# Chess and Matrices

Originally posted by Hurkyl
However, like tic-tac-toe, chess is a game with a finite number of board positions, and can, in principle, be brute forced from the starting position down through all possible combinations of moves (though, at present, that feat is too formidible to attempt)
Yes-yes, I completely agree with that part. It IS possible to brute force all possible combinations. My point was merely that this is not enough, or more likely - useless. Any game is infinitesimal subset of all possible combinations, and knowing all possible combinations "doesn't make a game".

From any state of board, you could find optimal solution. But that optimal solution depends on what opponent does. You assume that opponent takes also optimal solution, and contradiction occurs - one of players has to fail. Therefore, one opponent is going to make suboptimal move, sending whole 'optimal solution' way south. At this point, 'new optimal solution' must be found. Over and over again. And thats precisely what chess players are constantly doing. Single move on the board can change notion of optimal solution. When players recognise what 'optimal solution' opponent invented, they immediately try to ruin it. And its always possible, unless player has already failed.

From any state there are many equally optimal solutions, and player has to choose ONE to pursuit. When that choice is made, player is committed, he has spent some of his most precious resource - number of moves. And if that 'path' is ruined, player is left with shaky ground. Do you see what I mean?

chroot
Staff Emeritus
Gold Member
Originally posted by wimms
From any state of board, you could find optimal solution. But that optimal solution depends on what opponent does.
Of course -- and every time your opponent moves, you'd have to calculate the most optimal move.

The idea is a "dictionary." Because there are a finite number of board states, you could (in principle) solve every single one for the best move. Then playing perfect chess just boils down to looking up the board state in the dictionary and making the pre-computed best move.

Chess is a turn-based game, so all chess computers work one move at a time. We're not talking about designing forty-move strategies and then hoping the opponent does the right (wrong) things. Instead, you just imagine that every board state has a single optimal move. This is why you can walk up to a chess game in progress (having no information about the game other than its current state) and still know what move to make.

You are correct that there could be dozens of moves which are all ranked "optimal" when considering only the state of the board at one time. You can break the tie by considering the results of all possible opponent moves, and then all your moves after his. At this point, the likelihood of any two moves being equally good is very small.
You assume that opponent takes also optimal solution, and contradiction occurs - one of players has to fail.
Good chess is actually based around the concept of forced moves -- situations where, even when playing perfectly, a player must move into a situation where he loses a piece, or some development. Chess would be no fun without gambits and forced moves. Your idea that two perfect players would never lose to each other is, quite frankly, silly.

Of course, such perfect chess would be pretty boring -- perfect chess players would always make the same moves. In fact, given two perfect chess players, the very first move would determine the entire game. Since you can only move 18 ways in the start-game, there would be exactly 18 different chess games -- period.
At this point, 'new optimal solution' must be found. Over and over again. And thats precisely what chess players are constantly doing.
And not coincidentally, it's exactly what chess computers are doing.

- Warren

Hurkyl
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Gold Member
From any state there are many equally optimal solutions, and player has to choose ONE to pursuit. When that choice is made, player is committed, he has spent some of his most precious resource - number of moves. And if that 'path' is ruined, player is left with shaky ground. Do you see what I mean?
I know what you mean; you're only thinking one move at a time. When a game theorist says "optimal move", they are looking ahead for all possible moves.

Originally posted by Hurkyl
I know what you mean; you're only thinking one move at a time. When a game theorist says "optimal move", they are looking ahead for all possible moves.
No. I'm not thinking one move at a time. Lets maybe call what you mean 'optimal strategy' that consists of and is picked from ALL predicted possible moves of both players. After analysing all possible moves, we find strategy that with highest force leads to win. Now to implement that strategy, you have to make moves in its direction. Opponent has to reply with 'expected' moves. As long as this is the case, we are inside our optimal strategy. If opponent makes 'bad' move, we could win faster. maybe.

But, I'm saying that there exists no single 'optimal strategy'. There are many equal strategies, too many to count. Any single of them leads to win IF opponent replies as 'expected'. Situation you get into is this: I know 100 strategies that can lead to win. I can predict any possible move. But, if I pick strategy N and make first 3 moves in it, and opponent makes moves that doesn't fit into my strategy, after 4-5 moves my strategy is canceled and won't lead to win. I know that beforehand. So I pick another strategy. But, for any strategy whatever, there always exists set of opponent moves that cancels my strategy.
So, I am forced to cause my opponent to make a 'mistake'. Thats my only strategy.

Hurkyl
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Gold Member
But, if I pick strategy N and make first 3 moves in it, and opponent makes moves that doesn't fit into my strategy, after 4-5 moves my strategy is canceled and won't lead to win. I know that beforehand.
But you forget; the "optimal strategy" to which I'm referring considers all possible moves; in particular it's impossible for the opponent to make a move that doesn't fit into the strategy.

Originally posted by Hurkyl
But you forget; the "optimal strategy" to which I'm referring considers all possible moves; in particular it's impossible for the opponent to make a move that doesn't fit into the strategy.
Such strategy is impossibility in chess. If it were possible, they'd had found it already.
What you are saying is that there exists strategy within which whatever opponent does, we still achieve a win. Its not so. Success of your strategy always depends on what opponent does.

What I'm saying is that even if you know ALL possible moves, you HAVE TO select only 1 to make at a time. Then opponent makes his move, either one that fits your strategy or some that cancels it out. You'll never have ability to FORCE opponent into making always moves that suits you. You'll only have ability to know beforehand all opponent moves that destroys your strategy. You can't escape that, because any move on board is undoable 'decision'. When you have decided to take 1 path of many to your win, you can't take another path at the same time. And any action on board has counteraction.

Best you could achieve with knowing ALL possible moves, is this: in given state of board you'll find 100 possible paths to a win. Any of the paths needs different move after 2 turns and only few unique moves this turn. You have to choose. You know that for every move you make there exists genuine countermove that would lead to win of opponent. But You HAVE to choose.

What is your choice? Giving up? Gambling? Taking one move and after opponent makes a move that you know could lead to his win, would you capitulate? Or, would you consider facing situation like that as already lost game? But then, the only situation in which you could play is when every single opponent move is forced. There is not enough convergence in chess to offer such case even in principle.

The situation I described is uncertainty. Chess is full of states that have such Uncertainty, it consists of them! Such Uncertainty blocks your 'vision' of ALL possible Outcomes. Knowing All possible moves shows you equal chances to win and loose from any state of the board where strategic balance exists. Its the players who make decisions to win or loose.

See, there is difference between knowing the path, and walking the path.

Hurkyl
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Such strategy is impossibility in chess. If it were possible, they'd had found it already.
Chess has too many board positions to brute force. That's why I keep saying "In principle".

What you are saying is that there exists strategy within which whatever opponent does, we still achieve a win.
No I am not. I am saying there exists an optimal strategy. In chess, or any game with a finite number of states, this means that exactly one of the three situations is correct:

(1) There exists a strategy for white that guarantees a win, no matter what black does.

(2) There exists a strategy for black that guarantees a win, no matter what white does.

(3) There exist strategies for both white and black that guarantee you will not lose, no matter what your opponent does.

The rules of chess are fairly complex, to my knowledge nobody has proved which case is correct or eliminated a case. (But such analysis has been made for other interesting games; for instance there is a rigorous mathematical proof that there exists a strategy in go for the first player which guarantees the second player cannot win, and similarly, you can mathematically prove there is a strategy in hex for the first player which guarantees victory)

Originally posted by Hurkyl
No I am not. I am saying there exists an optimal strategy. In chess, or any game with a finite number of states, this means that exactly one of the three situations is correct:
I see. I was confusing optimal strategy to winning strategy. Interesting, all of the options include condition no matter what opponent does. I don't think this can be the case with chess. There, everything depends on what opponent does upto some point in game where one of your pointed situations is reached. Then players think about it, and agree on one of capulation or draw.

(1) There exists a strategy for white that guarantees a win, no matter what black does.
(2) There exists a strategy for black that guarantees a win, no matter what white does.
(3) There exist strategies for both white and black that guarantee you will not lose, no matter what your opponent does.
Chess game can take any of the above cases to be true. (1) Is the case of forced moves. There are states of board when black's moves can be fully forced. (2) is for case I showed above, when best strategy for white would be to skip the move but it can't and thus is forced to commit to one of his possible winning paths, and thus can be defeated. (3) is obviously also pretty usual in chess from states where win is not possible anymore.

So, for same game of finite states, there exists ANY of these options, depending on state of board. Not 'exactly one of'. For board in chess to reach any of above options needs previous moves. Thus fight in chess is not initially for a win, but for one of above options, strategic advantage. White wants to achive 1 without ever coming across 2. Black wants to force white into 2 without ever crossing 1. And option 3 is taken as goal in case all else goes too wrong.

Based on game theory its assumed that one of these 3 options is set in stone at the very start of game. But I question that. At start of game board is inert, no move has any immediate value. Its merely setting up the stage. Moves are not independant either. They are necessarity multimove opening strategies to build up confrontation of forces.
Initially board goes from low possible possibilities to large number possibilities. Each move contributes to combinatorial explosion. So, single move can create huge number of possibilities, not reducing them. Thus options 1-2 are excluded at this stage. Tension builds, possibilities to resolve it grow. Only when players start annihilating pieces happens reduction of possibilities, 'wave function collapse', that brings board to state that can be one of the 3 you mentioned, or still not defined.

If both players could see all possible moves, then probably 3rd option is the only viable. But, goal of game is to win not to draw. This forces both players to leave 'safe ground' and attack.

You know the game rock-paper-scissors? When its played as imperfect information game, there is no optimal strategy to it, but if its played as turnbased duell, its (2) opponent can always win. I took this as example to point that in chess there are very many branches of tree that have similarities. There is uncertainty about what opponent does. There are traps into which one can't step. And so game is most of the time in razorthin balance of forces in draw condition. Single even tiniest advatange in either pawn or moves can make a game. Such thin balance in face of so vast move space makes imo single optimal strategy very unlikely to exist.

Hurkyl
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Gold Member
Basically, the proof works like this:

First, you draw a big tree of every possible chess game history. (The three move repetition rule + finite number of board positions guarantee that there are only a finite number of game histories)

You do it as follows:

At the very top of the tree, you place the game history with no moves; i.e. the initial gamestate.

Then, you draw an edge for every move white can make, and you put the corresponding game history at the end of that edge.

Then from all of those nodes, you do the same for every possible move for black, then again for white, et cetera.

In principle, this gives is a big game tree that has the results of every possible combination moves the players can make.

Now, we work from the bottom up. The leaves of this tree represent finished games (they're leaves because there are no moves that can be made from them). We go through every leaf and label it W if white won, B if black won, and D if it was a draw.

Now, for every node in the tree such that we've labelled all of its children:

If it is white to move
:if there is a child node labelled W
::label this node W
:else if there is a child node labelled D
::label this node D
therwise label this node B
otherwise it is black to move
:if there is a child node labelled B
::label this node B
:else if there is a child node labelled D
::label this node D
therwise label this node W

The nodes labelled "W" are nodes from which white is guaranteed a win by perfect play. The nodes labelled "B" are nodes from which black is guaranteed a win by perfect play. The nodes labelled "D" are nodes where both players are guaranteed at least a draw through perfect play.

The proof that the above claim is correct is by induction:

It's obviously true when the node we consider is a leaf. White is clearly guaranteed a win if he's already won! et cetera

Now, choose some node in the tree, and assume the claim is correct for all of its descendants. It's easy to see that the claim is also true for this node!

For example, if it is white to play at this node, and one of the children is labelled "W", then white can play the move that leads to that node, and since that child node guarantees a win for white (through perfect play), the current node also guarantees a win for white, which is why we labelled it "W".

By induction and the fact the tree is finite, the above construction proves my claim that exactly one of those three possibilities is correct for any particular gamestate. In particular, there is an answer for the initial board position!

The only reason there is any uncertainty at all is because the game tree is far too large to compute (either by our brains or by computer). There is uncertainty because we run out of computing power and we're forced to guess at the labels instead of being able to continue the tree until its completed.