# Pancake race

This week’s Fiddler is a logic puzzle about getting home as fast as possible.

Alice, Bob, and Carey start together and each walk home at a different constant speed. Once all three get home, they can have pancakes! Alice can walk home in 10 minutes, Bob can do it in 20, and Carey in 30. Fortunately, any of them can carry any of the others on their back without reducing their own walking speed. Assume that they can pick someone up, set someone down, and change direction instantaneously. What is the fastest they can get to eat pancakes?

Extra Credit
There is now a fourth: Dee. Dee is the slowest, needing 60 minutes to walk home. As before, anyone can carry anyone else, and they won’t get pancakes until everyone gets home. What is the fastest this can happen?

My solution:
[Show Solution]

## 7 thoughts on “Pancake race”

1. tlk says:

There’s a better strategy for the first part, and by analogy probably for the extra credit too.
Instead of dropping C at home, A drops him earlier at point X. Then returns to fetch B, while C continues independently.
X is selected in such a way that all three arrive together.
I got 15 minutes with this approach.

The intuition here is that you don’t want any of them waiting at the destination when they could be walking.

1. You’re absolutely right! I had a feeling I was missing something… I’ll update my solution once I figure it out!

1. Done! I updated my solution. Thanks again for pointing out the error in my previous version!

2. Nice visualization. The 4-person problem can be optimized further, down to 17 minutes 5 seconds. Let the start point be at 0 and destination at 1. Let A carry C, and B carry D. Let A drop C off at some point x, and then A walks back. Since A moves twice as fast as B, A will walk back a length x/3 to meet B/D. Then A carries B, and walks forward (towards destination) a length of x/3 + y, where y is the length moved by C after being dropped off earlier by A. Since A moves thrice as fast as C, we have x/3 + x/3 + y = 3y, so x = 3y. When A/B meet C, B carries C to destination (a length of 1-x-y). A walks back to meet D. If D moved a length z between hopping off B earlier and meeting A again, then A walked 2(x/3 + y) -z = 4y -z during that time. Since A is 6 times as fast as D, we get 6z = 4y -z, so z = 4y/7. Then, A carries D, and they reach the destination at the same time as B/C (no point in anyone reaching early). During the time that B/C walked a length of 1-x-y to reach the destination, A walked 2(x/3 + y – z) + (1-x-y). Since A walks twice as fast as B, we get 2(x/3 + y – z) + (1-x-y) = 2(1-x-y), so 2(2y – 4y/7) = 1-4y, so y = 7/48. Hence x = 3y = 21/48. Since all 4 reach the destination simultaneously, time taken by them = time taken by C = x*10 + y*30 + (1-x-y)*20 = 30y + 30y + 20 – 80y = 20 -20y = 20 – 20*(7/48) = 205/12 = 17.0833… minutes, so 17 minutes 5 seconds. The same can be achieved with: A first carries and drops D at some point x. A walks back and picks up B, A carries B to D, and B carries D to the destination. A walks back to C, and carries C to the destination. All 4 reach the destination in the same 205/12 minutes.

1. Oh wow… Solving this puzzle has been humbling to say the least. Looks like I need to update my solution once again! Thanks!

I’ve been thinking about how I could model this as an optimization problem but I haven’t been able to come up with a good way to do it. It would be nice to have some assurance that this is indeed the optimal solution and there isn’t yet another even more clever way to solve the problem…