] filter ; : .queens ( n -- ) queens [ [ 1 "%d " printf ] each nl ] each ;variable solutionsvariable nodes : bits ( n -- mask ) 1 swap lshift 1- ;: low Bit ( mask -- bit ) dup negate and ;: low Bit- ( mask -- bits ) dup 1- and ; : next3 ( dl dr f files -- dl dr f dl' dr' f' ) invert low Bit- repeat else 1 solutions ! If using GCC, compile with gfortran -O2 -fopenmp queens.f90. nip ; : queens ( n -- l ) iota all-permutations [ solution? The original algorithm is slightly changed to start backtracking from column three.
n] do j := a[k]; p := i j - 1; q := i - j n; if up[p] and down[q] then if i = n then m := m 1; else up[p] := false; down[q] := false; a[k] := a[i]; a[i] := j; sub(i 1); up[p] := true; down[q] := true; a[i] := a[k]; a[k] := j; fi; fi; od; end; sub(1); return m;end; Queens := function(n) local a, up, down, v, sub; a := [1 ..
n]; up := List With Identical Entries(2*n - 1, true); down := List With Identical Entries(2*n - 1, true); m := 0; sub := function(i) local j, k, p, q; for k in [i ..
Therefore, we seek to eliminate all permutations specifying a pair of queens where ((change in y) ÷ (change in x)) ∊ 1 ¯1 , or (|change in y) = (|change in x)" Compare this to the Hettinger solution used in the first Python answer. 8 -- A path of the puzzle is a sequence of successive placements of-- queens on the board.
The logic is similar but the diagonal calculation is different and more expensive computationally (Both suffer from being unable to eliminate permutation prefixes that are invalid e.g. 8 -- A path of the puzzle is a sequence of successive placements of-- queens on the board. It is not explicitly defined as a type.-- A path is a potential solution in the making.
[$row]); def place(stream; i; j): # jq indexing is based on offsets but we are using the 1-based formulae: reduce stream as $s (.; setpath([-1 ($s|i), -1 ($s|j)]; q) ); def even(k): if ((k-2) % 6) ! $x) "\n"); single_solution_queens(8) | ppdef queens(n): def sums: . safe[q_List, n_] := With[, [email protected]@q == [email protected][q [email protected]] == [email protected][q - [email protected]] == l]n Queen[q_List: , n_] := If[safe[q, n], If[Length[q] == n, , Cases[n Queen[Append[q, #], n] & /@ Range[n], Except[ | ], ]], Null] Q .