 1. For every 1 ≤i ≤ d−1, let the set B define as follows:
i
B = {(I,J)|I,J ⊆ [d−i+1], I∩J = ∅, d−i+1 ∈I∪J, |I∪J|=6 d−i+1, |I∪J| ≤ k−1}.
i
Let for every pair (I,J) ∈ B ,Ri (x,y) denote the following polynomial
i (I,J)
Ri (x,y) = x y ( x + y −(d−i)).
(I,J) j j j j
Yj∈I jY∈J j6∈JX,j≤d−i j6∈IX,j≤d−i
5
Then
B = {Ri (x,y) | 1 ≤ i ≤ d−1, (I,J) ∈ B } (4)
(I,J) i
is a set of linearly independent polynomials.
Proof. To prove the assertion, assume this is false and let
γ1 R1 (x,y)+···+ γd−1 Rd−1 (x,y) = 0 (5)
(I,J) (I,J) (I,J) (I,J)
X X
(I,J)∈B1 (I,J)∈Bd−1
beanontriviallinearrelation. Supposethati isthegreatestsuperscriptand(I ,J )
0 0 0
is the subscript such that has minimum cardinality in the set B and γi0 6= 0.
i0 (I0,J0)
Substitute (v ,v ) for (x,y) in the linear relation 5. In view of the definition
I0 J0
of B all terms in the linear relation 5 but γi0 Ri0 (v ,v ) vanish. Since
i (I0,J0) (I0,J0) I0 J0
Ri0 (v ,v ) 6= 0, we have γi0 = 0. This is a contradiction which completes
(I0,J0) I0 J0 (I0,J0)
the proof. (cid:4)
Obviously, for every 1 ≤ i< d we have
k−22j+1 d−i i≤ d−k+1
j=0 j
|B |= P (cid:0) (cid:1) .
i
d−i−12j+1 d−i i≥ d−k+2
j=0 j
P (cid:0) (cid:1)
It is a well-known fact that
n
m+i m+n+1
= .
(cid:18) m (cid:19) (cid:18) m+1 (cid:19)
Xi=0
By this fact clearly, |B| = k−12j d − 2k + 2. Let A = {(M,N) | (M,N) ∈
j=1 j k
A, |M ∪ N| = k} and B isPdefined(cid:0)a(cid:1)s Lemma 1. We claim that {Pi(x,y) | 1 ≤
i ≤ n}∪{Q | (M,N) ∈ A } with all the polynomials Ri (x,y) ∈ B remain
(M,N) k (I,J)
linearly independent. Before prove the claim, we shall note that all polynomials in
the set B have this property that vanish in the point (v ,v ) for every 1 ≤ i ≤ n.
Bi Ai
Now, assume the claim is false and let
n d−1
α P (x,y)+ β Q (x,y)+ γi Ri (x,y) = 0 (6)
i i (M,N) (M,N) (I,J) (I,J)
Xi=1 (M,XN)∈Ak Xi=1(I,XJ)∈Bi
be a nontrivial linear relation.
Step 1. Let i be a subscript such that d 6∈ A and α 6= 0. Substitute
0 i0 i0
(v ,v ) for (x,y) in the linear relation 6. We know that for every i and every
Bi0 Ai0
(I,J) ∈ B , Ri (v ,v ) = 0. Also, d ∈ N so Q (v ,v ) = 0. Using
i (I,J) Bi0 Ai0 (M,N) Bi0 Ai0
these and by 1 all terms in the linear relation 6 but α P (v ,v ) vanish. Since
i0 i0 Bi0 Ai0
P (v ,v ) 6= 0, therefore α = 0.
i0 Bi0 Ai0 i0
6
Step 2. According to the step 1 we have
n d−1
α P (x,y)+ β Q (x,y)+ γi Ri (x,y) = 0.
i i (M,N) (M,N) (I,J) (I,J)
dX∈Ai (M,XN)∈Ak Xi=1(I,XJ)∈Bi
(7)
Let i be a subscript such that α 6= 0. We define v′ to be v −(0,...,0,1).
0 i0 Ai0 Ai0
Substitute (v ,v′ ) for (x,y) in the linear relation 7. For 1≤ i ≤ d−1 and every
Bi0 Ai0
pair (I,J) ∈ B , by the definition of Ri (x,y), it holds that Ri (v ,v ) =
i (I,J) (I,J) Bj Aj
Ri (v ,v′ ) for every 1 ≤ j ≤ n. Now, similar in the step 1 all terms in the
(I,J) Bj Aj
relation 7 but α P (v ,v′ ) vanish. Since P (v ,v′ )6= 0, therefore α = 0.
i0 i0 Bi0 Ai0 i0 Bi0 Ai0 i0
So, we have
d−1
β Q (x,y)+ γi Ri (x,y) = 0. (8)
(M,N) (M,N) (I,J) (I,J)
(M,XN)∈Ak Xi=1(I,XJ)∈Bi
Step 3. Since d ∈ N and d 6∈ J for all (I,J) ∈ B , i = 2,...,d − 1, if we
i
evaluate equality 8 in (v ,v ) then we conclude that γ = 0. Hence we have
I J (I,J)
β Q (x,y)+ γ1 R1 (x,y) = 0. (9)
(M,N) (M,N) (I,J) (I,J)
(M,XN)∈Ak (I,XJ)∈B1
Assume that (I ,J ) is a subscript such that has minimum cardinality in the set
0 0
A such that γ 6= 0. Substituting (v ,v ) in the equation 9 then since
1 (I0,J0) I0 J0
|I ∪J | ≤k−1 and |M ∪N|= k all terms in 9 but γ R1 (v ,v ) vanish.
0 0 (I0,J0) (I0,J0) I0 J0
So we have γ = 0
(I0,J0)
Step 4. By independence of the polynomials in the set P, each β = 0, there-
(M,N)
fore the claim is true.
So, we have n+2k−1 d−1 + k−12i d −2k +2 linearly independent polyno-
k−1 i=1 i
mials which, as the proof(cid:0)of T(cid:1)heoPrem 1,(cid:0)ar(cid:1)e in the space generated by k 2i d
i=0 i
monomials. Hence, P (cid:0) (cid:1)
d d−1
n(K,d) ≤ 2k −2k−1 +2k −1.
(cid:18)k(cid:19) (cid:18)k−1(cid:19)
That complete the proof of Theorem 2.
4 Proof of Theorem 3
Let K = {l ,l ,...,l }, and (A,B) be a cross K-intersection families with m blocks
1 2 k
on a set of d points. For every 1 ≤ j ≤d, let
def
X = {i| 1 ≤ i ≤ m,j ∈ A },
j i
7
def
Y = {i| 1 ≤ i≤ m,j ∈ B }.
j i
Now, for j = 1,2,...,d, we construct the complete bipartite graph G with vertex
j
set (X ,Y ), where X and Y were defined as above. Let ij be an arbitrary edge of
j j j j
K− , consider sets A and B . Without loss of generality assume that A ∩B =
m,m i j i j
{v ,v ,...,v },which1≤ l ≤ k. Itisnotdifficulttoseethattheedgeij wascovered
1 2 l
by the graphs G ,G ,...,G . So we have a biclique cover of type K and size d
v1 v2 vl
of K− . Conversely let {G ,G ,...,G } be a biclique cover of type K and size d
m,m 1 2 d
of the graph K− . Assume G has (X ,Y ) as the vertex set. For every 1 ≤ j ≤ m
m,m i i i
define
def
A = {i| 1≤ i ≤ d,j ∈ X },
j i
def
B = {i| 1 ≤ i ≤d,j ∈ Y }.
j i
LetA = {A ,...,A }andB ={B ,...,B }. SinceX ∩Y = ∅for every 1≤ i ≤ d
1 m 1 m i i
then A ∩B = ∅ for every 1 ≤ j ≤ m. Also, if {G ,G ,...,G } is the set of
j j v1 v2 vl
graphs that cover the edge ij then A ∩B = {v ,v ,...,v } where |A ∩B | ∈ K.
i j 1 2 l i j
Hence (A,B) is a cross K-intersection family.
Acknowledgment
The authors wish to express their tanks to Professor Hossein Hajiabolhassan for the
useful conversations during the preparation of the paper.
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