Below are three groups of three numbers each. Combine the numbers in each group using the standard binary operations (addition, subtraction, multiplication, division, and exponentiation) so that each group yields the same number.

1. 3, 15, 18
2. 10, 13, 36
3. 24, 27, 39

For an example, see the first problem.

The following program fails to find a solution. The program itself tries to produce all possible results and place them in a file with the result at the beginning of each record. That file is sorted and then another program looks for matches with three different equations producing them.

I had written a program for pluribus 1, but due to a bug (not permuting the numbers) did not find the answer. This program could have some other bug, or the answer is more exotic than allowed for here:

DEFDBL A-Z

op$ = "+-*/^"
DATA 3,5,18, 10,13,36, 24,27,29
FOR i = 1 TO 3
  FOR j = 1 TO 3
    READ opnd(i, j)
  NEXT
NEXT

OPEN "epluribs.txt" FOR OUTPUT AS #2

FOR eq = 1 TO 3
 DO
  FOR op1 = 1 TO 5
   err1 = 0
   SELECT CASE op1
     CASE 1
       rslt1a = opnd(eq, 1) + opnd(eq, 2)
     CASE 2
       rslt1a = opnd(eq, 1) - opnd(eq, 2)
     CASE 3
       rslt1a = opnd(eq, 1) * opnd(eq, 2)
     CASE 4
       rslt1a = opnd(eq, 1) / opnd(eq, 2)
     CASE 5
       IF LOG(opnd(eq, 1)) * opnd(eq, 2) < 300 * LOG(10) THEN
        rslt1a = opnd(eq, 1) ^ opnd(eq, 2)
       ELSE
        rslt1a = 1: err1 = 1
       END IF
   END SELECT
   IF err1 = 0 THEN
     FOR op2 = 1 TO 5
       err1 = 0
       err2 = 0
       SELECT CASE op2
         CASE 1
           rslt2a = opnd(eq, 2) + opnd(eq, 3)
           rslt1 = rslt1a + opnd(eq, 3)
         CASE 2
           rslt2a = opnd(eq, 2) - opnd(eq, 3)
           rslt1 = rslt1a - opnd(eq, 3)
         CASE 3
           rslt2a = opnd(eq, 2) * opnd(eq, 3)
           rslt1 = rslt1a * opnd(eq, 3)
         CASE 4
           rslt2a = opnd(eq, 2) / opnd(eq, 3)
           rslt1 = rslt1a / opnd(eq, 3)
         CASE 5
           IF LOG(opnd(eq, 2)) * opnd(eq, 3) < 300 * LOG(10) THEN
            rslt2a = opnd(eq, 2) ^ opnd(eq, 3)
           ELSE
            rslt2a = 1: err2 = 1
           END IF
           IF LOG(ABS(rslt1a)) * opnd(eq, 3) < 300 * LOG(10) THEN
            rslt1 = rslt1a ^ opnd(eq, 3)
           ELSE
            rslt1 = 0: err1 = 1
           END IF
       END SELECT
       SELECT CASE op1
         CASE 1
           rslt2 = opnd(eq, 1) + rslt2a
         CASE 2
           rslt2 = opnd(eq, 1) - rslt2a
         CASE 3
           rslt2 = opnd(eq, 1) * rslt2a
         CASE 4
           rslt2 = opnd(eq, 1) / rslt2a
         CASE 5
           IF LOG(opnd(eq, 1)) * rslt2a < 300 * LOG(10) THEN
            rslt2 = opnd(eq, 1) ^ rslt2a
           ELSE
            rslt2 = 1: err2 = 1
           END IF
       END SELECT

       IF err1 = 0 THEN
         PRINT #2, rslt1; TAB(30);
         PRINT #2, USING "## "; eq;
         PRINT #2, USING " # ## ! ## ! ##"; 1; opnd(eq, 1); MID$(op$, op1); opnd(eq, 2); MID$(op$, op2); opnd(eq, 3)
       END IF
       IF err2 = 0 THEN
         PRINT #2, rslt2; TAB(30);
         PRINT #2, USING "## "; eq;
         PRINT #2, USING " # ## ! ## ! ##"; 2; opnd(eq, 1); MID$(op$, op1); opnd(eq, 2); MID$(op$, op2); opnd(eq, 3)
       END IF

     NEXT
   END IF
  NEXT op1

  flag = 0
  IF opnd(eq, 1) < opnd(eq, 2) THEN
   IF opnd(eq, 2) < opnd(eq, 3) THEN
    SWAP opnd(eq, 2), opnd(eq, 3)
   ELSE
    IF opnd(eq, 3) < opnd(eq, 1) THEN
     h = opnd(eq, 1)
     opnd(eq, 1) = opnd(eq, 2)
     opnd(eq, 2) = opnd(eq, 3)
     opnd(eq, 3) = h
    ELSE
     h = opnd(eq, 3)
     opnd(eq, 3) = opnd(eq, 2)
     opnd(eq, 2) = opnd(eq, 1)
     opnd(eq, 1) = h
    END IF
   END IF
  ELSE
   IF opnd(eq, 2) < opnd(eq, 3) THEN
    SWAP opnd(eq, 2), opnd(eq, 3)
   ELSE
    SWAP opnd(eq, 1), opnd(eq, 3): flag = 1
   END IF
  END IF
  PRINT opnd(eq, 1), opnd(eq, 2), opnd(eq, 3)
 LOOP UNTIL flag

NEXT eq

Posted by Charlie on 2004-01-22 11:21:33