Presentaion on “MiniMax Algorithm and Water Jug Problem
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The document presents information on the Minimax algorithm and the water jug problem. It discusses how the Minimax algorithm works to determine the best possible move for a game by recursively searching a game tree. It also explains alpha-beta pruning which improves the efficiency of Minimax by pruning branches that don't affect the final decision. The water jug problem is then presented, where the goal is to get 2 gallons in the 4-gallon jug using two jugs and an unlimited water pump without volume markings. The state representation and possible state transitions are defined.
Presentaion on “MiniMax Algorithm and Water Jug Problem
- 1. Presentaion on “MiniMax Algorithm and Water Jug Problem Presented by Maruf Alom ID: 1256 World University of Bangladesh
- 2. Minimax and Alpha beta Reduction • Writing a machine player for a game, we need detraining the best possible to move. • Games such as Chess, tic-tac-toe etc. are interesting because they offer a pure abstraction of the competition between two armies. • Minimax is the recursive algorithm.
- 4. How Minimax Algorithm Works • Game Tree • Initial State • Successor function • Terminal State • Utility function
- 6. Problem with MiniMax • Too expensive • Need to prune tree
- 7. Alpha-beta pruning • If we apply alpha-beta pruning to a standard minimax algorithm, it returns the same move as the standard one, but it removes (prunes) all the nodes that are possibly not affecting the final decision. • Alpha: It is the best choice so far for the player MAX. We want to get the highest possible value here. • Beta: It is the best choice so far for MIN, and it has to be the lowest possible value.
- 8. Conclusion • Games are getting exciting. Game playing is to AI • Given a good implementation minimax algorithm can tough together.
- 9. Water Jug Problem • Consider the following problem: A Water Jug Problem: You are given two jugs, a 4-gallon one and a 3-gallon one, a pump which has unlimited water which you can use to fill the jug, and the ground on which water may be poured. Neither jug has any measuring markings on it. How can you get exactly 2 gallons of water in the 4- gallon jug? • State Representation and Initial State – We will represent a state of the problem as a tuple (x, y) where x represents the amount of water in the 4-gallon jug and y represents the amount of water in the 3-gallon jug. Goal state as (2,y).
- 10. • (x,y)If x<4 -> (4,y) • (x,y)If y<3 ->(x,3) • (x,y)If x>0 ->(x-d,y) • (x,y)If y>0 ->(x,y-d) • (x,y)If x>0 ->(0,y) • (x,y)If y>0 ->(x,0) • (x,y)If(x+y>=4 and y>0) ->(4,y-(4-x)) • (x,y)If (x+y>=3 and x>0) ->(x-(3-y),3) • (x,y)If(x+y<=4 and y>0) ->(x+y,0) • (x,y)If (x+y<=3 and x>0) ->(0,x+y) • (0,2)->(2,0)
- 11. 4 Gallon Jug 3 Gallon Jug 0 0 4 0 1 3 1 0 0 1 4 1 2 3 4 Gallon jug 3 Gallon jug pump