Lam, Gusfield, and Sridhar (2009) showed that a set of three-state characters
has a perfect phylogeny if and only if every subset of three characters has a
perfect phylogeny. They also gave a complete characterization of the sets of
three three-state characters that do not have a perfect phylogeny. However, it
is not clear from their characterization how to find a subset of three
characters that does not have a perfect phylogeny without testing all triples
of characters. In this note, we build upon their result by giving a simple
characterization of when a set of three-state characters does not have a
perfect phylogeny that can be inferred from testing all pairs of characters

Fernández-Baca, David

Shutters, Brad

English

arXiv

AbstractWe give a characterization of the minimal obstruction sets for the existence of a perfect phylogeny for a set of three-state characters that can be inferred by testing each pair of characters. This leads to a O(m2n+p) time algorithm for outputting all p minimal obstruction sets for a set of m three-state characters over a set of n taxa

Elsevier - Publisher Connector 

A simple characterization of the minimal obstruction sets for three-state perfect phylogenies 

Brad Shutters

David Fernandez-Baca

CiteSeerX

A Simple Characterization of the Minimal Obstruction Sets For Three-state Perfect Phylogenies

Applied Mathematics Letters 25 (2012) 1226–1229
Contents lists available at SciVerse ScienceDirect
Applied Mathematics Letters
journal homepage: www.elsevier.com/locate/aml
A simple characterization of the minimal obstruction sets for three-state
perfect phylogenies
Brad Shutters ∗, David Fernández-Baca
Department of Computer Science, Iowa State University, Ames, IA 50011, USA
a r t i c l e i n f o
Article history:
Received 18 July 2011
Received in revised form 23 February 2012
Accepted 24 February 2012
Keywords:
Computational biology
Phylogenetics
Perfect phylogeny
a b s t r a c t
We give a characterization of the minimal obstruction sets for the existence of a perfect
phylogeny for a set of three-state characters that can be inferred by testing each pair
of characters. This leads to a O(m2n + p) time algorithm for outputting all p minimal
obstruction sets for a set ofm three-state characters over a set of n taxa.
© 2012 Elsevier Ltd. All rights reserved.
1. Introduction
The k-state perfect phylogeny problem is a classic problem in computational biology [1]. The input is an n by m matrix
M of integers from the set K = {1, . . . , k}. We refer to each row of M as a taxon (plural taxa), each column of M as a
character, and each value that occurs in a column c ofM as a state of character c. A perfect phylogeny forM is a tree T with n
leaves such that each leaf is labeled by a distinct taxon ofM, each internal node is labeled by a vector in Km, and, for every
character c and every state i of c , the nodes labeled with state i for character c form a connected subtree of T . The problem
is to decide whether there exists a perfect phylogeny forM. If so, then the characters ofM are compatible. Otherwise, they
are incompatible. The general k-state perfect phylogeny problem is NP-complete [2,3]. However, for fixed k, the problem is
solvable in O(m2n) time [4,5], and in O(mn) time when k = 2 [6].
In this note, we focus on the three-state perfect phylogeny problem, and thus, fixM to be an n by m matrix of integers
from the set {1, 2, 3}. We remark that several specialized algorithms have been developed for the specific case where k = 3
that can construct a perfect phylogeny forM (when one exists) in O(m2n) time [7–11]. However, our main concern here
is the case where the characters ofM are incompatible. Since every subset of a compatible set of characters is compatible,
it follows that if the characters ofM are incompatible, there must be some minimal subset of the characters ofM that are
incompatible. A minimal obstruction set forM is a minimal subset of the characters ofM that are incompatible. A recent
breakthrough [12] showed that every minimal obstruction set forM has cardinality at most 3; implying an O(m3n) time
algorithm for outputting all minimal obstruction sets forM. Themain result of this work is a characterization of theminimal
obstruction sets forM that can be inferred by testing each pair of the characters ofM.We show that this leads to aO(m2n+p)
time algorithm for outputting all pminimal obstruction sets forM. Although there can be O(m3)minimal obstruction sets
forM, in practice we expect the number of minimal obstruction sets to be small.
We conclude with a theorem relating our characterization of the minimal obstruction sets for three-state perfect
phylogenies to monochromatic pairs of vertices of the partition intersection graph ofM with no legal minimal separator.
By the results of [10–12] the existence of such a pair of vertices certifies that no perfect phylogeny forM exists.
∗ Corresponding author. Tel.: +1 515 257 6855.
E-mail addresses: shutters@iastate.edu (B. Shutters), fernande@iastate.edu (D. Fernández-Baca).
0893-9659/$ – see front matter© 2012 Elsevier Ltd. All rights reserved.
doi:10.1016/j.aml.2012.02.060
B. Shutters, D. Fernández-Baca / Applied Mathematics Letters 25 (2012) 1226–1229 1227
Fig. 1. The ‘‘forbidden’’ sets of edges in the intersection graph of three three-state characters that admit a perfect phylogeny. In [12], four forbidden
subgraphs are given. However, one of the subgraphs given is a subgraph of another. For our purposes, the supergraph is not needed and so it is not shown
here. The colored edges are used in the proof of Theorem 6.
2. Preliminaries
We fixM to be an n bymmatrix of integers from the set {1, 2, 3}. For a subset C of the characters ofM, thematrixM|C is
obtained by restrictingM to the characters in C . G(M) is the intersection graph ofMwhich has a vertex ci for each character
c ofM and each state i of c , and an edge cidj precisely when there is a taxon ofM having state i for character c and state j
for character d. Note that G(M) cannot have an edge between vertices associated with different states of the same character
ofM.
In [7], a matrixM of two-state characters is derived fromM by adding, for each character c ofM, two-state characters
c(1), c(2), c(3) to M. All taxa having state i for c are given state 1 for c(i), and all other taxa are given state 2 for c(i).
Since every character ofM has two states, it follows from the splits equivalence theorem (also known as the four-gamete
condition) that two characters c(i) and d(j) ofM are incompatible if and only if the two columns corresponding to c(i) and
d(j) contain all four of the pairs (1, 1), (1, 2), (2, 1), and (2, 2); otherwise c(i) and d(j) are compatible [13]. Note that this implies
that c(i) and d(j) are compatible if and only if G(M|{c(i), d(j)}) is not a cycle. A set of two-state characters is compatible
if and only if each pair of characters in the set is compatible [13]. Theorem 1 shows that we can test for the existence of a
perfect phylogeny forM by finding a compatible subset of the characters inM.
Theorem 1 (See [7]). There is a perfect phylogeny for M if and only if there is a subset C of the characters inM such that both
of the following hold: (i) every pair of characters in C is compatible; and (ii) for every character c of M, C contains at least two
characters from the set {c(1), c(2), c(3)}.
Theorem 2 generalizes the splits equivalence theorem to three-state characters.
Theorem 2 (See [12]). There is a perfect phylogeny for M if and only if both of the following hold: (i) for every pair {a, b} of
characters of M, G(M|{a, b}) is acyclic; and (ii) for every triple {a, b, c} of characters of M, G(M|{a, b, c}) does not contain, up
to relabeling of characters and states, any of the subgraphs shown in Fig 1.
3. The main results
For a character c ofM and a state i of c , if there is a character d ofM and two states j and k of d such that c(i) is incompatible
with both d(j) and d(k), then we say that i is a dependent state of c with d as a witness. We give a complete characterization
of the obstruction sets forM in terms of the dependent states of its characters.
Lemma 3. Let c be a character of M and let i be a dependent state of c. No subset of the characters in M satisfying both
conditions (i) and (ii) of Theorem 1 contains c(i).
Proof. Let C be a subset of the characters of M containing c(i). Since i is a dependent state, there is a character d of M
and two states j, k of d such that c(i) is incompatible with both d(j) and d(k). It follows that either C contains at most one
of {d(1), d(2), d(3)}, or C contains a pair of incompatible characters. Thus, C cannot satisfy both conditions (i) and (ii) of
Theorem 1. 
Theorem 4. Let a, b, and c be three characters of Mwhere c has two dependent states i and j such that a is a witness that state i is
dependent, and b is a witness that state j is dependent. The set {a, b, c} is an obstruction set forM. Furthermore, if the characters
in {a, b, c} are pairwise compatible, then {a, b, c} is a minimal obstruction set for M.
Proof. Let M = M|{a, b, c} and suppose that C is a subset of the characters in M satisfying both conditions (i) and (ii) of
Theorem 1. By Lemma 3, c(i) ∉ C and c(j) ∉ C . Then C contains at most one of {c(1), c(2), c(3)}. This contradicts that
C satisfies condition (ii) of Theorem 1. Hence, no such C exists. So by Theorem 1, there is no perfect phylogeny for M and
{a, b, c} is an obstruction set forM. 
Wenowgive a characterization ofwhen a state is dependent using the intersection graph of two characters ofM, and then
show that every minimal obstruction set forM contains a character with two dependent states. For a path p : p1p2p3p4p5 of
length 4, we write mid(p) to denote p3, the ‘‘middle’’ vertex of p. We consider a 4-cycle to be a path of length 4 and mid(p)
is allowed to be any vertex on the cycle. Note that at least four vertices lie on any cycle in G(M), and for a pair {c, d} of
characters ofM, every state associated with a vertex on a cycle in G(M|{c, d}) is dependent.
1228 B. Shutters, D. Fernández-Baca / Applied Mathematics Letters 25 (2012) 1226–1229
Lemma 5. A state i of a character c of M is a dependent state if and only if there is a character d of M and a path p of length 4
in G(M|{c, d}) with ci = mid(p). Furthermore, if such a d exists, then d is a witness that i is a dependent state of c.
Proof. (⇒) Suppose that i is a dependent state of a character c ofM with d as witness. W.l.o.g. relabel the states of c and
d so that i = 1 and that c(1) is incompatible with both d(1) and d(2). Let G = G(M|{c, d}). Then, c1d1 and c1d2 are edges
of G. Since, c(1) is incompatible with d(1), either c2d1 or c3d1 is an edge of G. Since c(1) is incompatible with d(2), either
c2d2 or c3d2 is an edge of G. We show that in every case G contains a path p such that mid(p) = c1. If c2d1 and c2d2 are
edges of G, then c1d1c2d2c1 is a cycle containing c1. If c2d1 and c3d2 are edges of G, then c3d2c1d1c2 is the required path of
length 4. If c3d1 and c2d2 are edges of G, then c2d2c1d1c3 is the required path of length 4. If c3d1 and c3d2 are edges of G, then
c1d1c3d2c1 is a cycle containing c1. (⇐) Let d be a character ofM such that there is a path p of length 4 in G = G(M|{c, d})
with ci = mid(p). W.l.o.g. relabel the states of c and d so that i = 1. G cannot contain edges between two states of the same
character. So either p is a cycle containing c1, or the path c2d1c1d2c3 (up to possibly renaming states d1 and d2). In both cases,
the four-gamete condition shows that c(1) is incompatible with two of d(1), d(2), d(3). Hence, state 1 is a dependent state
of c with d as witness. 
Theorem 6. Let C be a minimal obstruction set for M. Then C contains a character with two dependent states.
Proof. By Theorem 2, the cardinality of C is either 2 or 3. Case 1. If the cardinality of C is 2, then it follows from Theorem 2
that G(M|C) contains a cycle. Since there cannot be an edge in G(M|C) between vertices associated with states of the same
character, it follows that any cycle G(M|C) has at least four vertices, and every state associated with a vertex on this cycle
is a dependent state. The theorem follows. Case 2. If the cardinality of C is 3, then it follows from Theorem 2 that G(M|C)
contains one of the graphs of Fig 1 as a subgraph (after possibly renaming the characters and states ofM|C). If Fig 1(a) is a
subgraph of G(M|C), then c3b1c1b2c2 (colored red) is a path showing that state 1 of c is dependent, and c3a1c2a3c1 (colored
blue) is a path showing that state 2 of c is dependent. If Fig 1(b) is a subgraph of G(M|C), then c3b1c1b2c2 (colored red) is a
path showing that state 1 of c is dependent, and c3a1c2a2c1 (colored blue) is a path showing that state 2 of c is dependent. If
Fig 1(c) is a subgraph ofG(M|C), then c3a2c1a1c2 (colored red) is a path showing that state 1 of c is dependent, and c3b3c2b1c1
(colored blue) is a path showing that state 2 of c is dependent. In every case,we have shown that states 1 and 2 are dependent
states of c. Thus,M contains a character with two dependent states. 
Theorems 4 and 6 together with Theorem 2 give us the following test for the existence of a perfect phylogeny forM.
Theorem 7. There is a perfect phylogeny for M if and only if there is at most one dependent state of each character c of M.
Proof. If there is a perfect phylogeny forM, then there is no obstruction set forM. Thus, by Theorem 4, there can be no
character ofM with more than one dependent state. If there is no perfect phylogeny forM, then there must exist some
minimal obstruction set forM. By Theorem 6, there is a character ofM with two or more dependent states. 
An immediate consequence of Theorem 7 is that every set C of three-state characters has a canonical subset that does
have a perfect phylogeny, namely the subset {c ∈ C : c has at most one dependent state}.
We now describe an algorithm, denoted by A, which outputs all of the minimal obstruction sets for M. Step 1 of A
computes for each character c ofM the following.
• A setB(c) of all characters d ofM such that G(M|{c, d}) contains a cycle.
• For each state i of c , a setD(c, i) of all characters d ofM such that d ∉ B(c) and there is a path p of length 4 inG(M|{c, d})
with ci = mid(p).
Step 2 ofA visits each character c ofM and outputs the following.
• For each character d inB(c), the set {c, d}.
• For each pair of states {i, j} of c with both D(c, i) and D(c, j) non-empty, each element of the set {{c, x, y} : x ∈
D(c, i), y ∈ D(c, j)}.
Theorem 8. A outputs all p minimal obstruction sets for M in O(m2n+ p) time.
Proof. We first establish the following claim.
Claim 1. For each character c of M, the setsB(c),D(c, 1),D(c, 2), andD(c, 3) are pairwise disjoint.
Proof of Claim 1. ClearlyB(c)∩(D(c, 1)∪D(c, 2)∪D(c, 3)) = ∅, so it suffices to show that for each character c ofM, the
setsD(c, 1),D(c, 2), andD(c, 3) are pairwise disjoint. W.l.o.g. let c be a character ofM and let d ∈ D(c, 1)∩D(c, 2). Let
G = G(M|{c, d}). Since d ∈ D(c, 1) ∩D(c, 2), d ∉ B(c), and there are paths p1 and p2 of length 4 in G with c1 = mid(p1)
and c2 = mid(p2). Since d ∉ B(c), G is acyclic. W.l.o.g. suppose that p1 is the path c2d1c1d2c3. Since mid(p2) = c2, there
must be two edges from c2 to vertices associated with states of d. We have that c2d1 is an edge of p1. If c2d2 is an edge of p2,
then we have a cycle in G. So c2d3 and either d3c3 or d3c1 are edges of p2. In either case, there is a cycle in G. 
B. Shutters, D. Fernández-Baca / Applied Mathematics Letters 25 (2012) 1226–1229 1229
By Lemma 5, A finds all dependent states, and hence, by Theorems 2, 4 and 6, outputs all of the minimal obstruction
sets for M. By Claim 1, every obstruction set output by A is minimal. We now establish the runtime. Step 1 of A takes
O(m2n) time to construct the intersection graphs of each pair of characters ofM. Since each intersection graph has exactly
six vertices and at most nine edges, it follows that it all cycles and paths of length 4 can be found in O(1) time. Hence
step 1 takes O(m2n) time. Step 2 ofA visits each of them characters ofM and takes O(1) time per set output. Any minimal
obstruction set of cardinality 2 will be output twice. If follows from Claim 1 that each minimal obstruction set of cardinality
3 will be output at most three times. Thus, step 2 takes O(m + p) time where p is the number of minimal obstruction sets.
Hence,A takes O(m2n+ p) time to complete both steps 1 and 2. 
Several approaches to determining the existence of a perfect phylogeny for M studied in the literature make use of
separating sets in G(M) [10,11]. For two vertices a and b of G(M), an a–b separator is a set of vertices whose removal
separates a from b. An a–b separator isminimal if no subset of it is an a–b separator. Aminimal separator is a separator that is
a minimal a–b separator for some pair a, b of vertices of G(M). A minimal separator S of G(M) is legal if, for each character
c ofM, S contains at most one vertex corresponding to a state of c . A pair of vertices of G(M) representing different states
of the same character ismonochromatic.
Theorem 9 (See [10–12]). There is a perfect phylogeny for M if and only if both of the following hold: (i) the characters of M
are pairwise compatible; and (ii) every monochromatic pair of vertices in G(M) is separated by a legal minimal separator.
We concludewith Theorem10 that relates dependent states to legalminimal separators. A consequence of Theorem10 is
that algorithmA can be easilymodified to outputmonochromatic pairs of vertices ofG(M)with no legalminimal separator.
Theorem 10. Suppose that the characters of M are pairwise compatible. Two states i and j of a character c of M are dependent
if and only if there is no legal minimal separator for ci and cj in G(M).
Proof. By Theorems 2 and 9, it suffices to show that the theorem holds for everyminimal obstruction set, i.e., for each graph
in Fig 1, amonochromatic pair of vertices has no legalminimal separator if and only if they correspond to a pair of dependent
states. This is verified by inspection. In the graph of Fig 1(a): {c1, c2} is the onlymonochromatic pair of vertices with no legal
minimal separator; 3 is the only dependent state of a; 2 is the only dependent state of b; and 1 and 2 are the only dependent
states of c . In the graph of Fig 1(b): (b1, b2) and (c1, c2) are the only monochromatic pairs of vertices with no legal minimal
separator; there are no dependent states of a; 1 and 2 are the only dependent states of b; and 1 and 2 are the only dependent
states of c. In the graph of Fig 1(c): (a1, a2), (b1, b3), (c1, c2), and are the only monochromatic pairs of vertices with no legal
minimal separator; 1 and 2 are the only dependent states of a; 1 and 3 are the only dependent states of b; and 1 and 2 are
the only dependent states of c. 
Acknowledgments
We thank an anonymous reviewer for several improvements to the exposition of our results. This work was supported
in part by the National Science Foundation under grants CCF-1017189 and DEB-0829674.
References
[1] D. Fernández-Baca, The perfect phylogeny problem, in: Steiner Trees in Industry, Kluwer, 2001, pp. 203–234.
[2] H. Bodlaender, M. Fellows, T. Warnow, Two strikes against perfect phylogeny, in: ICALP 1992, in: Lect. Notes Comput. Sci., vol. 623, Springer, 1992,
pp. 273–283.
[3] M. Steel, The complexity of reconstructing trees from qualitative characters and subtrees, J. Classif. 9 (1992) 91–116.
[4] R. Agarwala, D. Fernández-Baca, A polynomial-time algorithm for the perfect phylogeny problem when the number of character states is fixed, SIAM
J. Comput. 23 (6) (1994) 1216–1224.
[5] S. Kannan, T. Warnow, A fast algorithm for the computation and enumeration of perfect phylogenies, SIAM J. Comput. 26 (6) (1997) 1749–1763.
[6] D. Gusfield, Efficient algorithms for inferring evolutionary trees, Networks 21 (1) (1991) 19–28.
[7] A. Dress, M. Steel, Convex tree realizations of partitions, Appl. Math. Lett. 5 (3) (1992) 3–6.
[8] S. Kannan, T. Warnow, Inferring evolutionary history from DNA sequences, SIAM J. Comput. 23 (4) (1994) 713–737.
[9] D. Gusfield, Y. Wu, The three-state perfect phylogeny problem reduces to 2-SAT, Commun. Inf. Syst. 9 (4) (2009) 195–301.
[10] D. Gusfield, The multi-state perfect phylogeny problem with missing and removable data: solutions via integer-programming and chordal graph
theory, J. Comput. Biol. 17 (3) (2010) 383–399.
[11] R. Gysel, F. Lam, D. Gusfield, Constructing perfect phylogenies and proper triangulations for three-state characters, in: WABI 2011, in: Lect. Notes
Comput. Sc., vol. 6833, Springer, 2011, pp. 104–115.
[12] F. Lam, D. Gusfield, S. Sridhar, Generalizing the splits equivalence theorem and four gamete condition: perfect phylogeny on three-state characters,
SIAM J. Discrete Math. 25 (3) (2011) 1144–1175.
[13] C. Semple, M. Steel, Phylogenetics, Oxf. Lect. S. Math. Appl., Oxford University Press, 2003.


file:///data/core-remote/dit/data/elsevier/pdf/80d/aHR0cDovL2FwaS5lbHNldmllci5jb20vY29udGVudC9hcnRpY2xlL3BpaS9zMDg5Mzk2NTkxMjAwMTQ0OQ==.pdf

A Simple Characterization of the Minimal Obstruction Sets for Three-State Perfect Phylogenies

Abstract

Similar works

Full text

Available Versions

Elsevier - Publisher Connector

CiteSeerX

Elsevier - Publisher Connector