Cluemeister's Corner

Results for January

For January, the Cluemeister delved into the mysteries of sand timers, asking how eight minutes could be meted out with various combinations.

In part A, the two timers measured 13 and 6 minutes, respectively. Several readers sent in answers to this part. Most went like this: Flip the two timers at once. When the shorter one runs out, flip it again. When it runs out again, flip them both, leaving just one minute in the larger timer. When the larger one runs out, flip it, at which point there are five minutes left in the smaller one. When the smaller timer runs out, there will be eight minutes left in the larger timer, with which you can measure your task.

Bradley S. offered an algebraic abstraction of the problem in two variables, allowing one to generalize to any two timers. Jon S. pointed out that in an analog world, there is no such thing as measuring an exact period of time. A fair point! He also suggested that the error involved in flipping timers might cause us to lose the to-the-second precision the puzzle's set-up asks for. To an extent, however, this evens out, given that we are flipping both timers, and not just the same one over and over. It may indeed be a slight improvement to immediately stop both timers together whenever either one is to be flipped, and then turn them up together in the proper directions.

If one cares more about speed than accuracy, one can adopt Lisa S.'s solution: Run the two timers together until the short one runs out, and then start your task while flipping the short one. When the short one runs out again, flip it. When the long one runs out, flip the short timer again; as it has one minute of sand left, it will add to the seven elapsed minutes from the long timer to make eight in all. This solution saves us some time, but may be harder to perform in practice, depending on whether our task takes some setting up. It also suggests other, less efficient solutions taking advantage of the fact that the eight minutes need not be meted all at once by a single timer.

Part A was fairly straightforward, and indeed, by using the method of repeatedly flipping any timer that runs out, one can measure any whole number of minutes up to the length measured by the larger timer whenever one is given a pair of timers that measure M and N minutes, so long as M and N are relatively prime, meaning that they share no factors but 1.

This result from group theory, however, does not help us very much with parts B and C. In part B, where the timers measure 9 and 6 minutes, this method can only measure multiples of 3 minutes. In part C, where they measure 32 and 64 minutes, we are worse off, only being able to measure multiples of 32. In order to get to 8, we are therefore forced to get creative.

Tim L. suggested that as he is nerdy enough as to never be without a measuring device, his possession of one should be assumed for the purpose of the puzzle. He then proposed to use it to measure the relative heights of the sand in the two timers. Well, perhaps...unless their shape is tapered rather than cylindrical, as with some hourglasses....

Jon S. suggested that one timer could be used as a pendulum while the other is used to time its period, and then eight minutes worth of swings could be observed. Clever...so long as you have a string and can pay close enough attention to count the swings while performing your unnamed task.

Patricia Z. suggested smashing everything and melting the sand to make a bigger hourglass. Perhaps not eminently practical, but it is thinking outside the box!

The following are the Cluemeister's solutions:

Part B: Run both timers together until the 6-minute one runs out, and then lay them on their sides. Use some solid blunt object, such as the base of the 9-minute timer, to crack the glass or leverage open the plastic of the 6-minute timer on the end with sand in it. Carefully create an opening. Now break the 9-minute timer in half. One end will contain a third of its sand; the other end will contain two-thirds. Pour the sand from the end with less sand into the opening in the 6-minute timer. The 6-minute timer now contains a third again more sand than before. (Remember that the two timers are identical but for the widths of their waists, meaning that they contain equal amounts of sand to begin with.) Now, with its extra sand, the 6-minute timer is ready to measure 8 minutes when you turn it upright.

Part C follows a similar method, but is more elaborate. Run both timers and stop them when the smaller one runs out, as before. Break the 64-minute timer in half, or remove the lid. Remove the lid of the 32-minute timer (by cracking it on the floor if necessary). Pour out the sand. Now pour in half the sand from the other timer. It will take sixteen minutes to pass through into the bottom half of the former 32-minute timer. Flip this over while simultaneously running what remains of the former 64-minute timer, also now at half its original sand level. When the former 32-minute timer runs out, stop the other timer, which will now have a quarter of its original sand. Pour that sand into the former 32-minute timer. When you turn it over again, the sand will run for a quarter of 32 minutes, or 8 minutes.

Yes, it requires that the timers be breakable. Well, in such sticky cases, no solution is perfect! We make do with what we have.

For pointing out the world's sticky imperfections and offering a pendulum solution that might work for the other two parts, Jon S. wins this month's prize, to be collected at Minicon 46.