, If we assume an augmentation of type T , it is not difficult to see that an optimal choice for ? 1 and its corresponding sequence ? T are respectively ? and AAAC
, If we explore the different possibilities for an augmentation of type H, we find that the best choice is {? 1 , ? 2 } = {?, ?} with sequences AAAB and BBCC and with value ?
, Since µ < 1/3, we get that ?(?) > ?({?, ?}) and we opt for the augmentation N 0 = T ?1 (N, ?; {?}, ?) shown in Fig, p.17
, Taking any node y ? C different from ? we get that ?(y) < 0 and hence they are not good candidates. If we take y = ?, taking into account that µ < 1/3, we find an optimal sequence ?(?) = AACC. The value of ?(?) corresponds to three different evolution processes: two mutations, from AAAC to AACC and from AACC to AACC, and a hybridization of the sequences AACC and BBBB to BBCC, and hence we get ?(?) = (1 ? 2µ) 7 µ/2. Now, this value must be compared with the probability of evolution without this hybridization, i.e. the probability of speciations from BBBB to BBCC and from AAAC to AACC which is µ 3 (1 ? 2µ) 5 . Since (1 ? 2µ) 7 µ/2 > µ 3 (1 ? 2µ) 5 (assuming that µ < 0.2928, N 0 ) of ? and ? are in C
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