chessboards, hats, and chinese poetry : some rigorous and not-so-rigorous mathematical results c. l....
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Chessboards, Hats, and Chinese Poetry :Chessboards, Hats, and Chinese Poetry :
Some Rigorous and Not-So-RigorousSome Rigorous and Not-So-Rigorous
Mathematical ResultsMathematical Results
C. L. LiuC. L. Liu
詩裡有數
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List of books published
Introduction to Combinatorial Mathematics, McGraw-Hill Book Company, 1968, (Japanese translation, 1972, Chinese translation, 1982).
Topics in Combinatorial Mathematics, Mathematical Association of America, 1972.
Linear Systems Analysis, with J. W. S. Liu, McGraw-Hill Book Company, 1975.
Elements of Discrete Mathematics, McGraw-Hill Book Company, 1977, (Japanese translation, 1979, Chinese translation, 1981).
Pascal, with G. G. Belford, McGraw-Hill Book Company, 1984.
Elements of Discrete Mathematics, second edition, McGraw-Hill Book Company, 1985. (Chinese translation, 1993, Japanese translation, 1995, Indonesian translation, 1995, Greek, Spanish translation).
Solution of Design Automation Problems by the Method of Simulated Annealing, with D. F. Wong, and H. W. Leong, Kluwer Academic Publishers, 1988.
Fault Covering Problems in Reconfigurable VLSI Systems, with R. Libeskind-Hadas, N. Hasan, J. Cong, and P. McKinley, Kluwer Academic Publishers, 1992.
愛上層樓,天下遠見出版公司、國立清華大學出版社出版, 2002 年。
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Hats
ChinesePoetry
0cbxax 2Mathematics
Chessboards
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It all begins with a chessboard
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Covering a Chessboard
88 chessboard
21 domino
Cover the 88 chessboard with thirty-two 21 dominoes
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A Truncated Chessboard
21 domino
Cover the truncated 88 chessboard with thirty-one 21 dominoes
Truncated 88 chessboard
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Proof of Impossibility
21 domino
Truncated 88 chessboard
Impossible to cover the truncated 88 chessboard with thirty-one dominoes.
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Proof of Impossibility
Impossible to cover the truncated 88 chessboard with thirty-one dominoes. There are thirty-two white squares and thirty black squares. A 2 1 domino always covers a white and a black square.
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Proof of Impossibility
Impossible to cover the truncated 88 chessboard with thirty-one dominoes. There are thirty-two white squares and thirty black squares. A 2 1 domino always covers a white and a black square.
01 0
0 0 0 011 1 1
1 1
11 1 10 00
00 011
0
0 1110 0
0
0
0000
00 00
0
0 0
0
0
00
11
1
1
1
1
1
1
1
1
11
1 1
1 11
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Modulo-2 Arithmetic
1 2 3 4 5 6 …..
odd even odd even odd even…..
odd even
odd even odd
even odd even
0 1
0 0 1
1 1 0
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Coloring the Vertices of a Graph
vertex
edge
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2 - Colorability
A necessary and sufficient condition : No circuit of odd length
vertex
edge
vertex
edge
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3 - Colorability
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3 - Colorability
The problem of determining whether a graph is 3-colorable
is NP-complete. ( At the present time, there is no known
efficient algorithm for determining whether a graph is
3-colorable.)
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4 - Colorability : Planar Graphs
All planar graphs are 4-colorable.
How to characterize non-planar graphs ? Genus, Thickness, …
Kuratowski’s subgraphs
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A Defective Chessboard
Triomino
Any 88 defective chessboard can be covered with twenty-one triominoes
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Defective Chessboards
Any 2n2n defective chessboard can be covered with 1/3(2n2n -1) triominoes
Any 88 defective chessboard can be covered with twenty-one triominoes
Prove by mathematical induction
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The first domino falls.
If a domino falls, so will the next domino.
All dominoes will fall !
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Proof by Mathematical Induction
Basis : n = 1
Induction step :2 n+1
2 n+1
2 n 2 n
2 n
2 n
Any 2n2n defective chessboard can be covered with 1/3(2n2n -1) triominoes
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If there are n wise men wearing white hats, then at the nth hour allthe n wise men will raise their hands.
The Wise Men and the Hats
Basis : n =1 At the 1st hour. The only wise man wearing a white hat will raise his hand.
Induction step : Suppose there are n+1 wise men wearing white hats.
At the nth hour, no wise man raises his hand.
At the n+1th hour, all n+1 wise men raise their hands.
……
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Another Hat Problem
Design a strategy so that as few men will die as possible.
No strategy In the worst case, all men were shot.
Strategy 1 In the worst case, half of the men were shot.
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Another Hat Problem
x n x n-1 x n-2 x n-3 ……………… x1
………..
x n-1 x n-2 x n-3 ……… x1
x n-2 x n-3 ……… x1
x n-1 x n-3 ……… x1
x n-2
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Yet, Another Hat Problem
A person may say, 0, 1, or P(Pass)Winning : No body is wrong, at least one person is rightLosing : One or more is wrong
Strategy 1 : Everybody guesses Probability of winning = 1/8
Strategy 2 : First and second person always says P. Third person guesses Probability of winning = 1/2
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Strategy 3 :
observe call
00
01
10
11
1
P
P
0
pattern call
000001010011100101110111
111PP1P1P0PP1PPP0PPP0000
Probability of winning = 3/4
More people ?
Best possible ?
Generalization : 7 people, Probability of winning = 7/8
Application of Algebraic Coding Theory
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A Coin Weighing Problem
Twelve coins, possibly one of them is defective ( too heavyor too light ). Use a balance three times to pick out thedefective coin.
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1 2 3 4 5 6 7 8
G 9 10GG 11
12G 109
Step 1
Step 3
Step 2
Balance
Step 3Balance Imbalance
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7G
1 2 3 4 5 6 7 8
1 3 452 6
Step 1
Step 2
Imbalance
Step 3Balance
21
Step 3Imbalance
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1 2 3 4 5 6 7 8
1 3 452 6
Step 1
Step 2
Imbalance
43
Step 3Imbalance
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Another Coin Weighing Problem
Application of Algebraic Coding Theory
• Adaptive Algorithms• Non-adaptive Algorithms
Thirteen coins, possibly one of them is defective ( too heavyor too light ). Use a balance three times to pick out thedefective coin. However, an additional good coin is availablefor use as reference.
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Yet, Another Hat Problem
Hats are returned to 10 people at random, what is the probability that no one gets his own hat back ?
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Apples and Oranges
ApplesApples OrangesOrangesOrangesOranges
ApplesApples
Take out one fruit from one box to determine the contentsof all three boxes.
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Derangements
AA BB CC
a b c
a c b
b a c
b c a
c a b
c b a
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Derangement of 10 Objects
Number of derangements of n objects
]!
1)1(....
!3
1
!2
1
!1
11[!
nnd n
n
]!10
1)1(....
!3
1
!2
1
!1
11[!10 10
10 d
Probability !10
1)1(....
!3
1
!2
1
!1
11
!101010 d
36788.01 e
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Permutation
1 2 3 4
a
b
c
d
Positions
Objects
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Placement of Non-taking Rooks
1 2 3 4
a
b
c
d
Positions
Objects
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Permutation with Forbidden Positions
1 2 3 4
a
b
c
d
Positions
Objects1 2 3 4
a
b
c
d
Positions
Objects
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Placement of Non-taking Rooks
1 2 3 4
a
b
c
d
Positions
Objects1 2 3 4
a
b
c
d
Positions
Objects
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Placement of Non-taking Rooks
1 2 3 4
a
b
c
d
Positions
Objects
Rook Polynomial :
R (C) = r0 + r1 x + r2 x2 + …
ri = number of ways to place i non-taking rooks on chessboard C
R (C) = 1 + 6x + 10x2 + 4x3
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At Least One Way to Place Non-taking Rooks
1 2 3 4
a
b
c
d
Positions
Objects1 2 3 4
a
b
c
d
Positions
Objects
Theory of Matching !
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怎一個愁字了得
愁
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怎一個愁字了得
尋尋覓覓 冷冷清清 淒淒慘慘戚戚 乍暖還寒時候 最難將息 三杯兩盞淡酒 怎敵他晚來風急 雁過也 正傷心 卻是舊時相識 滿地黃花堆積 憔悴損 如今有誰堪摘 守著窗兒 獨自怎生得黑 梧桐更兼細雨 到黃昏點點滴滴這次第 怎一個愁字了得
李清照 <聲聲慢>
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只恐雙溪舴艋舟,載不動許多愁。
李清照 <武陵春>
一斛珠連萬斛愁,關山漂泊腰支細。
吳梅林 <圓圓曲>
舟載
斛量
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新妝宜面下朱樓,深鎖春光一院愁。劉錫禹 <春詞>
白髮三千丈、離愁似個長。 李白 <秋浦歌>
問君能有幾多愁,恰似一江春水向東流。李後主 <虞美人>
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一懷愁緒,幾年離索。陸游 <釵頭鳳>
凝眸處,從今又添,一段新愁。 李清照 <鳳凰台上憶吹簫>
一點芭蕉一點愁。徐再思 <雙調水仙子夜雨>
寂寞深閨,柔腸一寸愁千縷。李清照 <點絳唇>
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與爾同消萬古愁。
李白 <將進酒>
一種相思,兩處閒愁。
李清照 <一剪梅>
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怎一個愁字了得尋尋覓覓 冷冷清清 淒淒慘慘戚戚 乍暖還寒時候 最難將息 三杯兩盞淡酒 怎敵他晚來風急 雁過也 正傷心 卻是舊時相識 滿地黃花堆積 憔悴損 如今有誰堪摘 守著窗兒 獨自怎生得黑 梧桐更兼細雨 到黃昏點點滴滴這次第 怎一個愁字了得
2 > 1
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Conclusion
Mathematics is about finding connections, betweenspecific problems and more general results, and between one concept and another seemingly unrelatedconcept that really is related.
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有恆是信真君子無欲為剛大丈夫