For each node set N. At the end of the procedure

For each node set N. At the end of the procedure the set S contains every intersection sequence Step 0 (Initialisation): Let S = and let N = (-) Step 1: Move onto the next node PubMed ID:https://www.ncbi.nlm.nih.gov/pubmed/28497277 set N in the tree (as in Fig. 8). Step 2: For the node set N, apply Procedure 2 to identify partial state sequences P1, ..., Pk and sets of attractorsC1 ,..., Ck satisfyingA: For i = 1, ..., k, Pi involves the node set N (i.e.Pi = x iN ,..., x iN )0 q -B: For i = 1, ..., k, Pi occurs in every attractor A C i C: For i = 1, ..., PubMed ID:https://www.ncbi.nlm.nih.gov/pubmed/28551822 k, Pi does not occur in any attractor A C i D: For any i, j (1 KW-2478 i <j k), Ci C j = E: C1 ... Ck = A (the set of all attractors) F: Given the node set N, there are no other partial state sequences P' P1, ..., Pk that occur in any attractor AAStep 3: For i = 1, ..., k, add the pair Pi, C i to the set S Step 4: For i = 1, ..., k, check S to see if there is any pairM M Q = y 0 ,..., y r -1,D for which either of the followingare true (a) M N and D = Ci (b) M N and D = CiTherefore, when P3 and C 3 = A1, A2, A3 are considered in Step 4, it can never be removed from the set S. This is because when a larger node set M N is considered, we will never have a set of attractors C = C3 (or D C3 ).Page 19 of(page number not for citation purposes)BMC Bioinformatics 2007, 8:http://www.biomedcentral.com/1471-2105/8/Therefore, P3 satisfies the property 3 and is an intersection sequence. However, Px is not an intersection sequence since the corresponding set of attractors (A4) is still associated with a larger node set M N. Therefore it will be removed from the set S in Step 4 (either when N or M is analysed in the procedure). We now explain how the procedure can be made more efficientImproving efficiency(1)However, because all of the attractors are intersection sequence, the full node set V = n1, ..., nv should still be fully analysed in Steps 2 ?4 (possibly at the very end of the procedure).Improving efficiency(2) As can be seen in Fig. 8, some nodes appear less than others, with the least frequent nodes visited earlier in the tree. Therefore, it is likely to be advantageous to re-index nodes in the tree during the search. At any stage during the search, nodes along paths to the right (from a node set N) can be re-indexed without impairing our ability to search the tree. For example, once N = n1, n3 has been reached, re-indexing nodes n4, n5 to n5, n4 still allows us to reach the same node sets M = n1, n3, n4, M = n1, n3, n5 and M = n1, n3, n4, n5, as before. However, they would be visited in a different order (M = n1, n3, n5 then M = n1, n3, n4 then M = n1, n3, n4, n5).Consider node sets N M and two partial state sequencesM M N N P = x 0 ,..., x q -1 and P = y 0 ,..., y r -1 that both occurin some attractor A. Then, P' must contain P and hence only occur in an attractor whenever P does (proved in Lemma S1.12 of Additional file 1). Therefore, if N M V, and P only occurs in a single attractor A*, neither P nor P' can be intersection sequences. This is since the attractor A* is itself a partial state sequence (for a larger node set V) that occurs in exactly the same set of attractors (i.e. just A*), and so property 3 of Definition 4 fails. Therefore, if Step 2 of Procedure 3 identifies a partial stateN N sequence Pi = x 0 ,..., x q -1 that only occurs in a singleOnce a node set N has been analysed, re-indexing so that the next node nj to be visited maximises c (below) will speed up the search - For th.

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