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## Question

You have a fuse that burns at an uneven rate, but it's symmetric about its midpoint.

Starting from either end, it burns completely in one hour.

What's the shortest time interval you can measure using the fuse?

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a half-hour

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I want to say half an hour. From one endpoint to the midpoint. But that just seems too simple, and I'm sure there is something I'm missing

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one half hour

You have a fuse that burns at an uneven rate, but it's symmetric about its midpoint.

Starting from either end, it burns completely in one hour.

What's the shortest time interval you can measure using the fuse?

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1 hour...

nahhhh half hour

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You have a fuse that burns at an uneven rate, but it's symmetric about its midpoint.

Starting from either end, it burns completely in one hour.

What's the shortest time interval you can measure using the fuse?

15 minutes, light one end and then when it burns to midpoint light the other end.

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One hour, because you don't have a ruler to measure the where the midpoint is

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One hour, because you don't have a ruler to measure the where the midpoint is

Ah, well in that case the answer is zero as you do not have a match to light the fuse either.

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15 minutes, light one end and then when it burns to midpoint light the other end.

How could you be sure of 15 min? The symmetric point is in the middle, not the quarter.

I'm going with 30 minutes

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How could you be sure of 15 min? The symmetric point is in the middle, not the quarter.

I'm going with 30 minutes

Light the fuse at one end and the midpoint at the same time. The lit points will burn towards each other, and meet at 15 minutes

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It's 15 minutes. Light the fuse at one end AND the middle at the same time. When they meet, 15 minutes will have passed.

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Cut the fuse in half so that you have two identical pieces of fuse. Light both ends of one piece. Mark the spot on the other piece where the first fuse burns out. Light the second fuse at that spot and at both ends. The fuse will burn out in 7.5 minutes.

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what we know is a. the fuse burns at an uneven rate. b. the fuse will burn in 1 hour if lit from either end. Conclusion; if we light the fuse at both ends it will burn out in half an hour although it may not meet in the exact middle (burns at uneven rate). Hence the middle is an uncertain starting point to try and break it down further;therefore I say 1/2 an hour.

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How is it different from 2 unevenly burning ropes problems?

Ignite it in the middle and at the left end. When the left half has burned, ignite the right end.

Edited by Prime
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How is it different from 2 unevenly burning ropes problems?
Ignite it in the middle and at the left end. When the left half has burned, ignite the right end.

Even easier...

Once lit in the middle it should burn both ways, left and right. There's no need to light the second end, just wait for it to burn out.

Here's why the shorter answers work:

It doesn't matter that the fuse burns at different rates, you just need to know the total time to burn. The fact that it's symmetrical (same "shape" of burn times on both sides) helps you make a shorter time. If it wasn't symmetrical we wouldn't know where the "mid point" in burning time would be in the fuse. If you don't believe it works, try this. Take a string of numbers that adds up to, say, 10: 5 3 1 1. Let this string of numbers be the burn time for one side. So, 5 units, 3 units, etc. If we lit this at both sides our mid point in burning time will be between the 5 and the 3, but still half way to 10. You can arrange the numbers however you want, and if you light the fuse at both ends, they will meet in the middle (of burning time). The physical middle will shift depending on your arrangement of numbers.

Sorry for the long winded response.

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Even easier...

Once lit in the middle it should burn both ways, left and right. There's no need to light the second end, just wait for it to burn out.

I think what Prime meant was that the 7.5 min was time it would take to burn the remainder of the fuse. Lighting the fuse at the midpoint and one end, say the right end, the right side will have all burned in 15 min. Since the burning goes both ways from the midpoint as you said, it will have burned 15 min worth of the left side as well. Then you light this last piece on both sides and it will take 15/2=7.5 minutes to burn the remaining piece.

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It would depend on how many pieces you cut the fuse into. If you cut it in half and light both ends, 7.5 minutes. If you cut it in 4ths, 3.75 minutes. And so on until you can't cut it any more.

That aside, because the fuse burns at an uneven rate, is it even possible to know? The only thing we would know for certain is that the whole thing takes an hour. Unless no matter how you cut it, it'll always burn to where it's symmetric at its mid point. But yeah, the first bit is my answer.

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so, 30 minutes is pretty trivial

Light both ends.

Or, alternatively, light it anywhere along the fuse, other than at the end, (so that it burns both ways). When the two burning points meet, light the remaining unlit end - I think this should give a hint for the general solution....

There seems to be lots of suggestions for 15 minutes which just don't add up as far as I can see...

Light the middle and both ends. Except you don't know where the 'middle', with respect to time, is! For example

```start +------+------+

+----+  +-+

+--+

end	  ++```

This doesn't measure 15 mins because, at the end, there were only two bits of fuse still burning. The 30 minute solution relies on the fact that we burn at twice the rate. This burns at four times the rate until the first pair meet,from which point it only burns at twice the rate.

I think I have an answer leading on from this (and my solution to 30 mins) but will leave this up while I write it in case it helps anyone else.

The fact that it's symmetrical (same "shape" of burn times on both sides) helps you make a shorter time.

Who said it was symmetrical?

edit: correction to my incorrect alternative solution for 30 min!

Edited by foolonthehill
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Who said it was symmetrical?

The OP did.

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The OP did.

Sorry! My mistake - I thought we knew nothing about the rate of burning of the fuse.

Scratch my post which said that you couldn't get fifteen minutes by lighting the midpoint and two ends - you can!

Think I need to spend more time reading and less time posting!

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If the fuse burns unevenly, how could you be sure the mid and the end would meet at 15 min? Good Puzzle huh

edit: messed up spoiler

2nd edit for afterthought

I took symmetrical to mean the will burn to the middle at the same time if lit on both ends... maybe it doesn't?

Edited by Justsnapd8
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If you can move infinitely fast, you may be able to achieve any fraction of an half hour.

If we divide the fuse in half we'll have two 30 minute fuses. We light twelve flames on the first half and maintain twelve seperate flames. Meaning if one segment burns out, we would start a flame in the middle of a longer segment which would then divide into two. This would result in ten segments of fuse burning at both end at all times until the fuse is gone in about 1.5 minutes. I think this would work.

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It should be possible, I think, providing you are quick enough, to measure any time which is an even integer divisor of 60 minutes, by which I mean 60/2i for all integers, i. ALso, I don't think you need the fuse to be symmetrical because it adds nothing to the solution except a factor of two (and which I why I assumed that it wasn't symmetrical in my previous post!)

The solution is to ensure that there are 2i flames burning concurrently until the entire fuse is burnt, and this simply means lighting the the ends and i-1 points anywhere along the fuse.

As the fuse is burning from each end, the only way in which a flame can go out is by meeting another flame. Every time this happens you can replace them both by lighting anywhere except where a flame already is.

```1 +========================++=========++================+

2 -+======================+===+======+--+==============+

3 --+====================+======+===+----+============+

4 ---+==================+=========++------+==========+

5 ----+============++==+-------------------+========+

6 -----+==========+--++---------------------+======+

7 ------+===++===+---------------------------+====+

8-------+=+--+=+-----------------------------+==+

9--------+----+-------------------------------++```

On every row, there are six flames. The first meet at time 4 so a new pair is created. And the same happens at time 6. By time 9, all the pairs meet together and exactly 60/6, ie ten minutes has elapsed.

I can;t help feeling I've missed something because bonanova doesn't often put in extra information (the symmetry bit) unless it's necessary!

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I can;t help feeling I've missed something because bonanova doesn't often put in extra information (the symmetry bit) unless it's necessary!

It should be possible, I think, providing you are quick enough, to measure any time which is an even integer divisor of 60 minutes, by which I mean 60/2i for all integers, i. ALso, I don't think you need the fuse to be symmetrical because it adds nothing to the solution except a factor of two (and which I why I assumed that it wasn't symmetrical in my previous post!)

The solution is to ensure that there are 2i flames burning concurrently until the entire fuse is burnt, and this simply means lighting the the ends and i-1 points anywhere along the fuse.

As the fuse is burning from each end, the only way in which a flame can go out is by meeting another flame. Every time this happens you can replace them both by lighting anywhere except where a flame already is.

```1 +========================++=========++================+

2 -+======================+===+======+--+==============+

3 --+====================+======+===+----+============+

4 ---+==================+=========++------+==========+

5 ----+============++==+-------------------+========+

6 -----+==========+--++---------------------+======+

7 ------+===++===+---------------------------+====+

8-------+=+--+=+-----------------------------+==+

9--------+----+-------------------------------++```

On every row, there are six flames. The first meet at time 4 so a new pair is created. And the same happens at time 6. By time 9, all the pairs meet together and exactly 60/6, ie ten minutes has elapsed.

I think with both halves burning you can create more complex fractions, but I haven't thought it through all the way

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If you can move infinitely fast, you may be able to achieve any fraction of an half hour.
If we divide the fuse in half we'll have two 30 minute fuses. We light twelve flames on the first half and maintain twelve seperate flames. Meaning if one segment burns out, we would start a flame in the middle of a longer segment which would then divide into two. This would result in ten segments of fuse burning at both end at all times until the fuse is gone in about 1.5 minutes. I think this would work.

I don't think so.

So you have a piece of cord which takes time T to burn when ignited from one end.

Now you light it up in n+1 places having n segments burning from either end. When the fastest burning segment is gone. Extinguish all flames and stop to think about it.

It took x time (unknown) for the fastest segment to burn. What you have is x*n worth of ashes (timewise), and n-1 segments of cord remaining for the total time worth T - x*n. That does not help you to measure time.

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I don't think so.

So you have a piece of cord which takes time T to burn when ignited from one end.

Now you light it up in n+1 places having n segments burning from either end. When the fastest burning segment is gone. Extinguish all flames and stop to think about it.

It took x time (unknown) for the fastest segment to burn. What you have is x*n worth of ashes (timewise), and n-1 segments of cord remaining for the total time worth T - x*n. That does not help you to measure time.

Why would you stop to think about it? Moving infinitely fast is key.

by dividing the fuse into n segments and light just one end of each. When the fastest portion is done and burnt, chop the slowest into two and light one end of the new piece. You will always have n segments and n flames until they all burn out at the same instant. So the time is T/n.

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