# Category Archives: Game Theory

## A ball game and Konig’s lemma

In this post we will examine an application of Konig’s lemma to investigate a game proposed by Raymond Smullyan.

Imagine there is an unlimited supply of balls with us, with positive integers painted on each. It is very much possible that a particular positive integer is repeated many times on various balls. We do not know anything else about the distribution of numbers on the balls. Further imagine that there is a box with a ball initially within it.

To play the game we may do either of two things:

(1) We may remove a ball from the box and replace it with any finite number of balls, arbitrarily chosen, but constrained to have numbers painted on them which are less then the number of the ball removed. For example, if we have a number 1000 ball we may remove it and replace it by 6000 balls: 3000 balls on which 657 is painted, and 3000 balls on which 999 is painted.

(2) We may remove a ball from the box and not do any replacement.

After playing the game once we can continue playing provided there are balls left in the box.

The question is: Is it possible to play the ball game indefinitely? Can we ensure that the box will never be empty?

Smullyan himself answered this problem in the negative by using Konig’s lemma.

Theorem: The Smullyan game terminates after a finite number of steps.

Proof: For any play of the Smullyan game define a (rooted) tree as follows:
1. The vertices are the painted balls used during the game.
2. The root is the painted ball intially within the box.
3. The edges are given by the following criteria: There is an edge between the vertices $x$ and $y$ iff during some move of the game, the ball $y$ is among the balls which replaces the ball $x$ (or vice-versa).

By the definition of the game this is a rooted tree where each level is finite. Can there be an infinite path in the tree? No there can’t. This is because, by definition an infinite path would correspond to a infinite strictly decreasing sequence of positive integers which is absurd. By the contrapositive form of Konig’s lemma we can therefore conclude that the tree is finite. In other words only finitely many balls were used and so the game terminated after finitely many steps.

An alternate proof without using Konig’s lemma is available here.

Filed under Game Theory, Graph theory, Miscellaneous

## Mathematics and brainvita

In this post, I will analyse the single player game brainvita (also known as peg solitaire). The game is described by having a board with holes having the pattern:

Initially all holes except the central one are filled with pegs. The player may jump a peg horizontally or vertically over another peg into an empty hole. The jumped peg is then removed by the player. The objective is to finish the game with just one peg left.

The aim of this post is to show that there are (at most) $5$ ending board positions in brainvita. In other words, assuming the player wins, his last peg left on the board can be only among $5$ fixed board positions.

The proof uses abstract algebra. Suppose that the pegs correspond to coordinates in the plane in the natural fashion described below: The central peg is at $(0,0)$, and all others have integer coordinates as well. For convenience adjacent pegs may be taken a unit distance apart. Some coordinates are described below:

Now consider all possible ways of placing pegs on the board. There are $2^{33}$ of them. Many of these ways are impossible to reach in a valid game. Regardless of this fact, let $W$ be the set of all possible ways and for a given $w\in W$, let $C_w$ denote the set of all the coordinates where there is a peg in $w$. Now define a function $f:W\to \frac{\mathbb{Z}_2[x]}{(x^2+x+1)}$ by $f(w)=\sum_{(m,n)\in C_w}x^{m+n}$ where $\frac{\mathbb{Z}_2[x]}{(x^2+x+1)}$ is the finite field on $4$ elements. (If there are no pegs on the board we define $f$ to be $0$.)

The function $f$ has the property that for any given way $w$ of placing pegs on the board, if a legal move is performed to yield another way $w'$ then $f(w)=f(w')$. To see this, suppose that in the way $w$ there were pegs at $(m,n)$ and $(m+1,n)$ positions and the hole at the $(m+2,n)$ position was empty. A legal move may be performed by jumping the $(m,n)$ peg over the $(m+1,n)$ peg to yield a way $w'$ where the holes at $(m,n)$ and $(m+1,n)$ are empty and there is a peg at $(m+2,n)$. Clearly the only change between $f(w)$ and $f(w')$ occurs on account of these three coordinates and $f(w)-f(w')=x^{m+n}+x^{m+1+n}-x^{m+2+n}$. But as $x^{m+n}+x^{m+1+n}-x^{m+2+n}=x^{m+n}(1+x-x^2)=x^{m+n}.0=0$ in $\frac{\mathbb{Z}_2[x]}{(x^2+x+1)}$ so $f(w)-f(w')=0$ following which $f(w)=f(w')$. The same holds if instead of moving a peg to the right we had moved it above, below or to the left. Moreover all this is regardless of the fact whether the way $w$ can actually be reached by a valid sequence of moves in the game.

Now we define another function $g:W\to \frac{\mathbb{Z}_2[x]}{(x^2+x+1)}$ by $g(w)=\sum_{(m,n)\in C_w}x^{m-n}$ (the no pegs way means $g$ is $0$) and by a similar piece of reasoning conclude that $g$ also has the same property enjoyed by $f$. It is also easy to observe that both $f$ and $g$ are onto functions and so $f(W)\times g(W)$ has $16$ elements.

Next we partition the set $W$ as follows. Any $w\in W$ corresponds with a pair $(f(w),g(w))$ and so with exactly one among the $16$ elements of $f(W)\times g(W)$. In this way all the $2^{33}$ elements of $W$ are partitioned in $16$ parts. Further each part has the curious property that for any $w$ in that part, any sequence of legal moves will never change the value of $(f(w),g(w))$ and so the resultant $w'$ will also be in the same part. It is now obvious that among these $16$ parts one and only part consists of all the $w's$ which can actually occur in a game of brainvita. All other $w's$ correspond to ways which can never occur in a legal game.

What is the value of $(f(w),g(w))$ that this one special part corresponds to? The initial board (assuming it corresponds with $i\in W$) is set up so that $f(i)=g(i)=1$. So any way $w$ that arises during the game also has $f(w)=g(w)=1$. Assume now that the game has been successfully finished with a single peg at $(m,n)$, and that this corresponds to the way $w^*$. We therefore have $f(w^*)=g(w^*)=1$. But by definition $f(w^*)=x^{m+n}$ and $g(w^*)=x^{m-n}$. So we must have $x^{m+n}=1$. Since the multiplicative group $\{1,x,x^2\}$ is cyclic of order $3$ so as a consequence of Lagrange’s theorem we must have $3|(m+n)$. Similarly we have $3|(m-n)$. Combining these facts together we have $3$ dividing both $m$ and $n$. The only five coordinates which satisfy this condition are $(-3,0),(0,-3),(3,0),(0,3)$ and $(0,0)$. So the player has to end up at one among these $5$ positions. This completes the proof.

It should be noted that we have only found an upper bound on the number of positions the final peg can end up at. However by explicitly giving a sequence of moves for each of these $5$ coordinates it is possible to show that all these endings are indeed possible.

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Filed under Algebra, Game Theory