2012
DOI: 10.1109/tcbb.2011.137
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Algorithms for Reticulate Networks of Multiple Phylogenetic Trees

Abstract: Abstract-A reticulate network N of multiple phylogenetic trees may have nodes with two or more parents (called reticulation nodes). There are two ways to define the reticulation number of N. One way is to define it as the number of reticulation nodes in N [13]; in this case, a reticulate network with the smallest reticulation number is called an optimal type-I reticulate network of the trees. The better way is to define it as the total number of parents of reticulation nodes in N minus the number of reticulati… Show more

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Cited by 33 publications
(28 citation statements)
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“…[19] Suppose that C is a cycle of G F and r 1 , …, r ℓ are the nodes of C . Then, each r j ∈{ r 1 ,…, r ℓ } has two children u j and u’ j in F .…”
Section: Methodsmentioning
confidence: 99%
See 1 more Smart Citation
“…[19] Suppose that C is a cycle of G F and r 1 , …, r ℓ are the nodes of C . Then, each r j ∈{ r 1 ,…, r ℓ } has two children u j and u’ j in F .…”
Section: Methodsmentioning
confidence: 99%
“…In the supplementary material of [19], an O ( 3 k n )-time algorithm for solving GAF is detailed. The algorithm differs from Whidden et al ’s algorithm for enumerating all MAFs of T 1 and T 2 only in that we start with F 1 (as it is given) and F 2 = T 2 (instead of starting with F 1 = T 1 and F 2 = T 2 ) and then repeatedly modify F 1 and F 2 until either | F 1 |> k + k 0 or F 1 becomes a forest without edges, where k 0 is the original size of F 1 .…”
Section: Methodsmentioning
confidence: 99%
“…4 Mark each side s of G with |X N p (s)| ≤ 2 as finished and each other side as unfinished. 5 while there exists an unfinished side of G do 6 Let s be an unfinished side of G such that there is no unfinished side of G that is below s. 7 Let x + s and x − s be, respectively, the top and bottom leaf on side s in N p . 8 Let X s be the set of all x ∈ X that are not in N p and such that x + s → x, x − s .…”
Section: An Exponential-time Algorithmmentioning
confidence: 99%
“…For the case of multiple binary trees, there exists a kernel with at most 20k 2 leaves [15], various heuristics [7,8,29] and an exact approach without running-time bound [30].…”
Section: Introductionmentioning
confidence: 99%
“…. ‚ T K when K ‡ 3 (Wu, 2010;Park and Nakhleh, 2012;Chen and Wang, 2012). There are no existing methods for the exact computation of the hybridization number or reconstructing parsimonious networks with three or more trees.…”
Section: Wumentioning
confidence: 99%