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"Estimating Tree-Structured Covariance Matrices via Mixed-Integer Programming"

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セミナーで紹介したネタ論文.スライドはあわてて作ったので見映えは良くない.

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"Estimating Tree-Structured Covariance Matrices via Mixed-Integer Programming"

  1. 1. • – – – • – • – • • •
  2. 2. • – – – • – • – • • •
  3. 3. • •
  4. 4. • • • •
  5. 5. • •
  6. 6. • • • • •
  7. 7. • • • • • • • • • • • • • • •
  8. 8. • • • • • • • • • • • • • • •
  9. 9. • • • • • • • • • • • • • • •
  10. 10. • • • • • • • • • • • • • • •
  11. 11. • • • • • • • • • • • • • • •
  12. 12. • • • • • • • • • • • • • • •
  13. 13. • • • • • • • • • • • • • • •
  14. 14. • • • • • • • • • • • • • • •
  15. 15. • • • • • • • • • • • • • • •
  16. 16. • • • • • • • • • • • • • • •
  17. 17. • – – – • – • – • • •
  18. 18. T ;d • T 1 1 1 1 1 1
  19. 19. • B • Bij i j 1 1 1 1 1 1 T ;d
  20. 20. • B • Bij i j 1 1 1 1 1 1 T ;d
  21. 21. 1+1 1 1 1 1 1 1 T ;d • B • Bij i j
  22. 22. 1 1 1 1 1 1 T ;d • B • Bij i j
  23. 23. 1 11 1 1 1 1 1 T ;d • B • Bij i j
  24. 24. 1+1 1 0 0 1 1+1 0 0 0 0 1+1 1 0 0 1 1+1 1 1 1 1 1 1 T ;d • B • Bij i j
  25. 25. T • T 1 1 1 1 1 1 B T T 1+1 1 0 0 1 1+1 0 0 0 0 1+1 1 0 0 1 1+1
  26. 26. T • T 1 1 1 2 1 1 1+1 1 0 0 1 1+2 0 0 0 0 1+1 1 0 0 1 1+1 B T
  27. 27. • T 1 2 1 1 1 1 1+1 1 0 0 1 1+1 0 0 0 0 2+1 2 0 0 2 2+1 T B T
  28. 28. • T 0.9+2 0.9 0 0 0.9 0.9+3 0 0 0 0 2+1.1 2 0 0 2 2+1.5 T 0.9 2 2 3 1.1 1.5 B T
  29. 29. • T B. – B – B T
  30. 30. T T
  31. 31. • – – – • – • – • • •
  32. 32. B
  33. 33. B
  34. 34. B
  35. 35. B
  36. 36. B
  37. 37. B
  38. 38. B
  39. 39. B
  40. 40. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0
  41. 41. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1
  42. 42. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 2
  43. 43. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2
  44. 44. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 3
  45. 45. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1
  46. 46. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1
  47. 47. B 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 2 2 0 0 2 2 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
  48. 48. 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 2 2 0 0 2 2 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
  49. 49. • dk d0=0 d1=1 d2=2 d3=2 d4=3 d5=1 d6=1
  50. 50. • vk – p (2p -1) v0 v1 v2 v3 v4 v5 v6
  51. 51. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 VDVT d0=0 d1=1 d2=2 d3=2 d4=3 d5=1 d6=1
  52. 52. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 VkVk B d0=0 d1=1 d2=2 d3=2 d4=3 d5=1 d6=1
  53. 53. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 B = VDVT d0=0 d1=1 d2=2 d3=2 d4=3 d5=1 d6=1
  54. 54. 1+2 1 0 0 1 1+3 0 0 0 0 2+1 2 0 0 2 2+1 B = VDVT d0=0 d1=1 d2=2 d3=2 d4=3 d5=1 d6=1
  55. 55. 2.9 0.9 0 0 0.9 3.9 0 0 0 0 3.1 2 0 0 2 3.5 B T ? T B
  56. 56. T B 2.9 0.9 0 0 0.9 3.9 0 0 0 0 3.1 2 0 0 2 3.5 min (Bij) = 0 = d0 B
  57. 57. T B 2.9 0.9 0 0 0.9 3.9 0 0 0 0 3.1 2 0 0 2 3.5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 B 0
  58. 58. T B 2.9 0.9 0 0 0.9 3.9 0 0 0 0 3.1 2 0 0 2 3.5
  59. 59. T B
  60. 60. T B d0 = 0
  61. 61. T B 2.9 0.9 0.9 3.9 0.9 = d1
  62. 62. T B 2.9 0.9 0.9 3.9 d0 = 0 d1 = 0.9
  63. 63. T B 2.9 0.9 0.9 3.9 0.9 1 1 1 1
  64. 64. T B 2 0 0 3 0.9 1 1 1 1
  65. 65. T B d0 = 0 d1 = 0.9
  66. 66. T B d0 = 0 d1 = 0.9 2 0 0 3
  67. 67. T B 2 2 = d3
  68. 68. T B 3 3 = d4
  69. 69. T B d0 = 0 d1 = 0.9 d3 = 2 d4 = 3 2 3
  70. 70. T B d0 = 0 d1 = 0.9 d3 = 2 d4 = 3 2.9 0.9 0 0 0.9 3.9 0 0 0 0 3.1 2 0 0 2 3.5 d2 = 2 d5 = 1.1 d6 = 1.5 TB
  71. 71. V D VDV
  72. 72. D : a p p diagonal matrix All the entries are non-negative.
  73. 73. v0 V : a p (2p -1) Boolean matrix
  74. 74. v0 V has the ‘partition property’.
  75. 75. v0 V contains the vector of all ones as a column. V has the ‘partition property’.
  76. 76. v0 v1 v2 v3 v4 v5 v6 For every column vk with more than one non-zero entries, there exist exactly two columns vi and vj s.t. vi + vj = vk . V has the ‘partition property’.
  77. 77. v0 v1 v2 v3 v4 v5 v6 For every column vk with more than one non-zero entries, there exist exactly two columns vi and vj s.t. vi + vj = vk . V has the ‘partition property’.
  78. 78. v0 v1 v2 v3 v4 v5 v6 For every column vk with more than one non-zero entries, there exist exactly two columns vi and vj s.t. vi + vj = vk . V has the ‘partition property’.
  79. 79. v0 v1 v2 v3 v4 v5 v6 For every column vk with more than one non-zero entries, there exist exactly two columns vi and vj s.t. vi + vj = vk . V has the ‘partition property’.
  80. 80. For every column vk with more than one non-zero entries, there exist exactly two columns vi and vj s.t. vi + vj = vk . V has the ‘partition property’. v0 v1 v2 v3 v4 v5 v6
  81. 81. D A p p diagonal matrix with all entries non-negative V A p (2p -1) Boolean matrix with the ‘partition property‘ VDV BT B is a p p matrix spanned by vkvk in bijective correspondence with its rooted binary tree T with p leaves.
  82. 82. • – – – • – • – • • •
  83. 83. • S subject to minimise B (difference between B and S ) BT S Y (B is tree-structured)
  84. 84. subject to minimise D,V (difference between B and S ) (B is tree-structured) BT S Y
  85. 85. V B D
  86. 86. B S V
  87. 87. • • • • 0 0 0 0 0 0 0 0 d1 0 0 0 0 0 0 0 d2 0 0 0 0 0 0 0 d3 0 0 0 0 0 0 0 d4 0 0 0 0 0 0 0 d5 0 0 0 0 0 0 0 d6 d1 d2 d3 d4 d5 d6 B b
  88. 88. • •
  89. 89. • • • • subject to minimise d (B is tree-structured)
  90. 90. • •
  91. 91. • – – – • – • – • • •
  92. 92. B
  93. 93. ∵ ∵ ∵
  94. 94. F
  95. 95. F
  96. 96. • – – – • – • – • • •
  97. 97. ?
  98. 98. ?
  99. 99. ?
  100. 100. • – – – • – • – • • •
  101. 101. ∈ ℝ 𝑝 N
  102. 102. S
  103. 103. B
  104. 104. • – – •
  105. 105. • • •

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