## 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’, 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.

Unknown said...

I love this problem!

I was asked the inverse question (if you add x to the length how much will the string raise off the earth) years ago by physics professor. The method to arrive at the answer is deceptively simple and the actual value for the answer defies intuition. I have asked at least 100 people (mostly engineering and comp sci majors) this question over the last 25 or so years and only 2 or 3 arrived at the correct answer…most gave up before arriving at an answer. Some spent hours and filled up pages of paper before giving up.

With a basic understanding of differentials, the solution is even easier. The relationship between circumference and radius is linear (no squared or higher order terms). As a result, the change in circumference is directly related to the change in radius (in this case 4 inches) multiplied by 2pi.

This is shown below. In mathematical notation, a delta symbol (triangle) would represent the change.

C=2*pi*r
(change in)C=2*pi*(change in)r
(change in)C=6.28*4=25.12

Most people take the same approach that your son did…they make an assumption about the size of the earth and start from there. As you noted, the original circumference doesn’t matter. I always give a hint that the answer is the same whether the earth, a basketball or a golf ball is used for the problem. Even with that hint, most people do not arrive at the correct answer.

Cary Millsap said...

My kids (ages 12 and 14) are both doing interesting problems in school right now. "A rectangular prism has volume 42 cubic feet. What will be the volume of the prism if you multiply its side lengths by a factor of 1/2?" ...Stuff like that.

They're doing perimeters (with linear relationships, exactly as in your problem) areas (quadratic), and volumes (cubic). The problem is, though, that they're just following a mechanical process that their teacher has shown them. (If perimeter, do thing A; else, if area do thing B; else, if volume do thing C; otherwise, run!)

I like for them to see why there's a cubic relationship when there's a volume involved: because when you scale by some factor f, your V = L × H × W becomes V′ = fL × fH × fW, which is V′ = (f^3)(L × H × W) = (f^3)V.

I feel like the only way they'll actually be able to remember this stuff and apply it to their lives is to see why it works the way it does.

Fergal Taheny said...

Cary, I think you should pay up! If you hadn't confused him by giving him an actual value of 25000 for the diameter he would have got the answer

Hello, Cary!

It seems to me that your personal conclusions are more profound than the task by itself. This fact is very interesting that the difference between two lenghts of the strings disposed at the constant distance is the same. Let me note that it is true not only for spherical bodies but for every bodies with the similar type of its shape...

The main conclusion to my mind is the advantage of taking the final formula than numerical calculations.

So, if most programmers were powerful mathematicians, our (in the whole world) programms (including Oracle) would be much better. Do you agree with me?

Marcin Przepiorowski said...

Hello Cary,

I think that there a big change how math is taught at school. I remember that during my primary school years in Poland I have learned that math is not only numbers but it include formulas with all transformations plus basic algebra. Now when I work with my sons (12 and 14 years as well) in Ireland I can see that math is now about numbers only and I think that this is wrong approach as they can't solve problems described in abstract way like this "One brick is the one kilogram plus half a brick heavy What is the weight of one brick?"

Anyway I like your ending conclusion. I think that prove in IT is so important like prove in math.

regards,
Marcin

Dominic Delmolino said...

Excellent!

Wonder if it holds for other kinds of shapes?

http://www.oraclemusings.com/?p=168

Nice to see that you won't stoop to product placement Mr Millsap.

Chris said...

C'mon Cary, Everyone knows 2π is really τ [tau] so C=τr (C=tau*r)...

Get with the times ;)

Cary Millsap said...

Chris, I'm with you. I'm guessing that virtually anyone who reads my blog probably has a crush on Vi Hart. :-)

Chris said...

Cary - I had to go lookup Vi Hart. I had never heard of it :)

Interesting site.

I was just remembering about the "Tau" day versus "Pi" day and the relationship of Tau to Pi and the media spin on it a while back...

Chris said...

*sigh* Well, I just discovered Vi Hart is a female and not just a site LOL

Mahesh said...

Cary
Here is another interesting problem. With the numbers 1,2 and 3 used only once, get to 19 using the arithetic operators, sqrt and factorial.
Mahesh

asliwx said...

Cary,
My Dad is a Calculus/Math professor ... He loved it!
Would it be ok... if he uses this example in his class...
He proposed a similiar algebriac challenge to his students...

His response :

Assumptions and Variables

Perfect Circle with an exact perimeter = 25,000 miles

Then a second circle with a radius 4 inches longer than the first

Radius original = 25,000 divided by 2*pi gives the original radius in miles. This result hast o be multiplied by 5280 * 12 to convert to inches

25,000 * 5289 *12 / (2^pi). Lets call this number N. The length of the second (slightly longer circumfrance is given b y

(Old radius +4)* 2*pi is the total length

The incremental length would be the New minus Old

Since OLD is just a big number compared to the contribution of the additional radius, some simplications can be applied to prevent
errors related to subtracting two large and close to one another errors from creeping in

Cary Millsap said...

Andrew,

Certainly that would be ok.

—Cary

Jeremy said...
This comment has been removed by the author.
Jeremy said...

Thanks Cary for posting this problem. It was a lot of fun, and I had NO idea that the answer would end up looking like that! I posted it to my Facebook page (referencing your blog of course) to ask my friends to solve it too. :)

Cheers,

Jeremy

Unknown said...

Cary,

Did your second one solve the puzzle before the first?

Cary Millsap said...

Rahul,

Neither boy has taken a lot of interest in the puzzle since I posted this. I think they have too much money already to be interested for bribery's sake. :-)

Now that you've asked, my older boy is trying to save money for something he wants. I'll remind him that this offer still stands.

Cary

? said...

To give your boys a visual solution to this problem consider the following.

Start with a string with the ends tied up and shaped as a circle. The length of the string is 2 * π * Radius.

Twist the string to create a perfect 8 shape. You get two smaller, equal, tangential circles and it is easy to figure out that the radius for each of these two circles is half the size of the radius for the original circle.

Twist the string to create an imperfect 8 shape. You get two smaller, tangential circles and it is easy to figure out that the sum of radiuses for these two circles is equal to the radius of the original circle.

Now that one has the visual image of an imperfect 8 shape (that is, two tangential circles ... say smaller one at the top and big one at the bottom) consider the formula for calculating the "perimeter" of the 8 shape:

2 * π * R_top + 2 * π * R_bottom = 2 * π * (R_top + R_bottom)

But 2 * π * (R_top + R_bottom) is also the formula for calculating the circumference of a circle having the centre in the centre of the bottom circle in the 8 shape and a radius going through the tangential point to the centre of the top circle in the 8 shape.

Expressing this differently ... for any two concentric circles, the difference between the circumference of the outer one and the circumference of the inner one is equal to the circumference of a circle having the centre on any point on the outer circle and tangential to the inner circle.

In the image of the globe you have in your article consider having a small circle, right at the top, tangential to the earth circle and centered on the "suspended" circle. The sum of the circumferences for the two smaller circles (making up the 8 shape) is equal to the circumference of the "suspended" circle.

Doing the math in reverse … for any two concentric circles, the difference between the circumference of the outer one and the circumference of the inner one is equal to:

2 * π * (R_outer - R_inner)

For your particular case: 2 * π * 4 inches.

Gabe

Cary Millsap said...

Gabe, I'm sorry it's taken me so long to respond to your comment. I love it.

—Cary

Gary said...

about collective human intelligence and legacy... this problem and solution approach reminded me of line/method of eduction followed in vedic mathematics... though I never got a chance to learn that myself even though I like in India... but this approach is very akin to what is being discussed here (the heart and soul of that)...

Unknown said...

I love this problem. Never heard of it before.