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Figure 2 | Algorithms for Molecular Biology

Figure 2

From: Maximum Parsimony on Phylogenetic networks

Figure 2

Dynamic programming solution. The dynamic programming solution applied to a phylogenetic network. The states are 0, 1 and 2. The cost matrix used has all 1s except the diagonal elements which are all 0s. The tables shown on each vertex v are the costs, S v (i) (second row) of each state, i (first row) that are computed during the post-order traversal. Also shown at the vertices are the states of the child w, namely the t(v, w)(i) (third row) that correspond to the costs in the second row; when there are two children for a vertex, the entries in the third row are represented as a pair of states of the left child and the right child respectively. Each edge (v, w) is labelled with s(v, w)(i) for each state i. During the pre-order traversal, the states for each vertex are selected (shown in bold). The cost of 2 highlighted in bold at the root vertex gives a lower bound, S. The state assigned at each vertex is highlighted in bold. The algorithm finds a total of three substitutions (highlighted by bold edges). This is because the states assigned at the parent vertices of the reticulate vertex give conflicting assignments of 1 and 2 respectively, of which state 1 is assigned at the reticulate vertex. Thus with an extra cost of 1, we get the score of 3 (an upper bound of the optimal score) as the parsimony score corresponding to the assignment shown. Note that the optimum parsimony score on the network is 2 (equal to the lower bound), which can be found by exhaustive search and can be realized by changing the assignment from 1 to 2 for the left parent of the reticulate vertex and from 1 to 2 for the reticulate vertex. Thus the lower bound matches with the optimal score, although the assignment corresponding to the lower bound is not conflict-free and not the same as the assignment corresponding to the optimum.

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