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Chessboards, Hats, and Chinese Poetry : Some Rigorous and Not-So-Rigorous Mathematical Results C. L. Liu 詩裡有數.

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Presentation on theme: "Chessboards, Hats, and Chinese Poetry : Some Rigorous and Not-So-Rigorous Mathematical Results C. L. Liu 詩裡有數."— Presentation transcript:

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2 Chessboards, Hats, and Chinese Poetry : Some Rigorous and Not-So-Rigorous Mathematical Results C. L. Liu 詩裡有數

3 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, Linear Systems Analysis, with J. W. S. Liu, McGraw-Hill Book Company, Elements of Discrete Mathematics, McGraw-Hill Book Company, 1977, (Japanese translation, 1979, Chinese translation, 1981). Pascal, with G. G. Belford, McGraw-Hill Book Company, Elements of Discrete Mathematics, second edition, McGraw-Hill Book Company, (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, Fault Covering Problems in Reconfigurable VLSI Systems, with R. Libeskind-Hadas, N. Hasan, J. Cong, and P. McKinley, Kluwer Academic Publishers, 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。

4 Hats Chinese Poetry Mathematics Chessboards

5 It all begins with a chessboard

6 Covering a Chessboard 8  8 chessboard 2  1 domino Cover the 8  8 chessboard with thirty-two 2  1 dominoes

7 A Truncated Chessboard 2  1 domino Cover the truncated 8  8 chessboard with thirty-one 2  1 dominoes Truncated 8  8 chessboard

8 Proof of Impossibility 2  1 domino Truncated 8  8 chessboard Impossible to cover the truncated 8  8 chessboard with thirty-one dominoes.

9 Proof of Impossibility Impossible to cover the truncated 8  8 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.

10 Proof of Impossibility Impossible to cover the truncated 8  8 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

11 Modulo-2 Arithmetic ….. odd even odd even odd even….. oddeven oddevenodd evenoddeven

12 Coloring the Vertices of a Graph vertex edge

13 2 - Colorability A necessary and sufficient condition : No circuit of odd length vertex edge vertex edge

14 3 - Colorability

15 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.)

16 4 - Colorability : Planar Graphs All planar graphs are 4-colorable. How to characterize non-planar graphs ? Genus, Thickness, … Kuratowski’s subgraphs

17 A Defective Chessboard Triomino Any 8  8 defective chessboard can be covered with twenty-one triominoes

18 Defective Chessboards Any 2 n  2 n defective chessboard can be covered with 1/3(2 n  2 n -1) triominoes Any 8  8 defective chessboard can be covered with twenty-one triominoes Prove by mathematical induction

19 The first domino falls. If a domino falls, so will the next domino. All dominoes will fall !

20 Proof by Mathematical Induction Basis : n = 1 Induction step : 2 n+1 2 n Any 2 n  2 n defective chessboard can be covered with 1/3(2 n  2 n -1) triominoes

21 If there are n wise men wearing white hats, then at the n th hour all the n wise men will raise their hands. The Wise Men and the Hats Basis : n =1 At the 1 st 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 n th hour, no wise man raises his hand. At the n+1 th hour, all n+1 wise men raise their hands. ……

22 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.

23 Another Hat Problem x n x n-1 x n-2 x n-3 ……………… x 1 ……….. x n-1 x n-2 x n-3 ……… x 1 x n-2 x n-3 ……… x 1 x n-1 x n-3 ……… x 1 x n-2

24 Yet, Another Hat Problem A person may say, 0, 1, or P(Pass) Winning : No body is wrong, at least one person is right Losing : 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

25 Strategy 3 : observecall PP01PP0 patterncall PP1 P1P 0PP 1PP P0P PP0 000 Probability of winning = 3/4 More people ? Best possible ? Generalization : 7 people, Probability of winning = 7/8 Application of Algebraic Coding Theory

26 A Coin Weighing Problem Twelve coins, possibly one of them is defective ( too heavy or too light ). Use a balance three times to pick out the defective coin.

27 G 9 10 G G G 10 9 Step 1 Step 3 Step 2 Balance Step 3 BalanceImbalance

28 7 G Step 1 Step 2 Imbalance Step 3 Balance 2 1 Step 3 Imbalance

29 Step 1 Step 2 Imbalance 4 3 Step 3 Imbalance

30 Another Coin Weighing Problem Application of Algebraic Coding Theory Adaptive Algorithms Non-adaptive Algorithms Thirteen coins, possibly one of them is defective ( too heavy or too light ). Use a balance three times to pick out the defective coin. However, an additional good coin is available for use as reference.

31 Yet, Another Hat Problem Hats are returned to 10 people at random, what is the probability that no one gets his own hat back ?

32 Apples and Oranges ApplesOranges Oranges Apples Take out one fruit from one box to determine the contents of all three boxes.

33 Derangements ABC abc acb bac bca cab cba

34 Derangement of 10 Objects Number of derangements of n objects Probability

35 Permutation 1234 a b c d Positions Objects

36 Placement of Non-taking Rooks 1234 a b c d Positions Objects

37 Permutation with Forbidden Positions 1234 a b c d Positions Objects 1234 a b c d Positions Objects

38 Placement of Non-taking Rooks 1234 a b c d Positions Objects 1234 a b c d Positions Objects

39 Placement of Non-taking Rooks 1234 a b c d Positions Objects Rook Polynomial : R (C) = r 0 + r 1 x + r 2 x 2 + … r i = number of ways to place i non-taking rooks on chessboard C R (C) = 1 + 6x + 10x 2 + 4x 3

40 At Least One Way to Place Non-taking Rooks 1234 a b c d Positions Objects 1234 a b c d Positions Objects Theory of Matching !

41 怎一個愁字了得 愁

42 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>

43 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

44 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>

45 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>

46 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>

47 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

48 Conclusion Mathematics is about finding connections, between specific problems and more general results, and between one concept and another seemingly unrelated concept that really is related.

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50 有恆是信真君子 無欲為剛大丈夫


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