Since there is 1/2 left of the rope and we just lit the second end, it will burn for 1/4. Assuming we are able to measure 1/2, we do the following: light the rope at one end, wait 1/2, light the other end, and start the timer. So we have measured 5/8… and all we needed to do was be able to measure 1/4.Ģ) 1/4 comes from (1-1/2)/2, where 1/2 belongs to S1. Since there is 3/4 left of the rope and we just lit the second end, it will burn for 3/8. Assuming we are able to measure 1/4, we do the following: start the timer, light the rope at one end, wait 1/4, then light the other end. As you’ll see, it should be possible to write a recursive computer program that spits out the recipes for each possible number. This may seem like a lot of work but it’s very formulaic. We can construct the 5/8 solution by working in reverse. I interpreted “measure” to mean that it’s possible to use a stopwatch that you start and top at specified events and the time indicated on the stopwatch would be the “measured” time. Note: my solution is incomplete (see comments below!) Author Laurent Posted on FebruFebruCategories The Riddler Tags recursion, Riddler This measures $|\tau-1|$ or $|\tau-\tfrac$ hours. Starting at the first beep, light one side (or both sides) of the rope and measure the time elapsed between when the rope is fully burnt and the second beep.Measure the time elapsed between both beeps (don’t use the rope at all).Using a single rope, what new intervals of time can we now measure? There are several things we can do: We can imagine this as two beeps that occur exactly $\tau$ apart. Suppose we the ability to measure a time interval $\tau$ and nothing else. Let’s start by answering simple question that will give us the key insight needed to solve the problem: My solution uses a recursive approach: adding one rope at a time and seeing how the set of measurable time intervals increases.
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