An arborescence in a digraph is a tree directed away from its root. A classical theorem of Edmonds characterizes which digraphs have λ arc-disjoint arborescences rooted at r. A similar theorem of Menger guarantees that λ strongly arc disjoint rv-paths exist for every vertex v, where “strongly” means that no two paths contain a pair of symmetric arcs. We prove that if a directed graph D contains two arc-disjoint spanning arborescences rooted at r, then D contains two such arborences with the property that for every node v the paths from r to v in the two arborences satisfy Menger's theorem

Edmonds

Gabow

Giacomo Zambelli

Huck

Itai

Livio Colussi

Menger

Michele Conforti

Whitty

Colussi, Livio

Conforti, Michele

Zambelli, Giacomo

LSE Research Online

Disjoint paths in arborescences

An arborescence in a digraph is a tree directed away from its root.A classical theorem of Edmonds
characterizes which digraphs have \lambda arc-disjoint arborescences rooted at r. A similar theorem of
Menger guarantees that \lambda strongly arc disjoint rv-paths exist for every vertex v, where “strongly”
means that no two paths contain a pair of symmetric arcs.
We prove that if a directed graph D contains two arc-disjoint spanning arborescences rooted at r,
then D contains two such arborences with the property that for every node v the paths from r to v in
the two arborences satisfy Menger’s theorem

COLUSSI, LIVIO

CONFORTI, MICHELANGELO

ZAMBELLI, GIACOMO

Archivio istituzionale della ricerca - Università di Padova

Disjoint Paths in Arborescences

AbstractAn arborescence in a digraph is a tree directed away from its root. A classical theorem of Edmonds characterizes which digraphs have λ arc-disjoint arborescences rooted at r. A similar theorem of Menger guarantees that λ strongly arc disjoint rv-paths exist for every vertex v, where “strongly” means that no two paths contain a pair of symmetric arcs.We prove that if a directed graph D contains two arc-disjoint spanning arborescences rooted at r, then D contains two such arborences with the property that for every node v the paths from r to v in the two arborences satisfy Menger's theorem

Elsevier - Publisher Connector 

Discrete Mathematics 292 (2005) 187–191
www.elsevier.com/locate/disc
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Disjoint paths in arborescences
Livio Colussia, Michele Confortia, Giacomo Zambellib
aDipartimento di Matematica Pura ed Applicata, Università di Padova, Via Belzoni 7, 35131 Padova, Italy
bDepartment of Combinatorics and Optimization, University of Waterloo, 200 University Avenue West,
Waterloo, Ontario, Canada, N2L 3G1
Received 16 October 2002; received in revised form 29 November 2004; accepted 16 December 2004
Abstract
An arborescence in a digraph is a tree directed away from its root. A classical theorem of Edmonds
characterizes which digraphs have  arc-disjoint arborescences rooted at r. A similar theorem of
Menger guarantees that  strongly arc disjoint rv-paths exist for every vertex v, where “strongly”
means that no two paths contain a pair of symmetric arcs.
We prove that if a directed graph D contains two arc-disjoint spanning arborescences rooted at r,
then D contains two such arborences with the property that for every node v the paths from r to v in
the two arborences satisfy Menger’s theorem.
© 2005 Published by Elsevier B.V.
Keywords: Disjoint spanning arborescences
1. Introduction
Given a digraph D = (V ,A) and a subset S of V, deﬁne −D(S) to be the subset of A
with the head in S and the tail in V \S and −D(S) = |−D(S)|. Let +D(S) = −D(V \S),
+D(S)= |+D(S)|.
Let r be a node of D. An arborescence rooted at r is a subgraph F = (V (F ),E(F )) of
D which contains r, is connected and −F (r) = 0, while −F (v) = 1 for every other node of
V (F). The arborescence F is spanning if V (F)= V .
E-mail address: conforti@math.unipd.it (M. Conforti).
0012-365X/$ - see front matter © 2005 Published by Elsevier B.V.
doi:10.1016/j.disc.2004.12.005
188 L. Colussi et al. / Discrete Mathematics 292 (2005) 187–191
The following are two basic results on graph connectivity:
Theorem 1 (Edmonds [1]). A digraph D = (V ,A) with a speciﬁed node r contains 
pairwise arc-disjoint spanning arborescences rooted at r if and only if −D(S) for every∅ = S ⊆ V \r .
Two arcs are symmetric if they have the same endnodes but have opposite orientations. In
a digraph two paths are strongly arc-disjoint if they are arc-disjoint and they do not contain
a pair of symmetric arcs.
Theorem 2 (Menger [7]). A digraphD=(V ,A)with two speciﬁed nodes r and v contains
 pairwise strongly arc-disjoint paths from r to v if and only if −D(S) over all S ⊆ V \r
with v ∈ S.
The following conjecture, if true, provides a strengthening of both Theorems 1 and 2.
Conjecture 1. A digraph D = (V ,A) with a speciﬁed node r contains  pairwise arc-
disjoint spanning arborescences rooted at r such that, for every v ∈ V \r , the  paths from
r to v in each of these arborescences are strongly arc-disjoint if and only if −D(S) for
every ∅ = S ⊆ V \r .
Note that Conjecture 1 does not require the  arborescences to be strongly arc-disjoint.
Conjecture 1 obviously implies Theorem 1. That it implies Theorem 2 can be seen as
follows: Let D′ = (V ,A′) be obtained from D by adding  arcs from v to each node
x ∈ V \{r, v}. Then −D(S) over all S ⊆ V \r with v ∈ S if and only if −D′(S) over
all S ⊆ V \r and D contains  pairwise strongly arc-disjoint paths from r to v if and only if
D′ contains  pairwise arc-disjoint spanning arborescences rooted at r such that, for every
v ∈ V \r , the  paths from r to v in each of these arborescences are strongly arc-disjoint.
Although we cannot settle Conjecture 1 in the general case, we give below a proof when
= 2.
There is a known conjecture (see [2,6]) that is a undirected counterpart of Conjecture 1.
Given a undirected graphG= (V ,E) and a subset S = ∅ ofV, letG(S) be the set of edges
of E with one endnode in S and the other in V \S and G(S)= |G(S)|.
Conjecture 2. An undirected graphG=(V ,E)with a speciﬁed node r contains  spanning
trees such that, for every v ∈ V \r , the  paths from r to v in each of these trees are pairwise
edge-disjoint if and only if G(S) for every ∅ = SV \r .
Indeed,Conjecture 2 is a special case ofConjecture 1.To see this, given a graphG=(V ,E)
construct a digraph D = (V ,A) on the same node set by introducing a pair of symmetric
arcs (u, v), (v, u) for every edge uv of G. Given  spanning arborescences in D satisfying
Conjecture 1, the corresponding  spanning trees in G satisfy Conjecture 2. So Conjecture
1 implies Conjecture 2. In fact, the two conjectures are equivalent if all arcs in D come
in symmetric pairs. Again, Conjecture 2 has been proved only for  = 2 using depth ﬁrst
search [6].
L. Colussi et al. / Discrete Mathematics 292 (2005) 187–191 189
Similar results are known for the case where “strongly arc-disjoint paths” are replaced by
“internally disjoint paths” in Conjecture 1 (where two paths are internally disjoint if they
have no node in common, except possibly the ends).Whitty [8] proved the internally disjoint
version of the conjecture for = 2. A simpler proof is due to Huck [4]. Recently Huck [5]
found a counterexample to the internally disjoint version of the conjecture when > 2.
2. Proof of Conjecture 1 for = 2
If G contains two arc-disjoint spanning arborescences F1, F2 rooted at r, then, for all
S ⊆ V \r and i = 1, 2, |−D(S) ∩ A(Fi)|1; thus, −D(S)2.
For the converse, from Theorem 1 we may assume w.l.o.g. that the digraphD= (V ,A) is
the union of two arc-disjoint spanning arborescences rooted at r, that is, −D(r)=0, −D(v)=2
for every v ∈ V \r , and −D(S)2 for every S ⊆ V \r . So the arcs of D are partitioned in
pairs having the same head. Arcs in the same pair are mates. We may also assume w.l.o.g.
that +D(r) consists of two parallel arcs, say a and a′ with r ′ as head. If not, we may add a
new node r¯ and two parallel arcs from r¯ to r; one can easily verify that the case  = 2 of
Conjecture 1 holds for the new digraph D′ with speciﬁed node r¯ if and only if it holds for
D with speciﬁed node r.
Given an arborescence F = (V (F ),A(F )) of D, letD\F = (V ,A\A(F)). Assume now
that F satisﬁes the following.
Property 1. −D\F (S)1 for every S ⊆ V \r .
(That is, D\F contains a spanning arborescence.)
A subset of V \r is critical if it satisﬁes Property 1 with equality; the unique arc ofD\F
entering a critical set is called special. Since −D(v) = 2, every node v in V (F)\r belongs
to a critical set.
By submodularity of function −(·), if S and S′ are critical sets and S ∩ S′ = ∅, then
S ∩ S′ and S ∪ S′ are also critical. So if e is a special arc, there is a unique maximal critical
set Se(F ) entered by e.
Claim 1. Let e=(u, v)and e′=(u′, v′) be two special arcs. Ifu′∈Se(F ) thenSe′(F )Se(F ).
Proof. If u′ ∈ Se(F ) then Se(F ) ∪ Se′(F ) is critical and is entered by e. Since Se(F ) is
maximal, then Se(F )= Se(F ) ∪ Se′(F ). Since u′ /∈ Se′(F ), then Se′(F )Se(F ). 
A boundary node is a node v ∈ V (F) connected by an arc (v,w) to a nodew /∈V (F).
Let |V | = n and let F1, . . . , Fn−1 be arborescences rooted at r constructed as follows:
Let F1 be the arborescence with V (F1)= {r, r ′}, A(F1)= a and i = 1.
While i < n− 1, among all sets Se(Fi) that contain a boundary node v ∈ Se(Fi), pick
one which is inclusionwise minimal and let (v,w) be an arc such that w /∈V (Fi). Let
Fi+1 be obtained from Fi by adding node w and arc (v,w), set i = i + 1.
190 L. Colussi et al. / Discrete Mathematics 292 (2005) 187–191
We prove that Fn−1 can indeed be constructed by the above rule and that F = Fn−1 and
F ′ =D\F satisfy Conjecture 1. Note that by construction, F1 satisﬁes Property 1 and r is
not a boundary node.
Assume Fi , i < n − 1 satisﬁes Property 1. So Fi contains at least one boundary node.
Since every node in V (Fi)\r belongs to a critical set, the above procedure can be carried
out to construct Fi+1.
We now show that if Fi satisﬁes Property 1, then Fi+1 satisﬁes Property 1. This is
equivalent to showing that the arc (v,w) added to Fi by the above procedure is not
special.
Let Se(Fi) be the minimal critical set containing v. Assume (v,w) is special. Then by
Claim 1, S(v,w)(Fi)Se(Fi). Let SN = S(v,w)(Fi)\V (Fi) and SF = S(v,w)(Fi) ∩ V (Fi).
Both SN and SF are nonempty since w /∈V (Fi) and S(v,w)(Fi) is critical. Furthermore SN
is not a critical set, for it does not contain any node in V (Fi). So there exists one arc (y, z),
where y ∈ SF and z ∈ SN . Thus y is a boundary node in S(v,w)(Fi) and S(v,w)(Fi)Se(Fi),
contradicting the minimality of Se(Fi).
This shows that F and F ′ are arc-disjoint spanning arborescences of D.
We ﬁnally show that for every node z the two rz-paths in F and F ′ cannot contain a pair
of symmetric arcs.
Assume there exists a node z such that the rz-paths PFz and PF
′
z in F and F ′ contain
one of the arcs (u, v) and (v, u), respectively. Let (u′, v) be the mate of (u, v) (obviously
(u′, v) ∈ PF ′z ), let (v,w) be the arc in PFz with v as tail (possibly u′ = r or w = z) and
assume (v,w) ∈ A(Fi+1)\A(Fi).
Let u′ = z0, v = z1, u= z2, . . . , zm−1, zm = z the u′z-subpath of PF ′z . Since w /∈V (Fi)
both arcs entering z are in D\Fi and z /∈V (Fi). Since u ∈ V (Fi) there exist two nodes zk ,
zk+1 of lowest index such that zk is in V (Fi) and zk+1 is not (clearly, k2). Then zk is a
boundary node for Fi .
Since, for 1jk, all sets {zj } are critical, then all arcs (zj−1, zj ) are special, and
each set S(zj−1,zj )(Fi) contains the head zj of the next arc. By Claim 1, for 2jk,
S(zj−1,zj )(Fi)S(zj−2,zj−1)(Fi). So S(zk−1,zk)(Fi)Se(Fi) and contains the boundary node
zk , contradicting the minimality of Se(Fi). 
The construction in the proof can be implemented in polynomial time. Gabow [3] gave
an O(2n2) algorithm to ﬁnd  arc-disjoint arborescences in a digraph D; thus we may ﬁnd
two arc-disjoint spanning arborescences ofD in time O(n2), and assumeD is just the union
of such arborescences. We claim that our construction can be implemented, on such D, in
time O(n2) as well.
Notice that, at the ith iteration, if e = (u, v) is a special arc such that v is the unique
boundary node in Se(Fi), then Se(Fi) is inclusionwise minimal with such property; in fact,
if for some special arc e′, Se′(Fi) ⊆ Se(Fi) contains a boundary node, then v ∈ Se′(Fi) and
u /∈ Se′(Fi), so e′ = e.
Also, for any special arc e, if we denote by Ri(e) the set of nodes reachable from r in
D\(A(Fi) ∪ {e}), Se(Fi)= V \Ri(e).
In order to implement the construction in the proof, we need to show how to compute, at
the ith iteration, a minimal Se(Fi) containing a boundary node.
L. Colussi et al. / Discrete Mathematics 292 (2005) 187–191 191
Start from any boundary node v0, let (u0, v0) be the special arc entering v0, compute
Ri(u0, v0). Supposewehave computedRi(uj , vj ),wherevj is a boundary node and (uj , vj )
is a critical arc, 0j |V (Fi)|.
If S(uj ,vj )(Fi) = V \Ri(uj , vj ) does not contain any boundary node except vj , then
S(uj ,vj )(Fi) is minimal containing a boundary node.
Otherwise, choose a boundary node vj+1 = vj in V \Ri(uj , vj ), and let (uj+1, vj+1) be
the unique special arc entering vj+1. Compute the setR′ of nodes reachable fromRi(uj , vj )
in D\(A(Fi) ∪ {(uj+1, vj+1)}), and let Ri(uj+1, vj+1) := Ri(uj , vj ) ∪ R′.
Clearly, this procedure takes linear time at each iteration, and there are n− 1 iterations,
so the total running time is O(n2).
References
[1] J. Edmonds, Edge-disjoint branchings, in: R. Rustin (Ed.), Combinatorial Algorithms, Academic Press,
NewYork, 1973, pp. 91–96.
[2] A. Frank, Connectivity and Network Flows, Handbook of Combinatorics, 1995, pp. 111–177.
[3] H. Gabow, A matroid approach to ﬁnding edge connectivity and packing arborescences, J. Comput. System
Sci. 50 (1995) 259–273.
[4] A. Huck, Independent trees in graphs, Graphs Combin. 10 (1994) 29–45.
[5] A. Huck, Disproof of a conjecture about independent branchings in k-connected directed graphs, J. Graph
Theory 20 (1995) 235–239.
[6] A. Itai, M. Rodeh, The multi-tree approach to reliability in distributed networks, Inform. and Comput. 79
(1988) 43–59.
[7] K. Menger, Zur Allgemeinen Kurventheorie, Fund. Math. 10 (1927) 96–115.
[8] A. Whitty, Vertex-disjoint paths and edge-disjoint branchings in directed graphs, J. Graph Theory 11 (1987)
349–358.


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