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Mathematical Puzzles Mathematical Puzzles
and and
Not So Puzzling MathematicsNot So Puzzling Mathematics
C. L. LiuC. L. Liu
National 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 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 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
00
01
10
11
1
P
P
0
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 OrangesOrangesOrangesOranges
ApplesApples
Take out one fruit from one box to determine the contentsof all three boxes.
Derangements
AA BB CC
a 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[!
nnd n
n
]!10
1)1(....
!3
1
!2
1
!1
11[!10 10
10 d
Probability !10
1)1(....
!3
1
!2
1
!1
11
!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.