mathematical puzzles and not so puzzling mathematics

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Mathematical Puzzles and Not So Puzzling Mathematics. C. L. Liu National Tsing Hua University. It all begins with a chessboard. Covering a Chessboard. 21 domino. 8 8 chessboard. Cover the 8 8 chessboard with thirty-two 21 dominoes. 21 domino. A Truncated Chessboard. - PowerPoint PPT Presentation

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Mathematical Puzzles Mathematical Puzzles and and

Not So Puzzling MathematicsNot So Puzzling Mathematics

C. L. LiuC. L. LiuNational Tsing Hua UniversityNational Tsing Hua University

It all begins with a chessboard

Covering a Chessboard

88 chessboard

21 domino

Cover the 88 chessboard with thirty-two 21 dominoes

A Truncated Chessboard

21 domino

Cover the truncated 88 chessboard with thirty-one 21 dominoes

Truncated 88 chessboard

Proof of Impossibility

21 domino

Truncated 88 chessboard

Impossible to cover the truncated 88 chessboard with thirty-one dominoes.

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.

An Algebraic Proof

1 x x2 . . . . . . . . . . . . . . . . . . . . . . . x7

y7

............y2

y

1

(1+x) xi y j (1+y) x i y j

(1+x+x2+. . . x7) (1+y+y2+. . . y7) – 1 - x7y7

= (1+x) xi y j + (1+y) x i y j xi yj

Impossible !Let x = -1 y = -1 -2 = 0

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

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 knownefficient 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 88 defective chessboard can be covered with twenty-one triominoes

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

Principle of Mathematical Induction

To show that a statement p (n) is true

1. Basis : Show the statement is true for n = n0

2. Induction step : Assuming the statement is true for

n = k , ( k n0 ) , show the statement is true for n = k + 1

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

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.

……

Principle of Strong Mathematical Induction

To show that a statement p (n) is true1. Basis : Show the statement is true for n = n0

2. Induction step : Assuming the statement is true for n = k , ( k n0 ) , show the statement is true for n = k + 1 n0 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 ……………… 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

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

Strategy 3 : observe call

00011011

1PP0

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

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.

1 2 3 4 5 6 7 8

G 9 10GG 11

12G 109

Step 1

Step 3

Step 2

Balance

Step 3Balance Imbalance

7G

1 2 3 4 5 6 7 8

1 3 452 6

Step 1

Step 2

Imbalance

Step 3Balance

21

Step 3Imbalance

1 2 3 4 5 6 7 8

1 3 452 6

Step 1

Step 2

Imbalance

43

Step 3Imbalance

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.

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

ApplesApples OrangesOrangesOrangesOrangesApplesApples

Take out one fruit from one box to determine the contentsof all three boxes.

Derangements

AA BB CCa b c

a c b

b a c

b c a

c a b

c b a

Derangement of 10 Objects

Number of derangements of n objects

]!

1)1(....!3

1!2

1!1

11[!n

nd nn

]!10

1)1(....!3

1!2

1!1

11[!10 1010 d

Probability !101)1(....

!31

!21

!111

!101010 d

36788.01 e

Permutation

1 2 3 4

a

b

c

d

Positions

Objects

Placement of Non-taking Rooks

1 2 3 4

a

b

c

d

Positions

Objects

Permutation with Forbidden Positions

1 2 3 4

a

b

c

d

Positions

Objects1 2 3 4

a

b

c

d

Positions

Objects

Placement of Non-taking Rooks

1 2 3 4

a

b

c

d

Positions

Objects1 2 3 4

a

b

c

d

Positions

Objects

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

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 !

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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