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Ingy döt Net
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3887 changed files with 59894 additions and 7280 deletions
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@ -1,4 +1,4 @@
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Solve the [[WP:Eight_queens_puzzle|eight queens puzzle]]. You can extend the problem to solve the puzzle with a board of side NxN.
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Solve the [[WP:Eight_queens_puzzle|eight queens puzzle]]. You can extend the problem to solve the puzzle with a board of side NxN. Number of solutions for small values of N is [http://oeis.org/A000170 here].
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;Cf.
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* [[Knight's tour]]
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22
Task/N-queens-problem/Common-Lisp/n-queens-problem-1.lisp
Normal file
22
Task/N-queens-problem/Common-Lisp/n-queens-problem-1.lisp
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@ -0,0 +1,22 @@
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(defun n-queens (n m)
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(if (= n 1)
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(loop for x from 1 to m collect (list x))
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(loop for sol in (n-queens (1- n) m) nconc
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(loop for col from 1 to m when
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(loop for row from 0 to (length sol) for c in sol
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always (and (/= col c)
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(/= (abs (- c col)) (1+ row)))
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finally (return (cons col sol)))
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collect it))))
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(defun show-solution (b n)
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(loop for i in b do
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(format t "~{~A~^~}~%"
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(loop for x from 1 to n collect (if (= x i) "Q " ". "))))
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(terpri))
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(let ((i 0) (n 8))
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(mapc #'(lambda (s)
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(format t "Solution ~a:~%" (incf i))
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(show-solution s n))
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(n-queens n n)))
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45
Task/N-queens-problem/Common-Lisp/n-queens-problem-2.lisp
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45
Task/N-queens-problem/Common-Lisp/n-queens-problem-2.lisp
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@ -0,0 +1,45 @@
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(defun queens (nmax)
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(let ((a (make-array `(,nmax)))
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(s (make-array `(,nmax)))
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(u (make-array `(,(- (* 4 nmax) 2)) :initial-element 0))
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y z i j p q r m (v nil))
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(dotimes (i nmax) (setf (aref a i) i))
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(loop for n from 1 to nmax do
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(tagbody
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(setf m 0 i 0 r (1- (* 2 n)))
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(go L40)
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L30
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(setf (aref s i) j (aref u p) 1 (aref u (+ q r)) 1)
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(incf i)
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L40
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(if (>= i n) (go L80))
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(setf j i)
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L50
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(setf y (aref a j) z (aref a i))
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(setf p (+ (- i y) (1- n)) q (+ i y))
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(setf (aref a i) y (aref a j) z)
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(if (and (zerop (aref u p)) (zerop (aref u (+ q r)))) (go L30))
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L60
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(incf j)
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(if (< j n) (go L50))
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L70
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(decf j)
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(if (= j i) (go L90))
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(rotatef (aref a i) (aref a j))
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(go L70)
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L80
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(incf m)
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L90
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(decf i)
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(if (minusp i) (go L100))
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(setf p (+ (- i (aref a i)) (1- n)) q (+ i (aref a i)) j (aref s i))
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(setf (aref u p) 0 (aref u (+ q r)) 0)
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(go L60)
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L100
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;(princ n) (princ " ") (princ m) (terpri)
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(push (cons n m) v)
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)) (reverse v)))
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> (queens 14)
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((1 . 1) (2 . 0) (3 . 0) (4 . 2) (5 . 10) (6 . 4) (7 . 40) (8 . 92) (9 . 352)
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(10 . 724) (11 . 2680) (12 . 14200) (13 . 73712) (14 . 365596))
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@ -1,6 +1,6 @@
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import std.stdio, std.conv;
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uint nQueens(in uint nn) pure nothrow
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ulong nQueens(in uint nn) pure nothrow
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in {
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assert(nn > 0 && nn <= 27,
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"'side' value must be in 1 .. 27.");
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@ -12,9 +12,9 @@ in {
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immutable uint full = uint.max - ((1 << (ulen - nn)) - 1);
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immutable n = nn - 3;
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uint count;
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typeof(return) count;
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uint[32] l=void, r=void, c=void;
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uint[33] mm; // mm and mmi are a stack
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uint[33] mm; // mm and mmi are a stack.
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// Require second queen to be left of the first queen, so
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// we ever only test half of the possible solutions. This
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@ -24,22 +24,22 @@ in {
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uint d = n;
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// c: columns occupied by previous queens.
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c[n] = b0 | b1;
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// l: columns attacked by left diagonals
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// l: columns attacked by left diagonals.
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l[n] = (b0 << 2) | (b1 << 1);
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// r: by right diagnoals
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// r: by right diagnoals.
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r[n] = (b0 >> 2) | (b1 >> 1);
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// availabe columns on current row
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// Availabe columns on current row.
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uint bits = full & ~(l[n] | r[n] | c[n]);
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uint mmi = 1;
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mm[mmi] = bits;
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while (bits) {
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// d: depth, aka row. counting backwards
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// because !d is often faster than d != n
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// d: depth, aka row. counting backwards.
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// Because !d is often faster than d != n.
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while (d) {
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// pos is right most nonzero bit
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// immutable uint pos = 1U << bits.bsf; // Slower.
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immutable uint pos = -(cast(int)bits) & bits;
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// Mark bit used. Only put current bits on
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@ -53,9 +53,9 @@ in {
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}
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d--;
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l[d] = (l[d+1] | pos) << 1;
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r[d] = (r[d+1] | pos) >> 1;
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c[d] = c[d+1] | pos;
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l[d] = (l[d + 1] | pos) << 1;
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r[d] = (r[d + 1] | pos) >> 1;
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c[d] = c[d + 1] | pos;
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bits = full & ~(l[d] | r[d] | c[d]);
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@ -84,7 +84,7 @@ in {
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return count * 2;
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}
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void main(string[] args) {
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immutable int side = (args.length >= 2) ? to!int(args[1]) : 8;
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writefln("N-queens(%d) = %d solutions.", side, nQueens(side));
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void main(in string[] args) {
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immutable uint side = (args.length >= 2) ? args[1].to!uint : 8;
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writefln("N-queens(%d) = %d solutions.", side, side.nQueens);
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}
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@ -1,106 +1,122 @@
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MODULE QUEENS_MOD
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IMPLICIT NONE
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INTEGER, PARAMETER :: LONG=SELECTED_INT_KIND(17)
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CONTAINS
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FUNCTION PQUEENS(N,K1,K2) RESULT(M)
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IMPLICIT NONE
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INTEGER(KIND=LONG) :: M
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INTEGER, INTENT(IN) :: N,K1,K2
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INTEGER, PARAMETER :: L=20
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INTEGER :: A(L),S(L),U(4*L-2)
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INTEGER :: I,J,Y,Z,P,Q,R
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DO 10 I=1,N
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10 A(I)=I
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DO 20 I=1,4*N-2
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20 U(I)=0
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M=0
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R=2*N-1
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IF(K1.EQ.K2) RETURN
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P=1-K1+N
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Q=1+K1-1
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IF((U(P).NE.0).OR.(U(Q+R).NE.0)) RETURN
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U(P)=1
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U(Q+R)=1
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Z=A(1)
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A(1)=A(K1)
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A(K1)=Z
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P=2-K2+N
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Q=2+K2-1
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IF((U(P).NE.0).OR.(U(Q+R).NE.0)) RETURN
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U(P)=1
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U(Q+R)=1
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IF(K2.NE.1) THEN
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Z=A(2)
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A(2)=A(K2)
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A(K2)=Z
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ELSE
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Z=A(2)
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A(2)=A(K1)
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A(K1)=Z
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END IF
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I=3
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GO TO 40
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30 S(I)=J
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U(P)=1
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U(Q+R)=1
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I=I+1
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40 IF(I.GT.N) GO TO 80
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J=I
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50 Z=A(I)
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Y=A(J)
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P=I-Y+N
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Q=I+Y-1
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A(I)=Y
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A(J)=Z
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IF((U(P).EQ.0).AND.(U(Q+R).EQ.0)) GO TO 30
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60 J=J+1
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IF(J.LE.N) GO TO 50
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70 J=J-1
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IF(J.EQ.I) GO TO 90
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Z=A(I)
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A(I)=A(J)
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A(J)=Z
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GO TO 70
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80 M=M+1
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90 I=I-1
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IF(I.EQ.2) RETURN
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P=I-A(I)+N
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Q=I+A(I)-1
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J=S(I)
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U(P)=0
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U(Q+R)=0
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GO TO 60
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END FUNCTION
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END MODULE
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PROGRAM QUEENS
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USE OMP_LIB
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USE QUEENS_MOD
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IMPLICIT NONE
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INTEGER, PARAMETER :: L=20
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INTEGER :: N,I,J,A(L*L,2),K,P,Q
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INTEGER(KIND=LONG) :: S,B(L*L)
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DOUBLE PRECISION :: T1,T2
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DO N=6,18
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K=0
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P=N/2
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Q=MOD(N,2)*(P+1)
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DO I=1,N
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DO J=1,N
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IF((ABS(I-J).GT.1).AND.((I.LE.P).OR.((I.EQ.Q).AND.(J.LT.I)))) THEN
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K=K+1
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A(K,1)=I
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A(K,2)=J
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END IF
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END DO
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END DO
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S=0
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T1=OMP_GET_WTIME()
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C$OMP PARALLEL DO SCHEDULE(DYNAMIC)
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DO I=1,K
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B(I)=PQUEENS(N,A(I,1),A(I,2))
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END DO
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C$OMP END PARALLEL DO
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T2=OMP_GET_WTIME()
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PRINT '(I4,I12,F12.3)',N,2*SUM(B(1:K)),T2-T1
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END DO
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END PROGRAM
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program queens
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use omp_lib
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implicit none
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integer, parameter :: long = selected_int_kind(17)
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integer, parameter :: l = 18
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integer :: n, i, j, a(l*l, 2), k, p, q
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integer(long) :: s, b(l*l)
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real(kind(1d0)) :: t1, t2
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do n = 6, l
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k = 0
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p = n/2
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q = mod(n, 2)*(p + 1)
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do i = 1, n
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do j = 1, n
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if ((abs(i - j) > 1) .and. ((i <= p) .or. ((i == q) .and. (j < i)))) then
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k = k + 1
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a(k, 1) = i
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a(k, 2) = j
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end if
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end do
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end do
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s = 0
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t1 = omp_get_wtime()
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!$omp parallel do schedule(dynamic)
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do i = 1, k
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b(i) = pqueens(n, a(i, 1), a(i, 2))
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end do
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!$omp end parallel do
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t2 = omp_get_wtime()
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print "(I4, I12, F12.3)", n, 2*sum(b(1:k)), t2 - t1
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end do
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contains
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function pqueens(n, k1, k2) result(m)
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implicit none
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integer(long) :: m
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integer, intent(in) :: n, k1, k2
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integer, parameter :: l = 20
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integer :: a(l), s(l), u(4*l - 2)
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integer :: i, j, y, z, p, q, r
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do i = 1, n
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a(i) = i
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end do
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do i = 1, 4*n - 2
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u(i) = 0
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end do
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m = 0
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r = 2*n - 1
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if (k1 == k2) return
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p = 1 - k1 + n
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q = 1 + k1 - 1
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if ((u(p) /= 0) .or. (u(q + r) /= 0)) return
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u(p) = 1
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u(q + r) = 1
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z = a(1)
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a(1) = a(k1)
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a(k1) = z
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p = 2 - k2 + n
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q = 2 + k2 - 1
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if ((u(p) /= 0) .or. (u(q + r) /= 0)) return
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u(p) = 1
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u(q + r) = 1
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if (k2 /= 1) then
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z = a(2)
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a(2) = a(k2)
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a(k2) = z
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else
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z = a(2)
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a(2) = a(k1)
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a(k1) = z
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end if
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i = 3
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go to 40
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30 s(i) = j
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u(p) = 1
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u(q + r) = 1
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i = i + 1
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40 if (i > n) go to 80
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j = i
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50 z = a(i)
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y = a(j)
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p = i - y + n
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q = i + y - 1
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a(i) = y
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a(j) = z
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if ((u(p) == 0) .and. (u(q + r) == 0)) go to 30
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60 j = j + 1
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if (j <= n) go to 50
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70 j = j - 1
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if (j == i) go to 90
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z = a(i)
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a(i) = a(j)
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a(j) = z
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go to 70
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!valid queens position found
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80 m = m + 1
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90 i = i - 1
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if (i == 2) return
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p = i - a(i) + n
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q = i + a(i) - 1
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j = s(i)
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u(p) = 0
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u(q + r) = 0
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go to 60
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end function
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end program
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@ -6,7 +6,7 @@ file=1; rank=1; q=0 /*starting place, # of queens. */
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/*═════════════════════════════════════find solution: N queens problem.*/
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do while q<N /*keep placing queens until done.*/
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@.file.rank=1 /*place a queen on the chessboard*/
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if safe?(file,rank) then do; q=q+1 /*if not being attached, eureka! */
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if safe?(file,rank) then do; q=q+1 /*if not being attacked, eureka! */
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file=1 /*another attempt at file #1, */
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rank=rank+1 /*and also bump the rank pointer.*/
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end
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