May 2018
 

[URL="https://researchweb.watson.ibm.com/haifa/ponderthis/challenges/May2018.html"]May 2018[/URL]

Implicit assumption that all counterfeit coins have the same weight. Otherwise the shown solution can not solve the case N=7.

Solved for 11. Lot of waste, as \(C_{11}^2=55\) and you have \(3^4=81\) possibilities to arrange those "<", "=", and ">" cases.

(if you want to solve this, then a great help would be if you write down the solution for N=7 presented there, and see for all 21 cases if the solution indeed solves the puzzle, then you will get the idea how to solve the case for 11 without any calculus)

(edit: for N=7 there is also waste, as \(C_7^2=21\) and the space for the 3 weightings is \(3^3=27\), so if in their solution you replace "=" with "0", "<" with "1", and ">" with "2" and see each case as a number in base 3, between 0 and 26, then the cases 9, 14, 16, 18, 22, 26 are not used. However, N=7 is maximal for 3 weights, because \(C_8^2=28>27\) - end of edit)

In theory, one has to be able to solve this problem with 13 coins, still using 4 weightings. That is because \(C_{13}^2=78<81\).

"In theory, there is no difference between theory and practice. But, in practice, there is..." (which famous guy said this? I have seen it recently somewhere...)

(edit 2: If one of you finds a solution with 12 or 13, say here, so I will look further, maybe the guys saved some space for a "star" or even "two stars"? - as they usually do!)

I try to find a solution for 12, but I am not able to exceed 59 - will say, only 59 of the 66 combinations have different outcomes of the weighing process.

Update:
In my best approximation of a solution for 12 coins only 62 of the 66 combinations have different outcomes of the weighing process.
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Ok, so, according with you, there may be no solution for 12... this is a pity. No stars to hunt... (we believe what you say because we have no time to test that, and we can't think to an intelligent solution, beside brute force, where the search space for 12 and 13 would be 1.8e18 and respective 128e18, grrrr).

In the past we solved a different problem: we had one fake coin only and the rules allowed to take in consideration the former results of the weighting. We were asked to determine the maximum numbers of coins that could be tested with N weightings. We are curious how this changes in the case when only 2 coins are fake. Maybe we will tickle the problem if the time allows...

Probably I did not express myself well. What I wanted to say: The best solution I[COLOR=#0c0c0c]'v[/COLOR][COLOR=#181818]e fo[/COLOR][COLOR=#242424]und [/COLOR][COLOR=#303030]so[/COLOR] far has 62 different outcomes. I cannot exclude that there is a 12 solution. I continue to seek.

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