Bipartite graphs related to mutually disjoint S-permutation matrices

0.16 MB·English

by  Krasimir Yordzhev

Bipartite graphs related to mutually disjoint
S-permutation matrices
Krasimir Yordzhev
3
1
0 Faculty of Mathematics and Natural Sciences
2
South-West University, Blagoevgrad, Bulgaria

n E-mail: [email protected]
a
J Abstract

3
Some numerical characteristics of bipartite graphs in relation to the
2
problemoffindingalldisjointpairsofS-permutationmatricesinthegen-

] eral n2×n2 case are discussed in this paper. All bipartite graphs of the
O type g = hRg ∪Cg,Egi, where |Rg| = |Cg| = 2 or |Rg| = |Cg| = 3 are
C provided. The cardinality of the sets of mutually disjoint S-permutation
matrices in both the 4×4 and 9×9 cases are calculated.
.
h
t Keyword:Bipartite graph, Binary matrix, S-permutation matrix, Disjoint
a
matrices, Sudoku
m
MSC[2010] code: 05C30, 05B20, 05C50
[

2
1 Introduction
v
1
0 Let m be a positive integer. By [m] we denote the set
4
0 [m]={1,2,...,m}.
.
2
0 We let S denote the symmetric groupof order m i.e., the groupof all one-
m
2 to-one mappings of the set [m] to itself. If x ∈ [m], ρ ∈ S , then the image of
m
1
the element x in the mapping ρ we will denote by ρ(x).
:
v A bipartite graph is an ordered triple
i
X
g =hR ,C ,E i,
g g g
r
a
where R and C are non-empty sets such that R ∩C = ∅. The elements of
g g g g
R ∪C will be called vertices. The set of edges is E ⊆R ×C ={hr,ci|r ∈
g g g g g
R ,c∈C }. Multiple edges are not allowed in our considerations.
g g
Thesubjectofthe presentworkisbipartite graphsconsidereduptoisomor-
phism.
We refer to [3] or [6] for more details on graph theory.
Let n and k be two nonnegative integers and let 0≤k ≤n2. We denote by
G the set of all bipartite graphs of the type g =hR ,C ,E i, considered up
n,k g g g
to isomorphism, such that |R |=|C |=n and |E |=k.
g g g
LetP ,1≤i,j ≤n,ben2 squaren×nmatrices,whoseentriesareelements
ij
of the set [n2]={1,2,...,n2}. The n2×n2 matrix
1
P P ··· P
11 12 1n
P P ··· P
21 22 2n
P = ... ... ... ... 
 
 P P ··· P 
 n1 n2 nn 
 
is called a Sudoku matrix, if every row, every column and every submatrix
P , 1 ≤ i,j ≤ n comprise a permutation of the elements of set [n2], i.e., every
ij
numbers∈{1,2,...,n2}isfoundjustonceineachrow,column,andsubmatrix
P . Submatrices P are called blocks of P.
ij ij
Sudoku is a very popular game and Sudoku matrices are special cases of
Latin squares in the class of gerechte designs [1].
A matrix is called binary if all of its elements are equal to 0 or 1. A square
binary matrix is called permutation matrix, if in every row and every column
there is just one 1.
Let us denote by Σn2 the set of all n2 × n2 permutation matrices of the
following type:
A A ··· A
11 12 1n
A A ··· A
21 22 2n
A= ... ... ... ... ,
 
 A A ··· A 
 n1 n2 nn 
 
where for every s,t ∈ {1,2,...,n}, A is a square n × n binary submatrix
st
(block) with only one element equal to 1.
The elements of Σn2 will be called S-permutation matrices.
Two Σn2 matrices A = (aij) and B = (bij), 1 ≤ i,j ≤ n2 will be called
disjoint, if there are not elements a and b with the same indices such that
ij ij
a =b =1.
ij ij
The concept of S-permutation matrix was introduced by Geir Dahl [2] in
relation to the popular Sudoku puzzle.
Obviously,asquaren2×n2matrixP withentriesfrom[n2]={1,2,...,n2}is
a Sudoku matrix if andonly if there areΣn2 matrices A1,A2,...,An2, pairwise
disjoint, such that P can be written in the following way:
P =1·A1+2·A2+···+n2·An2 (1)
In[5]RobertoFontanaoffersanalgorithmwhichreturnsarandomfamilyof
n2×n2 mutuallydisjointS-permutationmatrices,wheren=2,3. Forn=3,he
ran the algorithm 1000 times and found 105 different families of nine mutually
disjoint S-permutation matrices. Then, applying (1), he decided that there are
at least 9! · 105 = 38 102 400 Sudoku matrices. This number is very small
comparedwith the exact number of 9×9 Sudoku matrices. In [4] it was shown
that there are exactly
9!·722·27·27704267971=6670903752021072936960
number of 9×9 Sudoku matrices.
To evaluate the effectiveness of Fontana’s algorithm, it is necessary to cal-
culate the probability of two randomly generatedmatrices being disjoint. As is
proved in [2], the number of S-permutation matrices is equal to
|Σn2|=(n!)2n.
2
Thus the question of finding a formula for counting disjoint pairs of S-
permutationmatricesnaturallyarises. Suchaformulaisintroducedandverified
in [8]. In this paper, we demonstrate this formula to compute the number of
disjoint pairs of S-permutation matrices in both the 4 × 4 and 9 × 9 cases.
2 A formula for counting disjoint pairs of S-
permutation matrices
Letg =hR ,C ,E i∈G forsomenaturalnumbersnandk andletv ∈V =
g g g n,k g
R ∪C .
g g
ByN(v)wedenotethesetofallverticesofV ,adjacentwithv,i.e.,u∈N(v)
g
if and only if there is an edge in E connecting u and v. If v is an isolated
g
vertex (i.e., there is no edge, incident with v), then by definition N(v)=∅ and
degree(v) = |N(v)| = 0. If v ∈ R , then obviously N(v) ⊆ C , and if v ∈ C ,
g g g
then N(v)⊆R .
g
Let g = hR ,C ,E i ∈ G and let u,v ∈ V = R ∪C . We will say that
g g g n,k g g g
u and v are equivalent and we will write u ∼v if N(u)=N(v). If u and v are
isolated,thenbydefinitionu∼vifandonlyifuandv belongsimultaneouslyto
R , or C . The above introduced relation is obviously an equivalence relation.
g g
ByV wedenotetheobtainedfactor-set(thesetoftheequivalenceclasses)
g/∼
according to relation ∼ and let
V ={∆ ,∆ ,...,∆ },
g/∼ 1 2 s
where ∆ ⊆R , or ∆ ⊆C , i=1,2,...s, 2≤s≤2n. We put
i g i g
δ =|∆ |, 1≤δ ≤n, i=1,2,...,s
i i i
and for every g ∈G we define multi-set (set with repetition)
n,k
[g]={δ ,δ ,...δ },
1 2 s
where δ ,δ ,...,δ are natural numbers, obtained by the above described way.
1 2 s
If z z ... z is a permutation of the elements of the set [n]={1,2,...,n}
1 2 n
and we shortly denote ρ this permutation, then in this case we denote by ρ(i)
the i-th element of this permutation, i.e., ρ(i)=z , i=1,2,...,n.
i
The following theorem is proved in [8]:
Theorem 1 [8] Let n ≥ 2 be a positive integer. Then the number Dn2 of all
disjoint ordered pairs of matrices in Σn2 is equal to
(n−|N(v)|)!
n2
Dn2 =(n!)4n+(n!)2(n+1) (−1)k v∈RYg∪Cg . (2)
δ!
kX=1 g∈XGn,k
δY∈[g]
The number dn2 of all non-ordered pairs of disjoint matrices in Σn2 is equal
to
1
dn2 = 2Dn2 (3)
(cid:3)
3
The proofofTheorem1 is describedin detailin [8] andherewe will miss it.
In order to apply Theorem 1 it is necessary to describe all bipartite graphs
up to isomorphism g =hR ,C ,E i, where |R |=|C |=n.
g g g g g
Let n and k are positive integers and let g ∈G . We examine the ordered
n,k
(n+1)-tuple
Ψ(g)=hψ (g),ψ (g),...,ψ (g)i, (4)
0 1 n
where ψ (g), i=0,1,...,n is equal to the number of vertices ofg incident with
i
n
exactly i number of edges. It is obvious that iψ (g) = 2k is true for all
i
i=1
g ∈G . Then formula (2) can be presented X
n,k
n
[(n−i)!]ψi(g)
n2
Dn2 =(n!)4n+(n!)2(n+1) (−1)k Yi=0 .
δ!
Xk=1 g∈XGn,k
δ∈[g]
Y
Since (n−n)!=0!=1 and [n−(n−1)]!=1!=1, then
n−2
[(n−i)!]ψi(g)
n2
Dn2 =(n!)4n+(n!)2(n+1) (−1)k Yi=0 . (5)
δ!
Xk=1 g∈XGn,k
δY∈[g]
Consequently,toapplyformula(5)foreachbipartitegraphg ∈G andfor
n,k
thesetG ofbipartitegraphs,itisnecessarytoobtainthefollowingnumerical
n,k
characteristics:
n−2
[(n−i)!]ψi(g)
ω(g)= iY=0 (6)
δ!
δY∈[g]
and
θ(n,k)= ω(g) (7)
g∈XGn,k
Using the numerical characteristics (6) and (7), we obtain the following
variety of Theorem 1:
Theorem 2
n2
Dn2 =(n!)4n+(n!)2(n+1) (−1)kθ(n,k), (8)
k=1
X
where θ(n,k) is described using formulas (7) and (6).
(cid:3)
4
3 Demonstrations in applying of Theorem 2
3.1 Counting the number D of all ordered pairs of disjoint
4
S-permutation matrices for n = 2
3.1.1 k =1
In n=2 and k =1, G consists of a single graph g shown in Figure 1.
2,1 1
g
1
(cid:23)(cid:20)(cid:23)(cid:20)
s c
s c
(cid:22)Rg1 (cid:21)(cid:22)Cg1 (cid:21)
Figure 1: n=2, k =1
For graph g ∈G we have:
1 2,1
[g ]={1,1,1,1}
1
Ψ(g )=hψ (g ),ψ (g ),ψ (g )i=h2,2,0i
1 0 1 1 1 2 1
Then we get:
[(2−0)!]2
ω(g )= =4
1
1!1!1!1!
and therefore
θ(2,1)= ω(g)=4. (9)
g∈XG2,1
3.1.2 k =2
The set G consists of three graphs g , g and g depicted in Figure 2.
2,2 2 3 4
g g g
2 3 4
(cid:23)(cid:20)(cid:23)(cid:20) (cid:23)(cid:20)(cid:23)(cid:20) (cid:23)(cid:20)(cid:23)(cid:20)
(cid:8) H
s c s (cid:8) c s H c
(cid:8) H
(cid:8) H
s c s c s c
(cid:22)Rg2 (cid:21)(cid:22)Cg2 (cid:21) (cid:22)Rg3 (cid:21)(cid:22)Cg3 (cid:21) (cid:22)Rg4 (cid:21)(cid:22)Cg4 (cid:21)
Figure 2: n=2, k =2
For graph g ∈G we have:
2 2,2
[g ]={1,1,1,1}
2
Ψ(g )=hψ (g ),ψ (g ),ψ (g )i=h0,4,0i
2 0 2 1 2 2 2
[(2−0)!]0
ω(g )= =1
1
1!1!1!1!
For graphs g ∈G and g ∈G we have:
3 2,2 4 2,2
[g ]=[g ]={2,1,1}
3 4
5
Ψ(g )=Ψ(g )=h1,2,1i
3 4
[(2−0)!]1
ω(g )=ω(g )= =1
3 4
2!1!1!
Then for the set G we get:
2,2
θ(2,2)= ω(g)=1+1+1=3. (10)
g∈XG2,2
3.1.3 k =3
In n=2 and k =3, G consists of a single graph g shown in Figure 3.
2,3 5
g
5
(cid:23)(cid:20)(cid:23)(cid:20)
H (cid:8)
s H(cid:8) c
(cid:8)H
(cid:8) H
s c
(cid:22)Rg5 (cid:21)(cid:22)Cg5 (cid:21)
Figure 3: n=2, k =3
For graph g ∈G we have:
5 2,3
[g ]={1,1,1,1}
5
Ψ(g )=hψ (g ),ψ (g ),ψ (g )i=h0,2,2i
5 0 5 1 5 2 5
Then we get:
[(2−0)!]0
ω(g )= =1
5
1!1!1!1!
and therefore
θ(2,3)= ω(g)=1. (11)
g∈XG2,3
3.1.4 k =4
Whenn=2andk =4thereisonlyonegraphandthisisthecompletebipartite
graph g which is shown in Figure 4.
6
g
6
(cid:23)(cid:20)(cid:23)(cid:20)
H (cid:8)
s H(cid:8) c
(cid:8)H
(cid:8) H
s c
(cid:22)Rg6 (cid:21)(cid:22)Cg6 (cid:21)
Figure 4: n=2, k =4
For graph g ∈G we have:
6 2,4
[g ]={2,2}
6
Ψ(g )=hψ (g ),ψ (g ),ψ (g )i=h0,0,4i
6 0 6 1 6 2 6
6
Then we get:
[(2−0)!]0 1
ω(g )= =
6
2!2! 4
and therefore
1
θ(2,4)= ω(g)= . (12)
4
g∈XG2,1
Havinginmindthe formulas(8), (9), (10), (11) and(12)forthe numberD
4
of all ordered pairs disjoint S-permutation matrices in n=2 we finally get:
D =(2!)8+(2!)6[−θ(2,1)+θ(2,2)−θ(2,3)+θ(2,4)]= (13)
4
1
=256+64 −4+3−1+ =144.
4
(cid:18) (cid:19)
Thenumberd ofallnon-orderedpairsdisjointmatricesfromΣ is equalto
4 4
1
d = D =72. (14)
4 4
2
3.2 Counting the number D of all ordered pairs of disjoint
9
S-permutation matrices for n = 3
3.2.1 k =1
Graph g , which is displayed in Figure 5 is the only bipartite graph belonging
7
to the set G .
3,1
g
7
(cid:23) (cid:20)(cid:23) (cid:20)
s c
s c
s c
(cid:22)Rg7 (cid:21)(cid:22)Cg7 (cid:21)
Figure 5: n=3, k =1
For graph g ∈G we have:
7 3,1
[g ]={1,1,2,2}
7
Ψ(g )=hψ (g ),ψ (g ),ψ (g ),ψ (g ),ψ (g )i=h4,2,0,0i
7 0 7 1 7 2 7 3 7 4 8
Then we get:
[(3−0)!]4[(3−1)!]2 64·22
ω(g )= = =1296
7
1!1!2!2! 1·1·2·2
and therefore
θ(3,1)= ω(g)=1296. (15)
g∈XG3,1
7
3.2.2 k =2
In this case G ={g ,g ,g }. The graphs g , g and g are shownin Figure
3,2 8 9 10 8 9 10
6.
g g g
8 9 10
(cid:23) (cid:20)(cid:23) (cid:20) (cid:23) (cid:20)(cid:23) (cid:20) (cid:23) (cid:20)(cid:23) (cid:20)
H (cid:8)
s c s H c s (cid:8) c
H (cid:8)
H (cid:8)
s c s c s c
s c s c s c
(cid:22)Rg8 (cid:21)(cid:22)Cg8 (cid:21) (cid:22)Rg9 (cid:21)(cid:22)Cg9 (cid:21) (cid:22)Rg10 (cid:21)(cid:22)Cg10 (cid:21)
Figure 6: n=3, k =2
For graph g ∈G we have:
8 3,2
[g ]={1,1,1,1,1,1}
8
Ψ(g )=hψ (g ),ψ (g ),ψ (g ),ψ (g ),ψ (g )i=h2,4,0,0i
8 0 8 1 8 2 8 3 8 4 8
[(3−0)!]2[(3−1)!]4
ω(g )= =62·24 =576
8
1!1!1!1!1!1!
For graphs g ∈G and g ∈G we have:
9 3,2 10 3,2
[g ]=[g ]={1,1,2,2}
9 10
Ψ(g )=Ψ(g )=h3,2,1,0i
9 10
[(3−0)!]3[(3−1)!]2 63·22
ω(g )=ω(g )= = =216
9 10
1!1!2!2! 1·1·2·2
Then for the set G we get:
3,2
θ(3,2)= ω(g)=576+216+216=1008. (16)
g∈XG3,2
3.2.3 k =3
When n=3 and k =3 the set G ={g ,g ,g ,g ,g ,g } consists of six
3,3 11 12 13 14 15 16
bipartite graphs, which are shown in Figure 7.
For graph g ∈G we have:
11 3,3
[g ]={1,1,1,1,1,1}
11
Ψ(g )=h0,6,0,0i
11
[(3−0)!]0[(3−1)!]6
ω(g )= =60·26 =64
11
1!1!1!1!1!1!
8
g g g
11 12 13
(cid:23) (cid:20)(cid:23) (cid:20) (cid:23) (cid:20)(cid:23) (cid:20) (cid:23) (cid:20)(cid:23) (cid:20)
H (cid:8)
s c s H c s (cid:8) c
H (cid:8)
H (cid:8)
s c s c s c
s c s c s c
(cid:22)Rg11 (cid:21)(cid:22)Cg11 (cid:21) (cid:22)Rg12 (cid:21)(cid:22)Cg12 (cid:21) (cid:22)Rg13 (cid:21)(cid:22)Cg13 (cid:21)
g g g
14 15 16
(cid:23) (cid:20)(cid:23) (cid:20) (cid:23) (cid:20)(cid:23) (cid:20) (cid:23) (cid:20)(cid:23) (cid:20)
H (cid:8) H
s H c s (cid:8) c s H c
H (cid:8) H
H (cid:8) H
H (cid:8)
s c s H c s (cid:8) c
H (cid:8)
H (cid:8)
s c s c s c
(cid:22)Rg14 (cid:21)(cid:22)Cg14 (cid:21) (cid:22)Rg15 (cid:21)(cid:22)Cg15 (cid:21) (cid:22)Rg16 (cid:21)(cid:22)Cg16 (cid:21)
Figure 7: n=3, k =3
For graphs g ,g ∈G we have:
12 13 3,3
[g ]=[g ]={1,1,1,1,2}
12 13
Ψ(g )=Ψ(g )=h1,4,1,0i
12 13
[(3−0)!]1[(3−1)!]4 61·24
ω(g )=ω(g )= = =48
12 13
1!1!1!1!2! 2
For graph g ∈G we have:
14 3,3
[g ]={1,1,1,1,1,1}
14
Ψ(g )=h2,2,2,0i
14
[(3−0)!]2[(3−1)!]2
ω(g )= =62·22 =144
14
1!1!1!1!1!1!
For graphs g ,g ∈G we have:
15 16 3,3
[g ]=[g ]={1,2,3}
15 16
Ψ(g )=Ψ(g )=h2,3,0,1i
15 16
[(3−0)!]2[(3−1)!]3 62·23
ω(g )=ω(g )= = =24
15 16
1!2!3! 2·6
Then for the set G we get:
3,3
θ(3,3)= ω(g)=64+48+48+144+24+24=352. (17)
g∈XG3,3
9
3.2.4 k =4
When n=3 and k =4 the set G ={g ,g ,g ,g ,g ,g ,g } consists of
3,4 17 18 19 20 21 22 23
seven bipartite graphs, which are shown in Figure 8.
g g g
17 18 19
(cid:23) (cid:20)(cid:23) (cid:20) (cid:23) (cid:20)(cid:23) (cid:20) (cid:23) (cid:20)(cid:23) (cid:20)
H (cid:8) H H
s H(cid:8) c s H c s H c
(cid:8)H H H
(cid:8) H H H
H
s c s H c s c
H
H
s c s c s c
(cid:22)Rg17 (cid:21)(cid:22)Cg17 (cid:21) (cid:22)Rg18 (cid:21)(cid:22)Cg18 (cid:21) (cid:22)Rg19 (cid:21)(cid:22)Cg19 (cid:21)
g g
20 21
(cid:23) (cid:20)(cid:23) (cid:20) (cid:23) (cid:20)(cid:23) (cid:20)
(cid:8) H
s (cid:8) c s H c
(cid:8) H
(cid:8) H
H (cid:8)
s H c s (cid:8) c
H (cid:8)
H (cid:8)
s c s c
(cid:22)Rg20 (cid:21)(cid:22)Cg20 (cid:21) (cid:22)Rg21 (cid:21)(cid:22)Cg21 (cid:21)
g g
22 23
(cid:23) (cid:20)(cid:23) (cid:20) (cid:23) (cid:20)(cid:23) (cid:20)
(cid:8) H
s (cid:8) c s H c
(cid:8) H
(cid:8) H
H (cid:8)
s H c s (cid:8) c
H (cid:8)
H (cid:8)
s c s c
(cid:22)Rg22 (cid:21)(cid:22)Cg22 (cid:21) (cid:22)Rg23 (cid:21)(cid:22)Cg23 (cid:21)
Figure 8: n=3, k =4
For graph g ∈G we have:
17 3,4
[g ]={1,1,2,2}
17
Ψ(g )=h2,0,4,0i
17
[(3−0)!]2[(3−1)!]0 62·20
ω(g )= = =9
17 1!1!2!2! 22
For graph g ∈G we have:
18 3,4
[g ]={1,1,2,2}
18
Ψ(g )=h0,4,2,0i
18
[(3−0)!]0[(3−1)!]4 60·24
ω(g )= = =4
18 1!1!2!2! 22
For graph g ∈G we have:
19 3,4
10