And I will be thrilled if someone comes with less weighings! (With coins, please. No mice!)

]]>Again, I didn’t see a straight answer to the problem, although I found colorblind’s solutions wuite exotic and entertaining!

Well, to avoid more mices eating from a same cake and that sort of variations, plus I don’t like mices at all, I take the coin version as set by Tanya, since I also love coin problems …

I found that the lighter coin can be found with a minimun of 3 weighings, which involves 6 pans:

Put 4 coins in one pan and 4 coins in another pan.

1) The pans balance: The coin is in the 4 coins remaining outside.

Put 2 of them in one pan and the other 2 ones in another pan.

1a) Take the 2 coins that weigh less and weigh them against each other ot find which is the lighter one.

And, if we want to generalize, based on this kind of thinking and using min 2 pans, we find that for:

n=2-3, 1 weighing is needed

n=4-7 -> 2 weighings are needed

n=8-15 -> 3 weighings are needed

…

That is, with k min weighings, we can find a lighter coin among n = 2^k to 2^(k+1)-1 coins.

Response 1 and 2 seem to work in the mice/cupcake universe but not the coin universe of course.

Notation: m = (x, y) where m is the most required weighings to solve the problem of a group x coins containing y heavier coins.

(12, 2) does have a solution in 5. Split the 12 into 4 groups of 3. Weigh group 1 vs. 2, 2 vs 3, and 3 vs 4. Either 1 or 2 of the 4 groups can be isolated as having heavier coins. Now there are needed at most (3,1) + (3, 1) = 2 additional weighings for the identified heavier groups.

I’m going to assert my (12, 2) solution is optimal and ask if anyone has the solution for the next step, (12,3).

If that’s too easy, here’s a variation to chew on: Let’s say your job is to take groups of 12 coins, all of which have exactly y heavier coins, and pick out those n heavier coins. What method will allow you to do your job the fastest? For example, in the above solution, there are instances where only m-1 weighings would be needed. Is it possible that a algorithm for which at most m+1 weighings might be needed would on average be faster than a solution for which only at most m weighings might be needed for some y?

]]>Alexander: yes, one can generalize this question to finding m faster mice among n; in fact, any “traditional” coin puzzle (n coins, of which some number m are known to be heavier, or lighter, or just not of the same weight (but not necessarily known to be heavier or lighter in advance); or questions like https://blog.tanyakhovanova.com/?p=148 , https://blog.tanyakhovanova.com/?p=179 or https://blog.tanyakhovanova.com/?p=217 (in alternate form, “what is the minimal number of weighings necessary?”)) can be rephrased in this context (asking for us to minimize the number of pans used rather than the number of weighings) — thus, every two-pan balance question has a “twin” question with a multi-pan balance (or, equivalently, with mice and cupcakes). It is this situation that made me attracted to the question in the first place.

By the way, if you happen to rediscover the solution for 12 and 2, it seems to me that it would be worth posting here.

]]>This week I came across your blog and this problem. I am presently with my relatives, the Zbarskys of Rockville, MD. The fellows came up with several ideas. First, there is an implicit assumption that the cupcakes consumption is uniform for all mice but one. Focusing on cupcakes, this can be interpreted as the uniform progress of cupcake disappearance once the mice have applied themselves to the food. The progress can be observed and a judgement passed whether two groups of mice progress with the same speed or not. If that is acceptable, then it may be not necessary to wait until a cupcake has disappeared in its entirety. The implication of this observation is that the cupcakes can be reused. For example, starting with three groups of 3 mice and 3 cupcakes, if the progress on all three cupcakes is the same, you can stop the competition midway ad then give any two cupcakes to the a pair of mice out of the remaining three. If the progress is not uniform, you still have time to stop the competition with two unfinished cupcakes of equal size. So it appears that 3 cupcakes will suffice to determine the fast eater.

They went further and posed a problem with 12 mice of which 2 eat faster than the rest. I missed the proposed solution but the answer they came up with was 6 cupcakes.

]]>Alternate solution 2: Pick the fat one. It’s doubtful the fast one stops eating once he’s had his share.

Alternate solution 3: Ask. A fantasy world with standard cupcakes may also have talking mice.

Alternate solution 4: Find a friend. Get them to give you their cupcakes …

]]>With a stopwatch it is a completely different problem.

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