Enigma 1780

Spoiler for New Scientist Enigma 1780 “Pure hedronism”  (Follow the link to see the puzzle.)

Not even with a bang…

Here’s the diagram with blanks labelled a through n. I also copied the 3 and 2 from the right to the left, and the 2 and 5 from the left to the right, to make the connections clearer.path3013 copy

Now it’s pretty much just reading it off. a must be 4 because we already have 1, 2, 3, and 5 along with a around a vertex. Then b and c must be 1 and 3, and b can’t be 1, so it’s 3 and is 1.

Likewise n is 1, and l and m must be 4 and 5, but m is not 4 so is 5 and l is 4.

d and e must be 2 and 5, but d is not 2, so is 5 and e is 2. That means h is 3.

Likewise j is 2 and k is 3. Then g is 5.

Finally f has to be 4 and i must be 1. ENIGMA is 324125.

Each of the four rows of triangles pointing in the same direction (there are two such rows interleaved in the middle) contains a cyclic permutation of 1-5-3-4-2.

And that’s it for Enigma.




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