You have a standard chess board, and as many queens as you need. What is the most queens that you can put on the board so that no two queens can attack each other? What is the formation to put them in?
Keep in mind that a chess board has 64 squares (8x8), and queens can go diagonal, up and down, and left and right.
(In reply to
solution by Charlie)
As for the programming exercise, this is what I used. The first part finds all 92 (as it turns out) solutions. Phase 2 then reads the file produced in part 1 and copies to a second file only those that are not trivial rotations/reflections of one that had come before.
DECLARE SUB placeQueen (col!)
DIM SHARED board(8, 8), ct
OPEN "8queens.txt" FOR OUTPUT AS #1
placeQueen 1
PRINT ct
CLOSE
phase2:
OPEN "8queens.txt" FOR BINARY AS #1
OPEN "8queensc.txt" FOR OUTPUT AS #2
row$ = SPACE$(16)
FOR soln = 1 TO ct
stPos = (soln  1) * 18 * 9 + 1
FOR row = 1 TO 8
GET #1, stPos + (row  1) * 18, row$
FOR col = 1 TO 8
IF MID$(row$, col * 2  1, 1) = "*" THEN
board(row, col) = 1
ELSE
board(row, col) = 0
END IF
NEXT
NEXT
good = 1
FOR solRef = 1 TO soln  1
stPos = (solRef  1) * 18 * 9 + 1
FOR row = 1 TO 8
GET #1, stPos + (row  1) * 18, row$
FOR col = 1 TO 8
IF MID$(row$, col * 2  1, 1) = "*" THEN
board2(row, col) = 1
ELSE
board2(row, col) = 0
END IF
NEXT
NEXT
FOR rotRef = 1 TO 7
bad = 1
FOR r = 1 TO 8
FOR c = 1 TO 8
SELECT CASE rotRef
CASE 1
r2 = 9  r
c2 = c
CASE 2
r2 = 9  r
c2 = 9  c
CASE 3
r2 = r
c2 = 9  c
CASE 4
r2 = 9  c
c2 = r
CASE 5
r2 = 9  c
c2 = 9  r
CASE 6
r2 = c
c2 = 9  r
CASE 7
r2 = c
c2 = r
END SELECT
IF board(r, c) <> board2(r2, c2) THEN bad = 0: EXIT FOR
NEXT
IF bad = 0 THEN EXIT FOR
NEXT
IF bad THEN good = 0: EXIT FOR
NEXT
IF good = 0 THEN EXIT FOR
NEXT
IF good THEN
stPos = (soln  1) * 18 * 9 + 1
FOR row = 1 TO 9
GET #1, stPos + (row  1) * 18, row$
PRINT #2, row$
NEXT
ct2 = ct2 + 1
END IF
NEXT
PRINT ct2
SUB placeQueen (col)
FOR row = 1 TO 8
good = 1
FOR c = 1 TO col  1
IF board(row, c) THEN good = 0: EXIT FOR
r = row  (col  c)
IF r > 0 THEN
IF board(r, c) THEN good = 0: EXIT FOR
END IF
r = row + (col  c)
IF r < 9 THEN
IF board(r, c) THEN good = 0: EXIT FOR
END IF
NEXT
IF good THEN
board(row, col) = 1
IF col = 8 THEN
ct = ct + 1
FOR r = 1 TO 8
FOR c = 1 TO 8
IF board(r, c) THEN
PRINT #1, "* ";
ELSE
PRINT #1, ". ";
END IF
NEXT
PRINT #1,
NEXT
PRINT #1, " "
ELSE
placeQueen col + 1
END IF
board(row, col) = 0
END IF
NEXT
END SUB

Posted by Charlie
on 20030603 08:30:47 