This is a variation of Puzzle #14 (https://alokgoyal1971.com/2013/04/28/puzzle-14-measuring-time/) that I found on the site fivethirtyeight.com. A fantastic puzzle!

Suppose you have four ropes and a lighter. Each rope burns at a non-constant rate but takes exactly one hour to burn completely from one end to the other. You can only light the ropes at either of their ends but can decide when to light each end as you see fit. If you’re strategic in how you burn the ropes, how many specific lengths of time can you measure? (For example, if you had one rope, you could measure two lengths of time: one hour, by simply burning the entire rope from one end, and half an hour, by burning the rope from both ends and marking when the flames meet.)

*Extra credit:* What if you had *N *ropes?

For younger children, try with two or three ropes.

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy burning!

P.S. – I am traveling over the next two weekends and hence there will be a delay in the next post- sorry!

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*Related*

The set for 1 rope is {0.5, 1}

Assume the set for n ropes is {t1, t2, t3… ti, ….}

The set for n+1 ropes is union of

{t1,…. ti…}

{t1+1, t2+1, ….. ti+1… }

{t1+0.5, t2+0.5, ti+0.5….. }

{1-t1, 1-t2, …. 1-ti…. }

{0.5-t1, 0.5-t2,…. }

{(1-t1)/2, (1-t2)/2,….}

{(1+t1)/2…..}

For 2 ropes, it would be {0.5, 1, 1.5, 2, 0.25, 0.75}

For 3 ropes, it would be {0.25, 0.5, 0.75, 1, 1.25, 1.5, 1.75, 2, 2.5, 3, 0.375, 0.125, 0.625, 0.875}

For n ropes it would be {i/2^n} where I varies from 1 to 2^n, UNION {1+i/2^(n-1)}where i varies from 1 to 2^(n-1), …..UNION {n-1+1/2, n}

So number of measurements = 2^n + 2^n-1+…..+ 2=(2^(n)-1)*2=2^(n+1)-2

Sorry, Can you explain how can we get 0.25 using 2 ropes?

My bad. I misunderstood the question and I got your solution now.

Thanks