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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20. T ;d
• T
1 1
1 1 1 1

21. • B
• Bij
i j
1 1
1 1 1 1
T ;d

22. • B
• Bij
i j
1 1
1 1 1 1
T ;d

23. 1+1
1 1
1 1 1 1
T ;d
• B
• Bij
i j

24. 1 1
1 1 1 1
T ;d
• B
• Bij
i j

25. 1
11 1
1 1 1 1
T ;d
• B
• Bij
i j

26. 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

27. 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

28. 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

29. • 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

30. • 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

31. • T B.
– B
– B T

33. T
T

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36. B

38. B

39. B

40. B

41. B

42. B

43. B

44. B

45. 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

46. 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

47. 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

48. 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

49. 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

50. 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

51. 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

52. 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

53. 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

54. • dk
d0=0
d1=1
d2=2
d3=2 d4=3
d5=1 d6=1

55. • vk
– p (2p -1)
v0
v1
v2
v3
v4 v5 v6

57. 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

58. 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

59. 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

60. 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

61. 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

62. 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

63. 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

64. T B
2.9 0.9 0 0
0.9 3.9 0 0
0 0 3.1 2
0 0 2 3.5

65. T B

66. T B
d0 = 0

67. T B
2.9 0.9
0.9 3.9
0.9 = d1

68. T B
2.9 0.9
0.9 3.9
d0 = 0
d1 = 0.9

69. T B
2.9 0.9
0.9 3.9
0.9
1 1
1 1

70. T B
2 0
0 3
0.9
1 1
1 1

71. T B
d0 = 0
d1 = 0.9

72. T B
d0 = 0
d1 = 0.9
2 0
0 3

73. T B
2 2 = d3

74. T B
3
3 = d4

75. T B
d0 = 0
d1 = 0.9
d3 = 2 d4 = 3
2
3

76. 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

77. V D
VDV

78. D : a p p diagonal matrix
All the entries are non-negative.

79. v0
V : a p (2p -1) Boolean matrix

80. v0
V has the ‘partition property’.

81. v0
V contains the vector of all ones as a column.
V has the ‘partition property’.

82. 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’.

83. 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’.

84. 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’.

85. 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’.

86. 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

87. 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.

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89. • S
subject to
minimise
B
(difference between B and S )
BT
S Y
(B is tree-structured)

90. subject to
minimise
D,V
(difference between B and S )
(B is tree-structured)
BT
S Y

91. V
B D

92. B S
V

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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

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subject to
minimise
d
(B is tree-structured)

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100. B

101. ∵
∵
∵

110. F

111. F

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124. ?

125. ?

126. ?

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134. ∈ ℝ 𝑝
N

135. S

136. B

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