1.解決6x6 與9x9數獨問題。
擴充套件原例子4x4的數獨，重複寫行、列、格仔太麻煩了，直接貼4x4的例子：
:- use_module(library(clpfd)).
valid().
valid([head|tail]) :-
all_different(head),
valid(tail).
sudoku(puzzle, solution) :-
solution = puzzle,
puzzle = [s11, s12, s13, s14,
s21, s22, s23, s24,
s31, s32, s33, s34,
s41, s42, s43, s44],
puzzle ins 1..4,
row1 = [s11, s12, s13, s14],
row2 = [s21, s22, s23, s24],
row3 = [s31, s32, s33, s34],
row4 = [s41, s42, s43, s44],
col1 = [s11, s21, s31, s41],
col2 = [s12, s22, s32, s42],
col3 = [s13, s23, s33, s43],
col4 = [s14, s24, s34, s44],
square1 = [s11, s12, s21, s22],
square2 = [s13, s14, s23, s24],
square3 = [s31, s32, s41, s42],
square4 = [s33, s34, s43, s44],
valid([row1, row2, row3, row4,
col1, col2, col3, col4,
square1, square2, square3, square4]),
writeln(row1),
writeln(row2),
writeln(row3),
writeln(row4).
2.採用乙個皇后列表的方式解決八皇后問題，使用1~8範圍內的數字代表每個皇后，通過皇后列表中的位置取其行號並通過其在列表的值取其列號。
ps.不知道怎麼取表中位置，用index代替下表
:- use_module(library(clpfd)).
valid_col().
valid_col([col|tail]) :-
member(col, [1, 2, 3, 4, 5, 6, 7, 8]),
valid_col(tail).
diags1(, , ).
diags1([row|rowstail], [col|colstail], [diagonal|diagonaltail]) :-
diagonal is row - col,
diags1(rowstail, colstail, diagonaltail).
diags2(, , ).
diags2([row|rowstail], [col|colstail], [diagonal|diagonaltail]) :-
diagonal is row + col,
diags2(rowstail, colstail, diagonaltail).
eight_queens(cols) :-
cols = [_, _, _, _, _, _, _, _],
valid_col(cols),
colindex = [1, 2, 3, 4, 5, 6, 7, 8],
diags1(colindex, cols, diags1),
diags2(colindex, cols, diags2),
all_different(cols),
all_different(diags1),
all_different(diags2).
2 第三天作業
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第三天python作業
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第三天，努力！！
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