On a Class of Binary Matrices

0.13 MB·English

by  Krasimir Yordzhev

ON A CLASS OF BINARY MATRICES
Krasimir Yankov Yordzhev
Abstract
The paper studies the set of all square binary matrices containing an exact numberof
1’s in each rows and in each column. A connection is established between the cardinal
2
1 number of this set and the cardinal number of its subset of matrices containing 1 in the
0 lowerrightcorner. WiththehelpofthisresultanewproofisadvancedoftheI.Goodand
2 J. Grook theorem. In connection with the firs result a classification has also been made
of square binary matrices containing three 1’s in each row and column and 1 in the lower
n
right corner.
a
J

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1 Introduction

]
O Abinary(orboolean,or(0,1)-matrix)isamatrixwhoseallelementsbelongtothesetB={0,1}.
C With B we will denote the set of all n×n binary matrices.
n
. Using the notationfrom [12], we will callΛk-matrices all n×n binary matrices in eachrow
h n
and each column of which there are exactly k in number 1’s.
t
a Let us accept the associative operations ”+” and ”·” in the set B = {0,1} defined as:
m
0+0 = 0, 1+0 = 0+1 = 1+1 = 1, 0·0 = 1·0 = 0·1 = 0, 1·1 = 1. Let the scalar
[ productoftwovectors(definedbythen-tuple)withelementsfromBbeoverdeterminedbythe

1 operations thus introduced. Let us accept that in Bn the common definition for the operation
v of matrix product as a scalar product of the corresponding row vector and column vector is
1 overdeterminedby the operationofscalarproductintroducedabove. Consideringallthis,B ,
n
1
alongwiththe operationmatrixproduct,is asemigroupthatis isomorphicwiththe semigroup
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B(M) comprising all binary relation in a set M, where |M|=n<∞ (see for example [5]).
1
Let us consider the following combinatorialproblem:
.
1
0
Problem 1 Find out the number of all binary relations ω ∈ B(M), |M| = n < ∞, such that
2
for all a∈M the equation
1
:
v |{x∈M |(a,x)∈ω}|=|{x∈M |(x,a)∈ω}|=k,
i
X
is correct, where n and k are natural numbers.
r
a
Using the language of graph theory (see for example [1] or [4]) problem 1 is equivalent to
Problem 2 Find out the number of all bipartite graphs G = (V ∪V ,E), such that |V | =
1 2 1
|V |=n and each vertex is incident with exactly k edges.
2
Clearly, problems 1 and 2 can be reduced to solving the following
1
Problem 3 Find out the number of all n×n matrices containing exactly k 1’s in each row
and each column, e.g. the number of all Λk-matrices.
n
The goal of this paper is to consider certain special cases of problem 3.
Let us denote the number of all Λk-matrices with λ .
n n,k
In [10] is offered the formula:
(n!)2
λ = (1)
n,2 n
2x2+3x3X+···+nxn=n x !(2r)xr
r
rY=2
One of the first recursive formulas for the calculation of λ appeared in [2] :
n,2
1
λn,2 = 2n(n−1)2 (2n−3)λ(n−2),2+(n−2)2λ(n−3),2 for n≥4 (2)
λ =0, λ =1(cid:2), λ =6 (cid:3)
1,2 2,2 3,2
Another recursive formula for the calculation of λ occurs in [3] :
n,2
(n−1)2n
λn,2 =(n−1)nλ(n−1),2+ 2 λ(n−2),2 for n≥3 (3)
λ =0, λ =1
1,2 2,2
The following recursive system for the calculation of λ is put forward in [9]:
n,2
λ(n+1),2 =n(2n−1)λn,2+n2λ(n−1),2−πn+1 for n≥2
(cid:12)(cid:12) πn+1 = n2(n4−1)2[8(n−2)(n−3)λ(n−2),2+(n−2)2λ(n−3),2−4πn−1] for n≥4 (4)
(cid:12) λ =0, λ =1, π =π =π =0, π =9
(cid:12) 1,2 2,2 1 2 3 4
(cid:12)
(cid:12)
where π identifies the number of a special class of Λ2-matrices.
n n
For the classificationof allnondefined concepts andnotations as wellas forcommonasser-
tion which have not been proved here, we recommend sources [1, 6, 8, 11].
2 On a partition of the set Λk
n
Let us introduce the notations:
Λk+ =|{A=(a )∈Λk|a =1}| (5)
n ij n nn
Λk− =|{A=(a )∈Λk|a =0}| (6)
n ij n nn
Obviously
Λk+∩Λk− =∅ and Λk+∪Λk− =Λk, (7)
n n n n n
in other words {Λk+,Λk−} represents a partition of the set Λk.
n n n
We set:
λ+ =|Λk+| (8)
n,k n
λ− =|Λk−| (9)
n,k n
2
Formula (3) occurs for the first time in [3] and it has been deduced in a manner applicable
only to the calculation of the number of the Λ2-matrices. The method for the obtaining of
n
the recursive relation (3) which we offer and which we will describe in section 3.1 will being
as closer to the discovery of the analogical formula for values greater than k. In this case k
represents the number of units in each row and each column of the respective square matrices.
The method is grounded in the following assertion:
Theorem 1 It is true the equation
n−k
λk− = λk+, (10)
n k n
where λ+ and λ− are set by formulas (8) and (9) respectively.
n,k n,k
Proof. Let us accept that A and B are Λk-matrices. We will say that A and B are ρ-
n
equivalent (AρB), if the removing of the columns ending in 1 results in equal n × (n − k)
matrices. Obviously, ρ is an equivalence relation. We use ρ to denote the set of elements to
A
which A is related by the equivalence relation ρ.
LetA=(a )be aΛk-matrix. Let’sdenotewithp+ the numberofallmatricesρ-equivalent
ij n
−
to A in which the element in the lower right corner is equal to 1 and with p the number
of all matrices ρ-equivalent to A in which the element in the lower right corner is equal to
0. Let K ,K ,...,K are the row-vectors of matrix A with 1 in final position. The set
j1 j2 jk
J ={j ,j ,...,j } is partitioned into subsets J , r =1,2,...s, such that j and j are part of
1 2 k r u v
s
the same subset if and only if K =K . It is easy to detect that J = J and J ∩J =∅
ju jv r u v
r[=1
for u6=v. We set k =|J |, r =1,2,...,s. Obviously
r r
k +k +···+k =k. (11)
1 2 s
LetCbethen×(n−k)matrixwhichcomesfromAbyremovingthecolumnsK ,K ,...,K .
j1 j2 jk
In this case, with the help of the different ways of adding new columns to C, we will obtain
all elements of the set ρ . Let us first we add k columns which equal to those columns of A
A 1
whosenumbersbelongtothesetJ . Thiscanbedoneusing n−k+k1 numberofways. Wecan
1 k1
then add k equal columns in n−k+k1+k2 possible ways. T(cid:0)hese equ(cid:1)al columns are also equal
2 k2
to the columns in A with numb(cid:0)er tags fro(cid:1)m J2, etc. Therefore
n−k+k n−k+k +k n−k+k +k +···+k
1 1 2 1 2 s
|ρ |= ··· =
A
(cid:18) k (cid:19)(cid:18) k (cid:19) (cid:18) k (cid:19)
1 2 s
n n−ks n−ks−ks−1 n−ks−ks−1−ks−2+···+k2
= ··· =
(cid:18)ks(cid:19)(cid:18) ks−1 (cid:19)(cid:18) ks−2 (cid:19) (cid:18) k1 (cid:19)
n!(n−ks)!(n−ks−ks−1)!···(n−ks−ks−1−···−k2)!
= =
ks!(n−ks)!ks−1!(n−ks−ks−1)!···k1!(n−ks−ks−1−···−k2−k1)!
n!
=
k !k !···k !(n−k)!
1 2 s
−
Analogically, for p we get
(n−1)!
−
p = ,
k !k !···k !(n−1−k)!
1 2 s
having the mind the fact that we cannot add new columns after the last column of matrix C.
3
For p+ we obtain the equation:
n! (n−1)!
p+ =|ρ |−p− = − =
A
k !k !···k !(n−k)! k !k !···k !(n−1−k)!
1 2 s 1 2 s
k(n−1)!
=
k !k !···k !(n−k)!
1 2 s
−
p n−k n−k
Then = , e.g. p− = p+. Summarize by equivalence classes we arrive a the
p+ k k
equation we were supposed to prove.
Considering equations (5) ÷ (9) and theorem 1, it is possible to formulate
Corollary 1
n−k n
λ =λ+ +λ− =λ+ + λ+ = λ+ (12)
n,k n,k n,k n,k k n,k k n,k
3 Some applications
Theorem 1 and corollary 1 are useful in that they facilitate the calculation Λk+ as compared
n
to that of all Λk-matrices.
n
3.1 A different proof of the I. Good and J. Grook theorem
Using corollary 1 in order to obtain a formula for the n×n binary matrices, it is enough to
find a formula for λ+ . We can this with the help of the following
n,2
Theorem 2 If n≥3, then
λ+n,2 =2(n−1)λ(n−1),2+(n−1)2λ(n−2),2
Proof. Let A = (a ) be a Λ2 -matrix. The matrix A can be give rise to the Λ2-matrix
ij n−1 n
B = (b ) following manner: We choose p,q such that a = 1. This can be accomplished
ij pq
in 2(n−1) ways. We set b = 0, b = b = b = 1, b = b = 0 and b = a for
pq pn nq nn in nj ij ij
1 ≤ i,j ≤ n−1, i 6= p, j 6= q. It is easy to see that B s Λ2-matrix with 1 in the lower right
n
corner. Besides p and q can be identified uniquely through B and matrix A can be restored.
Consequentlyλ+n,2 =2(n−1)λ(n−1),2+t,wheretis the numberofallΛ2n-matricescontaining1
in the lowerrightcorner,whichcannotbe generatedin the mannerdescribedabove. These are
Λ2-matrices, B = (b ) for which there are p and q such that b = b = b = b = 1 and
n ij pq nq pn nn
these are the only 1’s (2 ineachrowandcolumn)in rowswith number p andn andin columns
with number q and n. In this case, however, removing rows number p and n and columns
number q and n, we obviously obtain a Λ2 -matrix. On the contrary, each Λ2 -matrix can
n−2 n−2
give riseto a Λ2-matrix by insertingtwo new rows,their numbers willbe p andn andtwonew
n
columns, their numbers will be q n, with 0 everywhere except for the places of intersection.
Since p and q vary from 1 to n−1, then t=(n−1)2λ(n−2),2. This proves the theorem.
Applying theorems 1 and 2 directly we obtain:
Theorem 3 [3] The number of all n×n square binary matrices with exactly two 1’s in each
row and each column is given by the next formula:
(n−1)2n
λn,2 =(n−1)nλ(n−1),2+ 2 λ(n−2),2 for n≥3
λ =0, λ =1
1,2 2,2
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3.2 On the number of Λ3-matrices
n
The following formula in an explicit form for the calculation of λ (n) is offered in [3].
3
n!2 (−1)β(β+3γ)!2α3β
λ (n)= (13)
3 6n α!β!γ!26γ
X
where the sum is done as regard all (n+2)(n+1) solutions in nonnegative whole numbers of the
2
equation α+β+γ =n.
As it is noted in [7] formula (13) does not give us good opportunities to study behavior
of λ . The aim of the current consideration is to go one step closer to the obtaining of a
n,3
new recursiveformulaforthe calculationofλ ,whichcouldhelpavoidcertaininconveniences
n,3
resulting from use of formula (13).
Let X =(x )∈Λ3+ and all1’s in the last columns and in the lastrow are respectively the
ij n
elements x , x , x , x , x , where s,t,p,q ∈ {1,2,...,n−1}, s 6= t, p 6= q. X will be
sn tn np nq nn
denoted 2×2 submatrix
e
x x
X = sp sq (14)
(cid:18) xtp xtq (cid:19)
e
The set Λ3+ is partitioned into the following nonintersecting subsets:
n
1 1
A = X ∈Λ3+ X = (15)
n (cid:26) n (cid:12) (cid:18) 1 1 (cid:19)(cid:27)
(cid:12)
(cid:12)e
B = X ∈Λ3+(cid:12)X ∈ 0 1 , 1 0 , 1 1 , 1 1 (16)
n (cid:26) n (cid:12) (cid:26)(cid:18) 1 1 (cid:19) (cid:18) 1 1 (cid:19) (cid:18) 0 1 (cid:19) (cid:18) 1 0 (cid:19)(cid:27)(cid:27)
(cid:12)
(cid:12)e
Γ = X ∈Λ3+(cid:12)X ∈ 1 0 , 0 1 (17)
n (cid:26) n (cid:12) (cid:26)(cid:18) 1 0 (cid:19) (cid:18) 0 1 (cid:19)(cid:27)(cid:27)
(cid:12)
(cid:12)e
∆ = X ∈Λ3+(cid:12)X ∈ 1 1 , 0 0 (18)
n (cid:26) n (cid:12) (cid:26)(cid:18) 0 0 (cid:19) (cid:18) 1 1 (cid:19)(cid:27)(cid:27)
(cid:12)
(cid:12)e
E = X ∈Λ3+(cid:12)X ∈ 1 0 , 0 1 (19)
n (cid:26) n (cid:12) (cid:26)(cid:18) 0 1 (cid:19) (cid:18) 1 0 (cid:19)(cid:27)(cid:27)
(cid:12)
(cid:12)e
Z = X ∈Λ3+(cid:12)X ∈ 1 0 , 0 1 , 0 0 , 0 0 (20)
n (cid:26) n (cid:12) (cid:26)(cid:18) 0 0 (cid:19) (cid:18) 0 0 (cid:19) (cid:18) 1 0 (cid:19) (cid:18) 0 1 (cid:19)(cid:27)(cid:27)
(cid:12)
(cid:12)e
H = X ∈Λ3+(cid:12)X = 0 0 (21)
n (cid:26) n (cid:12) (cid:18) 0 0 (cid:19)(cid:27)
(cid:12)
(cid:12)e
We set (cid:12)
α =|A |, β =|B |, γ =|Γ |, δ =|∆ |,
n n n n n n n n (22)
ǫ =|E |, ζ =|Z |, η =|H |
n n n n n n
Obviously
γ =δ (23)
n n
and
λ+ =α +β +γ +δ +ǫ +ζ +η . (24)
n,3 n n n n n n n
Theorem 4
3(n−1)(3n−8)
λ+n,3 = 2 λ(n−1),3+αn+βn+2γn−ηn
5
Proof. Let Y = (yij) ∈ Λ3n−1. In Y we choose two 1’s not belonging to the same row or
column. Let then these be the elements y and y , s,t,p,q ∈ {1,2,...,n−1}, s 6= t, p 6= q.
sp tq
This can happen in 3(n−1)[3(n−1)−5] = 3(n−1)(3n−8) ways. The 1’s thus selected are turned
2 2
into 0’s and in Y in the last place one more column (number n) and one more row (number
n) are added, so that y = y = y = y = y = 1 и y = y = 0 for i ∈/ {s,t,n},
sn tn np nq nn in nj
j ∈/ {p,q,n}. Obviously, the matrix thus formed belongs to one of the set E , Z or H .
n n n
Onthe contrary,letX =(x )be amatrixfromE orZ . X thenhasuniquezerodiagonal
ij n n
whose elements we turn into 1’s and remove the last row (number n) and the last column
e
(number n). In this way a Λ3 -matrix is generated.
n−1
x x 0 0
Let X = (x ) ∈ H and X = sp sq = . We select one diagonal of X
ij n (cid:18) xtp xtq (cid:19) (cid:18) 0 0 (cid:19)
and turn 0’s of this diagonal ineto 1’s. Remove the last row n and the last column n. We getea
Λ3 -matrix. Consequently for each H -matrix correspond two Λ3 -matrices. Then,
n−1 n n−1
3(n−1)(3n−8)
λ+n,3 = 2 λ(n−1),3−ηn+t
where t=α +β +γ +δ . Considering (23) and (24) we proved the theorem.
n n n n
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Krasimir Yankov Yordzhev
South-West University ”N. Rilsky”
2700 Blagoevgrad
Bulgaria
e-mail: [email protected],[email protected]
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