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Mathematical Puzzles and Not So Puzzling Mathematics C. L. Liu National Tsing Hua University

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It all begins with a chessboard

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

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

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

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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.

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An Algebraic Proof 1 x x 2....................... x 7 y7............y2y1y7............y2y1 (1+x) x i y j (1+y) x i y j (1+x+x 2 +... x 7 ) (1+y+y 2 +... y 7 ) – 1 - x 7 y 7 = (1+x) x i y j + (1+y) x i y j x i y j Impossible ! Let x = -1 y = -1 -2 = 0

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Modulo-2 Arithmetic 1 2 3 4 5 6 ….. odd even odd even odd even….. oddeven oddevenodd evenoddeven 01 001 110

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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

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2 - Colorability Necessity : If there is a circuit of odd length, Sufficiency : If there is no circuit of odd length, use the “contagious” coloring algorithm.

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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 8 8 defective chessboard can be covered with twenty-one triominoes

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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

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Principle of Mathematical Induction To show that a statement p (n) is true 1.Basis : Show the statement is true for n = n 0 2.Induction step : Assuming the statement is true for n = k, ( k n 0 ), show the statement is true for n = k + 1

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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

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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. ……

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Principle of Strong Mathematical Induction To show that a statement p (n) is true 1.Basis : Show the statement is true for n = n 0 2.Induction step : Assuming the statement is true for n = k, ( k n 0 ), show the statement is true for n = k + 1 n 0 n k,

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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 ……………… 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

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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 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

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Strategy 3 : observecall 00 01 10 11 1PP01PP0 patterncall 000 001 010 011 100 101 110 111 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

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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.

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1 2 3 4 5 6 7 8 G 9 10 G G 11 12 G 10 9 Step 1 Step 3 Step 2 Balance Step 3 BalanceImbalance

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7 G 1 2 3 4 5 6 7 8 1 3 4 5 2 6 Step 1 Step 2 Imbalance Step 3 Balance 2 1 Step 3 Imbalance

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1 2 3 4 5 6 7 8 1 3 4 5 2 6 Step 1 Step 2 Imbalance 4 3 Step 3 Imbalance

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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 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.

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

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Derangements ABC abc acb bac bca cab cba

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Derangement of 10 Objects Number of derangements of n objects Probability

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Permutation 1234 a b c d Positions Objects

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

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

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

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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

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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 !

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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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