There are four 12-hour clocks arranged in a 2x2 grid, as shown in the diagram. The long minute handles of the left two clocks are touching, while the others are not. Two minute handles touch if they are both vertical (0 and 30 minutes) or both horizontal (15 and 45 minutes). You can set the time of each clock separately. How can you set the time of these clocks, so that there are as many touches between pairs of minute handles as possible in a 24 hour period?
1$\begingroup$ When you want to post two or more puzzles of the same kind, please make sure the answers are sufficiently interesting and different. If the same strategy or approach works, it is essentially the same puzzle. $\endgroup$– BubblerMar 17, 2021 at 4:24
$\begingroup$ @Bubbler Yeah I didn't realize the same strategy works. I deleted the other question. But now that I think about it, I am not so sure that the same strategy works. Can you show me the solution to the 3x3? $\endgroup$– Dmitry KamenetskyMar 17, 2021 at 4:29
$\begingroup$ Actually let me repost it. I think there is more to the 3x3 problem. $\endgroup$– Dmitry KamenetskyMar 17, 2021 at 4:35
1$\begingroup$ A more different variation might be if the clocks were instead arranged in an equilateral triangle. $\endgroup$– Chris SunamiMar 17, 2021 at 15:59
$\begingroup$ @ChrisSunamisupportsMonica ha I was going to put them in a triangle, before I decided to use a 2x2. Feel free to make a new puzzle with this arrangement. $\endgroup$– Dmitry KamenetskyMar 18, 2021 at 9:57
If we set the clocks like this (hour hands omitted, since we don't care about them)
The image of the clock face is from Wikimedia Commons and used under the CC BY 3.0 licence.
we get touching minute hands at every possible occasion, that is, whenever a minute hand points at another clock, there will always be a minute hand pointing back.
Since we cannot change how often a minute hand points at another clock, and there cannot be any touching when a minute hand points anywhere else, this must be optimal.
The solution is also unique in the sense that if we let the clocks run for an hour, they will visit all the possible initial solution states: the necessary condition (which is also sufficient) is that
the minute hands of any two neighbouring clocks must be offset by exactly 30 minutes,
and there is only one way of achieving those offsets, because we are ignoring the hour hands.
RE-EDIT after noticing that OP also posted a 3x3 version:
The method also generalises to any number of clocks placed on a rectangular grid in any pattern:
1: Calculate the sum of x and y coordinates of the clock's position in the grid
2: Point all minute hands of clocks with an even coordinate sum to the same direction
3: Point the minute hands of clocks with an odd coordinate sum to the opposite direction
This creates a kind of a checkerboard pattern: wherever there's a pair of clocks in two orthogonally adjacent squares in the grid, those two clocks will have different coordinate sum parity, meaning they will be offset by exactly the right amount, and they will touch minute hands every hour.
$\begingroup$ Nice. Is this the best we can do? Can we get more frequent touches? $\endgroup$ Mar 17, 2021 at 1:43
2$\begingroup$ @DmitryKamenetsky to me it looks like you just spent two entire sentences to ask if there's something more frequent than "every possible occasion". That seems a bit unlikely, so there must be a miscommunication here somewhere. $\endgroup$– BassMar 17, 2021 at 1:55
1$\begingroup$ @DmitryKamenetsky For every pair of touching clocks, the "minute hand touching" event may happen at most once per hour. It happens exactly once if the two clocks have 30-minutes difference, zero times otherwise. Bass's answer is optimal in that all four pairs satisfy the condition. $\endgroup$– BubblerMar 17, 2021 at 2:26
3$\begingroup$ Thanks everyone, I see now that there was indeed a miscommunication, and that it was caused by my being a bit too terse with the proof. I hope the edit makes it easier to follow. $\endgroup$– BassMar 17, 2021 at 8:44
2$\begingroup$ @DmitryKamenetsky You could get more frequent alignments if you put the clocks in a hexagonal grid pattern instead. Place 7 clocks in a honeycomb arrangement, arranged so that the center clock aligns with an outer clock every 20 minutes. Unfortunately, in this arrangement none of the outer clocks will ever align with each other. $\endgroup$ Mar 17, 2021 at 14:32