An upper bound for the number of perfect matchings in graphs

An upp er b ound fo r the n um b er of p erfect matc hings in g raphs Shm uel F riedland ∗ Departmen t of M a thematics, Statistics, and Computer Science, Univ ersit y of Illinois at Chicago Chicago, Illinois 60607- 7045, USA 6 Marc h, 2008 Abstract W e giv e an upp er b ound on the num b er of p erfect matchings in an undirected simple graph G with an even num b er of vertic es, in terms of the degrees of all the vertices in G . This b ound is sharp if G is a u nion of complete bipartite graphs. This b ound is a generalization of the upp er b ound on the num b er of p erfect matchings in bipartite graphs on n + n vertice s given b y the Bregman-Minc in eq ualit y for the p ermanents of (0 , 1) matrices. 2000 Mathematics Su b ject Clas sification: 05A15, 05C70. Keywo rd s and phrases: P erfect matc h ings, p ermanen ts, hafnians. 1 In tro duction Let G = ( V , E ) b e an undirected s imple g raph with the set of vertices V and edg es E . F or a vertex v ∈ V denote by deg v the degree of the vertex v . Assume that # V is even. Denote by pe r fmat G the num b er of p erfect matching in G . Our main re sult states that per fmat G ≤ Y v ∈ V ((deg v )!) 1 2 deg v , (1.1) W e assume here that 0 1 0 = 0. This result is sharp if G is a disjoint union of complete bipartite gr aphs. F or bipartite gra phs the a bov e ineq ua lit y fo llo ws from the Br e g man-Minc inequality for the per manen ts of (0 , 1) matrices, conjectur e d by Minc [4] and prov e d by Bregman [2]. In fact, the inequality (1.1 ) is the analo g of the Breg man-Minc inequality for the hafnians of (0 , 1) symmetric of even order with ze ro dia gonal. Our pr oof follows closely the pro of o f the Bregman-Minc inequality given by Schrijver [6]. 2 P ermanen ts and Hafnians If G is a bipar tite graph on n + n vertices then p erfmat G = p erm B ( G ), where B ( G ) = [ b ij ] ∈ { 0 , 1 } n × n is the incidence matrix of the bipar tite graph G . Thus V = V 1 ∪ V 2 and E ⊂ V 1 × V 2 , where V i = { v 1 ,i , . . . , v n,i } for i = 1 , 2 . Then b ij = 1 if and only if ( v i, 1 , v j, 2 ) ∈ E . Recall that the per manen t o f B ∈ R n × n is given by p erm B = P σ ∈S n Q n i =1 b iσ ( i ) , where S n is the symmetric group of all pe r m utations σ : h n i → h n i . Vice versa, given any (0 , 1) matrix B = [ a ij ] ∈ { 0 , 1 } n × n , then B is the incidence matrix of the induced G ( B ) = ( V 1 ∪ V 2 , E ). Deno te by h n i := { 1 , . . . , n } , m + h n i := { m + 1 , . . . , m + ∗ Visiting Professor, F all 2007 - Wint er 2008, Berlin Mathematical School, Berlin, Germany 1 n } for a ny tw o p ositive integers m, n . It is conv enient to identify V 1 = h n i , V 2 = n + h n i . Then r i := P n j =1 b ij is the i − th degree of i ∈ h n i . The celebrated Bregman-Minc inequalit y , conjectured by Minc [4] and prov ed by Breg man [2], states per m B ≤ n Y i =1 ( r i !) 1 r i . (2.1) A simple pro of Breg man-Minc inequality is given [6]. F urthermo r e the a bov e inequality is generalized to nonneg ativ e matrices. See [1, 5] for additiona l pr oofs of (2.1). Prop osition 2. 1 L et G = ( V 1 ∪ V 2 , E ) b e a bip artite gr aph with # V 1 = # V 2 . Then (1.1) holds. If G is a union of c omplete bip artite gr aphs then e quality holds in (1.1). Proof A s sume tha t # V 1 = # V 2 = n . Clearly , per fmat G = p erm B ( G ) = p erm B ( G ) ⊤ = p per m B ( G ) q per m B ( G ) ⊤ . Note that the i − th row sum of B ( G ) ⊤ is the degr ee of the vertex n + i ∈ V 2 . Apply the Bregman-Minc inequality to per m B ( G ) and p erm B ( G ) ⊤ to deduce (1.1). Assume that G is the c o mplete bipar tite g raph K r,r on r + r vertices. Then B ( K r,r ) = J r = { 1 } r × r . So p erfmat K r,r = r !. Hence equa lity holds in (1.1). Assume that G is a (disjoint ) union of G 1 , . . . G L . Since p erfmat G = Q L i =1 per fmat G i , we deduce (1.1) is sharp if each G i is a complete bipartite graph.  Let A ( G ) ∈ { 0 , 1 } m × m be the adjac ency matrix of an undirected simple gra ph G o n m vertices. Note that A ( G ) is a symmetric matr ix with zero diag o nal. Vice versa, an y symmetric (0 , 1) matrix with zero diagonal induces an indirected simple gr a ph G ( A ) = ( V , E ) on m vertices. Ident ify V with h m i . Then r i , the i − t h row sum o f A , is the degree of the vertex i ∈ h m i . Let K 2 n be the complete graph on 2 n vertices, a nd denote by M ( K 2 n ) the set of all p er- fect matches in K 2 n . T he n α ∈ M ( K 2 n ) ca n b e repres en ted as α = { ( i 1 , j 1 ) , ( i 2 , j 2 ) , .., ( i n , j n ) } with i k < j k for k ∈ h n i . It is convenien t to view ( i k , j k ) as an edge in K 2 n . W e ca n view α as an inv olutio n in S 2 n with no fixed p oin ts. So for l ∈ h 2 n i α ( l ) is seco nd vertex corre - sp onding to l in the per fect match g iv en by α . Vice versa, any fixed p oin t free inv olution of h 2 n i induces a p erfect matc h α ∈ M ( K 2 n ). Denote by S m the space of m × m real symmetric matrices. Assume that A = [ a ij ] ∈ S 2 n . Then the hafnian of A is defined as hafn A := X α = { ( i 1 ,j 1 ) , ( i 2 ,j 2 ) ,.., ( i n ,j n ) }∈M ( K 2 n ) n Y k =1 a i k j k . (2.2) Note that hafn A do es no t de p end o n the diagonal ent r ies of A . Let i 6 = j ∈ h 2 n i . Denote by A ( i, j ) ∈ S 2 n − 2 the symmetr ic matrix obtained fro m A by deleting the i, j rows a nd columns of A . The following prop osition is str aigh tfor w ard, and is known as the expans io n of the hafnian by the r o w, (column), i . Prop osition 2. 2 L et A ∈ S 2 n . Then for e ach i ∈ h 2 n i hafn A = X j ∈h 2 n i\{ i } a ij hafn A ( i, j ) (2.3) It is clear that p erfmat G = hafn A ( G ) for any G = ( h 2 n i , E ). Then (1.1) is equiv alent to the inequality hafn A ≤ 2 n Y i =1 ( r i !) 1 2 r i for all A ∈ { 0 , 1 } (2 n ) × (2 n ) ∩ S 2 n, 0 (2.4) Our pro of of the ab ov e inequa lity follows the pro of of the Bregman-Minc inequalit y given by A. Schrijv er [6]. 2 3 Preliminaries Recall that x lo g x is a s trict c o n vex function on R + = [0 , ∞ ), where 0 log 0 = 0. Hence P r j =1 t j r log P r j =1 t j r ≤ 1 r r X j =1 t j log t j , for t 1 , . . . , t r ∈ R + . (3.1) Clearly , the ab ov e ineq ualit y is equiv alent to the inequality ( r X j =1 t j ) P r j =1 t j ≤ r P r j =1 t j r Y j =1 t t j i for t 1 , . . . , t r ∈ R + . (3 .2) Here 0 0 = 1. Lemma 3. 1 L et A = [ a ij ] ∈ { 0 , 1 } (2 n ) × (2 n ) ∩ S 2 n, 0 . Then for e ach i ∈ h 2 n i (hafn A ) hafn A ≤ r hafn A i Y j,a ij =1 (hafn A ( i, j )) hafn A ( i,j ) . (3.3) Proof Let t j = hafn A ( i, j ) for a ij = 1. Use (2.3) and (3.2) to deduce (3.3).  T o prove our main r esult we nee d the following tw o lemmas. Lemma 3. 2 Th e s e qu enc e ( k !) 1 k , k = 1 , . . . , is an incr e asing se quenc e. Proof Clearly , the inequality ( k !) 1 k < (( k + 1)!) 1 k +1 is equiv alent to the inequality ( k !) k +1 < (( k + 1)!) k , which is in turn e q uiv a le n t to k ! < ( k + 1) k , which is obvious.  Lemma 3. 3 F or an inte ger r ≥ 3 the fol lowing ine quality holds. ( r !) 1 r (( r − 2)!) 1 r − 2 < (( r − 1)!) 2 r − 1 . (3.4) Proof Raise the b oth sides of (3.4) to the p ow er r ( r − 1 )( r − 2) to deduce that (3.4) is equiv alent to the inequa lity ( r !) ( r − 1)( r − 2) (( r − 2 )!) r ( r − 1) < (( r − 1)!) 2 r ( r − 2) . Use the identities r ! = r ( r − 1)! , ( r − 1 )! = ( r − 1 )( r − 2)! , 2 r ( r − 2) = ( r − 1 )( r − 2) + r ( r − 1) − 2 , r ( r − 1 ) − 2 = ( r + 1 )( r − 2) to deduce that the a bov e inequality is equiv alent to r ( r − 1)( r − 2) (( r − 2)!) 2 < ( r − 1) ( r +1)( r − 2) . T ake the logarithm of the ab ov e inequality , divide it by ( r − 2) deduce tha t (3.4) is equiv alent to the inequality ( r − 1) log r + 2 r − 2 log( r − 2)! − ( r + 1) log( r − 1) < 0 . This inequality is equiv alent to s r := ( r − 1) log r r − 1 + 2  1 r − 2 log( r − 2)! − log( r − 1 )  < 0 for r ≥ 3 . (3.5) Clearly ( r − 1 ) log r r − 1 = ( r − 1) log(1 + 1 r − 1 ) < ( r − 1) 1 r − 1 = 1 . 3 Hence (3.5) holds if 1 r − 2 log( r − 2)! − log ( r − 1) < − 1 2 . (3.6) Recall the Stirling’s for m ula [3, pp. 52] log k ! = 1 2 log(2 π k ) + k log k − k + θ k 12 k for some θ k ∈ (0 , 1) . (3.7) Hence log( r − 2)! r − 2 < log 2 π ( r − 2) 2( r − 2) + log ( r − 2 ) − 1 + 1 12( r − 2) 2 . Thu s 1 r − 2 log( r − 2)! − log( r − 1 ) < log 2 π ( r − 2) 2( r − 2) + log r − 2 r − 1 + 1 12( r − 2) 2 − 1 . Since e x is conv ex, it follows that 1 + x ≤ e x . Hence 1 r − 2 log( r − 2)! − lo g( r − 1) < log 2 π ( r − 2 ) 2( r − 2) − 1 r − 1 + 1 12( r − 2) 2 − 1 . Note that − 1 r − 1 + 1 12( r − 2) 2 < 0 for r ≥ 3 . Therefor e 1 r − 2 log( r − 2)! − log( r − 1 ) < log 2 π ( r − 2) 2( r − 2) − 1 . (3.8) Observe next that that the function log 2 π x 2 x is decrea sing for x > e 2 π . Hence the right-hand side o f (3.8) is a decrea s ing sequence for r = 3 , . . . , . Since log 2 π · 3 2 · 3 = 0 . 489 4, it follows that the r igh t-ha nd side of (3.8) is less than − 0 . 51 for r ≥ 5. Therefore (3.5) holds for r ≥ 5. Since s 3 = log 9 16 < 0 , s 4 = log 128 243 < 0 we deduce the lemma.  The arg umen ts of the Pro of of Lemma 3 .3 yield that s r , r = 3 , . . . , conv er ges to − 1. W e chec ked the v alues of this sequence for r = 3 , . . . , 100, and we found that this seq uence decreases in this r ange. W e c o njecture that the sequence s r , r = 3 , . . . decreas es. 4 Pro of of generalized Bregman-Minc inequalit y Theorem 4. 1 L et G = ( V , E ) b e un di re cte d simple gr aph on an even nu mb er of vertic es. Then t he ine quality (1.1) holds. Proof W e pr o ve (2.4). W e use the induction on n . F or n = 1 (2.4) is trivial. Assume that theorem ho lds for n = m − 1. Let n = m . It is enough to a ssume that hafn A > 0. In particular each r i ≥ 1. If r i = 1 for some i , then by expanding hafn A by the ro w i , using the induction hypo thesis and Lemma 3.2, we deduce ea sily the theorem in this c a se. Hence we assume that r i ≥ 2 for each i ∈ h 2 n i . Let G = G ( A ) = ( h 2 n i , E ) b e the g raph induced by A . Then hafn A > 0 is the n umber of per fect matchings in G . Denote by M := M ( G ) ⊂ M ( K 2 n ) the s e t of a ll p erfect matchings in G . Then # M = hafn A . W e now follow the a rgumen ts in the pr o of of the Bre g man-Minc theorem given in [6] with the corres p onding mo difications. 4 (hafn A ) 2 n hafn A (1) = 2 n Y i =1 (hafn A ) hafn A (2) ≤ 2 n Y i =1  r hafn A i Y j,a ij =1 (hafn A ( i, j )) hafn A ( i,j )  (3) = Y α ∈M  2 n Y i =1 r i  2 n Y i =1 hafn A ( i, α ( i )  (4) ≤ Y α ∈M  2 n Y i =1 r i  2 n Y i =1  Y j ∈h 2 n i\{ i,α ( i ) } ,a ij = a α ( i ) j =0 ( r j !) 1 2 r j   Y j ∈h 2 n i\{ i,α ( i ) } ,a ij + a α ( i ) j =1 (( r j − 1)!) 1 2( r j − 1)  Y j ∈h 2 n i\{ i,α ( i ) } ,a ij + a α ( i ) j =2 (( r j − 2)!) 1 2( r j − 2)  (5) = Y α ∈M  2 n Y i =1 r i  2 n Y j =1  Y i ∈h 2 n i\{ j,α ( j ) } ,a ij = a α ( i ) j =0 ( r j !) 1 2 r j   Y i ∈h 2 n i\{ j,α ( j ) } ,a ij + a α ( i ) j =1 (( r j − 1)!) 1 2( r j − 1)  Y i ∈h 2 n i\{ j,α ( j ) } ,a ij + a α ( i ) j =2 (( r j − 2)!) 1 2( r j − 2)  (6) ≤ Y α ∈M  2 n Y i =1 r i  2 n Y j =1  ( r j !) 2 n − 2 r j 2 r j  (( r j − 1)!) 2( r j − 1) 2( r j − 1)  (7) = Y α ∈M  2 n Y i =1 ( r i !) 2 n 2 r i  (8) =  2 n Y i =1 ( r i !) 1 2 r i  2 n hafn A . W e now expla in each step o f the pro of. 1. T r ivial. 2. Use (3.3). 3. The n umber of factors of r i is e q ual to ha fn A on b oth s ides, while the n umber of factors hafn A ( i, j ) equals to the num b er of α ∈ M such that α ( i ) = j . 4. Apply the induction hypothes is to each ha fn A ( i, α ( i )). Note that since the edge ( i, α ( i )) app ears in the p erfect matching α ∈ M , it follows that ha fn A ( i, α ( i )) ≥ 1. Hence if j ∈ h 2 n i\ { i, α ( i ) } and r j = 2 we must have tha t a ij + a α ( i ) j ≤ 1. 5. Chang e the or der of multiplication. 6. Fix α ∈ M and j ∈ h 2 n i . Then j is matched w ith α ( j ). Consider all other n − 1 edges ( i, α ( i )) in α . j is connected to r j − 1 vertices in h 2 n i\ { j, α ( j ) } . Assume there are s tria ngles formed by j and the s edges out o f n − 1 edg es in α \ ( j, α ( j )). Then j is connected to t = r j − 1 − 2 s edges vertices i ∈ h 2 n i\ { j, α ( j ) } such that j is no t connected to α ( i ). Hence there a re 2 n − 2 − (2 t + 2 s ) vertices k ∈ h 2 n i\{ j, α ( j ) } such that j is not connected to k and α ( k ). Therefore, for this α and j we hav e the following terms in (5):  Y i ∈h 2 n i\{ j,α ( j ) } ,a ij = a α ( i ) j =0 ( r j !) 1 2 r j  Y i ∈h 2 n i\{ j,α ( j ) } ,a ij + a α ( i ) j =1 (( r j − 1)!) 1 2( r j − 1)   Y i ∈h 2 n i\{ j,α ( j ) } ,a ij + a α ( i ) j =2 (( r j − 2)!) 1 2( r j − 2)  = ( r j !) 2 n − 2 − (2 s +2 t ) 2 r j (( r j − 1)!) 2 t 2( r j − 1) (( r j − 2)!) 2 s 2( r j − 2)! = ( r j !) 2 n − r j − 1 2 r j (( r j − 2)!) r j − 1 2( r j − 2)!  ( r j !) − 1 r j (( r j − 2)!) − 1 ( r j − 2) (( r j − 1)!) 2 ( r j − 1)  t 2 . (4.1) 5 In the last step we used the equa lit y r j − 1 = 2 s + t . Ass ume first tha t r j > 2. Use Lemma 3.3 to deduce that (4.1) increa ses in t . Hence the maximum v alue of (4.1) is achiev ed w hen s = 0 a nd t = r j − 1. Then (4.1) is equa l to ( r j !) 2 n − 2 r j 2 r j (( r j − 1)!) 2( r j − 1) 2( r j − 1) . If r j = 2 then, a s we expla ined ab ov e, s = 0. Hence (4 .1 ) is also eq ual to the ab ov e expression. Hence (6 ) holds . 7. T r ivial. 8. T r ivial. Thu s (hafn A ) 2 n hafn A ≤  2 n Y i =1 ( r i !) 1 2 r i  2 n hafn A . This establishes (2.4).  References [1] N. Alon and J.H. Spencer , The Pr ob abilistic Metho d , Wiley , New Y ork, 19 92. [2] L.M. Bre g man, Some pro perties of no nnegativ e matrices a nd their pe rmanen ts, Soviet Math. Dokl. 14 (197 3), 945-9 49. [3] W. F eller, An Int r o duction to Pr ob ability and Its Applic ations , vol I, J.Wiley , 1 958. [4] H. Minc, Upper b ounds for per manen ts of (0 , 1)-matrices , Bul l. Amer. Math. So c. 69 (1963 ), 7 89-791. [5] J. Radhakris hna n, An entropy pr o of of Bregma n’s theor em, J. Comb. The ory S er. A 77 (1997 ), 1 61-164. [6] A. Sc hrijver, A short pro of of Minc’s conjecture, J. Comb. The ory Ser. A 25 (1978), 80-83 . 6