Mathematical Puzzles and Not So Puzzling Mathematics

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

2. It all begins with a chessboard

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

4. 21 domino A Truncated Chessboard Truncated 88 chessboard Cover the truncated 88 chessboard with thirty-one 21 dominoes

5. 21 domino Proof of Impossibility Truncated 88 chessboard Impossible to cover the truncated 88 chessboard with thirty-one dominoes.

6. 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.

7. Impossible ! An Algebraic Proof y7 . . . . . . . . . . . . y2 y 1 (1+x+x2+. . . x7) (1+y+y2+. . . y7) – 1 - x7y7 = (1+x) xi y j + (1+y)  x i y j xi yj Let x = -1 y = -1 -2 = 0 1 x x2 . . . . . . . . . . . . . . . . . . . . . . . x7 (1+y) x i y j (1+x) xi y j

8. 1 2 3 4 5 6 ….. odd even odd even odd even….. Modulo-2 Arithmetic

9. Coloring the Vertices of a Graph vertex edge

10. 2 - Colorability A necessary and sufficient condition : No circuit of odd length vertex edge

11. 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.

12. 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.)

13. 4 - Colorability : Planar Graphs Kuratowski’s subgraphs All planar graphs are 4-colorable. How to characterize non-planar graphs ? Genus, Thickness, …

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

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

16. Principle of Mathematical Induction • To show that a statement p (n) is true • Basis : Show the statement is true for n = n0 • Induction step : Assuming the statement is true for • n = k , ( k  n0 ) , show the statement is true for n = k + 1

17. 2 n 2 n 2 n 2 n Proof by Mathematical Induction Any 2n2n defective chessboard can be covered with 1/3(2n2n -1) triominoes Basis : n = 1 Induction step : 2 n+1 2 n+1

18. …… The Wise Men and the Hats If there aren wise men wearing white hats, then at the nth hour all the n wise men will raise their hands. 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 thenth hour, no wise man raises his hand. At the n+1th hour, all n+1 wise men raise their hands.

19. Principle of Strong Mathematical Induction • To show that a statement p (n) is true • Basis : Show the statement is true for n = n0 • Induction step : Assuming the statement is true for • n = k , ( k  n0 ) , show the statement is true for n = k + 1 • n0 n  k,

20. Another Hat Problem No strategy In the worst case, all men were shot. Strategy 1 In the worst case, half of the men were shot. Design a strategy so that as few men will die as possible.

21. x n-1 x n-2 x n-3 ……… x1 x n-3 ……… x1 x n-2 x n-3 ……… x1 Another Hat Problem ……….. x n x n-1 x n-2 x n-3 ……………… x1 x n-1 x n-2

22. 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

23. Generalization : 7 people, Probability of winning = 7/8 More people ? Best possible ? Application of Algebraic Coding Theory Strategy 3 : Probability of winning = 3/4

24. 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.

25. 1 5 7 4 6 2 3 8 G G G G 9 11 12 10 10 9 Step 1 Balance Step 2 Balance Imbalance Step 3 Step 3

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

27. 1 5 7 4 6 2 3 8 1 5 3 6 2 4 3 4 Step 1 Imbalance Step 2 Imbalance Step 3

28. 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.

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

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

31. Derangements

32. Derangement of 10 Objects Number of derangements of n objects Probability

33. Permutation Positions Objects

34. Placement of Non-taking Rooks Positions Objects

35. Permutation with Forbidden Positions Positions Positions Objects Objects

36. Placement of Non-taking Rooks Positions Positions Objects Objects

37. Placement of Non-taking Rooks Positions ri = number of ways to place i non-taking rooks on chessboard C Objects Rook Polynomial : R (C) = r0 + r1 x + r2 x2 + … R (C) = 1 + 6x + 10x2 + 4x3

38. At Least One Way to Place Non-taking Rooks Positions Positions Objects Objects Theory of Matching !

39. 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.