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You are given some matches, a rope, and a pair of scissors. The rope burns irregularly and takes 60 minutes to burn end to end; however, we know that the burn rate at distance $x$ from the left end is the same as the burn rate at distance $x$ from the right.

What is the smallest time interval that you can measure?

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  • $\begingroup$ I know I can get it down to 3.75 minutes. But I'm not sure if there's a way to get it shorter. $\endgroup$
    – Justin
    Oct 4, 2014 at 4:21
  • $\begingroup$ Does the time interval have to be from the start of burning the rope, or can the timing start at any point? $\endgroup$
    – Rob Watts
    Oct 4, 2014 at 4:40
  • $\begingroup$ @RobWatts I'm fairly certain we can start at any point, because otherwise I think 15 minutes is the minimum. $\endgroup$
    – Justin
    Oct 4, 2014 at 4:42
  • $\begingroup$ You may want to specify the shortest length of time you can reliably measure. I'm not 100% sure, but I think it's possible to measure arbitrarily short lengths of time if you're lucky $\endgroup$
    – Ben Aaronson
    Oct 4, 2014 at 23:12

1 Answer 1

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Since you can cut it in half, this is equivalent to having two ropes of 30 minutes long. Lay them in rows and set fire to opposite ends, so you have this:

F-------
-------F

Eventually, the flames will meet, like this:

   F----
---F

Each is now a 15 minute rope. Set fire to the other end of the smaller rope, so that it burns down in 7.5 minutes. Then set fire to other end of the remaining rope, which will finish burning in 3.75 minutes.

I don't think you can do better than that. I can't think of a rigorous reason, but here's a unrigorous one: Say two ropes have "similar ends" if they burn at the same rate from each end. Given two ropes with different ends and burn time N, the best you can do is N/4. This is because in time N/2, for one rope you can either

  1. Destroy it (burn both ends)
  2. Reduce its length by N/2 (burn one end)

You need to destroy at least one rope in order to reduce the rest, because you have no other way of keeping time. So with two ropes you can destroy one to reduce the other to N/2 in N/2 time, then destroy the other in N/4 time.

That's with two ropes with different ends. With two ropes with similar ends, you can also produce two half-length ropes with different ends, so the best you can do is actually N/8. And one rope with both ends similar can be cut in half to produce two ropes with similar ends, so you can do N/16.

The bold part is the handwavey part, because I can't think of a strong reason why there isn't a more creative, better thing you can do there.

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  • $\begingroup$ That's exactly the solution I came up with. $\endgroup$
    – Justin
    Oct 4, 2014 at 5:05
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    $\begingroup$ @Quincunx Does "The rope burns irregularly" imply that cutting the rope in half doesn't result in two 30-minute segments? $\endgroup$
    – user20
    Oct 4, 2014 at 5:13
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    $\begingroup$ @Emrakul No. Note the special symmetry property of the rope, that the burn rate a distance x from the left end is the burn rate a distance x from the right end. $\endgroup$
    – Justin
    Oct 4, 2014 at 5:21
  • $\begingroup$ Instead of cutting when the flames reached the center, you would need to do it based off of time. There is no guarantee that the center of each half of the rope is the 'halfway consumed' point. One side could burn the first 50% of the rope in 5 minutes, then take 25 to burn the rest, while the other half may burn at a different proportional rate.. $\endgroup$
    – guildsbounty
    Oct 6, 2014 at 18:27
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    $\begingroup$ @guildsbounty: Actually, because of the symmetry, it still works (assuming you can accurately decide the fire is at the same place). The "half-time point" is going to be in the same spot on both points on the ropes, even if it's not in the physical center. At that point, there is 15 minutes worth of rope to the left and 15 to the right. Then, 15 minutes after lighting opposite sides, you will end up with the left 15 minutes of one and the right 15 minutes of the other, and both flames will be at the mutual half-time point. $\endgroup$
    – TheRubberDuck
    Oct 7, 2014 at 21:01

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