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8 Queens. Problem: Placing 8 queens on a chessboard such that they don’t attack each other Three different Prolog programs are suggessted as solutions.

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Presentation on theme: "8 Queens. Problem: Placing 8 queens on a chessboard such that they don’t attack each other Three different Prolog programs are suggessted as solutions."— Presentation transcript:

1 8 Queens

2 Problem: Placing 8 queens on a chessboard such that they don’t attack each other Three different Prolog programs are suggessted as solutions to this problem in Ivan Bratko’s book “Prolog Programming for Artificial Intelligence”. A chess-board is an 8x8 grid Y X

3 The problem is now finding such a list with the queens positioned on the board in such a way that they don’t attack one another. We will define a predicate: solution(Pos). We may represent the board as a list of eight elements. [X1:Y1, X2:Y2, X3:Y3, X4:Y4, X5:Y5, X6:Y6, X7:Y7, X8:Y8] Each X:Y pair represents the position of one queen on the board Representing the board

4 solution(Pos) returns a solution in a list “Pos”. Y X Y1 = 4, Y2 = 2, Y3 = 7, Y4 = 3, Y5 = 6, Y6 = 8, Y7 = 5, Y8 = 1Here is an example solution: [1:4, 2:2, 3:7, 4:3, 5:6, 6:8, 7:5, 8:1]

5 Q3 X=3 Y=4 Q4 X=8 Y=4 X Q1 X=1 Y=6Q2 X=5 Y=6 Q6 X=5 Y=2 Q5 X=1 Y=2 Y \= Y1, % different rows (Y1 - Y) \= (X1 - X), (Y1 - Y) \= (X - X1), % different diagonals

6 To avoid vertical attacks, queens have to be on different columns. We may fix the X coordinates to achieve this. [1:Y1, 2:Y2, 3:Y3, 4:Y4, 5:Y5, 6:Y6, 7:Y7, 8:Y8] Each of the Y’s will be a Y coordinate (a number between 1 to 8). Solution 1 We will put the queens on the board one by one making sure they do not attack one another until we have them all on the board. We may break the problem down to two cases of having an empty list or a list with a head and a tail as we usually do with problems involving lists.

7 Case 1 There are no queens on the board then the no attack condition holds and we have: solution([ ]). Case 2 first queen other queens There are some queens on the board in which case the list will be [X:Y|Sofar] In such a case, there will be a solution if 1. No attacks between the queens in Sofar 2. X and Y are integers between A queen at square X:Y must not attack any of the queens in the list Sofar solution([ ]). % Nothing to attack! solution([X:Y | Sofar]) :- % Add a new queen solution(Sofar), % Sofar is OK member(Y, [1, 2, 3, 4, 5, 6, 7, 8]), % Generate Y noattack(X:Y, Sofar). % Test

8 noattack(X:Y, [X1:Y1 | Rest]):- Y =\= Y1, % different rows (Y1 - Y) =\= (X1 - X), (Y1 - Y) =\= (X - X1), % different diagonals noattack(X:Y, Rest). % OK with the others. Now we must define the predicate noattack(Queen, List_of_Queens). This can again be broken into two cases. Case 1 If the List_of_Queens is empty then there are no queens to be attacked: noattack( _, [ ]). Case 2 If the List_of_Queens is not empty then it can be represented as [Queen1 | Rest_of_Queens] and we must check two conditions: 1. The queen at position Queen must not attack the one at position Queen1 2. The queen at position Queen must not attack the ones in the Rest_of_Queens

9 solution([ ]). % Nothing to attack! solution([X:Y | Sofar]) :- % Add a new queen solution(Sofar), % Sofar is OK member(Y, [1, 2, 3, 4, 5, 6, 7, 8]), % Generate Y (X is known) noattack(X:Y, Sofar). % Test % A solution template template([1:Y1, 2:Y2, 3:Y3, 4:Y4, 5:Y5, 6:Y6, 7:Y7, 8:Y8]) Putting it all together noattack(_, [ ]). noattack(X:Y, [X1:Y1 | Rest]):- Y =\= Y1, % different rows (Y1 - Y) =\= (X1 - X), (Y1 - Y) =\= (X - X1), % different diagonals noattack(X:Y, Rest). % OK with the others.

10 The query: ?-template(Pos),solution(Pos). generates Y1 = 4, Y2 = 2, Y3 = 7, Y4 = 3, Y5 = 6, Y6 = 8, Y7 = 5, Y8 = 1 as the first solution. There are 92 solutions.


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