# Burning ropes as timers - How many time intervals can be measured?

Note: This is a follow-up question to Burning ropes as timers. The following question and its answers may contain spoilers.

To sum up the puzzle, these are the rules (slightly modified):

• You have some number N of ropes, each with the following property.
• When you light one end of a rope, the fire will reach the other end after exactly one hour.
• You start the puzzle by lighting one or more ends of the ropes.
• You may light or extinguish any end of a rope after that.
• You can only light or extinguish a rope once another rope has completely burned out.
• You may only light the end of a rope, since lighting anywhere else is inaccurate.
• You may not organize the ropes so that they light each other or themselves.
• You must accurately measure some interval of time, between any two distinct times.

The question is, how many intervals of time (excluding 0) is it possible to measure with N ropes?

For example, when N = 1, the answer is 2.

• 1 hour (by burning the length of the rope)
• 1/2 hour (by lighting both ends of the rope at once)
• Do you know the answer? Can you measure 15 minutes with 2 ropes? (I mean is that allowed?) May 16 '14 at 19:01
• @martijnn2008 No, I haven't mathed it out yet. Yes, a 15-minute interval is possible, between the time the first and second rope burn out in the answer to the original puzzle. May 16 '14 at 19:05
• You might look at this paper on fusible numbers. The set of times that can be measured (in the paper, only intervals starting at zero are considered, so $1/4$ cannot be measured.) is extremely complicated. May 17 '14 at 0:45
• Do we need to start measuring right away, or can we measure 15 minutes, half an hour from now?
– SQB
May 21 '14 at 12:28
• @SQB See the last rule, and the second comment. May 21 '14 at 13:01

To get started on this problem, here is a general observation for any $n$:

You can only measure intervals of time of the form $$\frac{m}{2^n}\cdot 1\mbox{h}$$

where $m$ and $n$ are non-negative integers.

Proof (using the extremal principle):

Suppose there was a number of ropes you could use to measure an interval which is not of the proposed form above. Take the minimum number of ropes $n$ for which this is possible, then $n≥2$ (as for $n=1$ it is not possible as shown in the question).

There must be a way of lighting the $n$ ropes so that the interval $\ t = t_2-t_1$ between two "distinct times"1 $\ t_1$ and $t_2$, where $t_1<t_2$, cannot be described by the above form. As $t_2>0$, there must be a rope $r$ that has completely burned down (otherwise, $t_2$ would not be a distinct time - see the footnote). Due to $t_1<t_2$, you can measure $t_1$ without using $r$ (so $n-1$ ropes suffice), which means that $t_1$ has the form above due to the minimal choice of $n$.

Consider the very last time $t_3$ rope $r$ is lit. By the same argument as above (you do not need $r$ to measure this time), it follows that $t_3$ has the above form. Accordingly, $\left|t_3-t_1\right|$ is also of that form. As $t = t_2-t_1 = (t_2-t_3)+(t_3-t_1)$ is by assumption not of the above form (but $\left|t_3-t_1\right|$ is), it follows that $t_2-t_3$ is not of the form above either.

However, this cannot be true:

Look at amounts of time $t_{r,1},\ldots,t_{r,l}$ that $r$ actually burns. These satisfy the relation $$n_1\cdot t_{r,1} + \ldots + n_1\cdot t_{r,l} = 1\mbox{h}$$

where $n_i = 1,2$, depending on whether $r$ burns at one or both ends. All the amounts of time $t_{r,1},\ldots,t_{r,l-1}$ are of the above form, because they are measured without using $r$. So $t_{r,l}$ also has to be of the form above. But $t_{r,l} = t_2-t_3$, which is a contradiction to the conclusion of the last paragraph.

Therefore, the assumption that there is a number of ropes which can be used to measure an amount of time not of the form above has to be wrong.

1. The start of the experiment or the time a rope has burned out, as there are no other possible events.

• I'm interested, how do you measure 37.5 minutes (5/8 of an hour)?
– user88
May 20 '14 at 20:45
• @JoeZ. My answer does not imply that you can measure every interval of the form $\frac{m}{2^n}\cdot 1h$, only that you cannot measure other intervals (like $\frac{1}{3}h$). However it should indeed be possible to measure all such intervals given that you have enough ropes, which was addressed in another (now unfortunately deleted) answer to this question: May 25 '14 at 22:41
• To measure $\frac{5}{8}h$ using five ropes, you light rope 1 on one side and rope 2 on both sides. After 30min, rope 2 is burned, and you light rope 1 at the other side and rope 3 on one side. After 15min, rope 1 is burned, and you light rope 4 on one side. After 45min, rope 3 is burned, and you light rope 4 on the other side, starting the measurement of the 37.5 minutes. After 7.5min, rope 4 is burned, you light rope 5 on both sides so it is burned after 30min, ending the measurement of 37.5 minutes. May 25 '14 at 22:41
• @JoeZ You can solve this problem using 4 ropes. Here is how. Light both sides of rope one and one side of ropes two and three. After 30 minutes, rope one is burned through. Light the other end of rope two. After 15 minutes, rope two is burned through. Your 37.5 minutes starts here. Light the other end of rope three. In 7.5 minutes, rope three will burn through. Light both sides of rope four. in 30 minutes rope four will burn through. When this happens, 37.5 minutes will have passed since you lit the other end of rope three.
– user5559
Nov 17 '14 at 3:44
• @JoeZ. Its possible only with 3 ropes. Light 1st rope on both ends and 2nd rope on one end. Timer starts after 1st rope is completely burnt. Now 30 mins are remaining on 2nd rope. Now lit other end of 2nd rope and one end of 3rd rope. 2nd rope will burn in 15mins and 45mins will be remaining in 3rd rope. Now lit other end of 3rd rope which burn in 22.5mins. Total 37.5mins in 3 ropes. Nov 26 '14 at 6:05