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Probabilistic lower bounds on maximal determinants of binary matrices 6 1 0 2 Richard P. Brent Judy-anne H. Osborn t Australian National University The University of Newcastle c O Canberra, ACT 2600 Callaghan, NSW 2308 Australia Australia 5 2 Warren D. Smith ] Center for Range Voting O 21 Shore Oaks Drive C Stony Brook, NY 11790 . h USA t a m In memory of Mirka Miller 1949–2016 [ 7 Abstract v 5 Let (n) be the maximal determinant for n n 1 -matrices, 3 2 and (nD) = (n)/nn/2 be the ratio of (n) to th×e Ha{d±am}ard upper R D D 6 bound. Using the probabilistic method, we prove new lower bounds 0 on (n) and (n) in terms of d = n h, where h is the order of a . D R − 1 Hadamard matrix and h is maximal subject to h n. For example, ≤ 0 5 2 d/2 1 (n)> if 1 d 3, and R πe ≤ ≤ v: (cid:18) (cid:19) Xi (n)> 2 d/2 1 d2 π 1/2 if d>3. r R πe − 2h a (cid:18) (cid:19) (cid:18) (cid:16) (cid:17) (cid:19) By a recent result of Livinskyi, d2/h1/2 0 as n , so the second boundiscloseto(πe/2)−d/2forlargen. →Previousl→ow∞erboundstended tozeroasn withdfixed,exceptinthecasesd 0,1 . Ford 2, →∞ ∈{ } ≥ our bounds are better for all sufficiently large n. If the Hadamard conjecture is true, then d 3, so the first bound above shows that (n) is bounded below by ≤a positive constant (πe/2)−3/2 >0.1133. R 1 Introduction Let (n) be the maximal determinant possible for an n n matrix with ele- D × ments in 1 . Hadamard [15] proved that (n) nn/2, and the Hadamard {± } D ≤ conjecture isthatamatrixachievingthisupperboundexistsforeachpositive integer n divisible by four. The function (n) := (n)/nn/2 is a measure R D of the sharpness of the Hadamard bound. Clearly (n) = 1 if a Hadamard R matrix of order n exists; otherwise (n) < 1. In this paper we give lower R bounds on (n) and (n). D R Let be the set of orders of Hadamard matrices, and let h be H ∈ H maximal subject to h n. Then d = n h can be regarded as the “gap” ≤ − between n and the nearest (lower) Hadamard order. We are interested the case that n is not a Hadamard order, i.e. d > 0 and (n) < 1. R Except in the cases d 0,1 , previous lower bounds on (n) tended to ∈{ } R zero as n . For example, the well-known bound of Clements and Lind- → ∞ ström[11,CorollarytoThm.2]showsthat (n) > (3/4)n/2,and[5, Thm. 9] shows that (n) (ne/4)−d/2. In contrRast, our results imply that, for R ≥ fixedd, (n)isboundedbelowbyapositiveconstant(dependingonlyon d). R Our lower bound proof uses the probabilistic method pioneered by Erdős (see for example [1, 13]). This method does not appear to have been applied previously to the Hadamard maximal determinant problem, except in the case d = 1 (so n 1 mod4); in this case the concept of excess has been ≡ used [14], and lower bounds on the maximal excess were obtained by the probabilistic method [2, 9, 13, 14]. §2 describes our probabilistic construction and determines the mean µ and variance σ2 of elements in the Schur complement generated by the construction (see Lemmas 2.6 and 2.9). Informally, we adjoin d extra columns to an h h Hadamard matrix A, and fill their h d entries with random × × (uniformly and independently distributed) 1 values. Then we adjoin d ex- ± tra rows, andfill theird (h+d)entries with values chosen deterministically × in a way intended to approximately maximise the determinant of the final matrix A. To do so, we use the fact that this determinant can be expressed in terms of the d d Schur complement of A in A. × In thee case d = 1, this method is essentially the same as the known method involving the excess of matrices Hadameard-equivalent to A, and leads to the same bounds that can be obtained by bounding the excess in a probabilistic manner. In §3 we give lower bound results on both (n) and (n). Of course, a D R lower bound on (n) immediately gives an equivalent lower bound on (n). D R However, we use some elementary inequalities to obtain simpler (though 2 slightly weaker) bounds on (n). For example, if d 3 then Theorem 3.6 R ≤ states that (n) hh/2(µd η), where µ and η are certain functions of h and d. TheoDrem 3≥.6 also stat−es the (weaker) result that (n)> (πe/2)−d/2. R The lower bound on (n) clearly shows that the ratio of our bound to the Hadamard bound is aRt least (πe/2)−3/2 > 0.1133, whereas this conclusion is not immediately obvious from the lower bound on (n). D We outline the bounds on (n) here. Theorem 3.4 gives a lower bound R d/2 2 π 1/2 (n) > 1 d2 (1) R πe − 2h (cid:18) (cid:19) (cid:18) (cid:19) (cid:16) (cid:17) which is nontrivial whenever h > πd4/2. By the results of Livinskyi [20], d = O(h1/6) as h (see [7, §6] for details), so the condition h > πd4/2 holds for all suffici→ent∞ly large n. Also, as n , d2/h1/2 = O(n−1/6) 0, sothelowerbound(1)iscloseto(πe/2)−d/2.→Fo∞rfixedd> 1andlargen→,our lower bounds on (n) are better than previous bounds (see Table 1 in §4). R Theorem 3.6 applies only for d 3, but whenever it is applicable it gives ≤ sharperresultsthanTheorem3.4. Infact,Theorem3.6showsthatthefactor 1 O(d2/h1/2) in (1) can be omitted when d 3, giving (n)> (πe/2)−d/2. − ≤ R Theorem 3.6 is always applicable if the Hadamard conjecture is true, since this conjecture implies that d 3. ≤ In§4,wegivesomenumericalexamplestoillustrateTheorems3.4and3.6, and to compare our results with previous bounds on (n) and/or (n). D R Rokicki et al [23] showed, by extensive computation, that (n) 1/2 R ≥ for n 120, and conjectured that this inequality always holds. It seems difficul≤t to bridge the gap between the constants 1/2 and (πe/2)−3/2 by the probabilistic method. The best that we can do is to improve the term of order d2/h1/2 in the bound (1) at the expense of a more complicated proof – for details see [7]. 2 The probabilistic construction We now describe our probabilistic construction and prove some of its properties. In the case d = 1 our construction reduces to that of Best [2]. Let A be a Hadamard matrix of order h 4. We add a border of d rows ≥ and columns to give a larger (square) matrix A of order n. The border is defined by matrices B, C and D as shown: e A B A = . (2) C D (cid:20) (cid:21) e 3 The d d matrix D CA−1B is known as the Schur complement of A in A × − after Schur [24]. The Schur complement lemma (see for example [12]) gives e det(A) = det(A)det(D CA−1B). (3) − Inourconstructionthemeatrices A,B,andC haveentries in 1 . Weallow {± } the matrix D to have entries in 0, 1 , but each zero entry can be replaced { ± } byoneof+1or 1withoutdecreasing det(A),soanylowerboundsthatwe − | | obtain on max( det(A)) are valid lower bounds on maximal determinants | | of n n 1 -matrices. Note that the Schuer complement is not in general × {± } a 1 -matrix. e {± } In the proof of Lemma 3.2 we show that our choice of B, C and D gives a Schur complement D CA−1B that, with positive probability, has − sufficiently large determinant. From equation (3) and the fact that A is a Hadamard matrix, a large value of det(D CA−1B) implies a large value of − det(A). 2.1 eDetails of the probabilistic construction Let A be any Hadamard matrix of order h. B is allowed to range over the set of all h d 1 -matrices, chosen uniformly and independently from the × {± } 2hd possibilities. The d h matrix C = (c ) is a function of B. We choose ij × c = sgn(ATB) , ij ji where +1 if x 0, sgn(x) := ≥ ( 1 if x < 0. − To complete the construction, we choose D = I. As mentioned above, it − is inconsequential that D is not a 1 -matrix. {± } 2.2 Properties of the construction DefineF = CA−1B andG= F D = F+I (so GistheSchurcomplement defined above). Note that, sinc−e A is a Hadam−ard matrix, AT = hA−1, so hF = CATB. Since B is random, we expectthe elements of ATB to beusually of order h1/2. The definition of C ensures that there is no cancellation in the inner products defining the diagonal entries of hF = C (ATB). Thus, we expect · the diagonal entries f of F to be nonnegative and of order h1/2, but the ii off-diagonal entries f (i = j) to be of order unity with high probability. ij 6 4 Similarly for the elements of G. This intuition is justified by Lemmas 2.6 and 2.9. In the following we denote the expectation of a random variable X by E[X], and the variance by V[X] = E[X2] E[X]2. − 1 Lemmas 2.1–2.2 are essentially due to Best [2] and Lindsey. Lemma 2.1. If h 2 and F = (f ) is chosen as above, then ij ≥ h 2−hh if i= j, E[f ] = h/2 ij  (cid:18) (cid:19) 0 if i = j.  6 Proof. The case i = j followsas in Best [2, proof of Theorem 3]. The case i = j is easy, since B is chosen randomly. 6 Lemma2.2. IfF = (f )ischosenasabove, then f h1/2 for1 i,j d. ij ij | |≤ ≤ ≤ Proof. The matrix Q := h−1/2AT is orthogonal with rows and columns of unit length (in the Euclidean norm). Thus Qb = b = h1/2 for each 2 2 || || || || column b of B. Since h1/2F = C.QB, each element h1/2f of h1/2F is the ij inner product of a row of C (having length h1/2) and a column of QB (also having length h1/2). It follows from the Cauchy-Schwartz inequality that h1/2f h1/2 h1/2 = h, so f h1/2. ij ij | | ≤ · | | ≤ Lemma 2.3. If F is chosen as above and i,j k,ℓ = , then f and ij { }∩{ } ∅ f are independent. kℓ Proof. This follows from the fact that f depends only on the fixed matrix ij A and on columns i and j of B. Lemma 2.4. Let A 1 h×h be a Hadamard matrix, C 1 d×h, and U = CA−1. Then, fo∈r e{a±ch}i with 1 i d, ∈ {± } ≤ ≤ h u2 =1. ij j=1 X Proof. Since A is Hadamard, UUT = h−1CCT. Also, since c = 1, ij ± diag(CCT)= hI. Thus diag(UUT) = I. Lemma 2.5. If F = (f ) is chosen as above, then ij E[f2]= 1 for i = j. (4) ij 6 1See[13, footnote on pg. 88]. 5 Proof. We can assume, without loss of generality, that i = 1, j > 1. Write F = UB, where U = CA−1 = h−1CAT. Now f = u b , (5) 1j 1k kj k X where 1 u = c a , c = sgn b a . 1k 1ℓ kℓ 1ℓ m1 mℓ h ! ℓ m X X Observe that c and u depend only on the first column of B. Thus, f 1ℓ 1k 1j depends only on the first and j-th columns of B. If we fix the first column of B and take expectations over all choices of the other columns, we obtain E[f2 ]= E u u b b . 1j 1k 1ℓ kj ℓj " # k ℓ XX The expectation of the terms with k = ℓ vanishes, and the expectation of 6 the terms with k = ℓ is u2 . Thus, (4) follows from Lemma 2.4. k 1k Lemma 2.6. Let A bePa Hadamard matrix of order h 4 and B, C be 1 -matrices chosen as above. Let G = F +I where F =≥CA−1B. Then {± } h h E[g ] = 1+ , (6) ii 2h h/2 (cid:18) (cid:19) E[g ] = 0 for 1 i,j d, i = j, (7) ij ≤ ≤ 6 h(h 1) h/2 2 h2 h 2 V[g ] = 1+ − , (8) ii 2h+1 h/4 − 22h h/2 (cid:18) (cid:19) (cid:18) (cid:19) V[g ] = 1 for 1 i,j d, i = j. (9) ij ≤ ≤ 6 Proof. Since G = F +I, the results (6), (7) and (9) follow from Lemma 2.1 and Lemma 2.5 above. Thus, we only need to prove (8). Since g = f +1, ii ii it is sufficient to compute V[f ]. ii Since A is a Hadamard matrix, hF = CATB. We compute the second moment about the origin of the diagonal elements hf of hF. Since h is a ii Hadamard order and h 4, we can write h = 4k where k Z. Consider h ≥ ∈ independent random variables X 1 , 1 j h, where X = +1 with j j ∈ {± } ≤ ≤ probability 1/2. Define random variables S , S by 1 2 4k 2k 4k S = X , S = X X . 1 j 2 j j − j=1 j=1 j=2k+1 X X X 6 Consider a particular choice of X ,...,X and suppose that k + p of 1 h X ,...,X are +1, and that k+q of X ,...,X are +1. Then we have 1 2k 2k+1 4k S = 2(p + q) and S = 2(p q). Thus, taking expectations over all 24k 1 2 − possible (equally likely) choices, we see that 4 2k 2k E[S S ] = 4E[p2 q2 ] = p2 q2 | 1 2| | − | 24k k+p k+q | − | p q (cid:18) (cid:19)(cid:18) (cid:19) XX 4 2k 2 h2 2k 2 = 2k2 = . 24k · k 2h+1 k (cid:18) (cid:19) (cid:18) (cid:19) Here the closed form for the double sum is a special case of [4, Prop. 1.1]. By the definitions of B, C and F, we see that hf is a sum of the form ii Y +Y + +Y , where each Y is a random variable with the same distri- 1 2 h j ··· bution as S , and each product Y Y (for j = ℓ) has the same distribution 1 j ℓ | | 6 as S S . Also, Y2 has the same distribution as S 2 = S2. The random | 1 2| j | 1| 1 variables Y are not independent, but by linearity of expectations we obtain j h2 2k 2 h2E[f2] = hE[S2]+h(h 1)E[S S ] = h2+h(h 1) . ii 1 − | 1 2| − · 2h+1 k (cid:18) (cid:19) This gives 2 h(h 1) 2k E[f2]= 1+ − . ii 2h+1 k (cid:18) (cid:19) The result for V[g ] now follows from V[g ]= V[f ]= E[f2] E[f ]2. ii ii ii ii − ii For convenience we write µ(h) := E[g ] = E[f ]+1 and σ(h)2 := V[g ]. ii ii ii If h is understood from the context we write simply µ and σ2 respectively. To estimate µ and σ2 from Lemma 2.6, we need a sufficiently accurate estimate for a central binomial coefficient 2m (where m = h/2 or h/4). An m asymptotic expansion for ln 2m may bededuced from Stirling’s asymptotic m (cid:0) (cid:1) expansion of lnΓ(z), as in [16]. However, [16] does not give an error bound. (cid:0) (cid:1) We state such a bound in the following Lemma. Lemma 2.7. If k and m are positive integers, then 2m ln(πm) k−1 B (1 4−j) ln = mln4 2j − m1−2j +e (m), (10) k m − 2 − j(2j 1) (cid:18) (cid:19) j=1 − X where B e (m) < | 2k| m1−2k. (11) k | | k(2k 1) − 7 Proof. Using the facts that m is real and positive, and that the sign of the Bernoullinumber B is( 1)k−1,weobtainfromOlver[21,(4.03)and(4.05) 2k − of Ch. 8] that k−1 ln(2π) B lnΓ(m) = (m 1)lnm m+ + 2j m1−2j ( 1)kr (m), −2 − 2 2j(2j 1) − − k j=1 − X (12) where B 0 <r (m) < | 2k| m1−2k. (13) k 2k(2k 1) − Now 2m (2m)! 2 Γ(2m) = = , m m!m! mΓ(m)2 (cid:18) (cid:19) so from (12) and the same equation with m 2m we obtain (10) with 7→ e (m) = ( 1)k(2r (m) r (2m)). k k k − − Using the bound (13), this gives ( 1)k B e (m) = − | 2k|m1−2kθ, k k(2k 1) − where 2−2k < θ < 1. In particular, θ < 1, so we obtain the desired − | | bound (11). Remark 2.8. Lemma 2.7 can be sharpened. In fact, e (m) has the same k sign as the first omitted term (corresponding to j = k) and has smaller magnitude. This is proved in [3, Corollary 2]. We now show that µ(h) is of order h1/2, and that σ(h) is bounded. Lemma 2.9. For h 4Z, h 4, we have ∈ ≥ σ(h)2 < 1 (14) and 2h 2h +0.9 < µ(h) < +1. (15) π π r r Proof. From Lemma 2.7 with k = 2 and m a positive integer, we have 2m 4m 1 θ m = exp + , (16) m √πm −8m 180m3 (cid:18) (cid:19) (cid:20) (cid:21) 8 where θ < 1. m | | First consider the bounds (16) on µ(h). Taking m = h/2 and using the expression (6) for µ(h), the inequality (15) is equivalent to m 1 m 2m m < < . π − 20 4m m π r (cid:18) (cid:19) r The upper bound is immediate from (16), since 1 + 1 < 0. −8m 180m3 For the lower bound, a computation verifies the inequality for m = 2, since 2/π 1 < 3 = m 2m . Hence, we can assume that m 4. The − 20 4 4m m ≥ lower bound now follows from (16), since p (cid:0) (cid:1) m 2m m 1 1 m 1 1 > exp > 1 4m m π −8m − 180m3 π − 8m − 180m3 (cid:18) (cid:19) r (cid:20) (cid:21) r (cid:20) (cid:21) and m 1 1 1 + < . π 8m 180m3 20 r (cid:20) (cid:21) Now consider the upper bound (14) on σ(h)2. From (16) we have h/2 2 2h+2 1 32 < exp + h/4 πh −h 45h3 (cid:18) (cid:19) (cid:20) (cid:21) and h 2 22h+1 1 4 > exp . h/2 πh −2h − 45h3 (cid:18) (cid:19) (cid:20) (cid:21) Using these inequalities in (8) and simplifying gives 2h 1 32 1 4 σ(h)2 < 1+ exp + exp π −h 45h3 − −2h − 45h3 (cid:20) (cid:18) (cid:19) (cid:18) (cid:19)(cid:21) 2 1 32 exp + . (17) − π −h 45h3 (cid:18) (cid:19) Itiseasytoseethattheterminsquarebrackets isnegative forh 4, so(17) ≥ implies (14). Remark 2.10. We can show from (17) and a corresponding lower bound on σ(h)2 that σ(h+4)2 < σ(h)2, so σ(h)2 is monotonic decreasing and bounded above by σ(4)2 = 1. Also, for large h we have σ(h)2 = (1 3/π)+O(1/h). 4 − Since these results are not needed below, we omit the details. 9 3 A probabilistic lower bound We now prove lower bounds on (n) and (n) where, as usual, n = h+d D R and h is the order of a Hadamard matrix. The key result is Lemma 3.2. Theorem 3.4 simply converts the result of Lemma 3.2 into lower bounds on (n) and (n), giving away a little for the sake of simplicity in the latter D R case. For the proof of Lemma 3.2 we need the following bound on the determinant of a matrix which is “close” to the identity matrix. It is due to Ostrowski [22, eqn. (5,5)]; see also [8, Corollary 1]. Lemma 3.1 (Ostrowski). If M = I E Rd×d, e ε for 1 i,j d, ij − ∈ | | ≤ ≤ ≤ and dε 1, then ≤ det(M) 1 dε. ≥ − The idea of Lemma 3.2 is that we can, with positive probability, apply Lemma 3.1 to the matrix M = µ−1G, thus obtaining a lower bound on the maximum value attained by det(G). Lemma 3.2. Suppose d 1, 4 h , n = h+d, G as in 2.2. Then, ≥ ≤ ∈ H § with positive probability, detG d2 1 . (18) µd ≥ − µ Proof. Let λ be a positive parameter to be chosen later, and µ = µ(h). We say that G is good if the conditions of Lemma 3.1 apply with M = µ−1G and ε = λ/µ. Otherwise G is bad. Assume 1 i,j d. From Lemma 2.6, V[g ] = 1 for i = j; from ij ≤ ≤ 6 Lemma 2.9, V[g ] = σ2 < 1. It follows from Chebyshev’s inequality [10] ii that 1 P[g λ] for i= j, | ij| ≥ ≤ λ2 6 and σ2 P[g µ λ] . | ii − | ≥ ≤ λ2 Thus, d(d 1) dσ2 d2 P[G is bad] − + < . ≤ λ2 λ2 λ2 Taking λ = d gives P[G is bad] < 1, so P[G is good] is positive. When- ever G is good we can apply Lemma 3.1 to µ−1G, obtaining µ−ddet(G) = det(µ−1G) 1 dε = 1 dλ/µ = 1 d2/µ. ≥ − − − 10