For example, let sB = s1s2s3 and π(sB) = {s1, s2, s3}, where s1 = aaaa, s2 = aaat and s3 = aaa. We use score scheme I and c = -1. The valid value of i is 1, 2, ..., 7 since |s2s3| = 7. For i = 5 ≥ |s3|, pre(s2, 2) is optimally aligned with s3, s1 and suf(s2s3, 5) is scored as  (Figure 2(a)). So V(π(sB), c, 5) = μ(s1, s2) + μ(pre(s2, 2), s3) +  × (|s1| + 5) = . For i = 2 < |s3|, s2 is optimally aligned with s3, pre(s3, 1) is scored as μ(x, Δ) and suf(s3, 2) is scored as  (Figure 2(b)). In this case, V(π(sB), c, 2) = μ(s1, s2) + μ(s2, s3) + μ(x, Δ) × 1 +  × (|s1| + 2) = 0. For i = 4, at the right ends of the optimal self-alignment of π(sB) (Figure 2(c)), there are 4 letters that match spaces. The last letter t in s2 matches a space at the right end of the alignment. The assumption that  >μ(x, Δ) forces this column to have score  instead of μ(t, Δ) to maximize V(π, c). We have V(π(sB), c) = V(π(sB), c, 4) = 1. For i = 1, 3, 6, 7, the values are lower than V(π(sB), c) = V(π(sB), c, 4) = 1.