Proof that a common brain teaser is wrong (Burning Rope)

Question

There is a common brain teaser that goes like this:

You are given two ropes and a lighter. This is the only equipment you can use. You are told that each of the two ropes has the following property: if you light one end of the rope, it will take exactly one hour to burn all the way to the other end. But it doesn't have to burn at a uniform rate. In other words, half the rope may burn in the first five minutes, and then the other half would take 55 minutes. The rate at which the two ropes burn is not necessarily the same, so the second rope will also take an hour to burn from one end to the other, but may do it at some varying rate, which is not necessarily the same as the one for the first rope. Now you are asked to measure a period of 45 minutes. How will you do it?

Now I usually love brain teasers but this one frustrated me for a while because I could not prove that if a rope of non-uniform density is burned at both ends it burns in time $T/2$. I think I have sketched a proof by induction that shows that it's not actually true.

Given a rope of uniform density the burn rate at either end is equal so clearly it burns in time $T/2$. Now, consider a rope of non-uniform density, the total time T for this rope to burn is the linear combination of the times of the uniform density "chunks" to burn, i.e. $T = T_1 + T_2 + \ldots + T_n$. So consider, $T/2 = T_1/2+ T_2/2 + \ldots + T_n/2$. If we look at each $T_i/2$ this is precisely the time it takes to burn the uniform segment $T_i$ if lit at both ends. Therefore, in order to arrive at a rope that burns in time $T/2$, one would need to light each uniform segment on both ends, not simply the end of both ends of the total rope. What am I doing wrong?

Answer 1

Let's say yes half burns in 5 mins and half burns in 55 mins. We now lit fire in both the ends. Now whatever rate might be of burning, the total rope get burned out at 30 mins.

Now take the 2nd rope , lit it's two ends, and lit any two more portions of the rope. That makes the rope burn down in 15 mins. So now u have total of 30 + 15 = 45 mins

Answer 2

Light both ends of rope $A$, and one end of rope $B$. When rope $A$ is burnt out, after 30 minutes, light the other end of rope $B$. Rope $B$ will be burnt out after a further 15 minutes.

Answer 3

You say "Therefore, in order to arrive at a rope that burns in time $T/2$, one would need to light each uniform segment on both ends, not simply the end of both ends of the total rope. What am I doing wrong?"

If I understand what you mean here, the idea is that you cut the rope into short segments, each one of (approximately) uniform density. Then you light the first one at both ends. As soon as it's done burning, you light the next one at both ends, and so forth.

That's certainly one way to arrive at a rope that burns in time $T/2$. Your mistake is thinking that's the only way -- i.e., instead of "one would need to," it's "one could." But here's another way to do it: Suppose you knew, or somehow guessed, exactly how far the rope would burn in time $T/2$ if lit from one end (say the left). If you imagine cutting the rope at exactly that point, the other portion will also burn in time $T/2$. But you don't actually have to make the cut in order to start burning the other portion -- you can light it from the other end. So you don't need to know where to make the cut; you just light both ends. When they're done burning, at time $T/2$, you'll know where the cut would have been.

Answer 4

Here's my proof why the answer is wrong as well, if you assume nothing about the rope properties. Essentially things like feedback might affect the rope burning.

Rope = [ 1 | 2 | 3 | 4 | 5 ]

Imagine the rope split into 5 parts.

The rope’s units are not guaranteed to burn at the same rate depending on which side you light it from.

From left: 1 - 50min; 2 - 2min; 3 - 5min; 4 - 1min; 5 - 2min.
From right: 1 - 10min; 2 - 5min; 3 - 15min; 4 - 20min; 5 - 10min.
Both ways add up to 1hr if lit individually.

Light both ends: 50 minutes in and finally it’s done. Left side burnt only Unit 1 and right side burnt units 5, 4, 3, 2.

Or after 30 minutes, unit 1 is not burnt, and only unit 5 and 4 are burnt.

The fastest that lighting both sides can take is 0 minutes, and the slowest is 1 hour.

Answer 5

For the first rope, label one end 0 and the other end A. Starting from end 0, let derivative of time passed w.r.t. fire position x be f(x). The corresponding function starting from end A is f'(x).

Since rope A takes 1h to burn from either end, integrating f(x) from 0 to A, and integrating f'(x) from A to 0 gives time of an hour. Hence f(x)=-f'(x)

Now, if rope A burns from both ends, let time needed for complete burning be t hours and meeting point be a.

Since time passed for both processes of burning from each end is the same, integrating f(x) from 0 to a, and integrating f'(x) from A to a (which is the same as integrating f(x) from a to A, as established earlier) gives time t.

Hence 2t, which is twice the integral of f(x) from 0 to a = twice the integral of f(x) from a to A = integral of f(x) from 0 to a + integral of f(x) from a to A = integral of f(x) from 0 to A = 1 hour. Hence, t is 0.5 hour.

Next, for rope B, label one end 0 and the other end B and define the corresponding function for derivate of time w.r.t. position as g(x). After burning for 0.5 hour from one end, let position of fire be b.

We know that integrating g(x) from 0 to b gives half an hour, and integrating g(x) from 0 to B gives 1 hour. Then it can be seen that integrating g(x) from b to B, which is time needed to burn remaining the rope, which we can call rope C, is half an hour.

When we start burning the remaining rope, rope C from the other end, the same argument for rope A gives the time for complete burning as 0.25 hour. Hence we can get 0.75 hour.

Answer 6

If you light both ends of the rope, it will burn in 30 minutes. Think about it this way: if the left half of the rope takes 40 minutes to burn, then the right half should take 20 minutes, totaling 60 minutes. Now, let's light both ends. After 20 minutes, the right half of the rope is gone, but the left half has only burned half way.

However, the remaining 1/4 of rope, which would take 20 minutes to burn from one end, will burn in half the time because the flame that just burned the right half has now started on the left half as well. So the remaining 1/4 takes only 10 minutes. 10 + 20 = 30 minutes.

Short version: both flames will keep burning until the rope is gone (at which point the flames meet). And it doesn't matter how much rope is burnt, but rather how long the flames are burning it. Kind of like man-hours. Two men can do a job in half the time it takes one to do it (in theory).

Answer 7

Here is my proof that no matter how much the rate of burning varies over the time interval $[0, 1]$ as long as the time to burn when lit at one end is 1 hour then lighting the rope simultaneously at both ends will consume the rope after 1/2 hour.

Formally, the amount the rope (lit at one end) that burns in time $a$ is $\int_{t=0}^a r(t)dt$ where $r(t)$ is the rate of burning. The amount the rope that burns in time $a$ if lit from the other end is $\int_{t=0}^a r(1-t)dt$. The one hour restriction is the fact $\int_{t=0}^1 r(t)dt = L$ where $L$ is the length of the rope. Thus the amount of rope (lit at both ends) that burns in time $a$ is $\int_{t=0}^a r(t)dt + \int_{t=0}^a r(1-t)dt = \int_{t=0}^a r(t)dt + \int_{t=1-a}^1 r(t)dt$ (using substitution rule for definite integrals). When time $a$ reaches 1/2 it is easy to see that the rope is entirely consumed but not before. Although this is an easy result and fundamental to this type of physical process, it is glossed over in presentations of the problem to the general (non-mathematical) public.