Towards the end of term, Mr McIvor our maths teacher lightened up and cut us some slack. “Here’s a problem for the weekend – but only if you like. It’s not particularly mathematical. All you need is pure logic. It’s called the twelve cannon ball problem.”
You have twelve cannon balls. They look identical, and eleven of them are, but one has a different weight from the others; it could be heavier or it could be lighter. You have a simple set of scales – two pans on a fulcrum. Can you identify the unique cannon ball in three weighings, and say whether it is lighter or heavier?
I realise this blog might be aimed at a minority interest group. If you are allergic to numbers, it may be you are more interested in the qualitative rather than the quantitative aspects of cannon balls. Therefore I digress for a moment on your behalf. I had always thought the expression, “It was cold enough to freeze the balls off a brass monkey” had a somewhat vulgar connotation. Not so! It turns out that a brass monkey was a rack for holding cannon balls, and that in very cold weather the rack contracted more than the cannon balls and extruded them. Back to the twelve cannon ball problem – but, as Mr McIvor said, only if you like.
So, to return to the dim and distant past, I was up for it. I took the problem home with me and wrestled with it. It absorbed me completely. By Friday night I had realised that to put six balls on each pan was a squandered shot because it yielded too little information. I was pretty certain you needed to place four balls on each pan. All throughout Saturday I experimented with four ball patterns. Didn’t go to the swimming baths, didn’t go to badminton club. Didn’t watch telly. I became a recluse and an irritant to my family. By Saturday night I had solved the special case, when your first weighing, four x four, shows equilibrium. Therefore the rogue ball lies within the remaining four. You leave three of the balls you have already weighed on one pan to act as a control; you know they are all normal. You take three of the four untested balls and put them on the other pan. Suppose there is equilibrium. Then the remaining unweighed ball is the rogue ball. You use your third weighing to weigh it against any control ball in order to establish whether it is lighter or heavier.
Go back to the second weighing, three x three, however, and suppose there is disequilibrium.
(Are you still with me? Or have your eyes glazed over because you’re not that interested? Don’t fret.)
Well, now you know the rogue ball is one of three, and you know, depending on which way the scales have gone, whether it is lighter or heavier. So take two of the suspect three balls, clear the scales, and weigh one suspect ball against another. If there is equilibrium, the third ball is the rogue ball, and you already know whether it is lighter or heavier. If there is disequilibrium, you identify the rogue ball because you know whether you are looking for a lighter or heavier ball. Take note of this last paragraph; it is a recurring theme. If you know the rogue ball is one of a group of three, and you know the relative weight of the ball you are looking for, you have a scenario in which the ball can be identified with just one further weighing. Let us call this scenario “The Brass Monkey Scenario”.
Fine. Saturday midnight, and I have my special theory. But I’m miles away from my general theory. I still have no idea what to do if the first four v four weighing shows disequilibrium.
Sunday. Sunday is murder.
I struggle with combinations and permutations and seem to be getting nowhere. Moreover, I am dogged by a nagging sense of guilt that I’ve wasted a precious weekend in a futile gesture. But I can’t stop. It’s not fun any more, yet I can’t stop. Isn’t this what it is supposed to be like to be addicted to something? Am I in some kind of trouble?
Give it till Sunday teatime. I’m sure I’ve missed something. There’s something creative I need to do. Something I haven’t thought of. A little piece of lateral thinking.
I skipped Sunday evening youth orchestra.
Eureka! I worked on, with growing excitement. Yet still with that nagging sense of guilt, that would just not go away, even when I was on the brink of success.
Late at night, when Barry Alldis was doing his Top 20 countdown on Luxembourg 208, the dark metallic strangled voice of Gene Pitney recounted the tale of the man who lost everything twenty four hours from Tulsa. I had it. And yet that terrible sense of squandered energy just wouldn’t leave me. Why are you wasting your substance on this rubbish? This is a complete waste of time.
Maybe not completely. If this weekend has taught me anything, it has taught me that, even when a problem seems utterly insurmountable, when you have run out of rope, still with a little bit of creativity, and lateral thinking, there might just be a way out.
McIvor had me demonstrate my general theory on the blackboard before the class. The extra pressure of performance might have snarled up my logical circuits, but I was fired up, and my exposition was sound.
Let the balls be numbered 1,2,3,4,5,6,7,8,9,10,11,12. First weighing: balls 1,2,3,4 v balls 5,6,7,8 gives disequilibrium. You know the rogue ball is somewhere on the scales.
Second weighing is 1,2,3,5 v 9,10,11,4.
That’s the little creative leap, the little piece of lateral thinking. You swap two of the balls.
Consider now what might happen. The scales show equilibrium. Therefore the rogue ball is in the group of three you have removed, namely balls 6,7, and 8. You already know whether the rogue ball is lighter or heavier because of the disequilibrium of the first weighing. This is The Brass Monkey Scenario. All you need do now is weigh ball 6 against ball 7.
Consider now that in the second weighing a disequilibrium occurs but the scales tip over in the other direction. Now the rogue ball has to be one of the balls you have swapped – 4 or 5. If 4 is on the down scale and it turns out to be the rogue ball, then it is heavier. If 4 is on the up scale and it turns out to be the rogue ball, then it is lighter. Same scenario for ball 5.
Third weighing: ball 4 against any control ball. If there is equilibrium, then ball 5 is the rogue ball, and you already know whether it is lighter or heavier. If there is disequilibrium, then ball 4 is the rogue ball and it is lighter or heavier depending on how it now tips the scales.
Consider now that in the second weighing the same disequilibrium occurs as in the first weighing. Then the rogue ball is in the group 1, 2, 3, because balls 9, 10, 11 are control balls, and, had balls 4 or 5 been rogue, the scales would have tipped over. We are back to The Brass Monkey Scenario. We now know that the rogue ball (1,2, or 3) is lighter or heavier depending on how the scales tipped, and, as before, we clear the scales and weigh ball 1 against ball 2, knowing we are looking for a lighter or a heavier ball. If the scales show equilibrium, then ball 3 is the rogue ball, and we already know whether it is lighter or heavier.
That’s it. Every situation dealt with.
I sat down.
There was no applause from the class. They had long since lost interest. Mr McIvor said, “Well done!” And indeed there was admiration. Yet I could sense something else. I thought I could detect a sense of guilt that he had subjected me to this conundrum in the first place, and driven me to spend so much time on it. I was completely confused. It crossed my mind that he pitied me. He might have shaken his head and said, “You poor sad boy!”
jamescalumcampbell 