The Asymptotic Hadamard Conjecture

Transcription

1 p. 1/4 The Asymptotic Hadamard Conjecture Rod Canfield Department of Computer Science University of Georgia Atlanta Lecture Series in Combinatorics and Graph Theory November 13, 2010

2 p. 2/4 Thank You s Emory University, Georgia Tech, Georgia State University with support from the NSA and NSF Co-authors: Jason Gao, Catherine Greenhill, Brendan McKay, Bob Robinson Host institution: Australia National University Authors: Warwick de Launey & David A. Levin (hopefully they are future co-authors) NSA Mathematical Sciences Program

3 p. 3/4 Main Definitions An n t matrix over {±1} is a partial Hadamard matrix provided the rows are pairwise orthogonal. Definition: H nt := the # of n t partial Hadamard matrices Example. One of the matrices counted by H 58 : Example illustrates: n 3 & H nt 0 = 4 t

4 p. 4/4 The Problem Find an asymptotic formula for H nt valid certain infinite sequences of pairs (n, t).

5 p. 4/4 The Problem Find an asymptotic formula for H nt valid certain infinite sequences of pairs (n, t). Theorem (de Launey & Levin, 2010) Let (n,t) be an infinite sequence of pairs satisfying 4 t and t n 12+ɛ. Then along this sequence H nt 2nt+(n 1)2 (2πt) d/2, d = ( ) n. 2

6 p. 4/4 The Problem Find an asymptotic formula for H nt valid certain infinite sequences of pairs (n, t). Theorem (de Launey & Levin, 2010) Let (n,t) be an infinite sequence of pairs satisfying 4 t and t n 12+ɛ. Then along this sequence H nt 2nt+(n 1)2 (2πt) d/2, d = ( ) n. 2 A Fourier-analytic approach to counting partial Hadamard matrices Cryptography and Communications 2(2010) pp

7 p. 5/4 The Hadamard Conjecture The HADAMARD CONJECTURE states that there exist square Hadamard matrices of size n n for n {1, 2, 4, 8, 12, 16, 20,...}. Various constructions have been found n = 668 is the first undecided value n = 428 was decided in 2004

8 p. 6/4 Outline Circle Method Estimates Short digression: Latin rectangles The de Launey - Levin Theorem An extension

9 p. 7/4 The Circle Method a n = [z n ]f(z) = 1 2πi z =r f(z) z n+1dz δ +π = 1 f(re iθ ) 2πr n π e niθ dθ [ = 1 +δ 2πr n + δ θ π ]

10 p. 8/4 Example 1: Stirling s Formula 1 n! = [zn ]e z = 1 2πi z =r exp(z) 1 dz = zn+1 2π +π π exp(re iθ ) r n e niθ dθ = er 2πr n [ +δ δ exp ( (r n)iθ (1/2)rθ 2 + O(r θ 3 ) ) dθ +O(1) δ θ π exp ( crθ 2) ] dθ

11 p. 8/4 Example 1: Stirling s Formula 1 n! = [zn ]e z = 1 2πi z =r exp(z) 1 dz = zn+1 2π +π π exp(re iθ ) r n e niθ dθ = er 2πr n [ +δ δ exp ( (r n)iθ (1/2)rθ 2 + O(r θ 3 ) ) dθ +O(1) δ θ π exp ( crθ 2) ] dθ exp(re iθ ) = exp(r cosθ) = e r exp( r(1 cosθ)) e r exp( crθ 2 )

12 p. 8/4 Example 1: Stirling s Formula 1 n! = [zn ]e z = 1 2πi z =r exp(z) 1 dz = zn+1 2π +π π exp(re iθ ) r n e niθ dθ = er 2πr n [ +δ δ exp ( (r n)iθ (1/2)rθ 2 + O(r θ 3 ) ) dθ +O(1) δ θ π exp ( crθ 2) ] dθ r = n rδ 2 rδ 3 0

13 p. 8/4 Example 1: Stirling s Formula 1 n! = [zn ]e z = 1 2πi z =r exp(z) 1 dz = zn+1 2π +π π exp(re iθ ) r n e niθ dθ = er 2πr n [ +δ δ exp ( (r n)iθ (1/2)rθ 2 + O(r θ 3 ) ) dθ +O(1) δ θ π exp ( crθ 2) ] dθ = en [ ( 2πn n n 1/2 ) ] 2π + o(1) + O(e cnδ2 )

14 p. 9/4 W. K. Hayman A generalisation of Stirling s formula Journal für die reine und angewandte Mathematik vol 196 (1956) /***************************************************************/ Hayman s method can be used to prove p(n) = 1 4n 3 exp ( C = π 2/3 ) ( Cn 1/2 ( C C )n 1/2 + )

15 p. 10/4 Example 2: Binary Matrices Let s = s 1,s 2,...,s m, t = t 1,t 2,...,t n be integer sequences. Assume s j = t k. Let B(m,n;s,t) be the number of m n matrices over {0, 1} with row sums s and column sums are t. Then B(m,n;s,t) = [x s 1 1 xs m y t 1 1 yt n n ] 1 j m 1 k n (1 + x j y k )

16 p. 10/4 Example 2: Binary Matrices Let s = s 1,s 2,...,s m, t = t 1,t 2,...,t n be integer sequences. Assume s j = t k. Let B(m,n;s,t) be the number of m n matrices over Then {0, 1} with row sums s and column sums are t. B(m,n;s,t) = [x s 1 1 xs m y t 1 1 yt n n ] = 1 (2π) m+n 1 j m 1 k n (1 + xj y k ) x s y t dxdy n+1 n (1 + x j y k )

17 p. 11/4 The Integral x j = q j e iθ j y k = r k e iφ k 1 + x j y k = (1 + q j r k ) ( 1 + q ) jr k (e i(θ j+φ k ) 1) 1 + q j r k B(m,n;s,t) = m j=1 n k=1 (1 + q jr k ) (2π) m+n m j=1 qs j j n k=1 rt k k π π π π n k=1 (1 + λ jk(e i(θ j+φ k ) 1)) exp(i m j=1 s jθ j + i n k=1 t kφ k ) m j=1 dθdφ

18 p. 11/4 The Integral x j = q j e iθ j y k = r k e iφ k 1 + x j y k = (1 + q j r k ) ( 1 + q ) jr k (e i(θ j+φ k ) 1) 1 + q j r k B(m,n;s,t) = m j=1 n k=1 (1 + q jr k ) (2π) m+n m j=1 qs j j n k=1 rt k k π π π π n k=1 (1 + λ jk(e i(θ j+φ k ) 1)) exp(i m j=1 s jθ j + i n k=1 t kφ k ) m j=1 dθdφ λ jk = q jr k 1 + q j r k

19 p. 12/4 The Integrand m j=1 n k=1 (1 + λ jk(e i(θ j+φ k ) 1)) exp(i m j=1 s jθ j + i n k=1 t kφ k ) Get m + n saddlepoint equations

20 p. 12/4 The Integrand m j=1 n k=1 (1 + λ jk(e i(θ j+φ k ) 1)) exp(i m j=1 s jθ j + i n k=1 t kφ k ) Get m + n saddlepoint equations n k=1 λ jk = s j

21 p. 12/4 The Integrand m j=1 n k=1 (1 + λ jk(e i(θ j+φ k ) 1)) exp(i m j=1 s jθ j + i n k=1 t kφ k ) Get m + n saddlepoint equations n k=1 q j r k 1 + q j r k = s j

22 p. 12/4 The Integrand m j=1 n k=1 (1 + λ jk(e i(θ j+φ k ) 1)) exp(i m j=1 s jθ j + i n k=1 t kφ k ) Get m + n saddlepoint equations m j=1 q j r k 1 + q j r k = t k

23 p. 13/4 Comparison of the Two Examples 1/n! B(m,n;s,t) FirstExample SecondExample ========== ========== ========== primary exp( (1/2)Bθ 2 ) diagonalize quad. form beyond quadratic O( θ 3 ) c jkl θ j θ k φ l + d jk θj 2φ k secondary + + O(( ) 5 ) δ θ π complicated

24 p. 14/4 Consequence exp B(m,n;s,t) = ( 1 2 j ( 1 R 2Amn ( n s j ) ( mn λmn k ) ( m t k ) ) ( 1 C 2Amn ) ) + o(1)

25 p. 14/4 Consequence exp B(m,n;s,t) = ( 1 2 j ( 1 R 2Amn ( n s j ) ( mn λmn k ) ( m t k ) ) ( 1 C 2Amn ) ) + o(1) A = (1/2)λ(1 λ), λ = density m n R = (s j s) 2, C = (t k t) 2 j=1 k=1

26 p. 15/4 References Brendan McKay, Catherine Greenhill, & erc Asymptotic enumeration of dense 0-1 matrices with specified line sums Journal of Combinatorial Theory, series A vol 115 (2008) See also A. Barvinok & J.A. Hartigan arxiv: for treatment of non-negative integer matrices

27 p. 16/4 Latin Rectangles Another two-parameter asymptotic counting problem How many k n Latin rectangles are there? Erdos & Kaplansky 1946 k = O(log n) 3/2 ɛ ) Yamamoto 1951 k = o(n 1/3 ) Stein 1978 k = o(n 1/2 ) Godsil & McKay 1990 k = o(n 6/7 ) (n!) k ( (n)k n k ) n ( 1 k ) n/2 e k/2 n

28 p. 17/4 Main Symbols n t d δ the height of the phm the width = ( n 2) the dimension of an integral defines primary/secondary regions.

29 p. 18/4 y and Z(y) Given vector y of height n y =. y j. {±1} n, define the vector of inner products, Z(y), by Z(y) =. y j y k. {±1} d.

30 p. 19/4 Example of Z(y) Z

31 p. 20/4 H nt as a constant term Define M = M n = {Z(y) : y {±1} n }; M = 2 n 1 H nt = 2 t #{( m 1,..., m t ) : m k M & k m k = 0} = 2 t [x 0 12 x0 n 1n ] ( m M ) t jk xm jk jk

32 p. 21/4 Example of Previous Formula H 38 = 2 8 the constant term in the expansion of corresponding to Z(y) = y = u v w ( uvw + v uw + u vw + w ) 8 uv 1 1, 1 1, 1 1 = 1 1 1, , 1, ,

33 p. 22/4 Constant Term as Integral Let x jk = e iλ jk and define ψ(λ) = 1 M m M e iλ m. Then, H nt = 2 t [x 0 12 x 0 n 1n] m M jk x m jk jk t = 2nt (2π) d +π π +π π ψ(λ) t dλ.

34 p. 23/4 Primary/Secondary From now on, 4 t. H nt = 2nt (2π) d [2 (n 1)2 B δ ψ(λ) t dλ + ] ψ(λ) t dλ R δ B δ = {λ : λ δ} R δ = the R(est) of [ π, +π] d. Why the factor 2 (n 1)2?

35 p. 24/4 The Primary Region Where does ψ(λ) = 1? ψ(λ) = 1 M m M e iλ m Start with λ = 0; place π in 2 (n 2) ways; add π/2 in 2 (n 1 2 ) ways. ( ) n 2 + ( ) n 1 2 = (n 1) 2 Without 4 t, the factor (1 t + ( 1) t + (i) t + ( i) t ) vanishes.

36 p. 25/4 Taylor Series for Primary ψ(λ) = 1 M m M e iλ m ψ(λ) = G M even (n) jk λ µ jk(g) jk µ jk (G)!, M even (n) equals even multigraphs on [n] = {1, 2,...,n} Series is convergent for all λ, but we would like ψ(λ) = exp( ).

37 p. 26/4 Primary Integral Assuming nδ 0, Assuming tδ 2, ψ(λ) t = exp ( (t/2) λ 2 +O(tn 3 δ 3 ) ) +δ δ exp ( (t/2)x 2) = 2π t ( ) 1 + O(e tδ2 /2 ). Provided tn 3 δ 3 0, B δ ψ(λ) t = ( ) d/2 2π (1 + o(1)). t

38 p. 27/4 Secondary Integral de Launey and Levin prove that for any k, 1 k n, ψ(λ) n j=1 j k cos 2λ jk This gives ψ(λ) t (2π) d e ctδ2 R δ

39 p. 28/4 Combining the Two H nt = 2nt (2π) d H nt = 2nt (2π) d [ [2 (n 1)2 2 (n 1)2 ( 2π t B δ ψ(λ) t dλ + ] ψ(λ) t dλ R δ ) d/2 (1 + o(1)) + O(1)(2π) d e ctδ2 ] H nt = 2 nt [ 2 (n 1)2 (2πt) d/2 (1 + o(1)) + O(1)e ctδ2] The assumptions: tδ 2, tn 3 δ 3 0.

40 p. 29/4 Conditions on δ H nt = 2 nt [ 2 (n 1)2 (2πt) d/2 (1 + o(1)) + O(1)e ctδ2] We arrive at delauney and Levin s formula, H nt 2nt+(n 1)2 (2πt) d/2, obtained under assumption of δ > 0 such that tn 3 δ 3 0 and tδ 2 = Ω(n 2 log t).

41 p. 30/4 Setting δ H nt 2nt+(n 1)2 (2πt) d/2, Necessary conditions: tn 3 δ 3 0 and tδ 2 = Ω(n 2 log t). Can do for t n 12+ɛ. δ = C n 2 log t tδ 2 = Ω(n 2 log t) tn 3 δ 3 = O(n 6 log 3/2 (t)/t 1/2 ) 0 t

42 p. 31/4 Revisiting Secondary ψ(λ) n j=1 j k cos 2λ jk ψ(λ) 2 e cδ2, λ R δ Worst case: δ λ jk π for one pair jk In fact: δ λ jk π for cn 2 pairs jk = small measure Leads to ψ(λ) t dλ (2π) d e ctnδ2, R δ new factor of n in the exponent

43 p. 32/4 Resetting δ H nt = 2 nt [ 2 (n 1)2 (2πt) d/2(1 + ) + O(e ctnδ2 ) ] For tnδ 2 = Ω(n 2 log t), secondary is negligible vs primary δ = C n log t t previously n 2 in the numerator

44 p. 33/4 Back to Primary We ll assume t n 4+ɛ. (More than needed for nδ = o(1).) ψ(λ) t = exp (t/2) λ 2 it j<k<l λ jk λ jl λ kl + O(tn 2 jk λ 4 jk ) Integrate one variable at a time, starting with x λ 12. +δ δ exp ( (t/2)x 2 itxb + O(tn 2 x 4 ) ) dx B = n l=3 λ 1l λ 2l = O(nδ 2 ) 0

45 p. 34/4 Primary, continued +δ δ Since tn 2 δ 4 n 4 /t 0, exp ( (t/2)x 2 itxb + O(tn 2 x 4 ) ) dx +δ δ exp ( (t/2)x 2 itxb ) ( 1 + O(tn 2 x 4 ) ) dx + exp ( (t/2)x 2 itxb ) ( 1 + O(tn 2 x 4 ) ) dx + O(e ctδ2 )

46 p. 35/4 Complete Square + + e tb2 /2 exp ( (t/2)x 2 itxb ) ( 1 + O(tn 2 x 4 ) ) dx exp ( (t/2)x 2) ( 1 + O(tn 2 (x 4 + B 4 )) ) dx e tb2 /2 2π t ( 1 + O(n 2 /t) + O(tn 2 B 4 ) )

47 p. 36/4 Power-sum bounds for B 2,B 4 B = n λ 1l λ 2l (1/2) n (λ 2 1l + λ2 2l ) l=3 l=3 B 2 = O(n) B 4 = O(n 3 ) n (λ 4 1l + λ4 2l ) l=3 n (λ 8 1l + λ8 2l ) l=3

48 p. 37/4 Combining Previous e tb2 /2 2π t ( 1 + O(n 2 /t) + O(tn 2 B 4 ) ) equals 2π t exp ( O(n 2 /t) + O(tB 2 ) + O(tn 2 B 4 ) )

49 p. 38/4 Combining, continued Previous: 2π t exp ( O(n 2 /t) + O(tB 2 ) + O(tn 2 B 4 ) ) Thence, with the power-sum bounds for B 2,B 4 Primary ( exp O(tn) = exp ( O(n 2 /t) ) 2π B δ t ) n n (λ 4 1l + λ4 2l ) + O(tn5 ) (λ 8 1l + λ8 2l ) l=3 l=3 integrand with remaining d 1 variables

50 Later Variables When integrating with respect to x λ jk, +δ δ exp ( (t/2)x 2 itxb jk + O(tn 2 x 4 ) + O(tn 6 x 8 ) ) dx, because λ 4 jk,λ8 jk terms were inserted O(n) times. Inductively, after τ variables, Primary = exp ( τo(n 2 /t) + τo(n 6 /t 3 ) ) ( ) τ/2 2π B δ t ( ) n n exp O(tn) (λ 4 jl + λ4 kl ) + O(tn5 ) (λ 8 jl + λ8 kl ) l=k+1 l=k+1 integrand with remaining d τ variables p. 39/4

51 p. 40/4 Integration Complete When τ = d, Primary = exp ( do(n 2 /t) + do(n 6 /t 3 ) ) B δ ( 2π t ) d/2 Since, n 4 /t,n 8 /t 3 0, de Launey - Levin formula holds for t n 4+ɛ.

52 p. 41/4 Challenge Want H nt for t as small as can be Would at least like to know the limits of standard methods Current δ gives nδ = o(1) for t n 3+ɛ ; then must integrate ψ(λ) t = exp ( (t/2) λ 2 it λ jk λ jl λ kl + j<k<l + + O(tn ) 4 λ 6 jk ) jk