[bonanova]

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?

[Prime]

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.

You cannot know which segment is slowest until all others burned completely. Not that it would help if you did.

Divide rope into 3 segments and light each segment at both ends.

Say first one burned in 1 min. (but you don't know time). You performed whatever addigional lighting. The next one took 3 min to burn, the next one after that 50 min...

Where is 1/n here?

The only trick to these problems is the implication that, although uneven, cords burn at the same speed regardless of the direction in which flame moves. So lighting a 1-hour cord from 2 ends gives you 1/2 hour.

[bonanova]

The "standard" puzzle [uneven burn rate] is solved by lighting both ends.

Two continuously burning flames burn the fuse in 1/2 the time.

Prof T asks to generalize to n flames, which would burn the fuse in 1/n of the time.

But the difficulty seems to lie in having n flames complete their burn together. [With two flames, that part is automatic.]

So he says start a new flame whenever one goes out.

That's what introduces the infinitely fast requirement.

Sounds like juggling live cats, but it's true that n flames, if they finished together, would give 1/n hour burn time..

But I'd like to see a complete scenario for say n=4, just to be sure you have a free end to light whenever needed.

Because starting a new flame not at an end would produce two new flames.

Oh, we already have an 1/4 solution. Try n=8.

Prime says you don't have to move instantaneously fast - you could extinguish the flames and take your time relighting them.

But that introduces an unknown time interval to the process.

Are we in agreement yet for the best answer for a finitely fast person?

To add specificity, your available tools are scissors and matches.

No rulers or stopwatches.

[Guest]

Prof T asks to generalize to n flames, which would burn the fuse in 1/n of the time.

But the difficulty seems to lie in having n flames complete their burn together. [With two flames, that part is automatic.]

So he says start a new flame whenever one goes out.

That's what introduces the infinitely fast requirement.

Sounds like juggling live cats, but it's true that n flames, if they finished together, would give 1/n hour burn time..

But I'd like to see a complete scenario for say n=4, just to be sure you have a free end to light whenever needed.

Because starting a new flame not at an end would produce two new flames.

The n=4 (15 mins) solution is simply to light the middle and both ends, and no relighting is necessary because of your symmetric fuse.

The n=6 (10 mins) is the first solution which requires infinitely fast reactions because you can light both ends and two points anywhere along the fuse. When two flames meet, relight another 'midpoint' (but doesn't have to be in the middle!).

Actually, if you are also allowed to cut up the fuse, this is the same as cutting it into n/2 pieces and lighting all the ends.

I put a demonstration of n=6 in my last post (with the solution that I only realised afterwards Prof T had posted about ten minutes before!)

[Guest]

7.5 minutes.

Burn in the middle and one end at the start. when the fire meets from middle and one end(15 min) burn the other end and start measuring the time. When the fire meets it will be 7.5 min.

[bonanova]

7.5 minutes.

Burn in the middle and one end at the start. when the fire meets from middle and one end(15 min) burn the other end and start measuring the time. When the fire meets it will be 7.5 min.

I think that's the best finite-speed solution so far.

I think I can beat it.

[Prof. Templeton]

You cannot know which segment is slowest until all others burned completely. Not that it would help if you did.

Divide rope into 3 segments and light each segment at both ends.

Say first one burned in 1 min. (but you don't know time). You performed whatever addigional lighting. The next one took 3 min to burn, the next one after that 50 min...

Where is 1/n here?

The only trick to these problems is the implication that, although uneven, cords burn at the same speed regardless of the direction in which flame moves. So lighting a 1-hour cord from 2 ends gives you 1/2 hour.

You don't have to know which is the true slowest

You know which is the one(s) fastest because it burns out first. When one is gone you divide another into two and light one end of the new segment. It could be the second fastest or the slowest or any or the remaining segments. You would want to shoot for the slowest, so fewer operations were required, but it doesn't have to be the slowest only a remaining segment.

I'd like to get past this point, because I think I have a general equation that will work for any fraction and uses both halves of the fuse and I'd like some input on it.

[Guest]

I think that's the best finite-speed solution so far.

I think I can beat it.

Let the ends of fuses be named A,B and M be the middle. Cut the fuse in two parts at the middle. so now we have two fuses AM and BM. Now put the fuses in parallel to each other such that A of AM is in line with B of BM. Burn A side of AM and M of BM . Since they are lying parallel as soon as you see the flames are meeting each other extinguish the fire.

Now you have two fuses that burn in 15 minutes each. and one can use them to calculate 3.75 minutes. I guess that should be clear

[Guest]

Let the ends of fuses be named A,B and M be the middle. Cut the fuse in two parts at the middle. so now we have two fuses AM and BM. Now put the fuses in parallel to each other such that A of AM is in line with B of BM. Burn A side of AM and M of BM . Since they are lying parallel as soon as you see the flames are meeting each other extinguish the fire.

Now you have two fuses that burn in 15 minutes each. and one can use them to calculate 3.75 minutes. I guess that should be clear

Very nice. Could we not continue that process as far as "practical"? 30 - 15 - 7.5 - 3.75 - 1.875 - etc. And by practical I mean you couldn't really get a time out of burning both ends.

[Guest]

Very nice. Could we not continue that process as far as "practical"? 30 - 15 - 7.5 - 3.75 - 1.875 - etc. And by practical I mean you couldn't really get a time out of burning both ends.

I dont think so as the basic requirement here is that it burns symmetrically on both sides of the middle point. This cannot be said for the smaller fuses.

[bonanova]

Let the ends of fuses be named A,B and M be the middle. Cut the fuse in two parts at the middle. so now we have two fuses AM and BM. Now put the fuses in parallel to each other such that A of AM is in line with B of BM. Burn A side of AM and M of BM . Since they are lying parallel as soon as you see the flames are meeting each other extinguish the fire.

Now you have two fuses that burn in 15 minutes each. and one can use them to calculate 3.75 minutes. I guess that should be clear

Imram has it.

Even left the final step as an exercise. Kudos.

[Prof. Templeton]

So you can move infinitely fast and perform infinitely fast operations, but you left your stopwatch at home and all you have is 1 hour worth of this uneven burning fuse, a pair of scissors and a flame. To obtain any fraction of a half hour between zero and one

for a faction n/d, you would have to maintain d-n+1 flames on the first half of the fuse and d flames on the second half. If two flames go out at the same time because a segment was burning at both ends, another segment should be ignited in the middle which will cut the segment and produce two new flames. If a segment that was burning from one end is consumed another segment needs to be cut in the middle and just one of the new ends ignited. Once the first half of the fuse is gone, put out all but one of the remaining flames on the second fuse and then maintain one flame until the second fuse is all gone. When the fuse is all spent n/d of a half hour will have passed. In some cases both of the fuses will burn out at the same time. I hope I've explained this clear enough.

[Prime]

So you can move infinitely fast and perform infinitely fast operations, but you left your stopwatch at home and all you have is 1 hour worth of this uneven burning fuse, a pair of scissors and a flame. To obtain any fraction of a half hour between zero and one

for a faction n/d, you would have to maintain d-n+1 flames on the first half of the fuse and d flames on the second half. If two flames go out at the same time because a segment was burning at both ends, another segment should be ignited in the middle which will cut the segment and produce two new flames. If a segment that was burning from one end is consumed another segment needs to be cut in the middle and just one of the new ends ignited. Once the first half of the fuse is gone, put out all but one of the remaining flames on the second fuse and then maintain one flame until the second fuse is all gone. When the fuse is all spent n/d of a half hour will have passed. In some cases both of the fuses will burn out at the same time. I hope I've explained this clear enough.

Work out an example. Say you want to measure 10 min. Take one of the cut pieces burns in 2 min, second in 4, and the third in 24 (or one half of that time each when lighted from both ends.) Or use your own numbers. Remember, when you cut any piece in half lengthwise, it does not tell you how fast each half would burn.

[Prof. Templeton]

Work out an example. Say you want to measure 10 min. Take one of the cut pieces burns in 2 min, second in 4, and the third in 24 (or one half of that time each when lighted from both ends.) Or use your own numbers. Remember, when you cut any piece in half lengthwise, it does not tell you how fast each half would burn.

It doesn't matter how fast they burn. This is the third time I've said it. When the 2 min. piece is burned up you cut one of the ones that's left. Let's say you cut the segment that takes 24 min. and it gives a 10 min and a 12 min. when the 4 min segment is gone you cut a remaining segment in half and light it. Let's say it's the longer one which give us a 6 min and a 4 min (the 12 min segment is now 10 min.). By now 4 min. have passed and we have a 8 min. , a 6 min. and a 4 min. segment. Another 4 min. pass we cut another segment and and it gives us a 1min. segment and a 3 min. (the 8 min. segment is now 4 min. long). 8 min have passed and our segments are 3, 2, and 1 min. After another min. 9 have passed and we cut the longer segment in half and it gives us 1 min. and 1 min. now we have three 1 min. segments and when they are gone 10 minutes have passed. Use any numbers you like, if there is always 3 segments burning it will take ten minutes. The cuts are always widthwise never lengthwise.

[Prime]

It doesn't matter how fast they burn. This is the third time I've said it. When the 2 min. piece is burned up you cut one of the ones that's left. Let's say you cut the segment that takes 24 min. and it gives a 10 min and a 12 min. when the 4 min segment is gone you cut a remaining segment in half and light it. Let's say it's the longer one which give us a 6 min and a 4 min (the 12 min segment is now 10 min.). By now 4 min. have passed and we have a 8 min. , a 6 min. and a 4 min. segment. Another 4 min. pass we cut another segment and and it gives us a 1min. segment and a 3 min. (the 8 min. segment is now 4 min. long). 8 min have passed and our segments are 3, 2, and 1 min. After another min. 9 have passed and we cut the longer segment in half and it gives us 1 min. and 1 min. now we have three 1 min. segments and when they are gone 10 minutes have passed. Use any numbers you like, if there is always 3 segments burning it will take ten minutes. The cuts are always widthwise never lengthwise.

How did you know when to stop cutting and burning?

Although, I am beginning to see your point. What you get is as close approximation to 1/n of the total time as you can manage by moving fast. Unless, you got lucky at some point and divided remainders into n of equal time length.

Here is how I would express it in general terms.

Lets take an example of n=3. That is we want to measure T/3 time interval using uneven cord of total burning time T.

Cut the cord into 3 arbitrary pieces and ignite them at the same instance. Suppose the fastest took time x₁ to burn. Then the total worth of burned cord is 3x₁ at the instance it finished.

Cut quickly an end from the longer remaining piece and ignite it. Again you have 3 pieces burning simultaneously. Suppose the fastest took x₂ time to burn. Thus you have burned another 3x₂ worth of cord and thus far x₁ + x₂ total time elapsed.

Repeat the procedurre ad infinitum.

In the end you have burned the entire T = 3x₁ + 3x₂ + ... worth of the cord. While the total time elapsed is x₁ + x₂ + ... Which happens to be exactly 1/3 of the total time.

Interesting!

Here is the procedure for simultaneous ignition and division:

When you see one of the cords is about to expire, cut off the end of another cord and join it to the non-burning end of the expiring cord. This way you don't need to be all that fast.

[bonanova]

Here is the procedure for simultaneous ignition and division:

When you see one of the cords is about to expire, cut off the end of another cord and join it to the non-burning end of the expiring cord. This way you don't need to be all that fast.

Nice!

[Prof. Templeton]

How did you know when to stop cutting and burning?

Although, I am beginning to see your point. What you get is as close approximation to 1/n of the total time as you can manage by moving fast. Unless, you got lucky at some point and divided remainders into n of equal time length.

Edit

Here is the procedure for simultaneous ignition and division:

When you see one of the cords is about to expire, cut off the end of another cord and join it to the non-burning end of the expiring cord. This way you don't need to be all that fast.

You keep cutting as long as you have to to mantain the proper number of segments and flames. I like your idea of having a burning fuse ignite another. In some cases a fuse that is about to burn out will be burning from both ends and another segment which is already burning as well would have to be laid on top of it in a X formation, so you may have to handle burning fuses while trying to center them.

[Guest]

Here is the procedure for simultaneous ignition and division:

When you see one of the cords is about to expire, cut off the end of another cord and join it to the non-burning end of the expiring cord. This way you don't need to be all that fast.

Very nice idea. Great