0
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1
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just now, we have analysed

2
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corresponding to

3
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the regular tetrahedron

4
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how to colour the vertex

5
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we will be thinking

6
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just now, we have used ABCD this kind of

7
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symbols to represent its vertices

8
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but at the end, during the calculation

9
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are they still useful

10
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in Burnside's Lemma and

11
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Polya's Theorem

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what we are considering？

13
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actually, we are considering the permutation structure

14
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maybe, we do not even need

15
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to write how does each permutation look like

16
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we still take this kind of tetrahedron as an example

17
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we do have ABCD these 4 vertices

18
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but when we are calculating, we need to

19
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know for this kind of operations

20
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how many cycles are there

21
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we just need to write out the

22
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pattern of the cycles

23
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for example, when we are considering the vertex and the face centre

24
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performing positive/negative 120 degree rotation

25
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we will realize a vertex is fixed

26
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the other 3 vertices in one cycle

27
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therefore, we just need to write it as

28
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one 1 order cycle

29
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one 3 order cycle

30
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That is enough

31
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Besides the pattern of the permutations

32
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there is still a question like this

33
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how many this kind of permutations are there?

34
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we just need to find how many angles are there

35
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how many vertices

36
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there are total of 4 different vertices

37
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there are 2 different angles

38
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therefore, there are total of 8 permutation

39
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which are one 1 order cycle with

40
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one 3 order cycle

41
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besides, we look at middle points of two oppositing edges

42
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while performing exchange, it is nothing but just

43
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the exchange between 2 vertices

44
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there are two 2-order cycle

45
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how many corresponding edge pairs are there

46
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in total, there are 3

47
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next, fixed permutation is very simple

48
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total of 4 vertices

49
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each vertex is 1 order cycle

50
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therefore, there are four 1 order cycle

51
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at this time, we just need to simply

52
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list all the permutation pattern

53
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do not need to understand the actual details

54
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we just directly adopt the corresponding

55
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Polya's Theorem

56
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write in the corresponding several order cycle

57
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corresponding several colour power series into it

58
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it managed to get the answer directly .

59
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next, let us look at another question

60
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for 3 different types of beads

61
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for example,

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I have 3 kinds of beads with different colours

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I want to string a 4 beads necklace

64
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how many types of solutions are there

65
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It looks very similar with

66
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the circular permutation

67
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but, there are differences

68
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which is we have only

69
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3 different colours of beads

70
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but, if we need to make a 4 beads necklace

71
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1 colour must be repeated

72
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therefore, this a repeatable circle permutation

73
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at the same time, this arrangement can be rotated

74
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it is actually corresponding to

75
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perform rotation within a space

76
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therefore, we may need to consider

77
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its permutation group

78
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it should have a problem

79
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if you asked do I need to consider flipping?

80
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sometimes, it is a very realistic problem

81
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what does it mean by considering flipping

82
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if after flipping, it could be the

83
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same as the previous one in some cases

84
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then, we may need to consider this movement

85
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if after it was turned over and it would be changed for all solutions

86
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then, we do not need to consider this flipping

87
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therefore, everyone should think clearly

88
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in reality, do we actually need to consider about it

89
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it contains a practical significance

90
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besides, with or without flipping

91
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does it hold the same answer ?

92
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we can solve it separately

93
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for example, if we consider after the flipping

94
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the bead are split into half

95
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rounded beads

96
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after it was turned over, it then became flat

97
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it simply means

98
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that its looking was changed

99
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therefore, after flipping, it is treated as different

100
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we consider only rotation without flipping

101
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for this kind of 4 points on a flat surface

102
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how many different rotations are there

103
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means how many different rotation angles around the centre point

104
00:03:55,580 --> 00:03:58,610
for example,  positive/negative 90 degree

105
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each vertex may

106
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become the next vertex

107
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one 4 order cycle,

108
00:04:03,870 --> 00:04:05,200
how many angles are there

109
00:04:05,200 --> 00:04:07,070
there are total of 2 different angles

110
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positive/negative 90 degree

111
00:04:08,750 --> 00:04:09,970
other than those

112
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there is still

113
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another 180 degree rotation

114
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performing two-by-two exchange

115
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therefore, there are two 2 order cycle

116
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besides, there is still one fixed permutation

117
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directly apply Polya's Theorem

118
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corresponding to 3 colours colouring,

119
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therefore it is

120
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3 to the power of 1, added up with 3 to the power of 1

121
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because, there are 2 types of different

122
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positive and negative 90 degree

123
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then, added up with 3 square

124
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then, added up with 3 to the power of 4

125
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then, divided by all the number of permutation - 4

126
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the answer is exactly 24

127
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if we considered

128
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this bead is round shaped

129
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after we turned it over

130
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it is still a necklace

131
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then, we need to consider flipping as well

132
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at this time if we are considering flipping

133
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similarly, we are going to write

134
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rotation of positive and negative 90 degree around the centre axis

135
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rotation of 180 degree around the centre axis

136
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besides, we need to find its symmetry axis to perform the flipping

137
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at this time, we will realize

138
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there are 2 types of symmetry axes

139
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one type of the symmetry axes is over their

140
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corresponding centre of the necklace

141
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which is when the symmetry axis does not

142
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go through the bead

143
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we will realize actually it is

144
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performing two-by-two flipping exchange

145
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therefore, two 2 order cycles

146
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how many this kind of symmetry axes?

147
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it is nothing more that this angle and that angle

148
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there are total of 2

149
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we will also realize

150
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if we split the bead over its centre point

151
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it can still produce another symmetry axis

152
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whereas for this symmetry axis, its

153
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permutation is different

154
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because the bead over the axis is fixed

155
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perform exchange between 2 beads

156
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therefore, its permutation can be written as

157
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2 1 order cycles

158
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then, one 2 order cycle

159
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how many different axes are there

160
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similarly, there are 2 different choices

161
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then, there is still fixed one

162
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at this point, we can look at

163
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actually, in total it contains

164
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how many different permutations

165
00:06:03,710 --> 00:06:06,900
in total, there are 8 types of different permutations

166
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whereas, corresponding to

167
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applying 3 colours colouring into Polya's Theorem

168
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the answer is 21

169
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it can be seen that with

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without considering flipping

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the answers are different