Comments on: Help me understand a little logic problem.

Question: Help me understand a little logic problem.

TheHollowSeasThatRoar — Wed, 05 Sep 2012 04:39:47 -0800

Help me understand this. Math/logic puzzle follows...

Here are the puzzle and answer, as given by Futility Closet:

A worm crawls along an elastic band that's 1 meter long. It starts at one end and covers 1 centimeter per minute. Unfortunately, at the end of each minute the band is instantly and uniformly stretched by an additional meter. Heroically, the worm keeps its grip and continues crawling. Will it ever reach the far end?

Surprisingly, yes. At the end of the first minute the worm has crawled 1/100 of the way to the far end. At the end of the second minute, it's crawled (1/100 + 1/200) of the way. So it will reach the far end in t minutes when 1/100 (1/1 + 1/2 + 1/3 + ... + 1/t) equals or exceeds 1. The expression in parentheses is the harmonic series, which can be made as large as one desires, but it'll take a while: In this example the worm will reach its goal in t = 1.509269 × 1043 minutes, or about 286,961,000,000,000,000,000,000,000,000,000,000 centuries.

This makes zero sense to me. Won't the worm always be 1/100 of the way along? First minute: 1 cm of 1 metre. Second minute: 2 cm of 2 metres. Third minute: 3cm of 3 metres. And so on forever.

Am I mistaken somehow?

By: A Thousand Baited Hooks

A Thousand Baited Hooks — Wed, 05 Sep 2012 04:48:39 -0800

The thing you're missing is that the part of the band that the worm has already passed grows as well as the part ahead of the worm. So the worm is 1cm along after 1 minute, but after 2 minutes it's actually 3cm along because the 1cm of band it crawled over in the first minute has doubled in length. The gains the worm makes this way slow down very quickly as the minutes pass and the band gets longer, but it always stays slightly ahead.

By: pompomtom

pompomtom — Wed, 05 Sep 2012 04:50:31 -0800

First minute: 1 cm of 1 metre. Second minute: 2 cm of 2 metres. Third minute: 3cm of 3 metres. And so on forever.

The bit behind the worm is also stretched, so the second minute figure is 3cm of 2m.

By: TheHollowSeasThatRoar

TheHollowSeasThatRoar — Wed, 05 Sep 2012 05:17:33 -0800

Ah! Thank you very much.

On a side note, I read the words "stretch" and "uniform" and "elastic band", and yet visualized the worm on a ruler that is extended only from the far end. I find that a fascinating, self-created error -- my brain unconsciously simplified the problem to make it easier and quicker to "solve".

Thanks again.

By: Aquinas

Aquinas — Wed, 05 Sep 2012 08:50:42 -0800

As a side note, here's a way to calculate that value of t, since the quoted answer leaves that out.

What we need to do is solve the equation

100 = (1 + 1/2 + 1/3 + ... + 1/t)

for t. My first instinct here was to get an estimate by approximating the sum by the integral from 1 to t of dx/x, which is log(t). It's easy enough to argue that this will always underestimate the sum, so it will overestimate the time required to reach the end. Indeed, the solution we get this way is t = exp(100), or about 2.688x10^43, which is the right order of magnitude but too large.

Fortunately, the relationship between the sum and the integral is, in this particular case, very simple. The difference between the nth partial sum and log(n) converges to a known constant "gamma," called the Euler-Mascheroni constant. So, we can get a much better estimate by calculating exp(100 - gamma), and this produces the result in the answer above.

By: iotic

iotic — Wed, 05 Sep 2012 11:02:21 -0800

Also note, you must assume it is an ideal worm. Real worms cannot walk for 286,961,000,000,000,000,000,000,000,000,000,000 centuries.

By: yohko

yohko — Wed, 05 Sep 2012 11:17:57 -0800

And an ideal elastic band! For a real elastic band, it will eventually snap, catapulting the worm off into space, where it will cover half the distance to the next math/logic puzzle a la zeno's paradox.