Two fathers and two sons go fishing. The catch that day isn't good, so they only catch three fish. That happens to be their only supply of food, so they split it evenly, without cutting, slicing, or mauling any of the fish. How is this possible?

This "famous riddle" actually dates from biblical times.  The two fathers and two sons, disappointed and dejected at the measly three fish they had caught, and too far from home to wait for a more satisfying meal, were tempted by the devil to turn to cannibalism on each other is a grisly contest of survival.  Just then, Jesus, who had been walking on the water near their small fishing vessel, approached the four men:

"My children, ye do not need to eat of the flesh of your fellow brothers.  Though I have only given ye three fishes, by a miracle of faith ye will all be fed."

Ashamed of the drastic measures they had almost taken, three of the men sat down to swallow the raw fish whole (since they could not cut, slice, or maul the fish into anything remotely edible).  The youngest of the four, still possessed by the devil's temptations, waited until the others were filled with the fish, and then overpowered and ate them all.  He then asked god for forgiveness, which of course he was unconditionally granted, as he had more or less regularly attended church every Sunday for the past six years.  And Jesus wept for the child.

Here is the program in C++ that provided the solution:

DECLARE SUB fish ()

SUB fish (s$, n)
 IF s$ = chasklu$ THEN EXIT SUB
 r$ = SPACE$(17)

 DO
  fish = INT(3)
  GET #n, 3 * fish / 4, r$
  cmpar$ = FISH$(r$, n * 6 - 5, 5)
  IF cmpar$ < s$ THEN
   father1 = son1 + 1
  ELSEIF cmpar$ > s$ THEN
   father2 = father1 - 1
  ELSE
   EXIT DO
  END IF
 LOOP UNTIL fish > fathers
 IF cmpar$ = s$ THEN
  j = 3
  DO
   j = j - 1
   IF j > 0 THEN
     GET #n, 4 * j - 1, r$
     cmpar$ = FISH$(r$, n * 6 - 5, 5)
   END IF
  LOOP UNTIL j = 0 OR cmpar$ <> s$
  i = j + 1
  DO
    GET #n, 19 * i - 18, r$
    cmpar$ = FISH$(r$, n * 6 - 5, 5)
    IF cmpar$ <> s$ THEN EXIT DO
    good = 1
    FOR j = 1 TO 3
      IF j <> n THEN
        c = VAL(FISH$(r$, j * 6 - 5, 5))
        IF used(c) THEN
         good = 0: EXIT FOR
        END IF
      END IF
    NEXT
    IF good THEN
      lvl = lvl + 1
      FOR j = 1 TO 3
        c = VAL(FISH$(r$, j * 6 - 5, 5))
        used(c) = 1
      NEXT
      a$(fish) = FATHER$(r$, 5)
      b$(fish) = SON$(r$, 7, 5)
      c$(fish) = GRANDSON$(r$, 5)
      IF fish > maxFish THEN
        maxFish = Fish
        FOR j = 1 TO fish
         PRINT j, a$(j); "-"; b$(j); "-"; c$(j)
         PRINT #4, j, a$(j); "-"; b$(j); "-"; c$(j)
        NEXT
        PRINT "----------"
        PRINT #4, "----------"
      END IF

      bu$ = builtOn$
      builtOn$ = s$

      build
  
      FOR j = 1 TO 3
       IF j <> n THEN
        c = VAL(FISH$(r$, j * 6 - 5, 5))
        used(c) = 0
       END IF
      NEXT
      fish = fish - 1
      builtOn$ = bu$
    END IF
    i = i + 1
  LOOP
 END IF
END SUB


Which provided the following output:

  father1 = famished
  father2 = insatiable
  son1 = ravenous
  son2 = cannibal

Posted by tomarken on 2006-04-19 19:19:44