from zombies to fibonaccis: an introduction to the theory ......p 2 tic-tac-toe rectangle game...
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√2 Tic-Tac-Toe Rectangle Game Zombies I Love Rectangles Triangle Game Zeckendorf Games Games More Irrational
From Zombies to Fibonaccis: AnIntroduction to the Theory of Games
Steven J. Miller, Williams Collegehttp:
//www.williams.edu/Mathematics/sjmiller/public_html
New Jersey Math Camp: Summer 2018
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√2
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√2 Is Irrational
Standard Proof: Assume√
2 = a/b.
WLOG, assume b is the smallest denominator among allfractions that equal
√2.
2b2 = a2 thus a = 2m is even.
Then 2b2 = 4m2 so b2 = 2m2 so b = 2n is even.
Thus√
2 = a/b = 2m/2n = m/n, contradicts minimality of n.
(Could also do by contradiction from a,b relatively prime.)
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Tennenbaum’s Proof
Assume√
2 = a/b with b minimal.
b
a-b
ba-b
2b-a
Figure: 2b2 = a2 so (2b − a)2 = 2(a− b)2 and√
2 = 2b−aa−b .
As 0 < a− b < b (if not, a− b ≥ b so a ≥ 2b and√2 = a/b ≥ 2), contradicts minimality of b.
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Challenge
WHAT OTHER NUMBERS HAVE GEOMETRICIRRATIONALITY PROOFS?
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Tic-Tac-Toe
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Tic-Tac-Toe
Figure: How many opening moves? How many first two moves?
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Tic-Tac-Toe: First Move
Figure: Analyzing Opening Moves: Corners all equivalent.
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Tic-Tac-Toe: First Move
Figure: Analyzing Opening Moves: Middles all equivalent.
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Tic-Tac-Toe: First Move
Figure: Analyzing Opening Moves: Only one center: 3 classes ofmoves.
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Tic-Tac-Toe: Second Move
Figure: Analyzing Second Player Response.
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Tic-Tac-Toe: Second Move
Figure: Analyzing Second Player Response.
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Tic-Tac-Toe: Second Move
Figure: Analyzing Second Player Response.
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Tic-Tac-Toe: Second Move
Figure: Analyzing Second Player Response.
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Tic-Tac-Toe: Second Move
Figure: Analyzing Second Player Response.
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Tic-Tac-Toe: Second Move
Figure: Analyzing Second Player Response: Thus there are 12possible pairs of first two moves.
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Tic-Tac-Toe: Questions
Does either player have a winning strategy?
What happens if we play randomly? Chance of Player 1winning?
How can we make it interesting?
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Tic-Tac-Toe: Interesting Variants
Gobble Tic-Tac-Toe
Larger board (and handicaps)
Tic-Tac-Toe in Tic-Tac-Toe
Bidding Tic-Tac-Toe
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Tic-Tac-Toe in Tic-Tac-Toe
Figure: Rule: next move from position of previous.
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Tic-Tac-Toe in Tic-Tac-Toe
Figure: Rule: next move from position of previous.
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Tic-Tac-Toe in Tic-Tac-Toe
Figure: Rule: next move from position of previous.
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Tic-Tac-Toe in Tic-Tac-Toe
Figure: Rule: next move from position of previous.
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Tic-Tac-Toe in Tic-Tac-Toe
Figure: Rule: next move from position of previous.
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Tic-Tac-Toe in Tic-Tac-Toe
Figure: Rule: next move from position of previous.
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Tic-Tac-Toe in Tic-Tac-Toe
Figure: Rule: next move from position of previous.
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Tic-Tac-Toe in Tic-Tac-Toe
Figure: Rule: next move from position of previous.
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Tic-Tac-Toe in Tic-Tac-Toe
Figure: Rule: next move from position of previous.
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Bidding Tic-Tac-Toe
Figure: Rules: Start with $1000 and $1000, whomever bids moregets move and gives money to other. Opening bid?
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Bidding Tic-Tac-Toe
Figure: Rules: Start with $1000 and $1000, whomever bids moregets move and gives money to other. Opening bid?
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Bidding Tic-Tac-Toe
Figure: Rules: Start with $1000 and $1000, whomever bids moregets move and gives money to other. Opening bid?
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Bidding Tic-Tac-Toe
Figure: Rules: Start with $1000 and $1000, whomever bids moregets move and gives money to other. Opening bid?
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Bidding Tic-Tac-Toe
Figure: Rules: Start with $1000 and $1000, whomever bids moregets move and gives money to other. Opening bid?
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Bidding Tic-Tac-Toe
Figure: Rules: Start with $1000 and $1000, whomever bids moregets move and gives money to other. Opening bid?
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Bidding Tic-Tac-Toe: Overbidding for First Move
Question: If each have $1000, how much is too much to bid forthe first move?
Easy answer:
$1000.
Can we do better? Assume if tie Player 1 gets the move.
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Bidding Tic-Tac-Toe: Overbidding for First Move
Question: If each have $1000, how much is too much to bid forthe first move?
Easy answer: $1000.
Can we do better? Assume if tie Player 1 gets the move.
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Bidding Tic-Tac-Toe: Overbidding for First Move
Question: If each have $1000, how much is too much to bid forthe first move?
Easy answer: $1000.
Can we do better? Assume if tie Player 1 gets the move.
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Bidding Tic-Tac-Toe: Overbidding for First Move (cont)
Player 2 bids x and wins:(1000,1000) −→ (1000 + x ,1000− x).
Player 1 bids 1000− x and just wins:(1000 + x ,1000− x) −→ (2x ,2000− 2x).
Player 1 bids 2000− 2x and just wins:(2x ,2000− 2x) −→ (4x − 2000,4000− 4x).
Player 1 bids 4000− 4x and just wins:(4x − 2000,4000− 4x) −→ (8x − 6000,8000− 8x).
Need 4x − 2000 ≥ 4000− 4x or 8x ≥ 6000 or x ≥ 750.
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Bidding Tic-Tac-Toe: Overbidding for First Move (cont)
Player 2 bids x and wins:(1000,1000) −→ (1000 + x ,1000− x).
Player 1 bids 1000− x and just wins:(1000 + x ,1000− x) −→ (2x ,2000− 2x).
Player 1 bids 2000− 2x and just wins:(2x ,2000− 2x) −→ (4x − 2000,4000− 4x).
Player 1 bids 4000− 4x and just wins:(4x − 2000,4000− 4x) −→ (8x − 6000,8000− 8x).
Need 4x − 2000 ≥ 4000− 4x or 8x ≥ 6000 or x ≥ 750.
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Bidding Tic-Tac-Toe: Overbidding for First Move (cont)
Player 2 bids x and wins:(1000,1000) −→ (1000 + x ,1000− x).
Player 1 bids 1000− x and just wins:(1000 + x ,1000− x) −→ (2x ,2000− 2x).
Player 1 bids 2000− 2x and just wins:(2x ,2000− 2x) −→ (4x − 2000,4000− 4x).
Player 1 bids 4000− 4x and just wins:(4x − 2000,4000− 4x) −→ (8x − 6000,8000− 8x).
Need 4x − 2000 ≥ 4000− 4x or 8x ≥ 6000 or x ≥ 750.
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Bidding Tic-Tac-Toe: Overbidding for First Move (cont)
Player 2 bids x and wins:(1000,1000) −→ (1000 + x ,1000− x).
Player 1 bids 1000− x and just wins:(1000 + x ,1000− x) −→ (2x ,2000− 2x).
Player 1 bids 2000− 2x and just wins:(2x ,2000− 2x) −→ (4x − 2000,4000− 4x).
Player 1 bids 4000− 4x and just wins:(4x − 2000,4000− 4x) −→ (8x − 6000,8000− 8x).
Need 4x − 2000 ≥ 4000− 4x or 8x ≥ 6000 or x ≥ 750.
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Bidding Tic-Tac-Toe: Overbidding for First Move (cont)
Player 2 bids x and wins:(1000,1000) −→ (1000 + x ,1000− x).
Player 1 bids 1000− x and just wins:(1000 + x ,1000− x) −→ (2x ,2000− 2x).
Player 1 bids 2000− 2x and just wins:(2x ,2000− 2x) −→ (4x − 2000,4000− 4x).
Player 1 bids 4000− 4x and just wins:(4x − 2000,4000− 4x) −→ (8x − 6000,8000− 8x).
Need 4x − 2000 ≥ 4000− 4x or 8x ≥ 6000 or x ≥ 750.
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Bidding Tic-Tac-Toe: Overbidding for First Move (cont)
Player 2 bids 750 and wins: (1000,1000) −→ (1750,250).
Player 1 bids 250 and just wins: (1750,250) −→ (1500,500).
Player 1 bids 500 and just wins: (1500,500) −→ (1000,1000).
Player 1 bids 1000 and just wins: (1000,1000) −→ (0,2000).
Note: If Player 2 spends $750 to win the first two moves thenPlayer 1 can win!
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Bidding Tic-Tac-Toe: Overbidding for First Move (cont)
Player 2 bids 750 and wins: (1000,1000) −→ (1750,250).
Player 1 bids 250 and just wins: (1750,250) −→ (1500,500).
Player 1 bids 500 and just wins: (1500,500) −→ (1000,1000).
Player 1 bids 1000 and just wins: (1000,1000) −→ (0,2000).
Note: If Player 2 spends $750 to win the first two moves thenPlayer 1 can win!
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Rectangle Game
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Rectangle Game
Figure: Winning strategy? Function of board dimension?
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Rectangle Game
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Rectangle Game
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Rectangle Game
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Rectangle Game
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Rectangle Game
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Rectangle Game
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Rectangle Game
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Rectangle Game
Gather data! Try various sized boards, strategies.
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Rectangle Game: Data
Figure: Do you see a pattern?
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Mono-variant
A mono-variant is a quantity that moves on only one direction(either non-decreasing or non-increasing).
Idea: Associate a mono-variant to the rectangle game....
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Rectangle Game: Solution
Every time move, increase number of pieces by 1!
Figure: Move: 0; Pieces: 1.
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Rectangle Game: Solution
Every time move, increase number of pieces by 1!
Figure: Move: 1; Pieces: 2.
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Rectangle Game: Solution
Every time move, increase number of pieces by 1!
Figure: Move: 2; Pieces: 3.
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Rectangle Game: Solution
Every time move, increase number of pieces by 1!
Figure: Move: 3; Pieces: 4.59
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Rectangle Game: Solution
Every time move, increase number of pieces by 1!
Figure: Move: 4; Pieces: 5.
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Rectangle Game: Solution
Figure: Move: 5; Pieces: 5. Player 1 Wins.
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Rectangle Game: Solution (Continued)
Mono-variant is the number of pieces.
If board is m × n, game ends with mn pieces.
Thus takes mn − 1 moves.
If mn even then Player 1 wins else Player 2 wins.
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Zombies
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General Advice: What are your tools and how can they be used?
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Zombine Infection: Rules
If share walls with 2 or more infected, become infected.Once infected, always infected.
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Zombine Infection: Rules
If share walls with 2 or more infected, become infected.Once infected, always infected.
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Zombine Infection: Rules
If share walls with 2 or more infected, become infected.Once infected, always infected.
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Zombine Infection: Rules
If share walls with 2 or more infected, become infected.Once infected, always infected.
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Zombine Infection: Rules
If share walls with 2 or more infected, become infected.Once infected, always infected.
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Zombine Infection: Conquering The World
Easiest initial state that ensures all eventually infected is...?
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Zombine Infection: Conquering The World
Easiest initial state that ensures all eventually infected is...?
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Zombine Infection: Conquering The World
Next simplest initial state ensuring all eventually infected...?
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Zombine Infection: Conquering The World
Next simplest initial state ensuring all eventually infected...?
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Zombie Infection: Conquering The World
Fewest number of initial infections needed to get all...?
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Zombie Infection: Conquering The World
Fewest number of initial infections needed to get all...?
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Zombie Infection: Conquering The World
Fewest number of initial infections needed to get all...?
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Zombie Infection: Conquering The World
Fewest number of initial infections needed to get all...?
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Zombie Infection: Conquering The World
Fewest number of initial infections needed to get all...?
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Zombie Infection: Conquering The World
Fewest number of initial infections needed to get all...?
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Zombie Infection: Conquering The World
Fewest number of initial infections needed to get all...?
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Zombie Infection: Conquering The World
Fewest number of initial infections needed to get all...?
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Zombie Infection: Conquering The World
Fewest number of initial infections needed to get all...?
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Zombie Infection: Can n − 1 infect all?
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Zombie Infection: Can n − 1 infect all?
Perimeter of infection unchanged.
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Zombie Infection: Can n − 1 infect all?
Perimeter of infection unchanged.
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Zombie Infection: Can n − 1 infect all?
Perimeter of infection decreases by 2.
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Zombie Infection: Can n − 1 infect all?
Perimeter of infection decreases by 4.
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Zombie Infection: n − 1 cannot infect all
If n − 1 infected, maximum perimeter is 4(n − 1) = 4n − 4.
Mono-variant: As time passes, perimeter of infection neverincreases.
Perimeter of n × n square is 4n, so at least 1 square safe!
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Zombie Infection: n − 1 cannot infect all
If n − 1 infected, maximum perimeter is 4(n − 1) = 4n − 4.
Mono-variant: As time passes, perimeter of infection neverincreases.
Perimeter of n × n square is 4n, so at least 1 square safe!
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Zombie Infection: n − 1 cannot infect all
If n − 1 infected, maximum perimeter is 4(n − 1) = 4n − 4.
Mono-variant: As time passes, perimeter of infection neverincreases.
Perimeter of n × n square is 4n, so at least 1 square safe!
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I Love Rectangles
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Tiling the Plane with Squares
Have n× n square for each n, place one at a time so that shapeformed is always connected and a rectangle.
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Tiling the Plane with Squares
Have n×n square for each n, extra 1×1 square, place one at atime so that shape formed is always connected and a rectangle.
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Tiling the Plane with Squares: 1× 1, 1× 1, 2× 2, 3× 3, . . . .
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Tiling the Plane with Squares: 1× 1, 1× 1, 2× 2, 3× 3, . . . .
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Tiling the Plane with Squares: 1× 1, 1× 1, 2× 2, 3× 3, . . . .
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Tiling the Plane with Squares: 1× 1, 1× 1, 2× 2, 3× 3, . . . .
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Tiling the Plane with Squares: 1× 1, 1× 1, 2× 2, 3× 3, . . . .
1, 1, 2, 3, 5, . . . .
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Fibonacci Spiral:https://www.youtube.com/watch?v=kkGeOWYOFoA
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Fibonacci Spiral:https://www.youtube.com/watch?v=kkGeOWYOFoA
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Fibonacci Spiral:https://www.youtube.com/watch?v=kkGeOWYOFoA
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Triangle Game
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Rules for Triangle Game
Take an equilateral triangle, label corners 0, 1 and 2.
Subdivide however you wish into triangles.
Add labels, if a sub-triangle labeled 0–1–2 then Player 1 wins,else Player 2.
Take turns adding labels, subject to:On 0–1 boundary must use 0 or 1On 1–2 boundary must use 1 or 2On 0–2 boundary must use 0 or 2
Who has the winning strategy? What is it?
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Rules for Triangle Game
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Rules for Triangle Game
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The Line Game
Consider one-dimensional analogue: if have a 0–1 segmentPlayer 1 wins, else Player 2 wins.
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The Line Game
Consider one-dimensional analogue: if have a 0–1 segmentPlayer 1 wins, else Player 2 wins.
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The Line Game
Consider one-dimensional analogue: if have a 0–1 segmentPlayer 1 wins, else Player 2 wins.
Cannot prevent at least one 0–1 segment.
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The Line Game (cont)
Mono-variant: as add labels, number of 0–1 segments staysthe same or increases by 2.
Can also view this as a parity argument.109
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Zeckendorf Gameswith Cordwell, Epstein, Flynt, Hlavacek, Huynh, Peterson, Vu
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Introduction: Summand Minimality
Fibonaccis: F1 = 1,F2 = 2,F3 = 3,F4 = 5,Fn+2 = Fn+1 + Fn.
Zeckendorf’s TheoremEvery positive integer can be written uniquely as a sum of oneor more non-consecutive Fibonacci numbers.
Example:2018 = 1597 + 377 + 34 + 8 + 2 = F16 + F13 + F8 + F5 + F2.
Conversely, we can construct the Fibonacci sequence usingthis property:
1, 2, 3, 5, 8, 13. . .
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Introduction: Summand Minimality
Fibonaccis: F1 = 1,F2 = 2,F3 = 3,F4 = 5,Fn+2 = Fn+1 + Fn.
Zeckendorf’s TheoremEvery positive integer can be written uniquely as a sum of oneor more non-consecutive Fibonacci numbers.
Example:2018 = 1597 + 377 + 34 + 8 + 2 = F16 + F13 + F8 + F5 + F2.
Conversely, we can construct the Fibonacci sequence usingthis property:
1
, 2, 3, 5, 8, 13. . .
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Introduction: Summand Minimality
Fibonaccis: F1 = 1,F2 = 2,F3 = 3,F4 = 5,Fn+2 = Fn+1 + Fn.
Zeckendorf’s TheoremEvery positive integer can be written uniquely as a sum of oneor more non-consecutive Fibonacci numbers.
Example:2018 = 1597 + 377 + 34 + 8 + 2 = F16 + F13 + F8 + F5 + F2.
Conversely, we can construct the Fibonacci sequence usingthis property:
1, 2
, 3, 5, 8, 13. . .
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Introduction: Summand Minimality
Fibonaccis: F1 = 1,F2 = 2,F3 = 3,F4 = 5,Fn+2 = Fn+1 + Fn.
Zeckendorf’s TheoremEvery positive integer can be written uniquely as a sum of oneor more non-consecutive Fibonacci numbers.
Example:2018 = 1597 + 377 + 34 + 8 + 2 = F16 + F13 + F8 + F5 + F2.
Conversely, we can construct the Fibonacci sequence usingthis property:
1, 2, 3
, 5, 8, 13. . .
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Introduction: Summand Minimality
Fibonaccis: F1 = 1,F2 = 2,F3 = 3,F4 = 5,Fn+2 = Fn+1 + Fn.
Zeckendorf’s TheoremEvery positive integer can be written uniquely as a sum of oneor more non-consecutive Fibonacci numbers.
Example:2018 = 1597 + 377 + 34 + 8 + 2 = F16 + F13 + F8 + F5 + F2.
Conversely, we can construct the Fibonacci sequence usingthis property:
1, 2, 3, 5
, 8, 13. . .
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Introduction: Summand Minimality
Fibonaccis: F1 = 1,F2 = 2,F3 = 3,F4 = 5,Fn+2 = Fn+1 + Fn.
Zeckendorf’s TheoremEvery positive integer can be written uniquely as a sum of oneor more non-consecutive Fibonacci numbers.
Example:2018 = 1597 + 377 + 34 + 8 + 2 = F16 + F13 + F8 + F5 + F2.
Conversely, we can construct the Fibonacci sequence usingthis property:
1, 2, 3, 5, 8
, 13. . .
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Introduction: Summand Minimality
Fibonaccis: F1 = 1,F2 = 2,F3 = 3,F4 = 5,Fn+2 = Fn+1 + Fn.
Zeckendorf’s TheoremEvery positive integer can be written uniquely as a sum of oneor more non-consecutive Fibonacci numbers.
Example:2018 = 1597 + 377 + 34 + 8 + 2 = F16 + F13 + F8 + F5 + F2.
Conversely, we can construct the Fibonacci sequence usingthis property:
1, 2, 3, 5, 8, 13. . .
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Summand Minimality
Example18 = 13 + 5 = F6 + F4, legal decomposition, twosummands.18 = 13 + 3 + 2 = F6 + F3 + F2, non-legaldecomposition, three summands.
TheoremThe Zeckendorf decomposition is summand minimal.
Overall QuestionWhat other recurrences are summand minimal?
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Positive Linear Recurrence Sequences
DefinitionA positive linear recurrence sequence (PLRS) is thesequence given by a recurrence {an} with
an := c1an−1 + · · ·+ ctan−t
and each ci ≥ 0 and c1, ct > 0. We use ideal initial conditionsa−(n−1) = 0, . . . ,a−1 = 0,a0 = 1 and call (c1, . . . , ct) thesignature of the sequence.
Theorem (Cordwell, Hlavacek, Huynh, M., Peterson, Vu)For a PLRS with signature (c1, c2, . . . , ct), the GeneralizedZeckendorf Decompositions are summand minimal if and only if
c1 ≥ c2 ≥ · · · ≥ ct .119
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Proof for Fibonacci Case
Idea of proof:
D = b1F1 + · · ·+ bnFn decomposition of N, setInd(D) = b1 · 1 + · · ·+ bn · n.
Move to D′ by� 2Fk = Fk+1 + Fk−2 (and 2F2 = F3 + F1).� Fk + Fk+1 = Fk+2 (and F1 + F1 = F2).
Monovariant: Note Ind(D′) ≤ Ind(D).� 2Fk = Fk+1 + Fk−2: 2k vs 2k − 1.� Fk + Fk+1 = Fk+2: 2k + 1 vs k + 2.
If not at Zeckendorf decomposition can continue, if atZeckendorf cannot. Better: Ind′(D) = b1
√1 + · · ·+ bn
√n.
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Proof for Fibonacci Case
Idea of proof:
D = b1F1 + · · ·+ bnFn decomposition of N, setInd(D) = b1 · 1 + · · ·+ bn · n.
Move to D′ by� 2Fk = Fk+1 + Fk−2 (and 2F2 = F3 + F1).� Fk + Fk+1 = Fk+2 (and F1 + F1 = F2).
Monovariant: Note Ind(D′) ≤ Ind(D).� 2Fk = Fk+1 + Fk−2: 2k vs 2k − 1.� Fk + Fk+1 = Fk+2: 2k + 1 vs k + 2.
If not at Zeckendorf decomposition can continue, if atZeckendorf cannot. Better: Ind′(D) = b1
√1 + · · ·+ bn
√n.
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Rules
Two player game, alternate turns, last to move wins.
Bins F1, F2, F3, . . . , start with N pieces in F1 and othersempty.
A turn is one of the following moves:� If have two pieces on Fk can remove and put one
piece at Fk+1 and one at Fk−2(if k = 1 then 2F1 becomes 1F2)
� If pieces at Fk and Fk+1 remove and add one at Fk+2.
Questions:Does the game end? How long?For each N who has the winning strategy?What is the winning strategy?
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Rules
Two player game, alternate turns, last to move wins.
Bins F1, F2, F3, . . . , start with N pieces in F1 and othersempty.
A turn is one of the following moves:� If have two pieces on Fk can remove and put one
piece at Fk+1 and one at Fk−2(if k = 1 then 2F1 becomes 1F2)
� If pieces at Fk and Fk+1 remove and add one at Fk+2.
Questions:Does the game end? How long?For each N who has the winning strategy?What is the winning strategy?
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Rules
Two player game, alternate turns, last to move wins.
Bins F1, F2, F3, . . . , start with N pieces in F1 and othersempty.
A turn is one of the following moves:� If have two pieces on Fk can remove and put one
piece at Fk+1 and one at Fk−2(if k = 1 then 2F1 becomes 1F2)
� If pieces at Fk and Fk+1 remove and add one at Fk+2.
Questions:Does the game end? How long?For each N who has the winning strategy?What is the winning strategy?
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Rules
Two player game, alternate turns, last to move wins.
Bins F1, F2, F3, . . . , start with N pieces in F1 and othersempty.
A turn is one of the following moves:� If have two pieces on Fk can remove and put one
piece at Fk+1 and one at Fk−2(if k = 1 then 2F1 becomes 1F2)
� If pieces at Fk and Fk+1 remove and add one at Fk+2.
Questions:Does the game end? How long?For each N who has the winning strategy?What is the winning strategy?
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Sample Game
Start with 10 pieces at F1, rest empty.
10 0 0 0 0[F1 = 1] [F2 = 2] [F3 = 3] [F4 = 5] [F5 = 8]
Next move: Player 1: F1 + F1 = F2
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Sample Game
Start with 10 pieces at F1, rest empty.
8 1 0 0 0[F1 = 1] [F2 = 2] [F3 = 3] [F4 = 5] [F5 = 8]
Next move: Player 2: F1 + F1 = F2
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Sample Game
Start with 10 pieces at F1, rest empty.
6 2 0 0 0[F1 = 1] [F2 = 2] [F3 = 3] [F4 = 5] [F5 = 8]
Next move: Player 1: 2F2 = F3 + F1
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Sample Game
Start with 10 pieces at F1, rest empty.
7 0 1 0 0[F1 = 1] [F2 = 2] [F3 = 3] [F4 = 5] [F5 = 8]
Next move: Player 2: F1 + F1 = F2
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Sample Game
Start with 10 pieces at F1, rest empty.
5 1 1 0 0[F1 = 1] [F2 = 2] [F3 = 3] [F4 = 5] [F5 = 8]
Next move: Player 1: F2 + F3 = F4.
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Sample Game
Start with 10 pieces at F1, rest empty.
5 0 0 1 0[F1 = 1] [F2 = 2] [F3 = 3] [F4 = 5] [F5 = 8]
Next move: Player 2: F1 + F1 = F2.
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Sample Game
Start with 10 pieces at F1, rest empty.
3 1 0 1 0[F1 = 1] [F2 = 2] [F3 = 3] [F4 = 5] [F5 = 8]
Next move: Player 1: F1 + F1 = F2.
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Sample Game
Start with 10 pieces at F1, rest empty.
1 2 0 1 0[F1 = 1] [F2 = 2] [F3 = 3] [F4 = 5] [F5 = 8]
Next move: Player 2: F1 + F2 = F3.
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Sample Game
Start with 10 pieces at F1, rest empty.
0 1 1 1 0[F1 = 1] [F2 = 2] [F3 = 3] [F4 = 5] [F5 = 8]
Next move: Player 1: F3 + F4 = F5.
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Sample Game
Start with 10 pieces at F1, rest empty.
0 1 0 0 1[F1 = 1] [F2 = 2] [F3 = 3] [F4 = 5] [F5 = 8]
No moves left, Player One wins.
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Sample Game
Player One won in 9 moves.
10 0 0 0 08 1 0 0 06 2 0 0 07 0 1 0 05 1 1 0 05 0 0 1 03 1 0 1 01 2 0 1 0
0 1 1 1 00 1 0 0 1
[F1 = 1] [F2 = 2] [F3 = 3] [F4 = 5] [F5 = 8]136
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Sample Game
Player Two won in 10 moves.
10 0 0 0 08 1 0 0 06 2 0 0 07 0 1 0 05 1 1 0 05 0 0 1 03 1 0 1 01 2 0 1 02 0 1 1 00 1 1 1 00 1 0 0 1
[F1 = 1] [F2 = 2] [F3 = 3] [F4 = 5] [F5 = 8]137
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Games end
TheoremAll games end in finitely many moves.
Proof: The sum of the square roots of the indices is a strictmonovariant.
Adding consecutive terms:(√
k +√
k)−√
k + 2 < 0.
Splitting: 2√
k −(√
k + 1 +√
k + 1)< 0.
Adding 1’s: 2√
1−√
2 < 0.
Splitting 2’s: 2√
2−(√
3 +√
1)< 0.
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Games Lengths: I
Upper bound: At most n logφ
(n√
5 + 1/2)
moves.
Fastest game: n − Z (n) moves (Z (n) is the number ofsummands in n’s Zeckendorf decomposition).
From always moving on the largest summand possible(deterministic).
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Games Lengths: II
Figure: Frequency graph of the number of moves in 9,999simulations of the Zeckendorf Game with random moves whenn = 60 vs a Gaussian. Natural conjecture....
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Winning Strategy
TheoremPayer Two Has a Winning Strategy
Idea is to show if not, Player Two could steal Player One’sstrategy.
Non-constructive!
Will highlight idea with a simpler game.
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Winning Strategy: Intuition from Dot Game
Two players, alternate. Turn is choosing a dot at (i , j) andcoloring every dot (m,n) with i ≤ m and j ≤ n.
Once all dots colored game ends; whomever goes last loses.
Prove Player 1 has a winning strategy!
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Winning Strategy: Intuition from Dot Game
Two players, alternate. Turn is choosing a dot at (i , j) andcoloring every dot (m,n) with i ≤ m and j ≤ n.
Once all dots colored game ends; whomever goes last loses.
Proof Player 1 has a winning strategy. If have, play; if not, steal.
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Winning Strategy: Intuition from Dot Game
Two players, alternate. Turn is choosing a dot at (i , j) andcoloring every dot (m,n) with i ≤ m and j ≤ n.
Once all dots colored game ends; whomever goes last loses.
Proof Player 1 has a winning strategy. If have, play; if not, steal.
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Winning Strategy: Intuition from Dot Game
Two players, alternate. Turn is choosing a dot at (i , j) andcoloring every dot (m,n) with i ≤ m and j ≤ n.
Once all dots colored game ends; whomever goes last loses.
Proof Player 1 has a winning strategy. If have, play; if not, steal.
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Sketch of Proof for Player Two’s Winning Strategy
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Sketch of Proof for Player Two’s Winning Strategy
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Sketch of Proof for Player Two’s Winning Strategy
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Sketch of Proof for Player Two’s Winning Strategy
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Sketch of Proof for Player Two’s Winning Strategy
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Sketch of Proof for Player Two’s Winning Strategy
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Sketch of Proof for Player Two’s Winning Strategy
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Sketch of Proof for Player Two’s Winning Strategy
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Sketch of Proof for Player Two’s Winning Strategy
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Sketch of Proof for Player Two’s Winning Strategy
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Future Work
What if p ≥ 3 people play the Fibonacci game?
Does the number of moves in random games converge toa Gaussian?
Define k -nacci numbers by Si+1 = Si + Si−1 + · · ·+ Si−k ;game terminates but who has the winning strategy?
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Games
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Games: Coins on a line
You have 2N coins of varying denominations (each is anon-negative real number) in a line. Players A and B take turnschoosing one coin from either end. Does Player A or B have awinning strategy (i.e., a way to ensure they get at least as muchas the other?) If yes, who has it and find it if possible!
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Games: Devilish Coins
You die and the devil comes out to meet you. In the middle ofthe room is a giant circular table and next to the walls are manysacks of coins. The devil speaks. We’ll take turns putting coinsdown flat on the table. I’ll put down a coin and then you’ll putdown a coin, and so on. The coins cannot overlap and theycannot hang over the edge of the table. The last person to putdown a coin wins, or equivalently, the last person who can nolonger put a coin down on the table loses. You decide if youwant to go first.
Do you have a winning strategy for the game? If yes, what?
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Games: Prime Heaps
Alice and Bob play a game in which they take turns removingstones from a heap that initially has n stones. The number ofstones removed at each turn must be one less than a primenumber. The winner is the player who takes the last stone.Alice plays first. Prove that there are infinitely many such nsuch that Bob has a winning strategy. (For example, if n = 17,then Alice might take 6 leaving 11; then Bob might take 1leaving 10; then Alice can take the remaining stones to win.)
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More Irrationals
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√3
Assume√
3 = a/b with b minimal.
a
b
2b-a
Figure: Geometric proof of the irrationality of√
3. The whiteequilateral triangle in the middle has sides of length 2a− 3b.
Have 3(2b − a)2 = (2a− 3b)2 so√
3 = (2a− 3b)/(2b − a),note 2b − a < b (else b ≥ a), violates minimality.
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√5
Figure: Geometric proof of the irrationality of√
5.
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√5
Figure: Geometric proof of the irrationality of√
5: the kites, trianglesand the small pentagons.
This leaves five doubly covered pentagons, and one largerpentagon uncovered.
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√5
A straightforward analysis shows that the five doubly coveredpentagons are all regular, with side length a− 2b, and themiddle pentagon is also regular, with side lengthb − 2(a− 2b) = 5b − 2a.
We now have a smaller solution, with the five doubly countedregular pentagons having the same area as the omittedpentagon in the middle. Specifically, we have5(a− 2b)2 = (5b − 2a)2; as a = b
√5 and 2 <
√5 < 3, note
that a− 2b < b and thus we have our contradiction.
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√6
Figure: Geometric proof of the irrationality of√
6.166
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Closing Thoughts
Could try to do√
10 but eventually must break down. Note3,6,10 are triangular numbers (Tn = n(n + 1)/2).
T8 = 36 and thus√
T8 is an integer!
Can you get a cube-root?
What other numbers?
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