*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.

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 *c *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 *d *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.