Code spectrum and reliability function: Gaussian channel
aa r X i v : . [ c s . I T ] J un Problems of Information Transmission,vol. 43, no. 2, pp. 3-24, 2007. M. V. BurnashevCODE SPECTRUM AND RELIABILITY FUNCTION:GAUSSIAN CHANNEL 1A new approa h for upper bounding the hannel reliability fun tion usingthe ode spe trum is des ribed. It allows to treat both low and high rate asesin a uni(cid:28)ed way. In parti ular, the earlier known upper bounds are improved,and a new derivation of the sphere-pa king bound is presented.Ÿ 1. Introdu tion and main resultsWe onsider the dis rete time hannel with independent additive Gaussian noise, i.e. if x = ( x , . . . , x n ) is the input odeword then the re eived blo k y = ( y , . . . , y n ) is y i = x i + ξ i , i = 1 , . . . , n , where ( ξ , . . . , ξ n ) are independent Gaussian r.v.'s with E ξ i = 0 , E ξ i = 1 .For x , y ∈ R n denote ( x , y ) = n P i =1 x i y i , k x k = ( x , x ) , d ( x , y ) = k x − y k and S n − ( b ) = { x ∈ R n : k x k = b } . We assume that all odewords x satisfy the ondition k x k = An , where A > is a given onstant. A subset C = { x , . . . , x M } ⊂ S n − ( √ An ) ,M = e Rn , is alled a ( R, A, n ) - ode of rate R and length n . The minimum distan e of the ode C is d ( C ) = min { d ( x i , x j ) : i = j } .The hannel reliability fun tion [1, 2℄ is de(cid:28)ned as E ( R, A ) = lim sup n →∞ n ln 1 P e ( R, A, n ) , where P e ( R, A, n ) is the minimal possible de oding error probability for a ( R, A, n ) - ode.After the fundamental results of the paper [1℄, further improvements of various boundsfor E ( R, A ) have been obtained in [2(cid:21)9℄. In parti ular, on the exa t form of the fun tion E ( R, A ) it was known only that [1℄ E (0 , A ) = A , E ( R, A ) = E sp ( R, A ) , R crit ( A ) ≤ R ≤ C ( A ) , (1)1The resear h des ribed in this publi ation was made possible in part by the Russian Fund forFundamental Resear h (proje t number 06-01-00226).1here C = C ( A ) = 12 ln(1 + A ) , R crit ( A ) = 12 ln 2 + A + √ A + 44 , (2) E sp ( R, A ) = A − p A (1 − e − R ) g ( R, A )2 − ln g ( R, A ) +
R ,g ( R, A ) = 12 (cid:16)p A (1 − e − R ) + p A (1 − e − R ) + 4 (cid:17) . (3)Moreover, re ently [8℄ the exa t form of E ( R, A ) for a new region R ( A ) ≤ R ≤ R crit ( A ) was laimed under some restri tion on A . Similar to the ase of the binary symmetri hannel (BSC), that assertion follows from a useful observation that the tangent (it hasthe slope ( − ) to the fun tion E sp ( R, A ) at the point R = R crit ( A ) tou hes the previouslyknown upper bound for E ( R, A ) [5(cid:21)7℄. Sin e those results from [5(cid:21)7℄ were proved undersome restri tions on A , those restri tions were remaining in [8℄ as well. Sin e there aresome ina ura ies in the formulation of that result in [8℄ we do not expose orrespondingformulas from [8℄ (moreover, they have a di(cid:27)erent from ours form).From theorem 1 and the formula (9) (see below) the exa t form of E ( R, A ) followsfor the region R ( A ) ≤ R ≤ R crit ( A ) for any A > . Moreover, if A > A ≈ . (see(14)) then from theorem 2 below the exa t form of E ( R, A ) follows for a wider region R ( A ) ≤ R ≤ R crit ( A ) , where R ( A ) < R ( A ) and R ( A ) ≈ R crit ( A ) − . , A ≥ A .For < R < R ( A ) , < A ≤ A , or < R < R ( A ) , A > A , still only lower andupper bounds for E ( R, A ) are known [1(cid:21)9℄, and in this paper the most a urate of theupper bounds is improved.We begin by explaining what onstituted the di(cid:30) ulty in upper bounding the fun tion E ( R, A ) in the earlier papers [5(cid:21)9℄. Note that when testing only two odewords x i , x j withlarge distan e k x i − x j k = d we have the de oding error probability P e ∼ e − d/ . Let B ρn bethe average number of ea h odeword x i neighbors on the approximate distan e A (1 − ρ ) n .It was shown in [5℄ that for a ( R, A, n ) - ode there exists ρ su h that B ρn & b ( ρ ) n , where thefun tion b ( ρ ) > is des ribed below, and A (1 − ρ ) n does not ex eed the best upper bound(linear programming) for the minimal ode distan e d ( C ) . Therefore, if ea h odeword x i has approximately B ρn neighbors on the distan e A (1 − ρ ) n , then it is natural to expe tthat P e & B ρn e − A (1 − ρ ) n/ for large n (and not very small ρ ), i.e. a variant of an additivelower bound for the probability of the union of events holds.The (cid:28)rst variant of su h additive bound was obtained in [5℄ under rather severe onstraints on R and A . Those results of [5℄ have been strengthened in [6, 7℄, using themethod of [10(cid:21)12℄. However there were still ertain onstraints on R and A . It shouldbe noted that the investigation of E ( R, A ) for the Gaussian hannel is similar to theinvestigation of E ( R, A ) for the BSC. The di(cid:27)eren e is only that due to the dis retestru ture of a binary alphabet some expressions be ome simpler. For the BSC the methodof [6℄ was re ently [14, 15℄ further developed. Although the approa h of [14, 15℄ is still basedon [6℄, some additional arguments allowed the approa h to be essentially strengthened andsimpli(cid:28)ed. 2t should also be noted that until the papers [14, 15℄, all papers mentioned made useof various variants of the se ond order Bonferroni inequalities.The main aim of this paper is to prove an additive bound without any onstraints on R or A . For that purpose the method of [14, 15℄ is applied. It is also worth noting thatBonferroni inequalities are not used. This approa h allows us to treat both low and highrate R ases in a uni(cid:28)ed way. As an example, in Ÿ 2 a new derivation of the sphere-pa kingbound is presented.Introdu e some notations. For a ode C = { x , . . . , x M } ⊂ S n − ( √ An ) denote ρ ij = ( x i , x j ) An , d ij = k x i − x j k = 2 An (1 − ρ ij ) . (4)Below it will be onvenient to use the parametri representation of the transmission rate R = R ( t ) via the monotoni in reasing fun tion R ( t ) = (1 + t ) ln(1 + t ) − t ln t , t ≥ . (5)Consequently, for a rate R ≥ introdu e t R ≥ as the unique root of the equation R = R ( t R ) = (1 + t R ) ln(1 + t R ) − t R ln t R . (6)Introdu e also the fun tions τ ( t ) = 2 p t (1 + t )1 + 2 t , τ R = τ ( t R ) . (7)We shall need the values t ( A ) = p √ A − , τ ( A ) = τ ( t ( A )) = A √ A ,R ( A ) = R ( t ( A )) , (8)where the fun tions τ ( t ) , R ( t ) are de(cid:28)ned in (7) and (6). Sometimes below we shall omitthe argument A in t ( A ) , τ ( A ) , R ( A ) .One of the main results of the paper isT h e o r e m 1. For any A > the following relations hold: E ( R, A ) = (cid:26) E sp ( R crit , A ) + R crit − R , R ≤ R ≤ R crit ,E sp ( R, A ) , R crit ≤ R ≤ C , (9)and E ( R, A ) ≤ A (1 − τ R )4 + ln(1 + 2 t R ) − R , ≤ R ≤ R , (10)where R crit ( A ) , R ( A ) , τ R and t R are de(cid:28)ned in (2), (8), (7) and (6), respe tively.Remark 1. We have R ( A ) < R crit ( A ) , A > . Moreover, max A (cid:8) R crit ( A ) − R ( A ) (cid:9) ≈ . , and it is attained for A = A ≈ . .3emark 2. Note that (see the formulas (9) and (10) for R = R ) E sp ( R crit , A ) + R crit = A (1 − τ )4 + ln(1 + 2 t ) . (11)Validity of (11) an be he ked using the formulas (6), (7) and the relations t = r A τ , R crit = 12 ln 11 − τ ,A (cid:0) − e − R crit (cid:1) = Aτ = Aτ − , g ( R crit ) = (1 + τ ) √ A √ τ . (12)If A > A ≈ . (see (14)) then the upper bound (10) an be slightly improved, and,moreover, the validity region of the (cid:28)rst of formulas (9) an be enlarged to R ≤ R ≤ R crit ,where R ( A ) < R ( A ) (see (14)). To explain the possibility of su h an improvement onsiderthe problem of upper bounding the minimal ode distan e δ ( R, n ) of a spheri al ode. Thebest upper bound for δ ( R, n ) was obtained in [4℄ using the linear programming bound.It was also noti ed in [4, p. 20℄ that for R > . a better upper bound for δ ( R, n ) isobtained if the linear programming bound is applied not dire tly to the original spheri al ode, but to its sub ode on a spheri al ap. That observation was re ently used in [9℄ whenestimating the ode spe trum and the fun tion E ( R, A ) . Using the approa h of [6℄ an upperbound for E ( R, A ) was obtained in [9℄. But it is rather di(cid:30) ult to use that upper boundsin e it is expressed as an optimization problem over four parameters. In fa t, it is possibleto get a more a urate and rather simple bound that onstitutes theorem 2 below.Introdu e the fun tion D ( t ) = ln 1 + tt − p t (1 + t ) −
11 + 2 t , t > , (13)and denote t ≈ . the unique root of the equation D ( t ) = 0 . The equivalent equation(with a sign misprint) appeared earlier in [4, p. 20℄. Denote also R = R ( t ) ≈ . , τ = τ ( t ) ≈ . ,R ( A ) = R crit ( A ) + R + 12 ln(1 − τ ) ≈ R crit ( A ) − . ,A = min (cid:8) A : R ( A ) ≥ R (cid:9) ≈ . . (14)The next result strengthens theorem 1 when A > A .T h e o r e m 2. If A > A ≈ . then the following relations hold: E ( R, A ) = (cid:26) E sp ( R crit , A ) + R crit − R , R ≤ R ≤ R crit ,E sp ( R, A ) , R crit ≤ R ≤ C , (15)and E ( R, A ) ≤ A (1 − τ R ) + ln(1 + 2 t R ) − R , < R ≤ R , Aae − R −
12 ln(2 − ae − R ) −
12 ln a , R ≤ R ≤ R ( A ) , (16)4here a = (1 − τ ) e R ≈ . .For a omparison purpose we present also the best known lower bound for the fun tion E ( R, A ) [1;3, Theorem 7.4.4℄ E ( R, A ) ≥ A (cid:0) − √ − e − R (cid:1) / , ≤ R ≤ R low ,E sp ( R crit , A ) + R crit − R , R low ≤ R ≤ R crit ,E sp ( R, A ) , R crit ≤ R ≤ C ( A ) , (17)where R low ( A ) = 12 ln 2 + √ A + 44 . (18)Combining analyti al and numeri al methods it an be shown that for A > A we have R low ( A ) < R < R ( A ) < R ( A ) < R crit ( A ) . (19)On the (cid:28)gure the plots of upper (15),(16) and lower (17) bounds for E ( R, A ) with A = 4 are presented.The paper is organized as follows. In Ÿ2 the main analyti al tool (proposition 1) ispresented and, as an example, the sphere-pa king upper bound is derived. In Ÿ3 proposition1 and the ode spe trum are ombined in propositions 2(cid:21)3. In Ÿ4 (using results of Ÿ3 andthe known bound for the ode spe trum - theorem 3) theorem 1 is proved. In Ÿ5 theorem2 is proved. Proofs of some auxiliary results are presented in Appendix.Ÿ 2. New approa h and sphere-pa king exponentFor the onditional output probability distribution density p ( y | x ) of the input odeword x the formula holds ln p ( y | x ) = − d ( y , x ) − n π ) , x , y ∈ R n (in a similar formula in [6℄ there is a misprint - the minus sign is missing). To des ribe ourapproa h, we (cid:28)x a small δ = o (1) , n → ∞ , and s > and for an output y de(cid:28)ne the set: X s ( y ) = { x i ∈ C : | d ( y , x i ) − sn | ≤ δn } , y ∈ R n . (20)All odewords { x i } are assumed equiprobable. For a hosen de oding method denote P ( e | y , x i ) the onditional de oding error probability provided that x i was transmittedand y was re eived. Denote p e ( y ) the probability distribution density to get the output y and to make a de oding error. Then p e ( y ) = M − M X i =1 p ( y | x i ) P ( e | y , x i ) ≥ M − X x i ∈ X s ( y ) p ( y | x i ) P ( e | y , x i ) == M − (2 π ) − n/ X x i ∈ X s ( y ) e − d ( y , x i ) / P ( e | y , x i ) ≥≥ M − (2 πe s + δ ) − n/ X x i ∈ X s ( y ) P ( e | y , x i ) ≥ M − (2 πe s + δ ) − n/ [ | X s ( y ) | − + , [ z ] + = max { , z } and | A | (cid:21) the ardinality of the set A . For the de oding errorprobability P e we get P e = Z y ∈ R n p e ( y ) d y ≥ M − (2 πe s + δ ) − n/ Z y : | X s ( y ) | ≥ [ | X s ( y ) | − d y . Sin e ( a − ≥ a/ , a ≥ , we have P e ≥ (2 M ) − (2 πe s + δ ) − n/ Z y : | X s ( y ) | ≥ | X s ( y ) | d y , (21)where X s ( y ) is de(cid:28)ned in (20). To develop further the right-hand side of (21) we (cid:28)x some r > and for ea h x i introdu e the set Z s,r ( i ) = (cid:8) y : (cid:12)(cid:12) k y k − rn (cid:12)(cid:12) ≤ δn , | d ( y , x i ) − sn | ≤ δn, | X s ( y ) | ≥ (cid:9) == (cid:26) y : |k y k − rn | ≤ δn , | d ( y , x i ) − sn | ≤ δn andthere exists x j = x i with | d ( x j , y ) − sn | ≤ δn (cid:27) . (22)For a measurable set A ⊆ R n denote by m ( A ) its Lebesque measure. Then Z y : | X s ( y ) | ≥ | X s ( y ) | d y ≥ M X i =1 m ( Z s,r ( i )) and from (21) we getP r o p o s i t i o n 1. With any δ > for the de oding error probability P e the lowerbound holds P e ≥ M max s,r ( (2 πe s + δ ) − n/ M X i =1 m ( Z s,r ( i )) ) , (23)where Z s,r ( i ) is de(cid:28)ned in (22).Example: sphere-pa king upper bound. We show (cid:28)rst how to get the sphere-pa king upper bound E ( R, A ) ≤ E sp ( R, A ) from (23) ( f. [1;3, Chapter 7.4℄). To simplifyformulas we write below a ≈ b if | a − b | ≤ δ , where δ = o (1) , n → ∞ . Note that Z s,r ( i ) = Z (1) s,r ( i ) \ Z (2) s,r ( i ) , Z (1) s,r ( i ) = (cid:8) y : k y k /n ≈ r , d ( y , x i ) /n ≈ s (cid:9) , Z (2) s,r ( i ) = (cid:8) y : k y k /n ≈ r , d ( y , x i ) /n ≈ s, | X s ( y ) | = 1 (cid:9) == (cid:8) y : k y k /n ≈ r , d ( y , x i ) /n ≈ s and there is no x j = x i with d ( x j , y ) /n ≈ s (cid:9) . Then we have M [ i =1 Z (2) s,r ( i ) = Y s = (cid:8) y : k y k /n ≈ r , | X s ( y ) | = 1 (cid:9) == (cid:8) y : k y k /n ≈ r and there exists exa tly one x i with d ( y , x i ) /n ≈ s (cid:9) , Y s ⊆ Y ( r ) = (cid:8) y : k y k /n ≈ r (cid:9) , P e ≥ (2 M ) − (2 πe s + δ ) − n/ h M m (cid:16) | Z (1) s,r (1) | (cid:17) − m ( Y ( r )) i + . The surfa e area of a n -dimensional sphere of radius a is S n ( a ) = nπ n/ a n − / Γ( n/ ∼ (2 πea /n ) n/ . Then from a standard geometry we get m (cid:16) | Z (1) s,r (1) | (cid:17) ∼ (2 πer ) n/ , m ( Y ( r )) ∼ (2 πer ) n/ ,r = s − ( r − A − s ) A = r − ( r + A − s ) A .
Therefore the lower bound (23) takes the form P e & M − ( e s + δ − ) − n/ h M r n/ − r n/ i + . (24)We want to maximize the right-hand side of (24) over s, r . Sin e we are interested onlyin exponents in n , we may assume that M r n/ = r n/ , i.e. e R r = r . Then we shouldmaximize the fun tion f ( s, r ) = ln r − s provided s − ( r − A − s ) A − re − R = 0 . As usual, onsidering the fun tion g ( s, r ) = ln r − s + λ (cid:20) s − ( r − A − s ) A − re − R (cid:21) , and solving the equations g ′ s = g ′ r = 0 , we get r = 11 − λ (1 − e − R ) , s = r + A − Aλ , where λ satis(cid:28)es the equation (cid:0) − e − R (cid:1) λ + A (cid:0) − e − R (cid:1) λ − A = 0 . Therefore λ = √ Ag √ − e − R , where g = g ( R, A ) is de(cid:28)ned in (3). Note that g − g p A (1 − e − R ) , − λ (cid:0) − e − R (cid:1) = 1 g , ln r − s = 2 ln g − − A + g p A (1 − e − R ) . e R r = r , we get from (24) and (3) n ln 1 P e ≤ s − − ln r = s −
12 + R −
12 ln r == A − p A (1 − e − R ) g ( R, A )2 − ln g ( R, A ) + R = E sp ( R, A ) , whi h gives the sphere-pa king upper bound E ( R, A ) ≤ E sp ( R, A ) .Ÿ 3. Lower bound (23) and ode spe trumFor a ode C ⊂ S n − ( √ An ) introdu e the ode spe trum fun tion B ( s, t ) = 1 |C| (cid:12)(cid:12)(cid:12)(cid:12)(cid:26) u , v ∈ C : s ≤ ( u , v ) An < t (cid:27)(cid:12)(cid:12)(cid:12)(cid:12) , (25)and denote b ( ρ, ε ) = 1 n ln B ( ρ − ε, ρ + ε ) , < ε < ρ . To simplify notation we write below a ≈ b if | a − b | ≤ δ , where δ = 1 / √ An . For some r > we onsider only the set of outputs Y ( r ) = (cid:8) y : k y k /n ≈ r (cid:9) ⊆ R n . (26)To investigate the fun tion E ( R, A ) , R < R crit , we use a variant of the lower bound(23) P e ≥ (2 M ) − max s,r> max ρ ( (2 πe s + δ ) − n/ M X i =1 m ( Z s,r ( ρ, i )) ) , (27)where Z s,r ( ρ, i ) = (cid:26) y ∈ Y ( r ) : there exists x j with ρ ij ≈ ρ and d ( x i , y ) /n ≈ d ( x j , y ) /n ≈ s (cid:27) , (28)and ρ ij is de(cid:28)ned in (4). We develop the lower bound (27), relating it to the ode spe trum(25), i.e. to the distribution of the pairwise inner produ ts { ρ ij } .For odewords x i , x j with ρ ij ≈ ρ introdu e the set Z s,r ( ρ, i, j ) = (cid:8) y ∈ Y ( r ) : d ( x i , y ) /n ≈ d ( x j , y ) /n ≈ s (cid:9) . (29)Then for any i from (28) and (29) we have Z s,r ( ρ, i ) = [ j : ρ ij ≈ ρ Z s,r ( ρ, i, j ) . (30)Denoting Z ( s, r, ρ ) = m ( Z s,r ( ρ, i, j )) (31)8sin e the measure of that set does not depend on indi es ( i, j ) ), we have (see Appendix) n ln Z ( s, r, ρ ) = 12 ln [2 πez ( s, r, ρ )] + o (1) , n → ∞ , (32)where z ( s, r, ρ ) = r − ( A + r − s ) A (1 + ρ ) . (33)Note that due to (30), for the sum in the right-hand side of (27) for any ρ we have M X i =1 m ( Z s,r ( ρ, i )) ≤ X ( i,j ): ρ ij ≈ ρ m ( Z s,r ( ρ, i, j )) = Z ( s, r, ρ ) |{ ( i, j ) : ρ ij ≈ ρ }| == exp n n πez ( s, r, ρ )] + [ R + b ( ρ )] n + o ( n ) o , (34)sin e for b ( ρ ) = b ( ρ, δ ) the following formula holds (see (25)) |{ ( i, j ) : ρ ij ≈ ρ }| = e Rn B ( ρ − δ, ρ + δ ) = e ( R + b ( ρ )) n . Suppose that for some ρ = ρ in the relation (34) the following asymptoti equality holds: n ln " M X i =1 m ( Z s,r ( ρ , i )) = 12 ln [2 πez ( s, r, ρ )] + R + b ( ρ ) + o (1) , n → ∞ . (35)Using the fun tions s = s ( ρ ) , r = r ( ρ ) (they are hosen below), from (27), (35) and(33) for su h ρ we get n ln 1 P e ≤ s − −
12 ln (cid:20) r − ( A + r − s ) A (1 + ρ ) (cid:21) − b ( ρ ) + o (1) . (36)We set below s ( ρ ) = A (1 − ρ )2 + 1 , r ( ρ ) = A (1 + ρ )2 + 1 . (37)Su h hoi e of s ( ρ ) , r ( ρ ) minimizes (over s, r ) the right-hand side of (36). Optimality ofsu h s, r an also be dedu ed from the formulas (72) (see Appendix).For su h s ( ρ ) , r ( ρ ) we have r − ( A + r − s ) / [2 A (1 + ρ )] = 1 , and then (36) takes thesimple form n ln 1 P e ≤ A (1 − ρ )4 − b ( ρ ) + o (1) . (38)Note that b ( ρ ) ≥ if there exists a pair ( x i , x j ) with ρ ij ≈ ρ , and b ( ρ ) = −∞ if there isno any pair with ρ ij ≈ ρ .We formulate the result obtained as follows.P r o p o s i t i o n 2. If for some ρ the ondition (35) is ful(cid:28)lled, then the inequality(38) for the de oding error probability P e holds.9e show that as su h ρ we may hoose the value ρ , minimizing the right-hand sideof (38). In other words, de(cid:28)ne ρ as follows Aρ + 4 b ( ρ ) = max | ρ |≤ { Aρ + 4 b ( ρ ) } . (39)Remark 3. If there are several su h ρ , we may use any of them. It is not importantthat we do not know the fun tion b ( ρ ) . We may use as b ( ρ ) any lower bound for it (seeproofs of theorems 1 and 2).P r o p o s i t i o n 3. For ρ from (39) the ondition (35) holds and therefore theinequality (38) is valid.P r o o f. It is onvenient to (cid:16)quantize(cid:17) the range of possible values of the normalizedinner produ ts ρ ij . For that purpose we partition the whole range [ −
1; 1] of values ρ ij onsubintervals of the length δ = 1 / √ An . There will be n = 2 /δ of su h subintervals. Wemay assume that ρ ij takes values from the set {− ρ < . . . < ρ n = 1 } .We all ( x i , x j ) a ρ -pair if ( x i , x j ) / ( An ) ≈ ρ . Then M e nb ( ρ ) is the total number of ρ -pairs. We use s = s ( ρ ) , r = r ( ρ ) from (37) and onsider only outputs y ∈ Y ( r ) = Y ( r ( ρ )) . We say that su h a point y is ρ - overed if there exists a ρ -pair ( x i , x j ) su hthat d ( x i , y ) /n ≈ d ( x j , y ) /n ≈ s . Then the total (taking into a ount the overingmultipli ities) Lebesque measure of all ρ - overed points y equals M e nb ( ρ ) Z ( s, r, ρ ) .Introdu e the set Y ( ρ , ρ ) of all ρ - overed points yY ( ρ , ρ ) = { y ∈ Y ( r ) : y is ρ − overed } . We onsider the set Y ( ρ , ρ ) and perform its (cid:16) leaning(cid:17), ex luding from it all points y thatare also ρ - overed for any ρ su h that | ρ − ρ | ≥ δ , i.e. we onsider the set Y ′ ( ρ , ρ ) = Y ( ρ , ρ ) \ [ | ρ − ρ |≥ δ Y ( ρ , ρ ) == (cid:26) y ∈ Y ( r ) : y is ρ − overed and is not ρ − overedfor any ρ su h that | ρ − ρ | ≥ δ (cid:27) . (40)Ea h point y ∈ Y ′ ( ρ , ρ ) an be ρ - overed only if | ρ − ρ | < δ . We show that both sets Y ( ρ , ρ ) and Y ′ ( ρ , ρ ) have essentially the same Lebesque measures. Note that a ρ -pair ( x i , x j ) ρ - overs the set Z s,r ( ρ, i, j ) from (29) with the Lebesque measure Z ( s, r, ρ ) . We ompare the values P | ρ − ρ |≥ δ e nb ( ρ ) Z ( s, r, ρ ) and e nb ( ρ ) Z ( s, r, ρ ) (see (40)). For that purposewe onsider the fun tion g ( ρ ) = 1 n ln e nb ( ρ ) Z ( s, r, ρ ) e nb ( ρ ) Z ( s, r, ρ ) = b ( ρ ) − b ( ρ ) + 12 ln z ( s, r, ρ ) z ( s, r, ρ ) + o (1) , (41)where z ( s, r, ρ ) is de(cid:28)ned in (33). From (33) we also have z ( s, r, ρ ) = 1 + A (1 + ρ )( ρ − ρ )2(1 + ρ ) . b ( ρ ) ≤ b ( ρ ) − A ( ρ − ρ ) / (see (39)), for the fun tion g ( ρ ) from (41) we get g ( ρ ) ≤
12 ln (cid:20) A (1 + ρ )( ρ − ρ )2(1 + ρ ) (cid:21) − A ( ρ − ρ )4 ≤ − A ( ρ − ρ ) ρ ) . (42)Sin e ρ − ρ = iδ , | i | ≥ , after simple al ulations we have P | ρ − ρ |≥ δ e nb ( ρ ) Z ( s, r, ρ ) e nb ( ρ ) Z ( s, r, ρ ) = X | ρ − ρ |≥ δ e ng ( ρ ) ≤ X i ≥ exp (cid:26) − Anδ i (cid:27) = 2 X i ≥ e − i / < . Therefore we get e nb ( ρ ) Z ( s, r, ρ ) − X | ρ − ρ |≥ δ e nb ( ρ ) Z ( s, r, ρ ) > e nb ( ρ ) Z ( s, r, ρ ) . Then the total (taking into a ount the overing multipli ities) Lebesque measure of all ρ - overed points y ∈ Y ′ ( ρ , ρ ) ex eeds M e nb ( ρ ) Z ( s, r, ρ ) / . Remind that any point y ∈ Y ′ ( ρ , ρ ) an be ρ - overed only if | ρ − ρ | < δ .For ea h point y ∈ Y ′ ( ρ , ρ ) onsider the set X s ( y ) de(cid:28)ned in (20), i.e. the set ofall odewords { x i } su h that d ( x i , y ) /n ≈ s . The odewords from X s ( y ) satisfy also the ondition | ( x i , x j ) / ( An ) − ρ | < δ , i.e. the set { x i } onstitutes almost a simplex. It israther lear that the number | X s ( y ) | of su h odewords is not exponential on n , i.e. max y ∈ Y ′ ( ρ ,ρ ) (cid:26) n ln | X s ( y ) | (cid:27) = o (1) , n → ∞ . (43)Formally the validity of (43) follows from lemma 2 (see below).Note that if A , . . . , A N ⊂ R n are a measurable sets, and any point a ∈ S i A i is overedby the sets { A i } not more than K times, then m N [ i =1 A i ! ≥ K N X i =1 m ( A i ) . (44)For y ∈ Y ′ ( ρ , ρ ) denote X i ( y ) = (cid:8) x j : d ( x i , y ) /n ≈ d ( x j , y ) /n ≈ s, ρ ij ≈ ρ (cid:9) ,X max = max i, y ∈ Y ′ ( ρ ,ρ ) | X i ( y ) | . (45)Due to (43) we have n ln X max = o (1) , n → ∞ . (46)11in e any point y ∈ Y ′ ( ρ , ρ ) an be ρ - overed not more than X max times and Y ′ ( ρ , ρ ) ⊆ Y ( ρ , ρ ) , then from (43)(cid:21)(46) we get n ln " M X i =1 m ( Z s,r ( ρ , i )) ≥ n ln m ( Y ′ ( ρ , ρ )) ≥≥ n ln (cid:0) M e nb ( ρ ) Z ( s, r, ρ ) (cid:1) + o (1) == 12 ln [2 πez ( s, r, ρ )] + R + b ( ρ ) + o (1) , n → ∞ . (47)Therefore due to the inequalities (34) and (47), the ondition (35) is ful(cid:28)lled, and then therelation (38) holds.To omplete the proof of proposition 2 it remains to establish the formula (43). Weprove it (cid:28)rst for a simpler (but a more natural) ase ρ ∗ ≤ τ , and then onsider the general ase.C a s e ρ ≤ τ . In that ase the relation (43) follows from simple lemma (see proof inAppendix).L e m m a 1. Let y ∈ R n with k y k = rn . Let C = { x , . . . , x M } ⊂ S n − ( √ An ) be a ode with k x i − y k = sn, i = 1 , . . . , M , and max i = j ( x i , x j ) ≤ Anρ . If A + r − s ≥ p Arρ , (48)then M ≤ n .For s ( ρ ) , r ( ρ ) from (37) the ondition (48) holds, if ρ ≤ A √ A = τ ( A ) . (49)From lemma 1 and (49) the relation (43) follows.G e n e r a l a s e. Although a ode with ρ > τ an hardly de rease the de odingerror probability P e , its investigation needs a bit more e(cid:27)orts. The relation (43) followsfrom lemma (see proof in Appendix).L e m m a 2. Let for a ode C = { x , . . . , x M } ⊂ S n − ( √ An ) and some ρ < it holdsthat max i = j | ( x i , x j ) − Aρn | = o ( n ) , n → ∞ . Then ln M = o ( n ) , n → ∞ .It ompletes the proof of proposition 3. N Using proposition 3 and two lower bounds for b ( ρ ) we shall prove theorems 1 and 2.Ÿ 4. Proof of theorem 112irst we investigate the fun tion E ( R, A ) for < R ≤ R ( A ) and prove the upper bound(10). Then for R ( A ) < R < R crit ( A ) , using the (cid:16)straight-line bound(cid:17) [2℄, we will prove theformula (9). To apply proposition 3 we use the known bound for the ode spe trum. Thenext result is a slight re(cid:28)nement of [5, Theorem 9℄ (see also [6, Theorem 1℄).T h e o r e m 3. Let C ⊂ S n − ( √ An ) be a ode with |C| = e Rn , R > . Then for any ε = ε ( n ) > there exists ρ su h that ρ ≥ τ R and b ( ρ ) = 1 n ln B ( ρ − ε, ρ + ε ) ≥ R − J ( t R , ρ ) + ln εn + o (1) , n → ∞ ,J ( t, ρ ) = (1 + 2 t ) ln [2 tρ + q ( t, ρ )] − ln q ( t, ρ ) − t ln[4 t (1 + t )] ,q ( t, ρ ) = ρ + p (1 + 2 t ) ρ − t (1 + t ) , (50)where t R , τ R are de(cid:28)ned in (4) and (7), and o (1) does not depend on ε .Note that J ′ ρ ( t, ρ ) = 4 t (1 + t ) ρ + p (1 + 2 t ) ρ − t (1 + t ) ,J ′′ ρρ ( t, ρ ) = − t (1 + t )[ ρ + p (1 + 2 t ) ρ − t (1 + t )] " t ) ρ p (1 + 2 t ) ρ − t (1 + t ) ,J ′ t ( t, ρ ) = 2 ln [2 tρ + q ( t, ρ )] − ln[4 t (1 + t )] , [ R ( t ) − J ( t, ρ )] ′ t = 2 ln 2(1 + t )2 tρ + q > , J ( t R , τ R ) = ln(1 + 2 t R ) , J ( t R ,
1) =
R . (51)P r o p o s i t i o n 4. For the fun tion E ( R, A ) the upper bound (10) holds.P r o o f. Due to theorem 2 there exists ρ ≥ τ R su h that the inequality (50) holds.Denote ρ ∗ the largest of su h ρ . Sin e b ( ρ ) ≥ b ( ρ ∗ ) − A ( ρ − ρ ∗ ) / (ñì. (39)), from (38)and (50) we get n ln 1 P e ≤ A (1 − ρ )4 − b ( ρ ) + o (1) ≤ A (1 − ρ ∗ )4 − b ( ρ ∗ ) + o (1) ≤≤ A (1 − ρ ∗ )4 + J ( t R , ρ ∗ ) − R + o (1) . (52)Note that if τ R ≤ τ (i.e. if R ≤ R ( A ) ) then (see Appendix) [ J ( t R , ρ ) − Aρ/ ′ ρ ≤ , ρ ≥ τ R , (53)and therefore the fun tion J ( t R , ρ ) − Aρ/ monotone de reases on ρ ≥ τ R . Sin e ρ ∗ ≥ τ R then for τ R ≤ τ we an ontinue (52) as follows n ln 1 P e ≤ A (1 − τ R )4 + J ( t R , τ R ) − R + o (1) == A (1 − τ R )4 + ln(1 + 2 t R ) − R , < R ≤ R , (54)13hi h is the desired upper bound (10). N To prove the relation (9) note that the best upper bound for E ( R, A ) is a ombinationof the upper bound (10) and the sphere-pa king bound via the (cid:16)straight-line bound(cid:17) [2℄,whi h gives E ( R, A ) ≤ A (1 − τ )4 + ln(1 + 2 t ) − R , R ≤ R ≤ R crit . On the other hand, the random oding bound [1, 3℄ gives E ( R, A ) ≥ E sp ( R crit , A ) + R crit − R , R ≤ R crit , where E sp ( R, A ) is de(cid:28)ned in (3). Together with the formula (11) it ompletes the proof oftheorem 1. N Ÿ 5. Proof of theorem 2As was already mentioned in Ÿ 1, for
R > . the upper bounds for the minimal ode distan e [4, p. 20℄ of a spheri al ode and its spe trum [9℄ an be improved, if thelinear programming bound is not dire tly applied to the original spheri al ode, but to itssub odes on spheri al aps. The same approa h allows to improve the upper bound for E ( R, A ) as well. For that purpose we will need a bound for a ode spe trum better than(50). The bound obtained below (theorem 4), probably, is equivalent to the similar boundin [9, Theorem 3℄ (expressed in a di(cid:27)erent terms), but its derivation is simpler and a morea urate.Sin e we are interested only in angles between odewords x i , x j , for the formulassimpli(cid:28) ation we may set An = 1 , and onsider a ode C ⊂ S n − (1) = S n − . Let T nθ ( z ) bethe spheri al ap with half-angle ≤ θ ≤ π/ and enter z ∈ S n − , i.e. T nθ ( z ) = (cid:8) x ∈ S n − : ( x , z ) ≥ cos θ (cid:9) . It will be onvenient to onsider sub odes of C not on spheri al aps T nθ ( z ) , but on relatedwith them thin ring-shaped surfa es D nθ ( z ) . We set further δ = 1 /n , and denote D nθ ( z ) as D nθ ( z ) = T nθ ( z ) \ T nθ − δ ( z ) = (cid:8) x ∈ S n − : cos θ ≤ ( x , z ) ≤ cos( θ − δ ) (cid:9) . (55)Denote D n ( θ ) the surfa e area of D nθ ( z ) . Then [1, formula (21)℄ D n ( θ ) = ( n − π ( n − / Γ(( n + 1) / θ Z θ − δ sin n − u du , δ ≤ θ ≤ π/ . It is not di(cid:30) ult to show that − n sin θ ≤ D n ( θ )Γ(( n + 1) / n π ( n − / ( n −
1) sin n − θ ≤ . | S n − | of the sphere S n − equals nπ n/ / Γ( n/ , we have uniformlyover /n ≤ θ ≤ π/ n ln D n ( θ ) | S n − | = ln sin θ + o (1) , n → ∞ . For the ode
C ⊂ S n − and θ su h that max { arcsin e − R , /n } ≤ θ ≤ π/ , and z ∈ S n − we onsider the sub ode C ( θ, z ) = C ∩ D nθ ( z ) with |C ( θ, z ) | = e nr ( z ) odewords. Then m ( S n − ) Z z ∈ S n − |C ( θ, z ) | d z = |C| D n ( θ ) | S n − | = exp { ( R + ln sin θ ) n + o ( n ) } , i.e. in average (over z ∈ S n − ) a sub ode C ( θ, z ) has the rate r = R + ln sin θ + o (1) . Allits |C ( θ, z ) | odewords are lo ated in the ball B n (sin θ, z ′ ) of radius sin θ and entered at z ′ = z cos θ . Moreover, they are lo ated in a thin (of thi kness ∼ δ ) torus orthogonal to z .If x ∈ D nθ ( z ) , then we denote x ′ = x − z ′ the orresponding ve tor from B n (sin θ, z ′ ) . Theoriginal angle ϕ between two ve tors x , y ∈ D nθ ( z ) be omes the angle ϕ ′ + O ( δ ) betweenthe ve tors x ′ , y ′ ∈ B n (sin θ, z ′ ) , where sin( ϕ ′ /
2) = sin( ϕ/ / sin θ . The original value ρ = cos ϕ be omes the value ρ ′ + O ( δ ) , where ρ ′ = cos ϕ ′ is de(cid:28)ned by the formula − ρ = (1 − ρ ′ ) sin θ , (56)sin e ρ ′ = cos (cid:18) (cid:18) sin( ϕ/ θ (cid:19)(cid:19) = 1 − ( ϕ/ θ = 1 − (1 − ρ ) e R − r ) . The angle ϕ ′ and the value ρ ′ orrespond to the ase when the ve tors x ′ , y ′ are orthogonalto z . The ode C ( θ, z ) is then transferred to the ode C ′ ( z ) = C ′ ( θ, z ) ⊂ B n (sin θ, z ′ ) .To evaluate the average number e nb C ( ρ ) of ρ -neighbors in the ode C , we onsider anypair x i , x j with ( x i , x j ) = ρ and introdu e the sets Z ( x , a ) = (cid:8) z ∈ S n − : ( x , z ) ≥ a (cid:9) , Z ( x , y , a ) = (cid:8) z ∈ S n − : ( x , z ) ≥ a and ( y , z ) ≥ a (cid:9) . Denote by Ω n ( θ ) the surfa e area of the spheri al ap T nθ ( z ) . For ≤ θ < π/ we have Ω n ( θ ) = π ( n − / sin n − θ Γ(( n + 1) /
2) cos θ (1 + o (1)) , n → ∞ . Then for the Lebesque measure m ( a ) of the set Z ( x , a ) we have m ( a ) = m ( Z ( x , a )) = Ω n (arccos a ) .
15e evaluate the Lebesque measure m ( ρ, a ) of the set Z ( x , y , a ) provided ( x , y ) = ρ . Notethat if x , y ∈ S n − and ( x , y ) = ρ , then k x + y k = 2(1 + ρ ) . Therefore v =( x + y ) / p ρ ) ∈ S n − , and then Z ( x , y , a ) ⊆ (cid:8) z ∈ S n − : ( x + y , z ) ≥ a (cid:9) == n z ∈ S n − : ( v , z ) ≥ a p / (1 + ρ ) o = Z (cid:16) v , a p / (1 + ρ ) (cid:17) . Therefore we get m ( ρ, a ) = m ( Z ( x , y , a )) ≤ m (cid:16) Z (cid:16) v , a p / (1 + ρ ) (cid:17)(cid:17) = Ω n (cid:16) arccos (cid:16) a p / (1 + ρ ) (cid:17)(cid:17) . That upper bound for m ( ρ, a ) is logarithmi ally (as n → ∞ ) exa t. In parti ular, if a =cos θ and ( x , y ) = ρ , then n ln m (cos θ ) m ( ρ, cos θ ) ≥ ln sin θ − ln sin (cid:16) arccos (cid:16)p / (1 + ρ ) cos θ (cid:17)(cid:17) == ln sin θ − ln p − θ/ (1 + ρ ) . We use below the values ρ ′ = ρ ′ ( ρ, θ ) from and (56) and ε ′ = ε/ sin θ . Then denoting B C ( ρ ) = B C ( ρ − ε, ρ + ε ) , B C ′ ( z ) ( ρ ′ ) = B C ′ ( z ) ( ρ ′ − ε ′ , ρ ′ + ε ′ ) , for any ρ, ε we have B C ( ρ ) |C| = 1 m ( ρ, cos θ ) Z z ∈ S n − B C ′ ( z ) ( ρ ′ ) |C ′ ( z ) | d z . (57)Indeed, the value B C ( ρ ) |C| is the total number of pairs x i , x j ∈ C with | ( x i , x j ) − ρ | ≤ ε ,and B C ′ ( z ) ( ρ ′ ) |C ′ ( z ) | is the total number of similar pairs x ′ i , x ′ j ∈ C ′ ( z ) with | ( x ′ i , x ′ j ) / ( k x ′ i k·k x ′ j k ) − ρ ′ | ≤ ε ′ . Moreover, ea h pair x ′ i , x ′ j ∈ C ′ ( z ) gives the ontribution m ( ρ, cos θ ) tothe integral, from whi h the formula (57) follows. From (57) for any set A ⊆ S n − we have e nb C ( ρ ) ≥ m ( ρ, cos θ ) |C| Z z ∈A e nb C′ ( z ) ( ρ ′ ) |C ′ ( z ) | d z , (58)and also |C| = 1 m (cos θ ) Z z ∈ S n − |C ′ ( z ) | d z ≥ m (cos θ ) Z z ∈A |C ′ ( z ) | d z . The ode C ′ ( z ) has the rate r ( z ) = (ln |C ′ ( z ) | ) /n . Then there exists r su h that |C| = e o ( n ) m (cos θ ) max t (cid:8) e tn m (cid:0) z ∈ S n − : | r ( z ) − t | ≤ ε (cid:1)(cid:9) = e r n + o ( n ) m ( S ) m (cos θ ) ,S = (cid:8) z ∈ S n − : | r ( z ) − r | ≤ ε (cid:9) . (59)Sin e m ( S ) ≤ m ( S n − ) then r ≥ n ln |C| m (cos θ ) m ( S n − ) = R + ln sin θ + o (1) . (60)16e set A = S and ε = o (1) , n → ∞ . Then using the Jensen inequality, from (58) and(59) we have e nb C ( ρ ) ≥ m ( ρ, cos θ ) |C| Z z ∈ S e nb C′ ( z ) ( ρ ′ ) |C ′ ( z ) | d z ≥≥ m (cos θ ) e o ( n ) m ( ρ, cos θ ) m ( S ) Z z ∈ S e nb C′ ( z ) ( ρ ′ ) d z ≥≥ m (cos θ ) e o ( n ) m ( ρ, cos θ ) exp nm ( S ) Z z ∈ S b C ′ ( z ) ( ρ ′ ) d z , from whi h we get b C ( ρ ) ≥ n ln m (cos θ ) m ( ρ, cos θ ) + 1 m ( S ) Z z ∈ S b C ′ ( z ) ( ρ ′ ) d z + o (1) . (61)Due to theorem 3 for ea h ode C ′ ( z ) , z ∈ S , there exists ρ ′′ = ρ ′′ ( z ) su h that ρ ′′ ≥ τ r and b C ′ ( z ) ( ρ ′′ ) ≥ r − J ( t r , ρ ′′ ) + o (1) . Therefore there exists ρ ′ ≥ τ r and the orresponding ρ = ρ ( ρ ′ ) from (56) su h that fromthe inequality (61) we get b C ( ρ ) ≥ n ln m (cos θ ) m ( ρ, cos θ ) + r − J ( t r , ρ ′ ) + o (1) ≥ = 1 n ln m (cos θ ) m ( ρ, cos θ ) + R + ln sin θ − J ( t R +ln sin θ , ρ ′ ) + o (1) ≥≥ R + 2 ln sin θ − J ( t R +ln sin θ , ρ ′ ) − ln p − θ/ (1 + ρ ) + o (1) == R + ln sin θ − J ( t R +ln sin θ , ρ ′ ) + 12 ln (1 + ρ )(1 + ρ ′ ) + o (1) , (62)where we used the formula (60) and monotoni ity of the fun tion r − J ( t r , ρ ) on r (see(51)), and ρ ′ = ρ ′ ( ρ, θ ) is de(cid:28)ned in (56). After the variable hange sin θ = e r − R from (62)we getT h e o r e m 4. Let C ⊂ S n − (1) be a ode with |C| = e Rn , R > . Then for any r ≤ R there exists ρ ′ su h that ρ ′ ≥ τ r and for ρ = 1 − (1 − ρ ′ ) e r − R ) the following inequality holds b C ( ρ ) ≥ r − J ( t r , ρ ′ ) + 12 ln (1 + ρ )(1 + ρ ′ ) + o (1) . (63)Using the relation (63) in the inequality (38) we prove theorem 2. We have n ln 1 P e ≤ min r ≤ R max ρ ′ ≥ τ r (cid:26) A (1 − ρ )4 − b ( ρ ) (cid:27) + o (1) ≤≤ min r ≤ R max ρ ′ ≥ τ r (cid:26) A (1 − ρ ′ ) e r − R ) − r + J ( t r , ρ ′ ) + 12 ln 1 + ρ ′ ρ (cid:27) = min r ≤ R max ρ ≥ τ r f ( r, ρ ) , (64)17here f ( r, ρ ) = A (1 − ρ ) e r − R ) R − r + J ( t r , ρ ) + 12 ln 1 + ρ e R − r ) + ρ − . With t = t r and (1 − τ r ) e r − R ) = 2 z we have f ′ ρ = − Ae r − R ) − e R − r ) + ρ −
1) + 4 t (1 + t ) ρ + p (1 + 2 t ) ρ − t (1 + t ) + 12(1 + ρ ) ,f ′ ρ (cid:12)(cid:12) ρ = τ r = − Ae r − R ) − e R − r ) + τ r −
1) + 12(1 − τ r ) == Az − ( A + 2) z + 12(1 − z )(1 − τ r ) , f ′′ ρρ < . Sin
e f ′′ ρρ < then ρ = τ r is optimal if f ′ ρ (cid:12)(cid:12) ρ = τ r ≤ . Sin
e r ≤ R then z ≤ . Therefore f ′ ρ (cid:12)(cid:12) ρ = τ r ≤ if the following inequalities are ful(cid:28)lled: A + 2 + √ A + 4 ≤ z ≤ A + 2 + √ A + 42 A . (65)The right one of the inequalities (65) is always satis(cid:28)ed. The left one of the inequalities(65) is equivalent to the inequality f ( r ) = 2 r + ln(1 − τ r ) ≥ R − R crit ( A ) . (66)The next simple te
hni
al lemma
on
erns the fun
tion f ( r ) in the left-hand side of (66).L e m m a 3. The fun
tion f ( r ) from (66) monotone de
reases on ≤ r < R , andmonotone in
reases on r > R , where R is de(cid:28)ned in (14). Moreover, the formula holds ln (1 − τ ( A )) = − R crit ( A ) , A > . (67)Sin
e the fun
tion E ( R, A ) , R ≥ R ( A ) , is known exa
tly (see theorem 1), we
onsideronly the
ase R < R ( A ) . Then two
ases are possible: R ≤ min { R ( A ) , R } and R 12 ln[ v (2 − v )] , v ( r ) = (1 − τ r ) e r − R ) . (69)Note that for r = R the inequality (68) redu
es to the previous bound (10). We show thatsu
h r is optimal in (68). We have v (2 − v ) C ′ v = − Av + 2( A + 2) v − , C ′′ v > . Sin
e ≤ v ≤ , the equation C ′ v = 0 has the unique root v , where v = 4 A + 2 + √ A + 4 = e − R crit ( A ) . (70)The fun
tion C ( v ) , ≤ v ≤ , monotone de
reases on ≤ v < v and monotone in
reaseson v > v . Note that sin
e v ( r ) = e f ( r ) − R , then (see lemma 3) the fun
tion v ( r ) monotonede
reases on ≤ r < R and monotone in
reases on r > R .If now R ≤ min { R ( A ) , R } , then v ( r ) ≥ v for r ≤ R . Therefore r = R is optimal in(68), and then (68) redu
es to the previous bound (10).C a s e R < R < R ( A ) (i.e. A > A ). Then R < R ( A ) < R ( A ) , where R ( A ) isde(cid:28)ned in (14). Consider (cid:28)rst the
ase R ≤ R ≤ R ( A ) . It is simple to
he
k that then theinequality (66) is again satis(cid:28)ed (see (14)). Therefore ρ = τ r is optimal in the right-handside of (64), and (64) takes the form (68). Sin
e R ≤ R ( A ) , then v ( r ) ≥ v for r ≤ R .Sin
e R ≥ R then r = R is optimal in (68), and then from (68) the se
ond of bounds(16) follows.It remains to
onsider the
ase R ≤ R ( A ) ≤ R ≤ R ( A ) . Sin
e minimum of C ( v ) over ≤ v ≤ is attained for v = v (see (70)), then min ≤ v ≤ C ( v ) = C ( v ) = E sp ( R crit , A ) + R crit , (71)where the formula was used E sp ( R crit , A ) + R crit = Av − 12 ln v − 12 ln(2 − v ) . Now in the right-hand side of (64) we set r su
h that v ( r ) = v (it is possible when R ≥ R ).Then again the inequality (66) is ful(cid:28)lled and ρ = τ r is optimal in the right-hand side of(64). From (68) and (71) the (cid:28)rst of upper bounds (15) follows. The upper bound (15)
an also be proved applying the (cid:16)straight-line bound(cid:17) to the sphere-pa
king bound and these
ond of upper bounds (16) at R = R , and the formula E sp ( R crit , A ) + R crit − R = Aae − R − 12 ln(2 − ae − R ) − 12 ln a , whi
h is simple to
he
k using the relations (12). It
ompletes the proof of theorem 2. N x i , x j , y have the form x i = ( x , x , , . . . , , x j = ( − x , x , , . . . , , y = (0 , y , y , . . . , y n ) , from whi
h we have d ( x i , x j ) = 4 x = 2 An (1 − ρ ij ) ,d ( x i , y ) = x + ( y − x ) + n X k =3 y k = sn ,x + x = An , n X k =2 y k = rn . Solving those equations we get x = r An (1 − ρ ij )2 , x = r An (1 + ρ ij )2 , y = ( A + r − s ) n p An (1 + ρ ij ) , (72)and therefore n X k =3 y k = rn − y = rn − ( A + r − s ) n A (1 + ρ ij ) = r n , from whi
h the formula (32) follows. N Optimality of s ( ρ ) , r ( ρ ) from the formulas (37) also follows from (72).P r o o f o f f o r m u l a (53). For the fun
tion f ( ρ ) = J ( t R , ρ ) − Aρ/ from (51)we have f ′ = 4 t R (1 + t R ) ρ + p (1 + 2 t R ) ρ − t R (1 + t R ) − A , f ′′ ( t, ρ ) < . Then for ρ ≥ τ R we have f ′ ≤ f ′ (cid:12)(cid:12)(cid:12) ρ = τ R = 4 t R (1 + t R ) τ R − A τ R − τ R − A ≤ , if τ R ≤ τ ( A ) , whi
h proves the formula (53). N P r o o f o f l e m m a 1. Let { x , . . . , x M } ⊂ S n − ( √ An ) be a
ode su
h that max i = j ( x i , x j ) ≤ , i.e. min i = j k x i − x j k ≥ A . Then,
learly, M ≤ n .In lemma 1 for all i we have ( x i , y ) = ( A + r − s ) n/ . Consider M ve
tors { x ′ i = x i − a y } ,where a = ( A + r − s ) / (2 r ) . Then due to the
ondition (48) we have max i = j (cid:0) x ′ i , x ′ j (cid:1) ≤ (cid:2) Arρ − ( A + r − s ) (cid:3) n/ (4 r ) ≤ , M ≤ n . N P r o o f o f l e m m a 2. To prove lemma we redu
e it to the
ase ρ ≈ , and thenuse lemma 4 (see below). We set some integer m su
h that < m < M , and introdu
e theve
tor z = a m X k =1 x k , a = ρ m − ρ . After simple
al
ulations we get ρ − δ − m ≤ k z k ≤ ρ + δ ,ρ − δ − ρ ) / ( mρ ) ≤ ( x i , z ) ≤ ρ + δ − ρ ) / ( mρ ) , i = m + 1 , . . . , M. (73)Consider the normalized ve
tors u i = x i − z k x i − z k , i = m + 1 , . . . , M. Using the formulas (73), for any i, j ≥ m + 1 , i = j , we get ( u i , u j ) ≤ − ρ ) (cid:18) δ + 1 m (cid:19) = o (1) , n → ∞ , (74)if we set m → ∞ as n → ∞ . To upperbound the maximal possible number M − m ofve
tors { u i } satisfying the
ondition (74), we use a modi(cid:28)
ation of [16, Theorem 2℄.L e m m a 4. Let C = { x , . . . , x M } ⊂ S n − (1) be a
ode with ( x i , x j ) ≤ µ, i = j . Thenfor n ≥ the upper bound holds M ≤ n / (1 − µ ) − n/ , ≤ µ < . (75)P r o o f. Denote µ = cos(2 ϕ ) , and let M ( ϕ ) be the maximal
ardinality of su
h a
ode.For M ( ϕ ) the upper bound holds [16, Theorem 2℄ M ( ϕ ) ≤ ( n − √ π Γ (cid:18) n − (cid:19) sin β tan β (cid:16) n (cid:17) (cid:2) sin n − β − f ( β, n − 2) cos β (cid:3) , < ϕ < π , (76)where β = arcsin( √ ϕ ) and f ( β, n − 2) = ( n − β Z sin n − z dz . f ( β, n − we have f ( β, n − 2) = sin n − β cos β − sin n +1 β ( n + 1) cos β − n + 1) β Z sin n +2 z cos z dz ≥≥ sin n − β cos β − sin n +1 β ( n + 1) cos β − β ( n + 1) f ( β, n − , and therefore . (cid:20) βn − (cid:21) ≤ f ( β, n − . (cid:26) sin n − β cos β (cid:20) − tan βn + 1 (cid:21)(cid:27) ≤ , (77)if tan β < n + 1 , i.e. if ϕ < ( n + 1) / ( n + 2) . From (76) and (77) we get M ( ϕ ) ≤ √ π Γ (cid:18) n − (cid:19) ( n − 1) cos β (cid:16) n (cid:17) sin n − β < n p πn (1 − ϕ ) √ (cid:0) √ ϕ (cid:1) n − , (78)sin
e Γ (cid:18) z − (cid:19) ( z − (cid:30) Γ (cid:16) z (cid:17) < √ z / e /z , z ≥ . From (78) the inequality (75) follows provided ϕ < ( n +1) / ( n +2) , i.e. if µ > / ( n +2) .Sin
e the fun
tion M ( ϕ ) is
ontinuous on the left for ϕ ∈ (0 , π ] , the upper bound (78)remains valid for µ = 1 / ( n + 2) as well. For µ = 1 / ( n + 2) , n ≥ , the right-hand side of(78) does not ex
eed n p πe/ , whi
h in turn does not ex
eed the right-hand side of (75)for any µ ≥ , n ≥ . Sin
e M ( ϕ ) is a de
reasing fun
tion, it proves the inequality (75) forany µ ≥ , n ≥ . Clearly, (75) remains valid for n = 1 as well. N Now from (74) and (75) we get lemma 2. N The author thanks L.A.Bassalygo, G.A.Kabatyansky and V.V.Prelov for usefuldis
ussions and
onstru
tive
riti
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hInstitute for Information Transmission Problems RASburniitp.ru 24 R E ( R ) (15), (16)(17) A = 4 C = 0,8047 R crit = 0,4812 R = 0,4162 R = 0,4125 R crit R Figure. Upper (15),(16) and lower (17) bounds for E ( R, A ) and A = 4= 4