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Fri, 26 Sep 2008

Sprague-Grundy theory
I'm on a small mailing list for math geeks, and there's this one guy there, Richard Penn, who knows everything. Whenever I come up with some idle speculation, he has the answer. For example, back in 2003 I asked:

Let N be any positive integer. Does there necessarily exist a positive integer k such that the base-10 representation of kN contains only the digits 0 through 4?
M. Penn was right there with the answer.

Yesterday, M. Penn asked a question to which I happened to know the answer, and I was so pleased that I wrote up the whole theory in appalling detail. Since I haven't posted a math article in a while, and since the mailing list only has about twelve people on it, I thought I would squeeze a little more value out of it by posting it here.

Richard Penn asked:

N dots are placed in a circle. Players alternate moves, where a move consists of crossing out any one of the remaining dots, and the dots on each side of it (if they remain). The winner is the player who crosses out the last dot. What is the optimal strategy with 19 dots? with 20? Can you generalize?
M. Penn observed that there is a simple strategy for the 20-dot circle, but was not able to find one for the 19-dot circle. But solving such problems in general is made easy by the Sprague-Grundy theory, which I will explain in detail.

0. Short Spoilers

Both positions are wins for the second player to move.

The 20-dot case is trivial, since any first-player move leaves a row of 17 dots, from which the second player can leave two disconnected rows of 7 dots each. Then any first-player move in one of these rows can be effectively answered by the second player in the other row.

The 19-dot case is harder. The first player's move leaves a row of 16 dots. The second player can win by removing 3 dots to leave disconnected rows of 6 and 7 dots. After this, the strategy is complicated, but is easily found by the Sprague-Grundy theory. It's at the end of this article if you want to skip ahead.

Sprague-Grundy theory is a complete theory of all finite impartial games, which are games like this one where the two players have exactly the same moves from every position.

The theory says:

  1. Every such game position has a "value", which is a non-negative integer.
  2. A position is a second-player win if and only if its value is zero.
  3. The value of a position can be calculated from the values of the positions to which the players can move, in a simple way.
  4. The value of a collection of disjoint positions (such as two disconnected rows of dots) can be calculated from the values of its component positions in a simple way.
Order
Winning Ways for Your Mathematical Plays, Vol. 1
Winning Ways for Your Mathematical Plays, Vol. 1
with kickback
no kickback
Long details follow. They are also found in "Winning Ways", Vol I, by Berlekamp, Conway, and Guy.

1. Nim

In the game of Nim, one has some piles of beans, and a legal move is to remove some or all of the beans from any one pile. The winner is the player who takes the last bean. Equivalently, the winner is the last player who has a legal move.

Nim is important because every position in every impartial game is somehow equivalent to a position in Nim, as we will see. In fact, every position in every impartial game is equivalent to a Nim position with at most one heap of beans! Since single Nim-heaps are trivially analyzed, one can completely analyze any impartial game position by calculating the Nim-heap to which it is equivalent.

2. Disjoint sums of games

Definition: The "disjoint sum" A # B of two games A and B is a new game whose rules are as follows: a legal move in A # B is either a move in A or a move in B; the winner is the last player with a legal move.

Three easy exercises:

  1. # is commutative.
  2. # is associative.
  3. Let (a,b,c...) represent the Nim position with heaps a, b, c, etc. Then the game (a,b,c,...) is precisely (a) # (b) # (c) # ... .
Consider the trivial game with no legal moves for anyone. This game is called 0, because:

0 # a = a # 0 = a
for all games a. 0 is a win for the previous player: the next player to move has no legal moves, and loses.

We will call the next player to move "P1", and the player who just moved "P2".

Note that a Nim-heap of 0 beans is precisely the 0 game.

3. Sums of Nim-heaps

We usually represent a single Nim-heap with n beans as "∗n". I'll do that from now on.

We observed that ∗0 is a win for the second player. Observe now that when n is positive, ∗n is a win for the first player, by a trivial strategy.

From now on we will use the symbol "=" to mean a weaker relation on games than strict equality. Two games A and B will be equivalent if their outcomes are the same in a rather strong sense:

A = B means that for any game X, A # X is a winning position if and only if B # X is also.
Taking X = 0, the condition A = B implies that both games have the same outcome in isolation: if one is a first-player win, so is the other. But the condition is stronger than that. Both ∗1 and ∗2 are first-player wins, but ∗1 ≠ ∗2, because ∗1 # ∗1 is a second-player win, while ∗2 # ∗1 is a first-player win.

Exercise: ∗x = ∗y if and only if x = y.

It so happens that the disjoint sum of two Nim-heaps is equivalent to a single Nim-heap:

Nim-sum theorem:a # ∗b = ∗(ab), Where ⊕ is the bitwise exclusive-or operation.

I'll omit the proof, which is pretty easy to find. ⊕ is often described as "write a and b in binary, and add, ignoring all carries." For example 1 ⊕ 2 = 3, and 13 ⊕ 7 = 10. This implies that ∗1 # ∗2 = ∗3, and that ∗13 # ∗7 = ∗10.

Although I omitted the proof that # for Nim-heaps is essentially the ⊕ operation in disguise, there are many natural implications of this that you can use to verify that the claim is plausible. For example:

  1. The Nim-sum theorem implies that ∗0 is a neutral element for #, which we already knew.
  2. Since aa = 0, we have:
    a # ∗a = ∗0 for all a
    That is, ∗a # ∗a is a win for P2. And indeed, P2 has an obvious strategy: whatever P1 does in one pile, P2 does in the other pile. P2 never runs out of legal moves until after P1 does, and so must win.

  3. Since aa = 0, we have, more generally:
    a # ∗a # X = X for all a, X
    No matter what X is, its outcome is the same as that of ∗a # ∗a # X. Why?

    Suppose you are the player with a winning strategy for playing X alone. Then it is easy to see that you have a winning strategy in ∗a # ∗a # X, as follows: ignore the ∗a # ∗a component, until your opponent moves in it, when you should copy their move in the other half of that component. Eventually the ∗a # ∗a part will be used up (that is, reduced to ∗0 # ∗0 = 0) and your opponent will be forced to move in X, whereupon you can continue your winning strategy there until you win.

  4. According to the ⊕ operation, ∗1 # ∗2 = ∗3, and so ∗1 # ∗2 # ∗3 = ∗3 # ∗3 = 0, so P2 should have a winning strategy in ∗1 # ∗2 # ∗3. Which he does: If P1 removes any entire heap, P2 can win by equalizing the remaining heaps, leaving ∗1 # ∗1 = 0 or ∗2 # ∗2 = 0, which he wins easily. If P1 equalizes any two heaps, P2 can remove the third heap, winning the same way.

  5. Let's reconsider the game of the previous paragraph, but change the ∗1 to something else. 2 ⊕ 3 ⊕ x > 0 so if ∗x ≠ 1, ∗2 # ∗3 # ∗x = ∗y, where y>0. Since ∗y is a single nonempty Nim-heap, it is obviously a win for P1, and so ∗2 # ∗3 # ∗x should be equivalent, also a win for P1. What is P1's winning strategy in ∗2 # ∗3 # ∗x? It's easy. If x > 1, then P1 can reduce ∗x to ∗1, leaving ∗2 # ∗3 # ∗1, which we saw is a winning position. And if x = 0, then P1 can move to ∗2 # ∗2 and win.

4. The MEX rule

The important thing about disjoint sums is that they abstract away the strategy. If you have some complicated set of Nim-heaps ∗a # ∗b # ... # ∗z, you can ignore them and pretend instead that they are a single heap ∗(ab ⊕ ... ⊕ z). Your best move in the compound heap can be easily worked out from the corresponding best move in the fictitious single heap.

For example, how do you figure out how to play in ∗2 # ∗3 # ∗x? You consider it as (∗2 # ∗3) # ∗x = ∗1 # ∗x. That is, you pretend that the ∗2 and the ∗3 are actually a single heap of size 1. Then your strategy is to win in ∗1 # ∗x, which you obviously do by reducing ∗x to size 1, or, if ∗x is already ∗0, by changing ∗1 to ∗0.

Now, that is very facile, but ∗2 # ∗3 is not the same game as ∗1, because from ∗1 there is just one legal move, which is to ∗0. Whereas from ∗2 # ∗3 there are several moves. It might seem that your opponent could complicate the situation, say by moving from ∗2 # ∗3 to ∗3, which she could not do if it were really ∗1.

But actually this extra option can't possibly help your opponent, because you have an easy response to that move, which is to move right back to ∗1! If pretending that ∗2 # ∗3 was ∗1 was good before, it is certainly good after you make it ∗1 for real.

From ∗2 # ∗3 there are a whole bunch of moves:

Move to ∗3
Move to ∗2
Move to ∗1 # ∗3 = ∗2
Move to ∗2 # ∗1 = ∗3
Move to ∗2 # ∗2 = ∗0
But you can disregard the first four of these, because they are reversible: if some player X has a winning strategy that works by pretending that ∗2 # ∗3 is identical with ∗1, then the extra options of moving to ∗2 and ∗3 won't help X's opponent, because X can reverse those moves and turn the ∗2 # ∗3 component back into ∗1. So we can ignore these options, and say that there's just one move from ∗2 # ∗3 worth considering further, namely to ∗2 # ∗2 = 0. Since this is exactly the same set of moves that is available from ∗1, ∗2 # ∗3 behaves just like ∗1 in all situations, and have just proved that ∗2 # ∗3 = ∗1.

Unlike the other moves, the move from ∗2 # ∗3 to ∗0 is not reversible. Once someone turns ∗2 # ∗3 into ∗0, by equalizing the piles, it cannot then be turned back into ∗1, or anything else.

Considering this in more generality, suppose we have some game position P where the options are to move to one of several possible Nim-heaps, and M is the smallest Nim-heap that is not among the options. Then P = ∗M. Why? Because P has just the same options that ∗M has, namely the options of moving to one of ∗0 ... ∗(M-1). P also has some extra options, but we can ignore these because they're reversible. If you have a winning strategy in X # ∗M, then you have a winning strategy in X # P also, as follows:

  • If your opponent plays in X, then follow your strategy for X # ∗M, since the same move will also be available in X # P.

  • If your opponent makes P into ∗y, with y < M, then they've discarded their extra options, which are now irrelevant; play as you would if they had moved from X # ∗M to X # ∗y.

  • If your opponent makes P into ∗y, with y > M, then just move from ∗y to ∗M, leaving X + ∗M, which you can win.

MEX Theorem: If all the legal moves from a position P are equivalent to Nim-heaps of sizes {s1, ..., sk}, then P itself is equivalent to a nim-heap of size MEX(s1, ..., sk), where the MEX is the "Minimal EXcluded" element of the set: the smallest nonnegative integer that is not in the set.

For example, let's consider what happens if we augment Nim by adding a special token, called ♦. A player may, in lieu of a regular move, replace ♦ by a pile of beans of any positive size. What effect does this have on Nim?

Since the legal moves from ♦ are {∗1, ∗2, ∗3, ...} and the MEX is 0, ♦ should behave like ∗0. That is, adding a ♦ token to any position should leave the outcome unaffected. And indeed it does. If you have a winning strategy in game G, then you have a winning strategy in G # ♦ also, as follows: If your opponent plays in G, reply in G. If your opponent replaces ♦ with a pile of beans, remove it, leaving only G.

Exercise: Let G be a game where all the legal moves are to Nim-heaps. Then G is a win for P1 if and only if one of the legal moves from G is to ∗0, and a win for P2 if and only if none of the legal moves from G is to ∗0.

5. The Sprague-Grundy theory

An "impartial game" is one where both players have the same moves from every position.

Sprague-Grundy theorem: Any finite impartial game is equivalent to some Nim-heap ∗n, which is the "Nim-value" of the game.

Now let's consider Richard Penn's game, which is impartial. A legal move is to cross out any dot, and the adjacent dot or dots, if any.

The Sprague-Grundy theorem says that every row of dots in Penn's game is equivalent to some Nim-heap. Let's tabulate the size of this heap (the Nim-value) for each row of n dots. We'll represent a row of n dots as [οοοοο...ο]. Obviously, [] = ∗0 so the Nim-value of [] is 0. Also obviously, [ο] = ∗1, since they're exactly the same game.

[οο] = ∗1 also, since the only legal move from [2] is to [] = 0, and the MEX of {0} is 1.

The legal moves from [οοο] are to [] = ∗0 and [ο] = ∗1, so {∗0, ∗1}, and the MEX is 2. So [οοο] = ∗2.

Let's check that this is working. Since the Nim-value of [οοο] is 2, the theory predicts that [οοο] # ∗2 = 0 and so should be a win for P2. P2 should be able to pretend that [οοο] is actually ∗2.

Suppose P1 turns the ∗2 into ∗1, moving to [οοο] # ∗1. Then P2 should turn [οοο] into ∗1 also, which he can do by crossing out an end dot and the adjacent one, leaving [ο] # ∗1, which he easily wins. If P1 turns ∗2 into ∗0, moving to [οοο] # ∗0, then P2 should turn [οοο] into ∗0 also, which he can do by crossing out the middle and adjacent dots, leaving [] # ∗0, which he wins immediately.

If P1 plays in the [οοο] component, she must move to [] or to [ο], each equivalent to some Nim-heap of size x < 2, and P2 can answer by reducing the true Nim-heap ∗2 to contain x beans also.

Continuing our analysis of rows of dots: In Penn's game, the legal moves from [οοοο] are to [οο] and [ο]. Both of these have Nim-value ∗1, so the MEX is 0.

Easy exercise: Since [οοοο] is supposedly equivalent to ∗0, you should be able to show that a player who has a winning strategy in some game G also has a winning strategy in G + [οοοο].

The legal moves from [οοοοο] are to [οοο], [οο], and [ο] # [ο]. The Nim-values of these three games are ∗2, ∗1, and ∗0 respectively, so the MEX is 3 and [οοοοο] = ∗3.

The legal moves from [οοοοοο] are to [οοοο], [οοο], and [ο] # [οο]. The Nim-values of these three games are 0, 2, and 0, so [οοοοοο] = ∗1.

6. Richard Penn's game analyzed

Row of
n dots
Nim-
value
Winning
move
0 0  
1 1 []
2 1 []
3 2 []
4 0  
5 3 [ο] # [ο]
6 1 [ο] # [οο]
7 1 [οο] # [οο]
8 0  
9 3 [οοο] # [οοο]
10 3 [οοοοοοοο]
11 2 [οοοο] # [οοοο]
12 2 [οο] # [οοοοοοο]
13 4 [οοοοο] # [οοοοο]
14 0  
15 5 [οοοοοο] # [οοοοοο]
16 2 [ο × 14]
17 2 [οοοοοοο] # [οοοοοοο]
18 3 [οοο] # [ο × 12]
19 3 [οοοοοοοο] # [οοοοοοοο]
20 0  
Continuing in this way, we get the table of Nim-values that you see at left.

The table says that a row of 19 dots should be a win for P1, if she reduces the Nim-value from 3 to 0. And indeed, P1 has an easy winning strategy, which is to cross the 3 dots in the middle of the row, replacing [οοοοοοοοοοοοοοοοοοο] with [οοοοοοοο] # [οοοοοοοο]. But no such easy strategy obtains in a row of 20 dots, which, indeed, is a win for P2.

The original question involved circles of dots, not rows. But from a circle of n dots there is only one legal move, which is to a row of n-3 dots. From a circle of 20 dots, the only legal move is to [ο × 17] = ∗2, which should be a win for P1. P1 should win by changing ∗2 to ∗0, so should look for the move from [ο × 17] to ∗0. This is the obvious solution Richard Penn discovered: move to [οοοοοοο] # [οοοοοοο]. So the circle of 20 dots is an easy win for P2, the second player.

But for the circle of 19 dots the answer is the same, a win for the second player. The first player must move to [ο × 16] = ∗2, and then the second player should win by moving to a 0 position. [ο × 16] must have such a move, because if it didn't, the MEX rule would imply that its Nim-value was 0 instead of 2. So what's the second player's zero move here? There are actually two options. The second player can win by playing to [ο × 14], or by splitting the row into [οοοοοο] # [οοοοοοο].


7. Complete strategy for 19-bean circle

Just for completeness, let's follow one of these purportedly winning moves in detail. I claimed that the second player could win by moving to [οοοοοο] # [οοοοοοο]. But what next?

First recall that any isolated row of four dots, [οοοο], can be disregarded, because any first-player move in such a row can be answered by a second-player move that crosses out the rest of the row. And any pair of isolated rows of one or two dots, [ο] or [οο], can be similarly disregarded, because any move that crosses out one can be answered by a move that crosses out the other. So in what follows, positions like [οο] # [ο] # [οοοο] will be assumed to have been won by the second player, and we will say that the second player "has an easy win" if he has a move to such a position.

  • The first player has three possible moves in the left [οοοοοο] component, as follows:

    1. If the first player moves to [οοοο] # [οοοοοοο], the second player has an easy win by moving to [οοοο] # [οοοο].

    2. If the first player moves to [οοο] # [οοοοοοο] = ∗2 # ∗1, the second player should reduce the left component to ∗1, by moving to [ο] # [οοοοοοο]. Then no matter what the first player does, the second player has an easy win.

    3. If the first player moves to [ο] # [οο] # [οοοοοοο] = ∗1 # ∗1 # ∗1, the second player can disregard the [ο] # [οο] component. The second player instead plays to [ο] # [οο] # [οοοο] and wins.

  • The first player has four moves in the right [οοοοοοο] component, as follows:

    1. If the first player moves to [οοοοοο] # [οοοοο] = ∗1 # ∗3, the second player should move from ∗3 to ∗1. There must be a move in [οοοοο] to a position with Nim-value 1. (If there weren't, [οοοοο] would have Nim-value 1 instead of 3, by the MEX rule.) Indeed, the second player can move to [οοοοοο] # [οο]. Now whatever the first player does the second player has an easy win, either to [οοοο] or to X # X for some row X.

    2. If the first player moves to [οοοοοο] # [οοοο] = ∗1 # ∗0, the second player should move from ∗1 to ∗0. There must be a move in [οοοοοο] to a position with Nim-value 0, and indeed there is: the second player moves to [οοοο] # [οοοο] and wins.

    3. If the first player moves to [οοοοοο] # [ο] # [οοο] = ∗1 # ∗1 # ∗2, the second player can disregard the ∗1 # ∗1 component and should move in the ∗2 component, to ∗0, which he does by eliminating it entirely, leaving the first player with [οοοοοο] # [ο]. After any move by the first player the second player has an easy win.

    4. If the first player moves to [οοοοοο] # [οο] # [οο] = ∗1 # ∗1 # ∗1, the second player has a number of good choices. The simplest thing to do is to disregard the [οο] # [οο] component and move in the [οοοοοο] to some position with Nim-value 0. Moving to [οοοο] # [οο] # [οο] suffices.

So [ο × 17] is indeed a win for the next player to move, and a circle of 20 dots is therefore a win for the previous player, who is the second player.

But the important point here is not the strategy itself, which is hard to remember, and which could have been found by computer search. The important thing to notice is that computing the table of Nim-values for each row of n dots is easy, and once you have done this, the rest of the strategy almost takes care of itself. Do you need to find a good move from [οοοοοοο] # [οοοοοοοοο] # [οοοοοοοοοο]? There's no need to worry, because the table says that this can be viewed as ∗1 # ∗3 # ∗3, and so a good move is to reduce the ∗1 component, the [οοοοοοο], to ∗0, say by changing it to [οοοο] or to [οο] # [οο]. Whatever your opponent does next, calculating your reply will be similarly easy.


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