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

It all begins with a chessboard

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

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

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

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.

An Algebraic Proof 1 x x x 7 y y2y1y y2y1 (1+x) x i y j (1+y) x i y j (1+x+x x 7 ) (1+y+y 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 = = 0

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

Coloring the Vertices of a Graph vertex edge

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

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.

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.)

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

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

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

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

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

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. ……

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,

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.

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

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

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

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.

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

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

Step 1 Step 2 Imbalance 4 3 Step 3 Imbalance

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.

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

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

Derangements ABC abc acb bac bca cab cba

Derangement of 10 Objects Number of derangements of n objects Probability

Permutation 1234 a b c d Positions Objects

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

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

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

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

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 !

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.