3
$\begingroup$

By optimal I mean that:

  • If max has a winning strategy then minimax will return the strategy for max with the fewest number of moves to win.
  • If min has a winning strategy then minimax will return the strategy for max with the most number of moves to lose.
  • If neither has a winning strategy then minimax will return the strategy for max with the most number of moves to draw.

The idea is that you want to win in the fewest number of moves possible but if you can't win then you want to drag out the game for as long as possible so that the opponent has more chances of making mistakes.

So, how do you make minimax return the best strategy for max?

$\endgroup$
  • $\begingroup$ Why couldn't you just add a couple if else statements with the logic and get where you needed to go? $\endgroup$ – hisairnessag3 Dec 8 '17 at 23:22
2
$\begingroup$

Minimax deals with two kinds of values:

  1. Estimated values determined by a heuristic function.
  2. Actual values determined by a terminal state.

Commonly, we use the following denotational semantics for values:

  1. A range of values centered around 0 denote estimated values (e.g. -999 to 999).
  2. A value less than the smallest heuristic value denotes a loss for max (e.g. -1000).
  3. A value more than the biggest heuristic value denotes a win for max (e.g. 1000).
  4. The value 0 denotes either an estimated draw or an actual draw.

The advantage of this denotational semantics is that comparing values is the same as comparing numbers (i.e. you don't need a special comparison function). We can extend this denotational semantics to incorporate optimality of winning and losing as follows:

  1. A range of values centered around 0 denote estimated values (e.g. -999 to 999).
  2. A range of values less than the smallest heuristic value denote loss (e.g. -2000 to -1000).
  3. A range of values more than the biggest heuristic value denote win (e.g. 1000 to 2000).
  4. The value 0 denotes either an estimated draw or an actual draw.
  5. A loss in n moves is denoted as -(m - n) where m is a sufficiently large number (e.g. 2000).
  6. A win in n moves is denoted as (m - n) where m is a sufficiently large number (e.g. 2000).

Using this denotational semantics for values requires only a small change to the minimax algorithm:

function minimax(node, depth, max)
    if max
        return negamax(node, depth, 1)
    else
        return -negamax(node, depth, -1)

function negamax(node, depth, color)
    if terminal(node)
        return -2000

    if depth = 0
        return color * heuristic(node)

    value = -2000

    foreach child of node
        v = -negamax(child, depth - 1, -color)

        if v > 1000
            v -= 1

        if v > value
            value = v

    return value

Incorporating optimality for draws is a lot more difficult.

$\endgroup$
1
+150
$\begingroup$

Short version below.

When implementing a minimax algorithm the purpose is usually to find the best possible position of a game board for the player you call max after some amount of moves. In some games like tic-tac-toe, the game tree (a graph of all legal moves) is small enough that the minimax search can be applied exhaustively to look at the whole game tree. More complex games like chess have too large of a game tree to be feasibly searched exhaustively.

A simple version of minimax just travels through the game tree, evaluating every legal move for the position currently being evaluated before going further and evaluating the possible answers to those moves. To find an optimal winning move; minimax needs only to search until a winning state of the game has been found. If implemented using the aforementioned breadth first search minimax algorithm, it will have found the way to win in the least amount of moves.

In the case where min has a forced win the truly optimal move doesn't exist. If min is not an optimal player, the definition of optimal can be the move that is most likely to cause him to make an error, that enables max to force a win. That move isn't necessarily the move that leads to the most moves until loss. As an example, consider a position in some game where max has two moves, move A and move B, move A leads to a loss in 100 moves and move B to a loss in one move. Naively move A is better but in this game the only legal moves following move A lead to a loss and move B leads to a position where min has hundreds of legal moves but only one causes him to win. Albeit a bit extreme, this example demonstrates that optimality is hard to define in a losing position. Put simply, is a very complex loss in 6 moves worse than a obvious loss in 20?

You did define a version of optimality however and implementing it is possible. Since you are only considering optimal moves, an exhaustive search must be performed and thus, there is no reason to give a score to any positions but a win, loss, and draw. The method I would use is to assign each state a score, much larger than the maximum possible amount of moves, e.g. a loss is -100,000 a win is 100,000 and a draw is 0. Then you maintain a variable that is the depth of the search or number of moves that have to be performed to reach this state. Then I would add the number of moves to the large number. So a loss in 20 moves would have a score of -99,980 and a draw in 15 moves would have the score of 15. 100,000 is a bit excessive for most games but it just has to be large enough that a loss, win, and draw is never confused as a draw in 100,001 moves would look better than a win in 1. Note that this method should only be used for losses and draws since using this method for wins would result in a win in 10 having a score of 100,010 and a win in 20 a score of 100,020 and thus looking better.

Short version:

  • Use breadth first search.
  • For winning positions: terminate the minimax when a win is found.
  • For losses and draws: search the whole game tree and give the position a score of 0+MTP for draws and L+MTP for losses.

L is a large number and MTP is the number of moves to reach the position.

$\endgroup$
  • $\begingroup$ Welcome to AI! Props for the long & short versions. $\endgroup$ – DukeZhou Dec 13 '17 at 17:36
  • $\begingroup$ A win in 10 should have the value 99,990 and a win in 20 should have the value 99,980. $\endgroup$ – Aadit M Shah Dec 14 '17 at 22:31
  • $\begingroup$ Ah yes, by adjusting the algorithm using negamax, that superior behavior can be achieved $\endgroup$ – Arnaldur Dec 16 '17 at 16:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.