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17.
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18.
T ;d
• T
1 1
1 1 1 1

19.
• B
• Bij
i j
1 1
1 1 1 1
T ;d

20.
• B
• Bij
i j
1 1
1 1 1 1
T ;d

21.
1+1
1 1
1 1 1 1
T ;d
• B
• Bij
i j

22.
1 1
1 1 1 1
T ;d
• B
• Bij
i j

23.
1
11 1
1 1 1 1
T ;d
• B
• Bij
i j

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.
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.
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.
• 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.
• 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.
• T B.
– B
– B T

30.
T
T

31.
•
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–
•
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•
–
•
•
•

32.
B

33.
B

34.
B

35.
B

36.
B

37.
B

38.
B

39.
B

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

50.
• vk
– p (2p -1)
v0
v1
v2
v3
v4 v5 v6

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.
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.
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.
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.
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.
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.
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.
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.
T B

60.
T B
d0 = 0

61.
T B
2.9 0.9
0.9 3.9
0.9 = d1

62.
T B
2.9 0.9
0.9 3.9
d0 = 0
d1 = 0.9

63.
T B
2.9 0.9
0.9 3.9
0.9
1 1
1 1

64.
T B
2 0
0 3
0.9
1 1
1 1

65.
T B
d0 = 0
d1 = 0.9

66.
T B
d0 = 0
d1 = 0.9
2 0
0 3

67.
T B
2 2 = d3

68.
T B
3
3 = d4

69.
T B
d0 = 0
d1 = 0.9
d3 = 2 d4 = 3
2
3

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.
V D
VDV

72.
D : a p p diagonal matrix
All the entries are non-negative.

73.
v0
V : a p (2p -1) Boolean matrix

74.
v0
V has the ‘partition property’.

75.
v0
V contains the vector of all ones as a column.
V has the ‘partition property’.

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.
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.
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.
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.
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.
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.
•
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83.
• S
subject to
minimise
B
(difference between B and S )
BT
S Y
(B is tree-structured)

84.
subject to
minimise
D,V
(difference between B and S )
(B is tree-structured)
BT
S Y

85.
V
B D

86.
B S
V

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.
•
•

89.
•
•
•
•
subject to
minimise
d
(B is tree-structured)

90.
•
•

91.
•
–
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•
•
•

92.
B

93.
∵
∵
∵

94.
F

95.
F

96.
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•
•
•

97.
?

98.
?

99.
?

100.
•
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–
•
•
•

101.
∈ ℝ 𝑝
N

102.
S

103.
B

104.
•
–
–
•

105.
•
•
•