Le Monde puzzle [#879]

Here is the last week puzzle posted in Le Monde:

Given an alphabet with 26 symbols, is it possible to create 27 different three-symbol words such that

  1. all symbols within a word are different
  2. all triplets of symbols are different
  3. there is no pair of words with a single common symbol

Since there are

28x27x26/3×2=2925

such three-symbol words, it could be feasible to write an R code that builds the 27-uplet, assuming it exists. However, by breaking those words into primary words [that share no common symbols] and secondary words [that share two symbols with one primary word], it seems to me that there can be a maximum of 26 words under those three rules…

2 Responses to “Le Monde puzzle [#879]”

  1. Jean-Louis Foulley Says:

    Thanks Christian for your solution and discussion to this Le Monde puzzle.

    I should have missed something in my reasoning as I only found 24 triplets of such words with any pair of them being made of different triplets or triplets sharing two symbols in common.
    My reasoning is as follows: we need 4 letters to build 4 words with words in any pair sharing 2 letters in common: bcd, acd, abd, abc. Now we can repeat this procedure on disjoint subsets of letters ie efgh, ijkl,mnop, qrts. For the last one uvwx, we have to add, x and y so as to use all the 26 letters of the alphabet. But we cannot build more than 4 words out of these 6 letters eg uvw,vwx,vwy and vwz. In total that make 4 times 6=24 words. What are the two words I missed? Or where did I get wrong in this way of thinking?
    Sorry to bother you with that but I would have been curious to see the detailed list of your 26 words.
    Anyway I agree with you: we cannot build 27 words with the 3 conditions set up initially.

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