there is one bit representing a single bond in Q 0 that is not in S 0 . Thus, the A 1 → A 2 mapping used to compute Q 0 must ensure that A 1 → S 0 and A 2 → S 0 are respected. This can be done by composing the maps as A 1 → A 2 = A 1 → S 0 → A 2 noting that the inverse map S 0 → A 2 will always exist since T is always a bijection. Using this composition would give the correct pattern that matches S 0 as Q 0 =[#6;X3,X4;H1,H3]-[#6X3H1]=[#6;X2,X3;H0,H1].