# Puzzle from today's Futility Closet

An interesting puzzle at the Futility Closet page today. It first appeared in 1967 in Eureka the Cambridge math journal. No answer was ever published. I am guessing the answer is none (if they don’t have to all arrive together), but I am having trouble proving it. So am I right?

Charles, Donald, and Edward are triplets.

The triplets (whose abilities at walking, cycling, and donkey riding are identical) always leave home together at the last possible minute and arrive at school together on the last stroke of the bell.

They used to walk the 4 1/2 miles, and so had to set out at 8.00; then they acquired a bicycle and found that they did not have to leave home until 8.15 (Charles rode it for the first 1 1/2 miles, left it, and walked on; Donald walked 1 1/2 miles, cycled 1 1/2 miles, and walked again; Edward walked 3 miles and cycled the rest). More recently they have been given a donkey. After experiments to determine the donkey’s speed and to verify that it stood stock still when left, they found that — using the bicycle and the donkey — they did not need to leave home until 8.25. There were several schemes of changing over which they could use to do this, of course; but naturally they chose a scheme which involved the minimum number of changes. Going to school tomorrow Charles will start on foot and Edward will arrive on foot. How far will Donald walk?

The puzzle is clear about that: They always arrive at school together on the last stroke of the bell. And I think it’s implicit, also: If any triplet arrived earlier than necessary, then they could change the rota such that that triplet got less use of the vehicles, leaving more usage for the others.

And I think that the answer is simple:

A total of 13.5 person-miles must be traversed. The bicycle traverses a third of those, and the donkey traverses another third. Therefore a total of 4.5 person-miles must be traversed on foot. If the rota splits evenly, then each must, somehow or other, walk 1.5 miles.

However, I’m not completely certain of that, because

While all three must gain equal total benefit from the vehicles, it’s possible that one uses disproportionately more of both the fastest mode and the slowest mode, and less of the median mode.

I don’t know. It seems on the surface that everyone still has to walk a third of the way, but somehow I don’ think it is that simple.

I think the “minimum number of changes” constraint ensures a solution here.

Just playing around with numbers, I get walking=3 mph, donkey=4.5 mph, and bike=6 mph. Arrival time is 9:30.

Charles can walk for 2 miles and then bike for 2.5 miles and arrive at the right time, assuming the bike is there when he arrives.

This was a fun one.

Let’s label the boys C, D, and E and the transports w(alk), b(ike), d(onkey).

The most minimal changeover pattern possible in principle would be a sequence like this:

C:wwwwdddd
D:bbbbbbbb
E:ddddwwww

Three of our unknowns are (say) walking-only duration T, walking speed vw, and biking speed vb. The first scenario given in the problem (walking only) relates T and vw. The second scenario (walk+bike) then gives a relation between vw and vb. More to the point, it leads to a value for (1/vw + 1/vb).

Then, our hypothesized third scenario above also gives for boy D a relation between T (and thus vw) and vb, which can be rearranged trivially to be an expression again for (1/vw+1/vb).

These two expressions for the same quantity are numerically different, which is to say that we’ve proven that this is not a valid solution.

The next obvious thing to try might be a symmetric case, at least to rule it out. That is, if all three boys each take w, b, and d transports for some leg, then those durations must be equal in order to facilitate changeovers. However, this is impossible since someone will always “beat” their intended changeover target. For instance:

C:wwbbdd
D:ddwwbb
E:bbddww

fails because D would get to the second changeover point before the bike was ready for him (since his ddww legs are necessarily net faster than C’s wwbb legs.)

This leaves us trying the following arrangement:

C:wwddddddd
D:ddwwwwbbb
E:bbbbbbwww

where it’s clear that the legs are not all the same length. Here, D walks in the middle, carrying the donkey out first and bringing the bike in on the home stretch.

In principle the unknowns are T, vw, vb, vd, and the three leg lengths L1, L2, and L3.

Immediate constraints are L1 + L2 + L3 = 4.5 mi, and T is related to vw as before. So we go from seven to five unknowns. The walk+bike scenario and the three boys’ apparently independent time sums in the sequence above provide four equations for these five unknowns. Seems underconstrained! But…

As earlier, with a page or so of algebra, the numbers in the problem work out such that you can leave yourself with two equations relating the distance desired (L2) to identical combinations of vw and vd, allowing one to eliminate both of the latter in one fell swoop.

Doing so gives a walking distance for D of 1.75 mi (assuming I didn’t mess up the algebra anywhere. I did it all by hand, but it felt clean enough.)

Not yet looking at Pasta’s answer:

I get:
Don bikes 2 miles, walks 1.75, then rides the donkey 0.75 miles
Charles walks 2 miles, picks up the bike that Don left, then rides the remaining 2.5 miles
Ed rides the donkey 3.75 miles then walks the rest of the way.

Charles can obviously get the bike since he was walking. And Don can get the donkey from Ed, since Ed is walking the last segment and so must have gotten there first for the times to match up.

All of this assumes walking 3 mph, donkey 4.5, and bike 6 mph. Only 4 changeovers total.

…or if the donkey is slower than the bike as noted by Dr.Strangelove, a different changeover still has to fail given the required permutings.

I don’t think the speeds I came up with are uniquely defined, since we really only have two given constraints (the 15 and 25 minute savings) and there are three unknowns. But assuming that the problem actually has a solution, then any solution is good enough. I work better when I have some concrete solution to work with.

That the three speeds are at least different is constrained by the setup. As Pasta notes, the donkey could be faster than the bike, but if that’s the case then you can just swap them in my solution to ensure no timing conflicts.

I didn’t know there were any other Futility Closet fans here. I haven’t visited the site much since they stopped the podcast.

Yeah, that’s right. However, it can be shown (although I didn’t require it) that the bike is faster than the donkey. More generally, the speeds involved have to satisfy:
1/vd - 1/vb = 1/18 hr/mi
1/vw - 1/vb = 1/6 hr/mi

Thanks–I wasn’t sure about that. It seemed unlikely based on the savings, but I wasn’t sure if there was some non-linear effect going on. I suppose that might be possible if the donkey took away from time spent on the bike, but that’s not actually the case–it only takes away from walking time. So the fact that it saves only an additional 10 minutes vs. 15 for the bike guarantees that it’s slower.

Thank you to all you math whiz folks for helping to illuminate this.

The post on Futility Closet was updated with a reader-submitted solution which is similar to Dr. Strangelove’s as far as distance by each mode of transportation, but the order within an individual rider and who did each stage is somewhat different:

The author does mention my exact solution in the third-to-last paragraph. It’s just a mirrored version though, and Donald walks the same distance for each one. Edward and Charles have to swap roles, though.