On Hadamard s Maximal Determinant Problem

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1 Judy-anne Osborn MSI, ANU April 2009

2 m m 1 max det =?

3 A Naive Computer Search Order max det Time 1 1 fast 2 1 fast 3 2 fast 4 3 fast 5 5 fast 6 9 order of days 7 32 order of years 8 56 order of the age of the Universe

4 As well by hand? I found nested Max Dets }2}3}5}9 }1} 1

5 As well by hand? I found nested Max Dets }2}3}5}9 }1} 1

6 As well by hand? I found nested Max Dets }2}3}5}9 }1} 1

7 As well by hand? I found nested Max Dets }2}3}5}9 }1} 1

8 As well by hand? I found nested Max Dets }2}3}5}9 }1} 1 But no further!

9 The problem turns out to be famous Hadamard s Maximal Determinant Problem was posed in 1893 Jacques Hadamard

10 A little selected history on this century-old question...

11 Observe: 2D: ( ) 0 1 or ( ) 1 1 ( ) 1 0 ( ) 1 0 ( 0 ) 1 ( 1 )

12 Geometry: max det = max (hyper-)volume 3D:

13 An equivalent problem: {+1, 1} matrices ( ( ( ( m m matrix ( ( ( ( ( 2) border add row 1 ( ( column ops ( row ops (

14 An equivalent problem: {+1, 1} matrices ( ( ( ( m m matrix ( ( ( ( ( 2) border add row 1 det new = 2 m det old ( ( column ops ( row ops (

15 Volume interpretation upper bound on max det ±1 ±1 max.... }. n ±1 ±1 } n What is the upper bound?

16 Volume interpretation upper bound on max det ±1 ±1 max.... }. n ±1 ±1 } n What is the upper bound? ( (±1) (±1) 2 ) n = n n/2

17 Volume interpretation upper bound on max det ±1 ±1 max.... }. n ±1 ±1 } n What is the upper bound? Why? ( (±1) (±1) 2 ) n = n n/2

18 Volume interpretation upper bound on max det ±1 ±1 max.... }. n ±1 ±1 } n What is the upper bound? ( (±1) (±1) 2 ) n = n n/2 Why? (Columns/rows orthogonal)

19 When is the bound tight? Tight when {+1, 1} square matrix H of order n satisfies HH T = ni

20 When is the bound tight? Tight when {+1, 1} square matrix H of order n satisfies HH T = ni H is called a Hadamard Matrix.

21 When is the bound tight? Tight when {+1, 1} square matrix H of order n satisfies HH T = ni H is called a Hadamard Matrix. A necessary condition on existence of H is: n = 1, 2 or n 0 (mod 4)

22 When is the bound tight? Tight when {+1, 1} square matrix H of order n satisfies HH T = ni H is called a Hadamard Matrix. A necessary condition on existence of H is: n = 1, 2 or n 0 (mod 4) Hadamard Conjecture (Paley, 1933): this is also sufficient.

23 Evidence for Hadamard Conjecture Many constructions for infinite families, including Sylvester, 2 r First Paley, using finite fields, p r + 1, p prime Second Paley, using finite fields, 2p r + 2, p prime

24 Evidence for Hadamard Conjecture Many constructions for infinite families, including Sylvester, 2 r First Paley, using finite fields, p r + 1, p prime Second Paley, using finite fields, 2p r + 2, p prime Other constructions and ad hoc examples due to people including Williamson Jenny Seberry

25 Evidence for Hadamard Conjecture Many constructions for infinite families, including Sylvester, 2 r First Paley, using finite fields, p r + 1, p prime Second Paley, using finite fields, 2p r + 2, p prime Other constructions and ad hoc examples due to people including Williamson Jenny Seberry Smallest n 0 (mod 4) currently undecided: n = 668.

26 More evidence Order Number of inequivalent Hadamard matrices see Sloan s sequence A

27 Max Dets for non-hadamard orders?

28 Max Dets for non-hadamard orders? n max det 2 n n max det 2 n n max det 2 n

29 Max Dets for non-hadamard orders? n max det 2 n n max det 2 n n max det 2 n

30 Max Dets for non-hadamard orders? n max det 2 n n max det 2 n n max det 2 n

31 Tighter upper bounds? The Hadamard bound of n n/2 holds for all orders but is never tight for n 0 (mod 4) when n > 2.

32 Tighter upper bounds? The Hadamard bound of n n/2 holds for all orders but is never tight for n 0 (mod 4) when n > 2. Better upper bounds for non-hadamard orders were proved by Barba in 1933 Ehlich in 1962, 64 Wojtas in 1964 Cohn (proved a new bound tightness result) in 2000

33 The best known upper bounds are: The Barba-Ehlich bound holds for n 1 (mod 4): 2n 1(n 1) (n 1)/2

34 The best known upper bounds are: The Barba-Ehlich bound holds for n 1 (mod 4): 2n 1(n 1) (n 1)/2 The Ehlich-Wojtas bound holds for n 2 (mod 4): (2n 2)(n 2) (n/2) 1

35 The best known upper bounds are: The Ehlich bound holds for n 3 (mod 4): (n 3) n s u v ur 2 (n 3+4r) 2 (n+1+4r) 2 1 n 3 + 4r v(r + 1) n r where s = 3 for n = 3, s = 5 for n = 7, s = 5 or 6 for n = 11, s = 6 for n = 15, 19,..., 59, and s = 7 for n 63, r = n s, n = rs + v and u = s v.

36 Percentages of bounds met: summary from Will Orrick s: maxdet

37 The idea behind the bounds Let R be a maximal determinant square ±1 matrix of order n.

38 The idea behind the bounds Let R be a maximal determinant square ±1 matrix of order n. Consider gram matrix G := RR T

39 The idea behind the bounds Let R be a maximal determinant square ±1 matrix of order n. Consider gram matrix G := RR T 1. G has all n s on the diagonal

40 The idea behind the bounds Let R be a maximal determinant square ±1 matrix of order n. Consider gram matrix G := RR T 1. G has all n s on the diagonal 2. G is positive definite off-diagonal entries have size < n

41 The idea behind the bounds Let R be a maximal determinant square ±1 matrix of order n. Consider gram matrix G := RR T 1. G has all n s on the diagonal 2. G is positive definite off-diagonal entries have size < n 3. G has all off-diagonal entries n (mod 2)

42 The idea behind the bounds Let R be a maximal determinant square ±1 matrix of order n. Consider gram matrix G := RR T 1. G has all n s on the diagonal 2. G is positive definite off-diagonal entries have size < n 3. G has all off-diagonal entries n (mod 2) 4. G has all off-diagonal entries n (mod 4) with R normalized

43 The idea behind the bounds Let R be a maximal determinant square ±1 matrix of order n. Consider gram matrix G := RR T 1. G has all n s on the diagonal 2. G is positive definite off-diagonal entries have size < n 3. G has all off-diagonal entries n (mod 2) 4. G has all off-diagonal entries n (mod 4) with R normalized 5. G is symmetric

44 The idea behind the bounds Let R be a maximal determinant square ±1 matrix of order n. Consider gram matrix G := RR T 1. G has all n s on the diagonal 2. G is positive definite off-diagonal entries have size < n 3. G has all off-diagonal entries n (mod 2) 4. G has all off-diagonal entries n (mod 4) with R normalized 5. G is symmetric Let G n be the set of all gram matrices, G, of order n

45 The idea behind the bounds, continued Let G n be the set of matrices for which properties 1 5 hold.

46 The idea behind the bounds, continued Let G n be the set of matrices for which properties 1 5 hold. Then G n G n

47 The idea behind the bounds, continued Let G n be the set of matrices for which properties 1 5 hold. Then G n G n Hence max det ( Gn ) max det (G n )

48 Case: n 1 (mod 4) The matrix which was proven by Barba and Ehlich to have largest determinant in G n is n n

49 Case: n 2 (mod 4) The matrix which was proven by Ehlich and Wojtas to have largest determinant in G n is ( ) n 2 F 0, where F =... 0 F 2 n

50 Case: n 3 (mod 4) We expect a matrix with largest determinant in G n to be: n n In general, this is wrong Ehlich proved: the correct best determinant matrix in G n is a block form with off-diagonal entries from the set { 1, +3}.

51 Structure in the n 1 (mod 4) case What is a necessary condition for tightness of: Barba-Ehlich: 2n 1(n 1) (n 1)/2?

52 Structure in the n 1 (mod 4) case What is a necessary condition for tightness of: Barba-Ehlich: 2n 1(n 1) (n 1)/2? Lemma: The number 2n 1 is a perfect square iff q N such that n = q 2 + (q + 1) 2

53 In the literature, a conjecture on tightness: Conjecture: The Barba-Ehlich bound is tight whenever n is a sum of two consecutive squares: n = q 2 + (q + 1) 2

54 In the literature, a conjecture on tightness: Conjecture: The Barba-Ehlich bound is tight whenever n is a sum of two consecutive squares: n = q 2 + (q + 1) 2 Evidence: True for q = 2, 4

55 In the literature, a conjecture on tightness: Conjecture: The Barba-Ehlich bound is tight whenever n is a sum of two consecutive squares: n = q 2 + (q + 1) 2 Evidence: True for q = 2, 4 and when q = p r for p an odd prime proved by Brouwer s Construction.

56 What chance an exact maxdet formula n 1 (mod 4)? n max det 2 n Guess

57 What chance an exact maxdet formula n 1 (mod 4)? n max det 2 n Guess A guess/conjecture due to Will Orrick is that n = 4k + 1 is always divisible by max det 2 n 1 for k 2k 1,

58 What chance an exact maxdet formula n 1 (mod 4)? n max det 2 n Guess A guess/conjecture due to Will Orrick is that n = 4k + 1 is always divisible by max det 2 n 1 for k 2k 1, with coefficients growing quadratically between n s for which n = q 2 + (q + 1) 2.

59 What can we hope to compute? The ideas of Ehlich and Wojtas were reused by Chadjipantelis, Kounias and Moissiadis in the 1980 s to find max det matrices for n = 17 and n = 21.

60 What can we hope to compute? The ideas of Ehlich and Wojtas were reused by Chadjipantelis, Kounias and Moissiadis in the 1980 s to find max det matrices for n = 17 and n = 21. Will Orrick used similar ideas in the 2000 s to prove maximal an n = 15 matrix that had previously been found by Cohn; as well as filling in some gaps in CKM s published proofs.

61 What can we hope to compute? The ideas of Ehlich and Wojtas were reused by Chadjipantelis, Kounias and Moissiadis in the 1980 s to find max det matrices for n = 17 and n = 21. Will Orrick used similar ideas in the 2000 s to prove maximal an n = 15 matrix that had previously been found by Cohn; as well as filling in some gaps in CKM s published proofs. Are we within reach of n = 29?

62 Basic Idea Two steps: 1. Find candidate gram matrices in G n. 2. Check if candidates decompose in the form RR T.

63 Essentials of Step 1. There exist theorems which bound the determinants of candidate gram matrices in terms of their sub-matrices. So we can set a target determinant and build candidates: pruning too-small sub-matrices as we go. n n n n n

64 Computational considerations for Step 1. We must have efficient ways to calculate determinants! Rank-One Update Theorems: O(size 3 ) O(size 2 ), at the expense of some book-keeping.

65 Computational considerations for Step 1. ( Need to prune equivalent gram matrices: eg. ( ( ( under simultaneous row and column permutation

66 Computational considerations for Step 1. So need a (partial ordering) which prunes heavily enough, and is practical to compute on-the-fly?

67 Computational considerations for Step 1. So need a (partial ordering) which prunes heavily enough, and is practical to compute on-the-fly? Make sure we don t miss any valid candidates!

68 Essentials of Step 2. Consider gram candidates A and B whose determinants agree.

69 Essentials of Step 2. Consider gram candidates A and B whose determinants agree. These are candidates for RR T and R T R

70 Essentials of Step 2. Consider gram candidates A and B whose determinants agree. These are candidates for RR T and R T R Implement two kinds of constraints: Linear, Quadratic which use A and B simultaneously

71 Essentials of Step 2. the linear constraints Let A = (a ij ) and B = (b ij ).

72 Essentials of Step 2. the linear constraints Let A = (a ij ) and B = (b ij ). Assume r 0 R =. r n 1 = ( ) c 0 c n 1

73 Essentials of Step 2. the linear constraints Let A = (a ij ) and B = (b ij ). Assume r 0 R =. r n 1 = ( ) c 0 c n 1 Then a ij = r i.r j and b ij = c i.c j

74 Essentials of Step 2. the linear constraints Let A = (a ij ) and B = (b ij ). Assume r 0 R =. r n 1 = ( ) c 0 c n 1 Then a ij = r i.r j and b ij = c i.c j We need to implement an ordering to prune duplicates

75 Considerations for the linear constraints of Step 2. eg A = If we have already built r 0 = (1,,, 1, 1,,,,,, 1, 1, 1, 1, 1, 1, 1) r 1 = (,, 1,,,,, 1, 1, 1,,,, 1, 1, 1, 1) then r 2 breaks into blocks: r 2 = (a; b; c; d, e; f, g; h, i, j; k, l, m; n, o, p, q) and

76 Considerations for the linear constraints of Step 2. eg A = If we have already built r 0 = (1,,, 1, 1,,,,,, 1, 1, 1, 1, 1, 1, 1) r 1 = (,, 1,,,,, 1, 1, 1,,,, 1, 1, 1, 1) then r 2 breaks into blocks: r 2 = (a; b; c; d, e; f, g; h, i, j; k, l, m; n, o, p, q) and Adding more rows is a process of successive block refinement

77 Considerations for the linear constraints of Step 2. Because we work with both rows and columns, we need a way of refining row-blocks and column blocks simultaneously! ?

78 Essentials of Step 2. the quadratic constraints To derive the quadratic constraints, write key ingredients in block form: ( ) ( ) 1 y T R = x R, R T 1 x T = y R T, A = ( n a T a A ), B = ( ) n b T b B

79 Essentials of Step 2. the quadratic constraints To derive the quadratic constraints, write key ingredients in block form: ( ) ( ) 1 y T R = x R, R T 1 x T = y R T, A = ( n a T a A ), B = ( ) n b T b B This allow several quadratic constraints to be found, eg. det(a xx T ) = a perfect square = det(b yy T )

80 Computational Considerations for Step 2. We need to decide in which order to implement the various quadratic and linear constraints.

81 Computational Considerations for Step 2. We need to decide in which order to implement the various quadratic and linear constraints. How to decide?

82 THE END

Decomposing Gram matrices

Decomposing Gram matrices Richard P. Brent MSI & CECS ANU Describing joint work with Will Orrick, Judy-anne Osborn and Paul Zimmermann 14 May 2010 Abbreviations maxdet Hadamard maximal determinant problem or a matrix with maximal