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External Symmetry Numbers and Graph Automorphism Groups

So, there was a question on BioStar about calculating the 'external symmetry number' of a molecule - something I hadn't heard of, but turns out to be something like the subgroup of rotations and reflections of the automorphism group of a graph. Since I have some code to calculate the automorphism group, I naïvely thought it would be simple...

The questioner - Nick Vandewiele - kindly provided some test cases, which ended up as this code. Although many of these tests now pass, they only do so because I commented out the hydrogen adding! :)

On the one hand, there are some recent improvements that try to handle vertex and edge 'colors' - in other words, element symbols and bond orders. For example, consider the improbable molecule C1OCO1 :
These are the three permutations that leave the carbons and the oxygens in the same positions; when you include the identity, that makes 4. Cyclobutane (without hydrogens!) has a symmetry group of order 8. Similarly, cyclobutadiene now gives 4 instead of 8.

So what goes wrong when there are hydrogens? Well, it's a deeper problem than just hydrogens, but it starts there. Consider methane : it has an external symmetry number of 12, but my code gives 24 - why? Well the main answer is 'inversion', look:

The permutation (0)(1)(2, 3)(4) just swaps hydrogens 2 and 3. This effectively changes the chirality of the molecule ... sortof. It's not actually chiral, but its a reasonable description of the transformation. Apparently, this does happen (another thing I didn't know; there are lots more :) according to this document, but quite slowly compared to rotations - "slower than 1 cycle s-1".

This kind of pseudo-chirality will happen at any tetrahedral center. Or at any atom with 4 neighbours, I think - like XeF4, which is square planar. As an example, take this spira-fused ring system:

with a transform that swaps 7 and 9 but not the pairs (0, 5)(1, 4)(2, 3). Effectively this changes the parity at carbon 6. Somehow I doubt that this kind of 'movement' actually occurs in solution, but I could well be wrong. In any case, it seems likely that the external symmetry number is 2, and not 4.

In summary, it is probably not possible to calculate the external symmetry number correctly without 3D coordinates, or symmetry axes, or point groups. I have a feeling that the positional info could be recorded as a 3D combinatorial map which would give explicit orientations for atoms with four neighbours.

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