Lower bounds on the minimum average distance of binary codes

Lo w er b ounds on the minim um a v erage distance of binary co des Beniamin Mounits ∗ April 20, 2022 Abstract Let β ( n, M ) denote the min i mum a v erage Hammin g distance of a binary co de of length n and cardinalit y M . In this pap er we consider low er b ounds on β ( n, M ) . All the kno wn low er b ounds on β ( n, M ) are us e fu l when M is at least of size ab out 2 n − 1 /n. W e deriv e new lo we r b ounds whic h giv e go od estimations when size of M is ab out n. These b ounds are obtained using linear programming approac h. In particular, it is pro ved that lim n →∞ β ( n, 2 n ) = 5 / 2 . W e also give new r e cur siv e inequalit y f o r β ( n , M ) . Keyw ords: Binary co des, minim um av erage distance, linear progr a m ming ∗ CWI, Amsterdam, T h e Netherlands, e-mail: B. Mounits@cw i.nl . 1 1 In tro duct ion Let F 2 = { 0 , 1 } a nd let F n 2 denotes the set of all binary words of length n . F or x, y ∈ F n 2 , d ( x, y ) denotes the Hamming distance b et w een x and y and w t ( x ) = d ( x, 0 ) is the we ight of x , where 0 denotes all-zeros w ord. A binary co de C of length n is a nonempty subset of F n 2 . An ( n, M ) co de C is a binary co de of length n with cardinalit y M . In this pap er w e will consider only binary co des. The a verage Hamming distance o f an ( n, M ) co de C is defined b y d ( C ) = 1 M 2 X c ∈C X c ′ ∈C d ( c, c ′ ) . The minimum a v er age Hamming distanc e of an ( n, M ) co de is defined by β ( n, M ) = min { d ( C ) : C is an ( n, M ) co de } . An ( n, M ) co de C for whic h d ( C ) = β ( n, M ) will b e called e x t r emal co de. The problem o f determining β ( n, M ) w as prop osed b y Ahlsw ede and Katona in [2]. Upp er b ounds on β ( n, M ) are obtained b y constructions. F or surv ey on the know n upp er b ounds the reader is referred to [9]. In this pap er w e consider t he lo w er b ounds on β ( n, M ) . W e only ha v e to consider the case where 1 ≤ M ≤ 2 n − 1 b ecaus e of the f ollo wing result whic h w as prov ed in [6]. Lemma 1. F or 1 ≤ M ≤ 2 n β ( n, 2 n − M ) = n 2 − M 2 (2 n − M ) 2  n 2 − β ( n, M )  . First exact v alues o f β ( n, M ) w ere found by Jaeger et al. [7 ]. Theorem 1. [7] β ( n, 4) = 1 , β ( n, 8) = 3 / 2 , wher e as fo r M ≤ n + 1 , M 6 = 4 , 8 , we have β ( n, M ) = 2  M − 1 M  2 . Next, Alth¨ ofer and Sillk e [3] ga v e the following b ound. Theorem 2. [3] β ( n, M ) ≥ n + 1 2 − 2 n − 1 M , wher e e quality h o lds only for M = 2 n and M = 2 n − 1 . Xia and F u [10] improv ed Theorem 2 for o dd M . Theorem 3. [10] If M is o dd, then β ( n, M ) ≥ n + 1 2 − 2 n − 1 M + 2 n − n − 1 2 M 2 . 2 F urther, F u et al. [6] f ound the follo wing b ounds. Theorem 4. [6] β ( n, M ) ≥ n + 1 2 − 2 n − 1 M + 2 n − 2 n M 2 , if M ≡ 2( mod 4) , β ( n, M ) ≥ n 2 − 2 n − 2 M , for M ≤ 2 n − 1 , β ( n, M ) ≥ n 2 − 2 n − 2 M + 2 n − 1 − n 2 M 2 , if M is o dd and M ≤ 2 n − 1 − 1 . Using Lemma 1 and Theorems 3, 4 the following v alues of β ( n, M ) w ere determined: β ( n, 2 n − 1 ± 1) , β ( n, 2 n − 1 ± 2) , β ( n, 2 n − 2 ) , β ( n, 2 n − 2 ± 1) , β ( n, 2 n − 1 + 2 n − 2 ) , β ( n, 2 n − 1 + 2 n − 2 ± 1) . The b ounds in Theorems 3, 4 were o bta ine d by considering constrain ts on distance distribution o f co de s whic h w ere dev elop e d by Delsarte in [5]. W e will recall these constrain ts in the next section. Notice that the previous b ounds are only useful when M is at least of size ab out 2 n − 1 /n. Ahlsw ede and Alth¨ ofer determined β ( n, M ) asymptotically . Theorem 5. [1] L et { M n } ∞ n =1 b e a se quenc e of natur al numb ers with 0 ≤ M n ≤ 2 n for al l n and lim n →∞ inf  M n /  n ⌊ αn ⌋  > 0 for some c onstant α, 0 < α < 1 / 2 . T hen lim n →∞ inf β ( n, M n ) n ≥ 2 α (1 − α ) . The b ound of Theorem 5 is a s ymptotically ac hiev ed b y taking constan t w eigh t co de C = { x ∈ F n 2 : w t ( x ) = ⌊ α n ⌋ } . The rest of the pap er is orga nized as follows. In Section 2 w e giv e necessary bac kground in linear prog ramming approac h for deriving b ounds for co des . This includes Delsarte’s inequalities on distance distribution of a co de and some prop erties of binary Kr awtc houk p olynomials. In Section 3 w e obtain low er b ounds on β ( n, M ) whic h a re useful in case when M is relativ ely large. In par tic ular, w e sho w tha t the b ound of Theorem 2 is deriv ed via linear progra mming tec hnique. W e also impro ve some b ounds from Theorem 4 for M < 2 n − 2 . In Section 4, w e obta in new lo w er b ounds on β ( n, M ) whic h are useful when M is at least of size ab out n/ 3 . W e also pro ve that t hese b ounds ar e asymptotically tight for the case M = 2 n. Finally , in Section 5 , w e give new recursiv e inequalit y for β ( n, M ) . 2 Preliminaries The distance distribution of an ( n, M ) co de C is the ( n + 1)-tuple of rational nu mbers { A 0 , A 1 , · · · , A n } , where A i = |{ ( c, c ′ ) ∈ C × C : d ( c, c ′ ) = i }| M 3 is the a ve ra g e n umber of co dew ords whic h are at distance i from a n y giv en co dew ord c ∈ C . It is clear that A 0 = 1 , n X i =0 A i = M and A i ≥ 0 fo r 0 ≤ i ≤ n . (1) If C is an ( n, M ) co de with distance distribution { A i } n i =0 , the dual distance distribution { B i } n i =0 is defined b y B k = 1 M n X i =0 P n k ( i ) A i , (2) where P n k ( i ) = k X j =0 ( − 1) j  i j  n − i k − j  (3) is the binary Kra wtchouk p olynomial of degree k . It w as pro v ed b y Delsarte [5] that B k ≥ 0 fo r 0 ≤ k ≤ n . (4) Since the Kraw tchouk p olynomials satisfy the fo llowing o r thogonal relation n X k =0 P n k ( i ) P n j ( k ) = δ ij 2 n , (5) w e hav e n X k =0 P n j ( k ) B k = n X k =0 P n j ( k ) 1 M n X i =0 P n k ( i ) A i = 1 M n X i =0 A i n X k =0 P n j ( k ) P n k ( i ) = 2 n M A j . (6) It’s easy to see from (1),(2),(3), and ( 6) tha t B 0 = 1 and n X k =0 B k = 2 n M . (7) Before w e pro cee d, w e list some of the prop erties of binary Krawtc houk p olynomials (see for example [8]). • Some examples ar e: P n 0 ( x ) ≡ 1 , P n 1 ( x ) = n − 2 x , P n 2 ( x ) = ( n − 2 x ) 2 − n 2 , P n 3 ( x ) = ( n − 2 x )(( n − 2 x ) 2 − 3 n + 2) 6 . • F or any p olynomial f ( x ) of degree k t here is the unique Kra wtc houk expansion f ( x ) = k X i =0 f i P n i ( x ) , where the co efficien ts are f i = 1 2 n n X j =0 f ( j ) P n j ( i ) . 4 • Krawtc houk p olynomials satisfy the fo llo wing recurrent r elatio ns : P n k +1 ( x ) = ( n − 2 x ) P n k ( x ) − ( n − k + 1) P n k − 1 ( x ) k + 1 , (8) P n k ( x ) = P n − 1 k ( x ) + P n − 1 k − 1 ( x ) . (9) • Let i b e nonnegative integer, 0 ≤ i ≤ n. The follo wing symmetry relations hold:  n i  P n k ( i ) =  n k  P n i ( k ) , (10) P n k ( i ) = ( − 1) i P n n − k ( i ) . (11) 3 Bounds for “large” co des The k ey observ ation for obtaining the b ounds in Theorems 3, 4 is the following result. Lemma 2. [10] F or an a rbit r ary ( n, M ) c o de C the fol low ing holds: d ( C ) = 1 2 ( n − B 1 ) . F rom Lemma 2 follows t ha t any upp er b ound on B 1 will pro vide a lo we r b ound on β ( n, M ) . W e will obtain upp er b ounds on B 1 using linear programming tec hnique. Consider the follow ing linear programming problem: maximize B 1 sub ject to n X i =1 B i = 2 n M − 1 , n X i =1 P n k ( i ) B i ≥ − P k (0) , 1 ≤ k ≤ n , and B i ≥ 0 for 1 ≤ i ≤ n. Note that the constrain ts are obtained from (6) and (7). The next theorem follows from the dual linear program. W e will giv e an indep enden t pro of. 5 Theorem 6. L et C b e a n ( n, M ) c o de such that for 2 ≤ i ≤ n and 1 ≤ j ≤ n ther e holds that B i 6 = 0 ⇔ i ∈ I and A j 6 = 0 ⇔ j ∈ J. Supp ose a p olynomial λ ( x ) of de gr e e at most n c an b e found with the fol lowin g pr op erties. If the Kr awtchouk exp ansion of λ ( x ) is λ ( x ) = n X j =0 λ j P n j ( x ) , then λ ( x ) s hould satisfy λ (1) = − 1 , λ ( i ) ≤ 0 for i ∈ I , λ j ≥ 0 for j ∈ J . Then B 1 ≤ λ (0) − 2 n M λ 0 . (12) The e quality in ( 12 ) holds iff λ ( i ) = 0 for i ∈ I and λ j = 0 for j ∈ J. Pr o of. Let C b e an ( n, M ) co de whic h satisfies the ab o ve conditions. Thus , using (1), (2), (4) and (5), w e hav e − B 1 = λ (1) B 1 ≥ λ (1) B 1 + X i ∈ I λ ( i ) B i = n X i =1 λ ( i ) B i = n X i =1 λ ( i ) 1 M n X j =0 P n i ( j ) A j = 1 M n X j =0 A j n X i =1 λ ( i ) P n i ( j ) = 1 M n X j =0 A j n X i =1 n X k =0 λ k P n k ( i ) P n i ( j ) = 1 M n X j =0 A j n X k =0 λ k n X i =0 P n k ( i ) P n i ( j ) − P n k (0) P n 0 ( j ) ! = 1 M n X j =0 A j n X k =0 λ k δ k j 2 n − 1 M n X j =0 A j n X k =0 λ k P n k (0) = 2 n M n X j =0 λ j A j − λ (0 ) = 2 n M λ 0 A 0 + n X j ∈ J λ j A j ! − λ (0 ) ≥ 2 n M λ 0 A 0 − λ (0 ) = 2 n M λ 0 − λ (0) . 6 Corollary 1. If λ ( x ) = n X j =0 λ j P n j ( x ) satisfies 1. λ (1) = − 1 , λ ( i ) ≤ 0 for 2 ≤ i ≤ n, 2. λ j ≥ 0 for 1 ≤ j ≤ n, then β ( n, M ) ≥ 1 2  n − λ (0) + 2 n M λ 0  . Example 1. Consid er the fol lowing p olynomial: λ ( x ) ≡ − 1 . It is ob vious that the conditions of the Coro llary 1 are satisfied. Th us w e ha v e a b ound β ( n, M ) ≥ n + 1 2 − 2 n − 1 M whic h coincides with the o ne from Theorem 2. Example 2. [6, The or em 4] Consider the fol lowing p olynomial: λ ( x ) = − 1 2 + 1 2 P n n ( x ) . F rom (11) w e see that P n n ( i ) = ( − 1) i P n 0 ( i ) =  1 if i is ev en − 1 if i is o dd , and, therefore, λ ( i ) =  0 if i is ev en − 1 if i is o dd . F urthermore, λ j = 0 fo r 1 ≤ j ≤ n − 1 and λ n = 1 / 2 . Th us, the conditions o f the Corollary 1 are satisfied and w e obta in β ( n, M ) ≥ 1 2  n − 2 n − 1 M  = n 2 − 2 n − 2 M . This b ound was obtained in [6 , Theorem 4] and is tigh t for M = 2 n − 1 , 2 n − 2 . Other b ounds in Theorems 3, 4 we re obtained b y considering a dditio nal constraints on distance distribution co efficien ts give n in the next theorem. 7 Theorem 7. [4] L et C b e a n arbi tr ary binary ( n, M ) c o de. If M is o d d , then B i ≥ 1 M 2  n i  , 0 ≤ i ≤ n . If M ≡ 2( mod 4) , then ther e exists a n ℓ ∈ { 0 , 1 , · · · , n } such that B i ≥ 2 M 2  n i  + P n i ( ℓ )  , 0 ≤ i ≤ n . Next, w e will impro v e the b ound of Example 2 for M < 2 n − 2 . Theorem 8. F or n > 2 β ( n, M ) ≥        n 2 − 2 n − 2 M + 1 n − 2  2 n − 2 M − 1  if n is eve n n 2 − 2 n − 2 M + 1 n − 1  2 n − 2 M − 1  if n is o dd . Pr o of. W e distinguish b et w een t w o cases. • If n is ev en, n > 2 , consider the follow ing p olyn omial: λ ( x ) = 1 2( n − 2)  3 − n + P n n − 1 ( x ) + P n n ( x )  . Using (11), it’s easy to see that λ ( i ) =    2 − i n − 2 if i is ev en i +1 − n n − 2 if i is o dd . • If n is o dd, n > 1 , consider the followin g p olynomial: λ ( x ) = 1 2( n − 1)  2 − n + P n n − 1 ( x ) + 2 P n n ( x )  . Using (11), it’s easy to see that λ ( i ) =    2 − i n − 1 if i is ev en i − n n − 1 if i is o dd . In b oth cases, the claim of the theorem fo llo ws from Corollary 1. 8 4 Bounds for “small” co des W e will use the following lemma, whose pro of easily follo ws from (5). Lemma 3. L et λ ( x ) = n X i =0 λ i P n i ( x ) b e an arbitr ary p olynom ial. A p olynomial α ( x ) = n X i =0 α i P n i ( x ) satisfies α ( j ) = 2 n λ j iff α i = λ ( i ) . By substituting the p olynomial λ ( x ) from Theorem 6 in to Lemma 3 , we ha ve the following. Theorem 9. L et C b e an ( n, M ) c o de such that for 1 ≤ i ≤ n and 2 ≤ j ≤ n ther e hold s that A i 6 = 0 ⇔ i ∈ I and B j 6 = 0 ⇔ j ∈ J. Supp ose a p ol yno m ial α ( x ) of de gr e e at most n c an b e found with the fol lowi n g pr op erties. If the Kr awtchouk exp a n sion of α ( x ) is α ( x ) = n X j =0 α j P n j ( x ) , then α ( x ) should sa t isfy α 1 = 1 , α j ≥ 0 , for j ∈ J , α ( i ) ≤ 0 , for i ∈ I . Then B 1 ≤ α (0) M − α 0 . (13) The e quality in ( 13 ) holds iff α ( i ) = 0 for i ∈ I a n d α j = 0 for j ∈ J. Note that Theorem 9 follow s from the dual linear program of the follo wing o ne : maximize n X i =1 P n 1 ( i ) A i = M B 1 − n sub ject to n X i =1 A i = M − 1 , n X i =1 P n k ( i ) A i ≥ − P k (0) , 1 ≤ k ≤ n , and A i ≥ 0 for 1 ≤ i ≤ n, whose constrain ts are obtained fro m (1) and (4). 9 Corollary 2. If α ( x ) = n X j =0 α j P n j ( x ) satisfies 1. α 1 = 1 , α j ≥ 0 for 2 ≤ j ≤ n, 2. α ( i ) ≤ 0 fo r 1 ≤ i ≤ n, then β ( n, M ) ≥ 1 2  n + α 0 − α (0) M  . Example 3. Consid er α ( x ) = 2 − n + P n 1 ( x ) = 2( 1 − x ) . It’s ob vious that the conditions of t he Corollary 2 are satisfied and w e o bt a in Theorem 10. β ( n, M ) ≥ 1 − 1 M . Note that the b ound o f Theorem 10 is tight fo r M = 1 , 2 . Example 4. Consid er the fol lowing p olynomial: α ( x ) = 3 − n + P n 1 ( x ) + P n n ( x ) . F rom (11) w e obtain α ( i ) =  4 − 2 i if i is eve n 2 − 2 i if i is o dd . Th us, conditions of the Corolla ry 2 are satisfied and w e ha v e Theorem 11. β ( n, M ) ≥ 3 2 − 2 M . Note that the b ound o f Theorem 11 is tight fo r M = 2 , 4 . Example 5. L et n b e even inte ger. Consider the fol lo wing p olynomi a l: α ( x ) = n (4 − n ) n + 2 + P n 1 ( x ) + 4  n 2  ( n + 2)  n n 2 +1  P n n 2 +1 ( x ) . (14) 10 In this p olynomial α 1 = 1 and α j ≥ 0 for 2 ≤ j ≤ n . Thus, condition 1 in Corollary 2 is satisfied. F rom (10) we obtain that for nonnegative integer i, 0 ≤ i ≤ n, P n n 2 +1 ( i ) =  n n 2 +1   n i  P n i  n 2 + 1  and, therefore, α ( i ) = n (4 − n ) n + 2 + P n 1 ( i ) + 4  n 2  ( n + 2)  n i  P n i  n 2 + 1  . (15) It follow s fro m (8) that P n 1  n 2 + 1  = − 2 , P n 2  n 2 + 1  = 4 − n 2 , P n 3  n 2 + 1  = n − 2 , P n 4  n 2 + 1  = ( n − 2)( n − 8) 8 , P n 5  n 2 + 1  = ( n − 2)(4 − n ) 4 . (16) No w it’s easy to ve rify from (15) and (16) that α (1 ) = α (2) = α (3) = 0 . W e define e α ( i ) := n (4 − n ) n + 2 + P n 1 ( i ) + 4  n 2  ( n + 2)  n i     P n i  n 2 + 1     . It is clear that α ( i ) ≤ e α ( i ) for 0 ≤ i ≤ n. W e will prov e that e α ( i ) ≤ 0 for 4 ≤ i ≤ n. F rom (11) and (16) one can v erify t ha t e α ( n ) = 0 , e α ( n − 1) = e α ( n − 2) = 2 n (4 − n ) n + 2 , and e α ( n − 3) = 2(6 − n ) (17) whic h implies that e α ( n − j ) ≤ 0 for 0 ≤ j ≤ 3 (of course, w e are not in terested in v alues e α ( n − j ) , 0 ≤ j ≤ 3 , if n − j ∈ { 1 , 2 , 3 } ). So, it is left to pro ve tha t fo r every integer i, 4 ≤ i ≤ n − 4 , e α ( i ) ≤ 0 . No t e that for an in teger i, 4 ≤ i ≤ n/ 2 , e α ( n − i ) = n (4 − n ) n + 2 + P n 1 ( n − i ) + 4  n 2  ( n + 2)  n n − i     P n n − i  n 2 + 1     = n (4 − n ) n + 2 + (2 i − n ) + 4  n 2  ( n + 2)  n i     ( − 1) n 2 +1 P n i  n 2 + 1     ≤ n (4 − n ) n + 2 + ( n − 2 i ) + 4  n 2  ( n + 2)  n i     P n i  n 2 + 1     = e α ( i ) . Therefore, it is enough to ch eck that e α ( i ) ≤ 0 only for 4 ≤ i ≤ n/ 2 . F rom (16) w e obtain that e α (4) = − 2 − 6 n − 3 < 0 and e α (5) = − 4 − 12( n − 8) ( n + 2)( n − 3) < 0 , where, in view of (17), w e assume that n ≥ 8 . T o pr ov e that e α ( i ) ≤ 0 for 6 ≤ i ≤ n/ 2 w e will use the followin g lemma whose pro of is giv en in the App endix. 11 Lemma 4. If n is an even p os it ive inte ger and i is an arbitr ary inte ger numb er, 2 ≤ i ≤ n/ 2 , then    P n i  n 2 + 1     <  n ⌊ i 2 ⌋  . By Lemma 4, the following holds for 2 ≤ i ≤ n/ 2 . e α ( i ) = n (4 − n ) n + 2 + P n 1 ( i ) + 4  n 2  ( n + 2)  n i     P n i  n 2 + 1     < n (4 − n ) n + 2 + n − 2 i + 4  n 2  n ⌊ i 2 ⌋  ( n + 2)  n i  = 6 n n + 2 − 2 i + 4  n 2  n ⌊ i 2 ⌋  ( n + 2)  n i  = − 12 n + 2 − 2( i − 3) + 4  n 2  n ⌊ i 2 ⌋  ( n + 2)  n i  . Th us, to prov e t h at e α ( i ) ≤ 0 for 6 ≤ i ≤ n/ 2 , it’s enough to prov e that − 2( i − 3) + 4  n 2  n ⌊ i 2 ⌋  ( n + 2)  n i  < 0 for 6 ≤ i ≤ n/ 2 . Lemma 5. L et n b e an even i n t e ger. F or 6 ≤ i ≤ n/ 2 we have ( i − 3)  n i   n ⌊ i 2 ⌋  > n ( n − 1) n + 2 . The pro of of this lemma a pp ears in the App endix. W e ha v e pro v ed that the b oth conditions of the Corollary 2 are satisfied and, therefore, for ev en inte ger n, w e hav e β ( n, M ) ≥ 3 n n + 2 − n M . Once w e hav e a b ound for an ev en (o dd) n , it’s easy to deduce one for o dd (ev en) n due to the follo wing fa c t whic h follow s fro m (9). Lemma 6. L et α ( x ) = n X j =0 α j P n j ( x ) b e an a rb it r ary p olynomial. Then for a p olyno m ial µ ( x ) = n − 1 X j =0 µ j P n − 1 j ( x ) , wher e µ j = α j + α j +1 , 0 ≤ j ≤ n − 1 , the fol lowing h olds: µ ( x ) = α ( x ) for 0 ≤ x ≤ n − 1 . 12 Example 6. L et n b e o dd inte ger, n > 1 . Con s ider the fol lowing p olynomial: µ ( x ) = 6 + 3 n − n 2 n + 3 + P n 1 ( x ) + 4  n +1 2  ( n + 3)  n +1 n +3 2   P n n +1 2 ( x ) + P n n +3 2 ( x )  (18) whic h is obtained from α ( x ) g iv en in (14) b y the construction of Lemma 6 . Thus , b y Corollary 2, for o dd integer n, w e ha v e β ( n, M ) ≥ 3( n + 1) n + 3 − n + 1 M . W e summarize the b ounds from the Examples 5, 6 in the next theorem. Theorem 12. β ( n, M ) ≥    3 n n +2 − n M if n is eve n 3( n +1) n +3 − n +1 M if n is o dd . Example 7. F or n ≡ 1 ( mod 4) , n 6 = 1 , c on s ider α ( x ) = (1 − n )( n − 5) n + 1 + P n 1 ( x ) + 4 n ( n − 2) ( n + 1)  n n +1 2  P n n +1 2 ( x ) + P n n ( x ) . (19) One can v erify tha t α (0) = 4( n − 1) , α (1 ) = α (2) = α (3) = α (4 ) = 0 , α (5) = α (6) = 4(1 − n ) n − 4 , and α ( n ) = − 6 ( n − 1) 2 n + 1 , α ( n − 1) = α ( n − 2) = α ( n − 3) = α ( n − 4) = − 2 ( n − 5)( n − 1) n + 1 , α ( n − 5) = α ( n − 6) = − 2( n − 9)( n − 2)( n − 1) ( n + 1)( n − 4) . W e define e α ( i ) := (1 − n )( n − 5 ) n + 1 + P n 1 ( x ) + 4 n ( n − 2) ( n + 1)  n i      P n i  n + 1 2      + | P n n ( i ) | . As in the previous example, it’s easy to see that α ( i ) ≤ e α ( i ) for 0 ≤ i ≤ n and e α ( n − i ) ≤ e α ( i ) for 0 ≤ i ≤ ( n − 1) / 2 . Therefore, to pro ve that α ( i ) ≤ 0 for 1 ≤ i ≤ n, we only hav e to sho w that e α ( i ) ≤ 0 for 7 ≤ i ≤ ( n − 1) / 2 . It is follows f rom the next t w o lemmas. Lemma 7. I f n is o dd p ositive inte ger and i is an arbitr ary inte ger numb er, 2 ≤ i ≤ ( n − 1 ) / 2 , then     P n i  n + 1 2      <  n ⌊ i 2 ⌋  . 13 Lemma 8. L et n b e o dd in t e ger. F or 7 ≤ i ≤ ( n − 1) / 2 we have ( i − 4)  n i   n ⌊ i 2 ⌋  > 2 n ( n − 2) n + 1 . Pro ofs of the Lemmas 7 , 8 are v ery similar to those of Lemmas 4, 5 , resp ectiv ely , a nd they are omitted. Th us, we ha v e prov ed tha t the conditions of the Corollary 2 are satisfied and w e hav e the f o llo wing b ound. β ( n, M ) ≥ 7 n − 5 2( n + 1) − 2( n − 1) M , if n ≡ 1 ( mod 4) , n 6 = 1 . F rom Lemma 6, by c ho osing the follow ing p olyn omials: µ ( x ) = 2 + 5 n − n 2 n + 2 + P n 1 ( x ) + 4( n 2 − 1) ( n + 2)  n +1 n +2 2   P n n 2 ( x ) + P n n +2 2 ( x )  + P n n ( x ) , if n ≡ 0 ( mod 4) , e µ ( x ) = 9 + 4 n − n 2 n + 3 + P n 1 ( x ) + 4 n ( n + 2) ( n + 3)  n +2 n +3 2   P n n − 1 2 ( x ) + P n n +3 2 ( x )  + 8 n ( n + 2) ( n + 3)  n +2 n +3 2  P n n +1 2 ( x ) + P n n ( x ) , if n ≡ 3 ( mod 4) , n 6 = 3 , and b µ ( x ) = 16 + 3 n − n 2 n + 4 + P n 1 ( x ) + 4( n + 1)( n + 3) ( n + 4)  n +3 n +4 2   P n n − 2 2 ( x ) + P n n +4 2 ( x )  + 12( n + 1 )( n + 3) ( n + 4)  n +3 n +4 2   P n n 2 ( x ) + P n n +2 2 ( x )  + P n n ( x ) , if n ≡ 2 ( mod 4) , n 6 = 2 , w e obtain the b ounds whic h are summarized in the next theorem. Theorem 13. F or n > 3 β ( n, M ) ≥                        7 n +2 2( n +2) − 2 n M if n ≡ 0 ( mod 4) 7 n − 5 2( n +1) − 2( n − 1) M if n ≡ 1 ( mod 4) 7 n +16 2( n +4) − 2( n +2) M if n ≡ 2 ( mod 4) 7 n +9 2( n +3) − 2( n +1) M if n ≡ 3 ( mod 4) . It’s easy to see that the b ounds of Theorems 12 and 13 give similar estimations when the size of a co de is ab out 2 n. 14 Theorem 14. lim n →∞ β ( n, 2 n ) = 5 2 . Pr o of. Let C b e the follo wing ( n, 2 n ) co de: 000 · · · 00 100 · · · 00 010 · · · 00 . . . . . . . . . 000 · · · 01 110 · · · 00 101 · · · 00 . . . . . . . . . 100 · · · 01 One can ev aluate that β ( n, 2 n ) ≤ d ( C ) = 5 2 − 4 n − 2 n 2 . (20) On the other hand, Theorem 12 giv es β ( n, 2 n ) ≥    5 2 − 6 n +2 if n is ev en 5 2 − 13 n +3 2 n ( n +3) if n is o dd . (21) The claim of the theorem follows b y com bining (20) and (21). 5 Recursiv e inequalit y o n β ( n, M ) The follow ing recursiv e inequalit y w a s obtained in [10]: β ( n, M + 1) ≥ M 2 ( M + 1) 2 β ( n, M ) + M n ( M + 1) 2 1 − r 1 − 2 n β ( n, M ) ! . (22) In the next theorem w e give a new r ecursiv e inequalit y . Theorem 15. F or p ositive inte gers n and M , 2 ≤ M ≤ 2 n − 1 , β ( n, M + 1) ≥ M 2 M 2 − 1 β ( n, M ) . (23) Pr o of. Let C b e an extremal ( n, M + 1) co de, i.e., β ( n, M + 1 ) = d ( C ) = 1 ( M + 1) 2 X c ∈C X c ′ ∈C d ( c, c ′ ) . 15 Then there exists c 0 ∈ C suc h that X c ∈C d ( c 0 , c ) ≥ ( M + 1) β ( n, M + 1) . (24) Consider an ( n, M ) co de e C = C \ { c 0 } . Using ( 2 4) w e o bt a in β ( n, M ) ≤ d ( e C ) = 1 M 2 X c ∈ e C X c ′ ∈ e C d ( c, c ′ ) = 1 M 2 X c ∈C X c ′ ∈C d ( c, c ′ ) − 2 X c ∈C d ( c 0 , c ) ! ≤ 1 M 2  ( M + 1) 2 β ( n, M + 1 ) − 2( M + 1 ) β ( n, M + 1)  = M 2 − 1 M 2 β ( n, M + 1 ) . Lemma 9. F or p o sitive inte gers n and M , 2 ≤ M ≤ 2 n − 1 , the RHS of ( 23 ) is not s mal ler than RHS o f ( 22 ). Pr o of. One can v erify tha t RHS of (23) is not smaller than RHS of (22) iff β ( n, M ) ≤ M 2 − 1 M 2 · n 2 . By (23) w e hav e β ( n, M ) ≤ M 2 − 1 M 2 β ( n, M + 1) ≤ M 2 − 1 M 2 β ( n, 2 n ) = M 2 − 1 M 2 · n 2 , whic h completes the pro of. 6 App endix Pro of of Lemma 4: The pro of is b y induction. One can easily see from (16) that the claim is true for 2 ≤ i ≤ 5 , where i ≤ n/ 2 . Assume tha t we hav e pro ve d the claim for i, 4 ≤ i ≤ k ≤ n/ 2 − 1 . Th us    P n k +1  n 2 + 1     =      ( − 2) P n k  n 2 + 1  − ( n − k + 1) P n k − 1  n 2 + 1  k + 1      ≤ 2 k + 1    P n k  n 2 + 1     + n − k + 1 k + 1    P n k − 1  n 2 + 1     < 2 k + 1  n ⌊ k 2 ⌋  + n − k + 1 k + 1  n ⌊ k − 1 2 ⌋  = ( ∗ ) . 16 W e distinguish b et w een t w o cases. If k is o dd, then ( ∗ ) = 2 k + 1  n k − 1 2  + n − k + 1 k + 1  n k − 1 2  = 2 k + 1  n k − 1 2   1 + n − k + 1 2  = 1 n − k − 1 2 · n − k − 1 2 k +1 2  n k − 1 2  n − k + 3 2 = n − k + 3 2 n − k + 1  n k +1 2  <  n k +1 2  . Therefore, for o dd k , w e obtain    P k +1  n 2 + 1     <  n k +1 2  =  n ⌊ k +1 2 ⌋  . If k is ev en, then ( ∗ ) = 2 k + 1  n k 2  + n − k + 1 k + 1  n k 2 − 1  = 2 k + 1  n k 2  + n − k + 1 k + 1 · k 2 n − ( k 2 − 1) · n − ( k 2 − 1) k 2  n k 2 − 1  =  n k 2   2 k + 1 + n − k + 1 2 n − k + 2 · k k + 1  . Since k ≥ 4 , w e hav e ( ∗ ) =  n k 2       2 k + 1 + < 1 / 2 z }| { n − k + 1 2 n − k + 2 · < 1 z }| { k k + 1      <  n k 2   2 5 + 1 2  <  n k 2  . Therefore, for ev en k , w e obt a in    P k +1  n 2 + 1     <  n k 2  =  n ⌊ k +1 2 ⌋  . Pro of of Lemma 5: Denote a i = ( i − 3)  n i   n ⌊ i 2 ⌋  , 6 ≤ i ≤ n/ 2 . Th us, a 6 ( n + 2) n ( n − 1) = ( n + 2)( n − 3)( n − 4)( n − 5 ) 40 n ( n − 1) 17 = ( n − 2)( n − 7 ) 40 + 48 n − 120 40 n ( n − 1) n ≥ 12 z}|{ ≥ 5 4 + 48 · 12 − 120 40 n ( n − 1) > 5 4 and w e hav e prov ed that a 6 > n ( n − 1) n + 2 . Let’s see that a i ≥ a 6 for 6 ≤ i ≤ n/ 2 . Let i b e ev en inte ger suc h that 6 ≤ i ≤ n/ 2 − 2 . Then a i +2 a i = ( i − 1)( n − i − 1)( n − i ) ( i − 3)( i + 1)( n − 2 i ) i ≥ 6 z}|{ > ( i − 3)( n − 2 i )( n − i ) ( i − 3)( i + 1)( n − 2 i ) = n − i i + 1 i ≤ n/ 2 − 2 z}|{ > 1 . T ogether with a 6 > n ( n − 1) n + 2 , this implies that a i > n ( n − 1) n + 2 for ev ery eve n in teger i, 6 ≤ i ≤ n/ 2 . No w let i b e ev en intege r suc h that 6 ≤ i ≤ n/ 2 − 1 . Then a i +1 a i = ( i − 2)( n − i ) ( i − 3) ( i + 1) > n − i i + 1 i ≤ n/ 2 − 1 z}|{ > 1 , whic h completes the pro of. References [1] R. Ahlsw ede and I. Alth¨ ofer, “The asymptotic b eha viour of diameters in the a ve ra g e ”, in J. Comb i n . T he ory Ser. B 61, pp. 167–177, 1994. 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