P9 FACTORIZATION OF COMPLETE BIPARTITE GRAPH
U.S.RAJPUT AND BAL GOVIND SHUKLA
Abstract. Pk -factorization of a complete bipartite graph for k, an even
integer was studied by H.wang [1]. Further, Beiling Du [2] extended the work
of H.Wang, and studied the P2k -factorization of complete bipartite multigraph.
For odd value of k the work on factorization was done by a number
of researchers[3,4,5]. P3 -factorization of complete bipartite graph was studied by K.Ushio [3]. P5 -factorization of complete bipartite graph was studied
by J.Wang etal [4]. Further, P7 -factorization was studied by J.Wang [5] and
he gave necessary and su¢ cient conditions for its existence. In the present
paper we study the P9 -factorization of complete bipartite graph Km; n and
show that the necessary and su¢ cient conditions for its existence are:
(1)5m 4n;
(2)5n 4m;
(3)m + n 0(mod9) and
(4)9mn=8(m + n) is an integer.
1. Introduction
Let Km;n be a complete bipartite graph with two partite sets X and Y . The Pk
-factorization of Km;n is a partition of the set of Pk -factors of Km;n .Ushio[6] gave
the following necessary and su¢ cient conditions for the existence of Pk -factorization
of Km;n ,when k is odd and m and n are positive integers :
(1)
(2)
(3)
(4)
(k + 1)m (k 1)n;
(k + 1)n (k 1)m;
m + n 0(modk) and
kmn=[(k 1)(m + n)]is an integer.
The generalization was proved by J.Wang et al. [4]for k = 5 and J.Wang [5] for
k = 7.
In the present paper we will prove that this generalization is also true for k = 9.To
verify this result the necessary lemmas and theorem are developed.
2. Mathematical Analysis
Theorem 1. Let m and nbe positive integers. Then Km;n has P9 -factorization if
and only if the following are satis…ed :
(1)
(2)
(3)
(4)
5m 4n;
5n 4m;
m + n 0(mod9) and
9mn=[8(m + n)] is an integer.
1991 Mathematics Subject Classi…cation. 68R10,05C70.
Key words and phrases. Complete bipartite graph, factorization of graph, spaning graph.
1
2
U.S.RAJPUT AND BAL GOVIND SHUKLA
To prove this theorem the following well established number theoretic result is
used.
Lemma 1. Letg; h; p and q be positive integers. If gcd (p; q) = 1, then
gcd (p:q; p + g:q) = gcd (p; g). Similarly, if gcd (gp; hq) = 1 then gcd (gp + hq; pq)
= 1.
The following theorem will also be used in the proof.
Theorem 2. If Km;n has P9 -factorization then Ksm;sn has also P9 -factorization
for every positive integer s.
LetKs;s be 1-factorable[7], and {H1 , H2 . . . Hs } be a 1-factorization of it. For
each i with 1 i s, replace every edge of Hi with a Km;n to get spanning sub
graphs Gi’s ofKsm;sn such that
the Gi’s f1 i sg are pair wise edge disjoint and there sum is Ksm;sn .
Since Km;n is P9 -factorable, therefore it is obvious that each Gi is alsoP9 factorable, and hence; Ksm;sn is also P9 -factorable.
Now to prove the theorem (1), consider three cases:Case I (5m = 4n): In this case Km; n has aP9 -factorization, sinceK8;10 (trivial
case) has P9 - factorization
y1 x1 y2 x2 y3 x3 y4 x4 y5 ; y6 x5 y7 x6 y8 x7 y9 x8 y10 ;
y1 x2 y2 x3 y3 x4 y4 x5 y5 ; y6 x6 y7 x7 y8 x8 y9 x1 y10 ;
y1 x3 y2 x4 y3 x5 y4 x6 y5 ; y6 x7 y7 x8 y8 x1 y9 x2 y10 ;
y1 x4 y2 x5 y3 x6 y4 x7 y5 ; y6 x8 y7 x1 y8 x2 y9 x3 y10 ;
y1 x5 y2 x6 y3 x7 y4 x8 y5 , y6 x1 y7 x2 y8 x3 y9 x4 y10 .
Case II (5n = 4m, swapping the values of m; n): Obviously Km;n has P9 factorization.
and r
Case III (5m > 4n and 5n > 4m): Let a = 5m9 4n , b = 5n 94m , t = m+n
9
9mn
= 8(m+n)
.Where a; b; t and r must be integers, (for m; n satisfying theorem1). We
ab
have m = 4a + 5b,
n = 5a + 4b, r = 5(a+b)
+z, where z = 8(a+b)
.
2
Here,
t= the number of copies of P9 in any factor,
r = the number of P9 - factors in the factorization,
a = the number of copies of P9 with its endpoints in Y in a particular P9 - factor
(type M),
b = the number of copies of P9 with its endpoints in X in a particular P9 - factor
(type W),
c = the total number of copies of P9 in the whole factorization.
Let gcd (4a; 5b) = d and therefore 4a = dp, 5b = dq for some p; q;
dpq
where gcd (p; q) = 1. Consequently z = 8(5p+4q)
.
These equalities imply the following equalities:
d = 8(5p+4q)z
, m = 8(p+q)(5p+4q)z
, n = 2(25p+16q)(5p+4q)z
;
pq
pq
5pq
r = (p+q)(25p+16q)z
, a = 2p(5p+4q)z
and b = 8q(5p+4q)z
.
pq
pq
5pq
Now to compute the values of m; n; a; b and r, we established the following
lemma:
Case (1):
If gcd (p; 16) = 1and gcd (q; 25) = 1, then there exist a positive integer s such
that
m = 40(p+ q)(5p +4q)s, n = 2(25p + 16q)(5p + 4q)s,
P9 FACTORIZATION OF COM PLETE BIPARTITE GRAPH
3
a = 10p(5p + 4q)s; b = 8q(5p + 4q)s and r = 5(p + q)(25p + 16q)s.
Case (2):
If gcd (p; 16) = 1 and gcd (q; 25) = 5, let q = 5q1 . Then there exist a positive
integer s such that
m = 40(p + 5q1 )(p + 4q1 )s, n = 10(5p + 16q1 )(p + 4q1 )s,
a = 10p(p + q1 )s; b = 40q1 (p + 4q1 )s and r = 5(p + 5q1 )(5p + 16q1 )s.
Case (3):
If gcd (p; 16) = 1 and gcd (q; 25) = 25, let q = 25q2 . Then there exist a positive
integer s such that
m = 8(p + 25q2 )(p + 20q2 )s, n = 10(p + 16q2 )(p + 20q2 )s,
a = 2p(p + 20q2 )s; b = 40q2 (p + 20q2 )s and r = 5(p + 25q2 )(p + 16q2 )s.
Case (4):
If gcd (p; 16) = 2 and gcd(q; 25) = 1, let p = 2p1 . Then there exist a positive
integer s such that
m = 40(2p1 + q)(5p1 + 2q)s, n = 4(25p1 + 8q)(5p1 + 2q)s,
a = 20p1 (5p1 + 2q)s; b = 8q(5p1 + 2q)s and r = 5(2p1 + q)(25p1 + 8q)s.
Case (5):
If gcd (p; 16) = 2 and gcd(q; 25) = 5, let p = 2p1 , q = 5q1 . Then there exist a
positive integer
s such that
m = 40(2p1 + 5q1 )(p1 + 2q1 )s, n = 20(5p1 + 8q1 )(p1 + 2q1 )s,
a = 20p1 (p1 + 2q1 )s; b = 40q1 (p1 + 2q1 )s and r = (2p1 + 5q1 )(5p1 + 8q1 )s.
Case (6):
If gcd (p; 16) = 2, and gcd (q; 25) = 25, let p = 2p1 , q = 25q2 . Then there exist
a positive integer s such that
m = 40(2p1 + 25q2 )(p1 + 10q2 )s, n = 20(p1 + 8q2 )(p1 + 10q2 )s,
a = 40p1 (p1 + 10q2 )s; b = 40q2 (p1 + 10q2 )s and r = 5(2p1 + 25q2 )(p1 + 8q2 )s.
Case (7):
If gcd (p; 16) = 4 and gcd (q; 25) = 1, let p = 4p2 .Then there exist a positive
integer s such that
m = 40(4p2 + q)(5p2 + q)s; n = 8(25p2 + 4q)(5p2 + q)s;
a = 40p2 (5p2 + q)s; b = 8q(5p2 + q)s and r = 5(25p2 + 4q)(4p2 + q)s.
Case (8):
If gcd (p; 16) = 4, gcd (q; 25) = 5, let p = 4p2 , q = 5q1 . Then there exist a
positive integer s such that
m = 40(4p2 + 5q1 )(p2 + q1 )s; n = 40(5p2 + 4q1 )(p2 + q1 )s;
a = 40p2 (p2 + q1 )s; b = 40q1 (p2 + q1 )s and r = 5(4p2 + 5q1 )(5p2 + 4q1 )s:
Case (9):
If gcd (p; 16) = 4; gcd (q; 25) = 25; let p = 4p2 ; q = 25q2 : Then there exist a
positive integer s such that
m = 8(4p2 + 25q2 )(p2 + 5q2 )s; n = 40(p2 + 4q2 )(p2 + 5q2 )s;
a = 8p2 (p2 + 5q2 )s; b = 40q2 (p2 + 5q2 )s and r = 5(4p2 + 25q2 )(p2 + 4q2 )s:
Case (10):
If gcd (p; 16) = 8, gcd (q; 25) = 1, let p = 8p3 : Then there exist a positive integer
s such that
m = 20(8p3 + q)(10p3 + q)s, n = 8(25p3 + 2q)(10p3 + q)s,
a = 40p3 (10p3 + q)s; b = 4q(10p3 + q)s and r = 5(8p3 + q)(25p3 + 2q)s:
Case (11):
4
U.S.RAJPUT AND BAL GOVIND SHUKLA
If gcd (p; 16) = 8, gcd (q; 25) = 5, let p = 8p3 ; q = 5q1 . Then there exist a
positive integer s such that
m = 20(8p3 + 5q1 )(2p3 + q1 )s; n = 40(5p3 + 2q1 )(2p3 + q1 )s;
a = 40p3 (2p3 + q1 )s; b = 20q1 (2p3 + q1 )s and r = 5(8p3 + 5q1 )(5p3 + 2q1 )s:
Case (12):
If gcd (p; 16) = 8, gcd (q; 25) = 25, let p = 8p3 , q = 25q2 . Then there exist a
positive integer s such that
m = 4(8p3 + 25q2 )(2p3 + 5q2 )s; n = 40(p3 + 2p2 )(2p3 + 5q2 )s;
a = 8p3 (2p3 + 5q2 )s; b = 20q2 (2p3 + 5q2 )s and r = 5(8p3 + 25q2 )(p3 + 2q2 )s:
Case (13):
If gcd (p; 16) = 16; gcd (q; 25) = 1, let p = 16p4 . Then there exist a positive
integer s such that
m = 10(16p4 + q)(20p4 + q)s; n = 8(25p4 + q)(20p4 + q)s;
a = 40p4 (20p4 + q)s; b = 2q(10p4 + q)s and r = 5(16p4 + q)(25p4 + q)s:
Case (14):
If gcd (p; 16) = 16, gcd (q; 25) = 5, let p = 16p4 , q = 5q1 , Then there exist a
positive integer s such that
m = 10(16p4 + 5q1 )(4p4 + q1 )s; n = 40(5p4 + q1 )(4p4 + q1 )s;
a = 8p4 (4p4 + q1 )s; b = 10q1 (4p4 + q1 )s and r = 5(16p4 + 5q1 )(5p4 + q1 )s:
Case (15):
If gcd (p; 16) = 16, gcd (q; 25) = 25, let p = 16p4 , q = 25q2 . Then there exist a
positive integer s such that
m = 2(16p4 + 25q2 )(4p4 + 5q2 )s; n = 40(p4 + q2 )(4p4 + 5q2 )s;
a = 8p4 (4p4 + 5q2 )s; b = 10q2 (4p4 + 5q2 )s and r = 5(16p4 + 25q2 )(p4 + q2 )s:
Proof: We are giving the proof of case (1). Let gcd(p; q) = 1, gcd(p; 16) = 1and
gcd(q; 25) = 1,
then gcd(25p + 16q; 5) = 1 = gcd(5p + 4q; 5) and gcd(25p; 16q) = gcd(5p; 4q) = 1:
Hence, gcd(25p + 16q; pq) = gcd(5p + 4q; pq) = 1(lemma 1).
is an integer,
Since, n = 2(25p+16q)(5p+4q)z
5pq
hence we observe that z=5pq (call it s) will be an integer. Then the equalities in
(1) hold.
The proofs of other equalities in di¤erent cases are similar to (1). Now for each
case of lemma (2) we will establish the values of m and n for P9 - factorization.
We observe that cases (1) and (15), (2) and (14), (3) and (13), (4) and (12),
(5) and (11), (6) and (10) and (7) and (9) are symmetrical. Therefore we give the
direct construction of some cases the rest will be obvious.
Lemma 2. For any positive integer p and q let m = 40(p + q)(5p + 4q), and
n = 2(25p + 16q)(5p + 4q): Then Km;n has P9 -factorization.
Proof: Let a = 10p(5p + 4q) and b = 8q(5p + 4q). Hence t = a + b = 2(5p + 4q)2 ,
and r = r1 :r2 , where r1 = 5(p + q) and r2 = (25p + 16q):
Let X; Y be two partite sets of Km;n and set
X = fxij ; 1 i r1 , 1 j m0 },
and Y = fyij; 1
i
r2 ; 1
j
n0 }, where m0 = m=r1 = 8 (5p + 4q),and
n0 = n=r2 = 2(5p + 4q).
In this case there will be a = 10p(5p+4q), type M, P9 - factor and b = 8q(5p+4q)
type W, P9 –factor. Here type M denotes the P9 -factor with its end points in Y
and type W denote the P9 -factor with its end points in X.
P9 FACTORIZATION OF COM PLETE BIPARTITE GRAPH
5
Now for each 1 i 5p, let 1 j
2(5p + 4q), 0 u 1; 0 v; w 1; 0
s; t 1
Ei = fxi;j+(5p+4q)u+2(5p+4q)v+3(5p+4q)w y5(i 1)+s+t+u+v+1;j+2(i 1)+s+t g;
and for each 1 i q, let 1 j 2(5p + 4q); 1 u 3; 0 v; w; t 1;
E5p+i = fx5p+5(i
1)+2(u 1)+v;j+2(5p+4q)v+3(5p+4q)w+(5p+4q)t y25p+16(i 1)+4u+2v+w+1;j+10p+8(i 1)+2u+v+w g
Let F = U1 i 5p+q Ei. Obviously F contains t = a + b = 2(5p + 4q)2 =
(5p + 4q)n0 number of vertex disjoint and edge disjoint P9 component and span
Km;n :
!
Then the graph F is a P9 - factor of Km;n . De…ne a bijection : X [Y ontoX [Y
such that (xi;j ) = xi+1;j and (yi;j ) = yi+1;j for each i 2 (1; 2 : : : r1 ) and each
j 2 (1; 2 : : : r2 )
Let F ; = { (x) (y); x 2 X, y 2 Y; xy 2 F g,
It is shown that the graph, F ; {1
r1 ; 1
r2 },are edge disjoint P9 factor of Km;n and its union is also Km;n .
Thus fF ; : 1
r1 ; 1
r2 g is a P9 -factorization of Km;n .
The proofs of following lemmas (4–7) are similar to lemma 3, with only changes
in the values as given below.
Lemma 3. For any positive integer p and q, let m = 40(p + 5q)(p + 4q) and
n = 10(5p + 16q)(p + 4q) then Km;n has P9 -factorization.
Proof: Let a = 10p(p + 4q); b = 40q(p + 4q) and r = r1 :r2 where r1 = 5(p +
5q); r2 = 5p + 16q:
Let X = fxij ; 1 i r1 ; 1 j 8(p + 4q)g;
Y = fyi;j ; 1 i r2 ; 1 j 10(p + 4q)g;
For each 1 i 5p, let
Ei = fxi;j+3(p+4q)u yi;j+2(i 1)+(u 1)+v ; 1 j 2(p + 4q); 0 u 2; 0 v 1g:
Now for each 1 i 5q, let
E5p+i
1
= fx5p+5(i
j
1)+2u+v;j+3(p+4q)u y5p+3(i 1)+q+u+v;j+10p+8(i 1)+4u+v 1 ;
2(p + 4q); 0
u
2; 0
v
1g:
Lemma 4. For any positive integer p and q, let m = 40(2p + q)(5p + 2q), and
n = 4(25p + 8q)(25p + 2q) then Km;n has P9 - factorization.
Proof: Let r = r1 :r2 where r1 = 5(2p + q); r2 = (25p + 8q)
m0 = 8(5p + 2q); n0 = 4(5p + 2q) and
X = fxi;j ; 1 i r1 ; 1 j m0 g;
Y = fyi;j ; 1 i r2 ; 1 j n0 g:
For each 1 i 5p, let
Ei = fx2(i 1)+u;j+(5p+2q)u+2(5p+2q)v y5(i 1)+v+2u;j+4(i 1)+u+w ; 1
j
4(5p +
2q); 1 u 2; 0 v; w 1g:
Now for each 1 i q; let
E5p+i = fx10p+5(i 1)+2u+v;j+(5p+2q)(h 1)+2(5p+2q)w y25p+8(i 1)+3u+v+1;j+20p+8(i
1 j 20(p + 2q); 1 u 2; 0 v 1; 1 h 3g:
Lemma 5. For any positive integer p and q , let m = 40(2p + 5q)(p + 2q), and
n = 20(5p + 2q)(p + 2q) then Km;n has P9 - factorization.
Proof:: Let r = r1 :r2 , where r1 = (2p + 5q) and r2 = (5p + 8q), m0 = 40(p + 2q),
n0 = 20(p + 2q) , let
1)+h+2u+v ;
6
U.S.RAJPUT AND BAL GOVIND SHUKLA
X = fxi;j ; 1 i r1 ; 1 j m0 g;
Y = fyi;j ; 1 i r2 ; 1 j n0 g:
Now for each 1 i p, let
Ei = fx2(i 1)+u;j+(5p+2q)u+2(5p+2q)v y5(i 1)+u+3v;j+20(i
j 20(p + 2q); 1 u 2; 0 v 1; 0 w; h 1g:
Now for each 1 i q; let
Ep+i
1
= fx2p+5(i
j
1)+5u+5v+4w+2h 1 ; 1
1)+u+v;j+20(p+2q)w y5p+8(i 1)+2u+v;j+20p+40(i 1)+10u+4v+2w+h 1 ;
20(p + 2q); 1
u
3; 1
v
2; 0
w; h
1g:
Lemma 6. Consider m = 40(4p + 5q)(p + q) and n = 40(5p + 4q)(p + q) .Then
Km;n has P9 -factorization.
Proof: Let r = r1 :r2 where r1 = 5(4p + 5q), r2 = (5p + 4q) and m0 = 8(p + q),
n0 = 40(p + q), Now consider,
X = fxij ; 1 i r1 ; 1 j m0 g;
Y = fyij ; 1 i r2 ; 1 j n0 g:
Now for each 1 i p, let
Ei = fx20(i 1)+2h+u+v+5w;j y5(i 1)+u+v;j+40(i 1)+8h+w 1 ; 1 j
8(p + q); 1
h 5; 1 u 3; 1 v 2; 0 w 1g:
Now 1 i q, let
Ep+i = fx20p+5(i 1)+h;j y5p+4(i 1)+u+w;j+40p+40(i 1)+8h+w 1 ; 1
j
8(p +
q); 1 h 5; 1 u 3; 0 w 1g:
Proof: By applying lemmas (2) to (7) with theorem (1) along with theorem
(2), it can be seen that when the parameters m and n satisfy condition (1)-(4) in
theorem (1), the graph Km;n has P9 -factorization.
References
[1] Wang.H, P2k -factorization of a complete bipartite graph, Discrete math.120(1993)307-308:
[2] Du. B. P3 -factorization of complete multi graph. App. Math, j.Chinese univ. 14B (1999) 122124.
[3] Ushio.K. . P3 - factorization of complete bipartite graphs, Discrete math.72 (1988) 361-366.
[4] Wang.J, Du. B. P5 - factorization of complete bipartite graphs. Discrete math. 308 (2008) 1665
– 1673.
[5] Wang.J, P7 - factorization of complete bipartite graphs. Australasian Journal Of Combinatory,
volume 33 (2005), 129-137.
[6] Ushio.k.G-designs and related designs. Discrete maths. 116(1993) 299-311.
[7] Harary.F. Graph theory.Adison Wesley. Massachusetts, 1972.
Departmet of mathematics and astronomy,, Lucknow university Lucknow.
E-mail address : [email protected]
Departmet of mathematics and astronomy,, Lucknow university Lucknow.
E-mail address : [email protected]