An analog of Alspach’s problem on complete bipartite graph 高金美 淡江大學數學系 2005. 8. 5 2019/4/9

Outline Complete graph - m-cycle decomposition - Alspach’s conjecture Complete bipartite graph - known results - recent results Complete tripartite graph Further questions 2019/4/9

History Schoolgirls problem Is it possible for a schoolmistress to take 15 schoolgirls on a walk each day, walking with 5 rows of 3 girls each in such a way that each pair of girls walks together in the same row on exactly one day of each week? 2019/4/9

1847 Rev. T. P. Kirkman posed the problem 1847 Rev. T. P. Kirkman posed the problem. In 1850, he solved the existence problem of triple systems (3-cycle system). A Steiner triple system is an edge disjoint collection of 3-cycles which partition Kn. S(2,3,v)-design  v  1, 3 (mod 6) 2019/4/9

1960 Hanani A Steiner quadruple system is an edge disjoint collection of K4 which partition Kn. S(3,4,v)-design  v  2, 4 (mod 6) A Steiner pentagon system is an edge disjoint collection of K5 which partition Kn. Pentagon system C5|Kv  v  1, 5 (mod 10) 2019/4/9

An m-cycle system of order n is a pair (S,C), where C is an edge disjoint collection of m-cycles which partition Kn based on S. - Contemporary Design Theory(Chap.8) by J.H. Dinitz and D.R. Stinson (1992). - C.C.Lindner and C.A.Rodger, Decomposition into Cycles II: Cycle systems, 2019/4/9

m-cycle decomposition of Kn 2001 Alspach, Brian and Gavlas, Heather proved that An m-cycle decomposition of Kn exists iff (1) n  m； (2) n is odd； and (3) 2m divides n(n1). 2019/4/9

m-cycle decomposition of Kn – I An m-cycle decomposition of Kn – I exists iff (1) n  m； (2) n is even； and (3) 2m divides n(n2). 2019/4/9

Alspach’s Conjecture 1981 B. Alspach asked the following question: (1) If n is odd and a1 + a2 + … + am = n(n-1)/2 with 3  ai  n, can the edges of Kn be partitioned into m cycles C1, C2, …, Cm where Ci has length ai? (2) If n is even and a1 + a2 + … + am = n(n-2)/2, can the edges of Kn – F be partitioned into m cycles C1, C2, …, Cm where Ci has length ai? 2019/4/9

The Alspach’s conjecture is true, if n  10 ------------------- A. Rosa n  14 ---------------- P.N. Balister (2000) 2019/4/9

P. Adams, D.E. Bryant & A. Khodkar G = Kn or Kn – I G can be partitioned into r 3-cycles and s 5-cycles  3r + 5s = |E(G)| (1998) G can be partitioned into two even cycles a and b, when (a,b) = (4,10), (6,8), (6,10), or (8,10). (1998) G can be partitioned into two even cycles a and b, a < b and n < 7b.(2000) 2019/4/9

P. Adams, D.E. Bryant & etc G can be partitioned into h Hamilton cycles and t triangles  hn + 3t = |E(G)| (2004) K10n can be factored into  C5 –factors and  1-factors for 2 +  = 10n – 1. (2003) 2019/4/9

In 1989, K. Heinrich, P. Horak, and A In 1989, K. Heinrich, P. Horak, and A. Rosa: (1) a1, a2, …, am  {n – 2, n – 1, n} (2) a1, a2, …, am  {3, 4, 6} (3) a1, a2, …, am  {2k, 2k+1} 2019/4/9

In 1995, D.E. Bryant and P.Adams proved that, n is odd (1) a1,a2,…,am  {3, 3, 4, 5, …, n – 1}, as n  2 (2) a1,a2,…, am  {3,4,5,…, n – 4, n – 2, n – 1, n}, as n  7. 2019/4/9

P.N. Balister If the cycle lengths mi are bounded by some linear function of n and n is sufficiently large then this conjecture is true. 2019/4/9

1. Assume n  2 mod 144. If 72  mi  (n+37)/20 and m1+m2+ …+ mt = n(n-2)/2, then Kn  I can be packed with cycles of lengths m1,m2, …, mt. 2. Assume n  N1 and n  2 mod 144. If 3  mi  (n+37)/20 and m1+m2+ …+ mt = n(n-2)/2, then Kn  I can be packed with cycles of lengths m1,m2, …, mt. 3. Assume n  N2. If 3  mi  (n112)/20 and m1+m2+ …+ mt = n(n 1)/2 (n odd) or n(n-2)/2 (n even), then one Kn or Kn  I (n even) with cycles of lengths m1,m2, …, mt. 2019/4/9

Complete bipartite graphs 1981 D. Sotteau, (1) C2k|Km,n  m,n  k, 2k|mn (2) DC2k|K*m,n  m,n  k, k|mn. 2019/4/9

(1) Ka,b|Km,n (2) Ka,b|Kn,n,…,n, a  b 1981 Ushio, Kazuhiko (1) Ka,b|Km,n (2) Ka,b|Kn,n,…,n, a  b 1982 Little, Charles K4n,4n can be written as the union of n edge disjoint planar graphs and a 1-factor. 2019/4/9

1985 Truszczynski, Mirosaw Km,n (K 1985 Truszczynski, Mirosaw Km,n (K*m,n) can be partitioned into (1) path of length k (2) path of lengths k1, k2, …, kt, where k1 + k2 + …+ kt = mn 2019/4/9

Gavlas, Heather and etc m-cycle system of Kn,n – I exists  m  4 is even, n  3 is odd m 2n and m|n(n – 1) except m  0 (mod 4) with n < m < 2n (2004) 馬駿, 上海交大 2019/4/9

Alspach’s Conjecture!! (1) If n is even and a1 + a2 + … + am = n2, can the edges of Kn,n be partitioned into m cycles which have lengths a1, a2, …, am ? (2) If n is odd and a1 + a2 + … + am = n(n-1), can the edges of Kn,n – F be partitioned into m cycles which have lengths a1, a2, …, am ? 2019/4/9

K. Heinrich, P. Horak, and A. Rosa proved that (1) Let m and n be even. If Km,n = Ci1 + Ci2 + … + Cit then K2m,2n = Di1 + Di2 + … + Dit, where Dik = 4Cik or Dik = 2Cik, k = 1,2,…,t. (2) Let y and n be even and 0  y  n. Then K2n,2n = (2n  2y)C2n + yC4n 2019/4/9

Recent results: 2019/4/9

{4,8}-cycles decomposition of Km,n C.M.Fu, W.C.Huang (1998) (1) Let m and n be even and m  4, n  6. If 4p + 8q = mn, then Km,n = pC4 + qC8.. (2) Let n be odd. (i) If n 1(mod 4) and 4p + 8q = n(n1), then Kn,n  F = pC4 + qC8., and (ii) If n 3(mod 4) and 4p + 8q = (n3)(n+2), the Kn,n  (FC6) = pC4 + qC8.. 2019/4/9

How to do it? c1 c2 c3 c4 r1 1 1 2 2 r2 2 1 1 2 r3 2 2 1 1 r4 1 2 2 1 1 represents 8-cycle (r1, c1, r4, c4, r3, c3, r2, c2), 2 represents 8-cycle (r1, c3, r4, c2, r3, c1, r2, c4). r1 r2 r3 r4 c1 c2 c3 c4 2019/4/9

c1 c2 c3 c4 c5 r1 * 1 1 2 2 r2 2 * 1 1 2 r3 3 3 * 1 1 r4 3 1 3 * 1 r5 2 3 3 2 * 1 represents 8-cycle (r1, c2, r4, c5, r3, c4, r2, c3), 2 represents 6-cycle : (r1, c4, r5, c1, r2, c5), 3 represents 6-cycle : (r3, c1, r4, c3, r5, c2). 2019/4/9

--->K6,2t = K6,2t-4 + K6,4.  K4s+2,2t = K4s-4,2t + K6,2t. m and n are even: K4.4 = 4C4 = 2C8. K4,6, K4,8 ---> K4,2t  K4s,2t = s copies of K4,2t. --->K6,2t = K6,2t-4 + K6,4.  K4s+2,2t = K4s-4,2t + K6,2t. 2019/4/9

 K4s+1,4s+1 － F = (K4s-7,4s-7－F) + (K9,9－F) + 2 copies of (K8,4s-8). n is odd: K9,9 － F , K11,11 － (F  C6 )  K4s+1,4s+1 － F = (K4s-7,4s-7－F) + (K9,9－F) + 2 copies of (K8,4s-8).  K4s+3,4s+3 － (F  C6) = (K4s-7,4s-7－F) + (K11,11－(F  C6)) + 2 copies of K10,4s-8. 2019/4/9

{4,10}-cycles decomposition of Km,n 1. If 10r+4s = 4mn for all r, s ≥ 0, then K2m,2n = rC10+sC4 for m,n ≥ 3. 2. If 10r+4s = 2n(2n+1) for all r, s ≥ 0, then K2n+1,2n+1 – F = rC10+sC4, for n ≥ 2. 2019/4/9

Sketch Proof K2m,2n = K2m-10,2n-10+K2m-10,10+K2n-10,10+K10,10 10 2m-10 2019/4/9

Sketch Proof K2n+1,2n+1 – F = (K11,11 – F1)+(K2n-9,2n-9 – F2) +2K10,2n-10. 2n-9 10 2n-10 11 K10,2n-10 K2n-9,2n-9 – F1 K11,11 – F2 2019/4/9

Decomposition of Km,n into C4 or C12 If 12r+4s = 4mn for all r, s ≥ 0 then K2m,2n = rC12+sC4, for m, n ≥ 3. If 12r+4s = 4n(4n+1) for all r, s ≥ 0, then K4n+1,4n+1 – F = rC12+sC4 for n ≥ 2. If 12r+4s = 4n(4n+5) for all r, s ≥ 0, then K4n+3,4n+3 – (F ∪C6 )= rC12 + sC4 for n ≥ 1. 2019/4/9

Sketch Proof K4n+3,4n+3 – (F∪C6) = (K4n-9,4n-9 – (F1∪C6))+(K13,13 – F2)+2K12,4n-10 4n-9 12 4n-10 13 K13,13 - F K12,4n-10 K4n-9,4n-9 - (F∪C6) 2019/4/9

Decomposition of Km,n into C4 or C2t 1. Let t be odd. If K2u,2v can be decomposed into 4-cycles or 2t-cycles for t+1 ≤ 2u,2v ≤ 3t – 1 then K2m,2n can be decomposed into 4-cycles or 2t-cycles for 2m, 2n ≥ 3t+1. 2019/4/9

Proof : K2m,2n = K2m-t,2n-t + K2m-t,t + Kt,2n-t + Kt,t K2n-t,t 2019/4/9

t is odd 2. Let t be odd. If K2u+1,2u+1 - F can be decomposed into 4-cycles and 2t-cycles for t-1 ≤ 2u ≤ 3t-1 then K2n+1,2n+1 - F can be for 2n ≥ 3t+1. 2019/4/9

t is even 1. If K2u,2v can be decomposed into 4-cycles and 2t-cycles for t ≤ 2u,2v < 2t then K2m,2n can be decomposed into 4-cycles and 2t-cycles for m, n ≥ t. 2019/4/9

t is even 2. If K4u+1,4u+1 – F can be decomposed into 4-cycles or 2t-cycles for t ≤ 4u ≤ 3t – 4 then K4n+1,4n+1 – F can be decomposed into 4-cycles or 2t-cycles for 4n ≥ 3t. 2019/4/9

t is even 3. If K4u+3,4u+3 – (F∪C6) can be decomposed into 4-cycles or 2t-cycles for t – 2 ≤ 4u ≤ 3t – 6 then K4n+3,4n+3 – (F∪C6) can be decomposed into 4-cycles or 2t-cycles for 4n ≥ 3t – 2. 2019/4/9

{4,6,8}-cycles decomposition of Km,n C.M.Fu, W.C.Huang, C.C.Chou (2000) (1) Let m, n be even and m  4, n  6. If 4p + 6q + 8r = mn, then Km,n = pC4 + qC6 + rC8.. (2) Let n be odd. If 4p+6q+8r = n(n1), then Kn,n  F = pC4 + qC6 + rC8.. 2019/4/9

K4,4 = 4C4 = C4 +2C6 = 2C8 K2m,2n = K2m-6,2n + K6,2n K2n+1,2n+1 － F = (K2n-5,2n-5 － F) + (K7,7 － F) + 2 copies of K6,2n-6, or = (K2n-7,2n-7 － F) + (K9,9 － F) + 2 copies of K8,2n-8. 2019/4/9

{4,6,8}-cycles decomposition of 2Km,n Let n  m  4. If 4p + 6q + 8r = 2mn, then 2Km,n = pC4 + qC6 + rC8.. 2019/4/9

If p, q, r < m, then 4p + 6q + 8r < 18m 2mn < 18m  n < 9. If one of p, q, r > m, say p, then consider 2Km,n = 2Km,n-2 + 2K2,m 2019/4/9

Decomposition of Kn,n into cycles of different lengths What is the necessary condition? Is that the sufficient condition?.. 2019/4/9

What is the necessary condition? The degree of each vertex should be even. Consider the decomposition of K2n,2n and K2n+1,2n+1 – F . Since (4 + 6 + 8 + … + 4n) – (2n – 2) = 4n2 = |E(K2n,2n)| and [4 + 6 + 8 + … + (4n + 2) ] – (2n+2) – (2n-2) = 4n2 + 2n = |E(K2n+1,2n+1 – F )| 2019/4/9

For each n  3, Is that possible to decompose K2n,2n into C4, C6, C8, … , C4n except C2n-2 and K2n+1,2n+1 – F into C4, C6, C8, … , C4n+2 except C2n+2 and C2n-2? 2019/4/9

cycle decomposition of 2Kp,q,r Latin square of order n corresponds to the decomposition of Kn,n,n c1 c2 c3 c4 r1 e1 e2 e3 e4 r2 e2 e3 e4 e1 r3 e3 e4 e1 e2 r4 e4 e1 e2 e3 r1 r2 r3 r4 c1 c2 c3 c4 e1 e2 e3 e4 2019/4/9

Corresponding to cycle decomposition of 2Kp,q,r into most cycles, we will use Latin square of order r, where p  q  r. Example: p = 2, q = 3, r = 4. c1 c2 c3 c1 c2 c3 r1 e1 e2 e3 e4 r1 c3 c4 c2 c1 r2 e2 e3 e4 e1 r2 c2 e3 e1 e4 e3 e4 e1 e4 e1 e3 e4 e1 e2 e1 e2 e4 2019/4/9

result 2Kp,q,r can be decomposed into (1) 2pq triangles, and (pr+qr2pq)/2 4-cycles, when (p+q)r is even, (2) 2pq triangles, (pr+qr  2pq  3)/2 4-cycles and one 6-cycle, otherwise, where p  q  r. 2019/4/9

Further questions Is the Alspach’s conjecture true for Km,n? How about the cycle decompositions of the complete multipartite graph? 2019/4/9

The End 2019/4/9

謝謝 2019/4/9