New Upper Bounds on Sizes of Permutation Arrays

JOURNAL OF L A T E X CLASS FIL ES, V OL. 1, NO. 11, NO VE MBER 2002 1 Ne w Upper Bounds on S izes of Permutation Arrays Lizhen Y ang, Ling Dong, Kefei Chen Abstract A permutatio n arr ay(or code) of leng th n and d istance d , denoted by ( n, d ) P A, is a set of permutatio ns C from some fixed set of n elemen ts such th at the Hamming distan ce between d istinct members x , y ∈ C is at least d . Let P ( n, d ) den ote the maximu m size of an ( n, d ) P A. New u pper bound s on P ( n, d ) are given. F or co nstant α, β satisfying cer tain conditions, whenever d = β n α , the new upper bound s are asymptotically better than the previous ones. Index T erms permutatio n arrays (P As), permutation code, upp er bound. I . I N T RO D U C T I O N Let Ω be an arbitrary non empty in finite set. T wo disti nct permutatio ns x , y ov er Ω ha ve distance d if xy − 1 has exactly d unfixed poin ts. A permut ation array(permutation code, P A) of length n and distance d , denoted by ( n, d ) P A, is a set of permu tations C from some fixed set of n elements su ch that the distance betw een distinct mem bers x , y ∈ C is at least d . An ( n, d ) P A of size M is called an ( n, M , d ) P A. The m aximum size of an ( n, d ) P A is deno ted as P ( n, d ) . Manuscript receive d June 14, 2006. This work was su pported by NSFC under grants 90104 005 and 6057 3030. Lizhen Y ang is with the department of computer science and engineering, S hanghai Jiaotong Univ ersity , 800 DongChuan Road, Shan ghai, 200420 , R.P . Chin a (fax: 86-021-34204 221, email: l izhen yan g@msn.com). Ling Dong is with the department of computer science and enginee r ing, Shangh ai Jiaotong Uni versity , 800 Dong Chuan Road, Shanghai, 200420 , R. P . China (fax: 86-021-342 04221, email: l dong@sh16 3c.sta.net.cn). Ke f ei Chen is with the department of computer science and engin eering, Shanghai Jiaotong Uni versity , 800 DongChuan Road, Shanghai, 200420 , R. P . China (fax: 86-021-342 04221, email: C hen-kf@sjtu.edu.cn). Nov ember 16, 2021 DRAFT JOURNAL OF L A T E X CLASS FIL ES, V OL. 1, NO. 11, NO VE MBER 2002 2 P As are som ewhat studies in the 1970s . A recent appl ication by V inck [ ? ], [ ? ], [ ? ], [ ? ] of P As to a coding/m odulation schem e for comm unication over power lines has created renewed i nterest in P As. But there are still many problems unsolved in P As, e.g. one of the essential problem is to compute the values of P ( n, d ) . It’ s known th at determining the exactly va l ues o f P ( n, d ) is a diffic ul t task, except for special cases, it can be only to establish s ome lower boun ds and upper bounds on P ( n, d ) . In this correspondence, we give some n e w upper bounds o n P ( n, d ) , which are asym ptotically better t han the pre v ious ones. A. Concepts and Notations W e introduce concepts and n otations that will b e used t hroughout the correspondence. Since for two sets Ω , Ω ′ of t he same s ize, the s ymmetric groups S y m (Ω) and S y m (Ω ′ ) formed by the permutations ov er Ω and Ω ′ respectiv ely , under compos itions of mappings, are isomorphic, we need only to consider the P As over Z n = { 0 , 1 , . . . , n − 1 } and write S n to denote the special grou p S y m ( Z n ) . In the rest of the correspondence, withou t special p ointed out, we alwa ys assume that P As are over Z n . W e also writ e a permut ation a ∈ S n as an n − tuple ( a 0 , a 1 , . . . , a n − 1 ) , where a i is the i mage of i under a for each i . Especiall y , we writ e the identi cal permutation (0 , 1 , . . . , n − 1) as 1 for con venience. The Ham ming distance d ( a , b ) between two n − tuples a and b is the number of posit ions where they di f fer . Then th e distance between any two permutations x , y ∈ S n is equiv alent to their Ham ming distance. Let C be an ( n, d ) P A. For an arbitrary permutation x ∈ S n , d ( x , C ) stands for the Hamming distance between x and C , i.e., d ( x , C ) = min c ∈ C d ( x , c ) . A permut ation in C is also called a code word of C . For con venience for d iscussion, w ithout loss of generality , we always assum e that 1 ∈ C , and the indies of an n − tuple (vector , array) are started by 0 . The suppo rt o f a binary vector a = ( a 0 , a 1 , . . . , a n − 1 ) ∈ { 0 , 1 } n is defined as the set { i : a i = 1 , i ∈ Z n } , and the weight of a i s the size of i ts sup port, namely the numb er of ones in a . The support of a permutation x = ( x 0 , x 1 , . . . , x n − 1 ) ∈ S n is defined as th e set o f the p oints not fixed by x , namely { i ∈ Z n : x i 6 = i } = { i ∈ Z n : x ( i ) 6 = i } , and the weight of x , denoted as w t ( x ) , is defined as the s ize of it s support, namely the number of point s i n Z n not fixed by x . A derangement of order k is an element of S k with no fixed point s. Let D k be the number of derangement s of order k , with the conv entio n that D 0 = 1 . Then D k = k ! P k i =0 ( − 1) k k ! =  k ! e  , where [ x ] is t he nearest integer function, and e is the base of th e natural logarithm . The ball in Nov ember 16, 2021 DRAFT JOURNAL OF L A T E X CLASS FIL ES, V OL. 1, NO. 11, NO VE MBER 2002 3 S n of radius r wit h center x is t he set of all permutat ions of distance ≤ r from x . T he volume of such a ball is V ( n, r ) = r X i =0  n i  D i . (1) An ( n, d, w ) constant-weight bin ary code is a set of binary vectors of length n , such that each vector contains w o nes and n − w zeros, and any two vectors differ in at least d positions. The lar gest po ssible size of an ( n, d, w ) constant-weight bi nary code is denoted as A ( n, d, w ) . Similarly , we define an ( n, d, w ) constant -weight P A as an ( n, d ) P A such that each permutation is of weight w , and denote the l ar gest possi ble size of an ( n, d, w ) constant-weig ht P A as P ( n, d, w ) . The con cept o f P ( n, d ) can be further generalized. Let Ω ⊆ S n , t hen P Ω ( n, d ) denot es t he maximum size of an ( n, d ) P A C such t hat C ⊆ Ω . For trivial case Ω = S n , P ( n, d ) = P Ω ( n, d ) . B. Pr evious Resul ts The most b asic upp er bound on P ( n, d ) is g iv en by Deza and V anst one [ ? ]. Theor em 1: [ ? ]. P ( n, d ) ≤ n ! ( d − 1)! (2) W e call the P As which attain t he Deza-V ans tone bound perfect P As and the known perfect P As are • ( n, n, n ) P As for each n ≥ 1 ; • ( n, n ! , 2) P As for each n ≥ 1 ; • ( n, n ! / 2 , 3) P As for each n ≥ 1 [ ? ]; • ( q , q ( q − 1) , q − 1) P As for each prime power q [ ? ]; • ( q + 1 , ( q + 1) q ( q − 1) , q − 1) P As for each prime power q [ ? ]; • ( 1 1 , 11 · 10 · 9 · 8 , 8) P A [ ? ]; • ( 1 2 , 12 · 11 · 1 0 · 9 · 8 , 8) P A [ ? ]. The Deza-V anstone bound can be deriv ed by recursiv ely applying t he following inequ ality . Pr opositi on 1: [ ? ]. P ( n, d ) ≤ nP ( n − 1 , d ) . (3) Then for d ≤ m < n , if we k now P ( m, d ) ≤ M < m ! ( d − 1)! , we can get a s tronger upp er bo und on P ( n, d ) : P ( n, d ) ≤ n ! P ( m, d ) m ! ≤ n ! M m ! . Nov ember 16, 2021 DRAFT JOURNAL OF L A T E X CLASS FIL ES, V OL. 1, NO. 11, NO VE MBER 2002 4 Another nontrivial upper bo und on P ( n, d ) is the sphere packing bound obtained by considering the ball s of radi us ⌊ ( d − 1) / 2 ⌋ [ ? ]. Theor em 2: P ( n, d ) ≤ n ! V ( n, ⌊ ( d − 1) / 2 ⌋ ) . (4) For s mall values of n and d , still stron ger upper bo unds are founds in T arnanen [ ? ] by the method of linear programm ing. C. Or g anization and New Results The correspondence is or g anized as follows. In Section II, we first prove a relati on between P ( n, d ) and P Ω ( n, d ) that is the inequali ty P ( n, d ) ≤ n ! P Ω ( n, d ) | Ω | . Next, we give some elementary properties of P ( n, d, w ) , and then use them to show a new upper bound on P ( n, d ) for d is ev en and a new upper bound on P ( n, d ) for d is odd. They are given by the following inequ alities: P ( n, 2 k ) ≤ n ! V ( n, k − 1) + ( n k ) D k ⌊ n/k ⌋ , for 2 ≤ k ≤ ⌊ n/ 2 ⌋ ; P ( n, 2 k + 1) ≤ n ! V ( n, k ) + ( n k +1 ) D k +1 − A ( n − k , 2 k ,k +1) ( n k ) D k A ( n, 2 k ,k +1) , for 2 ≤ k ≤ ⌊ ( n − k − 1) / 2 ⌋ . In Section III, we compare the upper bou nds on P ( n, d ) and show for constant α , β s atisfying certain conditions , whene ver d = β n α , the new upper bou nds are asymptotically b etter th an th e pre v ious ones. I I . T H E N E W U P P E R B O U N D S Theor em 3: Let Ω be a su bset of S n . Then P ( n, d ) ≤ n ! P Ω ( n, d ) | Ω | . Pr oof: Suppose C is an ( n, P ( n, d ) , d ) P A. For any x ∈ S n , let x C = { xc : c ∈ C } . Then X x ∈ S n | x C ∩ Ω | = X c ∈ C X ω ∈ Ω |{ x ∈ S n : xc = ω }| = X c ∈ C X ω ∈ Ω |{ ω c − 1 }| = P ( n, d ) | Ω | . Nov ember 16, 2021 DRAFT JOURNAL OF L A T E X CLASS FIL ES, V OL. 1, NO. 11, NO VE MBER 2002 5 On the other hand, there must exist x ′ ∈ S n such that n ! | x ′ C ∩ Ω | ≥ P x ∈ S n | x C ∩ Ω | . Then n ! | x ′ C ∩ Ω | ≥ P ( n, d ) | Ω | , in other words, P ( n, d ) ≤ n ! | x ′ C ∩ Ω | | Ω | . This in conju nction wit h | x ′ C ∩ Ω | ≤ P Ω ( n, d ) results the theorem. QED. Since S d can be considered as a subset of S n for d ≤ n , Theorem 1 is als o a directly result of the above theorem, in fact P ( n, d ) ≤ | S n | P ( d, d ) | S d | = n ! d d ! = n ! ( d − 1)! . The following is also obtained im mediately by Theorem 3. Cor ollary 1: P ( n, d ) ≤ n ! P ( n, d, w )  n w  D w . The following are well-known elementary p roperties of A ( n, d, w ) , whi ch wil l be applied to the proof of the properties o f P ( n, d, w ) . Lemma 1: A ( n, d, w ) = 1 , if d > 2 w ; A ( n, 2 w , w ) = j n w k ; A ( n, 2 k , k + 1) ≤  n k + 1  n − 1 k  . Theor em 4: ( I ) P ( n, d, w ) ≤ A ( n, 2 d − 2 w , w ) , for d > w ; ( I I ) P ( n, d, w ) = 1 , for d > 2 w , w 6 = 1 , d ≥ 1; ( I I I ) P ( n, 2 k , k ) = ⌊ n k ⌋ , for 2 ≤ k ≤ ⌊ n/ 2 ⌋ ; ( I V ) P ( n, 2 k + 1 , k + 1) = A ( n, 2 k , k + 1) , for 1 ≤ k ≤ ⌊ ( n − 1) / 2 ⌋ ; ( V I ) P ( n, 4 , 3) ≤ 2 ( n 2 ) 3 , for n ≥ 4 . (5) Pr oof: Part ( I ) Let C be an ( n, d, w ) constant-weight P A wit h maxim al size P ( n, d, w ) , where d > w . Define f : S n 7→ { 0 , 1 } n such th at for any a = ( a 0 , a 1 , . . . , a n − 1 ) ∈ S n with support A , f ( a ) = a ′ = ( a ′ 0 , a ′ 1 , . . . , a ′ n − 1 ) ∈ { 0 , 1 } n , where a ′ i =    1 , for i ∈ A, 0 , for i 6∈ A. (6) Then C ′ = { f ( a ) : a ∈ C } is an ( n, 2 d − 2 w , w ) const ant-weight code with size P ( n, d, w ) and this means P ( n, d, w ) ≤ A ( n, 2 d − 2 w , w ) . T o prove thi s fact we need only t o prove that C ′ hav e mu tual distances ≥ 2 d − 2 w . Nov ember 16, 2021 DRAFT JOURNAL OF L A T E X CLASS FIL ES, V OL. 1, NO. 11, NO VE MBER 2002 6 Let a , b ∈ C , a 6 = b , and let A and B be the supports of a and b respectively . Suppose a ′ = f ( a ) , b ′ = f ( b ) . (6) im plies d ( a ′ , b ′ ) = | ( A/B ) ∪ ( B / A ) | = | A | + | B | − 2 | A ∩ B | = 2 w − 2 | A ∩ B | (7) On th e other hand , we have d ≤ d ( a , b ) ≤ | A ∪ B | = | A | + | B | − | A ∩ B | = 2 w − | A ∩ B | , namely | A ∩ B | ≤ 2 w − d . Putti ng this into (7) we obtain d ( a ′ , b ′ ) ≥ 2 d − 2 w . Since f is an ont o mapping , we comp lete the proof of Part ( I ) . Part ( I I ) For d > 2 w , w 6 = 1 and d ≥ 1 , since 2 d − 2 w > 2 · 2 w − 2 w = 2 w , (8) A ( n, 2 d − 2 w , w ) = 1 (by Lemma 1) . This i n conj unction with part ( I ) yields P ( n, d, w ) = 1 . Part ( I I I ) For 2 ≤ k ≤ ⌊ n/ 2 ⌋ , by part ( I ) and L emma 1 we have P ( n, 2 k , k ) ≤ A ( n, 2 k , k ) = ⌊ n/k ⌋ . On the other hand, we can construct an ( n, 2 k , k ) constant-weight P A as follows: C = { c i = ( c i, 0 , c i, 1 , . . . , c i,n − 1 ) | i = 0 , 1 , . . . , ⌊ n/k ⌋ − 1 } , where c i,j =          j + 1 , for j = ik , ik + 1 , . . . , ik + k − 2 ik , for j = ik + k − 1 j, others . Then we conclude P ( n, 2 k , k ) = ⌊ n/k ⌋ . Part ( I V ) For case 1 ≤ k ≤ ⌊ ( n − 1) / 2 ⌋ , by part ( I ) we hav e P ( n, 2 k + 1 , k + 1) ≤ A ( n, 2 k , k + 1) . Nov ember 16, 2021 DRAFT JOURNAL OF L A T E X CLASS FIL ES, V OL. 1, NO. 11, NO VE MBER 2002 7 Let C ′ be an ( n, 2 k , k + 1) constant-weight binary code with maxim al si ze A ( n, 2 k , k + 1) , then there exists C ⊆ S n such th at for each member of C there have one and o nly one member of C ′ with same support. W e wil l prove that C is an ( n, 2 k + 1 , k + 1) constant-weight P A, whi ch implies P ( n, 2 k + 1 , k + 1) ≥ A ( n, 2 k , k + 1) and then result s P ( n, 2 k + 1 , k + 1) = A ( n, 2 k , k + 1) . Let x , y ∈ C , x 6 = y with correspondi ng supports X and Y . For case X ∩ Y = Ø , d ( x , y ) = | X | + | Y | = 2 k + 2 . So we n eed o nly t o dis cuss th e case X ∩ Y 6 = Ø . L et x ′ , y ′ ∈ C ′ be the corresponding bi nary codew ords with supports X , Y . Since d ( x ′ , y ′ ) = | X | + | Y | − 2 | X ∩ Y | = 2( k + 1 ) − 2 | X ∩ Y | ≥ 2 k , | X ∩ Y | ≤ 1 . Therefore, if X ∩ Y 6 = Ø , then | X ∩ Y | = 1 . Suppose X ∩ Y = { a } . Then x ( a ) 6 = y ( a ) , ot herwise x ( a ) = y ( a ) = a and it lead to a contradictio n. Hence for this case, d ( x , y ) = | A/B ∪ B / A | + |{ a }| = | A/B | + | B / A | + 1 = 2 k + 1 . N ow we conclude that C is an ( n, 2 k + 1 , k + 1) constant-weig ht P A of size A ( n, 2 k , k + 1) , which completes the p roof of Part ( I V ) . Part ( V I ) Suppos e C is an ( n, 4 , 3) constant-weight P A. For any p air { i, j } ∈ Z n × Z n with i 6 = j , let C i,j ⊆ C be the maximal set such that for each x ∈ C i,j with support X , { i, j } ⊆ X . W e are now ready to prove | C i,j | ≤ 2 . Assume t he contrary , i .e., that | C i,j | ≥ 3 and x , y , z are distin ct elements of C i,j . W .l.o.g, ( x ( i ) , x ( j )) = ( k , i ) , where k 6 = i, j . Then ( y ( i ) , y ( j )) = ( j, k ′ ) , w here k ′ 6 = i, j, k , otherwise d ( x , y ) < 4 , which is a cont radiction. Similarly , ( z ( i ) , z ( j )) = ( j, k ′′ ) , where k ′′ 6 = i, j, k , k ′ . Thus d ( y , z ) < 4 , which is a contradictio n. T herefore | C i,j | ≤ 2 . Since there are  n 2  pairs of ( i, j ) ∈ Z n × Z n with i 6 = j , X i,j,i 6 = j | C i,j | ≤ 2  n 2  . (9) On th e other hand , for each mem ber of C , there are exactly 3 C i,j containing it, hence X i,j,i 6 = j | C i,j | = 3 | C | . (10) Substitutin g (10) into (9) yields | C | ≤ 2 ( n 2 ) 3 , this m eans P ( n, 4 , 3) ≤ 2 ( n 2 ) 3 . QED. Theor em 5: For 2 ≤ k ≤ ⌊ n/ 2 ⌋ , P ( n, 2 k ) ≤ n ! V ( n, k − 1) + ( n k ) D k ⌊ n/k ⌋ (11) Pr oof: Let t here be N k permutations in S n which ha ve di stance k to the ( n, M , d ) P A C . Then M V ( n, k − 1) + N k ≤ n ! (12) Nov ember 16, 2021 DRAFT JOURNAL OF L A T E X CLASS FIL ES, V OL. 1, NO. 11, NO VE MBER 2002 8 In order to estimate N k we consi der an arbitrary cod e word c whi ch we can take to be 1 (w .l.o.g.). Then all permutations o f weight k has distance k t o C . Since there are  n k  D k permutations of weight k , there must have  n k  D k permutations that have distance k to C . By varying c we thus count M  n k  D k permutations in S n that have di stance k to the P A. How often has each of th ese permutations been counted. T ake one of them; again w .l.o.g. we call it 1 . The codew ords with distance k to 1 form an ( n, 2 k , k ) const ant-weight P A since they hav e mutual di stances ≥ 2 k and weight k . Hence t here are at most P ( n, 2 k , k ) = ⌊ n/k ⌋ (by part ( I I I ) of Theorem 4) such code words. Thi s gives N k ≥ M ( n k ) D k ⌊ n/k ⌋ . Subs tituting this lower bounds on N k into (12 ) impli es the Theorem. QED. Theor em 6: For 2 ≤ k ≤ ⌊ ( n − k − 1) / 2 ⌋ , P ( n, 2 k + 1) ≤ n ! V ( n, k ) + ( n k +1 ) D k +1 − A ( n − k , 2 k ,k +1) ( n k ) D k A ( n, 2 k ,k +1) (13) Pr oof: Let C be an ( n, M , 2 k + 1) P A. For any x ∈ S n , let B i ( x ) = |{ c : c ∈ C, d ( c , x ) = i }| . The proof reli es on the foll owing lemma. Lemma 2: A ( n, 2 k , k + 1) P i 0 for any i < k , then B k ( x ) = B k +1 ( x ) = 0 and all the other s ummands are zeros, and there is not hing to prove. Assume, therefore, that B i ( x ) = 0 for all i < k . W e know that B k ( x ) ≤ P ( n, 2 k + 1 , k ) = 1 (by part ( I I ) of Theorem 4), in other words B k ( x ) is either 0 or 1: if it is 0, then th e claim becomes B k +1 ( x ) ≤ A ( n, 2 k , k + 1) = P ( n, 2 k + 1 , k + 1) (by part ( I V ) of Theorem 4), which is clear; if it is 1, t hen the claim becomes B k +1 ( x ) ≤ A ( n − k , 2 k , k + 1 ) = P ( n − k , 2 k + 1 , k + 1 ) , which is correct for there are no points moved by both cod e words of weight k and of weight k + 1 . QED. W e are now ready to compl ete the proof of th e theorem. It foll ows from Lemma 2 that P x ∈ S n  A ( n, 2 k , k + 1) P i 0 or α = 1 / 2 , 0 < β < e , whenev er d = β n α , lim n →∞ D V ( n, d ) S P ( n, d ) = ∞ . Pr oof: Let k = ⌊ ( d − 1) / 2 ⌋ . W e have lim n →∞ D V ( n, d ) S P ( n, d ) = lim n →∞ V ( n, k ) ( d − 1)! ≥ lim n →∞  n k  D k ( d − 1)! = lim n →∞ n ! k ! ek !( n − k )!( d − 1)! = lim n →∞ √ 2 π n ( n/e ) n e p 2 π ( n − k )(( n − k ) /e ) n − k p 2 π ( d − 1)(( d − 1) / e ) d − 1 (18) Nov ember 16, 2021 DRAFT JOURNAL OF L A T E X CLASS FIL ES, V OL. 1, NO. 11, NO VE MBER 2002 11 where the last equatio n is foll owed by Stirli ng’ s formula lim n →∞ n ! √ 2 π n ( n e ) n = 1 . By (18), lim n →∞ D V ( n, d ) S P ( n, d ) ≥ 1 √ 2 π lim n →∞ e d − k − 2 n n +1 / 2 ( n − k ) n − k +1 / 2 ( d − 1) d − 1 / 2 . (19) Let c be a constant such that c < 1 . Since lim n →∞  n n − k  n − k k = lim n →∞  1 + 1 n/k − 1  n/k − 1 = e, for n large enou gh,  n n − k  n − k k ≥ e c , i.e. ( n − k ) n − k ≤ e − ck n n − k . (20) Putting (20) into the right side of (19), and mul tiplying th e right side of (19) by lim n →∞ ( n − k ) 1 / 2 n 1 / 2 = 1 and lim n →∞ ( d − 1) d − 1 / 2 e − 1 ( β n α ) d − 1 / 2 = lim d →∞ e ( d − 1 d ) d − 1 / 2 = lim d →∞ e (1 − 1 /d ) d − 1 / 2 = 1 , we obtain lim n →∞ D V ( n, d ) S P ( n, d ) ≥ 1 √ 2 π lim n →∞ e d − k − 2 n n +1 / 2 e − ck n n − k +1 / 2 e − 1 ( β n α ) ( d − 1 / 2) = 1 √ 2 π lim n →∞ e d +( c − 1) k − 1 β − d +1 / 2 n k − αd + α/ 2 = 1 √ 2 π lim n →∞ e (1 − ln β ) d +( c − 1) k − 1+ln β / 2 n k − αd + α/ 2 ≥ 1 √ 2 π lim n →∞ e d ( 1+ c 2 − ln β ) − 1+ ln β 2 n (1 / 2 − α ) d − 1+ α/ 2 (21) where the last inequ ality follows from ( c − 1) k ≥ ( c − 1) d/ 2 and k − α d + α/ 2 ≥ ( d / 2 − 1) − αd + α/ 2 = (1 / 2 − α ) d − 1 + α / 2 . T o see the lim it of right side of (21), w e discuss in two cases: Case I:) 0 < α < 1 / 2 . Since the coeffic i ent 1 / 2 − α > 0 , the li mit is determined b y exponent n (1 / 2 − α ) d − 1+ α/ 2 , and t hen th e statement hol ds for this case. Case II:) α = 1 / 2 , 0 < β < e . The right si de of (21) is equal to 1 √ 2 π lim n →∞ e d ( 1+ c 2 − ln β ) − 1+ln β / 2 n − 3 / 4 . Nov ember 16, 2021 DRAFT JOURNAL OF L A T E X CLASS FIL ES, V OL. 1, NO. 11, NO VE MBER 2002 12 The statement holds also, s ince for 0 < β < e we can take c such that 2 ln β − 1 < c < 1 , i n other words, 0 < 1 + c 2 − ln β < 1 − ln β . QED. Lemma 5: For k ≥ 5 , S P ( n, 2 k ) − M E ( n, k ) > 2( n − k + 1)! n ( k − 1) . Pr oof: Since V ( n, k − 1) +  n k  D k ⌊ n/k ⌋ ≤ V ( n, k − 1) +  n k  D k = V ( n, k ) , S P ( n, 2 k ) − M E ( n, k ) = n ! V ( n, k − 1) − n ! V ( n, k − 1) + ( n k ) D k ⌊ n/k ⌋ = n ! · ( n k ) D k ⌊ n/k ⌋ V ( n, k − 1)  V ( n, k − 1) + ( n k ) D k ⌊ n/k ⌋  ≥ n !  n k  D k ⌊ n/k ⌋ V ( n, k − 1) V ( n, k ) . When k ≥ 5 , V ( n, k − 1) ≤ ( k − 1)  n k − 1  D k − 1 and V ( n, k ) ≤ k  n k  D k , thereby S P ( n, 2 k ) − M E ( n, k ) ≥ n !  n k  D k ⌊ n/k ⌋ k  n k  D k ( k − 1)  n k − 1  D k − 1 ≥ n ! n ( k − 1)  n k − 1  D k − 1 (22) When k ≥ 5 , D k − 1 = [( k − 1)! /e ] < ( k − 1)! 2 , put ting this i nto (22) we have S P ( n, 2 k ) − M E ( n, k ) > n ! n ( k − 1) · n ! ( n − k +1)!( k − 1)! · ( k − 1)! 2 = 2( n − k + 1)! n ( k − 1) . QED. Theor em 7: For constants α , β satis fying either 0 < α < 1 / 2 , β > 0 or α = 1 / 2 , 0 < β < e , whenev er 2 k = β n α , lim n →∞ (min { D V ( n, 2 k ) , S P ( n, 2 k ) } − M E ( n, k ) ) = ∞ . Nov ember 16, 2021 DRAFT JOURNAL OF L A T E X CLASS FIL ES, V OL. 1, NO. 11, NO VE MBER 2002 13 Pr oof: By Lemma 5, we have lim n →∞ S P ( n, 2 k ) − M E ( n, k ) ≥ lim n →∞ 2( n − k + 1)! n ( k − 1) = lim n →∞ 2( n − ( β n α ) / 2 + 1)! n (( β n α ) / 2 − 1) = ∞ . By Lemma 4, we hav e lim n →∞ ( D V ( n, 2 k ) − S P ( n, 2 k )) = lim n →∞ S P ( n, 2 k )  D V ( n, 2 k ) S P ( n, 2 k ) − 1  = ∞ , hence lim n →∞ ( D V ( n, 2 k ) − M E ( n, k )) = ∞ , and then follows the theorem. QED. As a simp le example of the superiority of the new bound M E ( n, k ) over D V ( n, 2 k ) and S P ( n, 2 k ) we can compare them for small va lues of d and n . Example 1: M E (20 , 4) < 0 . 218 · 10 15 , D V (20 , 8) > 0 . 48 2 · 10 15 , S P (20 , 8) > 0 . 98 4 · 10 15 , then M E (20 , 4) provides the best upper bo und on P (20 , 8) . Lemma 6: For k ≥ 4 , S P ( n, 2 k + 1) − M O ( n, k ) > 2( n − k )! ( k + 1) n ( n − 1)  1 + k − n − 1 k  . Pr oof: W e hav e S P ( n, 2 k + 1) − M O ( n, k ) = n ! V ( n, k ) − n ! V ( n, k ) + ( n k +1 ) D k +1 − A ( n − k , 2 k ,k +1) ( n k ) D k A ( n, 2 k ,k +1) = n !  ( n k +1 ) D k +1 − A ( n − k , 2 k ,k +1) ( n k ) D k A ( n, 2 k ,k +1)  V ( n, k )  V ( n, k ) + ( n k +1 ) D k +1 − A ( n − k , 2 k ,k +1) ( n k ) D k A ( n, 2 k ,k +1)  ≥ n !  ( n k +1 ) D k +1 − ( n − k )( n − k − 1) ( k +1) k ( n k ) D k n ( n − 1) ( k +1) k  V ( n, k ) V ( n, k + 1) (23) where the last inequality is followed by A ( n − k , 2 k , k + 1) ≤ ( n − k )( n − k − 1) ( k +1) k , A ( n, 2 k , k + 1) ≤ n ( n − 1) ( k +1) k (by Lemm a 1) and V ( n, k ) +  n k +1  D k +1 − A ( n − k , 2 k, k + 1)  n k  D k A ( n, 2 k , k + 1) ≤ V ( n, k ) +  n k + 1  D k +1 = V ( n, k + 1 ) . Nov ember 16, 2021 DRAFT JOURNAL OF L A T E X CLASS FIL ES, V OL. 1, NO. 11, NO VE MBER 2002 14 When k ≥ 4 , V ( n, k ) ≤ k  n k  D k and V ( n, k + 1) ≤ ( k + 1)  n k +1  D k +1 , then S P ( n, 2 k + 1) − M O ( n, k ) ≥ n !  ( n k +1 ) D k +1 − ( n − k )( n − k − 1) ( k +1) k ( n k ) D k n ( n − 1) ( k +1) k  k  n k  D k ( k + 1)  n k +1  D k +1 = ( n − 2)!  D k +1 D k − n − k − 1 k   n k  D k +1 Since for k ≥ 4 , D k +1 D k ≥ ( k +1)! /e − 1 k ! /e +1 = k + 1 − k +2 k ! /e +1 > k , and D k +1 ≤ ( k +1)! e + 1 < ( k + 1)! / 2 , S P ( n, 2 k + 1) − M O ( n, k ) > ( n − 2)!  k − n − k − 1 k   n k  ( k + 1)! / 2 = 2( n − k )! ( k + 1) n ( n − 1)  1 + k − n − 1 k  . QED. Theor em 8: For constant β s uch that 2 < β < e , whenever 2 k + 1 = β n 1 / 2 , lim n →∞ (min { D V ( n, 2 k + 1) , S P ( n, 2 k + 1) } − M O ( n, k ) ) = ∞ . Pr oof: Since 1 + k − n − 1 k ≥ 1 + 2 √ n − 1 2 − n − 1 2 √ n − 1 2 = 1 +  √ n − 1 2  −  √ n − 1 2 + √ n − 5 4 √ n − 1 2  = 3 4 √ n − 2 , by Lemma 6 we hav e lim n →∞ S P ( n, 2 k + 1) − M O ( n, k ) ≥ lim n →∞ 2( n − k )! ( k + 1) n ( n − 1)  1 + k − n − 1 k  ≥ lim n →∞ 2( n − k )! ( k + 1) n ( n − 1) · 3 4 √ n − 2 (24) = ∞ . By Lemma 4, we hav e lim n →∞ ( D V ( n, 2 k + 1 ) − S P ( n, 2 k + 1)) = lim n →∞ S P ( n, 2 k + 1)  D V ( n, 2 k + 1) S P ( n, 2 k + 1) − 1  = ∞ , hence lim n →∞ ( D V ( n, 2 k + 1 ) − M O ( n, k )) = ∞ , and then follows the theorem. QED. Nov ember 16, 2021 DRAFT JOURNAL OF L A T E X CLASS FIL ES, V OL. 1, NO. 11, NO VE MBER 2002 15 As a simple example of t he superiority of the ne w bound M O ( n, k ) over D V ( n, 2 k + 1) and S P ( n, 2 k + 1) we can compare them for small values of d and n . Example 2: M O (20 , 4) < 0 . 380 · 10 14 by Corrollary 2 , S P (2 0 , 9) > 0 . 528 · 10 14 , D V (20 , 9) > 0 . 603 · 10 14 , then M O (20 , 4) provide the best upp er bound on P (20 , 9) . mds November 18, 2 002 Nov ember 16, 2021 DRAFT