PROLOG

1. PROLOG 8 QUEENS PROBLEM

2. 8 QUEENS • Place 8 queens on an 8 X 8 chessboard so that: • Queens do not attack each other

3. 8 QUEENS • Solution will be programmed as a unary predicate • solution(Position) • Will be true if Position represents a valid chessboard position for the 8 queens and no 2 queens can attack each other

4. 8 QUEENS • Representation of the chess board position • Each queen will be represented by X and Y (row and column) • We will define a new operator, : • X:Y • Potential solution: • [X1:Y1, X2:Y2, X3:Y3, …, X8:Y8]

5. 8 QUEENS • Since every queen will have to be in a different row  • We can hard code all the Xis to be 1 to 8 • A potential solution can be templated as : • [1:Y1, 2:Y2, 3:Y3, .., 8:Y8]

6. 8 QUEENS • Recursive formulation of the problem • We will “think” about this problem in terms of placing N queens not just 8 •  recursive relation between N queens and (N – 1) queens • Use this to solve when number of queens is 8

7. 8 QUEENS • Recursive formulation of the problem • List of N queens • Base case: list of queens is empty • General case: list of queens looks like [X:Y | Others] and • Others itself is a solution • X and Y must be integers between 1 and 8 • The X:Y queen must not attack any queen in Others

8. 8 QUEENS • solution([]). % no queen  good • solution([X:Y | Others]) :- solution(Others), member(Y,[1,2,3,4,5,6,7,8]), noattack(X:Y,Others). % need to define % X will be hard coded 1 to 8

9. 8 QUEENS • Need to define predicate noattack • noattack takes 2 arguments: • A queen • A list of queens • Base case: list of queens is empty • The queen cannot attack the queens in the empty list (there is not any)

10. 8 QUEENS • Base case for noattack • noattack(_,[]). • A queen (1st argument) cannot be attacked by 0 queen

11. 8 QUEENS • General case • If the list of queens is not empty, then it can be written [X1:Y1 | Others] where X1:Y1 represents the first queen (head) and Others the other queens (tail) • noattack(X:Y,[X1:Y1 | Others]) :- ????

12. 8 QUEENS • noattack(X:Y,[X1:Y1 | Others]) :- ???? • That will be true if: • Queen X:Y cannot attack queen X1:Y1, and • Queen X:Y cannot attack any of the queens in the list of queens Others

13. 8 QUEENS • One queen vs another (X:Y vs X1:Y1) • X and X1 must be different but that is automatic (the Xis are hard coded) • Y and Y1 must be different (column) • Y1 – Y must be different from X1 – X (main diagonal) • Y1 – Y must be different from X – X1 (other diagonal)

14. 8 QUEENS • One queen vs another (X:Y vs X1:Y1) • Y1 and Y must be different • X1 and X are hard coded and automatically different • Y1 – Y different from X1 – X • Y1 – Y different from X – X1 • We will use the =\= operator , arithmetic not equal

15. 8 QUEENS • General case for noattack • noattack(X:Y,[X1:Y1 | Others]) :- Y =\= Y1, Y1 – Y =\= X1 – X, Y1 – Y =\= X – X1, noattack(X:Y, Others). % X and X1 are hard coded different

16. 8 QUEENS • solution([]). % no queen • solution([X:Y | Others]) :- solution(Others), member(Y,[1,2,3,4,5,6,7,8]), noattack(X:Y,Others).

17. 8 QUEENS • noattack(_,[]). • noattack(X:Y,[X1:Y1 | Others]) :- Y =\= Y1, Y1 – Y =\= X1 – X, Y1 – Y =\= X – X1, noattack(X:Y, Others).

18. 8 QUEENS • Hard code the Xis into 1 to 8 template([1:Y1, 2:Y2, 3:Y3, …, 8:Y8]). • To run the program: template(S), solution(S).