N-queens problem
You are encouraged to solve this task according to the task description, using any language you may know.
Solve the 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 here.
- Cf.
Ada
<lang Ada>with Ada.Text_IO; use Ada.Text_IO;
procedure Queens is
Board : array (1..8, 1..8) of Boolean := (others => (others => False));
function Test (Row, Column : Integer) return Boolean is
begin
for J in 1..Column - 1 loop
if ( Board (Row, J)
or else
(Row > J and then Board (Row - J, Column - J))
or else
(Row + J <= 8 and then Board (Row + J, Column - J))
) then
return False;
end if;
end loop;
return True;
end Test;
function Fill (Column : Integer) return Boolean is
begin
for Row in Board'Range (1) loop
if Test (Row, Column) then
Board (Row, Column) := True;
if Column = 8 or else Fill (Column + 1) then
return True;
end if;
Board (Row, Column) := False;
end if;
end loop;
return False;
end Fill;
begin
if not Fill (1) then
raise Program_Error;
end if;
for I in Board'Range (1) loop
Put (Integer'Image (9 - I));
for J in Board'Range (2) loop
if Board (I, J) then
Put ("|Q");
elsif (I + J) mod 2 = 1 then
Put ("|/");
else
Put ("| ");
end if;
end loop;
Put_Line ("|");
end loop;
Put_Line (" A B C D E F G H");
end Queens;</lang> Sample output:
8|Q|/| |/| |/| |/| 7|/| |/| |/| |Q| | 6| |/| |/|Q|/| |/| 5|/| |/| |/| |/|Q| 4| |Q| |/| |/| |/| 3|/| |/|Q|/| |/| | 2| |/| |/| |Q| |/| 1|/| |Q| |/| |/| | A B C D E F G H
ALGOL 68
<lang Algol68>INT ofs = 1, # Algol68 normally uses array offset of 1 #
dim = 8; # dim X dim chess board #
[ofs:dim+ofs-1]INT b;
PROC unsafe = (INT y)BOOL:(
INT i, t, x; x := b[y]; FOR i TO y - LWB b DO t := b[y - i]; IF t = x THEN break true ELIF t = x - i THEN break true ELIF t = x + i THEN break true FI OD; FALSE EXIT
break true:
TRUE
);
INT s := 0;
PROC print board = VOID:(
INT x, y;
print((new line, "Solution # ", s+:=1, new line));
FOR y FROM LWB b TO UPB b DO
FOR x FROM LWB b TO UPB b DO
print("|"+(b[y]=x|"Q"|: ODD(x+y)|"/"|" "))
OD;
print(("|", new line))
OD
);
main: (
INT y := LWB b;
b[LWB b] := LWB b - 1;
FOR i WHILE y >= LWB b DO
WHILE
b[y]+:=1;
# BREAK # IF b[y] <= UPB b THEN unsafe(y) ELSE FALSE FI
DO SKIP OD;
IF b[y] <= UPB b THEN
IF y < UPB b THEN
b[y+:=1] := LWB b - 1
ELSE
print board
FI
ELSE
y-:=1
FI
OD
)</lang>
AutoHotkey
Output to formatted Message box
<lang AutoHotkey>;
- Post
- http://www.autohotkey.com/forum/viewtopic.php?p=353059#353059
- Timestamp
- 05/may/2010
MsgBox % funcNQP(5) MsgBox % funcNQP(8)
Return
- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
- ** USED VARIABLES **
- Global
- All variables named Array[???]
- Function funcNPQ
- nQueens , OutText , qIndex
- Function Unsafe
- nIndex , Idx , Tmp , Aux
- Function PutBoard
- Output , QueensN , Stc , xxx , yyy
- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
funcNQP(nQueens) {
Global
Array[0] := -1
Local OutText , qIndex := 0
While ( qIndex >= 0 )
{
Array[%qIndex%]++
While ( (Array[%qIndex%] < nQueens) && Unsafe(qIndex) )
Array[%qIndex%]++
If ( Array[%qIndex%] < nQueens )
{
If ( qIndex < nQueens-1 )
qIndex++ , Array[%qIndex%] := -1
Else
PutBoard(OutText,nQueens)
}
Else
qIndex--
}
Return OutText
}
- ------------------------------------------
Unsafe(nIndex) {
Global
Local Idx := 1 , Tmp := 0 , Aux := Array[%nIndex%]
While ( Idx <= nIndex )
{
Tmp := "Array[" nIndex - Idx "]"
Tmp := % %Tmp%
If ( ( Tmp = Aux ) || ( Tmp = Aux-Idx ) || ( Tmp = Aux+Idx ) )
Return 1
Idx++
}
Return 0
}
- ------------------------------------------
PutBoard(ByRef Output,QueensN) {
Global
Static Stc = 0
Local xxx := 0 , yyy := 0
Output .= "`n`nSolution #" (++Stc) "`n"
While ( yyy < QueensN )
{
xxx := 0
While ( xxx < QueensN )
Output .= ( "|" ( ( Array[%yyy%] = xxx ) ? "Q" : "_" ) ) , xxx++
Output .= "|`n" , yyy++
}
}</lang>
Includes a solution browser GUI
This implementation supports N = 4..12 queens, and will find ALL solutions for each of the different sizes. The screenshot shows the first solution of 10 possible solutions for N = 5 queens. <lang AutoHotkey>N := 5 Number: ; main entrance for different # of queens
SI := 1 Progress b2 w250 zh0 fs9, Calculating all solutions for %N% Queens ... Gosub GuiCreate Result := SubStr(Queens(N),2) Progress Off Gui Show,,%N%-Queens StringSplit o, Result, `n
Fill: ; show solutions
GuiControl,,SI, %SI% / %o0%
Loop Parse, o%SI%, `,
{
C := A_Index
Loop %N%
GuiControl,,%C%_%A_Index% ; clear fields
GuiControl,,%C%_%A_LoopField%, r
}
Return ;-----------------------------------------------------------------------
Queens(N) { ; Size of the board
Local c, O ; global array r r1 := 1, c := 2, r2 := 3, O := "" ; init: r%c% = row of Queen in column c
Right: ; move to next column
If (c = N) { ; found solution
Loop %N% ; save row indices of Queens
O .= (A_Index = 1 ? "`n" : ",") r%A_Index%
GOTO % --c ? "Down" : "OUT" ; for ALL solutions
}
c++, r%c% := 1 ; next column, top row
GoTo % BAD(c) ? "Down" : "Right"
Down: ; move down to next row
If (r%c% = N)
GoTo % --c ? "Down" : "OUT"
r%c%++ ; row down
GoTo % BAD(c) ? "Down" : "Right"
OUT:
Return O
} ;----------------------------------------------------------------------------
BAD(c) { ; Check placed Queens against Queen in row r%c%, column c
Loop % c-1
If (r%A_Index% = r%c% || ABS(r%A_Index%-r%c%) = c-A_Index)
Return 1
} ;----------------------------------------------------------------------------
GuiCreate: ; Draw chess board
Gui Margin, 20, 15
Gui Font, s16, Marlett
Loop %N% {
C := A_Index
Loop %N% { ; fields
R := A_Index, X := 40*C-17, Y := 40*R-22
Gui Add, Progress, x%X% y%Y% w41 h41 Cdddddd, % 100*(R+C & 1) ;% shade fields
Gui Add, Text, x%X% y%Y% w41 h41 BackGroundTrans Border Center 0x200 v%C%_%R%
}
}
Gui Add, Button, x%x% w43 h25 gBF, 4 ; forth (default)
Gui Add, Button,xm yp w43 h25 gBF, 3 ; back
Gui Font, bold, Comic Sans MS Gui Add, Text,% "x62 yp hp Center 0x200 vSI w" 40*N-80
Menu FileMenu, Add, E&xit, GuiClose
Loop 9
Menu CalcMenu, Add, % "Calculate " A_Index+3 " Queens", Calculate ;%
Menu HelpMenu, Add, &About, AboutBox
Menu MainMenu, Add, &File, :FileMenu
Menu MainMenu, Add, &Calculate, :CalcMenu
Menu MainMenu, Add, &Help, :HelpMenu
Gui Menu, Mainmenu
Return ; ----------------------------------------------------------------------
AboutBox: ; message box with AboutText
Gui 1: +OwnDialogs MsgBox, 64, About N-Queens, Many thanks ...
Return
Calculate: ; menu handler for calculations
N := A_ThisMenuItemPos + 3 Gui Destroy GoTo Number ; -------------------------------------------------------------
BF:
SI := mod(SI+o0-2*(A_GuiControl=3), o0) + 1 ; left button text is "3" GoTo Fill ; ----------------------------------------------------------------
GuiClose:
ExitApp</lang>
BBC BASIC
The total number of solutions is displayed in the title bar and one solution is displayed. The code could be adapted to display a selected solution or multiple solutions.
<lang bbcbasic> Size% = 8
Cell% = 32
VDU 23,22,Size%*Cell%;Size%*Cell%;Cell%,Cell%,16,128+8,5
*font Arial Unicode MS,16
GCOL 3,11
FOR i% = 0 TO Size%-1 STEP 2
RECTANGLE FILL i%*Cell%*2,0,Cell%*2,Size%*Cell%*2
RECTANGLE FILL 0,i%*Cell%*2,Size%*Cell%*2,Cell%*2
NEXT
num% = FNqueens(Size%, Cell%)
SYS "SetWindowText", @hwnd%, "Total " + STR$(num%) + " solutions"
REPEAT : WAIT 1 : UNTIL FALSE
END
DEF FNqueens(n%, s%)
LOCAL i%, j%, m%, p%, q%, r%, a%(), b%(), c%()
DIM a%(n%), b%(n%), c%(4*n%-2)
FOR i% = 1 TO DIM(a%(),1) : a%(i%) = i% : NEXT
m% = 0
i% = 1
j% = 0
r% = 2*n%-1
REPEAT
i% -= 1
j% += 1
p% = 0
q% = -r%
REPEAT
i% += 1
c%(p%) = 1
c%(q%+r%) = 1
SWAP a%(i%),a%(j%)
p% = i% - a%(i%) + n%
q% = i% + a%(i%) - 1
b%(i%) = j%
j% = i% + 1
UNTIL j% > n% OR c%(p%) OR c%(q%+r%)
IF c%(p%)=0 IF c%(q%+r%)=0 THEN
IF m% = 0 THEN
FOR p% = 1 TO n%
MOVE 2*s%*(a%(p%)-1)+6, 2*s%*p%+6
PRINT "β";
NEXT
ENDIF
m% += 1
ENDIF
j% = b%(i%)
WHILE j% >= n% AND i% <> 0
REPEAT
SWAP a%(i%), a%(j%)
j% = j%-1
UNTIL j% < i%
i% -= 1
p% = i% - a%(i%) + n%
q% = i% + a%(i%) - 1
j% = b%(i%)
c%(p%) = 0
c%(q%+r%) = 0
ENDWHILE
UNTIL i% = 0
= m%</lang>
BCPL
<lang BCPL>// This can be run using Cintcode BCPL freely available from www.cl.cam.ac.uk/users/mr10.
GET "libhdr.h"
GLOBAL { count:ug; all }
LET try(ld, row, rd) BE TEST row=all
THEN count := count + 1
ELSE { LET poss = all & ~(ld | row | rd)
WHILE poss DO
{ LET p = poss & -poss
poss := poss - p
try(ld+p << 1, row+p, rd+p >> 1)
}
}
LET start() = VALOF { all := 1
FOR i = 1 TO 16 DO
{ count := 0
try(0, 0, 0)
writef("Number of solutions to %i2-queens is %i7*n", i, count)
all := 2*all + 1
}
RESULTIS 0
} </lang> The following is a re-implementation of the algorithm given above but using the MC package that allows machine independent runtime generation of native machine code (currently only available for i386 machines). It runs about 25 times faster that the version given above.
<lang BCPL> GET "libhdr.h" GET "mc.h"
MANIFEST {
lo=1; hi=16 dlevel=#b0000
// Register mnemonics ld = mc_a row = mc_b rd = mc_c poss = mc_d p = mc_e count = mc_f
}
LET start() = VALOF { // Load the dynamic code generation package
LET mcseg = globin(loadseg("mci386"))
LET mcb = 0
UNLESS mcseg DO
{ writef("Trouble with MC package: mci386*n")
GOTO fin
}
// Create an MC instance for hi functions with a data space // of 10 words and code space of 40000 mcb := mcInit(hi, 10, 40000)
UNLESS mcb DO
{ writef("Unable to create an mci386 instance*n")
GOTO fin
}
mc := 0 // Currently no selected MC instance mcSelect(mcb)
mcK(mc_debug, dlevel) // Set the debugging level
FOR n = lo TO hi DO
{ mcComment("*n*n// Code for a %nx%n board*n", n, n)
gencode(n) // Compile the code for an nxn board
}
mcF(mc_end) // End of code generation
writef("Code generation complete*n")
FOR n = lo TO hi DO
{ LET k = mcCall(n)
writef("Number of solutions to %i2-queens is %i9*n", n, k)
}
fin:
IF mc DO mcClose() IF mcseg DO unloadseg(mcseg)
writef("*n*nEnd of run*n")
}
AND gencode(n) BE { LET all = (1<<n) - 1
mcKKK(mc_entry, n, 3, 0)
mcRK(mc_mv, ld, 0) mcRK(mc_mv, row, 0) mcRK(mc_mv, rd, 0) mcRK(mc_mv, count, 0)
cmpltry(1, n, all) // Compile the outermost call of try
mcRR(mc_mv, mc_a, count) // return count mcF(mc_rtn) mcF(mc_endfn)
}
AND cmpltry(i, n, all) BE { LET L = mcNextlab()
mcComment("*n// Start of code from try(%n, %n, %n)*n", i, n, all)
mcRR(mc_mv, poss, ld) // LET poss = (~(ld | row | rd)) & all mcRR(mc_or, poss, row) mcRR(mc_or, poss, rd) mcR (mc_not, poss) mcRK(mc_and, poss, all)
mcRK(mc_cmp, poss, 0) // IF poss DO TEST n-i<=2 THEN mcJS(mc_jeq, L) // (use a short jump if near the last row) ELSE mcJL(mc_jeq, L)
TEST i=n
THEN { // We can place a queen in the final row.
mcR(mc_inc, count) // count := count+1
}
ELSE { // We can place queen(s) in a non final row.
LET M = mcNextlab()
mcL (mc_lab, M) // { Start of REPEATWHILE loop
mcRR(mc_mv, p, poss) // LET p = poss & -poss
mcR (mc_neg, p)
mcRR(mc_and, p, poss) // // p is a valid queens position
mcRR(mc_sub, poss, p) // poss := poss - p
mcR (mc_push, ld) // Save current state
mcR (mc_push, row)
mcR (mc_push, rd)
mcR (mc_push, poss)
// Call try((ld+p)<<1, row+p, (rd+p)>>1)
mcRR(mc_add, ld, p)
mcRK(mc_lsh, ld, 1) // ld := (ld+p)<<1
mcRR(mc_add, row, p) // row := row+p
mcRR(mc_add, rd, p)
mcRK(mc_rsh, rd, 1) // rd := (rd+p)>>1
cmpltry(i+1, n, all) // Compile code for row i+1
mcR (mc_pop, poss) // Restore the state
mcR (mc_pop, rd)
mcR (mc_pop, row)
mcR (mc_pop, ld)
mcRK(mc_cmp, poss, 0)
mcJL(mc_jne, M) // } REPEATWHILE poss
}
mcL(mc_lab, L)
mcComment("// End of code from try(%n, %n, %n)*n*n",
i, n, all)
} </lang>
C
C99, compiled with gcc -std=c99 -Wall. Take one commandline argument: size of board, or default to 8. Shows the board layout for each solution.<lang C>#include <stdio.h>
- include <stdlib.h>
int count = 0; void solve(int n, int col, int *hist) { if (col == n) { printf("\nNo. %d\n-----\n", ++count); for (int i = 0; i < n; i++, putchar('\n')) for (int j = 0; j < n; j++) putchar(j == hist[i] ? 'Q' : ((i + j) & 1) ? ' ' : '.');
return; }
- define attack(i, j) (hist[j] == i || abs(hist[j] - i) == col - j)
for (int i = 0, j = 0; i < n; i++) { for (j = 0; j < col && !attack(i, j); j++); if (j < col) continue;
hist[col] = i; solve(n, col + 1, hist); } }
int main(int n, char **argv) { if (n <= 1 || (n = atoi(argv[1])) <= 0) n = 8; int hist[n]; solve(n, 0, hist); }</lang>
Similiar to above, but using bits to save board configurations and quite a bit faster:<lang c>#include <stdio.h>
- include <stdlib.h>
- include <stdint.h>
typedef uint32_t uint; uint full, *qs, count = 0, nn;
void solve(uint d, uint c, uint l, uint r) { uint b, a, *s; if (!d) { count++;
- if 0
printf("\nNo. %d\n===========\n", count); for (a = 0; a < nn; a++, putchar('\n')) for (b = 0; b < nn; b++, putchar(' ')) putchar(" -QQ"[((b == qs[a])<<1)|((a + b)&1)]);
- endif
return; }
a = (c | (l <<= 1) | (r >>= 1)) & full; if (a != full) for (*(s = qs + --d) = 0, b = 1; b <= full; (*s)++, b <<= 1) if (!(b & a)) solve(d, b|c, b|l, b|r); }
int main(int n, char **argv) { if (n <= 1 || (nn = atoi(argv[1])) <= 0) nn = 8;
qs = calloc(nn, sizeof(int)); full = (1U << nn) - 1;
solve(nn, 0, 0, 0); printf("\nSolutions: %d\n", count); return 0; }</lang> Take that and unwrap the recursion, plus some heavy optimizations, and we have a very fast and very unreadable solution: <lang c>#include <stdio.h>
- include <stdlib.h>
typedef unsigned int uint; uint count = 0;
- define ulen sizeof(uint) * 8
/* could have defined as int solve(...), but void may have less
chance to confuse poor optimizer */
void solve(int n) { int cnt = 0; const uint full = -(int)(1 << (ulen - n)); register uint bits, pos, *m, d, e;
uint b0, b1, l[32], r[32], c[32], mm[33] = {0}; n -= 3; /* require second queen to be left of the first queen, so we ever only test half of the possible solutions. This is why we can't handle n=1 here */ for (b0 = 1U << (ulen - n - 3); b0; b0 <<= 1) { for (b1 = b0 << 2; b1; b1 <<= 1) { d = n; /* c: columns occupied by previous queens. l: columns attacked by left diagonals r: by right diagnoals */ c[n] = b0 | b1; l[n] = (b0 << 2) | (b1 << 1); r[n] = (b0 >> 2) | (b1 >> 1);
/* availabe columns on current row. m is stack */ bits = *(m = mm + 1) = full & ~(l[n] | r[n] | c[n]);
while (bits) { /* d: depth, aka row. counting backwards because !d is often faster than d != n */ while (d) { /* pos is right most nonzero bit */ pos = -(int)bits & bits;
/* mark bit used. only put current bits on stack if not zero, so backtracking will skip exhausted rows (because reading stack variable is sloooow compared to registers) */ if ((bits &= ~pos)) *m++ = bits | d;
/* faster than l[d+1] = l[d]... */ e = d--; l[d] = (l[e] | pos) << 1; r[d] = (r[e] | pos) >> 1; c[d] = c[e] | pos;
bits = full & ~(l[d] | r[d] | c[d]);
if (!bits) break; if (!d) { cnt++; break; } } /* Bottom of stack m is a zero'd field acting as sentinel. When saving to stack, left 27 bits are the available columns, while right 5 bits is the depth. Hence solution is limited to size 27 board -- not that it matters in foreseeable future. */ d = (bits = *--m) & 31U; bits &= ~31U; } } } count = cnt * 2; }
int main(int c, char **v) { int nn; if (c <= 1 || (nn = atoi(v[1])) <= 0) nn = 8;
if (nn > 27) { fprintf(stderr, "Value too large, abort\n"); exit(1); }
/* Can't solve size 1 board; might as well skip 2 and 3 */ if (nn < 4) count = nn == 1; else solve(nn);
printf("\nSolutions: %d\n", count); return 0; }</lang>
C++
<lang cpp>
- include <windows.h>
- include <iostream>
- include <string>
//-------------------------------------------------------------------------------------------------- using namespace std;
//-------------------------------------------------------------------------------------------------- class point { public:
int x, y;
point(){ x = y = 0; }
void set( int a, int b ){ x = a; y = b; }
}; //-------------------------------------------------------------------------------------------------- class nQueens { public:
void solve( int c )
{
_count = c; int len = ( c + 1 ) * ( c + 1 ); _queens = new bool[len]; memset( _queens, 0, len );
_cl = new bool[c]; memset( _cl, 0, c ); _ln = new bool[c]; memset( _ln, 0, c ); point pt; pt.set( rand() % c, rand() % c ); putQueens( pt, c ); displayBoard(); delete [] _queens; delete [] _ln; delete [] _cl;
}
private:
void displayBoard()
{
system( "cls" ); string t = "+---+", q = "| Q |", s = "| |"; COORD c = { 0, 0 }; HANDLE h = GetStdHandle( STD_OUTPUT_HANDLE ); for( int y = 0, cy = 0; y < _count; y++ ) { int yy = y * _count; for( int x = 0; x < _count; x++ ) { SetConsoleCursorPosition( h, c ); cout << t; c.Y++; SetConsoleCursorPosition( h, c ); if( _queens[x + yy] ) cout << q; else cout << s; c.Y++; SetConsoleCursorPosition( h, c ); cout << t; c.Y = cy; c.X += 4; } cy += 2; c.X = 0; c.Y = cy;
} }
bool checkD( int x, int y, int a, int b )
{
if( x < 0 || y < 0 || x >= _count || y >= _count ) return true; if( _queens[x + y * _count] ) return false; if( checkD( x + a, y + b, a, b ) ) return true; return false;
}
bool check( int x, int y )
{
if( _ln[y] || _cl[x] ) return false; if( !checkD( x, y, -1, -1 ) ) return false; if( !checkD( x, y, 1, -1 ) ) return false; if( !checkD( x, y, -1, 1 ) ) return false; if( !checkD( x, y, 1, 1 ) ) return false; return true;
}
bool putQueens( point pt, int cnt )
{
int it = _count; while( it ) { if( !cnt ) return true; if( check( pt.x, pt.y ) ) { _queens[pt.x + pt.y * _count] = _cl[pt.x] = _ln[pt.y] = true; point tmp = pt; if( ++tmp.x >= _count ) tmp.x = 0; if( ++tmp.y >= _count ) tmp.y = 0; if( putQueens( tmp, cnt - 1 ) ) return true; _queens[pt.x + pt.y * _count] = _cl[pt.x] = _ln[pt.y] = false; } if( ++pt.x >= _count ) pt.x = 0; it--; } return false;
}
int _count; bool* _queens, *_ln, *_cl;
}; //-------------------------------------------------------------------------------------------------- int main( int argc, char* argv[] ) {
nQueens n; int nq;
while( true )
{
system( "cls" ); cout << "Enter board size bigger than 3 (0 - 3 to QUIT): "; cin >> nq; if( nq < 4 ) return 0; n.solve( nq ); cout << endl << endl; system( "pause" );
} return 0;
} //-------------------------------------------------------------------------------------------------- </lang> Output:
+---+---+---+---+---+ +---+---+---+---+---+---+---+---+ +---+---+---+---+---+---+---+---+---+---+---+
| Q | | | | | | | | | Q | | | | | | | | | | | | | Q | | | |
+---+---+---+---+---+ +---+---+---+---+---+---+---+---+ +---+---+---+---+---+---+---+---+---+---+---+
| | | Q | | | | | Q | | | | | | | | | | | | | | | | | Q | |
+---+---+---+---+---+ +---+---+---+---+---+---+---+---+ +---+---+---+---+---+---+---+---+---+---+---+
| | | | | Q | | | | | | | | | Q | | | | | Q | | | | | | | |
+---+---+---+---+---+ +---+---+---+---+---+---+---+---+ +---+---+---+---+---+---+---+---+---+---+---+
| | Q | | | | | | | | | | Q | | | | | Q | | | | | | | | | |
+---+---+---+---+---+ +---+---+---+---+---+---+---+---+ +---+---+---+---+---+---+---+---+---+---+---+
| | | | Q | | | Q | | | | | | | | | | | | | Q | | | | | | |
+---+---+---+---+---+ +---+---+---+---+---+---+---+---+ +---+---+---+---+---+---+---+---+---+---+---+
| | | Q | | | | | | | | | | | | | | | | | Q |
+---+---+---+---+---+---+---+---+ +---+---+---+---+---+---+---+---+---+---+---+
| | | | | Q | | | | | Q | | | | | | | | | | |
+---+---+---+---+---+---+---+---+ +---+---+---+---+---+---+---+---+---+---+---+
| | | | | | | Q | | | | | Q | | | | | | | | |
+---+---+---+---+---+---+---+---+ +---+---+---+---+---+---+---+---+---+---+---+
| | | | | | Q | | | | | |
+---+---+---+---+---+---+---+---+---+---+---+
| | | | | | | | | Q | | |
+---+---+---+---+---+---+---+---+---+---+---+
| | | | | | | Q | | | | |
+---+---+---+---+---+---+---+---+---+---+---+
Version using Heuristics - explained here: Solution_construction <lang cpp>
- include <windows.h>
- include <iostream>
- include <string>
- include <vector>
- include <algorithm>
//-------------------------------------------------------------------------------------------------- using namespace std;
//-------------------------------------------------------------------------------------------------- typedef unsigned int uint;
//-------------------------------------------------------------------------------------------------- class nQueens_Heuristic { public:
void solve( uint n ) { makeList( n ); drawBoard( n ); }
private:
void drawBoard( uint n )
{
system( "cls" ); string t = "+---+", q = "| Q |", s = "| |"; COORD c = { 0, 0 }; HANDLE h = GetStdHandle( STD_OUTPUT_HANDLE ); uint w = 0; for( uint y = 0, cy = 0; y < n; y++ ) { for( uint x = 0; x < n; x++ ) { SetConsoleCursorPosition( h, c ); cout << t; c.Y++; SetConsoleCursorPosition( h, c ); if( x + 1 == solution[w] ) cout << q; else cout << s; c.Y++; SetConsoleCursorPosition( h, c ); cout << t; c.Y = cy; c.X += 4; } cy += 2; c.X = 0; c.Y = cy; w++; } solution.clear(); odd.clear(); evn.clear();
}
void makeList( uint n )
{
uint r = n % 6; for( uint x = 1; x <= n; x++ ) { if( x & 1 ) odd.push_back( x ); else evn.push_back( x ); } if( r == 2 ) { swap( odd[0], odd[1] ); odd.erase( find( odd.begin(), odd.end(), 5 ) ); odd.push_back( 5 ); } else if( r == 3 ) { odd.erase( odd.begin() ); odd.erase( odd.begin() ); odd.push_back( 1 ); odd.push_back( 3 ); evn.erase( evn.begin() ); evn.push_back( 2 ); } vector<uint>::iterator it = evn.begin(); while( it != evn.end() ) { solution.push_back( ( *it ) ); it++; } it = odd.begin(); while( it != odd.end() ) { solution.push_back( ( *it ) ); it++; }
}
vector<uint> odd, evn, solution;
}; //-------------------------------------------------------------------------------------------------- int main( int argc, char* argv[] ) {
uint n; nQueens_Heuristic nQH;
while( true )
{
cout << "Enter board size bigger than 3 (0 - 3 to QUIT): "; cin >> n; if( n < 4 ) return 0; nQH.solve( n ); cout << endl << endl;
} return 0;
} //-------------------------------------------------------------------------------------------------- </lang>
C#
<lang csharp>using System; using System.Collections.Generic; using System.Linq; using System.Text;
namespace NQueens {
class Program
{
const int N = 8;
static bool Allowed(bool[,] board, int x, int y)
{
for (int i=0; i<=x; i++)
{
if (board[i,y] || (i <= y && board[x-i,y-i]) || (y+i < N && board[x-i,y+i]))
{
return false;
}
}
return true;
}
static bool FindSolution(bool[,] board, int x)
{
for (int y = 0; y < N; y++)
{
if (Allowed(board, x, y))
{
board[x, y] = true;
if (x == N-1 || FindSolution(board, x + 1))
{
return true;
}
board[x, y] = false;
}
}
return false;
}
static void Main(string[] args)
{
bool[,] board = new bool[N, N];
if (FindSolution(board, 0))
{
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
{
Console.Write(board[i, j] ? "|Q" : "| ");
}
Console.WriteLine("|");
}
}
else
{
Console.WriteLine("No solution found for n = " + N + ".");
}
Console.ReadKey(true);
}
}
}</lang>
Clojure
This produces all solutions by essentially a backtracking algorithm. The heart is the extends? function, which takes a partial solution for the first k<size columns and sees if the solution can be extended by adding a queen at row n of column k+1. The extend function takes a list of all partial solutions for k columns and produces a list of all partial solutions for k+1 columns. The final list solutions is calculated by starting with the list of 0-column solutions (obviously this is the list [ [] ], and iterates extend for size times. <lang clojure>(def size 8)
(defn extends? [v n]
(let [k (count v)]
(not-any? true?
(for [i (range k) :let [vi (v i)]]
(or
(= vi n) ;check for shared row
(= (- k i) (Math/abs (- n vi)))))))) ;check for shared diagonal
(defn extend [vs]
(for [v vs
n (range 1 (inc size)) :when (extends? v n)]
(conj v n)))
(def solutions
(nth (iterate extend [[]]) size))
(doseq [s solutions]
(println s))
(println (count solutions) "solutions")</lang>
CoffeeScript
<lang coffeescript>
- Unlike traditional N-Queens solutions that use recursion, this
- program attempts to more closely model the "human" algorithm.
- In this algorithm, the function keeps placing queens on the board
- until there is no longer a safe square. If the 8th queen has been
- placed, the solution is noted. If fewer than 8th queens have been
- placed, then you are at a dead end. In either case, backtracking occurs.
- The LAST queen placed on the board gets pulled, then it gets moved
- to the next safe square. (We backtrack even after a "good" attempt in
- order to get to a new solution.) This backtracking may repeat itself
- several times until the original misplaced queen finally is proven to
- be a dead end.
- Many N-Queens solutions use lazy logic (along with geometry shortcuts)
- to determine whether a queen is under attack. In this algorithm, we
- are more proactive, essentially updating a sieve every time we lay a
- queen down. To make backtracking easier, the sieve uses ref-counts vs.
- a simple safe/unsafe boolean.
- We precompute the "attack graph" up front, and then we essentially ignore
- the geometry of the problem. This approach, while perhaps suboptimal for
- queens, probably is more flexible for general "coexistence" problems.
nqueens = (n) ->
neighbors = precompute_neighbors(n)
board = [] num_solutions = 0 num_backtracks = 0 queens = [] pos = 0
for p in [0...n*n]
board.push 0
attack = (pos, delta=1) ->
for neighbor in neighbors[pos]
board[neighbor] += delta
backtrack = ->
pos = queens.pop()
attack pos, -1 # unattack queen you just pulled
pos += 1
num_backtracks += 1
# The following loop finds all 92 solutions to
# the 8-queens problem (for n=8).
while true
if pos >= n*n
if queens.length == 0
break
backtrack()
continue
# If a square is empty
if board[pos] == 0
attack pos
queens.push pos
if queens.length == n
num_solutions += 1
show_queens queens, n
backtrack()
pos += 1
console.log "#{num_solutions} solutions"
console.log "#{num_backtracks} backtracks"
precompute_neighbors = (n) ->
# For each board position, build a list of all # the board positions that would be under attack if # you placed a queen on it. This assumes a 1d array # of squares. neighbors = []
find_neighbors = (pos) ->
arr = []
row = Math.floor pos / n
col = pos % n
for i in [0...n]
if i != col
arr.push row*n + i
r1 = row + col - i
r2 = row + i - col
if 0 <= r1 and r1 < n
arr.push r1*n + i
if 0 <= r2 and r2 < n
arr.push r2*n + i
if i != row
arr.push i*n + col
arr
for pos in [0...n*n] neighbors.push find_neighbors(pos) neighbors
show_queens = (queens, n) ->
# precondition: queens is a sorted array of integers,
# and each row is represented
console.log "\n------"
for q in queens
col = q % n
s =
for c in [0...n]
if c == col
s += "Q "
else
s += "* "
console.log s + "\n"
nqueens(8) </lang>
Common Lisp
<lang lisp>(defun n-queens (n m)
(if (= n 1) (loop for x from 1 to m collect (list x)) (loop for sol in (n-queens (1- n) m) nconc
(loop for col from 1 to m when (loop for row from 0 to (length sol) for c in sol always (and (/= col c) (/= (abs (- c col)) (1+ row))) finally (return (cons col sol))) collect it))))
(defun show-solution (b n)
(loop for i in b do
(format t "~{~A~^~}~%" (loop for x from 1 to n collect (if (= x i) "Q " ". "))))
(terpri))
(let ((i 0) (n 8))
(mapc #'(lambda (s)
(format t "Solution ~a:~%" (incf i)) (show-solution s n)) (n-queens n n)))</lang>
Alternate solution
Translation of Fortran 77 <lang lisp>(defun queens (nmax)
(let ((a (make-array `(,nmax)))
(s (make-array `(,nmax)))
(u (make-array `(,(- (* 4 nmax) 2)) :initial-element 0))
y z i j p q r m (v nil))
(dotimes (i nmax) (setf (aref a i) i))
(loop for n from 1 to nmax do
(tagbody
(setf m 0 i 0 r (1- (* 2 n)))
(go L40)
L30
(setf (aref s i) j (aref u p) 1 (aref u (+ q r)) 1)
(incf i)
L40
(if (>= i n) (go L80))
(setf j i)
L50
(setf y (aref a j) z (aref a i))
(setf p (+ (- i y) (1- n)) q (+ i y))
(setf (aref a i) y (aref a j) z)
(if (and (zerop (aref u p)) (zerop (aref u (+ q r)))) (go L30))
L60
(incf j)
(if (< j n) (go L50))
L70
(decf j)
(if (= j i) (go L90))
(rotatef (aref a i) (aref a j))
(go L70)
L80
(incf m)
L90
(decf i)
(if (minusp i) (go L100))
(setf p (+ (- i (aref a i)) (1- n)) q (+ i (aref a i)) j (aref s i))
(setf (aref u p) 0 (aref u (+ q r)) 0)
(go L60)
L100
;(princ n) (princ " ") (princ m) (terpri)
(push (cons n m) v)
)) (reverse v)))
> (queens 14) ((1 . 1) (2 . 0) (3 . 0) (4 . 2) (5 . 10) (6 . 4) (7 . 40) (8 . 92) (9 . 352)
(10 . 724) (11 . 2680) (12 . 14200) (13 . 73712) (14 . 365596))</lang>
Curry
Three different ways of attacking the same problem. All copied from A Catalog of Design Patterns in FLP <lang curry> -- 8-queens implementation with the Constrained Constructor pattern -- Sergio Antoy -- Fri Jul 13 07:05:32 PDT 2001
-- Place 8 queens on a chessboard so that no queen can capture -- (and be captured by) any other queen.
-- Non-deterministic choice operator
infixl 0 ! X ! _ = X _ ! Y = Y
-- A solution is represented by a list of integers. -- The i-th integer in the list is the column of the board -- in which the queen in the i-th row is placed. -- Rows and columns are numbered from 1 to 8. -- For example, [4,2,7,3,6,8,5,1] is a solution where the -- the queen in row 1 is in column 4, etc. -- Any solution must be a permutation of [1,2,...,8].
-- The state of a queen is its position, row and column, on the board. -- Operation column is a particularly simple instance -- of a Constrained Constructor pattern. -- When it is invoked, it produces only valid states.
column = 1 ! 2 ! 3 ! 4 ! 5 ! 6 ! 7 ! 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.
-- Constrained Constructor on a path -- Any path must be valid, i.e., any column must be in the range 1..8 -- and different from any other column in the path. -- Furthermore, the path must be safe for the queens. -- No queen in a path may capture any other queen in the path. -- Operation makePath add column n to path c or fails.
makePath c n | valid c && safe c 1 = n:c
where valid c | n =:= column = uniq c
where uniq [] = True
uniq (c:cs) = n /= c && uniq cs
safe [] _ = True
safe (c:cs) k = abs (n-c) /= k && safe cs (k+1)
where abs x = if x < 0 then -x else x
-- extend the path argument till all the queens are on the board -- see the Incremental Solution pattern
extend p = if (length p == 8)
then p
else extend (makePath p x)
where x free
-- solve the puzzle
main = extend [] </lang>
Another approach from the same source.
<lang curry> -- N-queens puzzle implemented with "Distinct Choices" pattern -- Sergio Antoy -- Tue Sep 4 13:16:20 PDT 2001 -- updated: Mon Sep 23 15:22:15 PDT 2002
import Integer
queens x | y =:= permute x & void (capture y) = y where y free
capture y = let l1,l2,l3,y1,y2 free in
l1 ++ [y1] ++ l2 ++ [y2] ++ l3 =:= y & abs (y1-y2) =:= length l2 + 1
-- negation as failure (implemented by encapsulated search): void c = (findall \_->c) =:= []
-- How does this permutation algorithm work? -- Only the elements [0,1,...,n-1] can be permuted. -- The reason is that each element is used as an index in a list. -- A list, called store, of free variables of length n is created. -- Then, the n iterations described below are executed. -- At the i-th iteration, an element, say s, -- of the initial list is non-deterministically selected. -- This element is used as index in the store. -- The s-th variable of the store is unified with i. -- At the end of the iterations, the elements of the store -- are a permutation of [0,1,...,n-1], i.e., the elements -- are unique since two iterations cannot select the same index.
permute n = result n
where result n = if n==0 then [] else pick n store : result (n-1)
pick i store | store !! k =:= i = k where k = range n
range n | n > 0 = range (n-1) ! (n-1)
store = free
-- end </lang>
Yet another approach, also from the same source.
<lang curry> -- 8-queens implementation with both the Constrained Constructor -- and the Fused Generate and Test patterns. -- Sergio Antoy -- Fri Jul 13 07:05:32 PDT 2001
-- Place 8 queens on a chessboard so that no queen can capture -- (and be captured by) any other queen.
-- Non-deterministic choice operator
infixl 0 ! X ! _ = X _ ! Y = Y
-- A solution is represented by a list of integers. -- The i-th integer in the list is the column of the board -- in which the queen in the i-th row is placed. -- Rows and columns are numbered from 1 to 8. -- For example, [4,2,7,3,6,8,5,1] is a solution where the -- the queen in row 1 is in column 4, etc. -- Any solution must be a permutation of [1,2,...,8].
-- The state of a queen is its position, row and column, on the board. -- Operation column is a particularly simple instance -- of a Constrained Constructor pattern. -- When it is invoked, it produces only valid states.
column = 1 ! 2 ! 3 ! 4 ! 5 ! 6 ! 7 ! 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.
-- Constrained Constructor on a path -- Any path must be valid, i.e., any column must be in the range 1..8 -- and different from any other column in the path. -- Furthermore, the path must be safe for the queens. -- No queen in a path may capture any other queen in the path. -- Operation makePath add column n to path c or fails.
makePath c n | valid c && safe c 1 = n:c
where valid c | n =:= column = uniq c
where uniq [] = True
uniq (c:cs) = n /= c && uniq cs
safe [] _ = True
safe (c:cs) k = abs (n-c) /= k && safe cs (k+1)
where abs x = if x < 0 then -x else x
-- extend the path argument till all the queens are on the board -- see the Incremental Solution pattern
extend p = if (length p == 8)
then p
else extend (makePath p x)
where x free
-- solve the puzzle
main = extend [] </lang>
D
Short Version
This high-level version uses the second solution of the Permutations task. <lang d>void main() {
import std.stdio, std.algorithm, std.range, permutations2;
enum n = 8;
n.iota.array.permutations.filter!(p =>
n.iota.map!(i => p[i] + i).array.sort().uniq.count == n &&
n.iota.map!(i => p[i] - i).array.sort().uniq.count == n)
.count.writeln;
}</lang>
- Output:
92
Intermediate Version
This version shows all the solutions.
<lang d>enum side = 8; __gshared int[side] board;
bool isUnsafe(in int y) nothrow {
immutable int x = board[y];
foreach (immutable i; 1 .. y + 1) {
immutable int t = board[y - i];
if (t == x || t == x - i || t == x + i)
return true;
}
return false;
}
void showBoard() nothrow {
import core.stdc.stdio;
static int s = 1;
printf("\nSolution #%d:\n", s++);
foreach (immutable y; 0 .. side) {
foreach (immutable x; 0 .. side)
putchar(board[y] == x ? 'Q' : '.');
putchar('\n');
}
}
void main() nothrow {
int y = 0; board[0] = -1;
while (y >= 0) {
do {
board[y]++;
} while (board[y] < side && y.isUnsafe);
if (board[y] < side) {
if (y < (side - 1))
board[++y] = -1;
else
showBoard;
} else
y--;
}
}</lang>
- Output:
Solution #1: Q....... ....Q... .......Q .....Q.. ..Q..... ......Q. .Q...... ...Q.... [...] Solution #91: .......Q ..Q..... Q....... .....Q.. .Q...... ....Q... ......Q. ...Q.... Solution #92: .......Q ...Q.... Q....... ..Q..... .....Q.. .Q...... ......Q. ....Q...
Fast Version
<lang d>import std.stdio, std.conv;
ulong nQueens(in uint nn) pure nothrow in {
assert(nn > 0 && nn <= 27,
"'side' value must be in 1 .. 27.");
} body {
if (nn < 4)
return nn == 1;
enum uint ulen = uint.sizeof * 8; immutable uint full = uint.max - ((1 << (ulen - nn)) - 1); immutable n = nn - 3;
typeof(return) count; uint[32] l=void, r=void, c=void; uint[33] mm; // mm and mmi are a stack.
// Require second queen to be left of the first queen, so
// we ever only test half of the possible solutions. This
// is why we can't handle n=1 here.
for (uint b0 = 1U << (ulen - n - 3); b0; b0 <<= 1) {
for (uint b1 = b0 << 2; b1; b1 <<= 1) {
uint d = n;
// c: columns occupied by previous queens.
c[n] = b0 | b1;
// l: columns attacked by left diagonals.
l[n] = (b0 << 2) | (b1 << 1);
// r: by right diagnoals.
r[n] = (b0 >> 2) | (b1 >> 1);
// Availabe columns on current row.
uint bits = full & ~(l[n] | r[n] | c[n]);
uint mmi = 1;
mm[mmi] = bits;
while (bits) {
// d: depth, aka row. counting backwards.
// Because !d is often faster than d != n.
while (d) {
// immutable uint pos = 1U << bits.bsf; // Slower.
immutable uint pos = -(cast(int)bits) & bits;
// Mark bit used. Only put current bits on
// stack if not zero, so backtracking will
// skip exhausted rows (because reading stack
// variable is slow compared to registers).
bits &= ~pos;
if (bits) {
mm[mmi] = bits | d;
mmi++;
}
d--;
l[d] = (l[d + 1] | pos) << 1;
r[d] = (r[d + 1] | pos) >> 1;
c[d] = c[d + 1] | pos;
bits = full & ~(l[d] | r[d] | c[d]);
if (!bits)
break;
if (!d) {
count++;
break;
}
}
// Bottom of stack m is a zero'd field acting as
// sentinel. When saving to stack, left 27 bits
// are the available columns, while right 5 bits
// is the depth. Hence solution is limited to size
// 27 board -- not that it matters in foreseeable
// future.
mmi--;
bits = mm[mmi];
d = bits & 31U;
bits &= ~31U;
}
}
}
return count * 2;
}
void main(in string[] args) {
immutable uint side = (args.length >= 2) ? args[1].to!uint : 8;
writefln("N-queens(%d) = %d solutions.", side, side.nQueens);
}</lang>
- Output:
N-queens(8) = 92 solutions.
With side = 17:
N-queens(17) = 95815104 solutions.
Run-time for side = 17 compiled with ldc2 is about 49.5 seconds.
N-queens(19) = 4968057848 solutions.
Dart
<lang dart>/** Return true if queen placement q[n] does not conflict with other queens q[0] through q[n-1]
- /
isConsistent(List q, int n) {
for (int i=0; i<n; i++) {
if (q[i] == q[n]) {
return false; // Same column
}
if ((q[i] - q[n]) == (n - i)) {
return false; // Same major diagonal
}
if ((q[n] - q[i]) == (n - i)) {
return false; // Same minor diagonal
}
}
return true;
}
/** Print out N-by-N placement of queens from permutation q in ASCII.
- /
printQueens(List q) {
int N = q.length;
for (int i=0; i<N; i++) {
StringBuffer sb = new StringBuffer();
for (int j=0; j<N; j++) {
if (q[i] == j) {
sb.add("Q ");
} else {
sb.add("* ");
}
}
print(sb.toString());
}
print("");
}
/** Try all permutations using backtracking
- /
enumerate(int N) {
var a = new List(N); _enumerate(a, 0);
}
_enumerate(List q, int n) {
if (n == q.length) {
printQueens(q);
} else {
for (int i = 0; i < q.length; i++) {
q[n] = i;
if (isConsistent(q, n)){
_enumerate(q, n+1);
}
}
}
}
void main() {
enumerate(4);
}</lang> Output:
* Q * * * * * Q Q * * * * * Q * * * Q * Q * * * * * * Q * Q * *
Erlang
Instead of spawning a new process to search for each possible solution I backtrack. <lang Erlang> -module( n_queens ).
-export( [display/1, solve/1, task/0] ).
display( Board ) -> %% Queens are in the positions in the Board list. %% Top left corner is {1, 1}, Bottom right is {N, N}. There is a queen in the max column. N = lists:max( [X || {X, _Y} <- Board] ), [display_row(Y, N, Board) || Y <- lists:seq(1, N)].
solve( N ) ->
Positions = [{X, Y} || X <- lists:seq(1, N), Y <- lists:seq(1, N)],
try
bt( N, Positions, [] )
catch
_:{ok, Board} -> Board
end.
task() ->
task( 4 ), task( 8 ).
bt( N, Positions, Board ) -> bt_reject( is_not_allowed_queen_placement(N, Board), N, Positions, Board ).
bt_accept( true, _N, _Positions, Board ) -> erlang:throw( {ok, Board} ); bt_accept( false, N, Positions, Board ) -> bt_loop( N, Positions, [], Board ).
bt_loop( _N, [], _Rejects, _Board ) -> failed; bt_loop( N, [Position | T], Rejects, Board ) -> bt( N, T ++ Rejects, [Position | Board] ), bt_loop( N, T, [Position | Rejects], Board ).
bt_reject( true, _N, _Positions, _Board ) -> backtrack; bt_reject( false, N, Positions, Board ) -> bt_accept( is_all_queens(N, Board), N, Positions, Board ).
diagonals( N, {X, Y} ) -> D1 = diagonals( N, X + 1, fun diagonals_add1/1, Y + 1, fun diagonals_add1/1 ), D2 = diagonals( N, X + 1, fun diagonals_add1/1, Y - 1, fun diagonals_subtract1/1 ), D3 = diagonals( N, X - 1, fun diagonals_subtract1/1, Y + 1, fun diagonals_add1/1 ), D4 = diagonals( N, X - 1, fun diagonals_subtract1/1, Y - 1, fun diagonals_subtract1/1 ), D1 ++ D2 ++ D3 ++ D4.
diagonals( _N, 0, _Change_x, _Y, _Change_y ) -> []; diagonals( _N, _X, _Change_x, 0, _Change_y ) -> []; diagonals( N, X, _Change_x, _Y, _Change_y ) when X > N -> []; diagonals( N, _X, _Change_x, Y, _Change_y ) when Y > N -> []; diagonals( N, X, Change_x, Y, Change_y ) -> [{X, Y} | diagonals( N, Change_x(X), Change_x, Change_y(Y), Change_y )].
diagonals_add1( N ) -> N + 1.
diagonals_subtract1( N ) -> N - 1.
display_row( Row, N, Board ) -> [io:fwrite("~s", [display_queen(X, Row, Board)]) || X <- lists:seq(1, N)], io:nl().
display_queen( X, Y, Board ) -> display_queen( lists:member({X, Y}, Board) ). display_queen( true ) -> " Q"; display_queen( false ) -> " .".
is_all_queens( N, Board ) -> N =:= erlang:length( Board ).
is_diagonal( _N, [] ) -> false; is_diagonal( N, [Position | T] ) -> Diagonals = diagonals( N, Position ), T =/= (T -- Diagonals) orelse is_diagonal( N, T ).
is_not_allowed_queen_placement( N, Board ) -> Pieces = erlang:length( Board ), {Xs, Ys} = lists:unzip( Board ), Pieces =/= erlang:length( lists:usort(Xs) ) orelse Pieces =/= erlang:length( lists:usort(Ys) ) orelse is_diagonal( N, Board ).
task( N ) ->
io:fwrite( "N = ~p. One solution.~n", [N] ), Board = solve( N ), display( Board ).
</lang>
- Output:
22> n_queens:task(). N = 4. One solution. . . Q . Q . . . . . . Q . Q . . N = 8. One solution. Q . . . . . . . . . . . . . Q . . . . . Q . . . . . . . . . . Q . Q . . . . . . . . . Q . . . . . . . . . Q . . . . Q . . . . .
Factor
<lang factor>USING: kernel sequences math math.combinatorics formatting io locals ; IN: queens
- /= ( x y -- ? ) = not ; inline
- safe? ( board q -- ? )
[let q board nth :> x
q iota [
x swap
[ board nth ] keep
q swap -
[ + /= ]
[ - /= ] 3bi and
] all?
] ;
- solution? ( board -- ? )
dup length iota [ dupd safe? ] all? nip ;
- queens ( n -- l )
iota all-permutations [ solution? ] filter ;
- .queens ( n -- )
queens
[
[ 1 + "%d " printf ] each nl
] each ;</lang>
Forth
<lang forth>variable solutions variable nodes
- bits ( n -- mask ) 1 swap lshift 1- ;
- lowBit ( mask -- bit ) dup negate and ;
- lowBit- ( mask -- bits ) dup 1- and ;
- next3 ( dl dr f files -- dl dr f dl' dr' f' )
invert >r 2 pick r@ and 2* 1+ 2 pick r@ and 2/ 2 pick r> and ;
- try ( dl dr f -- )
dup if
1 nodes +!
dup 2over and and
begin ?dup while
dup >r lowBit next3 recurse r> lowBit-
repeat
else 1 solutions +! then
drop 2drop ;
- queens ( n -- )
0 solutions ! 0 nodes ! -1 -1 rot bits try solutions @ . ." solutions, " nodes @ . ." nodes" ;
8 queens \ 92 solutions, 1965 nodes</lang>
Fortran
Using a back tracking method to find one solution <lang fortran>program Nqueens
implicit none
integer, parameter :: n = 8 ! size of board integer :: file = 1, rank = 1, queens = 0 integer :: i logical :: board(n,n) = .false.
do while (queens < n)
board(file, rank) = .true.
if(is_safe(board, file, rank)) then
queens = queens + 1
file = 1
rank = rank + 1
else
board(file, rank) = .false.
file = file + 1
do while(file > n)
rank = rank - 1
if (rank < 1) then
write(*, "(a,i0)") "No solution for n = ", n
stop
end if
do i = 1, n
if (board(i, rank)) then
file = i
board(file, rank) = .false.
queens = queens - 1
file = i + 1
exit
end if
end do
end do
end if
end do
call Printboard(board)
contains
function is_safe(board, file, rank)
logical :: is_safe
logical, intent(in) :: board(:,:)
integer, intent(in) :: file, rank
integer :: i, f, r
is_safe = .true.
do i = rank-1, 1, -1
if(board(file, i)) then
is_safe = .false.
return
end if
end do
f = file - 1
r = rank - 1
do while(f > 0 .and. r > 0)
if(board(f, r)) then
is_safe = .false.
return
end if
f = f - 1
r = r - 1
end do
f = file + 1
r = rank - 1
do while(f <= n .and. r > 0)
if(board(f, r)) then
is_safe = .false.
return
end if
f = f + 1
r = r - 1
end do
end function
subroutine Printboard(board)
logical, intent(in) :: board(:,:)
character(n*4+1) :: line
integer :: f, r
write(*, "(a, i0)") "n = ", n
line = repeat("+---", n) // "+"
do r = 1, n
write(*, "(a)") line
do f = 1, n
write(*, "(a)", advance="no") "|"
if(board(f, r)) then
write(*, "(a)", advance="no") " Q "
else if(mod(f+r, 2) == 0) then
write(*, "(a)", advance="no") " "
else
write(*, "(a)", advance="no") "###"
end if
end do
write(*, "(a)") "|"
end do
write(*, "(a)") line
end subroutine end program</lang> Output for 8, 16 and 32 queens
n = 8 +---+---+---+---+---+---+---+---+ | Q |###| |###| |###| |###| +---+---+---+---+---+---+---+---+ |###| |###| | Q | |###| | +---+---+---+---+---+---+---+---+ | |###| |###| |###| | Q | +---+---+---+---+---+---+---+---+ |###| |###| |###| Q |###| | +---+---+---+---+---+---+---+---+ | |###| Q |###| |###| |###| +---+---+---+---+---+---+---+---+ |###| |###| |###| | Q | | +---+---+---+---+---+---+---+---+ | | Q | |###| |###| |###| +---+---+---+---+---+---+---+---+ |###| |###| Q |###| |###| | +---+---+---+---+---+---+---+---+ n = 16 +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ | Q |###| |###| |###| |###| |###| |###| |###| |###| +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ |###| | Q | |###| |###| |###| |###| |###| |###| | +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ | |###| |###| Q |###| |###| |###| |###| |###| |###| +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ |###| Q |###| |###| |###| |###| |###| |###| |###| | +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ | |###| |###| |###| |###| |###| |###| Q |###| |###| +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ |###| |###| |###| |###| | Q | |###| |###| |###| | +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ | |###| |###| |###| |###| |###| |###| | Q | |###| +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ |###| |###| |###| |###| |###| |###| Q |###| |###| | +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ | |###| |###| |###| |###| |###| |###| |###| Q |###| +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ |###| |###| |###| Q |###| |###| |###| |###| |###| | +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ | |###| |###| |###| |###| |###| |###| |###| | Q | +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ |###| |###| |###| | Q | |###| |###| |###| |###| | +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ | |###| | Q | |###| |###| |###| |###| |###| |###| +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ |###| |###| |###| |###| |###| | Q | |###| |###| | +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ | |###| |###| |###| | Q | |###| |###| |###| |###| +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ |###| |###| |###| |###| |###| Q |###| |###| |###| | +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ n = 32 +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ | Q |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ |###| | Q | |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| | +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ | |###| |###| Q |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ |###| Q |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| | +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ | |###| | Q | |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ |###| |###| |###| |###| | Q | |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| | +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ | |###| |###| |###| |###| |###| Q |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ |###| |###| |###| |###| |###| |###| | Q | |###| |###| |###| |###| |###| |###| |###| |###| |###| | +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ | |###| |###| |###| |###| |###| |###| |###| Q |###| |###| |###| |###| |###| |###| |###| |###| |###| +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ |###| |###| |###| Q |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| | +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ | |###| |###| |###| |###| |###| |###| |###| |###| | Q | |###| |###| |###| |###| |###| |###| |###| +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| Q |###| |###| |###| |###| | +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ | |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| | Q | |###| |###| |###| +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| Q |###| | 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+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ | |###| |###| |###| |###| |###| |###| |###| |###| |###| | Q | |###| |###| |###| |###| |###| |###| +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| |###| Q |###| |###| |###| |###| |###| | +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
Alternate Fortran 77 solution
<lang fortran>C This one implements depth-first backtracking. C See the 2nd program for Scheme on the "Permutations" page for the C main idea. C As is, the program only prints the number of n-queens configurations. C To print also the configurations, uncomment the line after label 80.
program queens
implicit integer(a-z)
parameter(l=18)
dimension a(l),s(l),u(4*l-2)
do 10 i=1,l
10 a(i)=i
do 20 i=1,4*l-2
20 u(i)=0
do 110 n=1,l
m=0
i=1
r=2*n-1
go to 40
30 s(i)=j
u(p)=1
u(q+r)=1
i=i+1
40 if(i.gt.n) go to 80
j=i
50 z=a(i)
y=a(j)
p=i-y+n
q=i+y-1
a(i)=y
a(j)=z
if((u(p).eq.0).and.(u(q+r).eq.0)) goto 30
60 j=j+1
if(j.le.n) go to 50
70 j=j-1
if(j.eq.i) go to 90
z=a(i)
a(i)=a(j)
a(j)=z
go to 70
80 m=m+1
C print *,(a(k),k=1,n)
90 i=i-1
if(i.eq.0) go to 100
p=i-a(i)+n
q=i+a(i)-1
j=s(i)
u(p)=0
u(q+r)=0
go to 60
100 print *,n,m
110 continue
end
C Output C 1 1 C 2 0 C 3 0 C 4 2 C 5 10 C 6 4 C 7 40 C 8 92 C 9 352 C 10 724 C 11 2680 C 12 14200 C 13 73712 C 14 365596 C 15 2279184 C 16 14772512 C 17 95815104 C 18 666090624</lang>
Alternate Fortran 95 solution with OpenMP
This code is useful mainly for counting solutions. Here we use the same algorithm as with Fortran 77, with an optimization: because of symmetry of the chess board, computations are divided by two. The remaining is parallelized with OpenMP. The loop is done on the valid combinations of queens in the first two columns. The original algorithm is slightly changed to start backtracking from column three.
If using GCC, compile with gfortran -Wall -std=f95 -pedantic -O3 -fopenmp queens.f. With Absoft Pro Fortran, af90 -openmp -O3 queens.f.
With some versions of GCC the function OMP_GET_WTIME is not known, which seems to be a bug. Then it's enough to comment out the two calls, and the program won't display timings.
<lang fortran>program queens
use omp_lib implicit none integer, parameter :: long = selected_int_kind(17) integer, parameter :: l = 18 integer :: n, i, j, a(l*l, 2), k, p, q integer(long) :: s, b(l*l) real(kind(1d0)) :: t1, t2
do n = 6, l
k = 0
p = n/2
q = mod(n, 2)*(p + 1)
do i = 1, n
do j = 1, n
if ((abs(i - j) > 1) .and. ((i <= p) .or. ((i == q) .and. (j < i)))) then
k = k + 1
a(k, 1) = i
a(k, 2) = j
end if
end do
end do
s = 0
t1 = omp_get_wtime()
!$omp parallel do schedule(dynamic)
do i = 1, k
b(i) = pqueens(n, a(i, 1), a(i, 2))
end do
!$omp end parallel do
t2 = omp_get_wtime()
print "(I4, I12, F12.3)", n, 2*sum(b(1:k)), t2 - t1
end do
contains
function pqueens(n, k1, k2) result(m)
implicit none
integer(long) :: m
integer, intent(in) :: n, k1, k2
integer, parameter :: l = 20
integer :: a(l), s(l), u(4*l - 2)
integer :: i, j, y, z, p, q, r
do i = 1, n
a(i) = i
end do
do i = 1, 4*n - 2
u(i) = 0
end do
m = 0
r = 2*n - 1
if (k1 == k2) return
p = 1 - k1 + n
q = 1 + k1 - 1
if ((u(p) /= 0) .or. (u(q + r) /= 0)) return
u(p) = 1
u(q + r) = 1
z = a(1)
a(1) = a(k1)
a(k1) = z
p = 2 - k2 + n
q = 2 + k2 - 1
if ((u(p) /= 0) .or. (u(q + r) /= 0)) return
u(p) = 1
u(q + r) = 1
if (k2 /= 1) then
z = a(2)
a(2) = a(k2)
a(k2) = z
else
z = a(2)
a(2) = a(k1)
a(k1) = z
end if
i = 3
go to 40
30 s(i) = j
u(p) = 1
u(q + r) = 1
i = i + 1
40 if (i > n) go to 80
j = i
50 z = a(i)
y = a(j)
p = i - y + n
q = i + y - 1
a(i) = y
a(j) = z
if ((u(p) == 0) .and. (u(q + r) == 0)) go to 30
60 j = j + 1
if (j <= n) go to 50
70 j = j - 1
if (j == i) go to 90
z = a(i)
a(i) = a(j)
a(j) = z
go to 70
!valid queens position found
80 m = m + 1
90 i = i - 1
if (i == 2) return
p = i - a(i) + n
q = i + a(i) - 1
j = s(i)
u(p) = 0
u(q + r) = 0
go to 60
end function
end program</lang>
GAP
<lang gap># Quick and dirty solution, checking all permutations without backtracking (thus it's slow) IsSafe := function(a)
local n, i, j;
n := Length(a);
for i in [1 .. n - 1] do
for j in [i + 1 .. n] do
if AbsInt(a[j] - a[i]) = j - i then
return false;
fi;
od;
od;
return true;
end;
Queens := function(n)
local p, a, v;
v := [];
for p in SymmetricGroup(n) do
a := List([1 .. n], i -> i^p);
if IsSafe(a) then
Add(v, a);
fi;
od;
return v;
end;
v := Queens(8);; Length(v); PrintArray(PermutationMat(PermListList([1 .. 8], v[1]), 8)); [ [ 0, 0, 1, 0, 0, 0, 0, 0 ],
[ 0, 0, 0, 0, 0, 1, 0, 0 ], [ 0, 1, 0, 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 1, 0, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 0, 1 ], [ 1, 0, 0, 0, 0, 0, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 1, 0 ], [ 0, 0, 0, 1, 0, 0, 0, 0 ] ]</lang>
Go
<lang go>// A fairly literal translation of the example program on the referenced // WP page. Well, it happened to be the example program the day I completed // the task. It seems from the WP history that there has been some churn // in the posted example program. The example program of the day was in // Pascal and was credited to Niklaus Wirth, from his "Algorithms + // Data Structures = Programs." package main
import "fmt"
var (
i int q bool a [9]bool b [17]bool c [15]bool // offset by 7 relative to the Pascal version x [9]int
)
func try(i int) {
for j := 1; ; j++ {
q = false
if a[j] && b[i+j] && c[i-j+7] {
x[i] = j
a[j] = false
b[i+j] = false
c[i-j+7] = false
if i < 8 {
try(i + 1)
if !q {
a[j] = true
b[i+j] = true
c[i-j+7] = true
}
} else {
q = true
}
}
if q || j == 8 {
break
}
}
}
func main() {
for i := 1; i <= 8; i++ {
a[i] = true
}
for i := 2; i <= 16; i++ {
b[i] = true
}
for i := 0; i <= 14; i++ {
c[i] = true
}
try(1)
if q {
for i := 1; i <= 8; i++ {
fmt.Println(i, x[i])
}
}
}</lang> Output:
1 1 2 5 3 8 4 6 5 3 6 7 7 2 8 4
Groovy
Distinct Solutions
This solver starts with the N! distinct solutions to the N-Rooks problem and then keeps only the candidates in which all Queens are mutually diagonal-safe. <lang groovy>def listOrder = { a, b ->
def k = [a.size(), b.size()].min()
def i = (0..<k).find { a[it] != b[it] }
(i != null) ? a[i] <=> b[i] : a.size() <=> b.size()
}
def orderedPermutations = { list ->
def n = list.size() (0..<n).permutations().sort(listOrder)
}
def diagonalSafe = { list ->
def n = list.size()
n == 1 || (0..<(n-1)).every{ i ->
((i+1)..<n).every{ j ->
!([list[i]+j-i, list[i]+i-j].contains(list[j]))
}
}
}
def queensDistinctSolutions = { n ->
// each permutation is an N-Rooks solution orderedPermutations((0..<n)).findAll (diagonalSafe)
}</lang>
Unique Solutions
Unique solutions are equivalence classes of distinct solutions, factoring out all reflections and rotations of a given solution. See the Wikipedia page for more details. <lang groovy>class Reflect {
public static final diag = { list ->
final n = list.size()
def tList = [0] * n
(0..<n).each { tList[list[it]] = it }
tList
}
public static final vert = { list ->
list.reverse()
}
public static final horiz = { list ->
final n = list.size()
list.collect { n - it - 1 }
}
}
enum Rotations {
r0([]),
r90([Reflect.vert, Reflect.diag]),
r180([Reflect.vert, Reflect.diag, Reflect.vert, Reflect.diag]),
r270([Reflect.diag, Reflect.vert]);
private final List operations
private Rotations(List ops) {
operations = ops ?: []
}
public static void eliminateDups(primary, solutions) {
(r0..r270).each { rot -> rot.eliminateDuplicates(primary, solutions) }
}
private void eliminateDuplicates(primary, solutions) {
def rotated = [] + primary
operations.each { rotated = it(rotated) }
solutions.removeAll([rotated, Reflect.vert(rotated)])
}
}
def queensUniqueSolutions = { start ->
assert start instanceof Number || start instanceof List
def qus = (start instanceof Number) \
? queensDistinctSolutions(start) \
: [] + start
for (def i = 0; i < qus.size()-1; i++) {
Rotations.eliminateDups(qus[i], qus[(i+1)..<(qus.size())])
}
qus
}</lang>
Test and Results
This script tests both distinct and unique solution lists. <lang groovy>(1..9).each { n ->
def qds = queensDistinctSolutions(n)
def qus = queensUniqueSolutions(qds)
println ([boardSize:n, "number of distinct solutions":qds.size(), "number of unique solutions":qus.size()])
if(n < 9) { qus.each { println it } }
else { println "first:${qus[0]}"; println "last:${qus[-1]}" }
println()
}</lang>
Interpreting the Results:
Each individual result is given as a list of N numbers. Each number represents a column number within the list-indexed row. So, the following 4-queens solution:
[1, 3, 0, 2]
should be interpreted as follows:
row 0 has a queen in column 1 row 1 has a queen in column 3 row 2 has a queen in column 0 row 3 has a queen in column 2
In other words, this:
|///| Q |///| | --- --- --- --- | |///| |/Q/| --- --- --- --- |/Q/| |///| | --- --- --- --- | |///| Q |///|
Results:
[boardSize:1, number of distinct solutions:1, number of unique solutions:1] [0] [boardSize:2, number of distinct solutions:0, number of unique solutions:0] [boardSize:3, number of distinct solutions:0, number of unique solutions:0] [boardSize:4, number of distinct solutions:2, number of unique solutions:1] [1, 3, 0, 2] [boardSize:5, number of distinct solutions:10, number of unique solutions:2] [0, 2, 4, 1, 3] [1, 4, 2, 0, 3] [boardSize:6, number of distinct solutions:4, number of unique solutions:1] [1, 3, 5, 0, 2, 4] [boardSize:7, number of distinct solutions:40, number of unique solutions:6] [0, 2, 4, 6, 1, 3, 5] [0, 3, 6, 2, 5, 1, 4] [1, 3, 0, 6, 4, 2, 5] [1, 4, 0, 3, 6, 2, 5] [1, 4, 6, 3, 0, 2, 5] [1, 5, 2, 6, 3, 0, 4] [boardSize:8, number of distinct solutions:92, number of unique solutions:12] [0, 4, 7, 5, 2, 6, 1, 3] [0, 5, 7, 2, 6, 3, 1, 4] [1, 3, 5, 7, 2, 0, 6, 4] [1, 4, 6, 0, 2, 7, 5, 3] [1, 4, 6, 3, 0, 7, 5, 2] [1, 5, 0, 6, 3, 7, 2, 4] [1, 5, 7, 2, 0, 3, 6, 4] [1, 6, 2, 5, 7, 4, 0, 3] [1, 6, 4, 7, 0, 3, 5, 2] [2, 4, 1, 7, 0, 6, 3, 5] [2, 4, 7, 3, 0, 6, 1, 5] [2, 5, 1, 4, 7, 0, 6, 3] [boardSize:9, number of distinct solutions:352, number of unique solutions:46] first:[0, 2, 5, 7, 1, 3, 8, 6, 4] last:[3, 1, 6, 8, 0, 7, 4, 2, 5]
Haskell
<lang haskell>import Control.Monad import Data.List
-- given n, "queens n" solves the n-queens problem, returning a list of all the -- safe arrangements. each solution is a list of the columns where the queens are -- located for each row queens :: Int -> Int queens n = map fst $ foldM oneMoreQueen ([],[1..n]) [1..n] where
-- foldM :: (Monad m) => (a -> b -> m a) -> a -> [b] -> m a -- foldM folds (from left to right) in the list monad, which is convenient for -- "nondeterminstically" finding "all possible solutions" of something. the -- initial value [] corresponds to the only safe arrangement of queens in 0 rows
-- given a safe arrangement y of queens in the first i rows, and a list of -- possible choices, "oneMoreQueen y _" returns a list of all the safe -- arrangements of queens in the first (i+1) rows along with remaining choices oneMoreQueen (y,d) _ = [ (x:y, d\\[x]) | x <- d, safe x y]
-- "safe x y" tests whether a queen at column x is safe from previous -- queens as recorded in y safe x y = and [ x /= c && x /= c + n && x /= c - n | (n,c) <- zip [1..] y]
-- prints what the board looks like for a solution; with an extra newline printSolution y = do
let n = length y
mapM_ (\x -> putStrLn [if z == x then 'Q' else '.' | z <- [1..n]]) y
putStrLn ""
-- prints all the solutions for 6 queens main = mapM_ printSolution $ queens 6</lang>
If you just want one solution, simply take the head of the result of queens n; since Haskell is lazy, it will only do as much work as needed to find one solution and stop.
Alternative version
<lang haskell>import Control.Monad (foldM) import Data.List ((\\))
main :: IO () main = mapM_ print $ queens 8
queens :: Int -> Int queens n = foldM f [] [1..n]
where
f qs k = [q:qs | q <- [1..n] \\ qs, q `notDiag` qs]
q `notDiag` qs = and [abs (q - qi) /= i | (qi,i) <- qs `zip` [1..]]</lang>
Using permutations
This version uses permutations to generate unique horizontal and vertical position for each queen. Thus, we only need to check diagonals. However, it is less efficient than the previous version because it does not prune out prefixes that are found to be unsuitable. <lang haskell>import Data.List (nub, permutations)
-- checks if queens are on the same diagonal -- with [0..] we place each queen on her own row check f = length . nub . zipWith f [0..]
-- filters out results where 2 or more queens are on the same diagonal -- with [0..n-1] we place each queeen on her own column generate n = filter (\x -> check (+) x == n && check (-) x == n) $ permutations [0..n-1]
-- 8 is for "8 queens" main = print $ generate 8</lang>
Heron
<lang heron>module NQueens {
inherits {
Heron.Windows.Console;
}
fields {
n : Int = 4;
sols : List = new List();
}
methods {
PosToString(row : Int, col : Int) : String {
return "row " + row.ToString() + ", col " + col.ToString();
}
AddQueen(b : Board, row : Int, col : Int)
{
if (!b.TryAddQueen(row, col))
return;
if (row < n - 1)
foreach (i in 0..n-1)
AddQueen(new Board(b), row + 1, i);
else
sols.Add(b);
}
Main() {
foreach (i in 0..n-1)
AddQueen(new Board(), 0, i);
foreach (b in sols) {
b.Output();
WriteLine("");
}
WriteLine("Found " + sols.Count().ToString() + " solutions");
}
}
}
class Board {
fields {
rows = new List();
}
methods {
Constructor() {
foreach (r in 0..n-1) {
var col = new List();
foreach (c in 0..n-1)
col.Add(false);
rows.Add(col);
}
}
Constructor(b : Board) {
Constructor();
foreach (r in 0..n-1)
foreach (c in 0..n-1)
SetSpaceOccupied(r, c, b.SpaceOccupied(r, c));
}
SpaceOccupied(row : Int, col : Int) : Bool {
return rows[row][col];
}
SetSpaceOccupied(row : Int, col : Int, b : Bool) {
rows[row][col] = b;
}
ValidPos(row : Int, col : Int) : Bool {
return ((row >= 0) && (row < n)) && ((col >= 0) && (col < n));
}
VectorOccupied(row : Int, col : Int, rowDir : Int, colDir : Int) : Bool {
var nextRow = row + rowDir;
var nextCol = col + colDir;
if (!ValidPos(nextRow, nextCol))
return false;
if (SpaceOccupied(nextRow, nextCol))
return true;
return VectorOccupied(nextRow, nextCol, rowDir, colDir);
}
TryAddQueen(row : Int, col : Int) : Bool {
foreach (rowDir in -1..1)
foreach (colDir in -1..1)
if (rowDir != 0 || colDir != 0)
if (VectorOccupied(row, col, rowDir, colDir))
return false;
SetSpaceOccupied(row, col, true);
return true;
}
Output() {
foreach (row in 0..n-1) {
foreach (col in 0..n-1) {
if (SpaceOccupied(row, col)) {
Write("Q");
}
else {
Write(".");
}
}
WriteLine("");
}
}
}
}</lang>
Icon and Unicon
Here's a solution to the n = 8 case: <lang icon>procedure main()
write(q(1), " ", q(2), " ", q(3), " ", q(4), " ", q(5), " ", q(6), " ", q(7), " ", q(8))
end
procedure q(c)
static udiag, ddiag, row
initial {
udiag := list(15, 0)
ddiag := list(15, 0)
row := list(8, 0)
}
every 0 = row[r := 1 to 8] = ddiag[r + c - 1] = udiag[8 + r - c] do # test if free
suspend row[r] <- ddiag[r + c - 1] <- udiag[8 + r - c] <- r # place and yield
end</lang>
Notes:
- Solution assumes attempting to place 8 queens on a standard chessboard, and is a simplification of a program in the The Icon Programming Library (IPL) which is in the public domain.
- There are 15 left-side-down-diagonals and 15 left-side-up-diagonals represented in the lists. An unfilled row or diagonal has value 0, otherwise the row number is stored to indicate placement.
- The numeric equality operator =, like all the comparators in Icon, yields the right argument as its solution, or fails. The chain of 0 = A = B = C therefore tests each of A B and C for equality with 0; these semantics read very naturally.
- every drives the chain of = tests to yield every possible result; the iterable component is the generator 1 to 8 which is progressively stored into r and will be backtracked if any of the equality tests fail. If all the placements are zero, the chain of equalities suceeds, and the suspend is invoked for that iteration.
- <- is the "reversible assignment" operator. It restores the original value and fails if it is resumed by backtracking. The suspend will use it to temporarily consume the placements and then it will yield the value of the chosen row r.
- procedure q() attempts to place the c-th column queen into row 1 to 8 in turn, suspending only if that queen can be placed at [c,r]
- As the calls to q() are evaluated in main, each one will suspend a possible row, thereby allowing the next q(n) in main to be evaluated. If any of the q() fails to yield a row for the nth queen (or runs out of solutions) the previous, suspended calls to q() are backtracked progressively. If the final q(8) yields a row, the write() will be called with the row positions of each queen. Note that even the final q(8) will be suspended along with the other 7 calls to q(). Unless the write() is driven to produce more solutions (see next point) the suspended procedures will be closed at the "end of statement" ie after the write has "succeeded".
- If you want to derive all possible solutions, main() can be embellished with the every keyword:
<lang icon> procedure main()
every write(q(1), " ", q(2), " ", q(3), " ", q(4), " ", q(5), " ", q(6), " ", q(7), " ", q(8))
end </lang> This drives the backtracking to find more solutions.
The following is a general N-queens solution, adapted from a solution placed into the public domain by Peter A. Bigot in 1990. The program produces a solution for a specified value of N. The comment explains how to modify the program to produce all solutions for a given N. <lang icon>global n, rw, dd, ud
procedure main(args)
n := integer(args[1]) | 8 rw := list(n) dd := list(2*n-1) ud := list(2*n-1) solvequeen(1)
end
procedure solvequeen(c)
if (c > n) then return show() else suspend placequeen(c) & solvequeen(c+1)
end
procedure placequeen(c)
suspend (/rw[r := 1 to n] <- /dd[r+c-1] <- /ud[n+r-c] <- c)
end
procedure show()
static count, line, border
initial {
count := 0
line := repl("| ",n) || "|"
border := repl("----",n) || "-"
}
write("solution: ", count+:=1)
write(" ", border)
every line[4*(!rw - 1) + 3] <- "Q" do {
write(" ", line)
write(" ", border)
}
write()
return # Comment out to see all possible solutions
end</lang>
A sample run for N = 6:
->nq 6 solution: 1 ------------------------- | | | | Q | | | ------------------------- | Q | | | | | | ------------------------- | | | | | Q | | ------------------------- | | Q | | | | | ------------------------- | | | | | | Q | ------------------------- | | | Q | | | | ------------------------- ->
Two solutions are in the IPL queens and genqueen.
J
This is one of several J solutions shown and explained on this J wiki page
<lang j>perm =: ! A.&i. ] NB. all permutations of integers 0 to y comb2 =: (, #: I.@,@(</)&i.)~ NB. all size 2 combinations of integers 0 to y mask =: [ */@:~:&(|@-/) { queenst=: comb2 (] #"1~ mask)&.|: perm</lang>
Note that the Roger Hui's approach (used here) matches the description attributed to Raymond Hettinger (in the Python implementation). (Both were posted years ago: 2008 for Hui's version which was used here, and 2009 for Hettinger's.)
Java
<lang java>public class NQueens {
private static int[] b = new int[8]; private static int s = 0;
static boolean unsafe(int y) {
int x = b[y];
for (int i = 1; i <= y; i++) {
int t = b[y - i];
if (t == x ||
t == x - i ||
t == x + i) {
return true;
}
}
return false; }
public static void putboard() {
System.out.println("\n\nSolution " + (++s));
for (int y = 0; y < 8; y++) {
for (int x = 0; x < 8; x++) {
System.out.print((b[y] == x) ? "|Q" : "|_");
}
System.out.println("|");
}
}
public static void main(String[] args) {
int y = 0;
b[0] = -1;
while (y >= 0) {
do {
b[y]++;
} while ((b[y] < 8) && unsafe(y));
if (b[y] < 8) {
if (y < 7) {
b[++y] = -1;
} else {
putboard();
}
} else {
y--;
}
}
}
}</lang>
Liberty BASIC
Program uses permutation generator (stores all permutations) and solves tasks 4x4 to 9x9. It prints all the solutions. <lang lb> 'N queens '>10 would not work due to way permutations used 'anyway, 10 doesn't fit in memory Input "Input N for N queens puzzle (4..9) ";N if N<4 or N>9 then print "N out of range - quitting": end
ABC$= " " dash$ = "" for i = 0 to N-1
ABC$=ABC$+" "+chr$(asc("a")+i)
dash$ = dash$+"--"
next
dim q(N) t0=time$("ms")
fact = 1 for i = 1 to N
fact = fact*i
next
dim anagram$(fact) global nPerms print "Filling permutations array" t0=time$("ms")
res$=permutation$("", left$("0123456789", N))
t1=time$("ms") print "Created all possible permutations ";t1-t0
t0=time$("ms") 'actually fact = nPerms for k=1 to nPerms
for i=0 to N-1
q(i)=val(mid$(anagram$(k),i+1,1))
'print q(i);
next
'print
fail = 0
for i=0 to N-1
for j=i+1 to N-1
'check rows are different
if q(i)=q(j) then fail = 1: exit for
'check diagonals are different
if i+q(i)=j+q(j) then fail = 1: exit for
'check other diagonals are different
if i-q(i)=j-q(j) then fail = 1: exit for
next
if fail then exit for
next
if not(fail) then
num=num+1
print " ";dash$
for i=0 to N-1
print N-i; space$(2*q(i));" *"
next
print " ";dash$
print ABC$
end if
next
t1=time$("ms") print "Time taken ";t1-t0 print "Number of solutions ";num
'---------------------------------- 'from 'http://babek.info/libertybasicfiles/lbnews/nl124/wordgames.htm 'Programming a Word Game by Janet Terra, 'The Liberty Basic Newsletter - Issue #124 - September 2004 Function permutation$(pre$, post$) 'Note the variable nPerms must first be stated as a global variable.
lgth = Len(post$)
If lgth < 2 Then
nPerms = nPerms + 1
anagram$(nPerms) = pre$;post$
Else
For i = 1 To lgth
tmp$=permutation$(pre$+Mid$(post$,i,1),Left$(post$,i-1)+Right$(post$,lgth-i))
Next i
End If
End Function
</lang>
Locomotive Basic
Uses the heuristic from the Wikipedia article to get one solution.
<lang locobasic>10 mode 1:defint a-z 20 while n<4:input "How many queens (N>=4)";n:wend 30 dim q(n),e(n),o(n) 40 r=n mod 6 50 if r<>2 and r<>3 then gosub 320:goto 220 60 for i=1 to int(n/2) 70 e(i)=2*i 80 next 90 for i=1 to round(n/2) 100 o(i)=2*i-1 110 next 120 if r=2 then gosub 410 130 if r=3 then gosub 460 140 s=1 150 for i=1 to n 160 if e(i)>0 then q(s)=e(i):s=s+1 170 next 180 for i=1 to n 190 if o(i)>0 then q(s)=o(i):s=s+1 200 next 210 ' print board 220 cls 230 for i=1 to n 240 locate i,26-q(i):print chr$(238); 250 locate i,24-n :print chr$(96+i); 260 locate n+1,26-i :print i; 270 next 280 locate 1,1 290 call &bb06 300 end 310 ' the simple case 320 p=1 330 for i=1 to n 340 if i mod 2=0 then q(p)=i:p=p+1 350 next 360 for i=1 to n 370 if i mod 2 then q(p)=i:p=p+1 380 next 390 return 400 ' edit list when remainder is 2 410 for i=1 to n 420 if o(i)=3 then o(i)=1 else if o(i)=1 then o(i)=3 430 if o(i)=5 then o(i)=-1 else if o(i)=0 then o(i)=5:return 440 next 450 ' edit list when remainder is 3 460 for i=1 to n 470 if e(i)=2 then e(i)=-1 else if e(i)=0 then e(i)=2:goto 500 480 next 490 ' edit list some more 500 for i=1 to n 510 if o(i)=1 or o(i)=3 then o(i)=-1 else if o(i)=0 then o(i)=1:o(i+1)=3:return 520 next</lang>
Logo
<lang logo>to try :files :diag1 :diag2 :tried
if :files = 0 [make "solutions :solutions+1 show :tried stop] localmake "safe (bitand :files :diag1 :diag2) until [:safe = 0] [ localmake "f bitnot bitand :safe minus :safe try bitand :files :f ashift bitand :diag1 :f -1 (ashift bitand :diag2 :f 1)+1 fput bitnot :f :tried localmake "safe bitand :safe :safe-1 ]
end
to queens :n
make "solutions 0 try (lshift 1 :n)-1 -1 -1 [] output :solutions
end
print queens 8 ; 92</lang>
Lua
<lang Lua>N = 8
board = {} for i = 1, N do
board[i] = {}
for j = 1, N do
board[i][j] = false
end
end
function Allowed( x, y )
for i = 1, x-1 do
if ( board[i][y] ) or ( i <= y and board[x-i][y-i] ) or ( y+i <= N and board[x-i][y+i] ) then
return false
end
end return true
end
function Find_Solution( x )
for y = 1, N do
if Allowed( x, y ) then
board[x][y] = true
if x == N or Find_Solution( x+1 ) then return true end board[x][y] = false end
end return false
end
if Find_Solution( 1 ) then
for i = 1, N do for j = 1, N do if board[i][j] then
io.write( "|Q" ) else io.write( "| " ) end end print( "|" )
end
else
print( string.format( "No solution for %d queens.\n", N ) )
end </lang>
Mathematica
This code recurses through the possibilities, using the "safe" method to check if the current set is allowed. The recursive method has the advantage that finding all possibilities is about as hard (code-wise, not computation-wise) as finding just one. <lang Mathematica>safe[q_List, n_] :=
With[{l = Length@q},
Length@Union@q == Length@Union[q + Range@l] ==
Length@Union[q - Range@l] == l]
nQueen[q_List:{}, n_] :=
If[safe[q, n],
If[Length[q] == n, q,
Cases[Flatten[{nQueen[Append[q, #], n]}, 2] & /@ Range[n],
Except[{Null} | {}]]], Null]</lang>
This returns a list of valid permutations by giving the queen's column number for each row. It can be displayed in a list of chess-board tables like this: <lang Mathematica>matrixView[n_] :=
Grid[Normal@
SparseArray[MapIndexed[{#, First@#2} -> "Q" &, #], {n, n}, "."],
Frame -> All] & /@ nQueen[n]
matrixView[6] // OutputForm</lang>
{. . . Q . ., . . . . Q ., . Q . . . ., . . Q . . .}
Q . . . . . . . Q . . . . . . Q . . . . . . . Q
. . . . Q . Q . . . . . . . . . . Q . Q . . . .
. Q . . . . . . . . . Q Q . . . . . . . . . Q .
. . . . . Q . . . Q . . . . Q . . . Q . . . . .
. . Q . . . . Q . . . . . . . . Q . . . . Q . .
Alternate Solution This solution uses Permutations and subsets, also prints out a board representation.
<lang Mathematica>n=8;cnt=1;per=Permutations[Range[n],{n}];(* All Permutations of length n *) Do[perq=Partition[Riffle[Reverse[Range[n]],perq],2],{q,1,Length[per]}];(* Riffled in the reverse of [range n] partitioned into pairs*) Do[w=Subsets[pert,{2}];(* This is a full subset of the previous set of pairs taken 2 at a time *) tot=0; Do[y=Abs[wq,1,1-wq,2,1];x=Abs[wq,1,2-wq,2,2];If[x==y,tot++],{q,1,Length[w]}];(* x and y are the abs values of x1-y1 and x2-y2 if equal they are on same diagonal *) If[tot==0,g=Grid[Table[" ",{n},{n}],Alignment->Center,Frame->All,Spacings->{1.2,1}];(* If no clashing diagonals setup an array and print the permutation and the grid*) Do[g[[1,pert,w,1,pert,w,2]]="Q",{w,1,n}]; Print[cnt," ",pert," ",g];cnt++],{t,1,Length[per]}]</lang>
Maxima
<lang maxima>/* translation of Fortran 77, return solutions as permutations */
queens(n) := block([a, i, j, m, p, q, r, s, u, v, w, y, z], a: makelist(i, i, 1, n), s: a*0, u: makelist(0, i, 1, 4*n - 2), m: 0, i: 1, r: 2*n - 1, w: [ ], go(L40), L30, s[i]: j, u[p]: 1, u[q + r]: 1, i: i + 1, L40, if i > n then go(L80), j: i, L50, z: a[i], y: a[j], p: i - y + n, q: i + y - 1, a[i]: y, a[j]: z, if u[p] = 0 and u[q + r] = 0 then go(L30), L60, j: j + 1, if j <= n then go(L50), L70, j: j - 1, if j = i then go(L90), z: a[i], a[i]: a[j], a[j]: z, go(L70), L80, m: m + 1, w: endcons(copylist(a), w), L90, i: i - 1, if i = 0 then go(L100), p: i - a[i] + n, q: i + a[i] - 1, j: s[i], u[p]: 0, u[q + r]: 0, go(L60), L100, w)$
queens(8); /* [[1, 5, 8, 6, 3, 7, 2, 4],
[1, 6, 8, 3, 7, 4, 2, 5],
...]] */
length(%); /* 92 */</lang>
MUMPS
<lang MUMPS>Queens New count,flip,row,sol Set sol=0 For row(1)=1:1:4 Do try(2) ; Not 8, the other 4 are symmetric... ; ; Remove symmetric solutions Set sol="" For Set sol=$Order(sol(sol)) Quit:sol="" Do . New xx,yy . Kill sol($Translate(sol,12345678,87654321)) ; Vertical flip . Kill sol($Reverse(sol)) ; Horizontal flip . Set flip="--------" for xx=1:1:8 Do ; Flip over top left to bottom right diagonal . . New nx,ny . . Set yy=$Extract(sol,xx),nx=8+1-xx,ny=8+1-yy . . Set $Extract(flip,ny)=nx . . Quit . Kill sol(flip) . Set flip="--------" for xx=1:1:8 Do ; Flip over top right to bottom left diagonal . . New nx,ny . . Set yy=$Extract(sol,xx),nx=xx,ny=yy . . Set $Extract(flip,ny)=nx . . Quit . Kill sol(flip) . Quit ; ; Display remaining solutions Set count=0,sol="" For Set sol=$Order(sol(sol)) Quit:sol="" Do Quit:sol="" . New s1,s2,s3,txt,x,y . Set s1=sol,s2=$Order(sol(s1)),s3="" Set:s2'="" s3=$Order(sol(s2)) . Set txt="+--+--+--+--+--+--+--+--+" . Write !," ",txt Write:s2'="" " ",txt Write:s3'="" " ",txt . For y=8:-1:1 Do . . Write !,y," |" . . For x=1:1:8 Write $Select($Extract(s1,x)=y:" Q",x+y#2:" ",1:"##"),"|" . . If s2'="" Write " |" . . If s2'="" For x=1:1:8 Write $Select($Extract(s2,x)=y:" Q",x+y#2:" ",1:"##"),"|" . . If s3'="" Write " |" . . If s3'="" For x=1:1:8 Write $Select($Extract(s3,x)=y:" Q",x+y#2:" ",1:"##"),"|" . . Write !," ",txt Write:s2'="" " ",txt Write:s3'="" " ",txt . . Quit . Set txt=" A B C D E F G H" . Write !," ",txt Write:s2'="" " ",txt Write:s3'="" " ",txt Write ! . Set sol=s3 . Quit Quit try(col) New ok,pcol If col>8 Do Quit . New out,x . Set out="" For x=1:1:8 Set out=out_row(x) . Set sol(out)=1 . Quit For row(col)=1:1:8 Do . Set ok=1 . For pcol=1:1:col-1 If row(pcol)=row(col) Set ok=0 Quit . Quit:'ok . For pcol=1:1:col-1 If col-pcol=$Translate(row(pcol)-row(col),"-") Set ok=0 Quit . Quit:'ok . Do try(col+1) . Quit Quit Do Queens
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
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+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
7 |##| |##| |##| Q|##| | |##| |##| | Q| |##| | |##| |##| |##| | Q| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
6 | |##| | Q| |##| |##| | | Q| |##| |##| |##| | |##| Q|##| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
5 |##| Q|##| |##| |##| | |##| |##| |##| |##| Q| |##| |##| |##| |##| Q|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
4 | |##| |##| |##| | Q| | |##| |##| | Q| |##| | | Q| |##| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
3 |##| |##| | Q| |##| | |##| |##| Q|##| |##| | |##| |##| | Q| |##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
2 | |##| |##| |##| Q|##| | |##| |##| |##| Q|##| | Q|##| |##| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
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+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ A B C D E F G H A B C D E F G H A B C D E F G H +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
8 | |##| |##| | Q| |##| | |##| |##| | Q| |##| | |##| |##| | Q| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
7 |##| | Q| |##| |##| | |##| | Q| |##| |##| | |##| |##| Q|##| |##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
6 | |##| |##| |##| Q|##| | |##| |##| |##| Q|##| | | Q| |##| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
5 |##| Q|##| |##| |##| | |##| Q|##| |##| |##| | |##| |##| |##| |##| Q|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
4 | |##| |##| |##| | Q| | |##| | Q| |##| |##| | |##| |##| Q|##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
3 |##| |##| | Q| |##| | |##| |##| |##| |##| Q| |##| |##| |##| | Q| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
2 | Q|##| |##| |##| |##| | Q|##| |##| |##| |##| | Q|##| |##| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
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+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
7 |##| |##| |##| | Q| | |##| Q|##| |##| |##| | |##| Q|##| |##| |##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
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+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
5 |##| |##| |##| |##| Q| |##| |##| |##| Q|##| | |##| | Q| |##| |##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
4 | |##| |##| | Q| |##| | |##| |##| |##| | Q| | |##| |##| | Q| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
3 |##| |##| Q|##| |##| | |##| | Q| |##| |##| | |##| |##| |##| |##| Q|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
2 | Q|##| |##| |##| |##| | Q|##| |##| |##| |##| | Q|##| |##| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
1 |##| |##| | Q| |##| | |##| |##| |##| | Q| | |##| |##| | Q| |##| |
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6 | |##| Q|##| |##| |##| | |##| |##| |##| | Q| | |##| |##| Q|##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
5 |##| |##| |##| Q|##| | |##| Q|##| |##| |##| | |##| Q|##| |##| |##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
4 | |##| |##| |##| | Q| | |##| |##| |##| Q|##| | |##| |##| | Q| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
3 |##| |##| | Q| |##| | | Q| |##| |##| |##| | | Q| |##| |##| |##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
2 | Q|##| |##| |##| |##| | |##| Q|##| |##| |##| | |##| Q|##| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
1 |##| |##| Q|##| |##| | |##| |##| | Q| |##| | |##| |##| |##| |##| Q|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ A B C D E F G H A B C D E F G H A B C D E F G H +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
8 | |##| Q|##| |##| |##| | |##| |##| Q|##| |##| | |##| |##| Q|##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
7 |##| |##| |##| Q|##| | |##| |##| |##| | Q| | |##| |##| |##| | Q| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
6 | |##| |##| |##| | Q| | | Q| |##| |##| |##| | | Q| |##| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
5 |##| Q|##| |##| |##| | |##| |##| Q|##| |##| | |##| |##| |##| Q|##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
4 | |##| | Q| |##| |##| | |##| |##| |##| | Q| | |##| Q|##| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
3 | Q| |##| |##| |##| | | Q| |##| |##| |##| | | Q| |##| |##| |##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
2 | |##| |##| |##| Q|##| | |##| Q|##| |##| |##| | |##| | Q| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
1 |##| |##| | Q| |##| | |##| |##| |##| Q|##| | |##| |##| |##| |##| Q|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ A B C D E F G H A B C D E F G H A B C D E F G H +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
8 | |##| Q|##| |##| |##| | |##| Q|##| |##| |##| | |##| | Q| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
7 |##| |##| |##| Q|##| | |##| |##| |##| | Q| | |##| Q|##| |##| |##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
6 | | Q| |##| |##| |##| | | Q| |##| |##| |##| | |##| |##| |##| | Q|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
5 |##| |##| | Q| |##| | |##| |##| |##| |##| Q| |##| |##| | Q| |##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
4 | |##| |##| |##| | Q| | |##| |##| Q|##| |##| | |##| |##| |##| Q|##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
3 | Q| |##| |##| |##| | | Q| |##| |##| |##| | | Q| |##| |##| |##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
2 | |##| |##| |##| Q|##| | |##| | Q| |##| |##| | |##| Q|##| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
1 |##| |##| Q|##| |##| | |##| |##| |##| Q|##| | |##| |##| |##| Q|##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ A B C D E F G H A B C D E F G H A B C D E F G H +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
8 | |##| | Q| |##| |##| | | Q| |##| |##| |##| | |##| | Q| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
7 |##| Q|##| |##| |##| | |##| |##| Q|##| |##| | |##| |##| |##| |##| Q|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
6 | |##| |##| Q|##| |##| | |##| |##| | Q| |##| | |##| |##| Q|##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
5 |##| |##| |##| |##| Q| |##| |##| |##| |##| Q| |##| | Q| |##| |##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
4 | |##| |##| | Q| |##| | |##| Q|##| |##| |##| | Q|##| |##| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
3 | Q| |##| |##| |##| | | Q| |##| |##| |##| | |##| |##| |##| | Q| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
2 | |##| Q|##| |##| |##| | |##| |##| |##| Q|##| | | Q| |##| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
1 |##| |##| |##| | Q| | |##| |##| | Q| |##| | |##| |##| |##| Q|##| |
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8 | |##| |##| | Q| |##| | |##| Q|##| |##| |##| | |##| Q|##| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
7 |##| | Q| |##| |##| | |##| |##| | Q| |##| | |##| |##| |##| |##| Q|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
6 | |##| |##| Q|##| |##| | |##| |##| |##| | Q| | |##| | Q| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
5 |##| |##| |##| |##| Q| |##| |##| Q|##| |##| | |##| |##| |##| | Q| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
4 | Q|##| |##| |##| |##| | Q|##| |##| |##| |##| | Q|##| |##| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
3 |##| |##| Q|##| |##| | |##| |##| |##| | Q| | |##| |##| |##| Q|##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
2 | | Q| |##| |##| |##| | | Q| |##| |##| |##| | | Q| |##| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
1 |##| |##| |##| | Q| | |##| |##| |##| Q|##| | |##| |##| | Q| |##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ A B C D E F G H A B C D E F G H A B C D E F G H +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
8 | |##| |##| | Q| |##| | |##| Q|##| |##| |##| | |##| | Q| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
7 |##| |##| |##| |##| Q| |##| |##| | Q| |##| | |##| Q|##| |##| |##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
6 | | Q| |##| |##| |##| | | Q| |##| |##| |##| | |##| |##| |##| | Q|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
5 |##| |##| Q|##| |##| | |##| |##| |##| |##| Q| |##| |##| |##| Q|##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
4 | Q|##| |##| |##| |##| | Q|##| |##| |##| |##| | Q|##| |##| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
3 |##| |##| |##| | Q| | |##| |##| |##| | Q| | |##| | Q| |##| |##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
2 | |##| |##| Q|##| |##| | |##| | Q| |##| |##| | |##| |##| Q|##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
1 |##| | Q| |##| |##| | |##| |##| |##| Q|##| | |##| |##| |##| | Q| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ A B C D E F G H A B C D E F G H A B C D E F G H +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
8 | |##| |##| |##| | Q| | | Q| |##| |##| |##| | | Q| |##| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
7 |##| Q|##| |##| |##| | |##| |##| |##| | Q| | |##| |##| | Q| |##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
6 | |##| |##| Q|##| |##| | |##| |##| Q|##| |##| | |##| |##| |##| Q|##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
5 |##| | Q| |##| |##| | |##| |##| |##| |##| Q| |##| |##| Q|##| |##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
4 | Q|##| |##| |##| |##| | Q|##| |##| |##| |##| | Q|##| |##| |##| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
3 |##| |##| |##| | Q| | |##| |##| Q|##| |##| | |##| |##| |##| |##| Q|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
2 | |##| | Q| |##| |##| | |##| |##| | Q| |##| | |##| |##| | Q| |##|
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
1 |##| |##| |##| Q|##| | |##| | Q| |##| |##| | |##| | Q| |##| |##| |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+ A B C D E F G H A B C D E F G H A B C D E F G H +--+--+--+--+--+--+--+--+
8 | | Q| |##| |##| |##|
+--+--+--+--+--+--+--+--+
7 |##| |##| |##| Q|##| |
+--+--+--+--+--+--+--+--+
6 | |##| |##| |##| | Q|
+--+--+--+--+--+--+--+--+
5 |##| | Q| |##| |##| |
+--+--+--+--+--+--+--+--+
4 | Q|##| |##| |##| |##|
+--+--+--+--+--+--+--+--+
3 |##| |##| Q|##| |##| |
+--+--+--+--+--+--+--+--+
2 | |##| |##| |##| Q|##|
+--+--+--+--+--+--+--+--+
1 |##| |##| | Q| |##| |
+--+--+--+--+--+--+--+--+ A B C D E F G H</lang>
Objeck
<lang objeck>ο»Ώbundle Default {
class NQueens {
b : static : Int[];
s : static : Int;
function : Main(args : String[]) ~ Nil {
b := Int->New[8];
s := 0;
y := 0;
b[0] := -1;
while (y >= 0) {
do {
b[y]+=1;
}
while((b[y] < 8) & Unsafe(y));
if(b[y] < 8) {
if (y < 7) {
b[y + 1] := -1;
y += 1;
}
else {
PutBoard();
};
}
else {
y-=1;
};
};
}
function : Unsafe(y : Int) ~ Bool {
x := b[y];
for(i := 1; i <= y; i+=1;) {
t := b[y - i];
if(t = x | t = x - i | t = x + i) {
return true;
};
};
return false;
}
function : PutBoard() ~ Nil {
IO.Console->Print("\n\nSolution ")->PrintLine(s + 1);
s += 1;
for(y := 0; y < 8; y+=1;) {
for(x := 0; x < 8; x+=1;) {
IO.Console->Print((b[y] = x) ? "|Q" : "|_");
};
"|"->PrintLine();
};
}
}
} </lang>
OCaml
<lang ocaml>(* Authors: Nicolas Barnier, Pascal Brisset
Copyright 2004 CENA. All rights reserved. This code is distributed under the terms of the GNU LGPL *)
open Facile open Easy
(* Print a solution *) let print queens =
let n = Array.length queens in
if n <= 10 then (* Pretty printing *)
for i = 0 to n - 1 do
let c = Fd.int_value queens.(i) in (* queens.(i) is bound *)
for j = 0 to n - 1 do
Printf.printf "%c " (if j = c then '*' else '-')
done;
print_newline ()
done
else (* Short print *)
for i = 0 to n-1 do
Printf.printf "line %d : col %a\n" i Fd.fprint queens.(i)
done;
flush stdout;
(* Solve the n-queens problem *) let queens n =
(* n decision variables in 0..n-1 *) let queens = Fd.array n 0 (n-1) in
(* 2n auxiliary variables for diagonals *) let shift op = Array.mapi (fun i qi -> Arith.e2fd (op (fd2e qi) (i2e i))) queens in let diag1 = shift (+~) and diag2 = shift (-~) in
(* Global constraints *) Cstr.post (Alldiff.cstr queens); Cstr.post (Alldiff.cstr diag1); Cstr.post (Alldiff.cstr diag2);
(* Heuristic Min Size, Min Value *) let h a = (Var.Attr.size a, Var.Attr.min a) in let min_min = Goals.Array.choose_index (fun a1 a2 -> h a1 < h a2) in
(* Search goal *) let labeling = Goals.Array.forall ~select:min_min Goals.indomain in
(* Solve *) let bt = ref 0 in if Goals.solve ~control:(fun b -> bt := b) (labeling queens) then begin Printf.printf "%d backtracks\n" !bt; print queens end else prerr_endline "No solution"
let _ =
if Array.length Sys.argv <> 2
then raise (Failure "Usage: queens <nb of queens>");
Gc.set ({(Gc.get ()) with Gc.space_overhead = 500}); (* May help except with an underRAMed system *)
queens (int_of_string Sys.argv.(1));;</lang>
A stand-alone OCaml solution
<lang ocaml>let solutions n =
let show board =
let pr v =
for i = 1 to n do
print_string (if i=v then " q" else " _");
done;
print_newline() in
List.iter pr board;
print_newline() in
let rec safe i j k = function | [] -> true | h::t -> h<>i && h<>j && h<>k && safe i (j+1) (k-1) t in
let rec loop col p =
for i = 1 to n
do
if safe i (i+1) (i-1) p then
let p' = i::p in
if col = n then show p'
else loop (col+1) p'
done in
loop 1 [] in
let n =
if Array.length Sys.argv > 1 then int_of_string Sys.argv.(1) else 8 in
solutions n</lang> With output
$ ocaml queens.ml 6 _ _ _ _ q _ _ _ q _ _ _ q _ _ _ _ _ _ _ _ _ _ q _ _ _ q _ _ _ q _ _ _ _ _ _ _ q _ _ q _ _ _ _ _ _ _ _ _ q _ _ q _ _ _ _ _ _ _ _ _ q _ _ q _ _ _ _ _ q _ _ _ _ _ _ _ _ q _ q _ _ _ _ _ _ _ _ q _ q _ _ _ _ _ _ _ _ q _ _ _ q _ _ _ _ _ _ _ q _ _ _ _ _ _ _ q q _ _ _ _ _ _ _ q _ _ _ _ _ _ _ q _
Oz
A pretty naive solution, using constraint programming: <lang oz>declare
fun {Queens N}
proc {$ Board}
%% a board is a N-tuple of rows
Board = {MakeTuple queens N}
for Y in 1..N do
%% a row is a N-tuple of values in [0,1]
%% (0: no queen, 1: queen)
Board.Y = {FD.tuple row N 0#1}
end
{ForAll {Rows Board} SumIs1}
{ForAll {Columns Board} SumIs1}
%% for every two points on a diagonal
for [X1#Y1 X2#Y2] in {DiagonalPairs N} do
%$ at most one of them has a queen
Board.Y1.X1 + Board.Y2.X2 =<: 1
end
%% enumerate all such boards
{FD.distribute naive {FlatBoard Board}}
end
end
fun {Rows Board}
{Record.toList Board}
end
fun {Columns Board}
for X in {Arity Board.1} collect:C1 do
{C1
for Y in {Arity Board} collect:C2 do
{C2 Board.Y.X}
end}
end
end
proc {SumIs1 Xs}
{FD.sum Xs '=:' 1}
end
fun {DiagonalPairs N}
proc {Coords Root}
[X1#Y1 X2#Y2] = Root
Diff
in
X1::1#N Y1::1#N
X2::1#N Y2::1#N
%% (X1,Y1) and (X2,Y2) are on a diagonal if {Abs X2-X1} = {Abs Y2-Y1}
Diff::1#N-1
{FD.distance X2 X1 '=:' Diff}
{FD.distance Y2 Y1 '=:' Diff}
%% enumerate all such coordinates
{FD.distribute naive [X1 Y1 X2 Y2]}
end
in
{SearchAll Coords}
end
fun {FlatBoard Board}
{Flatten {Record.toList {Record.map Board Record.toList}}}
end
Solutions = {SearchAll {Queens 8}}
in
{Length Solutions} = 92 %% assert
{Inspect {List.take Solutions 3}}</lang>
There is a more concise and much more efficient solution in the Mozart documentation.
Pascal
<lang pascal>program queens;
const l=16;
var i,j,k,m,n,p,q,r,y,z: integer;
a,s: array[1..l] of integer; u: array[1..4*l-2] of integer;
label L3,L4,L5,L6,L7,L8,L9,L10;
begin
for i:=1 to l do a[i]:=i;
for i:=1 to 4*l-2 do u[i]:=0;
for n:=1 to l do
begin
m:=0;
i:=1;
r:=2*n-1;
goto L4;
L3:
s[i]:=j;
u[p]:=1;
u[q+r]:=1;
i:=i+1;
L4:
if i>n then goto L8;
j:=i;
L5:
z:=a[i];
y:=a[j];
p:=i-y+n;
q:=i+y-1;
a[i]:=y;
a[j]:=z;
if (u[p]=0) and (u[q+r]=0) then goto L3;
L6:
j:=j+1;
if j<=n then goto L5;
L7:
j:=j-1;
if j=i then goto L9;
z:=a[i];
a[i]:=a[j];
a[j]:=z;
goto L7;
L8:
m:=m+1;
{ uncomment the following to print solutions }
{ write(n,' ',m,':');
for k:=1 to n do write(' ',a[k]);
writeln; }
L9:
i:=i-1;
if i=0 then goto L10;
p:=i-a[i]+n;
q:=i+a[i]-1;
j:=s[i];
u[p]:=0;
u[q+r]:=0;
goto L6;
L10:
writeln(n,' ',m); end;
end.
{ 1 1
2 0 3 0 4 2 5 10 6 4 7 40 8 92 9 352 10 724 11 2680 12 14200 13 73712 14 365596 15 2279184 16 14772512 }</lang>
Perl
<lang perl>my ($board_size, @occupied, @past, @solutions);
sub try_column {
my ($depth, @diag) = shift;
if ($depth == $board_size) {
push @solutions, "@past\n";
return;
}
# @diag: marks cells diagonally attackable by any previous queens.
# Here it's pre-allocated to double size just so we don't need
# to worry about negative indices.
$#diag = 2 * $board_size;
for (0 .. $#past) {
$diag[ $past[$_] + $depth - $_ ] = 1;
$diag[ $past[$_] - $depth + $_ ] = 1;
}
for my $row (0 .. $board_size - 1) {
next if $occupied[$row] || $diag[$row];
# @past: row numbers of previous queens
# @occupied: rows already used. This gets inherited by each
# recursion so we don't need to repeatedly look them up
push @past, $row;
$occupied[$row] = 1;
try_column($depth + 1);
# clean up, for next recursion
$occupied[$row] = 0;
pop @past;
}
}
$board_size = 12; # takes a minute or so, 14,200 solutions try_column(0);
local $" = "\n"; print @solutions; print "total ", scalar(@solutions), " solutions\n";</lang>
Perl 6
Neither pretty nor efficient, a simple backtracking solution
<lang perl6>sub MAIN($N = 8) {
sub collision(@field, $row) {
for ^$row -> $i {
my $distance = @field[$i] - @field[$row];
return 1 if $distance == any(0, $row - $i, $i - $row);
}
0;
}
sub search(@field is rw, $row) {
if $row == $N {
return @field;
} else {
for ^$N -> $i {
@field[$row] = $i;
if !collision(@field, $row) {
my @r = search(@field, $row + 1) and return @r;
}
}
}
Nil;
}
for 0 .. $N / 2 {
if my @f = search [$_], 1 {
say ~@f;
last;
}
}
}
- output:
0 4 7 5 2 6 1 3</lang>
PHP
Probably not a great solution given this is one of my first forays into PHP. First solves the n rooks problem and then finds solutions for n-queens, disregarding any rotations/reflections. Checked up to n=10.
<lang PHP> <html> <head> <title> n x n Queen solving program </title> </head> <body> <?php
echo "
n x n Queen solving program
";
//Get the size of the board $boardX = $_POST['boardX']; $boardY = $_POST['boardX'];
// Function to rotate a board 90 degrees function rotateBoard($p, $boardX) { $a=0; while ($a < count($p)) { $b = strlen(decbin($p[$a]))-1; $tmp[$b] = 1 << ($boardX - $a - 1); ++$a; } ksort($tmp); return $tmp; }
// This function will find rotations of a solution function findRotation($p, $boardX,$solutions){ $tmp = rotateBoard($p,$boardX); // Rotated 90 if (in_array($tmp,$solutions)) {} else {$solutions[] = $tmp;}
$tmp = rotateBoard($tmp,$boardX); // Rotated 180 if (in_array($tmp,$solutions)){} else {$solutions[] = $tmp;}
$tmp = rotateBoard($tmp,$boardX); // Rotated 270 if (in_array($tmp,$solutions)){} else {$solutions[] = $tmp;}
// Reflected $tmp = array_reverse($p); if (in_array($tmp,$solutions)){} else {$solutions[] = $tmp;}
$tmp = rotateBoard($tmp,$boardX); // Reflected and Rotated 90 if (in_array($tmp,$solutions)){} else {$solutions[] = $tmp;}
$tmp = rotateBoard($tmp,$boardX); // Reflected and Rotated 180 if (in_array($tmp,$solutions)){} else {$solutions[] = $tmp;}
$tmp = rotateBoard($tmp,$boardX); // Reflected and Rotated 270 if (in_array($tmp,$solutions)){} else {$solutions[] = $tmp;} return $solutions; }
// This is a function which will render the board function renderBoard($p,$boardX) {
echo "
"; for ($y = 0; $y < $boardX; ++$y) { echo ''; for ($x = 0; $x < $boardX; ++$x){ if (($x+$y) & 1) { $cellCol = '#9C661F';} else {$cellCol = '#FCE6C9';} if ($p[$y] == 1 << $x) { echo "";} else { echo "";}}
echo ''; } echo '| <img width=30 height=30 src='./images/blackqueen.png'> | |
 ';
}
//This function allows me to generate the next order of rows. function pc_next_permutation($p) { $size = count($p) - 1; // slide down the array looking for where we're smaller than the next guy
for ($i = $size - 1; $p[$i] >= $p[$i+1]; --$i) { }
// if this doesn't occur, we've finished our permutations // the array is reversed: (1, 2, 3, 4) => (4, 3, 2, 1) if ($i == -1) { return false; }
// slide down the array looking for a bigger number than what we found before for ($j = $size; $p[$j] <= $p[$i]; --$j) { } // swap them $tmp = $p[$i]; $p[$i] = $p[$j]; $p[$j] = $tmp; // now reverse the elements in between by swapping the ends for (++$i, $j = $size; $i < $j; ++$i, --$j) { $tmp = $p[$i]; $p[$i] = $p[$j]; $p[$j] = $tmp; } return $p; }
//This function needs to check the current state to see if there are any function checkBoard($p,$boardX) { $a = 0; //this is the row being checked while ($a < count($p)) { $b = 1; while ($b < ($boardX - $a)){ $x = $p[$a+$b] << $b; $y = $p[$a+$b] >> $b; if ($p[$a] == $x | $p[$a] == $y) { return false;} ++$b; } ++$a; } return true; }
if (isset($_POST['process']) && isset($_POST['boardX']))
{
//Within here is the code that needs to be run if process is clicked.
//First I need to create the different possible rows
for ($x = 0; $x < $boardX; ++$x){
$row[$x] = 1 << $x;
}
//Now I need to create all the possible orders of rows, will be equal to [boardY]! $solcount = 0; $solutions = array(); while ($row != false) { if (checkBoard($row,$boardX)){ if(!in_array($row,$solutions)){ $solutions[] = $row; renderBoard($row,$boardX); $solutions = findRotation($row,$boardX,$solutions); ++$solcount; }
} $row = pc_next_permutation($row);
}
echo "
    Rows/Columns: ".$boardX."
    Unique Solutions: ".$solcount."
    Total Solutions: ".count($solutions)." - Note: This includes symmetrical solutions
";
//print_r($solutions);
}
//This code collects the starting parameters echo <<<_END <form name="input" action="queens.php" method="post">     Number of columns/rows <select name="boardX" /> <option value="1">One</option> <option value="2">Two</option> <option value="3">Three</option> <option value="4" >Four</option> <option value="5">Five</option> <option value="6">Six</option> <option value="7">Seven</option> <option value="8" selected="selected">Eight</option> <option value="9">Nine</option> <option value="10">Ten</option> </select>
<input type="hidden" name="process" value="yes" />
 <input type="submit" value="Process" /> </form>
_END;
?> </body> </html></lang>
PicoLisp
Calling 'permute'
<lang PicoLisp>(load "@lib/simul.l") # for 'permute'
(de queens (N)
(let (R (range 1 N) Cnt 0)
(for L (permute (range 1 N))
(when
(= N # from the Python solution
(length (uniq (mapcar + L R)))
(length (uniq (mapcar - L R))) )
(inc 'Cnt) ) )
Cnt ) )</lang>
Permuting inline
This alternative version does not first pre-generate all permutations with 'permute', but creates them recursively. Also, it directly checks for duplicates, instead of calling 'uniq' and 'length'. This is much faster. <lang PicoLisp>(de queens (N)
(let (R (range 1 N) L (copy R) X L Cnt 0)
(recur (X) # Permute
(if (cdr X)
(do (length X)
(recurse (cdr X))
(rot X) )
(or
(seek # Direct check for duplicates
'((L) (member (car L) (cdr L)))
(mapcar + L R) )
(seek
'((L) (member (car L) (cdr L)))
(mapcar - L R) )
(inc 'Cnt) ) ) )
Cnt ) )</lang>
Output in both cases:
: (queens 8) -> 92
PowerBASIC
<lang powerbasic> defint a-z
option base 1 input "n=",n dim a(n), s(n), u(4*n-2) for i=1 to n: a(i)=i: next for i=1 to 4*n-2: u(i)=0: next m=0 i=1 r=2*n-1 goto 20
10 s(i)=j
u(p)=1 u(q+r)=1 incr i
20 if i>n goto 60
j=i
30 z=a(i)
y=a(j) p=i-y+n q=i+y-1 a(i)=y a(j)=z if u(p)=0 and u(q+r)=0 goto 10
40 incr j
if j<=n goto 30
50 decr j
if j=i goto 70 swap a(i),a(j) goto 50
60 incr m
for k=1 to n: print a(k);: next: print
70 decr i
if i=0 goto 80 p=i-a(i)+n q=i+a(i)-1 j=s(i) u(p)=0 u(q+r)=0 goto 40
80 print m</lang>
Prolog
The code for these samples is taken from [1].
Solution #1: <lang Prolog>solution([]).
solution([X/Y|Others]) :-
solution(Others), member(Y, [1,2,3,4,5,6,7,8]), noattack(X/Y, Others).
noattack(_,[]).
noattack(X/Y,[X1/Y1|Others]) :-
Y =\= Y1, Y1 - Y =\= X1 - X, Y1 - Y =\= X - X1, noattack(X/Y,Others).
member(Item,[Item|Rest]).
member(Item,[First|Rest]) :-
member(Item,Rest).
template([1/Y1,2/Y2,3/Y3,4/Y4,5/Y5,6/Y6,7/Y7,8/Y8]).</lang>
Solution #2: <lang Prolog>solution(Queens) :-
permutation([1,2,3,4,5,6,7,8], Queens), safe(Queens).
permutation([],[]).
permutation([Head|Tail],PermList) :-
permutation(Tail,PermTail), del(Head,PermList,PermTail).
del(Item,[Item|List],List).
del(Item,[First|List],[First|List1]) :-
del(Item,List,List1).
safe([]).
safe([Queen|Others]) :-
safe(Others), noattack(Queen,Others,1).
noattack(_,[],_).
noattack(Y,[Y1|Ylist],Xdist) :-
Y1-Y=\=Xdist, Y-Y1=\=Xdist, Dist1 is Xdist + 1, noattack(Y,Ylist,Dist1).</lang>
Solution #3: <lang Prolog>solution(Ylist) :-
sol(Ylist,[1,2,3,4,5,6,7,8], [1,2,3,4,5,6,7,8], [-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7], [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]).
sol([],[],[],Du,Dv).
sol([Y|Ylist],[X|Dx1],Dy,Du,Dv) :-
del(Y,Dy,Dy1), U is X-Y, del(U,Du,Du1), V is X+Y, del(V,Dv,Dv1), sol(Ylist,Dx1, Dy1,Du1,Dv1).
del(Item,[Item|List],List).
del(Item,[First|List],[First|List1]) :-
del(Item,List,List1).</lang>
?- findall(S, solution(S), LS), length(LS,N), write(N). 92
Alternative Solution
Uses backtracking- a highly efficient mechanism in Prolog to find all solutions.
<lang prolog>% 8 queens problem. % q(Row) represents a queen, allocated one per row. No rows ever clash. % The columns are chosen iteratively from available columns held in a % list, reduced with each allocation, so we need never check verticals. % For diagonals, we check prior to allocation whether each newly placed % queen will clash with any of the prior placements. This prevents % most invalid permutations from ever being attempted. can_place(_, []) :- !. % success for empty board can_place(q(R,C),Board) :- % check diagonals against allocated queens member(q(Ra,Ca), Board), abs(Ra-R) =:= abs(Ca-C), !, fail. can_place(_,_). % succeed if no diagonals failed
queens([], [], Board, Board). % found a solution queens([q(R)|Queens], Columns, Board, Solution) :- nth0(_,Columns,C,Free), can_place(q(R,C),Board), % find all solutions queens(Queens,Free,[q(R,C)|Board], Solution). % recursively
queens :-
findall(q(N), between(0,7,N), Queens), findall(N, between(0,7,N), Columns),
findall(B, queens(Queens, Columns, [], B), Boards), % backtrack over all
length(Boards, Len), writef('%w solutions:\n', [Len]), % Output solutions
member(R,Boards), reverse(R,Board), writef(' - %w\n', [Board]), fail.
queens.</lang>
- Output:
?- queens. 92 solutions: - [q(0,0),q(1,4),q(2,7),q(3,5),q(4,2),q(5,6),q(6,1),q(7,3)] - [q(0,0),q(1,5),q(2,7),q(3,2),q(4,6),q(5,3),q(6,1),q(7,4)] - [q(0,0),q(1,6),q(2,3),q(3,5),q(4,7),q(5,1),q(6,4),q(7,2)] - [q(0,0),q(1,6),q(2,4),q(3,7),q(4,1),q(5,3),q(6,5),q(7,2)] ... - [q(0,7),q(1,1),q(2,4),q(3,2),q(4,0),q(5,6),q(6,3),q(7,5)] - [q(0,7),q(1,2),q(2,0),q(3,5),q(4,1),q(5,4),q(6,6),q(7,3)] - [q(0,7),q(1,3),q(2,0),q(3,2),q(4,5),q(5,1),q(6,6),q(7,4)] true.
PureBasic
A recursive approach is taken. A queen is placed in an unused column for each new row. An array keeps track if a queen has already been placed in a given column so that no duplicate columns result. That handles the Rook attacks. Bishop attacks are handled by checking the diagonal alignments of each new placement against the previously placed queens and if an attack is possible the solution backtracks. The solutions are kept track of in a global variable and the routine queens(n) is called with the required number of queens specified. <lang PureBasic>Global solutions
Procedure showBoard(Array queenCol(1))
Protected row, column, n = ArraySize(queenCol())
PrintN(" Solution " + Str(solutions))
For row = 0 To n
For column = 0 To n
If queenCol(row) = column
Print("|Q")
Else
Print("| ")
EndIf
Next
PrintN("|")
Next
EndProcedure
Macro advanceIfPossible()
x + 1 While x <= n And columns(x): x + 1: Wend If x > n ProcedureReturn #False ;backtrack EndIf
EndMacro
Procedure placeQueens(Array queenCol(1), Array columns(1), row = 0)
Protected n = ArraySize(queenCol())
If row > n
solutions + 1
showBoard(queenCol())
ProcedureReturn #False ;backtrack
EndIf
Protected x, queen, passed
While columns(x): x + 1: Wend
;place a new queen in one of the available columns
Repeat
passed = #True
For queen = 0 To row - 1
If ((queenCol(queen) - x) = (queen - row)) Or ((queenCol(queen) - x) = -(queen - row))
advanceIfPossible()
passed = #False
Break ;ForNext loop
EndIf
Next
If passed
queenCol(row) = x: columns(x) = 1
If Not placeQueens(queenCol(), columns(), row + 1)
columns(x) = 0
advanceIfPossible()
EndIf
EndIf
ForEver
EndProcedure
Procedure queens(n)
If n > 0 Dim queenCol(n - 1) Dim columns(n - 1) placeQueens(queenCol(), columns()) EndIf
EndProcedure
If OpenConsole()
Define i For i = 1 To 12 solutions = 0 queens(i) PrintN(#CRLF$ + Str(solutions) + " solutions found for " + Str(i) + "-queens.") Input() Next Print(#CRLF$ + "Press ENTER to exit") Input() CloseConsole()
EndIf</lang> Sample output showing the last solution (all are actually displayed) for 1 - 12 queens:
Solution 1
|Q|
1 solutions found for 1-queens. {Press ENTER}
0 solutions found for 2-queens. {Press ENTER}
0 solutions found for 3-queens. {Press ENTER}
Solution 2
| | |Q| |
|Q| | | |
| | | |Q|
| |Q| | |
2 solutions found for 4-queens. {Press ENTER}
Solution 10
| | | | |Q|
| | |Q| | |
|Q| | | | |
| | | |Q| |
| |Q| | | |
10 solutions found for 5-queens. {Press ENTER}
Solution 4
| | | | |Q| |
| | |Q| | | |
|Q| | | | | |
| | | | | |Q|
| | | |Q| | |
| |Q| | | | |
4 solutions found for 6-queens. {Press ENTER}
Solution 40
| | | | | | |Q|
| | | | |Q| | |
| | |Q| | | | |
|Q| | | | | | |
| | | | | |Q| |
| | | |Q| | | |
| |Q| | | | | |
40 solutions found for 7-queens. {Press ENTER}
Solution 92
| | | | | | | |Q|
| | | |Q| | | | |
|Q| | | | | | | |
| | |Q| | | | | |
| | | | | |Q| | |
| |Q| | | | | | |
| | | | | | |Q| |
| | | | |Q| | | |
92 solutions found for 8-queens. {Press ENTER}
Solution 352
| | | | | | | | |Q|
| | | | | | |Q| | |
| | | |Q| | | | | |
| |Q| | | | | | | |
| | | | | | | |Q| |
| | | | | |Q| | | |
|Q| | | | | | | | |
| | |Q| | | | | | |
| | | | |Q| | | | |
352 solutions found for 9-queens. {Press ENTER}
Solution 724
| | | | | | | | | |Q|
| | | | | | | |Q| | |
| | | | |Q| | | | | |
| | |Q| | | | | | | |
|Q| | | | | | | | | |
| | | | | |Q| | | | |
| |Q| | | | | | | | |
| | | | | | | | |Q| |
| | | | | | |Q| | | |
| | | |Q| | | | | | |
724 solutions found for 10-queens. {Press ENTER}
Solution 2680
| | | | | | | | | | |Q|
| | | | | | | | |Q| | |
| | | | | | |Q| | | | |
| | | | |Q| | | | | | |
| | |Q| | | | | | | | |
|Q| | | | | | | | | | |
| | | | | | | | | |Q| |
| | | | | | | |Q| | | |
| | | | | |Q| | | | | |
| | | |Q| | | | | | | |
| |Q| | | | | | | | | |
2680 solutions found for 11-queens. {Press ENTER}
Solution 14200
| | | | | | | | | | | |Q|
| | | | | | | | | |Q| | |
| | | | | | | |Q| | | | |
| | | | |Q| | | | | | | |
| | |Q| | | | | | | | | |
|Q| | | | | | | | | | | |
| | | | | | |Q| | | | | |
| |Q| | | | | | | | | | |
| | | | | | | | | | |Q| |
| | | | | |Q| | | | | | |
| | | |Q| | | | | | | | |
| | | | | | | | |Q| | | |
14200 solutions found for 12-queens. {Press ENTER}
Python
Raymond Hettingers permutations based solution
This solution, originally by Raymond Hettinger for demonstrating the power of the itertools module, generates all solutions.
<lang python>from itertools import permutations
n = 8 cols = range(n) for vec in permutations(cols):
if n == len(set(vec[i]+i for i in cols)) \
== len(set(vec[i]-i for i in cols)):
print ( vec )</lang>
The output is presented in vector form (each number represents the column position of a queen on consecutive rows). The vector can be pretty printed by substituting a call to board instead of print, with the same argument, and where board is pre-defined as:
<lang python>def board(vec):
print ("\n".join('.' * i + 'Q' + '.' * (n-i-1) for i in vec) + "\n===\n")</lang>
Raymond's description is:
- With the solution represented as a vector with one queen in each row, we don't have to check to see if two queens are on the same row. By using a permutation generator, we know that no value in the vector is repeated, so we don't have to check to see if two queens are on the same column. Since rook moves don't need to be checked, we only need to check bishop moves.
- The technique for checking the diagonals is to add or subtract the column number from each entry, so any two entries on the same diagonal will have the same value (in other words, the sum or difference is unique for each diagonal). Now all we have to do is make sure that the diagonals for each of the eight queens are distinct. So, we put them in a set (which eliminates duplicates) and check that the set length is eight (no duplicates were removed).
- Any permutation with non-overlapping diagonals is a solution. So, we print it and continue checking other permutations.
One disadvantage with this solution is that we can't simply "skip" all the permutations that start with a certain prefix, after discovering that that prefix is incompatible. For example, it is easy to verify that no permutation of the form (1,2,...) could ever be a solution, but since we don't have control over the generation of the permutations, we can't just tell it to "skip" all the ones that start with (1,2).
Alternative Solution
<lang python># From: http://wiki.python.org/moin/SimplePrograms, with permission from the author, Steve Howell BOARD_SIZE = 8
def under_attack(col, queens):
return col in queens or \
any(abs(col - x) == len(queens)-i for i,x in enumerate(queens))
def solve(n):
solutions = [[]]
for row in range(n):
solutions = [solution+[i+1]
for solution in solutions
for i in range(BOARD_SIZE)
if not under_attack(i+1, solution)]
return solutions
for answer in solve(BOARD_SIZE): print(list(enumerate(answer, start=1)))</lang>
Simple Backtracking Solution
A surprisingly simple change to the above code (changing the list comprehension to a generator expression) produces a backtracking solution:
<lang python>BOARD_SIZE = 8
def under_attack(col, queens):
return col in queens or \
any(abs(col - x) == len(queens)-i for i,x in enumerate(queens))
def solve(n):
solutions = [[]]
for row in range(n):
solutions = (solution+[i+1]
for solution in solutions # first for clause is evaluated immediately,
# so "solutions" is correctly captured
for i in range(BOARD_SIZE)
if not under_attack(i+1, solution))
return solutions
answers = solve(BOARD_SIZE) first_answer = next(answers) print(list(enumerate(first_answer, start=1)))</lang>
R
<lang r># Brute force, see the "Permutations" page for the next.perm function safe <- function(p) { n <- length(p) for(i in 1:(n-1)) { for(j in (i+1):n) { if(abs(p[j] - p[i]) == abs(j - i)) return(FALSE) } } return(TRUE) }
queens <- function(n) { p <- 1:n k <- 0 while(!is.null(p)) { if(safe(p)) { cat(p,"\n") k <- k + 1 } p <- next.perm(p) } return(k) }
queens(8)
- 1 5 8 6 3 7 2 4
- ...
- 92</lang>
Racket
Backtracking algorithm; returns one solution
<lang racket>
- lang racket
(struct Q (x y) #:transparent)
- returns true if given q1 and q2 do not conflict
(define (safe? q1 q2)
(match* (q1 q2)
[((Q x1 y1) (Q x2 y2))
(not (or (= x1 x2) (= y1 y2)
(= (abs (- x1 x2)) (abs (- y1 y2)))))]))
- returns true if given q doesn't conflict with anything in given list of qs
(define (safe-lst? q qs) (for/and ([q2 qs]) (safe? q q2)))
(define (nqueens n)
;; qs is partial solution; x y is current position to try
(let loop ([qs null] [x 0] [y 0])
(cond [(= (length qs) n) qs] ; found a solution
[(>= x n) (loop qs 0 (add1 y))] ; go to next row
[(>= y n) #f] ; current solution is invalid
[else
(define q (Q x y))
(if (safe-lst? q qs) ; is current position safe?
(or (loop (cons q qs) 0 (add1 y)) ; optimistically place a queen
; (and move pos to next row)
(loop qs (add1 x) y)) ; backtrack if it fails
(loop qs (add1 x) y))])))
(nqueens 8)
- => (list (Q 3 7) (Q 1 6) (Q 6 5) (Q 2 4) (Q 5 3) (Q 7 2) (Q 4 1) (Q 0 0))
</lang>
Show result with "How to Design Programs" GUI. <lang racket> (require htdp/show-queen)
(define (show-nqueens n)
(define qs (time (nqueens n)))
(show-queen
(for/list ([row n])
(for/list ([col n])
(if (member (Q row col) qs) #t #f)))))
(show-nqueens 8) </lang>
When hovering mouse, GUI also displays conflicts for potential additional queens.
Lazy-style solution, ie, generate all solutions, then filter out invalid ones.
Computes all solutions.
<lang racket>
- lang racket
(struct Q (x y) #:transparent)
(define-syntax-rule (lcons x y) (cons x (lazy y)))
(define (lazy-filter p? lst)
(define flst (force lst))
(if (null? flst) '()
(let ([x (car flst)])
(if (p? x)
(lcons x (lazy-filter p? (cdr flst)))
(lazy-filter p? (cdr flst))))))
(define (lazy-foldr f base lst)
(define flst (force lst))
(if (null? flst) base
(f (car flst) (lazy (lazy-foldr f base (cdr flst))))))
(define (tails lst)
(if (null? lst) '(())
(cons lst (tails (cdr lst)))))
(define (safe? q1 q2)
(match* (q1 q2)
[((Q x1 y1) (Q x2 y2))
(not (or (= x1 x2) (= y1 y2)
(= (abs (- x1 x2)) (abs (- y1 y2)))))]))
(define (safe-lst? lst)
(or (null? lst)
(let ([q1 (car lst)])
(for/and ([q2 (cdr lst)]) (safe? q1 q2)))))
(define (valid? lst) (andmap safe-lst? (tails lst)))
(define (nqueens n)
(define all-possible-solutions
(for/fold ([qss-so-far '(())]) ([row (in-range n)])
(lazy-foldr
(Ξ» (qs new-qss)
(append (for/list ([col (in-range n)]) (cons (Q row col) qs))
new-qss))
'() qss-so-far)))
(lazy-filter valid? all-possible-solutions))
</lang>
Taking the first solution does not compute the other solutions:
<lang racket> (car (nqueens 8))
- => (list (Q 7 3) (Q 6 1) (Q 5 6) (Q 4 2) (Q 3 5) (Q 2 7) (Q 1 4) (Q 0 0))
</lang>
Computing all solutions is also possible:
<lang racket> (define (force-and-print qs)
(define forced (force qs)) (unless (null? forced) (printf "~v\n" (car forced)) (force-and-print (cdr forced))))
(force-and-print (nqueens 8))
- =>
- (list (Q 7 3) (Q 6 1) (Q 5 6) (Q 4 2) (Q 3 5) (Q 2 7) (Q 1 4) (Q 0 0))
- (list (Q 7 4) (Q 6 1) (Q 5 3) (Q 4 6) (Q 3 2) (Q 2 7) (Q 1 5) (Q 0 0))
- (list (Q 7 2) (Q 6 4) (Q 5 1) (Q 4 7) (Q 3 5) (Q 2 3) (Q 1 6) (Q 0 0))
- (list (Q 7 2) (Q 6 5) (Q 5 3) (Q 4 1) (Q 3 7) (Q 2 4) (Q 1 6) (Q 0 0))
- (list (Q 7 4) (Q 6 6) (Q 5 0) (Q 4 2) (Q 3 7) (Q 2 5) (Q 1 3) (Q 0 1))
- (list (Q 7 3) (Q 6 5) (Q 5 7) (Q 4 2) (Q 3 0) (Q 2 6) (Q 1 4) (Q 0 1))
- (list (Q 7 2) (Q 6 5) (Q 5 7) (Q 4 0) (Q 3 3) (Q 2 6) (Q 1 4) (Q 0 1))
- (list (Q 7 4) (Q 6 2) (Q 5 7) (Q 4 3) (Q 3 6) (Q 2 0) (Q 1 5) (Q 0 1))
- (list (Q 7 4) (Q 6 6) (Q 5 3) (Q 4 0) (Q 3 2) (Q 2 7) (Q 1 5) (Q 0 1))
- (list (Q 7 3) (Q 6 0) (Q 5 4) (Q 4 7) (Q 3 5) (Q 2 2) (Q 1 6) (Q 0 1))
- (list (Q 7 2) (Q 6 5) (Q 5 3) (Q 4 0) (Q 3 7) (Q 2 4) (Q 1 6) (Q 0 1))
- (list (Q 7 3) (Q 6 6) (Q 5 4) (Q 4 2) (Q 3 0) (Q 2 5) (Q 1 7) (Q 0 1))
- (list (Q 7 5) (Q 6 3) (Q 5 1) (Q 4 7) (Q 3 4) (Q 2 6) (Q 1 0) (Q 0 2))
- (list (Q 7 5) (Q 6 3) (Q 5 6) (Q 4 0) (Q 3 7) (Q 2 1) (Q 1 4) (Q 0 2))
- (list (Q 7 0) (Q 6 6) (Q 5 3) (Q 4 5) (Q 3 7) (Q 2 1) (Q 1 4) (Q 0 2))
- (list (Q 7 5) (Q 6 7) (Q 5 1) (Q 4 3) (Q 3 0) (Q 2 6) (Q 1 4) (Q 0 2))
- (list (Q 7 5) (Q 6 1) (Q 5 6) (Q 4 0) (Q 3 3) (Q 2 7) (Q 1 4) (Q 0 2))
- (list (Q 7 3) (Q 6 6) (Q 5 0) (Q 4 7) (Q 3 4) (Q 2 1) (Q 1 5) (Q 0 2))
- (list (Q 7 4) (Q 6 7) (Q 5 3) (Q 4 0) (Q 3 6) (Q 2 1) (Q 1 5) (Q 0 2))
- (list (Q 7 3) (Q 6 7) (Q 5 0) (Q 4 4) (Q 3 6) (Q 2 1) (Q 1 5) (Q 0 2))
- (list (Q 7 1) (Q 6 6) (Q 5 4) (Q 4 7) (Q 3 0) (Q 2 3) (Q 1 5) (Q 0 2))
- (list (Q 7 0) (Q 6 6) (Q 5 4) (Q 4 7) (Q 3 1) (Q 2 3) (Q 1 5) (Q 0 2))
- (list (Q 7 1) (Q 6 4) (Q 5 6) (Q 4 3) (Q 3 0) (Q 2 7) (Q 1 5) (Q 0 2))
- (list (Q 7 3) (Q 6 1) (Q 5 6) (Q 4 4) (Q 3 0) (Q 2 7) (Q 1 5) (Q 0 2))
- (list (Q 7 4) (Q 6 6) (Q 5 0) (Q 4 3) (Q 3 1) (Q 2 7) (Q 1 5) (Q 0 2))
- (list (Q 7 5) (Q 6 3) (Q 5 0) (Q 4 4) (Q 3 7) (Q 2 1) (Q 1 6) (Q 0 2))
- (list (Q 7 4) (Q 6 0) (Q 5 3) (Q 4 5) (Q 3 7) (Q 2 1) (Q 1 6) (Q 0 2))
- (list (Q 7 4) (Q 6 1) (Q 5 5) (Q 4 0) (Q 3 6) (Q 2 3) (Q 1 7) (Q 0 2))
- (list (Q 7 5) (Q 6 2) (Q 5 6) (Q 4 1) (Q 3 7) (Q 2 4) (Q 1 0) (Q 0 3))
- (list (Q 7 1) (Q 6 6) (Q 5 2) (Q 4 5) (Q 3 7) (Q 2 4) (Q 1 0) (Q 0 3))
- (list (Q 7 6) (Q 6 2) (Q 5 0) (Q 4 5) (Q 3 7) (Q 2 4) (Q 1 1) (Q 0 3))
- (list (Q 7 4) (Q 6 0) (Q 5 7) (Q 4 5) (Q 3 2) (Q 2 6) (Q 1 1) (Q 0 3))
- (list (Q 7 0) (Q 6 4) (Q 5 7) (Q 4 5) (Q 3 2) (Q 2 6) (Q 1 1) (Q 0 3))
- (list (Q 7 2) (Q 6 5) (Q 5 7) (Q 4 0) (Q 3 4) (Q 2 6) (Q 1 1) (Q 0 3))
- (list (Q 7 5) (Q 6 2) (Q 5 0) (Q 4 6) (Q 3 4) (Q 2 7) (Q 1 1) (Q 0 3))
- (list (Q 7 6) (Q 6 4) (Q 5 2) (Q 4 0) (Q 3 5) (Q 2 7) (Q 1 1) (Q 0 3))
- (list (Q 7 6) (Q 6 2) (Q 5 7) (Q 4 1) (Q 3 4) (Q 2 0) (Q 1 5) (Q 0 3))
- (list (Q 7 4) (Q 6 2) (Q 5 0) (Q 4 6) (Q 3 1) (Q 2 7) (Q 1 5) (Q 0 3))
- (list (Q 7 1) (Q 6 4) (Q 5 6) (Q 4 0) (Q 3 2) (Q 2 7) (Q 1 5) (Q 0 3))
- (list (Q 7 2) (Q 6 5) (Q 5 1) (Q 4 4) (Q 3 7) (Q 2 0) (Q 1 6) (Q 0 3))
- (list (Q 7 5) (Q 6 0) (Q 5 4) (Q 4 1) (Q 3 7) (Q 2 2) (Q 1 6) (Q 0 3))
- (list (Q 7 7) (Q 6 2) (Q 5 0) (Q 4 5) (Q 3 1) (Q 2 4) (Q 1 6) (Q 0 3))
- (list (Q 7 1) (Q 6 7) (Q 5 5) (Q 4 0) (Q 3 2) (Q 2 4) (Q 1 6) (Q 0 3))
- (list (Q 7 4) (Q 6 6) (Q 5 1) (Q 4 5) (Q 3 2) (Q 2 0) (Q 1 7) (Q 0 3))
- (list (Q 7 2) (Q 6 5) (Q 5 1) (Q 4 6) (Q 3 4) (Q 2 0) (Q 1 7) (Q 0 3))
- (list (Q 7 5) (Q 6 1) (Q 5 6) (Q 4 0) (Q 3 2) (Q 2 4) (Q 1 7) (Q 0 3))
- (list (Q 7 2) (Q 6 6) (Q 5 1) (Q 4 7) (Q 3 5) (Q 2 3) (Q 1 0) (Q 0 4))
- (list (Q 7 5) (Q 6 2) (Q 5 6) (Q 4 1) (Q 3 3) (Q 2 7) (Q 1 0) (Q 0 4))
- (list (Q 7 3) (Q 6 1) (Q 5 6) (Q 4 2) (Q 3 5) (Q 2 7) (Q 1 0) (Q 0 4))
- (list (Q 7 6) (Q 6 0) (Q 5 2) (Q 4 7) (Q 3 5) (Q 2 3) (Q 1 1) (Q 0 4))
- (list (Q 7 0) (Q 6 5) (Q 5 7) (Q 4 2) (Q 3 6) (Q 2 3) (Q 1 1) (Q 0 4))
- (list (Q 7 2) (Q 6 7) (Q 5 3) (Q 4 6) (Q 3 0) (Q 2 5) (Q 1 1) (Q 0 4))
- (list (Q 7 5) (Q 6 2) (Q 5 6) (Q 4 3) (Q 3 0) (Q 2 7) (Q 1 1) (Q 0 4))
- (list (Q 7 6) (Q 6 3) (Q 5 1) (Q 4 7) (Q 3 5) (Q 2 0) (Q 1 2) (Q 0 4))
- (list (Q 7 3) (Q 6 5) (Q 5 7) (Q 4 1) (Q 3 6) (Q 2 0) (Q 1 2) (Q 0 4))
- (list (Q 7 1) (Q 6 5) (Q 5 0) (Q 4 6) (Q 3 3) (Q 2 7) (Q 1 2) (Q 0 4))
- (list (Q 7 1) (Q 6 3) (Q 5 5) (Q 4 7) (Q 3 2) (Q 2 0) (Q 1 6) (Q 0 4))
- (list (Q 7 2) (Q 6 5) (Q 5 7) (Q 4 1) (Q 3 3) (Q 2 0) (Q 1 6) (Q 0 4))
- (list (Q 7 5) (Q 6 2) (Q 5 0) (Q 4 7) (Q 3 3) (Q 2 1) (Q 1 6) (Q 0 4))
- (list (Q 7 7) (Q 6 3) (Q 5 0) (Q 4 2) (Q 3 5) (Q 2 1) (Q 1 6) (Q 0 4))
- (list (Q 7 3) (Q 6 7) (Q 5 0) (Q 4 2) (Q 3 5) (Q 2 1) (Q 1 6) (Q 0 4))
- (list (Q 7 1) (Q 6 5) (Q 5 7) (Q 4 2) (Q 3 0) (Q 2 3) (Q 1 6) (Q 0 4))
- (list (Q 7 6) (Q 6 1) (Q 5 5) (Q 4 2) (Q 3 0) (Q 2 3) (Q 1 7) (Q 0 4))
- (list (Q 7 2) (Q 6 5) (Q 5 1) (Q 4 6) (Q 3 0) (Q 2 3) (Q 1 7) (Q 0 4))
- (list (Q 7 3) (Q 6 6) (Q 5 2) (Q 4 7) (Q 3 1) (Q 2 4) (Q 1 0) (Q 0 5))
- (list (Q 7 3) (Q 6 7) (Q 5 4) (Q 4 2) (Q 3 0) (Q 2 6) (Q 1 1) (Q 0 5))
- (list (Q 7 2) (Q 6 4) (Q 5 7) (Q 4 3) (Q 3 0) (Q 2 6) (Q 1 1) (Q 0 5))
- (list (Q 7 3) (Q 6 1) (Q 5 7) (Q 4 4) (Q 3 6) (Q 2 0) (Q 1 2) (Q 0 5))
- (list (Q 7 4) (Q 6 6) (Q 5 1) (Q 4 3) (Q 3 7) (Q 2 0) (Q 1 2) (Q 0 5))
- (list (Q 7 6) (Q 6 3) (Q 5 1) (Q 4 4) (Q 3 7) (Q 2 0) (Q 1 2) (Q 0 5))
- (list (Q 7 7) (Q 6 1) (Q 5 3) (Q 4 0) (Q 3 6) (Q 2 4) (Q 1 2) (Q 0 5))
- (list (Q 7 6) (Q 6 1) (Q 5 3) (Q 4 0) (Q 3 7) (Q 2 4) (Q 1 2) (Q 0 5))
- (list (Q 7 4) (Q 6 0) (Q 5 7) (Q 4 3) (Q 3 1) (Q 2 6) (Q 1 2) (Q 0 5))
- (list (Q 7 3) (Q 6 0) (Q 5 4) (Q 4 7) (Q 3 1) (Q 2 6) (Q 1 2) (Q 0 5))
- (list (Q 7 4) (Q 6 1) (Q 5 7) (Q 4 0) (Q 3 3) (Q 2 6) (Q 1 2) (Q 0 5))
- (list (Q 7 2) (Q 6 6) (Q 5 1) (Q 4 7) (Q 3 4) (Q 2 0) (Q 1 3) (Q 0 5))
- (list (Q 7 2) (Q 6 0) (Q 5 6) (Q 4 4) (Q 3 7) (Q 2 1) (Q 1 3) (Q 0 5))
- (list (Q 7 7) (Q 6 1) (Q 5 4) (Q 4 2) (Q 3 0) (Q 2 6) (Q 1 3) (Q 0 5))
- (list (Q 7 2) (Q 6 4) (Q 5 1) (Q 4 7) (Q 3 0) (Q 2 6) (Q 1 3) (Q 0 5))
- (list (Q 7 2) (Q 6 4) (Q 5 6) (Q 4 0) (Q 3 3) (Q 2 1) (Q 1 7) (Q 0 5))
- (list (Q 7 4) (Q 6 1) (Q 5 3) (Q 4 5) (Q 3 7) (Q 2 2) (Q 1 0) (Q 0 6))
- (list (Q 7 5) (Q 6 2) (Q 5 4) (Q 4 7) (Q 3 0) (Q 2 3) (Q 1 1) (Q 0 6))
- (list (Q 7 4) (Q 6 7) (Q 5 3) (Q 4 0) (Q 3 2) (Q 2 5) (Q 1 1) (Q 0 6))
- (list (Q 7 3) (Q 6 1) (Q 5 4) (Q 4 7) (Q 3 5) (Q 2 0) (Q 1 2) (Q 0 6))
- (list (Q 7 3) (Q 6 5) (Q 5 0) (Q 4 4) (Q 3 1) (Q 2 7) (Q 1 2) (Q 0 6))
- (list (Q 7 5) (Q 6 2) (Q 5 0) (Q 4 7) (Q 3 4) (Q 2 1) (Q 1 3) (Q 0 6))
- (list (Q 7 4) (Q 6 2) (Q 5 0) (Q 4 5) (Q 3 7) (Q 2 1) (Q 1 3) (Q 0 6))
- (list (Q 7 3) (Q 6 1) (Q 5 7) (Q 4 5) (Q 3 0) (Q 2 2) (Q 1 4) (Q 0 6))
- (list (Q 7 5) (Q 6 2) (Q 5 4) (Q 4 6) (Q 3 0) (Q 2 3) (Q 1 1) (Q 0 7))
- (list (Q 7 5) (Q 6 3) (Q 5 6) (Q 4 0) (Q 3 2) (Q 2 4) (Q 1 1) (Q 0 7))
- (list (Q 7 3) (Q 6 6) (Q 5 4) (Q 4 1) (Q 3 5) (Q 2 0) (Q 1 2) (Q 0 7))
- (list (Q 7 4) (Q 6 6) (Q 5 1) (Q 4 5) (Q 3 2) (Q 2 0) (Q 1 3) (Q 0 7))
</lang>
Logic borrowed from the Ruby example <lang racket>
- lang racket
(define (remove x lst)
(for/list ([i (in-range (length lst))]
#:when (not (= x i)))
(list-ref lst i)))
(define (switch-pairs lst)
(cond [(null? lst) '()]
[(null? (cdr lst)) (list '() (car lst))]
[else (append (list (cadr lst) (car lst))
(switch-pairs (cddr lst)))]))
(define (switch-places a1 a2 lst)
(for/list ([i (length lst)]) (list-ref lst (cond [(= a1 i) a2] [(= a2 i) a1] [else i]))))
(define (position-queens n)
(cond [(= 1 n) (list (list 1))]
[(> 4 n) #f]
[else (possible-queens n)]))
(define (possible-queens n)
(define rem (remainder n 12))
(define lst (build-list n add1))
(define evens (filter even? lst))
(define odds (filter odd? lst))
(cond [(or (= rem 9) (= rem 3)) (case3or9 evens odds)]
[(= rem 8) (case8 evens odds)]
[(= rem 2) (case2 evens odds)]
[else (append evens odds)]))
(define (case3or9 evens odds)
(for/fold ([acum (append (cdr evens) (list (car evens)) odds)])
([i (in-list '(1 3))])
(append (remove (list-ref acum i) acum) (list i))))
(define (case8 evens odds)
(append evens (switch-pairs odds)))
(define (case2 evens odds)
(define nums (append evens odds))
(define idx (map (Ξ»(i) (list-ref nums i)) '(1 3 5)))
(append (remove (caddr idx)
(switch-places (car idx) (cadr idx) nums))
'(5)))
(define (queens n)
(define position-numbers (position-queens n))
(define positions-on-board
(for/list ([i n]) (cons i (sub1 (list-ref position-numbers i)))))
(for/list ([x n])
(for/list ([y n])
(if (member (cons x y) positions-on-board) "Q" "."))))
(define (print-queens n)
(for ([x (queens n)]) (displayln (string-join x))))
</lang>
Rascal
<lang Rascal>import Prelude;
public set[list[int]] Nqueens(int n){ cols = upTill(n); result = {}; for (vector <- permutations(cols)){ if (n == size({vector[j] + j |j <- cols}) && n == size({vector[j] - j |j <- cols})) result += vector;} return result; }</lang>
REXX
The logic was borrowed from the Fortran example and modified for speed; the display of the chessboard was
also changed to allow for the aspect ratio of display terminals to make the chessboard appear square.
Logic was added to the REXX program to preserve the color for a black square when a queen is on it.
<lang rexx>/*REXX program place N queens on a NxN chessboard (the 8 queens problem)*/
parse arg N . /*get board size arg (if any). */
if N== then N=8 /*No argument? Use the default.*/
file=1; rank=1; q=0 /*starting place, # of queens. */
@.=0; !=left(, 9* (N<18)) /*define empty board, indentation*/
/*βββββββββββββββββββββββββββββββββββββfind solution: N queens problem.*/
do while q<N /*keep placing queens until done.*/
@.file.rank=1 /*place a queen on the chessboard*/
if safe?(file,rank) then do; q=q+1 /*if not being attacked, eureka! */
file=1 /*another attempt at file #1, */
rank=rank+1 /*and also bump the rank pointer.*/
end
else do /*Β¬ safe, so it's a bad placement*/
@.file.rank=0 /* So, remove this queen. */
file=file+1 /*try the next file then. */
do while file>N; rank=rank-1
if rank==0 then call noSol
do j=1 for N
if @.j.rank then do; file=j; @.file.rank=0
q=q-1; file=j+1
leave /*j*/
end
end /*j*/
end /*do while file>N*/
end /*else do*/
end /*while q<N*/
/*ββββββββββββββββββββββββββββββββββββββshow chessboard with a solution.*/ say 'A solution for' N "queens:"; _ = substr( copies("βΌβββ", N) ,2) lineT = 'β'_"β"; say; say ! translate(lineT,'β¬',"βΌ") lineB = 'β'_"β"; lineB = translate(lineB, 'β΄', "βΌ") line = 'β'_"β€" /*define a line for cell boundry.*/ bar = 'β' /*kinds: horizonal/vertical/salad*/ Bqueen = 'βββ' /*glyph befitting the black queen*/ Wqueen = ' β ' /* " " " white " */ /*βββββββββββββββββββββββ==ββββββββββββββplace the queens on chessboard.*/
do r=1 for N; if r\==1 then say ! line; _= /*process the rank &*/
do f=1 for N; black=(f+r)//2 /*the file; is it a black square?*/
qgylph=Wqueen; if black then Qgylph=Bqueen /*use a black queen.*/
/*is it black sqare?*/
if @.f.r then _=_ || bar || Qgylph /*use the 3-char symbol for queen*/
else if black then _=_ || bar'βββ' /*ΒΌ dithering char. */
else _=_ || bar' ' /*three blanks. */
end /*f*/ /* [β] preserve square chessboard*/
say ! _ || bar
end /*r*/ /*80 cols can view 19x19 chessbrd*/
say ! lineB; say /*show last line, + a blank line.*/ exit 1 /*stick a fork in it, we're done.*/ /*ββββββββββββββββββββββββββββββββββNOSOL subroutineββββββββββββββββββββ*/ noSol: say "No solution for" N 'queens.'; exit 0 /*ββββββββββββββββββββββββββββββββββSAFE? subroutineββββββββββββββββββββ*/ safe?: procedure expose @.; parse arg file,rank
do k=rank-1 to 1 by -1; if @.file.k then return 0
end
f=file-1; r=rank-1
do while f\==0 & r\==0; if @.f.r then return 0
f=f-1; r=r-1
end
f=file+1; r=rank-1
do while f<=n & r\==0; if @.f.r then return 0
f=f+1; r=r-1
end
return 1</lang> output (when using the default of an 8x8 chessboard):
A solution for 8 queens:
βββββ¬ββββ¬ββββ¬ββββ¬ββββ¬ββββ¬ββββ¬ββββ
β β βββββ βββββ βββββ βββββ
βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€
βββββ βββββ βββββ βββββ β
βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€
β βββββ βββββ βββββ βββββ
βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€
βββββ βββββ βββββ β βββββ β
βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€
β βββββ β βββββ βββββ βββββ
βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€
βββββ βββββ βββββ βββββ β
βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€
β βββββ βββββ βββββ βββββ
βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€
βββββ βββββ β βββββ βββββ β
βββββ΄ββββ΄ββββ΄ββββ΄ββββ΄ββββ΄ββββ΄ββββ
output when using the input of: 20
A solution for 20 queens: βββββ¬ββββ¬ββββ¬ββββ¬ββββ¬ββββ¬ββββ¬ββββ¬ββββ¬ββββ¬ββββ¬ββββ¬ββββ¬ββββ¬ββββ¬ββββ¬ββββ¬ββββ¬ββββ¬ββββ β β βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ β βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€ β βββββ βββββ β βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€ βββββ β βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ β βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€ β βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ β βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€ β βββββ βββββ βββββ βββββ βββββ βββββ βββββ β βββββ βββββ βββββ βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€ βββββ βββββ βββββ βββββ βββββ βββββ β βββββ βββββ βββββ βββββ β βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€ β βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ β β βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€ β βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ β βββββ βββββ βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ β βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€ β βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ β βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€ β βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€ βββββ βββββ βββββ βββββ βββββ β βββββ βββββ βββββ βββββ βββββ β βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€ β βββββ βββββ βββββ β βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€ βββββ βββββ βββββ βββββ βββββ βββββ βββββ β βββββ βββββ βββββ β βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€ β βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββΌββββ€ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ βββββ β βββββ΄ββββ΄ββββ΄ββββ΄ββββ΄ββββ΄ββββ΄ββββ΄ββββ΄ββββ΄ββββ΄ββββ΄ββββ΄ββββ΄ββββ΄ββββ΄ββββ΄ββββ΄ββββ΄ββββ
Ruby
This implements the heuristics found on the wikipedia page to return just one solution <lang ruby># 1. Divide n by 12. Remember the remainder (n is 8 for the eight queens
- puzzle).
- 2. Write a list of the even numbers from 2 to n in order.
- 3. If the remainder is 3 or 9, move 2 to the end of the list.
- 4. Append the odd numbers from 1 to n in order, but, if the remainder is 8,
- switch pairs (i.e. 3, 1, 7, 5, 11, 9, β¦).
- 5. If the remainder is 2, switch the places of 1 and 3, then move 5 to the
- end of the list.
- 6. If the remainder is 3 or 9, move 1 and 3 to the end of the list.
- 7. Place the first-column queen in the row with the first number in the
- list, place the second-column queen in the row with the second number in
- the list, etc.
def n_queens(n)
if n == 1
return "Q"
elsif n < 4
puts "no solutions for n=#{n}"
return ""
end
evens = (2..n).step(2).to_a odds = (1..n).step(2).to_a
rem = n % 12 # (1) nums = evens # (2)
nums.push(nums.shift) if rem == 3 or rem == 9 # (3)
# (4)
if rem == 8
odds = odds.each_slice(2).inject([]) {|ary, (a,b)| ary += [b,a]}
end
nums.concat(odds)
# (5)
if rem == 2
idx = []
[1,3,5].each {|i| idx[i] = nums.index(i)}
nums[idx[1]], nums[idx[3]] = nums[idx[3]], nums[idx[1]]
nums.slice!(idx[5])
nums.push(5)
end
# (6)
if rem == 3 or rem == 9
[1,3].each do |i|
nums.slice!( nums.index(i) )
nums.push(i)
end
end
# (7)
board = Array.new(n) {Array.new(n) {"."}}
n.times {|i| board[i][nums[i] - 1] = "Q"}
board.inject("") {|str, row| str << row.join(" ") << "\n"}
end
(1 .. 15).each {|n| puts "n=#{n}"; puts n_queens(n); puts}</lang>
Output:
n=1 Q n=2 no solutions for n=2 n=3 no solutions for n=3 n=4 . Q . . . . . Q Q . . . . . Q . n=5 . Q . . . . . . Q . Q . . . . . . Q . . . . . . Q n=6 . Q . . . . . . . Q . . . . . . . Q Q . . . . . . . Q . . . . . . . Q . n=7 . Q . . . . . . . . Q . . . . . . . . Q . Q . . . . . . . . Q . . . . . . . . Q . . . . . . . . Q n=8 . Q . . . . . . . . . Q . . . . . . . . . Q . . . . . . . . . Q . . Q . . . . . Q . . . . . . . . . . . . . Q . . . . . Q . . . n=9 . . . Q . . . . . . . . . . Q . . . . . . . . . . Q . . Q . . . . . . . . . . . Q . . . . . . . . . . Q . . . . . . . . . . Q Q . . . . . . . . . . Q . . . . . . n=10 . Q . . . . . . . . . . . Q . . . . . . . . . . . Q . . . . . . . . . . . Q . . . . . . . . . . . Q Q . . . . . . . . . . . Q . . . . . . . . . . . Q . . . . . . . . . . . Q . . . . . . . . . . . Q . n=11 . Q . . . . . . . . . . . . Q . . . . . . . . . . . . Q . . . . . . . . . . . . Q . . . . . . . . . . . . Q . Q . . . . . . . . . . . . Q . . . . . . . . . . . . Q . . . . . . . . . . . . Q . . . . . . . . . . . . Q . . . . . . . . . . . . Q n=12 . Q . . . . . . . . . . . . . Q . . . . . . . . . . . . . Q . . . . . . . . . . . . . Q . . . . . . . . . . . . . Q . . . . . . . . . . . . . Q Q . . . . . . . . . . . . . Q . . . . . . . . . . . . . Q . . . . . . . . . . . . . Q . . . . . . . . . . . . . Q . . . . . . . . . . . . . Q . n=13 . Q . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . Q . Q . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . Q n=14 . Q . . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . . Q . . Q . . . . . . . . . . . Q . . . . . . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . . Q . . . . . Q . . . . . . . . . n=15 . . . Q . . . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . . . Q . . Q . . . . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . . . Q . . . . . . . . . . . . . . . . Q Q . . . . . . . . . . . . . . . . Q . . . . . . . . . . . .
Run BASIC
<lang runbasic>[loop] input "How many queens (N>=4)";n if n < 4 then
print "Must be greater than 4" goto [loop]
end if
dim plot$(100,100) dim q(n+20) dim e(n+20) dim o(n+20) r=n mod 6 if r<>2 and r<>3 then
gosub [samp] goto [shoBoard]
end if for i=1 to int(n/2)
e(i) = 2 * i
next for i=1 to int((n/2)+.5)
o(i) = 2 *i-1
next if r = 2 then gosub [edt2] if r = 3 then gosub [edt3] s = 1 for i=1 to n
if e(i)>0 then q(s) = e(i) s = s+1 end if
next for i=1 to n
if o(i) > 0 then q(s) = o(i) s = s + 1 end if
next ' print board [shoBoard] cls for i = 1 to n
plot$(i,26-q(i)) = "*" plot$(i,24-n) = chr$(96+i) plot$(n+1,26-i) = str$(i)
next i for ii = 1 to 100
for jj = 1 to 100 print left$(plot$(jj,ii)+" ",1); next jj
print next ii end
' the simple case [samp] p = 1 for i = 1 to n
if i mod 2=0 then q(p) = i p = p + 1 end if
next i for i = 1 to n
if i mod 2 then q(p) = i p = p + 1 end if
next return ' edit list when remainder is 2 [edt2] for i=1 to n
if o(i) = 3 then
o(i) = 1
else
if o(i)=1 then o(i) = 3
end if
if o(i) = 5 then
o(i)= o(i) -1
else
if o(i) = 0 then
o(i) = 5
return
end if
end if
next
' edit list when remainder is 3 [edt3] for i = 1 to n
if e(i) = 2 then
e(i) = e(i)-1
else
if e(i) = 0 then
e(i) = 2
goto [more]
end if
end if
next i ' edit list some more [more] for i = 1 to n
if (o(i)=1 or o(i)=3) then
o(i) = o(i)-1
else
if o(i) = 0 then
o(i) = 1
o(i+1) = 3
return
end if
end if
next</lang>
abcdefgh
* 8
*7
* 6
5
* * 4
* 3
* 2
* 1
SAS
<lang sas>/* Store all 92 permutations in a SAS dataset. Translation of Fortran 77 */ data queens; array a{8} p1-p8; array s{8}; array u{30}; n=8; do i=1 to n; a(i)=i; end; do i=1 to 4*n-2; u(i)=0; end; m=0; i=1; r=2*n-1; goto L40; L30: s(i)=j; u(p)=1; u(q+r)=1; i=i+1; L40: if i>n then goto L80; j=i; L50: z=a(i); y=a(j); p=i-y+n; q=i+y-1; a(i)=y; a(j)=z; if u(p)=0 and u(q+r)=0 then goto L30; L60: j=j+1; if j<=n then goto L50; L70: j=j-1; if j=i then goto L90; z=a(i); a(i)=a(j); a(j)=z; goto L70; L80: m=m+1; output; L90: i=i-1; if i=0 then goto L100; p=i-a(i)+n; q=i+a(i)-1; j=s(i); u(p)=0; u(q+r)=0; goto L60; L100: put n m; keep p1-p8; run;</lang>
Scala
The algorithm below is lazy. It returns an iterator, and each solution is computed as you ask for the next element of the iterator. If you ask for one element, it will only compute one solution.
The test for legal moves is a bit redundant, as the algorithm can never generate two positions in the same row.
<lang scala>case class Pos(row: Int, column: Int) {
def sameRow(p: Pos) = row == p.row def sameColumn(p: Pos) = column == p.column def sameDiag(p: Pos) = (p.column - column).abs == (p.row - row).abs def illegal(p: Pos) = sameRow(p) || sameColumn(p) || sameDiag(p) def legal(p: Pos) = !illegal(p)
}
def rowSet(size: Int, row: Int) = Iterator.tabulate(size)(column => Pos(row, column))
def expand(solutions: Iterator[List[Pos]], size: Int, row: Int) =
for {
solution <- solutions
pos <- rowSet(size, row)
if solution forall (_ legal pos)
} yield pos :: solution
def seed(size: Int) = rowSet(size, 0) map (sol => List(sol))
def solve(size: Int) = (1 until size).foldLeft(seed(size)) (expand(_, size, _))</lang>
Seed7
<lang seed7>$ include "seed7_05.s7i";
var array integer: board is 8 times 0; var integer: solutionNum is 0;
const func boolean: safe (in integer: y) is func
result
var boolean: safe is TRUE;
local
var integer: i is 1;
begin
while i < y and safe do
safe := board[y - i] <> board[y] and
board[y - i] <> board[y] - i and
board[y - i] <> board[y] + i;
incr(i);
end while;
end func;
const proc: putBoard is func
local
var integer: y is 0;
begin
incr(solutionNum);
writeln;
writeln("Solution " <& solutionNum);
for y range 1 to 8 do
writeln("|_" mult pred(board[y]) <& "|Q" <& "|_" mult (8 - board[y]) <& "|");
end for;
end func;
const proc: main is func
local
var integer: y is 1;
begin
while y >= 1 do
repeat
incr(board[y]);
until board[y] > 8 or safe(y);
if board[y] <= 8 then
if y < 8 then
incr(y);
board[y] := 0;
else
putBoard;
end if;
else
decr(y);
end if;
end while;
end func;</lang>
SNOBOL4
<lang SNOBOL4>
- N queens problem
- Set N to the desired number. The program prints out all solution boards.
N = 5 NM1 = N - 1; NP1 = N + 1; NSZ = N * NP1; &STLIMIT = 10 ** 9; &ANCHOR = 1 DEFINE('SOLVE(B)I')
- This pattern tests if the first queen attacks any of the others:
TEST = BREAK('Q') 'Q' (ARBNO(LEN(N) '-') LEN(N) 'Q' + | ARBNO(LEN(NP1) '-') LEN(NP1) 'Q' + | ARBNO(LEN(NM1) '-') LEN(NM1) 'Q') P = LEN(NM1) . X LEN(1); L = 'Q' DUPL('-',NM1) ' ' SOLVE() :(END) SOLVE EQ(SIZE(B),NSZ) :S(PRINT)
- Add another row with a queen:
B = L B LOOP I = LT(I,N) I + 1 :F(RETURN) B TEST :S(NEXT) SOLVE(B)
- Try queen in next square:
NEXT B P = '-' X :(LOOP) PRINT SOLUTION = SOLUTION + 1 OUTPUT = 'Solution number ' SOLUTION ' is:' PRTLOOP B LEN(NP1) . OUTPUT = :S(PRTLOOP)F(RETURN) END </lang>
Standard ML
This implementation uses failure continuations for backtracking. <lang Standard ML> (*
* val threat : (int * int) -> (int * int) -> bool * Returns true iff the queens at the given positions threaten each other *)
fun threat (x, y) (x', y') =
x = x' orelse y = y' orelse abs(x - x') = abs(y - y');
(*
* val conflict : (int * int) -> (int * int) list -> bool * Returns true if there exists a conflict with the position and the list of queens. *)
fun conflict pos = List.exists (threat pos);
(*
* val addqueen : (int * int * (int * int) list * (unit -> (int * int) list option)) -> (int * int) list option * Returns either NONE in the case that no solution exists or SOME(l) where l is a list of positions making up the solution. *)
fun addqueen(i, n, qs, fc) =
let
fun try j =
if j > n then fc()
else if (conflict (i, j) qs) then try (j + 1)
else if i = n then SOME((i, j)::qs)
else addqueen(i + 1, n, (i,j)::qs, fn() => try (j + 1))
in
try 1
end;
(*
* val queens : int -> (int * int) list option * Given the board dimension n, returns a solution for the n-queens problem. *)
fun queens(n) = addqueen(1, n, [], fn () => NONE);
(* SOME [(8,4),(7,2),(6,7),(5,3),(4,6),(3,8),(2,5),(1,1)] *) queens(8);
(* NONE *) queens(2); </lang>
SystemVerilog
Create a random board configuration, with the 8-queens as a constraint <lang SystemVerilog>program N_queens;
parameter SIZE_LOG2 = 3; parameter SIZE = 1 << SIZE_LOG2;
`define ABS_DIFF(a,b) (a>b?a-b:b-a)
class board; rand bit [SIZE_LOG2-1:0] row[SIZE];
constraint rook_moves {
foreach (row[i]) foreach (row[j]) if (i < j) {
row[i] != row[j];
}
}
constraint diagonal_moves {
foreach (row[i]) foreach (row[j]) if (i < j) {
`ABS_DIFF(row[i], row[j]) != `ABS_DIFF(i,j);
}
}
function void next;
randomize;
foreach (row[i]) begin
automatic bit [SIZE-1:0] x = 1 << row[i];
$display( " %b", x );
end
$display("--");
endfunction
endclass
board b = new; initial repeat(1) b.next;
endprogram </lang>
Tcl
This solution is based on the C version on wikipedia. By default it solves the 8-queen case; to solve for any other number, pass N as an extra argument on the script's command line (see the example for the N=6 case, which has anomalously few solutions).
<lang tcl>package require Tcl 8.5
proc unsafe {y} {
global b
set x [lindex $b $y]
for {set i 1} {$i <= $y} {incr i} {
set t [lindex $b [expr {$y - $i}]] if {$t==$x || $t==$x-$i || $t==$x+$i} { return 1 }
} return 0
}
proc putboard {} {
global b s N
puts "\n\nSolution #[incr s]"
for {set y 0} {$y < $N} {incr y} {
for {set x 0} {$x < $N} {incr x} { puts -nonewline [expr {[lindex $b $y] == $x ? "|Q" : "|_"}] } puts "|"
}
}
proc main {n} {
global b N
set N $n
set b [lrepeat $N 0]
set y 0
lset b 0 -1
while {$y >= 0} {
lset b $y [expr {[lindex $b $y] + 1}] while {[lindex $b $y] < $N && [unsafe $y]} { lset b $y [expr {[lindex $b $y] + 1}] } if {[lindex $b $y] >= $N} { incr y -1 } elseif {$y < $N-1} { lset b [incr y] -1; } else { putboard }
}
}
main [expr {$argc ? int(0+[lindex $argv 0]) : 8}]</lang> Sample output:
$ tclsh8.5 8queens.tcl 6 Solution #1 |_|Q|_|_|_|_| |_|_|_|Q|_|_| |_|_|_|_|_|Q| |Q|_|_|_|_|_| |_|_|Q|_|_|_| |_|_|_|_|Q|_| Solution #2 |_|_|Q|_|_|_| |_|_|_|_|_|Q| |_|Q|_|_|_|_| |_|_|_|_|Q|_| |Q|_|_|_|_|_| |_|_|_|Q|_|_| Solution #3 |_|_|_|Q|_|_| |Q|_|_|_|_|_| |_|_|_|_|Q|_| |_|Q|_|_|_|_| |_|_|_|_|_|Q| |_|_|Q|_|_|_| Solution #4 |_|_|_|_|Q|_| |_|_|Q|_|_|_| |Q|_|_|_|_|_| |_|_|_|_|_|Q| |_|_|_|Q|_|_| |_|Q|_|_|_|_|
Ursala
This is invoked as a command line application by queens -n, where n is a number greater than 3. Multiple solutions may be reported but reflections and rotations thereof are omitted. <lang Ursala>#import std
- import nat
remove_reflections = ^D(length@ht,~&); ~&K2hlPS+ * ^lrNCCs/~&r difference*D remove_rotations = ~&K2hlrS2S+ * num; ~&srlXSsPNCCs
- executable <'par',>
- optimize+
queens =
%np+~command.options.&h.keyword.&iNC; -+
~&iNC+ file$[contents: --<>+ mat` *+ %nP*=*],
remove_rotations+ remove_reflections+ ~&rSSs+ nleq-<&l*rFlhthPXPSPS,
~&i&& ~&lNrNCXX; ~&rr->rl ^/~&l ~&lrrhrSiF4E?/~&rrlPlCrtPX @r ^|/~& ^|T\~& -+
-<&l^|*DlrTS/~& ~&iiDlSzyCK9hlPNNXXtCS,
^jrX/~& @rZK20lrpblPOlrEkPK13lhPK2 ~&i&& nleq$-&lh+-,
^/~&NNXS+iota -<&l+ ~&plll2llr2lrPrNCCCCNXS*=irSxPSp+ ^H/block iota; *iiK0 ^/~& sum+-</lang>
The output shows one solution on each line. A solution is reported as a sequence of numbers with the -th number being the index of the occupied row in the -th column.
$ queens -4 2 3 0 1 $ queens -5 0 2 1 3 4 2 4 3 0 1 1 3 2 4 0 $ queens 6 4 3 0 2 1 5
Xanadu
Copied from http://www.cs.bu.edu/~hwxi/Xanadu/Examples/ <lang Xanadu> int abs(i: int) {
if (i >= 0) return i; else return -i;
}
unit print_dots(n: int) {
while (n > 0) { print_string("."); n = n - 1; }
return;
}
{size:int | 0 < size} unit print_board (board[size]: int, size: int(size)) {
var: int n, row;;
invariant: [i:nat] (row: int(i))
for (row = 0; row < size; row = row + 1) {
n = board[row];
print_dots(n-1);
print_string("Q");
print_dots(size - n);
print_newline();
}
print_newline();
return;
}
{size:int, j:int | 0 <= j < size} bool test (j: int(j), board[size]: int) {
var: int diff, i, qi, qj;;
qj = board[j];
invariant: [i:nat] (i: int(i))
for (i = 0; i < j; i = i + 1) {
qi = board[i]; diff = abs (qi - qj);
if (diff == 0) { return false; }
else { if (diff == j - i) return false; }
}
return true;
}
{size:int | 0 < size} nat queen(size: int(size)) {
var: int board[], next, row; nat count;;
count = 0; row = 0; board = alloc(size, 0);
invariant: [n:nat | n < size] (row: int(n))
while (true) {
next = board[row]; next = next + 1;
if (next > size) {
if (row == 0) break; else { board[row] = 0; row = row - 1; }
} else {
board[row] = next;
if (test(row, board)) {
row = row + 1;
if (row == size) {
count = count + 1;
print_board(board, size);
row = row - 1;
}
}
}
}
return count;
}
int main () {
return queen (8);
}</lang>
XSLT
Below simple stylesheet does produce this output (either by XSLT processors saxon-6.5.5, xsltproc, xalan, or any of the big5 browsers): <lang> 15863724 16837425 ... 88 lines omitted ... 83162574 84136275 </lang>
You can view the results directly in your browser (Chrome/FF/IE/Opera/Safari) here: [[2]]
This stylesheet is in category XSLT because it makes use or EXSLT [[3]] exslt:node-set() extension function not available in XSLT 1.0
It is extracted from a bigger solution described in this blog posting: [[4]]
- determine all 500 n-queens solutions for 4<=n<=9
- determine distict solutions and totals
- display solutions graphically nicely
- with references to external .gif images [[5]]
- with internal "data:..." .gif images [[6]]
This is the initial part of a screenshot from browser output:
Here is stylesheet 8-queens.xsl.xml which produces the (simple) output by having itself as input: [[7]]
<lang xml>
<?xml-stylesheet href="#" type="text/xsl"?>
<!DOCTYPE xsl:stylesheet [
<!ENTITY N "8">
]>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:exslt="http://exslt.org/common" xmlns:n-queens="urn:n-queens" exclude-result-prefixes="n-queens exslt"
>
<xsl:output omit-xml-declaration="yes"/>
<xsl:template match="/xsl:stylesheet">
<xsl:variable name="row0">
<xsl:call-template name="n-queens:row">
<xsl:with-param name="n" select="&N;"/>
</xsl:call-template>
</xsl:variable>
<xsl:variable name="row" select="exslt:node-set($row0)"/>
<xsl:variable name="rows0">
<xsl:for-each select="$row/*">
<r><xsl:copy-of select="$row"/></r>
</xsl:for-each>
</xsl:variable>
<xsl:variable name="rows" select="exslt:node-set($rows0)"/>
<html>
<!-- determine all solutions of $N queens problem -->
<xsl:call-template name="n-queens:search">
<xsl:with-param name="b" select="$rows/*"/>
</xsl:call-template>
</html>
</xsl:template>
<xsl:template name="n-queens:search"> <xsl:param name="b"/> <xsl:param name="s"/>
<xsl:if test="not($b)">
<xsl:value-of select="$s"/><xsl:text>
</xsl:text>
</xsl:if>
<xsl:for-each select="$b[1]/*">
<xsl:variable name="sieved0">
<xsl:call-template name="n-queens:sieve">
<xsl:with-param name="c" select="."/>
<xsl:with-param name="b" select="$b[position()>1]"/>
</xsl:call-template>
</xsl:variable>
<xsl:variable name="sieved" select="exslt:node-set($sieved0)"/>
<xsl:call-template name="n-queens:search">
<xsl:with-param name="b" select="$sieved/*"/>
<xsl:with-param name="s" select="concat($s, .)"/>
</xsl:call-template>
</xsl:for-each>
</xsl:template>
<xsl:template name="n-queens:sieve"> <xsl:param name="c"/> <xsl:param name="b"/>
<xsl:for-each select="$b">
<xsl:variable name="r" select="position()"/>
<r><xsl:copy-of select="*[. != $c][. - $r != $c][. + $r != $c]"/></r> </xsl:for-each> </xsl:template>
<xsl:template name="n-queens:row"> <xsl:param name="n"/>
<xsl:if test="$n>0">
<xsl:call-template name="n-queens:row">
<xsl:with-param name="n" select="$n - 1"/>
</xsl:call-template>
<f><xsl:value-of select="$n"/></f> </xsl:if> </xsl:template>
<msxsl:script xmlns:msxsl="urn:schemas-microsoft-com:xslt"
language="JScript" implements-prefix="exslt"
>
this['node-set'] = function (x) {
return x;
}
</msxsl:script>
</xsl:stylesheet> </lang>
XPL0
<lang XPL0>def N=8; \board size (NxN) int R, C; \row and column of board char B(N,N); \board include c:\cxpl\codes;
proc Try; \Try adding a queen to the board int R; \row, for each level of recursion
func Okay;
\Returns 'true' if no row, column, or diagonal from square R,C has a queen
int I;
[for I:= 0 to N-1 do
[if B(I,C) then return false; \row is occupied
if B(R,I) then return false; \column is occupied
if R+I<N & C+I<N then
if B(R+I, C+I) then return false; \diagonal down right
if R-I>=0 & C-I>=0 then
if B(R-I, C-I) then return false; \diagonal up left
if R-I>=0 & C+I<N then
if B(R-I, C+I) then return false; \diagonal up right
if R+I<N & C-I>=0 then
if B(R+I, C-I) then return false; \diagonal down left
];
return true;
]; \Okay
[ \Try if C>=N then
[for R:= 0 to N-1 do \display solution
[ChOut(0, ^ ); \(avoids scrolling up a color)
for C:= 0 to N-1 do
[Attrib(if (R|C)&1 then $0F else $4F); \checkerboard pattern
ChOut(6, if B(R,C) then $F2 else ^ ); \cute queen symbol
ChOut(6, if B(R,C) then $F3 else ^ );
];
CrLf(0);
];
exit; \one solution is enough
];
for R:= 0 to N-1 do
[if Okay(R,C) then \a queen can be placed here
[B(R,C):= true; \ so do it
C:= C+1; \move to next column
Try; \ and try from there
C:= C-1; \didn't work: backup
B(R,C):= false; \undo queen placement
];
];
]; \Try
[for R:= 0 to N-1 do \clear the board
for C:= 0 to N-1 do
B(R,C):= false;
C:= 0; \start at left column Try; ]</lang>
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