Thursday, March 29, 2012

The String Puzzle

I gave my two boys an old puzzle to solve yesterday. I told them that I’d give them each \$10 if they could solve it for me. It’s one of the ways we do the “allowance” thing around the house sometimes.

Here’s the puzzle. A piece of string is stretched tightly around the Earth along its equator. Imagine that this string along the equator forms a perfect circle, and imagine that to reach around that perfect circle, the string has to be exactly 25,000 miles long. Now imagine that you wanted to suspend this string 4 inches above the surface of the Earth, all the way around it. How much longer would the string have to be do do this?

Before you read any further, guess the answer. How much longer would the string have to be? A few inches? Several miles? What do you think?

Now, my older son Alex was more interested in the problem than I thought he would be. He knows the formula for computing the circumference of a circle as a function of its diameter, and he knew that raising the string 4 inches above the surface constituted a diameter change. So the kernel of a solution had begun to formulate in his head. And he had a calculator handy, which he loves to use.

We were at Chipotle for dinner. The rest of the family went in to order, and Alex waited in the truck to solve the problem “where he could have some peace and quiet.” He came into the restaurant in time to order, and he gave me a number that he had cooked up on his calculator in the truck. I had no idea whether it was correct or not (I haven’t worked the problem in many years), so I told him to explain to me how he got it.

When he explained to me what he had done, he pretty quickly discovered that he had made a unit conversion error. He had manipulated the ‘25,000’ and the ‘4’ as if they had been expressed in the same units, so his answer was wrong, but it sounded like conceptually he got what he needed to do to solve the problem. So I had him write it down. On a napkin, of course:

The first thing he did was draw a sphere (top center) and tell me that the diameter of this sphere is 25,000 miles divided by 3.14 (the approximation of π that they use at school). He started dividing that out on his calculator when I pulled the “Whoa, wait” thing where I asked him why he was dividing those two quantities, which caused him, grudgingly, to write out that C = 25,000 mi, that C = πd, and that therefore d = C/π. So I let him figure out that d ≈ 7,961 mi. There’s loss of precision there, because of the 3.14 approximation, and because there are lots of digits to the right of the decimal point after ‘7961’, and of course, the ‘25,000’ wasn’t precise to begin with, but more about that later.

I told him to call the length of the original string C (for circumference) and to call the 4-inch suspension distance of the string h (for height), and then write me the formula for the length of the 4-inch high string, without worrying about any unit conversion issues. He got the formula pretty close on the first shot. He added 4 inches to the diameter of the circle instead of adding 4 inches to the radius (you can see the ‘4’ scratched out and replaced with an ‘8’ in the “8 in/63360 in” expression in the middle of the napkin. Where did the ‘63360’ come from, I asked? He explained that this is the number of inches in a mile (5,280 × 12). Good.

But I asked him to hold off on the unit conversion stuff until the very end. He wrote the correct formula for the length of the new string, which is [(C/π) + 2h]·π (bottom left). Then I let him run the formula out on his calculator. It came out to something bigger than exactly 25,000; I didn’t even look at what he got. This number he had produced minus 25,000 would be the answer we were looking for, but I knew there would be at least two problems with getting the answer this way:
• The value of π is approximately 3.14, but it’s not exactly 3.14.
• Whenever he had to transfer a precise number from one calculation to the next, I knew Alex was either rounding or truncating liberally.
So, I told him we were going to work this problem out completely symbolically, and only plug the numbers in at the very end. It turns out that doing the problem this way yields a very nice little surprise.

Here’s my half of the napkin:

I called the new string length cʹ and the old string length c. The answer to the puzzle is the value of cʹ − c.

The new circumference cʹ will be π times the new diameter, which is c/π + 2h, as Alex figured out. The second step distributes the π factor through the addition, resulting in cʹ − c = πc/π + 2πh − c. The πc/π term simplifies to just c, and it’s the final step where the magic happens: cʹ − c = c + 2πhc reduces simply to cʹ − c = 2πh. The difference between the new string length and the old one is 2πh, which in our case (where h = 4 inches) is roughly 25.133 inches.

So, problem solved. The string will have to be about 25.133 inches longer if we want to suspend it 4 inches above the surface.

Notice how simple the solution is: the only error we have to worry about is how precisely we want to represent π in our calculation.

Here’s the even cooler part, though: there is no ‘c’ in the formula for the answer. Did you notice that? What does that mean?

It means that the original circumference doesn’t matter. It means that if we have a string around the Moon that we want to raise 4 inches off the surface, we just need another 25.133 inches. How about a string stretched around Jupiter? just 25.133 more inches. Betelgeuse, a star whose diameter is about the same size as Jupiter’s orbit? Just 25.133 more inches. The whole solar system? Just 25.133 more inches. The entire Milky Way galaxy? Just 25.133 more inches. A golf ball? Again, 25.133 more inches. A single electron? Still 25.133 inches.

This is the kind of insight that solving a problem symbolically provides. A numerical solution tends to answer a question and halt the conversation. A symbolic formula answers our question and invites us to ask more.

The calculator answer is just a fish (pardon the analogy, but a potentially tainted fish at that). The symbolic answer is a fishing pole with a stock pond.

So, did I pay Alex for his answer? No. Giving two or three different answers doesn’t close the deal, even if one of the answers is correct. He doesn’t get paid for blurting out possible answers. He doesn’t even get paid for answering the question correctly; he gets paid for convincing me that he has created a correct answer. In the professional world, that is the key: the convincing.

Imagine that a consultant or a salesman told you that you needed to execute a \$250,000 procedure to make your computer application run faster. Would you do it? Under what circumstances? If you just trusted him and did it, but it didn’t do what you had hoped, would you ever trust him again? I would argue that you shouldn’t trust an answer without a compelling rationale, and that the recommender’s reputation alone is not a compelling rationale.

The deal is, whenever Alex can show me the right answer and convince me that he’s done the problem correctly, that’s when I’ll give him the \$10. I’m guessing it’ll happen within the next three days or so. The interesting bet is going to be whether his little brother beats him to it.