LNCS 2950 - Aspects of Molecular Computing (Frontmatter Pages)

208 Tero Harju, Ion Petre, and Grzegorz Rozenberg
5 The Overlap Equivalence Problem
The mapping w ↦→ γw of legal strings to overlap graphs is not injective. Indeed,
for each legal string w = w1w2, wehave
γw1w2 = γw2w1 and γw = γ c(w) ,
where c is any morphism that selects one element c(p) fromp = {p, p} for each
p (and then obviously c(p) =c(p)). In particular, all conjugates of a legal string
w have the same overlap graph. Also, the reversal w R and the complementation
w C of a legal string w define the same overlap graph as w does.
Example 10. The following eight legal strings of pointers (in ∆ � )havethesame
overlap graph: 2323, 3232, 2323, 3232, 2323, 3232, 2323, 3232. ✷
Example 11. All string representations of an overlap graph γ need not be obtained
from a fixed representation string w by using the operations of conjugation,
inversion and reversing (and by considering isomorphic strings equal). For
instance, the strings v1 = 23342554 and v2 = 35242453 define the same overlap
graph (see Fig. 3), while v1 is not isomorphic to any signed string obtained from
v2 by the above operations. However, we will demonstrate that if strings u and
v are realistic, then γu = γv implies that u and v can be obtained from each
other by using the operations of conjugation, inversion and reversing. ✷
✎☞
✍✌
4 −
✎☞
✍✌
2 −
✎☞
✍✌
5 −
✎☞
✍✌
3 −
Fig. 3. The common overlap graph of legal strings v1 = 23342554 and v2 =
35242453. The signed string v1 and v2 are not obtainable from each other by the
operations of conjugation, reversal and complementation
Examples 10 and 11 lead naturally to the overlap equivalence problem for
realistic legal strings: whendotworealistic legal strings u and v have the same
overlap graph, i.e., when γu = γv? To solve this problem, we begin by characterizing
legal substrings of realistic legal strings.
For a signed string v ∈ ∆ � ,let
vmin =min(dom(v)) and vmax = max(dom(v)) .
Thus, dom(v) ⊆ [vmin,vmax].