Fill a 3x3 square with the numbers from 1 to 9, so the four sides and the two diagonals share the same sum, which is to be as large as possible.
(In reply to
re: Solution by Bryan)
Having myself written a program to solve this, I disagree with the wrecking ball analogy. Of course my program is bare bones and utilizes code from prior puzzles, but it assures that no solutions would be left unchecked. I agree that many of the bells and whistles attached to the previously posted program lend credence to the wrecking ball analogy, but to my mind the bare bones version is worthwhile, though I didn't previously think it was worth posting. But for comparison, and to dispel the wrecking ball analogy:
DECLARE SUB permute (a$)
OPEN "almmagic.txt" FOR OUTPUT AS #2
a$ = "123456789": h$ = a$
FOR i = 1 TO 362880
n1 = VAL(MID$(a$, 1, 1))
n2 = VAL(MID$(a$, 2, 1))
n3 = VAL(MID$(a$, 3, 1))
n4 = VAL(MID$(a$, 4, 1))
n5 = VAL(MID$(a$, 5, 1))
n6 = VAL(MID$(a$, 6, 1))
n7 = VAL(MID$(a$, 7, 1))
n8 = VAL(MID$(a$, 8, 1))
n9 = VAL(MID$(a$, 9, 1))
n10 = VAL(MID$(a$, 10, 1))
t1 = n1 + n4 + n7
t2 = n2 + n5 + n8
t3 = n3 + n6 + n9
t4 = n1 + n2 + n3
t5 = n4 + n5 + n6
t6 = n7 + n8 + n9
t7 = n1 + n5 + n9
t8 = n3 + n5 + n7
IF t1 = t3 AND t1 = t4 AND t1 = t6 AND t1 = t7 AND t1 = t8 THEN
PRINT #2, MID$(a$, 1, 3), t4
PRINT #2, MID$(a$, 4, 3)
PRINT #2, MID$(a$, 7, 3), t6
PRINT #2, t1; t3
PRINT #2, , , , t7, t8
PRINT #2,
END IF
nextPerm:
permute a$
IF LEFT$(a$, 2) <> pVal$ THEN pVal$ = LEFT$(a$, 2): PRINT pVal$
NEXT
CLOSE
The permute subroutine is the same as has been used elsewhere.

Posted by Charlie
on 20040622 14:34:29 