Number Theory Games - Lecture 10 · 10/12/2018 · Justin Stevens Number Theory Games (Lecture 10)...

Number Theory GamesLecture 10

Justin Stevens

Justin Stevens Number Theory Games (Lecture 10) 1 / 30

Outline

1 Number Theory GamesTAG5Divisors of 60 GameNimFibonacci Nim

2 Euclidean Algorithm Revisited

3 Parting Shots


TAG5 Rules

Consider the take-away game TAG5 below:Let there be n chips on the table and A,B be two players, Amaking the first move. Each player may in turn take-away either1, 2, 3, 4, or 5 chips from the pile. The winner is the player whoremoves the final chip(s), leaving none for the other player.

Play TAG5 for n = 42. Who has the winning strategy?Play TAG5 for n = 50. Who has the winning strategy?For a general n, does player A or player B have the winning strategy?


TAG5 Analysis

12

9 10 1187

6

3 4 521

0


Reducing by Divisors Game

Consider the variation of the take-away game below:Let there be 60 chips on the table and A,B be two players, Amaking the first move. Each player in turn generates newnumbers by subtracting a positive proper divisor from the currentsum. The winner is the player who makes the new number 1.

Does player A or player B have the winning strategy?


Divisors Game Analysis

23456789


Nim

Consider the rules for the following game, known as Nim:There are four piles of stones of size 1, 3, 5, and 7. Players A andB alternate choosing a pile and taking at least one stone fromthat pile. The game continues until there are no stones remaining.

There are two variations on the winning condition:Normal Play: The player to take the last stone wins.Misère Play: The player to take the last stone loses.


Nim-Sum Definition

Definition. The nim-sum of two numbers a and b is denoted a ⊕ b. Toform this, write the two numbers in binary and add without carry:

0⊕ 0 = 1⊕ 1 = 0 and 0⊕ 1 = 1⊕ 0 = 1.

The operation is also known as “exclusive or" (xor).For piles of size 1, 3, 5, and 7, consider the table below:

22 21 20

1 0 0 13 0 1 15 1 0 17 1 1 1

0 0 0

Therefore, 1 ⊕ 3 ⊕ 5 ⊕ 7 = 0. Whenthe nim-sum is 0, we call a positionbalanced. Otherwise, it is unbalanced.

Example. Compute 5⊕ 6⊕ 7 and 9⊕ 10⊕ 15.


Nim-Sum Examples

Example. Compute 5⊕ 6⊕ 7 and 9⊕ 10⊕ 15.

22 21 20

5 1 0 16 1 1 07 1 1 1

1 0 0

Therefore, 5⊕6⊕7 = 4. To make the positionbalanced, we remove 4 from any pile. Then,

1⊕ 6⊕ 7 = 5⊕ 2⊕ 7 = 5⊕ 6⊕ 3 = 0.

23 22 21 20

9 1 0 0 110 1 0 1 015 1 1 1 1

1 1 0 0

Therefore, 9 ⊕ 10 ⊕ 15 = 10. To make theposition balanced, we remove 4 from one ofthe first two piles, or 12 from the last pile:5⊕ 10⊕ 15 = 9⊕ 6⊕ 15 = 9⊕ 10⊕ 3 = 0.


Winning Strategy

The practical strategy to win at the game of Nim is for a player to get theother into one of the following positions:

2 Heaps 3 Heaps 4 Heaps2 2 1 2 3 1 1 n n3 3 1 4 5 1 2 4 74 4 1 6 7 1 2 5 65 5 1 8 9 1 3 4 66 6 2 4 6 1 3 5 77 7 2 5 7 2 3 4 5

For Misére Nim, only the last move changes. In particular,Play exactly like you would in normal play until your opponent leavesone pile of size greater than one.At this point, reduce this pile to size 1 or 0, whichever leaves an oddnumber of piles with only one object.


Misére Nim Example

Figure 1: Mathologer


Proof of Winning Strategy in Nim

Theorem. (Charles Bouton) In a normal Nim game, the player making thefirst move has a winning strategy if and only if the nim-sum of the sizes ofthe heaps is nonzero. Otherwise, the second player has a winning strategy.Proof. Notice that the nim-sum (⊕) obeys the usual associative andcommutative laws of addition and also satisfies the property x ⊕ x = 0.Let x1, · · · , xn be the size of the heaps before a move, and y1, · · · , yn be thecorresponding sizes after. Let s = x1 ⊕ · · · ⊕ xn and t = y1 ⊕ · · · ⊕ yn. Ifthe move was in heap k, we have xi = yi for all i 6= k, and xk > yk . Then,

t = 0⊕ t= s ⊕ s ⊕ t= s ⊕ (x1 ⊕ · · · ⊕ xn)⊕ (y1 ⊕ · · · ⊕ yn)= s ⊕ (x1 ⊕ y1)⊕ · · · ⊕ (xn ⊕ yn)= s ⊕ 0⊕ · · · ⊕ 0⊕ (xk ⊕ yk)⊕ 0⊕ · · · ⊕ 0= s ⊕ xk ⊕ yk .


Proof of Winning Strategy in Nim Continued

We’ve proven that if there is a move in heap k, then t = s ⊕ xk ⊕ yk .Lemma. If s = 0, then t 6= 0 no matter what move is made.

Proof.If there are no moves, then the lemma is vacuously true. Otherwise, anymove in heap k produces t = xk ⊕ yk , which is nonzero since xk 6= yk .

Lemma. If s 6= 0, then it is possible to make a move so that t = 0.

Proof.Let d be the position of the leftmost nonzero bit in the binary form of s,and choose k such that the dth bit of xk is also nonzero. Let yk = s ⊕ xk .

Since bit d decreases from 1 to 0 and any change in the remaining bits is atmost 2d − 1, the first player can take xk − yk stones from heap k. Then,

t = s ⊕ xk ⊕ yk = s ⊕ xk ⊕ (s ⊕ xk) = 0.


Fibonacci Nim

Consider the rules for the following game, known as Fibonacci Nim:Let there be n chips on the table and A,B be two players whoalternate removing chips from the pile. On the first move, a playeris not allowed to take all of the coins, and on each subsequentmove, the number of coins removed can be any number that is atmost twice the previous move. The winner is the player whoremoves the final chip(s), leaving none for the other player.

Play Fibonacci nim for n = 140. Who has the winning strategy?Play Fibonacci nim for n = 144. Who has the winning strategy?For a general n, does player A or player B have the winning strategy?


Zeckendorf’s Theorem

Example. (Zeckendorf’s Theorem) Prove that every positive integer N canbe represented as the unique sum of non-consecutive Fibonacci numbers. Inother words, there exists a unique {aj}mj=0 such that

N =m∑

j=0Faj , a0 ≥ 2 and aj+1 > aj + 1.


Fibonacci Sum Existence

Proof by Strong Induction.For the base case N = 1, the unique representation is 1 = F2. Now, assumethat every every integer up to K can be written as the unique sum ofdistinct non-consecutive Fibonacci numbers. Let Fa be the largest Fibonaccinumber such that Fa ≤ K + 1. If Fa = K + 1, then we are clearly done.Otherwise, Fa < K + 1 < Fa+1, therefore

0 < (K + 1)− Fa < Fa+1 − Fa = Fa−1. (?)

By our hypothesis, there exists a sequence {aj}mj=0 with aj+1 > aj + 1 and

K + 1 = Fa +m∑

j=0Faj .

Since m > a + 1 by (?), we have found a representation for K + 1.


Fibonacci Sum Uniqueness

LemmaThe sum of any set of distinct, non-consecutive Fibonacci numbers whoselargest member is Fj is strictly less than the next Fibonacci number Fj+1.

For the sake of contradiction, assume we have two representations:

K + 1 = Fa1 + Fa2 + · · ·+ Fam = Fb1 + Fb2 + · · ·+ Fbl .

WLOG assume that am ≥ bl . If am > bl , by our Lemma,

K + 1 = Fb1 + Fb2 + · · ·+ Fbl < Fbl +1 − 1≤ Fam − 1< Fa1 + Fa2 + · · ·+ Fam

= K + 1.

Therefore am = bl . By our hypothesis, K + 1− Fam = K + 1− Fbl has aunique representation, therefore, K + 1 also has a unique representation.


Fibonacci Nim Strategy

Theorem. When there is a Fibonacci number of chips in the starting pile,the position is losing for the first player. Otherwise, the first player can win.

Proof. We describe the optimal strategy in Fibonacci nim in terms of the“quota" q (the maximum number of coins that can currently be removed)and the Zeckendorf representation. In particular,

A given position is losing when q is less than the smallest Fibonaccinumber in the representation.In a winning position, it is always a winning move to remove all thecoins (if this is allowed) or otherwise to remove a number of coinsequal to the smallest Fibonacci number in the representation.

When this is possible, the opposing player will necessarily be faced with alosing position, because the new quota will be smaller than the smallestFibonacci number in the Zeckendorf representation of the remaining chips.


Sprague-Grundy theorem

Figure 2: In combinatorial game theory, the Sprague-Grundy theorem states thatevery impartial game under the normal play convention is equivalent to a nimber.


Outline

1 Number Theory Games

2 Euclidean Algorithm RevisitedThe Game of EuclidMatrices

3 Parting Shots


The Game of Euclid

A third variation is a two-player game called The Game of Euclid :Let (p, q) be a pair of positive integers satisfying p > q and letA,B be two players, A making the first move. Each player may inturn subtract as many times the smaller of the remaining integersfrom the larger without making the result negative. The winner isthe player who discovers the highest common factor of p and q,that is, by the Euclidean algorithm, the player who reduces one ofthe factors to zero. [Cole and Davie, 1969]

The game has a winning strategy utilizing the golden ratio, ϕ = 1.618 . . ..

(162, 100) A→ (100, 62) B→ (62, 38) A→ (38, 24) B→ (24, 14) A→ (14, 10) B→ (10, 4) A→ (6, 4) B→ (4, 2) A→ (2, 0)

(161, 100) A→ (100, 61) B→ (61, 39) A→ (39, 22) B→ (22, 17) A→ (17, 5) B→ (7, 5) A→ (5, 2) B→ (3, 2) A→ (2, 1) B→ (1, 0).

A wins the first game, while B wins the second. Observe 1.62 > ϕ > 1.61.


Properties of the Game of Euclid

Example. For starting position (n,m), the Game of Euclid has properties:(1) The ending state of the game is (gcd(n,m), 0).(2) For 1 < a/b < ϕ, the only move is (a, b)→ (b, a′), where b/a′ > ϕ.

Proof.(1) For the move (n,m)→ (n − sm,m), by the Euclidean algorithm,gcd(n,m) = gcd(n − sm,m). Therefore, each move preserves the gcd.Furthermore, each move reduces one of the two numbers. Since negativeintegers are not permitted, the ending state of the game is (gcd(n,m), 0).

(2) Since b < a < bϕ < 2b, the only possible move is (a, b)→ (b, a − b).Hence a′ = a − b. Furthermore, ϕ2 − ϕ = 1, so

ba′ = b

a − b = 1a/b − 1 >

1ϕ− 1 = ϕ.


Winning Strategy of the Game of Euclid

Theorem. Player A may force a win if n/m = 1 or if n/m > ϕ.

Proof by Infinite Descent.When ϕ < n/m < 2, player A moves to (m, n −m) since

mn −m = 1

n/m − 1 <1

ϕ− 1 = ϕ.

When n/m > 2 and n ≡ r (mod m) for 0 ≤ r < m, there are two moves:

(n,m)→ (m, r) or (n,m)→ (m + r ,m).

If r = 0, player A wins. Otherwise, ϕ is between m/r and (m + r)/m.Player A moves to the position whose ratio lies strictly between 1 and ϕ.

Player B is left in position (a, b) where 1 < a/b < ϕ. Player B must thenmove to (b, a′) where b/a′ > ϕ, from which the process is repeated.


2× 2 Matrices

A 2× 2 matrix stores 4 pieces of information: A =[a11 a12a21 a22

].

To add two 2× 2 matrices,

A + B =[a11 a12a21 a22

]+[b11 b12b21 b22

]=[a11 + b11 a12 + b12a21 + b21 a22 + b22

].

To multiply a matrix A by a vector ~x ,

A~x =[a11 a12a21 a22

] [x1x2

]=[a11x1 + a12x2a21x1 + a22x2

].

To multiply two 2× 2 matrices,

AB =[a11 a12a21 a22

] [b11 b12b21 b22

]=[a11b11 + a12b21 a11b12 + a12b22a21b11 + a22b21 a21b12 + a22b22

].


Bézout’s Identity Example

Example. Find integers m and n such that 107m + 44n = 1.Solution. The first method uses the Euclidean algorithm in reverse:

107 = 44 · 2 + 19 1 = 7 · 107− 17 · 44 ↑44 = 19 · 2 + 6 1 = −3 · 44 + 7 · 19 ↑19 = 6 · 3 + 1 =⇒ 1 = 19− 3 · 6 ↑

Another method rewrites the first equation as[10744

]=[2 11 0

] [4419

].

Continuing in this manner by looking at the quotients,[10744

]=[2 11 0

] [2 11 0

] [3 11 0

] [6 11 0

] [10

].

Let M represent the product of all the quotient matrices. We compute

M =[107 1744 7

].


Inverses and Determinants

The inverse of a 2× 2 matrix A is a matrix A−1 such that

AA−1 = A−1A = I =[1 00 1

].

We can compute an explicit formula for the inverse:[a bc d

]−1

= 1ad − bc

[d −b−c a

].

The determinant of a 2× 2 matrix is defined by

det[a bc d

]=∣∣∣∣∣a bc d

∣∣∣∣∣ = ad − bc.

Determinants are distributive, therefore detAB = detA detB.


Matrix Form of Euclidean Algorithm

Theorem. The gcd of two integers a and b can be computed by:

a = q0b + r0b = q1r0 + r1· · ·

rn−2 = qnrn−1 + 0.

We rewrite this as a product of quotient matrices and a remainder vector:[ab

]=[q0 11 0

] [br0

]=[q0 11 0

] [q1 11 0

] [r0r1

]= · · · =

n∏i=0

[qi 11 0

] [rn−10

].

Let M =[m11 m12m21 m22

]represent the product of all the quotient matrices.


Matrix Form of Bézout’s Identity

If M =[m11 m12m21 m22

]=

n∏i=0

[qi 11 0

], then the Euclidean algorithm is

[ab

]= M

[rn−10

]= M

[g0

].

To express g as a linear combination of a and b, we must compute M−1.We see detM = (−1)n+1, since it equals the product of the determinants ofthe quotient matrices, each of which is negative one. By the inverse formula,[

g0

]= M−1

[ab

]= (−1)n+1

[m22 −m12−m21 m11

] [ab

].

The top equation is g = (−1)n+1 (m22a −m12b), therefore integersolutions to ax + by = g are x = (−1)n+1 m22 and y = (−1)n m12.


Outline

1 Number Theory Games

2 Euclidean Algorithm Revisited

3 Parting Shots


Relevant Links

Matt Parker: The Unbeatable Game from the 60s: Dr NIMMathologer: NIM, or, always WIN with mathArchimedes’ Lab: Play Nim against your computer!CL Bouton: Nim, a Game with Complete Mathematical Theory.Steven Vajda: Mathematical Games and How to Play ThemA Game Based on the Euclidean Algorithm and a Winning Strategy for It


https://www.youtube.com/watch?v=9KABcmczPdg

https://www.youtube.com/watch?v=niMjxNtiuu8

http://www.archimedes-lab.org/game_nim/play_nim_game.html

https://paradise.caltech.edu/ist4/lectures/Bouton1901.pdf

https://www.amazon.com/Mathematical-Games-How-Play-Them/dp/0486462773

https://www.jstor.org/stable/3612461

Number Theory Games - Lecture 10 · 10/12/2018 · Justin Stevens Number Theory Games (Lecture 10)...

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Transcript of Number Theory Games - Lecture 10 · 10/12/2018 · Justin Stevens Number Theory Games (Lecture 10)...