COMP 221 Tutorial 6

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COMP 221 Tutorial 6

• Game Playing

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Factors affecting the goodness of game playing

• Searching algorithm
• Time is limited gt cannot expand the whole search
  tree
• Expand which node first?
• When to stop?
• Alpha-Beta pruning
• Prune away nodes that are useless
• Effectiveness depends on the ordering of node
  examination
• Evaluation function
• Evaluate the goodness of a game position

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Alpha-Beta Pruning
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• Exercise
• Perform alpha-beta pruning with
• DFS, expand the leftmost node first
• DFS, expand the rightmost node first
• Solution
• Effectiveness depends on order of node
  examination
• Can we control the ordering by introducing an
  utility of node expansion? Expand the node that
  is believed to be better first.

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Evaluation Function

• Considering Tic-Tac-Toe, (Me- , Opponent- )
• Its evaluation function could be
• 1. (row, columns, diagonals open for me)
• - (row, columns, diagonals open for
  opponent)
• 2. 8 if I win, -8 if opponent win
• 1. If two collinear board places and a third
  space in the
• line is empty, award the player 200 points.
• 2. If the player has two nearly complete lines
  then award
• the player 300 points.
• 3. If the player has a complete line award the
  player 600
• points.
• 4. Award the player one addition point for each
  line that
• could be completed from his/her current
  position.
• Eval 3X2 X1 - (3O2 O1) example from next
  exercise

F(n) 6-4 2
Me 201
Opponent 302
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Exercise 5.1 (Tic-Tac-Toe Me-X, Opponent-O)

• Define
• Xn - rows, columns, diagonals with n Xs and no
  Os
• On - rows, columns, diagonals with n Os and no
  Xs
• Eval 3X2 X1 - (3O2 O1)
• You need to
• Show the whole game tree, start from empty board,
  down to depth 2 (1 X and 1 O on the board).
  Taking symmetry into account.
• Mark down all the Eval for positions at level 2
• Mark down all the values for positions at level 0
  and 1, using minmax, and choose the best move
• Circle nodes at level 2 that would be pruned away
  by alpha-beta pruning (assuming nodes are
  generated in the optimal order)

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• Solution

1
-2
1
-1
-1
-2
0
-1
0
1
-1
0
0
1
1
2