13. Combinatorial Games
Robert Snapp
[email protected]
Department of Computer Science
University of Vermont
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
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1
Two-Player Games of Perfect Information
2
Charles Bouton’s Nim
Nim Strategies
Nimtrees
The Secret of Nim
Parity Sums
3
Northcott’s Game
4
Wythoff’s Game
5
Kayles
6
Sprague-Grundy Numbers
Analysis of Nim
Analysis of Whythoff’s Game
Analysis of Kayles
7
Appendix
Binary Numbers
Powers of Two
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Two-player Games of Perfect Information
Players alternate turns.
Each player has complete information about the current state.
No elements of luck (e.g., no dice).
No hidden information (e.g., no unexposed cards).
Examples?
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Two-player Games of Perfect Information
Players alternate turns.
Each player has complete information about the current state.
No elements of luck (e.g., no dice).
No hidden information (e.g., no unexposed cards).
Examples?
Robert R. Snapp © 2010, 2012
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Two-player Games of Perfect Information
Players alternate turns.
Each player has complete information about the current state.
No elements of luck (e.g., no dice).
No hidden information (e.g., no unexposed cards).
Examples?
Robert R. Snapp © 2010, 2012
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Two-player Games of Perfect Information
Players alternate turns.
Each player has complete information about the current state.
No elements of luck (e.g., no dice).
No hidden information (e.g., no unexposed cards).
Examples?
Robert R. Snapp © 2010, 2012
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Two-player Games of Perfect Information
Players alternate turns.
Each player has complete information about the current state.
No elements of luck (e.g., no dice).
No hidden information (e.g., no unexposed cards).
Examples?
Robert R. Snapp © 2010, 2012
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Two-player Games of Perfect Information
Players alternate turns.
Each player has complete information about the current state.
No elements of luck (e.g., no dice).
No hidden information (e.g., no unexposed cards).
Examples?
I
Tic-Tac-Toe
I
Dots and Boxes
I
I
Checkers (or Draughts)
I
Chinese Checkers
I
I
I
Hex (or Nash)
I
Brigde-It
I
Twixt
I
Three-dimensional Chess
Go
Nim
I
Northcott’s Game
I
Othello (or Reversi)
Chess
I
I
Konane (Hawaiian Checkers)
I
Sprouts
I
I
Chinese Chess (Xiangqi)
I
Go-Moku (5 in a row)
I
Connect Four
Whytoff’s Game
Kayles
Robert R. Snapp © 2010, 2012
Mancala & Bao
I
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Charles Bouton’s Nim
A two-player game of perfect information, possibly of Chinese origin. It was analyzed by
Charles Leonard Bouton in 1901.
A number of tokens are placed into a number of heaps (or rows).
Players alternate turns. During a turn, a player may remove an arbitrary number of
tokens from a single heap.
The player who removes the last token from the board, wins.
In the related game, Nim misère, the player who removes the last token, loses.
For further reading, see
1
2
Martin Gardner’s, Hexaflexagons and Other Mathematical Diversions, University of
Chicago Press, 1988, pp. 151–161.
Elwyn R. Berlekamp, John H. Conway, and Richard W. Guy, Winning Ways for your
mathematical plays, Vol. 1. Second Edition, A. K. Peters, Natick, MA, 2001,
pp. 53–79.
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Sample Game: Alice vs. Bob
Consider an initial configuration having three heaps of size 3, 6, and 6, as shown.
Alice’s turn: Which tokens should she take?
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Sample Game: Alice vs. Bob
Consider an initial configuration having three heaps of size 3, 6, and 6, as shown.
Alice’s turn: Which tokens should she take?
Answer: The three red tokens (from the first heap).
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Sample Game: Alice vs. Bob
Consider an initial configuration having three heaps of size 3, 6, and 6, as shown.
Bob’s turn: (He’s going down!) What might he do?
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Sample Game: Alice vs. Bob
Consider an initial configuration having three heaps of size 3, 6, and 6, as shown.
Bob’s turn: (He’s going down!) What might he do?
Answer: Maybe he removes four blue tokens (from the second heap).
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Sample Game: Alice vs. Bob
Consider an initial configuration having three heaps of size 3, 6, and 6, as shown.
Alice’s turn: How does she respond?
Robert R. Snapp © 2010, 2012
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Sample Game: Alice vs. Bob
Consider an initial configuration having three heaps of size 3, 6, and 6, as shown.
Alice’s turn: How does she respond?
Answer: Smart Alice balances the remaining heaps by taking four tokens from the
third heap.
Robert R. Snapp © 2010, 2012
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Sample Game: Alice vs. Bob
Consider an initial configuration having three heaps of size 3, 6, and 6, as shown.
Bob’s turn: Poor Bob is not sure what he can do?
Robert R. Snapp © 2010, 2012
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Sample Game: Alice vs. Bob
Consider an initial configuration having three heaps of size 3, 6, and 6, as shown.
Bob’s turn: Poor Bob is not sure what he can do?
Answer: He decides to take one token from the third heap.
Robert R. Snapp © 2010, 2012
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Sample Game: Alice vs. Bob
Consider an initial configuration having three heaps of size 3, 6, and 6, as shown.
Alice’s turn: What should she do now?
Robert R. Snapp © 2010, 2012
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Sample Game: Alice vs. Bob
Consider an initial configuration having three heaps of size 3, 6, and 6, as shown.
Alice’s turn: What should she do now?
Answer: She takes one token from the second heal: the copy-cat strategy.
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Sample Game: Alice vs. Bob
Consider an initial configuration having three heaps of size 3, 6, and 6, as shown.
Bob’s turn: Only two options exist. Both leave one token on the board for Alice to
take. Thus, Bob resigns.
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Nim Strategies
Before describing an optimal strategy for this game, let’s look at a few interesting
cases.
1
What would you do, if it were your turn and all of the remaining tokens belong to
a single heap? For example,
Answer: Remove the entire heap.
2
What would you do if it were your turn, and there were two unequal heaps?
Answer: Remove the difference from the larger, making them equal.
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Nim Strategies
Before describing an optimal strategy for this game, let’s look at a few interesting
cases.
1
What would you do, if it were your turn and all of the remaining tokens belong to
a single heap? For example,
2
What would you do if it were your turn, and there were two unequal heaps?
Answer: Remove the entire heap.
Answer: Remove the difference from the larger, making them equal.
Robert R. Snapp © 2010, 2012
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Nim Strategies
Before describing an optimal strategy for this game, let’s look at a few interesting
cases.
1
What would you do, if it were your turn and all of the remaining tokens belong to
a single heap? For example,
2
What would you do if it were your turn, and there were two unequal heaps?
Answer: Remove the entire heap.
Answer: Remove the difference from the larger, making them equal.
Robert R. Snapp © 2010, 2012
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Nim Strategies
Before describing an optimal strategy for this game, let’s look at a few interesting
cases.
1
What would you do, if it were your turn and all of the remaining tokens belong to
a single heap? For example,
2
What would you do if it were your turn, and there were two unequal heaps?
Answer: Remove the entire heap.
Answer: Remove the difference from the larger, making them equal.
Robert R. Snapp © 2010, 2012
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List Representation
We adopting a list representation for each state. Let (1 2 3), correspond to the state
and let (2 2) correspond to the state
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Nimtrees
Earlier in the course we learned how trees can be used to enumerate (or count) the
number of permutations (using a factorial tree), the number of sequences of coin
tosses (using a binary tree), and the number of different rolls of a pair of dice (using a
hexary tree).
We now extend that idea to nim, and will enumerate every possible game that can be
played from the state (2 2).
We first draw a root node and label it with the state (2 2). We shall call the first player
Max, and consider every possible move that Max can make. Max can
1
2
3
4
Remove one token from the first heap, yielding (1 2), or
Remove two tokens from the first heap, yielding (2), or
Remove one token from the second heap, yielding (2 1), or
Remove two tokens from the second heap, yielding (2).
Since Case 3 is symmetric to Case 1, and Case 4 is symmetric to Case 2, we only
consider the first two cases.
Robert R. Snapp © 2010, 2012
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Nimtrees (cont.)
(2 2)
1
(1 2)
1
(2)
C1 (1 1)
1
(1)
C1
(1)
C1
C1
(1)
()
1
C1
()
C1
()
(1)
1
1
()
()
(2)
1
()
1
1
We thus insert nodes for states (1 2) and (2) below the root node, (2 2), and connect
each to the root node with a green edge.
Next, we will consider the moves that the second player, called Min, can make from
each of these two states.
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Nimtrees (cont.)
(2 2)
1
(1 2)
1
(2)
C1 (1 1)
1
(1)
C1
(1)
C1
C1
(1)
()
1
C1
()
C1
()
(1)
1
1
()
()
(2)
1
()
1
1
There are three moves that Min can make from the state (1 2): removing 1 from the
first heap yields that state (2), 1 from the second yields (1 1), and 2 from the second
yields (1). There are only two moves that Min can make from the state (2): removing
one token yields (1); removing 2, yields ( ): victory for Min.
Robert R. Snapp © 2010, 2012
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Nimtrees (cont.)
(2 2)
1
(1 2)
1
(2)
C1 (1 1)
1
(1)
C1
(1)
C1
C1
(1)
()
1
C1
()
C1
()
(1)
1
1
()
()
(2)
1
()
1
1
We continue to analyze each of the four active states in turn. Max can convert (2)
into either (1) or ( ) (a win for Max). Max can convert (1 1) only to (1) (why?), and (1),
on either subtree, to ( ), other wins for Max.
Robert R. Snapp © 2010, 2012
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Nimtrees (cont.)
(2 2)
1
(1 2)
1
(2)
C1 (1 1)
1
(1)
C1
(1)
C1
C1
(1)
()
1
C1
()
C1
()
(1)
1
1
()
()
(2)
1
()
1
1
Only two active nodes are left. Min can convert (1), on either subtree, to ( ), wins for
Min. Note that in this full expanded nimtree, the root node is the initial state, and
every leaf node is an empty list, which we have colored to indicate the player who
removed the last token.
Robert R. Snapp © 2010, 2012
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Nimtrees (cont.)
(2 2)
1
(1 2)
1
(2)
C1 (1 1)
1
(1)
C1
(1)
C1
C1
(1)
()
1
C1
()
C1
()
(1)
1
1
()
()
(2)
1
()
1
1
We will now prove that Min (the second player) can always win from the initial
position, (2 2), no matter how Max plays.
First we assign a score of C1 to each game that Max wins, and a score of 1 to each
game that Min wins. These values can be assigned to the leaf nodes, as shown.
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Nimtrees (cont.)
(2 2)
1
(1 2)
1
(2)
C1 (1 1)
1
(1)
C1
(1)
C1
C1
(1)
()
1
C1
()
C1
()
(1)
1
1
()
()
(2)
1
()
1
1
Next, we assign a score to node of the tree that reflects the outcome of the game
assuming that both sides play optimally. We will work from the bottom of the tree (leaf
nodes) to the top (root).
Given the state (1), Min can only remove the last token. Thus the score of any state
(1) given to Min is 1, as shown.
Robert R. Snapp © 2010, 2012
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Nimtrees (cont.)
(2 2)
1
(1 2)
1
(2)
C1 (1 1)
1
(1)
C1
(1)
C1
C1
(1)
()
1
C1
()
C1
()
(1)
1
1
()
()
(2)
1
()
1
1
We continue to assign scores to nodes one step up the tree. Since Max wants to
maximize the score, given the state (2), Max would select () over (1). Thus the score
of (2) should equal the maximum score of its descendants. This same rule applies if
no choice exists, as is the case for states (1 1) and (1).
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Nimtrees (cont.)
(2 2)
1
(1 2)
1
(2)
C1 (1 1)
1
(1)
C1
(1)
C1
C1
(1)
()
1
C1
()
C1
()
(1)
1
1
()
()
(2)
1
()
1
1
Since Min wants to minimize the score, given the state (1 2), Min would select (1 1)
over (2) or (1). Thus the score of (1 2) should equal the minimum score of its
descendants. This same rule applies to the state (2) in the right subtree.
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Nimtrees (cont.)
(2 2)
1
(1 2)
1
(2)
C1 (1 1)
1
(1)
C1
(1)
C1
C1
(1)
()
1
C1
()
C1
()
(1)
1
1
()
()
(2)
1
()
1
1
All that is left is to assign a score to the root node, (2 2). Note that it is Max’s turn, so
the score of (2 2) should be the maximum score of its descendants. However, the
latter are equal to 1. Consequently, the score of the root node is also 1, so Min is
destined to win.
In more general contexts, Nimtrees are called Max-Min trees.
Robert R. Snapp © 2010, 2012
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Nimtrees (cont.)
(2 2)
1
(1 2)
1
(2)
C1 (1 1)
1
(1)
C1
(1)
C1
C1
(1)
()
1
C1
()
C1
()
(1)
1
1
()
()
(2)
1
()
1
1
All that is left is to assign a score to the root node, (2 2). Note that it is Max’s turn, so
the score of (2 2) should be the maximum score of its descendants. However, the
latter are equal to 1. Consequently, the score of the root node is also 1, so Min is
destined to win.
In more general contexts, Nimtrees are called Max-Min trees.
Robert R. Snapp © 2010, 2012
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Now you try!
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Now you try!
Hint: Look for powers of 2 (1, 2, 4, 8, 16, 32, : : :).
Robert R. Snapp © 2010, 2012
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The Secret of Nim
Before you make a move, group each heap into unique powers of 2. For example, the
state (5 9 11 14) can be partitioned as,
Note that there are 3 groups of 8, 2 groups of 4, 2 groups of 2, and 3 groups of 1.
Identify the powers of 2 with an odd multiplicity (here 8 and 1).
Then find a move that leaves each power of 2 with an even multiplicity. Either,
1
remove 9 tokens from the second heap, or
2
remove 9 tokens from the third heap, or
3
remove 7 tokens from the fourth heap.
Robert R. Snapp © 2010, 2012
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The Secret of Nim
Before you make a move, group each heap into unique powers of 2. For example, the
state (5 9 11 14) can be partitioned as,
Note that there are 3 groups of 8, 2 groups of 4, 2 groups of 2, and 3 groups of 1.
Identify the powers of 2 with an odd multiplicity (here 8 and 1).
Then find a move that leaves each power of 2 with an even multiplicity. Either,
1
remove 9 tokens from the second heap, or
2
remove 9 tokens from the third heap, or
3
remove 7 tokens from the fourth heap.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
11 / 39
The Secret of Nim
Before you make a move, group each heap into unique powers of 2. For example, the
state (5 9 11 14) can be partitioned as,
Note that there are 3 groups of 8, 2 groups of 4, 2 groups of 2, and 3 groups of 1.
Identify the powers of 2 with an odd multiplicity (here 8 and 1).
Then find a move that leaves each power of 2 with an even multiplicity. Either,
1
remove 9 tokens from the second heap, or
2
remove 9 tokens from the third heap, or
3
remove 7 tokens from the fourth heap.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
11 / 39
The Secret of Nim
Before you make a move, group each heap into unique powers of 2. For example, the
state (5 9 11 14) can be partitioned as,
Note that there are 3 groups of 8, 2 groups of 4, 2 groups of 2, and 3 groups of 1.
Identify the powers of 2 with an odd multiplicity (here 8 and 1).
Then find a move that leaves each power of 2 with an even multiplicity. Either,
1
remove 9 tokens from the second heap, or
2
remove 9 tokens from the third heap, or
3
remove 7 tokens from the fourth heap.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
11 / 39
The Secret of Nim
Before you make a move, group each heap into unique powers of 2. For example, the
state (5 9 11 14) can be partitioned as,
Note that there are 3 groups of 8, 2 groups of 4, 2 groups of 2, and 3 groups of 1.
Identify the powers of 2 with an odd multiplicity (here 8 and 1).
Then find a move that leaves each power of 2 with an even multiplicity. Either,
1
remove 9 tokens from the second heap, or
2
remove 9 tokens from the third heap, or
3
remove 7 tokens from the fourth heap.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
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The Secret of Nim
Before you make a move, group each heap into unique powers of 2. For example, the
state (5 9 11 14) can be partitioned as,
Note that there are 3 groups of 8, 2 groups of 4, 2 groups of 2, and 3 groups of 1.
Identify the powers of 2 with an odd multiplicity (here 8 and 1).
Then find a move that leaves each power of 2 with an even multiplicity. Either,
1
remove 9 tokens from the second heap, or
2
remove 9 tokens from the third heap, or
3
remove 7 tokens from the fourth heap.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
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The Secret of Nim
Before you make a move, group each heap into unique powers of 2. For example, the
state (5 9 11 14) can be partitioned as,
Note that there are 3 groups of 8, 2 groups of 4, 2 groups of 2, and 3 groups of 1.
Identify the powers of 2 with an odd multiplicity (here 8 and 1).
Then find a move that leaves each power of 2 with an even multiplicity. Either,
1
remove 9 tokens from the second heap, or
2
remove 9 tokens from the third heap, or
3
remove 7 tokens from the fourth heap.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
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The Secret of Nim
Before you make a move, group each heap into unique powers of 2. For example, the
state (5 9 11 14) can be partitioned as,
Note that there are 3 groups of 8, 2 groups of 4, 2 groups of 2, and 3 groups of 1.
Identify the powers of 2 with an odd multiplicity (here 8 and 1).
Then find a move that leaves each power of 2 with an even multiplicity. Either,
1
remove 9 tokens from the second heap, or
2
remove 9 tokens from the third heap, or
3
remove 7 tokens from the fourth heap.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
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The Secret of Nim
Before you make a move, group each heap into unique powers of 2. For example, the
state (5 9 11 14) can be partitioned as,
Note that there are 3 groups of 8, 2 groups of 4, 2 groups of 2, and 3 groups of 1.
Identify the powers of 2 with an odd multiplicity (here 8 and 1).
Then find a move that leaves each power of 2 with an even multiplicity. Either,
1
remove 9 tokens from the second heap, or
2
remove 9 tokens from the third heap, or
3
remove 7 tokens from the fourth heap.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
11 / 39
The Secret of Nim
Before you make a move, group each heap into unique powers of 2. For example, the
state (5 9 11 14) can be partitioned as,
Note that there are 3 groups of 8, 2 groups of 4, 2 groups of 2, and 3 groups of 1.
Identify the powers of 2 with an odd multiplicity (here 8 and 1).
Then find a move that leaves each power of 2 with an even multiplicity. Either,
1
remove 9 tokens from the second heap, or
2
remove 9 tokens from the third heap, or
3
remove 7 tokens from the fourth heap.
Robert R. Snapp © 2010, 2012
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The Secret of Nim
The first step is to resolve each heap into a sum of different powers of 2.
1
2
3
Find the largest power of 2 in the heap.
Subtract this power of 2 from the heap size to obtain the remainder.
Reapply steps 1 and 2, until the the remainder equals a power of 2.
;; Returns the largest power of 2 not greater than num.
(define (largest-binary-power num)
(letrec ((pot-aux (lambda (pot)
(if (> (* 2 pot) num)
pot
(pot-aux (* 2 pot))))))
(pot-aux 1)))
;; Decompose num into a list of powers of 2.
(define (binary-rep num)
(if (= num 0)
’()
(let ((pot (largest-binary-power num)))
(cons pot (binary-rep (- num pot))))))
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The Secret of Nim
For example, if a certain heap contained 328 tokens, we
1
2
3
4
5
6
Identify 256 D 28 as the largest power of 2 contained in 328.
Subtract 328 256 D 72. Thus, 328 D 256 C 72.
The remainder, 72 is not a power of 2, so we continue.
Identify 64 D 26 as the largest power of 2 contained in 72.
Subtract 72 64 D 8. Thus, 328 D 256 C 64 C 8.
Since the remainder 8 D 23 is a power of 2, we can stop. Thus,
328 D 256 C 64 C 8 D 28 C 26 C 23 D 101001000
.base 2/:
The rightmost number is written in binary notation. The 0 in the rightmost column
indicates that 20 does not appear in the sum. Neither does 21 nor 22 . The three 1s
correspond to 23 , 26 , and 28 which add up to 328. Thus, 101001000 is called the
binary representation of 328.
Robert R. Snapp © 2010, 2012
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The Secret of Nim
The powers of 2 are said to be “balanced” if each power of 2 occurs an even number
of times after the heaps have been grouped as sums of powers of two.
The configuration (1 2 3) is balanced because,
1 D 20 ;
2 D 21 ;
3 D 21 C 20 ;
i.e., both 20 and 21 occur twice: an even number of times.
Robert R. Snapp © 2010, 2012
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Parity Sums
Another way to demonstrate that a state is balanced is to
1
2
Represent each heap size as a binary (base 2) number.
Compute the parity sum of each column (or place).
I
I
3
The parity sum equals 1 if there are an odd number of 1s in that column,
otherwise it is 0.
If the parity sums in every colomun zero, then the state is balanced.
Thus, for the state (1 2 3)
Decimal
1
2
3
Parity Sum
Robert R. Snapp © 2010, 2012
Binary
01
10
11
00
D 0 ) state is balanced.
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Parity Sums (cont.)
Here is another example. Consider the state (5 9 11 14). By the first method we
decompose each heap size as a sum of powers of two
5
9
11
14
D
4
C 1 D
22
C 20
3
D 8
C 1 D 2
C 20
3
1
D 8
C 2C 1 D 2
C 2 C 20
3
2
D 8 C4 C 2
D 2 C 2 C 21
Note that 8 D 23 occurs three times, 4 D 22 occurs twice, 2 D 21 occurs twice, and
1 D 20 occurs three times. Since some powers of 2 occur an odd number of times
this state is unbalanced.
By the other procedure, we would write
Decimal
5
9
11
14
Parity Sum
Robert R. Snapp © 2010, 2012
Binary
0101
1001
1011
1110
1001
= 9 ¤ 0 ) state is unbalanced
13. Combinatorial Games
CS 32, Fall 2012
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Philosophical Remark
What do you think is the difference between these two procedures? “Very little,”
perhaps: both methods determine the oddness or evenness of each power of 2.
Humans players likely prefer the first method, as it is easy to imagine each heap as
consisting of groups of powers of 2. For example the heaps for the state (5 9 11 14)
can be horizontally regrouped as
The second method, however, is easier for a computer. The resulting parity sum is
often called the Sprague-Grundy number G.s/ of the current state s . The
Sprague-Grundy number is a useful concept as it can be extended to other games of
perfect information.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
17 / 39
The Secret of Nim
Bouton, who first analyzed Nim, called the balanced states safe, and the unbalanced
states unsafe.
The three essential facts of Nim, and Nim misère, are
1
2
3
From every safe state, one must move to an unsafe one.
From any unsafe state, one can move to at least one safe state.
Both Nim and Nim misère end in a safe state: ( ).
If a game of Nim begins in a safe state (e.g., (1 2 3)), then the second player can
always win: the first player must move to an unsafe state, and thus the second player
can rebalance the powers of two. For example, if Max removes the 20 from the first
heap, Min would remove 20 from the third, etc.
If a Nim game begins in an unsafe state, then the first player can win by moving to a
safe state, and returning to a safe state in each subsequent turn.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
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18 / 39
Northcott’s Game
5
0
1
1
4
2
4
1
Rules:
Game begins with exactly one red and one black token in each row of a rectangular grid.
One player choses red; the other player takes black and moves first.
Players alternate turns. During a turn, a player moves one of his or her tokens to any
vacant square in the same row, so long as it does not cross the opponent’s token in that
row.
The first player unable to move, because his or her pieces are blocked, loses.
Robert R. Snapp © 2010, 2012
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19 / 39
Northcott’s Game
5
0
1
1
4
2
4
1
It turns out the Northcott’s Game is essentially the same game as Nim, where the heap sizes
correspond to the number of empty spaces that lie between two tokens on the same row.
One slight difference, is that by moving away from one’s opponent, one can add tokens,
instead of removing them, during one’s turn. However, this does not change the outcome, as
the following player can remove the new empty spaces on his or her next turn.
How should black move in the state above?
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
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19 / 39
Wythoff’s Game
Wythoff’s Game is played like Nim, except
1
2
There are initially only two heaps; and
Each turn a player may either remove some tokens from a single heap, or the
same number of tokens from both heaps.
Can you obtain an optimal strategy?
Note that Whytoff’s game can also be played with a sufficiently large chessboard. If
the size of the first heap is x , and the size of the second is y , then place the queen
on the square that belongs to both column x C 1 (from the left side), and row y C 1
(from the top). Players alternate turns moving the queen one or more squares either
to the left, up, or along the upper-left diagonal, so long as the queen never leaves the
board. The first player who moves the queen to the upper-left-most square is the
winner.
Robert R. Snapp © 2010, 2012
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Whythoff’s Queen in State (6 8)
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
Robert R. Snapp © 2010, 2012
q
13. Combinatorial Games
CS 32, Fall 2012
21 / 39
Constrained Nim
Consider a game of Nim in which players can only remove one, two, or three tokens
from a single heap during a turn.
Can you obtain an optimal strategy for this variation?
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
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22 / 39
The Game of Kayles
This game begins with an arbitrary number of tokens in a single row:
Two players alternate turns. During a turn a player may remove either one or two
adjacent tokens from a single heap. Note that if tokens are removed from the middle
of a heap, then their removal creates a gap that divides that heap into two.
As in Nim, the player who removes the last token wins. Find an optimal strategy.
The game defined above is actually quite transparent. A more challenging version,
as defined by H. E. Dudeney, begins with the initial state that consists of the following
two heaps,
(See, Henry Ernest Dudeney, The Canterbury Puzzles, Dover, NY.)
Robert R. Snapp © 2010, 2012
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Minimal Excluded Number
For every finite list of integers greater than or equal to zero, there exists a minimum
integer that is not contained within the list. Berlekamp, Conway, and Guy (2001) call
this integer the minimal excluded number, or mex for short.
For example,
(mex ’(0 1 2 4 5)) ) 3
(mex ’(2 4 5))
) 0
) 0
(mex ’())
Can you implement mex in racket?
The Minimum Excluded Number is also called the Sprague-Grundy Number, in the
context of combinatorial games.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
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Racket definition of mex
;;; (mexaux i loi) returns the smallest integer not less
;;; than i that is absent from the list of integers loi.
;;; mexaux is an auxiliary function used by mex.
(define (mexaux i loi)
(if (memq i loi)
(mexaux (+ i 1) loi)
i))
;;; (mex loi) returns the smallest non-negative integer that is
;;; absent from the list of integers loi. mex stands for the
;;; minimum excluded integer: the Sprague-Grundy number of games
;;; of perfect information. Thus (mex ’(0 3 1 3)) => 2, and
;;; (mex ’()) => 0.
(define (mex loi)
(mexaux 0 loi))
Robert R. Snapp © 2010, 2012
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More efficient definition of mex
;; mex evaluates the minimal excluded (non-negative) number from
;; the list of integers that is bound to the argument loi.
;; Thus, (mex ’(0 3 1 3)) => 2, and (mex ’()) => 0.
(define (mex loi)
(letrec ((mexaux (lambda (i)
(if (memq i loi)
(mexaux (+ i 1))
i))))
(mexaux 0)))
This function uses the special built-in keyword letrec to define a local recursive
function called aux, that does the heavy lifting.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
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26 / 39
Minimal Excluded Numbers in Nim
(2 2)
0
(1 2)
3
(2)
2
(1)
1
0
()
0
(1 1) 0
()
(1) 1
(2)
2
(1)
(1)
1
1
0
()
0
()
0
()
0
()
Here we have redrawn the nimtree, but have relabeled each vertex by the parity sum
(or Sprague-Grundy number) of that state. Note that value of each node equals the
mex of its immediate descendants (child nodes). This property can be used to find
optimal strategies for other games of perfect information.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
27 / 39
Minimal Excluded Numbers in Nim
(2 2)
0
(1 2)
3
(2)
2
(1)
1
0
()
0
(1 1) 0
()
(1) 1
(2)
2
(1)
(1)
1
1
0
()
0
()
0
()
0
()
Here we have redrawn the nimtree, but have relabeled each vertex by the parity sum
(or Sprague-Grundy number) of that state. Note that value of each node equals the
mex of its immediate descendants (child nodes). This property can be used to find
optimal strategies for other games of perfect information.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
27 / 39
Minimal Excluded Numbers in Nim
(2 2)
0
(1 2)
3
(2)
2
(1)
1
0
()
0
(1 1) 0
()
(1) 1
(2)
2
(1)
(1)
1
1
0
()
0
()
0
()
0
()
Here we have redrawn the nimtree, but have relabeled each vertex by the parity sum
(or Sprague-Grundy number) of that state. Note that value of each node equals the
mex of its immediate descendants (child nodes). This property can be used to find
optimal strategies for other games of perfect information.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
27 / 39
Minimal Excluded Numbers in Nim
(2 2)
0
(1 2)
3
(2)
2
(1)
1
0
()
0
(1 1) 0
()
(1) 1
(2)
2
(1)
(1)
1
1
0
()
0
()
0
()
0
()
Here we have redrawn the nimtree, but have relabeled each vertex by the parity sum
(or Sprague-Grundy number) of that state. Note that value of each node equals the
mex of its immediate descendants (child nodes). This property can be used to find
optimal strategies for other games of perfect information.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
27 / 39
Minimal Excluded Numbers in Nim
(2 2)
0
(1 2)
3
(2)
2
(1)
1
0
()
0
(1 1) 0
()
(1) 1
(2)
2
(1)
(1)
1
1
0
()
0
()
0
()
0
()
Here we have redrawn the nimtree, but have relabeled each vertex by the parity sum
(or Sprague-Grundy number) of that state. Note that value of each node equals the
mex of its immediate descendants (child nodes). This property can be used to find
optimal strategies for other games of perfect information.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
27 / 39
Sprague Grundy Numbers for Whytoff’s Game
Let .i j / denote an arbitrary state in Whytoff’s Game, where i; j 2 f0; 1; : : :g.
1
First we assign the value 0 to the final winning state (0 0). Thus the
Sprague-Grundy number for (0 0) is,
G.0 0/ D 0:
2
(Sprague-Grundy numbers for safe states will be colored green.)
Next, we consider the states (0 1) and (1 0). Each of these states has only one
possible subsequent state: (0 0). Thus,
G.0 1/ D G.1 0/ D mexfG.0 0/g D mexf0g D 1:
3
(Sprague-Grundy numbers for unsafe states will be colored red.)
Now we consider the state (1 1). By the rules of the game (1 1) has three
possible successors: (1 0), (0 1), and (0 0). Thus,
G.1 1/ D mexfG.1 0/; G.0 1/; G.0 0/g
D mexf1; 1; 0g
D 2:
Robert R. Snapp © 2010, 2012
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28 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0 1 2 3 4 5 6 7 8 9
0 0
1
2
3
4
5
6
7
8
9
The cell in the i -th row of the j -th column will contain the Sprague-Grundy number
for the state with i tokens in the first heap, and j tokens in the second. Let’s fill in the
four values found so far. Can you complete the table?
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
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29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0 1 2 3 4 5 6 7 8 9
0 0 1
1
2
3
4
5
6
7
8
9
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
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29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0 1 2 3 4 5 6 7 8 9
0 0 1 2
1
2
3
4
5
6
7
8
9
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0 1 2 3 4 5 6 7 8 9
0 0 1 2 3
1
2
3
4
5
6
7
8
9
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4
1
2
3
4
5
6
7
8
9
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5
1
2
3
4
5
6
7
8
9
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5 6
1
2
3
4
5
6
7
8
9
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5 6 7
1
2
3
4
5
6
7
8
9
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5 6 7 8
1
2
3
4
5
6
7
8
9
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5 6 7 8 9
1
2
3
4
5
6
7
8
9
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9
1
2
3
4
5
6
7
8
9
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9
1 2
2
3
4
5
6
7
8
9
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9
1 2 0
2
3
4
5
6
7
8
9
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9
1 2 0 4
2
3
4
5
6
7
8
9
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9
1 2 0 4 5
2
3
4
5
6
7
8
9
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9
1 2 0 4 5 3
2
3
4
5
6
7
8
9
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9
1 2 0 4 5 3 7
2
3
4
5
6
7
8
9
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9
1 2 0 4 5 3 7 8
2
3
4
5
6
7
8
9
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9
1 2 0 4 5 3 7 8 6
2
3
4
5
6
7
8
9
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9
1 2 0 4 5 3 7 8 6 10
2
3
4
5
6
7
8
9
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0
0
1
2
3
4
5
6
7
8
9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
2 0 4 5 3 7 8 6 10
0
4
5
3
7
8
6
10
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0
0
1
2
3
4
5
6
7
8
9
1
1
2
0
4
5
3
7
8
6
10
2 3 4 5 6 7 8 9
2 3 4 5 6 7 8 9
0 4 5 3 7 8 6 10
1
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0
0
1
2
3
4
5
6
7
8
9
1
1
2
0
4
5
3
7
8
6
10
2
2
0
1
3 4 5 6 7 8 9
3 4 5 6 7 8 9
4 5 3 7 8 6 10
5
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0
0
1
2
3
4
5
6
7
8
9
1
1
2
0
4
5
3
7
8
6
10
2
2
0
1
3
3
4
5
4 5 6 7 8 9
4 5 6 7 8 9
5 3 7 8 6 10
3
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0
0
1
2
3
4
5
6
7
8
9
1
1
2
0
4
5
3
7
8
6
10
2
2
0
1
3
3
4
5
4
4
5
3
5 6 7 8 9
5 6 7 8 9
3 7 8 6 10
4
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0
0
1
2
3
4
5
6
7
8
9
1
1
2
0
4
5
3
7
8
6
10
2
2
0
1
3
3
4
5
4
4
5
3
5
5
3
4
6 7 8 9
6 7 8 9
7 8 6 10
8
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0
0
1
2
3
4
5
6
7
8
9
1
1
2
0
4
5
3
7
8
6
10
2
2
0
1
3
3
4
5
4
4
5
3
5
5
3
4
6
6
7
8
7 8 9
7 8 9
8 6 10
6
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0
0
1
2
3
4
5
6
7
8
9
1
1
2
0
4
5
3
7
8
6
10
2
2
0
1
3
3
4
5
4
4
5
3
5
5
3
4
6
6
7
8
7
7
8
6
8 9
8 9
6 10
7
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0
0
1
2
3
4
5
6
7
8
9
1
1
2
0
4
5
3
7
8
6
10
2
2
0
1
3
3
4
5
4
4
5
3
5
5
3
4
6
6
7
8
7
7
8
6
8
8
6
7
9
9
10
11
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0
0
1
2
3
4
5
6
7
8
9
1
1
2
0
4
5
3
7
8
6
10
2
2
0
1
5
3
4
8
6
7
11
3
3
4
5
4
4
5
3
5
5
3
4
6
6
7
8
7
7
8
6
8
8
6
7
9
9
10
11
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Whytoff’s Game (cont.)
0
1
2
3
4
5
6
7
8
9
0
0
1
2
3
4
5
6
7
8
9
1
1
2
0
4
5
3
7
8
6
10
2
2
0
1
5
3
4
8
6
7
11
3
3
4
5
6
2
0
1
9
10
12
4
4
5
3
2
7
6
9
0
1
8
5
5
3
4
0
6
8
10
1
2
7
6
6
7
8
1
9
10
3
4
5
13
7
7
8
6
9
0
1
4
5
3
14
8
8
6
7
10
1
2
5
3
4
15
9
9
10
11
12
8
7
13
14
15
16
As in nim, players only need to know which states are safe. Each is indicated by a 0.
The optimal strategy is to move always to a safe state, if possible. If the game starts
in an unsafe state, then a skillful first player will win. Otherwise, the second player,
playing optimally, will win.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
29 / 39
Sprague-Grundy Numbers for Kayles
(5)
(4)
(3)
(2)
(1)
(1)
()
(2)
(1 1)
(3)
(1 2)
(2)
(1 1)
(1)
(1 1)
(1 1)
(1 3)
(2 2)
(1 2) (1 1 1)
(3)
(1 2)
(1 2)
(2)
(1 1)
(1)
()
The state tree for a game of kayles with initially five contiguous tokens. Note that
redundant states are not expanded.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
30 / 39
Sprague-Grundy Numbers for Kayles (cont.)
(5)
(4)
(3)
(2)
(1)
(1)
0()
(2)
(1 1)
(3)
(1 2)
(2)
(1 1)
(1)
(1 1)
(1 1)
(1 3)
(2 2)
(1 2) (1 1 1)
(3)
(1 2)
(1 2)
(2)
(1 1)
(1)
0 ()
First we assign a Sprague-Grundy number (SGN) of 0 to each terminal state ().
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
31 / 39
Sprague-Grundy Numbers for Kayles (cont.)
(5)
(4)
(3)
(2)
1 (1)
1 (1)
0()
(2)
(1 1)
(3)
(1 2)
(2)
(1 1)
1 (1)
(1 1)
(1 1)
(1 3)
(2 2)
(1 2) (1 1 1)
(3)
(1 2)
(1 2)
(2)
(1 1)
1(1)
0 ()
Since the SGN of each state is the mex of SGNs of all of its descendants, we can
assign an SGN of 1 to the state (1), everywhere it appears.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
32 / 39
Sprague-Grundy Numbers for Kayles (cont.)
(5)
(4)
(3)
2 (2)
1 (1)
2(2)
1 (1) 0(1 1)
0()
(3)
(1 2) 0(1 1)
2(2)
0(1 1)
1 (1) 0(1 1)
(1 3)
(1 2) (1 1 1)
(2 2)
(3)
(1 2)
(1 2)
2(2)
0 (1 1)
1(1)
0 ()
The state (2) has two descendants: (1) and (), with SGNs of 1 and 0, respectively.
The SGN of state (2) is thus (mex ’(1 0)) ) 2. The state (1 1) has a single
descendant of (1) with an SGN of 1. Thus, its SGN is (mex ’(1)) ) 0.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
33 / 39
Sprague-Grundy Numbers for Kayles (cont.)
(5)
3(3)
(4)
3(3)
2 (2)
1 (1)
2(2)
1 (1) 0(1 1)
0()
3(1 2) 0(1 1)
2(2)
(1 3)
0(1 1) 3(1 2) 1(1 1 1) 3(3)
1 (1) 0(1 1)
(2 2) 3(1 2)
3(1 2)
2(2)
0 (1 1)
1(1)
0 ()
States (3) and (1 2) both share the same set of descendants: (2), (1), and (1 1),
which have SGNs of 2, 1, and 0, respectively. Thus, they share an SGN of (mex ’(2
1 0)) ) 3.
State (1 1 1) has only one descendant, (1 1), which has an SGN of 0. Consequently,
(1 1 1) is assigned an SGN of 1.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
34 / 39
Sprague-Grundy Numbers for Kayles (cont.)
(5)
3(3) 2(1 3)
1(4)
3(3)
2 (2)
1 (1)
2(2)
1 (1) 0(1 1)
0()
3(1 2) 0(1 1)
2(2)
0(1 1) 3(1 2) 1(1 1 1) 3(3)
1 (1) 0(1 1)
0(2 2) 3(1 2)
3(1 2)
2(2)
0 (1 1)
1(1)
0 ()
Since the descendants of state (4) have SGNs of 3, 2, and 0, the state (4) is assigned
a SGN of (mex ’(3 2 0)) ) 1. Similarly, the state (1 3) is assigned an SGN of
(mex ’(0 1 3)) )2, while (2 2) is assigned the value (mex ’(2 3)) ) 0.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
35 / 39
Sprague-Grundy Numbers for Kayles (cont.)
4 (5)
3(3) 2(1 3)
1(4)
3(3)
2 (2)
1 (1)
2(2)
1 (1) 0(1 1)
0()
3(1 2) 0(1 1)
2(2)
0(1 1) 3(1 2) 1(1 1 1) 3(3)
1 (1) 0(1 1)
0(2 2) 3(1 2)
3(1 2)
2(2)
0 (1 1)
1(1)
0 ()
Finally the initial state (5), the root node is assigned a SGN of (mex ’(0 1 2 3))
) 4.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
36 / 39
Sprague-Grundy Numbers for Kayles (cont.)
4 (5)
3(3) 2(1 3)
1(4)
3(3)
2 (2)
1 (1)
2(2)
1 (1) 0(1 1)
0()
3(1 2) 0(1 1)
2(2)
0(1 1) 3(1 2) 1(1 1 1) 3(3)
1 (1) 0(1 1)
0(2 2) 3(1 2)
3(1 2)
2(2)
0 (1 1)
1(1)
0 ()
Given the initial state (5), the optimal move for the first player is to remove the center
token so that the state becomes (2 2). As this state has a Sprague-Grundy number of
0, what ever moves the second play makes, the first player will always have a move
to another state with a Sprague-Grundy number of 0. By continuing in this manner,
the first player is insured an eventual victory.
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
37 / 39
Table of binary numbers
For reference, here is a table of the first 64 binary numbers:
Decimal
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Binary
000000
000001
000010
000011
000100
000101
000110
000111
001000
001001
001010
001011
001100
001101
001110
001111
Robert R. Snapp © 2010, 2012
Decimal
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
Binary
010000
010001
010010
010011
010100
010101
010110
010111
011000
011001
011010
011011
011100
011101
011110
011111
Decimal
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
13. Combinatorial Games
Binary
100000
100001
100010
100011
100100
100101
100110
100111
101000
101001
101010
101011
101100
101101
101110
101111
Decimal
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
Binary
110000
110001
110010
110011
110100
110101
110110
110111
111000
111001
111010
111011
111100
111101
111110
111111
CS 32, Fall 2012
38 / 39
Table of powers of two
For reference, here are the first 48 powers of 2.
n
2n
n
2n
n
2n
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
1
2
4
8
16
32
64
128
256
512
1024
2048
4096
8192
16384
32768
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
65536
131072
262144
524288
1048576
2097152
4194304
8388608
16777216
33554432
67108854
134217728
268435456
536870912
1073741824
2147483648
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
4294967296
8589934592
17179869184
34359738368
68719476736
137438953472
274877906944
549755813888
1099511627776
2199023255552
4398046511104
8796093022208
17592186044416
35184372088832
70368744177664
140737488355328
Robert R. Snapp © 2010, 2012
13. Combinatorial Games
CS 32, Fall 2012
39 / 39