In this paper, we have shown that Hill et al's conjecture [

1] and the underlying divide-and-conquer approach cannot be used to calculate the rSPR distance between two phylogenies exactly. To provide some intuition why this conjecture fails, consider the following. Let

be a cluster sequence with respect to Θ =

*solvable* for two rooted binary phylogenetic trees

and

. Calculating a maximum-agreement forest for each tree pair (

), taking their union, and, for each

*i* ∈; {1, 2, ...,

*t*}, joining the element containing

*a*
_{
i
} with the element containing

*ρ*
_{
i
} can potentially result in a set, say

, which contains an element that is a subset of {

*a*
_{1},

*a*
_{2}, ...,

*a*
_{
t
} ,

*ρ*
_{1},

*ρ*
_{2}, ...,

*ρ*
_{
t
}}. In the case of our counterexample,

contains one such element. Trivially, this element is not part of any agreement forest for
and
while
- {{*a*
_{1}, *a*
_{2}, *ρ*
_{1}, *ρ*
_{2}}} is precisely a maximum-agreement forest for
and
. Consequently, a divide-and-conquer approach that exactly calculates
needs to take into account the number of elements in
that are subsets of {*a*
_{1}, *a*
_{2}, ..., *a*
_{
t
} , *ρ*
_{1}, *ρ*
_{2}, ..., *ρ*
_{
t
}}; otherwise, the result may be an overestimate of the exact solution. Alternatively, one can approach the problem by finding a strategy which guarantees that no element in
is a subset of {*a*
_{1}, *a*
_{2}, ..., *a*
_{
t
} , *ρ*
_{1}, *ρ*
_{2}, ..., *ρ*
_{
t
} }. This is the underlying idea of Theorem 4 which uses a slightly more restricted version of Hill et al's conjecture and finally gives the desired outcome. Hence, decomposing
and
into a cluster sequence with respect to Θ = *subtree-like* can be used to speed up the exact calculation of
.

However, for practical problem instances, it may be unlikely to find many subtree-like clusters. For example, the two phylogenies shown in Figure 1 do not have any common subtree-like cluster. This is due to the restricted definition of such a cluster which requires that a vertex whose set of descendants is a common cluster of two rooted binary phylogenetic *X*-trees
and
has the same parent vertex than a common subtree of
and
. To lessen this problem, an alternative approach--that has recently been published by Linz and Semple [6]--can be applied. This paper describes a more general divide-and-conquer approach that exactly computes the rSPR distance between
and
for when a cluster sequence
with respect to Θ = *minimal* for
and
is given. Loosely speaking, the authors calculate a so-called minimum-weight partition
of *X* ∪ {*ρ*} ∪ {*a*
_{1}, *a*
_{2}, ..., *a*
_{
t
} , *ρ*
_{1}, *ρ*
_{2}, ..., *ρ*
_{
t
}} such that
contains an agreement forest (not necessarily a maximum-agreement forest) for each tree pair (
). To compute
, it has been shown that applying a 'bottom-up' approach which locally works on subtrees of each tree pair (
) guarantees that the number of elements in
that are subsets of {*a*
_{1}, *a*
_{2}, ..., *a*
_{
t
} , *ρ*
_{1}, *ρ*
_{2}, ..., *ρ*
_{
t
}} is maximized while |
| is minimized.