The Cophylogeny Reconstruction Problem is NP-Complete∗
Y. Ovadia and D. Fielder and C. Conow and R. Libeskind-Hadas
September 7, 2009
1
Introduction
The cophylogeny reconstruction problem arises in the study of host-parasite relationships. Specifically, we are given a host tree H, a parasite tree P , and a function ϕ mapping the leaves (extant
taxa) of P to the leaves of H. Four biologically plausible operations are considered: cospeciation,
duplication, host switching, and loss (Figure 1). A host switch is permitted in conjunction with a
duplication event but not with a cospeciation event [1].
A feasible solution is an extension of ϕ that maps each internal node of the parasite tree to a
vertex or edge of the host tree and can be constructed using the four types of events. The objective
of the cophylogeny reconstruction problem is to find one or more “optimal” solutions that reconcile
the parasite tree and the host tree with respect to these operations. One notion of optimality
simply assigns a cost to each of the four types of events and then seeks to minimize the total cost.
Since it is often difficult to estimate appropriate costs for each of the four types of operations,
an alternative approach is to find a Pareto optimal set of solutions [1]. In this case, a vector is
associated with each solution where the entries indicate the number of cospeciation, duplication,
loss, and host switch events, respectively. Let v = (c, d, `, h) and v 0 = (c0 , d0 , `0 , h0 ) be the cost
vectors of two solutions. We say that v is strictly less than v 0 if c ≤ c0 , d ≤ d0 , ` ≤ `0 , and h ≤ h0 ,
and at least one of these relationships is a strict inequality. A solution is said to be in the Pareto
optimal set if its cost vector is some v such that there is no solution with cost vector strictly less
than v. The optimization version of the cophylogeny reconstruction problem is to find the maximal
Pareto optimal set. The corresponding decision version of this problem, henceforth denoted CRDP
(Cophylogeny Reconstruction Decision Problem) asks: For a given cost vector v, is there a solution
whose cost vector is strictly less than v?
While practitioners have generally assumed for over a decade that this problem is computationally intractable, no rigorous proof of this conjecture has been found. Recently, Libeskind-Hadas
and Charleston have made progress in this direction by showing that a slightly more general version
of this problem is NP-complete [2]. Their proof assumes that the host phylogeny may be reticulate,
meaning that it can contain cycles (corresponding to biological hybridization events) rather than
a true tree. In addition, the proof assumes that ranges can be stipulated for the relative times of
events in the two trees. These two assumptions are necessary in the proof in order to construct
appropriate gadgetry. The problem of whether the classical cophylogeny reconstruction problem is
NP-complete was, therefore, left open.
∗ This
work was supported by the U.S. National Science Foundation under grant 0753306.
1
a
e
b
d
c
a
a
loss
loss
duplication
e
e
loss
loss
c
cospeciation
d
duplication
with host switch
c
d
cospeciation
b
b
Figure 1: A simple tanglegram with host tree in black at left and parasite tree in gray on right.
The associations ϕ between tips is shown in dotted lines. Below are two possible reconstructions
that explain the relationship between H and P with events labeled.
In this paper, we prove that the decision version of the cophylogeny reconstruction problem is indeed NP-complete. The proof relies on some ideas introduced in Libeskind-Hadas and Charleston’s
proof for the reticulate timed problem, but is substantially more involved due to the fact that we
can no longer exploit reticulation or timing.
2
Terms and Definitions
The following terms and definitions will be used throughout the remainder of the paper.
• V (T ) denotes the vertices in rooted tree T ,
• L(T ) denotes the leaves or tips of rooted tree T ,
• A host tree is a rooted regular binary tree with an additional edge (d, r) where d is called the
“dummy root” and r is the original root of the regular binary tree. This is required in order
to account for events in the parasite phylogeny that may predate the most recent common
ancestor in the host phylogeny.
We now formally state the decision problem.
Definition 1 An instance of the Cophylogeny Reconstruction Decision Problem (CRDP) is a 4-tuple
(H, P, ϕ, B) where
• H and P are the rooted host and parasite trees,
2
• ϕ : L(P ) → L(H) maps the tips of P to the tips of H,
• B is a 4-tuple (BC , BD , BS , BL ) of upper bounds on the number of cospeciation, duplication,
loss, and host switch events respectively.
The decision question is: Does there exist a mapping Φ that extends ϕ and whose cost is strictly
less than B?
A natural generalization of this problem allows an extant parasite to be mapped to some arbitrary number of tips in the host tree. This problem, called the Generalized Cophylogeny Reconstruction Decision Problem (GCRDP), is stated as follows:
Definition 2 An instance of the Generalized Cophylogeny Reconstruction Decision Problem (GCRDP)
is a 4-tuple (H, P, ϕ, B) where
• H and P are the rooted host and parasite trees,
• ψ : L(P ) → 2L(H) maps the tips of P to sets of tips of H,
• B is a 4-tuple (BC , BD , BS , BL ) of upper bounds on the number of cospeciation, duplication,
loss, and host switch events respectively.
The decision question is: Does there exist a mapping Ψ that extends ψ and whose cost is strictly
less than B?
We first prove that GCRDP is NP-complete via a reduction from 3SAT. We then show an instance of GCRDP constructed in the reduction from 3SAT can be transformed into a corresponding
instance of CRDP with the same answer, thus implying that CRDP is also NP-complete.
3
Polynomial Time Reduction of 3SAT to GCRDP
Given an instance of 3SAT with n variables x1 , . . . , xn and m clauses C1 , . . . , Cm where n is a power
of 2 and each clause contains at most a single instance of each variable1 , we construct an instance
of GCRDP for which a solution exists if and only if a solution exists to the 3SAT instance.
A basic gadget in the reduction is the k-thorn gadget, illustrated in Figure 2, consisting of a path
of length k where each vertex on the path, except the last, has one tip child. The last vertex on
the path may have one tip child and a second child in another gadget or may have two tip children.
The value of k will be determined later.
If y and z are two k-thorn gadgets where y is in the parasite tree and z is in the host tree, then
we say that y associates with z if ϕ maps each tip of y to the corresponding tip of z or, in the case
of GCRDP, if φ maps each tip of y to a set which contains the corresponding tip of z. Similarly,
we say that a solution mapping Ψ maps thorn y to thorn z to mean the solution maps each vertex
of y to the corresponding vertex in z.
In our reduction, variables in the 3SAT instance will be represented in the host tree while clauses
will be represented in the parasite tree. The tip mapping function will be used to encode the literals
that appear in each clause.
1 It is easily seen that any instance of 3SAT can be expressed as an equivalent instance with these two properties
using a simple padding of variables and clauses.
3
...
k
Figure 2: The k-thorn gadget is drawn to the left with filled-in tip vertices. Henceforth, we represent
this symbolically as shown to the right.
α0
α1
α2
α3
α4
α0
α0
α1
0
α2 α
α31
α
α42
α
α3
α4
α1 0
α
α21
α
α32
α
α43
α
α4
Figure 3: The assignment gadget is set to true if α1 occurs prior to α3 and is set to false otherwise.
4
λ0
λ1
λ2
λ3
λ4
Figure 4: A generic literal gadget.
The primary gadget in the host tree is called an assignment gadget. There will be four such
gadgets in the host tree for each variable in the 3SAT instance (only one of which determines the
variable’s assignment). This gadget, shown in Figure 3, begins with a k-thorn gadget α0 . This
thorn’s non-tip child has two descendant edges; one to thorn gadget α1 followed by thorn gadget
α2 , and the other which leads to α3 followed by α4 . The final thorns of α2 and α4 have two tip
children. This host tree gadget will represent the value true if α1 occurs before α3 and false
otherwise. These relative times will be induced by the mapping of corresponding gadgets in the
parasite tree, which we describe below.
The host tree, depicted
in Figure 5, begins with a dummy root whose single child is a k-thorn
P
0
τ 0 where k = BD + l∈L(P ) (|ψ(l)| − 1) + 1 (this value will be defined as BD
+ 1 later). This
k-thorn leads to a vertex which initiates a sequence of bifurcations forming a balanced binary tree
culminating with n vertices, one for each variable in the 3SAT instance. Each of these n vertices
is the root of a k-thorn gadget τi1 of size k = BL + 1. Each such thorn’s non-tip child leads to a
balanced binary tree with four children which serve as roots for assignment gadgets αi , βi , γi , and
δi .
The primary gadget in the parasite tree is called a literal gadget and is depicted in Figure 4
Each clause in a 3SAT instance contains exactly three literals and thus there will be three literal
gadgets per clause in the parasite tree. Let λ denote a literal gadget. λ begins with k-thorn gadget
λ0 which shares its root, followed by an additional thorn λ1 . The non-tip child of λ1 is incident to
one edge which leads to a k-thorn λ2 followed by a tip, and another edge which leads to k-thorns
λ3 and λ4 in series followed by another tip vertex.
The literal gadgets in the parasite tree will be mapped to assignment gadgets in the host tree
using the tip mapping function ψ. In particular, this mapping will depend on whether a given
literal, represented by a literal gadget in the parasite tree, appears unnegated or negated in a given
clause. In other words, the tip mapping function will encode whether a given literal satisfies its
clause by being true or false.
Henceforth, we will refer to a literal gadget from clause Cj that is the negated or unnegated
form of variable xi as λi,j .
5
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Figure 5: A sample hostβββtree
derived
from
The balloon shows the group of four
1 γi a 33SATδiinstance.
1
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λ4
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λii!! ,j,j L
D
S
1
1
Figure 6: A sample parasite tree derived from a 3SAT instance. On the left is the gadgetry
corresponding to clause Cj containing literals of variables xi , xi0 , and xi00 .
3.0.1
Budget Tuple
1
1
1
1
The budget tuple B = (BC , B , B , B ) is defined as follows
11
1
4
λ
λi4i!!!! ,j,j
BC = ∞
BD = 4m + 8n − 1
BS = 3m + 8n − 2
1
1
1
11 + 8n + 4
BL = (m + 2) log(n) + 4m
3.0.2
Parasite Tree
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
11
11
1
1
1
1
1
1
1
The parasite tree, illustrated in Figure 6, begins with a path (p1 , . . . pm+1 ) where p1 is the root.
Henceforth, for 1 ≤ j ≤ m, the jth parasite subtree refers to the subtree containing descendants
of pj ’s non-path edge. We arbitrarily label the subtrees containing descendants of pm+1 ’s edges as
the (m + 1)th and (m + 2)th parasite subtrees.
Each of the first m parasite subtrees correspond to one of the m clauses. Consider an arbitrary
clause Cj and its corresponding subtree. The jth subtree begins with a thorn gadget of size
0
k = BD
+ 1 labeled σj0 followed by an additional k-thorn labeled σj1 of size k = BL + 1. A path of
7
length 2 follows the non-tip child of σj1 which is followed by three edges followed by literal gadgets
labeled λi,j , λi0 ,j , and λi00 ,j each representing a literal contained in Cj .
The last two parasite subtrees do not correspond to clauses. Instead they serve to force an
order between assignment gadgets as we will describe later. The branches lead to the two thorns
0
1
0
1
0
σm+1
, σm+1
and σm+2
, σm+2
respectively where the 0th thorns are of size k = BD
+ 1, and the 1st
is of size k = BL + 1. These branches are then followed by n k-thorn triples (k = BL + 1) in series
πi0 , πi1 , πi2 and ρ0i , ρ1i , ρ2i respectively for 1 ≤ i ≤ n, and each triple except the last is followed by a
vertex with one tip child.
3.0.3
Tip Mappings
We begin by providing definitions of how literal gadgets are mapped to assignment gadgets in
negated and unnegated forms. This, in turn, will be used to describe the mapping function ψ.
If we wish to associate λ with α in unnegated form, then the tips are mapped as follows where
each parasite thorn, host thorn pair indicates that the parasite thorn is associated with the host
thorn.
(λ0 , α0 ), (λ1 , α1 ), (λ2 , α2 ), (λ3 , α3 ), (λ4 , α4 )
If, instead, we wish to associate λ with an assignment gadget α in negated form, then we use
the following set of associations.
(λ0 , α0 ), (λ1 , α3 ), (λ2 , α4 ), (λ3 , α1 ), (λ4 , α2 )
Figure 7, which depicts the intended solution mapping of a literal gadget onto an assignment
gadget, also serves to depict the tip mapping between these gadgets.
The parasite tip mappings are defined by the following:
• σj0 is associated with τ 0 for 1 ≤ j ≤ m.
• σj1 is associated with τi1 for all i, j where xi ∈ Cj or xi ∈ Cj .
• The (m + 1)th branch maps its sequence of thorns to the 0th , 1st, and 2nd thorn of each
assignment gadget as described by the following parasite thorn, host thorn associations (for
1 ≤ i ≤ n):
0
1
2
(π4i−3
, αi0 ), (π4i−3
, αi1 ), (π4i−3
, αi2 )
0
1
2
(π4i−2
, βi0 ), (π4i−2
, βi1 ), (π4i−2
, βi2 )
0
1
2
(π4i−1
, γi0 ), (π4i−1
, γi1 ), (π4i−1
, γi2 )
0
1
2
(π4i
, δi0 ), (π4i
, δi1 ), (π4i
, δi2 )
• Similarly, the (m + 2)th branch’s thorn tips map to the 0th , 1st , and 2nd thorn of each
assignment gadget (for 1 ≤ i ≤ n):
(ρ04i−3 , αi0 ), (ρ14i−3 , αi3 ), (ρ24i−3 , αi4 )
(ρ04i−2 , βi0 ), (ρ14i−2 , βi3 ), (ρ24i−2 , βi4 )
(ρ04i−1 , γi0 ), (ρ14i−1 , γi3 ), (ρ24i−1 , γi4 )
(ρ04i , δi0 ), (ρ14i , δi3 ), (ρ24i , δi4 )
8
0
1
2
• Recall that every k-thorn triple (except the last) of form π4i−3
, π4i−3
, π4i−3
(1 ≤ i ≤ n) is
followed by a vertex with exactly one tip child. This tip maps to a singleton set containing
2
the tip child of αi2 which is not already mapped to by the last thorn in π4i−3
.
• Similarly, the tips following thorn triples of form π4i−2 , π4i−1 , and π4i (1 ≤ i ≤ n) map to
the tip children of β, δ, and γ which are not already mapped to by the last thorn of the triple
respectively.
• The prior two mappings apply similarly for the (m + 2)th parasite subtree except the assignment gadget thorns use superscripts 0, 3, 4 in place of 0, 1, 2.
• For all 1 ≤ i ≤ n and 1 ≤ j ≤ m, where xi ∈ Cj we associate λi,j ’s tips with αi ’s tips in its
unnegated form as defined earlier.
• For all 1 ≤ i ≤ n and 1 ≤ j ≤ m, where xi ∈ Cj we associate λi,j with αi in its negated form
as defined earlier.
• For 1 ≤ i ≤ n and 1 ≤ j ≤ m, if λi,j ’s parent is the first vertex of the length 2 path following
σj1 , then λi0 ,j ’s and λi00 ,j ’s tips map to γi ’s and δi ’s tips respectively, where λi,j , λi0 ,j , and
λi00 ,j are distinct literal gadgets of the jth parasite subtree, and i0 < i00 .
• For 1 ≤ i ≤ n and 1 ≤ j ≤ m, if λi,j ’s parent is not the first vertex of the path following σj1 ,
then its sibling literal gadget λi0 ,j ’s tips map to βi , and the tips of λi00 ,j map to γi .
3.1
3.1.1
Proof of Correctness
Forward Direction
Assume that the given 3SAT instance is satisfiable. Let C1 , . . . , Cm denote the m clauses, and let
x1 , . . . , xn denote the n variables. We show the existence of a solution to the constructed GCRDP
instance by describing a solution Ψ which incurs a cost tuple less than B. For convenience, define
s(j) ∈ {1, . . . , n} (1 ≤ j ≤ m + 2) to be the index of a variable which satisfies clause Cj . If multiple
variables satisfy Cj , let s(j) be the least such index. We arbitrarily assign s(m + 1) = s(m + 2) = 1.
First we describe the mapping of a literal gadget onto an assignment gadget as shown in Figure
7. When we write that a literal gadget λ associates with an assignment gadget α, and ψ has been
defined such that λ’s tips are mapped to α’s tips in unnegated form, the following is intended:
• For each of the following parasite thorn, host thorn pairs, we map the parasite gadget to the
host gadget each at cost of BL + 1 cospeciations.
(λ0 , α0 ), (λ1 , α1 ), (λ2 , α2 ), (λ3 , α3 ), (λ4 , α4 )
• The edge between λ0 and λ1 incurs a loss as it passes over the non-tip child of α0 .
• The parent vertex of thorns λ2 and λ3 is mapped to the edge between α1 and α2 and incurs
a duplication followed by a host switch onto the edge preceding α3 .
If ψ maps λ’s tips to α’s tips in negated form, then we apply the following modifications to the
mapping above.
9
0
λ0
0
λ
α1
1
α
λ1
1
λ
α2
2
α
λ2
23
λ
α
3
α
λ3
3
λ
α4
4
α
λ4
λ4
α
λ0
1
α
α0
λ10 λ0 α0
0
α
α02 λ
α
α11
λ2 α
λ11
α3 λ
α2
3 α2
λ λ2
2
α4 λ
α33
λ4 α
λ33
λ4
α4
α4
λ4
λ
α0
λ0
α1
λ1
α2
λ2
α3
α0
λ3
4 λ0
α
0
α01 α
λ4 λ
λ11
α
α12
λ
λ22
α
α23
λ
λ33
α
α34
λ
λ4
α4
λ4
0
α0
λ0 0
α01 α
λ
λ11
α
α12
λ
λ22
λ0
α
α23
λα01
λ
3
λ3
αλ11
α
α34
λα12
λ
λ44
αλ22
α
λα23
4
λ
αλ33
λα34
αλ44
λ4
α
λ0
1
α
λ1
α2
λ2
α0
α3
0
λ3 λ0 α
0
α1
α4 λ
1
λ1
λ4 α
12
λ
α
α22
α00 λ2
α λ
α3
3
α
λ3
3
λ
α4
4
α
λ4
λ4
α0
λ0
α1
λ1
α2 α0
λ2 λ0 0
α
α3α01
λ
λ3 λ11
α
α4α12
λ
4 2
λ λ2
α3
α
λ23
λ
α34
α
λ34
λ
α4
λ4
α0
λ0 0
α10 α
λ1
λ
α21
α
λ21
λ
α32
α
λ32
λ
α43
α
λ43
λ
α4
λ4
Figure 7: On the left, we show a literal gadget λ mapping onto an assignment gadget α in unnegated
form. The mapping for negated form is shown to the right.
• The gadgets are mapped to each other according to the pairs
(λ0 , α0 )(λ1 , α3 )(λ2 , α4 )(λ3 , α1 )(λ4 , α2 )
• The parasite tree incurs a loss by passing over the other child of the non-tip child of α0 .
• The duplication and host switch events occur on the edge between α3 and α4 , and the switch
lands on the edge prior to thorn α1 .
1
Regardless of form, this mapping costs 5(BL + 1) cospeciations, 11 duplication, 1 loss, and
1 host
1
1
1
switch.
0
1
First, let all vertices of the path (p1 , . . . , pm+2 ) map to the host edge which precedes τ incurring
1
1
1
a cost of m + 1 duplications. For the jth parasite subtree
(1 ≤ j ≤ m + 2), we map the k-thorn σj0
1
1
0
1
1
to τ , assign the child edge to the path through the binary subtree between τ 0 and τs(j)
, and
map
1
1
1
0
1
m
+
1
duplication,
and
σj to τs(j) . This accumulates (m + 2) × (BD + 1 + BL + 1) cospeciation,
1
1
1
(m + 2) × (log2 (n)) loss events.
1
For 1 ≤ j ≤ m, the length two path following σj1 codiverges at vertices of the balanced binary
1
subtree following τs(j)
as depicted in Figure 8, incurring two cospeciations and one loss per parasite
1
1 assignment gadget
subtree associated with a 3SAT clause. The literal gadget λs(j),j then maps to
αs(j) . Since the given 3SAT instance is known to be satisfiable, no two values j and j 0 exist such
that i = s(j) = s(j 0 ) and xi ∈ Cj and xi ∈ Cj 0 . Thus we may apply the mapping defined earlier
between literal and assignment gadgets and incur a cost tuple of (5(BL + 1), 1, 1, 1) for each of the
m clauses (recall that this is the tuple we derived after describing how to map a literal gadget to
an assignment gadget). The other two literal gadgets of the jth parasite subtree each map to one
of βs(j) , γs(j) , or δs(j) in unnegated form accumulating an additional cost of (10(BL + 1), 2, 2, 2) for
each clause.
10
22
00
11
The tip
tip
child
of the
the vertex
vertex which
which follows
follows aa thorn
thorn triple
triple ππ4i−3
•• The
child
of
4i−3,,ππii
4i−3,,ππ4i−3
00
11
π
,π
••The
ππ4i−3
, πi2 i2 of
, π4i−3
Thetip
tipchild
childofofthe
thevertex
vertexwhich
whichfollows
followsaathorn
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triple
4i−3
4i−3
which isis not
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mapped to
to by
by the
the last
last thorn
thorn in
in
maps
to
the
tip,child
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of ααi2i2 which
maps
to
the
tip
22
maps
whichisisnot
notmapped
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the
lastthorn
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mapstotothe
thetip
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childofofααi i which
22 the
0
1
ππby
.. last
4i−3
, πi2
, π4i−3
• The tip child of the vertex which
follows
a thorn triple π4i−3
4i−3
22
2
0
1
ππ4i−3
• The tip child
of. . the vertex which follows
a thorn
4i−3
4i−3 , πis
i not mapped to by the last thorn in
4i−3
maps to
the tiptriple
childπof
αi2, πwhich
• Similarly,
Similarly,
we map
map the
the tips
tips following
following thorn
thorn triples
triples of
of form
form ππ4i−2
•
we
4i−2,, ππ4i−1
4i−1,,
2
mapped
to
by
the
last
thorn
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maps to the tip child of αi2 which is πnot
.
4i−3
••Similarly,
following
thorn
form
, ,πtip
,,
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wemap
mapthe
thetips
tips
following
thorntriples
triplesofof
form
π4i−1
4i−2
2
4i−2
4i−1
and
to
the
tip
children
ofβ,
β,δ,δ,and
andγγ which
whichare
arenot
notalready
alreadymapped
mappedto
to
and
ππ4i4iππto
the
children
of
π4i−3
.
and
ofofβ,β,δ,δ,and
γγwhich
already
mapped
andππ4i4itotothe
thetip
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children
andmap
which
arenot
not
already
mapped
to
• Similarly,
we
the are
tips
following
triplesto
ofthe
form
π4i−2
, π4i−1 ,
by
thethorn
last
thorn
of
the
triple
respectively.
by
the
last
thorn
of
triple
respectively.
• Similarly,by
we
map
thethorn
tips following
thorn
triples of form π4i−2
, γ which are not already mapped to
ofofthe
bythe
thelast
last
thorn
thetriple
triple
respectively.
4i−1
and
πrespectively.
of ,β,πδ,
and
4i to the tip children
and π4i to the tip children of β, δ, andbyγ the
which
not of
already
mapped
Thetoprior
prior two
two mappings
mappings apply
apply similarly
similarly for
for the
the m
m++22 lineage
lineage except
except the
the
•• respectively.
The
lastare
thorn
the triple
prior
two
mappings
apply
the
•The
The
prior
two
mappings
applysimilarly
similarlyfor
forthe
themm++2assignment
2lineage
lineageexcept
except
the
by the •last
thorn
of the
triple
respectively.
assignment
gadget
thorns use
use indices
indices 0,0,3,3,44 in
in place
place of
of 0,0,1,1,2.2.
gadget
thorns
assignment
use
indices
0,0,1,1,2.2. for the m + 2 lineage except the
• The
two0,mappings
applyofof
similarly
assignmentgadget
gadgetthorns
thorns
useprior
indices
0,3,3,44ininplace
place
• The prior two mappings apply similarly
for the m
+ 2 lineage
assignment
gadget
thorns except
use
in
place
0, 1, 2. λλi,j
Cjj4,, then
then
weofassociate
associate
’s tips
tips with
with ααii’s’s tips
tips in
in its
its unnegated
unnegated
•• indices
IfIf the
xxii ∈∈0,C3,
we
i,j’s
assignment
usewe
indices
0, 3, 4λin
place
of with
0,with
1, 2.αα
••IfIfgadget
xxi i∈∈Cthorns
associate
’s’stips
ininas
its
unnegated
Cj ,j ,then
then
we
associate
λi,ji,j
tips
tips
unnegated
i ’s
i ’stips
form
asits
defined
earlier.
form
defined
earlier.
form
formasasdefined
definedearlier.
earlier.• If xi ∈ Cj , then we associate λi,j ’s tips with αi ’s tips in its unnegated
• If xi ∈ Cj , then we associate λi,j ’s form
tips with
αi ’s tips
in its unnegated
as defined
earlier.
then we
we associate
associate λλi,j
with ααii in
in its
its negated
negated form
form as
as defined
defined
•• IfIf xxii ∈∈ CCjj,, then
i,j with
form as••
defined
IfIfxxi i∈earlier.
form
∈CCj ,j ,then
thenwe
weassociate
associateλλi,ji,jwith
withααi iininits
itsnegated
negated
formasasdefined
defined
earlier.
• If xi ∈ Cj , then we associate λearlier.
negated form as defined 0
i,j with αi in
2
1
0
1
βi its
The
tip child
the
vertex
which
follows
a thorn triple π4i−3a, πthorn
, πi2
, π4i−3
The
tip
π4i−3
• If xi ∈ Cjearlier.
,earlier.
then we associate λi,j with
αi in• its
negated
form of
as •
defined
4i−3 , πtriple
i
2 vertex which follows
0 child
1 of the
2
0
1
earlier.
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tipofchild
the vertex
whichmaps
follows
a triple
thorn
11 then
,,ππ tipisis,child
,i,j
ππ’s
• The •tipThe
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which follows
a thorn
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i λλ
4i−3
’s
the
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i i4i−3
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22 path
following
which
not
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to of
by
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notlength
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maps
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i parent
i which
11
2
2 first
•
If
λ
’s
parent
is
the
vertex
of
the
length
2
path
following
σ
,
then
which
is
not
mapped
to
by
the
last
thorn
in
maps
to
the
tip
child
of
α
•
If
λ
’s
parent
is
the
first
vertex
of
the
length
2
path
following
σ
,
then
α
not
tofirst
by the
last
thorn
in
maps to the
βπimapped
2
11
i,j
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i,jtip child of αi which
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! ,j,,
i λλ
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’s
and
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and
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whereλλi,j
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to
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0 γγii’s
•• IfThe
’s
parent
is
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the
2 tips
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following
σπand
, then
. the
,j’s
i!!j!!,j
i,j,,λλii! ,j
πii!2,j
.and
ia
ii’s
i,jtip
j4i−3
4i−3of
,’s
child
therespectively,
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which
follows
thorn
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2! ! ’s and λ !! !! ’s tips map
2
1 04i−3
i
4i−3
!i,j
! ,j
γ
λ
to
γ
’s
and
δ
’s
tips
where
,
2, λ
1δi λλ
"
""
α
π
.
λ
’s
and
λ
’s
tips
map
to
γ
’s
and
δ
’s
tips
respectively,
where
,
λ
,
"
""
• If π
λi,j
’s
parent
is
the
first
vertex
of
the
length
2
path
following
σ
,
then
.
i
i
,j
i
,j
i
i
i,j
i
i
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i
,j
i
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i
i
i,j
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4i−3
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, πrespectively,
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’s
where
,
are
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and
ithorn
and
are
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.,. λi ,jin
iwhich
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i ’s
,j,jtips
4i−3
idistinct,
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isλ
mapped
to by
the
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distinct,
and
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. are
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are
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irespectively,
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we
following
triples
of form
π4i−2
, π4i−1of, form π4i−2 , π4i−1 ,
•<
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wein map thorn
the tips
following
thorn
triples
i!! ,jλ’s
,jthe
,jare
is
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and
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iand
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•and
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the
of
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4i−2
4i−1
2 distinct,
4i−2
4i−1
δthorn
λi!!•
are
and
i"tips
<the
i""following
. tips4i−3
’sπof
parent
not
the first
first
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of the
the
path following
following
then its
its
i
γπi4i to triples
If
λλi,j
parent
isis not
the
vertex
of
path
σσ1j1,, then
,j π
andtriples
theform
tip••of
children
β, ,δ,πand
γ, which
are not
already
mapped
to
i,j
and
π,’s
to
to
4i
4i−3 .
1 1 the tip children of 1β, δ, and γ which are not already jmapped
•
If
λ
’s
parent
is
not
the
first
vertex
of
the
path
following
σ
,
then
its
π
toparent
the
tipischildren
of
β,’sγδ,vertex
and
which
are
not
already
to
Iftoλi,j
not
of
the
path
following
σjmapped
,respectively.
then
its
and π•4iand
the
children
of β,
δ,
and
which
are
not
already
mapped
to
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i
Figure 9: Here we depict the mapping of the (m + 1)th parasite subtree along the k-thorns with
superscripts 0, 1, and 2 of every assignment gadget. The expanded assignment gadget shows the
mapping details onto γi . This is similar to the mapping for the (m + 2)th parasite subtree which
codiverges along k-thorns of superscripts 0, 3, and 4.
11
1
1
1
1
1
1
For the (m + 1)th and (m + 2)th parasite subtrees, we map the edge following σj1 to the path
between τ11 and α10 at a cost of 2 loss events for each subtree. The first k-thorn triple of the (m+1)th
parasite subtree maps to the 0, 1, and 2 superscripted thorns of α1 , and the vertex which follows this
triple host switches to the edge preceding assignment gadget β1 . This continues through assignment
gadgets β1 , γ1 , δ1 , α2 , . . . , δn as depicted in Figure 9. Since each of the two parasite subtrees has 4n
k-thorn triples, and 4n − 1 vertices to following the triples in order to perform the host switches,
an additional cost of 24n(BL + 1) cospeciation, 8n − 2 duplication, 8n loss, and 8n − 2 host switch
events is incurred.
The total cost of this mapping is
0
(m + 2)(BD
+ 1) + (m + 2)(BL + 1) + 15m(BL + 1) + 2m + 24nk cospeciations
4m + 8n − 1 duplications
(m + 2) log n + 4m + 8n + 4 losses
3m + 8n − 2 host switches
Since this cost is within the budget tuple B, it follows that any instance of GCRDP constructed
from a satisfiable 3SAT instance has a valid solution.
3.1.2
Reverse Direction
Now we will show that the existence of a valid solution to an instance of GCRDP describing an
instance of 3SAT implies that the 3SAT instance is satisfiable.
First note that every parasite tree k-thorn must codiverge at least once since every such thorn
is made up of at least BL + 1 > BD + 1 vertices. It follows that for every 1 ≤ j ≤ m + 2, Ψ must
map at least one vertex of σj0 to its corresponding vertex on τ 0 yielding a cospeciation event, and
since the vertices, p1 , . . . , pm+1 are ancestors of σ10 , they must incur m + 1 duplications.
Consider the (m + 1)th and (m + 2)th parasite subtrees of the parasite tree. In order for a
solution to map any non-k-thorn vertex of these branches as a cospeciation event, we must map it
to a common ancestor of two assignment gadgets. However, such a mapping would exceed the loss
budget as the edge from this vertex to its tip child must work through one side of an assignment
gadget at a cost of 3 × (BL + 1) losses. Thus each of these 2 × (4n − 1) vertices must duplicate and
host switch. The purpose of this mapping is to force an ordering between all assignment gadgets
such that given any pair of assignment gadgets in the host tree, the k-thorns of one must occur
before all k-thorns of the other.
Note that we have now accounted for all but 3m duplications, and all but 3m host switches.
We claim that any solution must incur at least one duplication and host switch as it maps each
literal gadget to the host tree. Any mapping of a literal gadget λi,j which incurs no duplication
or host switch events must map the k-thorn λ0i,j as a cospeciation event to αi00 , βi00 , γi00 , or δi00 for
some i0 possibly equal to i. Without loss of generality, we will assume that the assignment gadget
is labeled αi00 . In order to avoid host switches we must map all vertices of λi,j to vertices of αi0 .
Now consider the parent of λ2i,j and λ3i,j which we will refer to as w. A solution which maps this
vertex as a cospeciation must map it to the parent of αi10 and αi20 , but such a solution requires that
w’s child lineage which contains λ2i,j incur at least BL + 1 loss events. Thus any valid solution
must duplicate and host switch at least one vertex of each λi,j , and since we have used all but 3m
duplications and host switches, each literal gadget duplicates and host switches exactly once.
12
!!
!!1
!!2
!!ki −1
!!ki
...
!
!
! !!
!!
!!1
!!1
!!2
!!2
!!ki −1
!!ki −1 !!ki
!!ki
!
!!
!!1
!!2
!
!!ki −1!!
!
!
!ki !1
!!2
!!ki −1
!!ki
!
!!
!
!1
!!2
!!ki −1
!!ki
Figure 10: Here we show a sample transformation of a parasite tip vertex to a disjunctive tip gadget.
We have now accounted for all host switches and duplications, so Φ must map all remaining
parasite vertices as cospeciation events. This implies that each vertex of a σj1 k-thorn gadget must
map to its corresponding vertex on some τi1 as a cospeciation, and that the length 2 path which
follows σj1 must codiverge with vertices of the balanced tree following τi1 .
Consider some λi,j which, without loss of generality, maps to an assignment gadget αi0 . Recall
that any attempt to codiverge the parent vertex of λ2i,j and λ3i,j will exceed the loss budget, so this
vertex must duplicate and host switch while all other vertices of the literal gadget codiverge with
vertices of αi .
Furthermore, we claim that this host switch cannot land outside of the assignment gadget αi0 .
Recall that the solution mapping of the (m + 1)th and (m + 2)th parasite subtrees imply that for
any pair of assignment gadgets in the host tree, all non-tip vertices of one gadget must occur before
all non-tip vertices of the other. Thus a host switch which lands outside of αi will place a switched
lineage with 2 × (BL + 1) internal vertices onto an edge that either precedes an assignment gadget
where any mapping will incur at least BL + 1 loss events or onto an edge which precedes a tip
vertex resulting in an excess of duplication events. Since the switched parasite lineage will contain
2 k-thorns of size k = BL + 1, the host switch must land on the edge preceding either αi10 or αi30
depending on ψ’s definition.
Note that this host switch requires that either all vertices of αi10 occur prior to all vertices of αi30
or that the reverse ordering occurs. If this condition cannot be met, it follows that no1 satisfactory
mapping exists to solve the GCRDP instance. Thus for any
1 solvable instance, a solution exists for
1 to
the associated 3SAT instance which we can obtain by assigning xi to true if αi1 occurs prior
1
αi3 and to false if we find the reverse ordering. To prove that this solution will
satisfy
the
3SAT
1
clauses, consider a clause Cj . Note that ψ is defined such that the property1 noted earlier where the
literal gadgets of a single clause must map to the assignment gadgets of the same variable implies
that some gadget λi,j exists which maps to αi . Since the solution mapping is valid, an ordering is
imposed between αi1 and αi3 which assigns the variable xi appropriately such that Cj is satisfied.
This is the case for all Cj and therefore a solution to the GCRDP instance implies the existence of
an assignment which satisfies the 3SAT clauses.
4
Polynomial Time Reduction of GCRDP to CRDP
We now show that an instance of GCRDP constructed from a 3SAT instance, as described above,
can be transformed into an instance of CRDP for which a solution exists if and only if the DCRP
instance is solvable.
13
Given an instance of GCRDP (H, P, ψ, B) corresponding to a 3SAT instance, we construct a
CRDP instance (H 0 , P 0 , ϕ, B 0 ) as follows. The transformation leaves the host tree unchanged so
H 0 = H. The parasite tree is modified, as illustrated in Figure 10 such that each tip ` ∈ L(P ) is
replaced with a disjunctive tip gadget. This gadget is similar in form to a k-thorn of size |ψ(`)| − 1
all of whose children are tips labeled `01 , `02 , . . . , `0|ψ(`)| in order of decreasing depth, and whose root
we label `0 . The tips are mapped by ϕ to a unique element in the set ψ(`), and we stipulate that
ϕ(`01 ) and ϕ(`02 ) are the most distant pair of host vertices in ψ(`) with respect to distance in the
host tree.
Finally, we define
X
X
B 0 = (BC , BD +
(|ψ(`)| − 1), BS +
(|ψ(`)| − 1), BL ).
`∈L(P )
4.1
`∈L(P )
Proof of Correctness
Given a solution, Ψ, for a GCRDP instance (H, P, ψ, B), we construct a solution to the CRDP
instance (H 0 , P 0 , ϕ, B 0 ) as follows. Let Φ(p) = Ψ(p) for all p ∈ P ∩ P 0 . For every disjunctive tip
gadget constructed to replace a tip ` ∈ L(P ), we map the vertices of the unique path from `0 to `0i
to the edge preceding hi where hi = Ψ(`), ϕ(li0 ) = hi , and every internal vertex v of the disjunctive
tip gadget host switches to the edge preceding some tip hi0 , where some descendant of v maps to
hi0 . The mapping of verticesPp ∈ P ∩ P 0 accumulates a cost of at most (BC , BD , BS , BL ) while the
disjunctive tip gadgets add `∈L(P ) (|ψ(`)| − 1) duplications and the same number of host switches.
The total cost is upper bounded by B 0 , and thus there exists a solution to the CRDP instance.
Now we demonstrate that the transformation above yields a solvable instance of CRDP only if
the transformed GCRDP instance is solvable. First we show that every internal vertex of a disjunctive tip gadget, which takes the place of a tip other than a thorn tip of some σj1 , must be mapped
as a duplication and host switch event. It will then be demonstrated that the vertices p1 , . . . , pm+1
must be mapped to the edge child of the host dummy root at a cost of m + 1 duplications, and
finally, that the solution must map each σj1 to some τi1 as cospeciation events. With these observations, we then show that defining Ψ(p) = Φ(p) for p ∈ P ∩ P 0 and Ψ(`) = Φ(`0 ) for ` ∈ L(P ), will
yield a valid solution to the GCRDP instance.
Consider an internal vertex v of a disjunctive tip gadget, which replaces a tip that is not part
of some σj1 . Note that in our definition of ψ, no two vertices exist together in some ψ(`) which are
both descendants of the same τi1 k-thorn gadget, so in order to map v such that only a cospeciation
is incurred, it must be mapped to an ancestor of two gadgets of form τi1 which would incur BL + 1
losses yielding an invalid solution. Thus it follows that these vertices, of which there are
X
(|ψ(`)| − 1)
l∈L(P )−{thorn tips of some σj1 }
must each incur a duplication and host switch. Furthermore, for any such mapping, it is always
desirable to perform these host switches on edges which precede tips in order to avoid loss events,
which yields a mapping similar to that described earlier.
We then note that in order to map a vertex of the initial parasite tree path, pj , to a descendant
of τ 0 , σj0 must duplicate and host switch all of its thorns in order to ensure that its tips properly
0
map with the tips of τ 0 . Since we specified that k = BD
+ 1 for the k-thorns, τ 0 and all σj0
14
(1 ≤ j ≤ m + 2), host switching all thorns of any σj0 will incur too many duplications for even the
CRDP instance, and therefore the path (p1 , . . . , pm+1 ) must accumulate m + 1 duplications.
Subtracting the host switch and duplication events due to the disjunctive tip gadgets below
thorns of form σj1 and the length m + 1 path, the remaining cost budget for both host switches and
duplications is
X
(|ψ(`)| − 1).
3m + 8n − 2 +
`∈{thorn tips of some σj1 }
Since every host switch requires an accompanying duplication, any additional duplication which
does not incur a host switch is no less expensive than a duplication followed by a host switch.
If we map each σj1 to some τi1 (1 ≤ i ≤ n and 1 ≤ j ≤ m + 2) and host switch the disjunctive
tip gadgets as described earlier, then each thorn contributes a cost of two duplications and host
switches. Now if we consider any mapping of a thorn vertex in σj1 which avoids this host switch,
we will show that such a mapping incurs an additional duplication and some number of loss events.
Let u and v be the two internal vertices of the disjunctive tip gadget associated with a thorn tip
of σj1 where v is a descendant of u, let `01 , `02 , and `03 be the disjunctive gadget’s tips in order of
increasing depth, and label ϕ(`0k ) = hk (k = 1, 2, 3). In order to avoid a single host switch at u
or v, Φ must map this vertex to a common ancestor vertex or edge of h1 and h2 . Mapping to the
latest common ancestor would yield a cospeciation, but the parent of u (an internal vertex of the
k-thorn gadget) would have to incur a duplication (else exceed BL losses) while an earlier common
ancestor would yield an extra duplication. In addition to this, the edges (v, `01 ) and (v, `02 ) would
each incur at least one loss at the parents of h1 and h2 . To avoid a host switch at u, the vertex
must map to a common ancestor of h1 , h2 , and h3 , but recall that `01 and `02 are defined such that
h1 and h2 are pairwise the most distant vertices in the image of this disjunctive tip gadget’s tips.
It follows that the common ancestor of h1 , h2 , and h3 is the same as the common ancestor of h1
and h2 , so if we wish to save host switches at u and v, we cannot codiverge at u, and must incur
an extra duplication in addition to those accumulated by avoiding the host switch at v.
Thus, the proposed solution, Ψ, to the GCRDP instance solution, where Ψ(p) = Φ(p) for
p ∈ P ∩ P 0 and Ψ(`) = Φ(`0 ) for ` ∈ L(P ) will incur a cost within the budget tuple B. Furthermore,
a valid CRDP instance solution requires that Φ(p) = h implies that there exists a tip descendant
p0 of p and a tip descendant h0 of h such that ϕ(p0 ) = h0 , so it follows that Ψ is valid in that it
properly extends the tip mapping function ψ.
References
[1] Michael Charleston. Jungles: A new solution to the hostparasite phylogeny reconciliation problem. Mathematical Biosciences, 149:191–223, 1998.
[2] Ran Libeskind-Hadas and Michael Charleston. On the computational complexity of the reticulate cophylogeny reconstruction problem. Journal of Computational Biology, 16(1):105–117,
2009.
15