Graph optimization problems of common root
(Parts I and II)
Kristóf Bérczi⇤
András Frank⇤
MTA-ELTE Egerváry Research Group
Department of Operations Research
Eötvös Loránd University
Budapest, Hungary
[email protected]
MTA-ELTE Egerváry Research Group
Department of Operations Research
Eötvös Loránd University
Budapest, Hungary
[email protected]
Abstract: The paper considers the following three graph optimization problems: (A) the
existence of k disjoint arborescences in a digraph each having a specified number of root-edges,
(B) Ryser’s maximal term rank formula and its matroidal extension, (C) characterization of
degree-sequences of simple k-node-connected digraphs. It is shown that these apparently
independent problems have one common root and the answers can be obtained from the
general min-max result on supermodular function of [8].
Keywords: covering supermodular functions, packing arborescences, graphical degree sequences, term rank
1
Introduction
There are several frameworks relying on sub- and supermodular functions in combinatorial optimization.
Matroids, polymatroids, submodular flows provide general tools to solve optimization problems in which
typically the weighted versions are also nicely tractable. The paper [6] of the second author introduced
a model for supermodular functions in which cardinality optimization can be done but the weighted
versions are NP-complete. Special cases include, for example, characterizations of degree-sequences of
k-edge-connected digraphs. Frank and Jordán [8] extended this framework further and they derived a
min-max theorem on minimal coverings of supermodular bi-set functions by a digraph. Special cases
include, for example, characterizations of degree-sequences of k-node-connected digraphs. It should be
noted that the analogous problems of simple digraphs remained open. It turned out that the general
min-max theorem of [8] has several applications distinct from connectivity augmentation. For example,
it implied an extension of a deep theorem of Győri on optimal covering of polyominos. It also implies
min-max theorems on maximal Kt,t -free t-matchings of a bipartite graph. Recently, Soto and Telha
[21, 22] found an elegant generalization of Győri’s theorem.
The main goal of the present work is to exhibit new applications of [8]. First we prove a minimax
theorem on covering a supermodular function with a degree-constrained simple bipartite graph. As
an application, we derive a necessary and sufficient condition for the existence of k disjoint spanning
arborescences F1 , . . . , Fk of root r0 in a digraph D so that the number of roof-edges of Fi is mi for
i = 1, . . . , k, where m1 , . . . , mk are k given positive integers. Another consequence is a derivation of
the maximum term rank formula of Ryser. Ford and Fulkerson [4] remark in their classic book (p. 89)
that this theorem is not amenable to a flow approach and recently Pálvölgyi [15] verified this remark
by showing that the weighted maximal term rank problem is already NP-complete. Still, we are not
only able to show that Ryser’s theorem follows from our general supermodular covering theorem but a
matroidal extension of it could also be derived.
⇤ The
authors were supported by the Hungarian Scientific Research Fund - OTKA, K109240.
242
As a third application of the framework of [8], we generalize an important recent result of Hong,
Liu, and Lai [11] on the characterization of degree-sequences of simple strongly connected digraphs.
Furthermore, a characterization of degree-sequences of simple k-node-connected digraphs is also exhibited
while the analogous problem for k-edge-connected digraphs remains open. Algorithmic aspects and
further extensions will be discussed in the journal version of the present work.
All the notions and notation not mentioned explicitly in the paper can be found in the book of the
second author [7].
2
Preliminaries
2.1
Partially ordered sets
Let P = (P, ) be a partially ordered set, or poset for short. A chain or an antichain is a set of pairwise
comparable or incomparable elements of P , respectively. Let C and A denote the sets of chains and
antichains. An antichain of maximum size is called a D-antichain.
Mirsky verified the following connection between ⌫P and ⌧P [14].
Theorem 1 (Mirsky) In a partially ordered set P = (P, ), the maximum size of a chain equals the
minimum number of antichains covering P .
Assume now that a weight function w : P ! R is given. An antichain ofPmaximum weight is called
a WD-antichain. An antichain cover is a function c : A ! R such that A3p c(A) w(p) for each
p 2 P . The weight of the antichain cover is c(A).
The weighted counterpart of Theorem 1 is as follows.
Theorem 2 (Mirsky, weighted version) For a partially ordered set P = (P, ) and weight function
w : P ! Z+ , the maximum weight of a chain equals the minimum weight of an antichain cover.
The following lemma will play an important role later on.
Lemma 3 Let P = (P, ) be a partially ordered set and w : P ! R a weight function. Assume
that p1 , . . . , pk are minimal elements of P such that for each i = 1, . . . , k there exists a WD-antichain
containing pi . Then there exists a WD-antichain containing all of the pi ’s simultaneously.
Proof: For antichains A1 , A2 2 A, let A1 ^ A2 denote the set of minimal elements of A1 [ A2 , while
A1 _ A2 denotes the set of maximal elements. We claim that w(A1 ) + w(A2 ) = w(A1 ^ A2 ) + w(A1 _ A2 ).
To prove this, we check the contribution of a 2 P to the two sides.
If a 2
/ A1 [ A2 or w(a) = 0 then the contribution of a to both sides is 0.
Let a 2 A1 [ A2 be such that w(a) > 0. If a 2 A1 \ A2 then a is both minimal and maximal in
A1 [ A2 , hence its contribution to both sides is 2w(a). Assume now that a 2 A1 , a 2
/ A2 . As A2 is a
WD-antichain, there is an element a0 2 A2 that is comparable with a, say, a0 < a. As A1 [ A2 is the
union of two antichains, its longest chain may contain at most two elements, hence a is maximal, a0 is
minimal in A1 [ A2 .
By w(A1 ) + w(A2 ) = w(A1 ^ A2 ) + w(A1 _ A2 ), A1 ^ A2 and A1 _ A2 are WD-antichains. Let Ai
be a WD-antichain containing pi . Then A1 ^ A2 ^ · · · ^ Ak is a WD-antichain that, by their minimality,
contains all of the pi ’s, thus concluding the proof. ⇤
2.2
Covering supermodular functions
Let D = (S, T ; A⇤ ) be a complete bipartite digraph in which each edge is oriented from S to T . We denote
S [ T by V . Subsets X, Y ✓ V are comparable if X ✓ Y or Y ✓ X. A family of pairwise comparable
sets is called a chain. The sets are ST -intersecting if X \ Y \ T 6= ; and properly ST -intersecting
243
if, in addition, they are not comparable. The sets are ST -crossing if X \ Y \ T 6= ; and S 6✓ X [ Y and
properly ST -crossing if they are not comparable. A set is trivial if X \ T = ; or S ✓ X. A function
p : 2V ! R is called positively ST -crossing supermodular if p(X) + p(Y )  p(X \ Y ) + p(X [ Y ) for
all ST -crossing pair X, Y ✓ V with p(X) > 0, p(Y ) > 0. X and Y are ST -independent if they are not
ST -crossing, that is, X \ Y \ T = ; or S ✓ X [ Y . A family of sets is ST -independent if its members
are pairwise ST -independent.
Let p : 2V ! Z+ be a positively ST -crossing supermodular function which is 0 on trivial sets. We
say that a vector z : A⇤ ! Z+ covers p if %z (X) p(X) for each X ✓ V . As p is 0 on trivial sets, there
exists z covering p. If z corresponds to the incidence vector of a subset of edges A ✓ A⇤ , then we say
that A covers p.
P
For a set F ✓ 2V we call X2F p(X) the p-value of F and denote it by pe(F ). Let ⌧p be the minimum
value of a cover of p, that is,
⌧p = min{e
z (A⇤ ) : z : A⇤ ! Z+ covers p}.
Meanwhile, the maximum p-value of an ST -independent family is denoted by ⌫p , so
⌫p = max{e
p(I) : I ✓ 2V ST -independent}.
The following min-max theorem appeared in [8].
Theorem 4 (Frank, Jordán) Let D = (S, T ; A⇤ ) be a complete bipartite digraph in which each edge is
oriented from S to T . Let V = S [T and assume that p : 2V ! R is a positively ST -crossing supermodular
function that is non-positive on trivial sets. Then ⌧p = ⌫p .
It is worth mentioning that Theorem 4 can be reformulated in terms of covering positively crossing
bi-set functions. Given a ground-set V , we call a pair X = (XO , XI ) of subsets a bi-set if XI ✓ XO ✓ V
where XO is the outer member and XI is the inner member of X. The set of bi-sets on ground-set
V is denoted by P2 (V ) = P2 . A bi-set X = (XO , XI ) with XO = XI can be naturally identified with
the normal set XI , hence the following notation can be also interpreted for sets. The intersection and
the union of bi-sets X, Y is defined as X \ Y = (XO \ YO , XI \ YI ) and X [ Y = (XO [ YO , XI [ YI ),
respectively. We write X ✓ Y if XO ✓ YO , XI ✓ YI . Note that this relation is a partial order on P2 .
Accordingly, when X ✓ Y or Y ✓ X, we call X and Y comparable. A family of pairwise comparable
bi-sets is called a chain. Two bi-sets X and Y are independent if XI \ YI = ; or V = XO [ YO . A
family of bi-sets is independent if its members are pairwise independent. Two bi-sets are intersecting
if XI \ YI 6= ; and properly intersecting if, in addition, they are not comparable. Two bi-sets are
crossing if XI \ YI 6= ; and XO [ YO 6= V and properly crossing if they are not comparable. A bi-set
(XO , XI ) is trivial if XI = ; or XO = V .
A directed edge enters or covers X if its head is in XI and its tail is outside XO . An edge covers
a family of bi-sets if it covers each member of the family. For a bi-set function p, a digraph D = (V, A)
is said to cover p if %D (X) p(X) for every X 2 P2 (V
P) where %D (X) denotes the number of edges of
D covering X. For a vector z : A ! R, let %z (X) := [z(a) : a 2 A, a covers X]. A vector z : A ! R
covers p if %z (X) p(X) for every X 2 P2 (V ).
A bi-set function p is said to satisfy the supermodular inequality on X, Y 2 P2 if
p(X) + p(Y )  p(X \ Y ) + p(X [ Y ).
(1)
If the reverse inequality holds, we speak of the submodular inequality. p is said to be fully supermodular (or supermodular, for short) if it satisfies the supermodular inequality for every pair of bi-sets
X, Y . If (1) holds for intersecting (crossing) pairs, we speak of intersecting (crossing) supermodular functions. Analogous notions can be introduced for submodularity. Sometimes (1) is required only
for pairs with p(X) > 0 and p(Y ) > 0 in which case we speak of positively supermodular functions.
Positively intersecting or crossing supermodular functions are defined analogously.
Although we will only use the result of [8], we state the theorem here in a more general form. Let
D = (V, A) be a directed graph. Assume that p : P2 (V ) ! R is a bi-set function that is non-positive on
trivial bi-sets, and
244
(a) for any pair of non-trivial bi-sets X ✓ Y ✓ V with p(X), p(Y ) > 0, there exists an arc a 2 A covering
X and Y simultaneously, and
(b) for any pair of bi-sets X, Y ✓ V with p(X), p(Y ) > 0, if there exists an arc a 2 A covering X and Y
simultaneously, then p(X) + p(Y )  p(X \ Y ) + p(X [ Y ).
Condition (a) altogether with p being non-positive on trivial bi-sets imply that there exists z covering
p. Let ⌧p be the minimum value of a cover of p, while the maximum p-value of an independent bi-set
family is denoted by ⌫p .
Theorem 5 Let D = (V, A) be a bipartite digraph, p : P2 (V ) ! R a bi-set function that is non-positive
on trivial bi-sets and satisfies (a) and (b). Then ⌧p = ⌫p .
Proof:
Let z : A !PZ+ be a cover of p and let I ✓ P2 be an independent family. Then ze(A)
P
[%z (X) : X 2 I]
[p(X) : X 2 I] = pe(I), hence ⌧p ⌫p .
To see equality, for each a 2 A let pa denote the function
(
(p(X) 1)+ , if a covers X
pa (X) =
p(X)
otherwise.
As pa  p, we have ⌫p
1  ⌫pa  ⌫p . We call an arc reducing if ⌫pa < ⌫p .
Claim 6 If ⌫p > 0, then there exists a reducing arc.
Proof: Suppose indirectly that ⌫pa = ⌫p for each a 2 A. That means that for each a 2 A, there
exists an independent family Ia with p(Ia ) = ⌫p and a does not cover any member of Ia . Let F be the
multi-union of sets Ia , that is, F contains as many copies of a bi-set X 2 P2 as the number of Ia ’s
containing X. Then pe(F) = |A|⌫p and each arc a enters at most |A| 1 members of F.
We may assume that each member of F has positive p-value as bi-sets of zero p-values can be left
out. Apply the uncrossing procedure as long as possible: if there are two bi-sets of positive p-values that
can be covered by an arc a 2 A, replace them with their union and intersection (onlyP
add those sets that
have positive p-value). The process will terminate after a finite number of steps as X2X |X|2 strictly
increases. Let F 0 denote the family thus obtained. By (b), pe(F 0 )
pe(F). Clearly, each arc covers at
most |A| 1 members of F 0 .
For each X 2 F 0 , let s(X) denote the number of occurrences of X in F 0 . We claim that in the poset
P = (F 0 , ✓) each chain has s-weight at most |A| 1. Indeed, if C is a chain of P with s(C)
|A|,
then F 0 contains |A| -not necessarily distinct- bi-sets that form a chain X1 ✓ · · · ✓ X|A| . By (a), there
exists an arc a 2 A covering the members of the chain simultaneously, contradicting the fact that each
arc covers at most |A| 1 members of F 0 .
By Theorem 2, there exist |A| 1 antichains in F 0 that together contains s(X) copies of X for each
X 2 F 0 . As pe(F 0 ) |A|⌫p , at least one of these antichains has p-value strictly larger than ⌫p . But the
members of an antichain of F 0 are independent, contradicting the definition of ⌫p . This completes the
proof of the claim. ⇤
We prove ⌫p
⌧p by induction on ⌫p . If ⌫p = 0, then p  0 and z ⌘ 0 covers p, hence ⌧p = 0.
Assume that ⌫p > 0. By Claim 6, there exists a reducing arc a 2 A. If we increase the value of z 0 (a) by
one in any cover z 0 of pa , then we get a cover of p, that is, ⌧p  ⌧pa + 1. By induction, ⌧pa = ⌫pa , so
⌧p 1  ⌧pa = ⌫p a  ⌫p 1, implying ⌧p  ⌫p . Hence the theorem follows. ⇤
245
3
Coverings by simple bipartite graphs
3.1
Base case
Theorem 7 Let S and T be non-empty disjoint sets, p a positively intersecting supermodular set-function
e S (S) = . There exists a simple bipartite
on T and mS : S ! Z+ a degree specification on S for which m
graph G = (S, T ; E) for which
|
G (Y
)|
p(Y ) holds for every Y ✓ T
(2)
and
dG (s) = mS (s) for every s 2 S
(3)
mS (s)  |T | for every s 2 S
(4)
if and only if
and
m
e S (X) +
q
X
[p(Ti )
|X|] 
(5)
i=1
holds for every sub-partition T = {T1 , . . . , Tq } of T and subset X of S. Inequality (5) is equivalent to
q
X
i=1
p(Ti )  m(S
e
X) + q|X|.
(6)
Proof: First we prove necessity. Suppose that G is is a graph with the requested properties. By the
simplicity of G, mS (s) = dG (s)  |T | holds for every s 2 S, that is, (3) is indeed necessary.
G has m
e S (X) edges ending in X. Furthermore, each Ti has at least p(Ti ) |X| neighbours in S X
implying that there are at least
Pqp(Ti ) |X| edges between Ti and S X. Therefore the total number
of edges is at least m
e S (X) + i=1 [p(Ti ) |X|], that is, (5) is also necessary.
Now we prove sufficiency. Recall that V = S [ T . A set Z ✓ V is called large if Z = V s for some
s 2 S, otherwise it is small. Consider the following set function on V .
(
if Z = V s is a large set for s 2 S,
mS (s),
0
p (Z) =
p(Z \ T ) |Z \ S|, if Z is small.
Condition (5), when applied to T = {T } and X = S s, implies that mS (s) p(T ) |S s| for each
s 2 S, hence p0 (Z) = mS (s) p(Z \ T ) |Z \ S| for every large set Z = V s. By choosing subpartition
{T 0 } together with X = S in (5), we get p(T 0 )  |S| for every T 0 ✓ T . If Z is a trivial set, then either
Z \ T = ; or S ✓ Z. In both cases, p0 (Z)  0 follows from the observation and p(;) = 0.
Let D = (S, T ; A⇤ ) be the complete bipartite digraph in which each arc is oriented from S to T . Note
that ⌧p0
as the set of every large sets is ST -independent with total p0 -value .
Claim 8 If ⌧p0 = , then there exists a – not-necessarily simple – bipartite graph G satisfying (2) and
(3).
Proof: Let z : A⇤ ! Z+ be a vector covering p0 such that ze(A⇤ ) = ⌧p0 = and let G = (S, T ; E) be a
bipartite graph in which there are z(uv) parallel edges between u 2 S and v 2 T . Note that ze(A⇤ ) =
implies dG (s) = z (s) = mS (s) for each s 2 S, hence (3) is satisfied.
Suppose indirectly that | G (Y )| < p(Y ) for some set Y ✓ T . Then 0 = %z (Y [ G (Y ))
p(Y )
| G (Y )| > 0, a contradiction, so (2) is also satisfied and the claim follows. ⇤
By Theorem 4 and Claim 8, it suffices to show that (5) implies ⌫p0 = .
Lemma 9 ⌫p0 = .
246
Proof: Theorem 4 and ⌧p0
implies ⌫p0
. Suppose indirectly that ⌫p0 > . Let I ✓ 2V be an ST 0
e
independent family with p (I) = ⌫p0 , modulo this let |I| be minimal. Define I1 = {Z 2 I : Z is large}
and I2 = {Z 2 I : Z is small}.
We claim that {Z \T : Z 2 I2 } is a subpartition of T . Indeed, if Z1 \Z2 \T 6= ; for some Z1 , Z2 2 I2 ,
then
p0 (Z1 ) + p0 (Z2 ) = p(Z1 \ T )
|Z1 \ S| + p(Z2 \ T )
 p((Z1 \ Z2 ) \ T )
|Z2 \ S|
|(Z1 \ Z2 ) \ S| + p((Z1 [ Z2 ) \ T )
|(Z1 [ Z2 ) \ S|
0
 p (Z1 \ Z2 ),
where the last inequality follows from p0 (Z1 [Z2 )  0 as Z1 [Z2 is trivial. Hence replacing Z1 and Z2 with
0
e0 0
e0
their intersection results in an independent family I 0 with
S p (I ) p (I) and |I | < |I|, a contradiction.
Recall that |S Z| = 1 for each Z 2 I1 . Let X = Z2I1 (S Z). As I is ST -independent, X ✓ Z
for each Z 2 I2 .
By the above observations, we have
X
< pe0 (I) =
p0 (Z)
Z2I
=
X
p0 (Z) +
Z2I1
 m(X)
e
+
contradicting (5), thus proving the lemma.
X
X
p0 (Z)
Z2I2
[p(Z \ T )
|X|],
Z2I2
⇤
Take an arbitrary bipartite graph G = (S, T ; E) satisfying (2) and (3). If there exist parallel edges
between nodes u 2 S and v 2 T , then, by mS (u)  |T | and dG (u) = mS (u), there exists a node w 2 T
such that uw 2
/ E. Delete an edge between u and v and add an edge uw to the graph. By repeating the
above step we get a simple graph G0 = (S, T ; E 0 ) still satisfying (2) and (3). This completes the proof of
the theorem. ⇤
3.2
Covering p by degree-specified bipartite graphs
Let S and T be two disjoint sets and V := S [ T . We are given a non-negative integer-valued degree
specification m : V ! Z+ whose restrictions to S and to T are denoted by mS and mT , respectively. We
e T (V ) and this common value will be denoted by . We say that
assume throughout that m
e S (V ) = m
the pair m or (mS , mT ) is a degree specification and that a bipartite graph G = (S, T ; E) fits this
degree specification if dG (v) = m(v) holds for every node v 2 V . Let G(mS , mT ) denote the set of simple
bipartite graphs fitting (mS , mT ).
Gale [10] and Ryser [19] found in an equivalent form the following characterization.
Theorem 10 (Gale and Ryser) There is a bipartite graph G fitting the degree specification, that is,
G(mS , mT ) is non-empty if and only if
m
e S (X) + m
e T (Z)
|X||Z| 
for every X ✓ S, Z ✓ T.
(7)
Moreover, (7) follows from its special case when X consists of the i largest values of mS and Z consists
of the j largest values of mT (i = 1, . . . , |S|, j = 1, . . . , |T |).
247
Theorem 11 Let p be a positively intersecting supermodular function p on T and let (mS , mT ) be a
degree specification. There is a simple bipartite graph G = (S, T ; E) fitting (mS , mT ) (that is, a member
of G(mS , mT )) wich covers p if and only if
e T (Z)
m
e S (X) + m
|Z||X| +
q
X
[p(Ti )
|X|] 
for every X ✓ S, Z ✓ T
(8)
i=1
where {Z, T1 , . . . , Tq } is a subpartition of T in which Z may be empty and q may be zero. The inequality
in (8) is equivalent to
q
X
m
e T (Z) +
p(Ti )  m
e S (S X) + (q + |Z|)|X|.
(9)
i=1
Proof: Necessity. Suppose that G is a member of G(mS , mT ) covering p. Furthermore each Ti has at
least p(Ti ) |X| neighbours in S X from which there are at least p(Ti ) |X| edges
Ti and
Pbetween
q
S X. Therefore the total number of edges is at least m
e S (X) + m
e T (Z) |X||Z| + i=1 [p(Ti ) |X|]
that is, (8) is indeed necessary.
Sufficiency. Observe first that (8) in the special case when q = 0 gives back (7), and hence G(mS , mT )
is non-empty by Theorem 10. Also, for any element t 2 T , by applying (8) to q = 1, T1 = T t, X = ;,
we obtain that m
e ( T t) + p(t)  , that is, p(t)  mT (t) for every t 2 T .
For each t 2 T , increase the value p(t) to mT (t) and let p1 denote the resulting function. Then p1 is
also intersecting supermodular, and (8) is equivalent to the inequality obtained from (5) by applying it
to p1 .
Observe that (8) when applied to X = {s}, Z = T and q = 0 requires that mS (s) + m
e T (T ) |T |  ,
that is, mS (s)  |T |. By applying Theorem 7 to p1 in place of p, we obtain that there isPa simple
bipartite P
graph G covering p for which
for every s 2 S. Furthermore, |E| = [dG (s) :
P dG (s) = mS (s) P
s 2 S] = [mS (s) : s 2 S] = = [mT (t) : t 2 T ]  [dG (t) : t 2 t] = |E| from which we must have
dG (t) = mT (t) for every t 2 T . ⇤
3.3
Upper and lower bounds on T
In Theorem 7 the graph in question could be chosen to be simple by greedily replacing parallel edges
by new ones having no parallel copies without destroying (2) and (3). Finding a simple graph becomes
significantly more difficult if the degree of each node in T is bounded. Note that lower bounds on the
degrees of nodes in T can be built in the values of p on singletons, hence only upper bounds are considered
explicitly.
Theorem 12 Let S and T be non-empty disjoint sets, p a positively intersecting supermodular setfunction on T , mS : S ! Z+ a degree specification on S for which m
e S (S) = and g : T ! Z+ an upper
bound function for which p(t)  g(t) for each t 2 T . There exists a simple bipartite graph G = (S, T ; E)
for which
| G (Y )| p(Y ) holds for every Y ✓ T ,
(10)
dG (s) = mS (s) for every s 2 S
(11)
dG (t)  g(t) for every t 2 T
(12)
and
if and only if
m
e S (X) +
q
X
[p(Ti )
i=1
248
|X|] 
(13)
holds for every sub-partition T = {T1 , . . . , Tq } of T and subset X of S, and
ge(T
Y)
[m
e S (X)
|Y ||X|] + [
q
X
p(Yi )
q|X|]
(14)
i=1
Y with |Yi |
holds for every Y ✓ T , subpartition {Y1 , . . . , Yq } of T
Inequality (13) is equivalent to
q
X
p(Ti )  m
e S (S X) + q|X|.
2 (i = 1, . . . , q) and X ✓ S.
(15)
i=1
Proof: Suppose that G is is a graph with the requested properties. Necessity of (13) was already proved
in Theorem 7. G has m
e S (X) edges ending in X. As G is simple, at least m
e S (X) |Y ||X| of these edges
have its other end in T Y . Furthermore, each Yi has at least p(Yi ) |X| neighbours in S X implying
Yi and S X. Therefore the total number of edges
that there are at least p(Yi ) |X| edges between P
q
ending in T Y is at least [m
e S (X) |Y ||X|] + [ i=1 [p(Yi ) |X|], that is, (14) is also necessary. It
is worth mentioning that (14) is also valid for a subpartition {Y1 , . . . , Yq } of T Y possibly containing
singletons. If Yi = {t} for some t 2 T , then deleting t from T Y decreases the left hand side of (14) by
g(t) while the right size decreases by p(t)
P and, as g(t) p(t), we get a stronger inequality.
Now we prove sufficiency. By (13), t2T p(t)  . We discuss two cases.
P
Case 1 t2T p(t) = .
By choosing Y = ; and X = {s}, (14) gives m(s)  |T |. By Theorem
P 7, there exists a simple bipartite
graph G = (S, T ; E) satisfying conditions (10) and (11). However, t2T p(t) = implies dG (t) = p(t) 
g(t) for each t 2 T , hence (12) also holds.
P
Case 2 t2T p(t) < .
For a subpartition T = {T1 , . . . , Tq } of T and X ✓ S the pair (T , X) is called tight if (13) holds with
equality, that is,
X
[p(Z) |X|] = .
m
e S (X) +
Z2T
0
Note that p (Z + X) 0 for each Z 2 T as otherwise (T {Z}, X) would violate (13).
A node t 2 T is called loose if p(t) < g(t). The set of loose nodes is denoted by L. P
By choosing
Y = ; and X = S, (14) gives ge(T ) m(S)
e
= . Hence L is not empty, as otherwise ge(T ) = t2T p(t) <
=m
e S (S)  ge(T ), a contradiction.
As increasing the values of p on singletons does not a↵ect its positively intersecting supermodularity,
we may assume that p is as large on singletons as possible so that (13) and (14) are met. Note that
increasing the value of p(t) cannot destroy (14) at all since p occurs in (14) only in p(Yi ) where |Yi | 2.
This means that for each t 2 T , either t is loose or there exists a tight pair (Tt , Xt ) such that T contains t
as a singleton. Such a pair is called a certificate for t, and shows that p(t) cannot be increased without
destroying (13).
The content of the next lemma is that there exists a universal tight pair (T , X) in the sense that it
is a certificate for each t 2 L.
Lemma 13 There exists a tight pair (T , X) such that {t} 2 T for every t 2 L.
Proof: For each t 2 L, let (Tt , Xt ) be a tight pair that is a certificate for t, modulo this choose |Tt | to
be minimal.
Claim 14 p(U ) > 0 for each U 2 Tt , U 6= {t}.
249
Proof: Suppose indirectly that p(U ) = 0 for some U 2 Tt , U 6= {t}. Then (Tt
for t, contradicting the minimality of |Tt |. ⇤
{U }, Xt ) is a certificate
Let
It = {U + Xt : U 2 Tt } [ {V
s : s 2 Xt }.
It is easy to see that It is an ST -independent family and pe0 (It ) = .
Let F be the multi-union of sets It (t 2 L). Clearly, pe0 (F) = |L| and
each arc a 2 A⇤ covers at most |Y | members of F.
(16)
While there exist two properly ST -intersecting sets in F for which p0 satisfies the supermodular inequality,
replace them with their intersectionP
and union. Only keep a set from the new ones if it has positive p0 value. The procedure is finite as
[|Z|2 : Z 2 F] strictly increases in each step. Let F 0 denote
0 for each Z 2 F 0 and p0 (Z) = 0 may only hold if
the family thus obtained. By Claim 14, p0 (Z)
|Z \ T | = 1, Z \ T 2 L.
Claim 15 F 0 does not contain a pair of properly ST -intersecting small sets. Specially, F 0 does not
contain a pair of properly ST -crossing sets.
Proof: Suppose indirectly that Z1 , Z2 2 F 0 are properly ST -intersecting small sets. Then p0 (Zi ) =
p(Zi \ T ) |Zi \ S| for i = 1, 2. By definition, p0 (Zi ) > 0 implies p(Zi ) > 0. If p0 (Zi ) = 0, then
|Zi \ T | = 1. As Zi 6✓ Z3 i , necessarily Zi \ S 6= ; holds, which in turn implies p(Zi \ T ) > 0. By
these observations, Z1 \ T and Z2 \ T are intersecting subsets of T having positive p values. Thus the
positively intersecting supermodularity of p implies
p0 (Z1 ) + p0 (Z2 ) = p(Z1 \ T )
|Z1 \ S| + p(Z2 \ T )
 p((Z1 \ Z2 ) \ T )
|Z2 \ S|
|(Z1 \ Z2 ) \ S| + p((Z1 [ Z2 ) \ T )
|(Z1 [ Z2 ) \ S|
 p0 (Z1 \ Z2 ) + p0 (Z1 [ Z2 ),
where the last inequality holds by p0 (Z) p(Z\T ) |Z\S| for large sets Z ✓ V . This is a contradiction, as
the procedure stops only if there exists no pair of properly ST -intersecting sets for which the supermodular
inequality holds. This finishes the proof of the claim. ⇤
Claim 16 For each t 2 L, there exists a set Zt in F 0 such that p0 (Zt )
inclusionwise minimal in F 0 .
0, Zt \ T = {t} and Zt is
Proof: For each t 2 L, Xt + t is a set in F such that p0 (Xt + t)
0. If a step of the procedure
a↵ects Xt + t, then a new set Xt0 + t gets into the family with Xt0 ✓ Xt . The definition of p0 implies
p0 (Xt + t)
0. Hence when the procedure stops, there exists a set Zt in F 0 such that
p0 (Xt0 + t)
0
p (Zt ) 0 and Zt \ T = {t} for each t 2 L. Choosing Zt to be inclusionwise minimal among such sets
satisfies the conditions of the claim. ⇤
Let Z = {Zt : t 2 L} be the family of sets provided by Claim 16. The next claim shows that each
member of Z is contained in an antichain, that is, in an independent family of maximum pe0 -value.
Claim 17 F 0 can be partitioned into independent families of pe0 -value .
Proof: The number of occurrences of Z in F 0 is denoted by s(Z). Observe that (16) remains valid
throughout the procedure. Hence each chain in F 0 has s-weight at most |L| as the members of any chain
can be covered by a single arc a 2 A⇤ . By Theorem 2, there exist |L| antichains of F 0 that together
contains s(Z) copies of Z for each Z 2 F 0 . By Claim 15, an antichain corresponds to a set of pairwise
250
independent sets. Recall that ⌫p0 = by Lemma 9. As pe0 (F 0 )
partition of F 0 into |L| independent families of pe0 -value . ⇤
pe0 (F) = |L| , these antichains give a
By Claim 17 and Lemma 3, there exists an ST -independent family I of pe0 -value containing all
members of Z. Partition I into two sets: I1 = {Z 2 I : Z is large} and I2 = {Z 2 I : Z is small}. By
Claim 15, {Z \ T : Z 2 I2 } is a subpartition of T . Moreover,
T contains the elements of L as singletons.
S
Recall that |S Z| = 1 for each Z 2 I1 . Let X = Z2I1 (S Z). As I is ST -independent, X ✓ Z
for each Z 2 I2 . Hence
X
= pe0 (I) =
p0 (Z)
Z2I
=
X
p0 (Z) +
X
p0 (Z)
Z2I2
Z2I10
m
e S (X) +
=m
e S (X) +
X
[p(Z \ T )
|X|]
Z2I2
X
|X|].
[p(U )
U 2T
By (13), equality holds, concluding the proof of the lemma.
⇤
Take a tight pair (T , X) provided by Lemma 13. Let Y ✓ T be the set of nodesPappearing in T as
singletons
and {Y1 , . . . , Yq } be the remaining partition classes. That is, m
e S (X) + t2Y [p(t) |X|] +
Pq
[p(Y
)
|X|]
=
.
i
i=1
We get
X
ge(T Y ) =
p(t)
t2T
=
X
t2T
<
Y
p(t)
X
X
p(t)
t2Y
p(t)
t2Y
= [m
e S (X)
|Y ||X|] + [
q
X
[p(Yi )
|X|],
i=1
contradicting (14), and this contradiction shows that Case 2 cannot occur.
4
⇤
Packing arborescences
Let D = (V, A) be a digraph. For a root-node r0 2 V , an arborescence rooted at r0 or an r0 arborescence (U, F ) is a directed tree in which each node in U is reachable from r0 . An arborescence
is spanning if its node-set is V . We sometimes identify an arborescence (U, F ) with its edge set F and
say that F spans U . The node-set of an r0 -arborescence F ✓ A is denoted by V (F ). An r0 -arborescence
may consist of the single node {r0 } and no edge but we always assume that an arborescence has at least
one node. Edges leaving the root-node r are called a root-arcs and their set is denoted by A0 , while
A⇤ = A A0 . For a set X ✓ V r0 , the set of root-arcs entering X is denoted by 0 (X). The tail
node and head node of an arc a are denoted by t(a) and h(a), respectively. The out-neighbours of
r0 , that is, the set {h(a) : a 2 A0 } is denoted by R.
A branching is set of node-disjoint arborescences, or in other words, a directed forest in which the
in-degree of each node is at most one. The set of nodes having in-degree 0 is called the root-set of the
branching. Note that a branching with root-set R is the union of |R| node-disjoint arborescences (where
251
an arborescence may consist of a single node and no edge). We sometimes identify a branching (U, B)
with its edge set B and say that B spans U . For a digraph D = (V, A) and root-set ; ⇢ R ✓ V a
branching (V, B) is called a spanning R-branching of D if its root-set is R. In particular, if R is a
singleton consisting of an element r0 , then a spanning branching is a spanning r0 -arborescence.
4.1
4.1.1
Motivation
Packing arborescences under matroid constraints
In his fundamental paper [2], Edmonds gave the following characterization of the existence of k pairwise
disjoint spanning arborescences rooted at the same node r0 .
Theorem 18 (Edmonds) In a digraph D = (V, A), there exist k pairwise disjoint spanning r0 -arborescences if and only if
%(X) k for all ; 6= X ✓ V r0 .
Edmonds actually proved a stronger form of the theorem.
Theorem 19 (Edmonds) Let D = (V, A) be a digraph, r0 2 V a root-node and {F10 , . . . , Fk0 } a family
of disjoint –not necessarily spanning– arborescences rooted at r0 . There are pairwise disjoint spanning
r0 -arborescences F1 , . . . , Fk such that Fi0 ✓ Fi if and only if
%(X)
p(X) for all ; 6= X ✓ V
r0 ,
(17)
where p(X) = |{i : V (Fi0 ) \ X = ;}|.
In the past four decades, great many extensions of the theorem have been proved. Recently, Durand de Gevigney, Nguyen and Szigeti considered the problem of packing arborescences under matroid
constraints [1]. They assumed that a matroid is given on A0 and were seeking for a packing of arborescences (not necessarily spanning the whole node set) such that the starting edges of the paths in the
arborescences going from r0 to a given node v form a base of the matroid.
However, the extra condition meant by the presence of a matroid on the set of root-arcs naturally
implies another question.
Problem 20 Given a matroid M on A0 , give a characterization of the existence of k pairwise disjoint
spanning r0 -arborescences F1 , . . . , Fk such that Fi \A0 form an independent set of M for each i = 1, . . . , k.
This problem can be further generalized by considering k matroids M1 , . . . , Mk with ground set A0 .
Problem 21 Given k matroids M1 , . . . , Mk on A0 , give a characterization of the existence of k pairwise
disjoint spanning r0 -arborescences F1 , . . . , Fk such that Fi \ A0 form an independent set of Mi for each
i = 1, . . . , k.
Unfortunately, Problem 21 turns out to be difficult even in the following very special case.
Problem 22
INPUT: A digraph D = (V, A), root-node r0 2 V and matroids M1 , M2 on A0 .
QUESTION: Are there pairwise disjoint spanning r0 -arborescences F1 , F2 in D such that Fi \ A0 forms
an independent set of Mi for i = 1, 2?
Theorem 23 Problem 22 is NP-complete.
Proof: The reduction is from the following NP-complete problem.
252
Problem 24
INPUT: A set S of q pairs and matroids M1 and M2 of rank q with ground set S.
QUESTION: Let M3 be the partition matroid in which a set I ✓ S is independent if it contains at most
one member of each pair. Do M1 , M2 and M3 have a common base?
Let S, M1 , M2 , M3 be an instance of Problem 24. Construct a directed graph as follows. The node
set of the graph consists of a root r0 and 2q nodes v11 , v12 , v21 , v22 , . . . , vq1 , vq2 while the set of arcs is A =
{r0 vij |i = 1, . . . , q, j = 1, 2} [ {vi1 vi2 , vi2 vi1 |i = 1, . . . , q}. Matroids M1 and M2 can naturally be projected
to A0 . Note that if F1 and F2 are disjoint spanning r0 -arborescences than both of them contains exactly
one edge from {r0 vi1 , r0 vi2 } for each i = 1, . . . , q. Hence the existence of two r0 -arborescences F1 , F2 such
that Fi \ A0 is independent in Mi is equivalent to M1 , M2 and M3 having a common base, concluding
the proof. ⇤
Although Problem 21 is difficult in general, its special cases are still interesting. By choosing Mi to
be the uniform matroid of rank mi and by requiring Fi \ A0 to be a base of Mi we get the following.
Problem 25 Given a digraph D = (V, A), a root-node r0 2 V and positive integers m1 , . . . , mk , find k
pairwise disjoint spanning arborescences F1 , . . . , Fk such that |Fi \ A0 | = mi for each i = 1, . . . , k.
In Section 4.2, we will show that this problem fits well in the framework presented in Section 3.
4.1.2
Matchings
Another motivation of our investigations was an earlier result of Folkman and Fulkerson on edge-colourings
in bipartite graphs [3]. They considered the problem of finding disjoint matchings of given sizes in a
bipartite graph and gave characterized when can the edge set of a bipartite graph partitioned into k1
matchings of size m1 and k2 matchings of size m2 .
Somewhat surprisingly, the problem becomes difficult when the matchings do not necessarily give a
partition of the edge set. Pálvölgyi recently showed the followings [15].
Theorem 26 (Pálvölgyi) The following problems are NP-complete.
(i) INPUT: A simple bipartite graph G = (S, T ; E) with maximum degree 3 and a positive integer k.
QUESTION: Does G contain a pair of disjoint matchings P and M such that P is a perfect matching
and M is a matching of size k?
(ii) INPUT: A simple bipartite graph G = (S, T ; E) with maximum degree 4.
QUESTION: Does G contain a pair of disjoint matchings P and M such that P is a perfect matching
and M covers every node of degree at least 2?
(iii) INPUT: A simple bipartite graph G = (S, T ; E) and positive integers k1 , k2 .
QUESTION: Does G contain three matchings P , M1 and M2 such that P is a perfect matching and
Mi is a matching of size ki for i = 1, 2?
It is worth mentioning that Rizzi and Cariolaro recently gave an algorithm for partitioning the edge
set of an arbitrary undirected graph into k pairwise disjoint matchings each having size at most B where
B is a given constant [17].
Similar questions can be asked about branchings, too.
Problem 27 Given a digraph D = (V, A) and positive integers m1 , . . . , mk , find k pairwise disjoint
branchings B1 , . . . , Bk such that |Bi | = mi for each i = 1, . . . , k.
Theorem 28 Problem 27 is a special case of Problem 25.
253
Pk
Proof: Take an instance of Problem 27. Add a new node r0 to the digraph and add i=1 mi parallel
arcs from r0 to v for each v 2 V . Let D 0 denote the digraph thus obtained. It is easy to see that there
are k pairwise disjoint branchings B1 , . . . , Bk in D with |Bi | = mi for each i = 1, . . . , k if and only if
there are k pairwise disjoint spanning arborescences F1 , . . . , Fk in D0 such that |Fi \ A0 | = n mi for
each i = 1, . . . , k where n = |V |. ⇤
4.2
Bounded number of root arcs
There are special cases when Problem 25 can be solved by using the following exchange property of
branchings [20].
Theorem 29 (Schrijver) Let B1 and B2 be branchings partitioning the arc set A of a digraph D =
(V, A) with root-set R1 and R2 , respectively. Let R10 and R20 be sets with R10 [ R20 = R1 [ R2 and
R10 \ R20 = R1 \ R2 . Then A can be split into branchings B10 and B20 such that the root-set of Bi0 is Ri0 if
and only if each strongly connected component K of D with %(K) = 0 intersects both R10 and R20 .
By applying Theorem 29 to two branchings obtained by the deletion of the common root-node of two
spanning r0 -arborescences, the next corollary easily follows.
Corollary 30 Let D = (V, F1 [ F2 ) be a digraph such that F1 and F2 are disjoint spanning r0 -arborescences. There exist disjoint spanning r0 -arborescences F10 and F20 such that F10 [ F20 = F1 [ F2 and
||F10 \ A0 | |F20 \ A0 ||  1.
The union of k pairwise disjoint r0 -arborescences is called a k-arborescence.
Corollary 31 Assume that F is a k-arborescence. There exist k pairwise disjoint r0 -arborescences
F1 , . . . , Fk ✓ F such that ||A0 \ Fi | |A0 \ Fj ||  1 for each i, j 2 {1, . . . , k}.
Proof: By Edmonds’ theorem (Theorem 18), F can be decomposed into k pairwise disjoint spanning
r0 -arborescences F1 , . . . , Fk . Choose indices i and j such that |A0 \ Fi | |A0 \ Fj | is maximum. If this
di↵erence is at least 2, apply Corollary 30 to the two arborescences. After a finite number of steps the
maximum di↵erence between the |A0 \ Fi |’s becomes 1, concluding the proof. ⇤
Consider the special case of Problem 25 when m1 = · · · = mk1 = p and mk1 +1 = · · · = mk = p + 1.
As k-arborescences form the common bases of two matroids, the weighted matroid intersection algorithm
can be used to find one using a minimum number of root-arcs. Let F denote such a k-arborescence.
If |A0 \ F | > pk1 + (p + 1)(k k1 ) or |A0 | < pk1 + (p + 1)(k k1 ), then there are no arborescences
satisfying the conditions of the problem. Otherwise we may add further root-arcs to F and delete some
of its non-root arcs as to get a k-arborescence that uses exactly pk1 + (p + 1)(k k1 ) root-arcs. By
Corollary 31, the k-arborescence thus obtained can be decomposed into k arborescences F1 , . . . , Fk with
|A0 \ Fi | = mi .
In [5], the special case of Problem 25 when m1 = · · · = mk = µ was considered.
Theorem 32 Let D = (V, A) be a digraph, r0 2 V a root-node and µ 2 Z+ . Assume that the maximum
number of parallel root-arcs is at most k. There exist k pairwise disjoint r0 -arborescences F1 , . . . , Fk such
that |Fi \ A0 | = µ if and only if
q
X
[k %A⇤ (Vj )]  µk  |A0 |
(18)
i=1
for every subpartition {V1 , . . . , Vq } of V
r0 .
In order to solve Problem 25 in general, we show that it fits well in the framework of Section 3. We start
with a simpler case which can be answered using Theorem 7. Recall that R denotes the out-neighbours
of r0 .
254
Theorem 33 Let D = (V, A) be a rooted k-edge-connected digraph with root-node r0 2 V and m1 , . . . , mk
Pk
positive integers such that
i=1 mi = |A0 |. There exist k pairwise disjoint spanning r0 -arborescences
F1 , . . . , Fk such that |Fi \ A0 | = mi for i = 1, . . . , k if and only if
%A0 (V0 ) +
q
X
[k
%A⇤ (Vi )] 
i=1
k
X
min{mj , |V0 | + q}
(19)
j=1
for every subpartition {V0 , V1 , . . . , Vq } of V
r0 where V0 ✓ R may be empty.
Proof: To see necessity, let F1 , . . . , Fk be arborescences satisfying |Fi \ A0 | = mi (i = 1, . . . , k) and
{V0 , V1 , . . . , Vq } a subpartition of V r0 where V0 ✓ R. Define P = {{v} : v 2 V0 } [ {Vi : i = 1, . . . , q}.
Pk
As i=1 mi = |A0 |, the arborescences give a partition of the root-arcs, so
%A0 (V0 ) +
q
X
[k
%A⇤ (Vi )] 
i=1
X
|{i : Fi \
0 (v)
6= ;}| +
k
X
|{i : Fi \
0 (Vi )
6= ;}|
i=1
v2V0
=
q
X
|{X 2 P : Fj \
0 (X)
6= ;}|
j=1

k
X
min{|V0 | + q, mj },
j=1
showing necessity of (19).
To see sufficiency, construct a bipartite graph as follows. Let S = {s1 , . . . , sk } and T be a node set
corresponding to R. Let m(si ) = mi and define set function p : 2T ! Z as
8
>
<max{k %A⇤ (Z) : ; 6= Z ✓ V, Y = R \ Z}, if |Y | 2,
p(Y ) = %A0 (v),
if Y = {v},
>
:
0,
if Y = ;.
Claim 34 p is an intersecting supermodular function on 2T .
2 and p(Ti ) = k %A⇤ (Zi )
Proof: Let T1 , T2 be properly intersecting subsets of T . Then |T1 |, |T2 |
for some Zi ✓ V attaining the minimum in the definition of p(Ti ). Note that the rooted k-connectivity
of D implies %A0 (v) max{k %A⇤ (Z) : ; 6= Z ✓ V, v = R \ Z}. By this and the submodularity of the
in-degree function, we have
p(T1 ) + p(T2 ) = k
k
%A⇤ (Z1 ) + k
%A⇤ (Z2 )
%A⇤ (Z1 \ Z2 ) + k
%A⇤ (Z1 [ Z2 )
 p(T1 \ T2 ) + p(T1 [ T2 ),
where the last step follows from T1 \ T2 = R \ (Z1 \ Z2 ), T1 [ T2 = R \ (Z1 [ Z2 ).
⇤
We claim that conditions of Theorem 7 are satisfied. Let {T1 , . . . , Tq } be a subpartition of T . For
each i = 1, . . . , q, let Zi be a subset of V attaining the minimum in the definition of p(Ti ). Assume that
Ti and Tj are such that for the corresponding sets Zi , Zj ✓ V r the intersection Zi \ Zj is non-empty.
As Ti \ Tj = ;, we have %A⇤ (Zi \ Zj ) = %D (Zi \ Zj ) k, so
p(Ti ) + p(Tj ) = k
k
%A⇤ (Zi ) + k
%A⇤ (Zj )
%A⇤ (Zi \ Zj ) + k
0+k
%A⇤ (Zi [ Zj )
 p(Ti [ Tj ),
255
%A⇤ (Zi [ Zj )
where the last step follows from Ti [ Tj = R \ (Zi [ Zj ). That is, replacing Ti and Tj with their union may
only increase the left hand side of (6), and may only decrease the right hand side. Hence (19) implies
(6).
P
e
s)+|T | < m(S),
e
a contradiction.
If m(s) > |T | for some s 2 S, then m(S)
e
= |A0 | = t2T p(t)  m(S
Hence m(s)  |T | for each s 2 S. By Theorem 7, there exists a simple bipartite graph P
G = (S, T ; E) such
that dG (si ) = m(si ) = mi for i = 1, . . . , k and | G (Y )| p(Y ) for each Y ✓ T . As t2T p(t) = m(S),
e
dG (t) = p(t) for each t 2 T .
Let Fi0 be the arborescence consisting of arcs {r0 v : v 2 R, si v 2 E} for i = 1, . . . , k. As dG (v) =
p(v) = %A0 (v) for each v 2 T , there are exactly %A0 (v) arborescences using an arc from r0 to v. | G (Y )|
p(Y ) implies that the arborescences F10 , . . . , Fk0 thus obtained satisfy the conditions of Edmonds strong
theorem (Theorem 19), hence they can be extended to spanning r0 -arborescences F1 , . . . , Fk such that
|A0 \ Fi | = mi for i = 1, . . . , k. ⇤
Pk
Theorem 7 sufficed to solve the above case due to extra condition i=1 mi = |A0 |. Without this
assumption the more general Theorem 12 can be applied.
Theorem 35 Let D = (V, A) be a digraph, r0 2 V a root-node and m1 , . . . , mk positive integers. There
exist k pairwise disjoint spanning arborescences F1 , . . . , Fk such that |Fi \ A0 | = mi for each i = 1, . . . , k
if and only if
q
k
X
X
[k %A⇤ (Vi )] 
min{mj , q}
(20)
i=1
for every subpartition {V1 , . . . , Vq } of V
%A0 (V0 )
[
X
mi
j=1
r0 , and
|I||R
V0 |] + [
[k
%A⇤ (Vj )]
q|I|]
(21)
j=1
i2I
for each V0 ✓ R, subpartition {V1 , . . . , Vq } of V
I ✓ {1, . . . , k}.
q
X
r0 with (
Sq
j=1
Vj ) \ R ✓ V0 and |R \ Vj |
2, and
Proof: The theorem follows from Theorem 12 (similarly as Theorem 33 followed from Theorem 7) by
setting g(v) = %A0 (v) for v 2 R and
(
max{k %A⇤ (Z) : ; 6= Z ✓ V, Y = R \ Z}, if Y 6= ;
p(Y ) =
0,
if Y = ;.
⇤
5
Maximal term rank
Let G = (S, T ; E) be a bipartite graph. The deficiency h(Y ) of a subset Y ✓ T is defined by h(Y ) :=
|Y | | (Y )| where (Y ) = G (Y ) denotes the set of neighbours of Y . Let µ = µ(G, T ) denote the
maximum deficiency of subsets of T while ⌫ = ⌫(G) is the maximum cardinality of a matching of G. The
defect form of Hall’s theorem (which is equivalent to Kőnig’s theorem) is as follows.
Theorem 36 In a bipartite graph G = (S, T ; E), there is a matching of ` edges if and only if the
deficiency of every subset of T is at most |T | `. Equivalently, ⌫(G) = |T | µ(G, T ).
256
5.1
Term rank of matrices
Graphs in G(mS , mT ) can be identified with (0, 1)-matrices of size |S| ⇥ |T | in which the row sum vector
is mS and the column sum vector is mT . Let M(mS , mT ) denote the set of these matrices.
Ryser [18] defined the term rank of a (0, 1)-matrix A by the maximum number of independent 1’-s
which is the matching number of the corresponding bipartite graph. Ryser developed a formula for the
maximum term rank of matrices in M(mS , mT ).
The maximum term rank problem is equivalent to finding the maximum matching number of bipartite
graphs in G(mS , mT ) which, in turn, is equivalent to the following.
Theorem 37 (Ryser) Let `  |T | be an integer. Suppose that G(mS , mT ) is non-empty, that is, (7)
holds. For an integer `  |T |, G(mS , mT ) has a member G with ⌫(G) ` if and only if
m
e S (X) + m
e T (Z)
|X||Z| + (`
|X [ Z|) 
for every X ✓ S, Z ✓ T.
(22)
Moreover, (22) follows from its special case when X consists of the i largest values of mS and Z consists
of the j largest values of mT (i = 1, . . . , |S|, j = 1, . . . , |T |).
Proof: Necessity. Let G be a bipartite graph with the requested properties. Since G is simple, it has
at least m
e S (X) + m
e T (Z) |X||Z| edges having at least one end-node in X [ Z. Moreover, since G has
a matching of ` edges, there are at least (|X [ Z| `) edges connecting S |X| and S Z. Therefore
e T (Z) |X||Z| + ` |X [ Z|, that is, (22) is indeed
the total number of edges is at least m
e S (X) + m
necessary.
To prove sufficiency, define a set-function p on T by p(Y ) := |Y | (|T | `) if Y is non-empty and
p(;) = 0. Then p is fully supermodular. If there is a simple bipartite graph G = (S, T ; E) fitting (mS , mT )
(that is, a member of G(mS , mT ) that covers p, then G has a matching of size ` by Theorem 36, and we
are done. If no such a G exists, then Theorem 11 implies that there is a subpartition T = {Z, T1 , . . . , Tq }
of T and a subset X of S for which
m
e T (Z) +
q
X
i=1
p(Ti ) > m
e S (S
X) + (q + |Z|)|X|.
(23)
Since in the present case p is fully supermodular, we can assume that q  1. If q = 0, then
X) + |Z||X|,
m
e T (Z) > m
e S (S
that is,
m
e T (Z) + m
e S (X)
|Z||X| > ,
contradicting (7).
If q = 1, then (23) is equivalent to
m
e T (Z) + p(T1 ) > m
e S (S
X) + (1 + |Z|)|X|.
Since p is monotone non-decreasing, we may assume that T1 = V
|T Z| (|T | `) = ` |Z| from which
m
e T (Z) + `
that is,
contradicting (22).
⇤
|Z| >
m
e T (Z) + m
e S (X)
Z. Furthermore p(T
m
e S (X) + (1 + |Z|)|X|,
|X||Z| + `
257
|X [ Z| > ,
Z) =
It is interesting to notice that Theorem 10 is well-known to be treatable by network flow techniques
and therefore its weighted extension is also tractable as well as the so-called b-matching problem in an
arbitrary bipartite graph H, when one is to find a subgraph G of H with specified degrees.
Such an extension for the max term rank problem is not known and Ford and Fulkerson remark in
their book ([4], p. 89) that:
”Neither term rank problem appears amenable to flow approach”.
Recently, Pálvölgyi [15] pointed out that there is a deeper explanation of this sentence, namely, he
proved that the problem of finding a subgraph G of a given bipartite graph H so that G is fitting the
degree specification (mS , mT ) and G has a matching of ` elements is NP-complete.
On the other hand, our approach above permits a natural matroidal extension of the max term rank
problem. We continue to use graph-theoretic terminology.
5.2
Matroidal term rank
Let M = (T, r) be a matroid on T with rank function r. The co-rank function t of M is defined by
t(X) := min{B \ X : B is a basis}. Clearly, t(X) = r(T ) r(T X), and hence t is supermodular. We
call a matching basis-covering if it covers a basis of M .
Recall the classic theorem of Rado [16] on independent transversals. Its defect form is as follows.
Theorem 38 Given a bipartite graph G = (S, T ; E) and a matroid M on T with co-rank function t, G
admits a basis-covering matching if and only if
|
G (Y
)|
t(Y ) for every Y ✓ T.
(24)
The following result is a straight extension of Theorem 37.
Theorem 39 Let M be a matroid on T with co-rank function t. Suppose that G(mS , mT ) is non-empty,
that is, (7) holds. There is a member of G(mS , mT ) admitting a basis-covering matching if and only if
e T (Z)
m
e S (X) + m
|X||Z| + (t(T
Z)
|X|) 
for every X ✓ S, Z ✓ T.
(25)
Proof: Necessity. Let G 2 G(mS , mT ) be a graph admitting a basis-covering matching N . Since G is
simple, it has at least m
e S (X)+ m
e T (Z) |X||Z| edges having at least one end-node in X [Z. Moreover, N
covers at least t(T Z) elements of T Z and hence N has at least t(T Z) |X| elements connecting T Z
e T (Z) |X||Z| + (t(T Z) |X|),
and S X. Therefore the total number of edges is at least m
e S (X) + m
that is, (25) is indeed necessary.
To prove sufficiency, define a set-function p to be t. Then p is fully supermodular. If there is a simple
bipartite graph G = (S, T ; E) fitting (mS , mT ) (that is, a member of G(mS , mT ) that covers p, then G
has a basis-covering matching by the Theorem 38, and we are done. If no such a G exists, then Theorem
11 implies that there is a subpartition T = {Z, T1 , . . . , Tq } of T and a subset X of S for which
m
e T (Z) +
q
X
i=1
p(Ti ) > m
e S (S
X) + (q + |Z|)|X|.
Since p is now fully supermodular, we can assume that q  1. If q = 0, then
that is,
contradicting (7).
m
e T (Z) > m
e S (S
m
e S (X) + m
e T (Z)
258
X) + |Z||X|,
|Z||X| > ,
(26)
If q = 1, then (23) is equivalent to
X) + (1 + |Z|)|X|.
m
e T (Z) + p(T1 ) > m
e S (S
Since p is monotone non-decreasing, we may assume that T1 = V
m
e T (Z) + p(T
that is,
contradicting (25).
⇤
m
e S (X) + (1 + |Z|)|X|,
Z) >
m
e S (X) + m
e T (Z)
Z. Hence
|X||Z| + p(T
Z)
|X| > ,
Theorem 37 can be obtained by applying Theorem 39 to the `-uniform matroid on T .
5.3
Wooded hypergraphs
In [13], Lovász gave the following characterization of bipartite graphs in which the Hall condition holds
with strict inequality.
Theorem 40 (Lovász) In a bipartite graph G = (S, T ; E), there exists a forest for which the degree of
every node t 2 T is exactly 2 if and only if
|
G (X)|
|X| + 1
(27)
for every non-empty subset X ✓ S.
Lovász’s theorem can be reformulated in terms of hypergraphs. A hypergraph H = (V, E) is called
a wooded if it can be trimmed to a graph which is a forest, that is, it is possible to select two distinct
elements from each hyperedge in such a way that the selected pairs, as graph edges, form a forest.
Theorem 40 is equivalent to the following.
Theorem 41 A hypergraph H = (V, E) is wooded if and only if the union of every j hyperedges has at
least j + 1 elements (j 1).
Theorem 7 allows us to characterize the existence of a bipartite graph G 2 G(mS , mT ) satisfying the
Hall condition with strict inequality. By Theorem 41, this gives the characterization of the existence of
a wooded hypergraph in which the degree of each node and the size of each hyperedge is prescribed.
Theorem 42 Let V be a set of n nodes, mV : V ! Z+ a degree prescription, I = {1, . . . , m} and
e E (I) = m
e V (V ) = . There exists
mE : I ! Z+ a size prescription such that mE (i) 2 for i 2 I and m
a wooded hypergraph H = (V, E) such that such that dH (v) = mV (v) for v 2 V , E = {e1 , . . . , em } and
|ej | = mE (j) for j 2 I if and only if
for every X ✓ V, Z ✓ I, and
m
e V (X) + m
e E (Z)
m
e V (X) + m
e E (Z) + |I| + 1
for every ; 6= X ✓ V, Z ⇢ I.
259
|X||Z| 
|X|
|Z|
(28)
|X||Z| 
(29)
Proof: Necessity of (28) follows from the Gale-Ryser theorem. Let H = (V, E) be a hypergraph with
the requested properties, and let G = (V, I; E) be the associated bipartite graph. As G is simple, it has
e E (Z) |X||Z| edges having at least one end-node in X [ Z. As I Z 6= ; and H is
at least m
e V (X) + m
wooded, I Z has at least |I| |Z|+1 neighbours, hence there are at least |I| |Z| |X|+1 edges between
e E (Z)+|I|+1 |X| |Z| |X||Z|  ,
I Z and V X. So the total number of edges is at least m
e V (X)+ m
showing the necessity of (29).
Now we prove sufficiency. Define a set function p : 2I ! Z+ as
8
>
<|Y | + 1, if |Y | 2,
p(Y ) = mE (i),
if Y = {i},
>
:
0,
if Y = ;.
2 (i 2 I), p is intersecting supermodular. Moreover, mV (v)  |I| follows from m
e E (I) =
As mE (i)
m
e V (V ) and (28).
We will show that (5) is satisfied, that is,
m
e S (X) +
q
X
[p(Ti )
|X|] 
(30)
i=1
for every subpartition {T1 , . . . , Tq } of T and X ✓ S.
Let Z ✓ T denote the set of nodes that appear as singletons in the subpartition. If |Ti | = 1 for
i = 1, . . . , q, then (30) is equivalent to (28).
If |X| 1, then replacing two sets Ti , Tj of cardinality at least 2 by their union may only increase the
left hand side. Hence we may assume that there is exactly one such set in the subpartition. Moreover,
we may assume that this set is T Z as adding further nodes to it may only increase the left hand side
of (30) while the right hand side does not change. In this case (30) is equivalent to (29).
If |X| = 0, then, by mE (i) 2 for i 2 I, the left hand side is maximal if the subpartition is in fact
e V (V ).
the partition {{v} : v 2 V }. In this case (30) follows from m
e E (I) = m
By Theorem 7, there exists a simple bipartite graph G = (V, I; E) satisfying | G (Z)|
|Z| + 1 for
mE (i) for i 2 I. As m
e V (V ) = m
e E (I), necessarily
Z ✓ I, dG (v) = mV (v) for v 2 V and dG (i)
dG (i) = mE (i) for i 2 I, so G corresponds to a hypergraph satisfying the degree and the hyperedge-size
prescriptions. By Theorem 41, the hypergraph is wooded, thus concluding the proof. ⇤
6
Connectivity augmentation with degree-specified simple digraphs
There is an extensive literature of problems concerning degree sequences of graphs or digraphs with
specific properties. In what follows, we have an out-degree specification mo : V ! Z and an in-degree
specification mi : V ! Z for which we assume throughout that m
e o (V ) = m
e i (V ) and this common value
will be denoted by . Here V is an n-element set. We say that the pair (mo , mi ) is a degree-specification
and that a digraph H = (V, F ) fits this degree-specification if %H (v) = mi (v) and H (v) = mo (v) holds
for every node v 2 V .
The following characterization (in a simpler but equivalent form) is due to Fulkerson [9].
Theorem 43 Let mo : V ! Z and mi : V ! Z be out- and in-degree specifications.
(A) There exists a loopless digraph fitting (mo , mi ) if and only if
mi (v) + mo (v) 
for every v 2 V .
(31)
(B) There exists a simple digraph fitting (mo , mi ) if and only if
m
e i (X) + m
e o (Z)
|X||Z| + |X \ Z| 
260
for every X, Z ✓ V.
(32)
Note that (31) follows from (32) by taking X = {v} and Z = {v}. We also remark that (32) is
redundant in the sense that it follows from its special case when X consists of the j largest values of mi
and Z consists of the i largest values of mo (i = 1, . . . , n, j = 1, . . . , n).
6.1
Edge-connectivity
In [6], a min-max theorem was derived on covering supermodular functions by digraphs. A special case of
this result provides a characterization of degree specifications of loopless k-edge-connected digraphs. The
more general abstract result of Frank and Jordán [8] gives rise to characterizations of degree sequences of
(loopless) k-node-connected digraphs. However, the general problem remained open for simple digraphs.
For a survey of the existing results, see the book [7] which also contains the notions and notation used
here.
The approach of [8] is actually allows one to characterize degree sequences of digraphs H whose
addition to a starting digraph D results in a k-edge-connected digraph D + H. Jordán [12] pointed out
that in this case we arrive at an NP-complete problem if the simplicity of the resulting D +H is requested.
In this section, we only require the simplicity of the augmenting digraph H.
Recently, Hong, Liu, and Lai [11] found a characterization for strongly connected digraphs. In order
to generalize conveniently their result, we formulate it in a slightly more redundant though equivalent
form.
Theorem 44 (Hong, Liu, and Lai) Suppose that there is a simple digraph fitting the degree-specification (mo , mi ). There is a strongly connected simple digraph fitting (mo , mi ) if and only if
e i (X) + 1
m
e o (Z) + m
|X||Z| 
(33)
holds for every pair of disjoint subsets X, Z ✓ V .
We are going to extend this result to make a k-edge-connected digraph D (k + 1)-edge-connected by
adding a simple digraph H that fits a degree specification. Since the subsets of nodes with in-degree k in
a k-edge-connected digraph D form a crossing family, the following result indeed solves this augmentation
problem.
Theorem 45 Let K be a crossing family of non-empty proper subsets of V . Suppose that there is a
simple digraph fitting the degree specification (mo , mi ), that is, (32) holds. There is a simple digraph
fitting (mo , mi ) which covers K if and only if
e i (X)
m
e o (Z) + m
|X||Z| + 1 
(34)
holds for every pair of disjoint subsets X, Z ✓ V for which there is a member K 2 K with Z ✓ K ✓ V
X.
Proof: Suppose that there is a requested digraph H. By the simplicity of H there are at most |X||Z|
arcs from Z to X. Therefore the total number of arcs with tail in Z or with head in X is at least
m
e o (Z) + m
e i (X) |X||Z|. Moreover, at least one arc enters K and such an arc neither leaves an element
of Z nor enters an element of X, from which we obtain that m
e o (Z) + mi (X) |X||Z| + 1  , that is,
(34) is necessary.
To prove sufficiency, we need some observations.
Claim 46
holds for every K 2 K.
m
e i (K)
1 and m
e o (V
K)
1
Proof: m
e i (K) 1 follows by applying (34) to Z = ; and X = V
with the choice X = ; and Z = K. ⇤
This claim immediately implies the following.
261
(35)
K, while m
e o (V
K)
1 follows
Claim 47
K has at most
pairwise disjoint and at most
pairwise co-disjoint members.
(36)
Two sets are said to be co-disjoint if their complements are disjoint.
Claim 48
mo (v)  n
1 and mi (v)  n
1 for every v 2 V .
(37)
Proof: By applying (32) to X = {v} and Z = V , one gets mi (v) + m
e o (V ) 1|V | + |{v}|, that is,
mi (v)  n 1, and mo (v)  n 1 is obtained analogously by choosing X = V and Z = {v}. ⇤
Define a bi-set function p as follows.
Type 1: For K 2 K, let p(K, K) = 1.
Type 2: For u 2 V , let p(V
u, V
u) = mo (u).
Type 3: For v 2 X ⇢ V , let p(X, v) = mi (v)
(|X|
1).
If a bi-set belongs to more than one of these types, then we take the maximum of the values in
question. For example, if K = V
u 2 K, then p(K, K) = max{1, mo (v)}, or if K = {v} 2 K, then
p(K, K) = max{1, mi (v)}.
Note that the role of mo and mi is not symmetric in the definition of p.
Claim 49 The bi-set function p defined above is positively crossing supermodular.
It follows from the definition of p that every digraph covering p must have at least
arcs.
Case 1 There is a digraph H = (V, F ) with arcs covering p.
P
P
Now = |F | = [%F (v) : v 2 V ]
[mi (v) : v 2 V ] = from which %F (v) = mi (v) follows for
every v 2 V . Analogously, we get F (v) = mo (v) for every v 2 V . By the definition of p, it also follows
that H covers K.
Claim 50 H contains no loop.
Proof: Suppose indirectly that there is a loop e = vv 2 F at node v. By the hypothesis that there is
a simple digraph fitting (mo , mi ), we know that mi (v) + mo (v)  for every v 2 V , that is, mi (v) 
m
e o (V v). Hence there must be an arc f = xy 2 F with x 6= v, y 6= v. By replacing the arcs e and f by
the arcs xv and vy, we obtain a digraph H 0 of arcs that covers p as well and H 0 has less loops than H
has. ⇤
Claim 51 H is a simple digraph.
Proof: Suppose indirectly that H has two parallel arcs e and f from u to v. Consider the bi-set (X, v)
for X = {u, v}. We have %(X, v)  %(v) 2 = mi (v) 2  p(X, v) 1, a contradiction. ⇤
We can conclude that in Case 1 the digraph requested by the theorem is indeed available.
Case 2. The minimum number or arcs of a digraph covering p is larger than .
By the main min-max theorem of [8], there is an independent family I of bi-sets for which pe(I) > .
Then I partitions into three parts according to the three possibly types its members belong to. Therefore
we have a subset F ✓ K, a subset Z ✓ V and a family B = {(X, v)} of bi-sets so that
I = {(K, K) : K 2 F} [ {(V
z, V
z) : z 2 Z} [ B
and
|F| + m
e o (Z) +
X
[mi (v)
(|(Kv , v)|
262
1) : (Kv , v) 2 B] = pe(I) > .
(38)
Claim 52 There are no two members (X, v) and (Y, v) of B with the same inner set {v}.
Proof: If indirectly we have two such members, then X [ Y = V by the independence of I. If we replace
the two members (X, v) and (Y, v) of I by the single bi-set (X \ Y, v), then the resulting family I 0 is also
independent since any arc covering (X \ Y, v) covers at least one of (X, v) and (Y, v). Furthermore,
p(X, v) + p(Y, v) = mi (v)
(|X|
1) + mi (v)
(|Y |
1)
(39)
and
p(X \ Y, v) = mi (v)
1)
(|X \ Y |
1).
We claim that p(X, v) + p(Y, v)  p(X \ Y, v). Indeed, this is equivalent to mi (v)  (|X|
(|X \ Y | 1), that is, mi (v)  |V | 1, but this holds by (37). ⇤
Let X := {v : there is a bi-set (Kv , v) 2 B}. Now (38) transforms to
X
|F| + m
e o (Z) +
[mi (v) (|(Kv , v)| 1) : v 2 X] = pe(I) > .
(40)
1) + (|Y |
(41)
We may assume that pe(I) is as large as possible, and modulo this, |F| is minimal.
Claim 53 For v 2 X
Z, one has Kv = Z + v.
Proof: By the independence of I, Z ✓ Kv . Suppose, indirectly, that there is an element u 2 Kv (Z +v).
Replace the member (Kv , v) of I by (Kv u, v). Since p(Kv u, v) = p(Kv ) + 1, the resulting system
I 0 is not independent, therefore there exists a member of I that is covered by arc uv. Since u 62 Z, this
member must be in F, that is, this member is of form (K, K) for some K 2 K. By leaving out (K, K)
from I 0 , we obtain an independent I 00 for which pe(I 00 ) = pe(I), contradicting the minimal choice of F.
⇤
Claim 54 For v 2 Z \ X, one has Kv = Z.
Proof: The independence of I implies that Z ✓ Kv . If indirectly there is an element u 2 Kv z, then
replace Kv by Kv u. The maximality of pe(I) implies that there must exist a member (K, K) of F
which is entered by uv. Therefore if we replace in I the member Kv by Kv u and remove (K, K), then
the resulting I 0 contradicts the minimal choice of F. ⇤
e i (X) (|X \Z|(|Z| 1)+(|X Z||Z|) >
Due to these claims, condition (38) reduces to |F|+ m
e o (Z)+ m
which is equivalent to
Claim 55 X [ Z 6= ;.
|F| + m
e o (Z) + m
e i (X)
|X||Z| + |X \ Z| > .
(42)
Proof: If X [ Z = ;, then (42) reduces to |F| > . It is an easy observation that the members of the
independent F are either pairwise disjoint or pairwise co-disjoint, contradicting (36).
Claim 56 Z 6= ;.
263
Proof: If indirectly Z = ;, then X 6= ; in which case (42) reduces to
|F| + m
e i (X) > .
(43)
By Claim 53 we have Kv = {v} for every element v 2 X. The independence of I implies that
K \ X = ; for every K 2 F. Hence F is a sub-partition {K1 , . . . , Kq } and X is disjoint from [j Kj .
Furthermore we can assume that X = V [j Kj . Now we have
X
e i ([j Kj ) =
[m
e i (Kj ) : j = 1, . . . , q] q,
q>m
e i (V X) = m
a contradiction.
⇤
Claim 57 X 6= ;.
Proof: Suppose, indirectly, that X = ;. Then (42) reduces to
|F| + m
e o (Z) > .
(44)
By the independence of I, we have Z ✓ K for every K 2 K = {K1 , . . . , Kq }. Hence {K 1 , . . . , K q } is a
subpartition and Z \ ([j K j ) = ;. We may assume that Z = V [j K j (= \j Kj ). Now (44) is equivalent
to
X
e o ([j K j )
[m
e o (Kj ) : k = 1, . . . , q] q,
q>
m
e o (Z) = m
a contradiction.
⇤
We have concluded that X 6= ; and Z 6= ;.
Case 1 X \ Z = ;.
Hence (32) reduces to m
e i (X) + m
e o (Z)
On the other hand (42) reduces to
|X||Z|  .
|F| + m
e o (Z) + m
e i (X)
|X||Z| > .
(45)
This implies that |F| 1. For K 2 F, the independence of I implies that Z ✓ K and X ✓ V K.
The independence of I also implies that F cannot have more than one member, that is, |F| = 1, and
hence K, X, Z violate (34).
Case 2 X \ Z 6= ;.
If K 2 F , then Z ✓ K. Moreover, for an element v 2 X \ Z, the independence of (K, K) and (Z, v)
implies that v 2 X \ Z and Z [ K = V , that is K = V , which is not possible. Hence F = ; and (42)
reduces to m
e o (Z) + m
e i (X) |X||Z| + |X \ Z| > contradicting (32). ⇤ ⇤
6.2
Node-connectivity
Let D = (V, A) be a starting digraph on n k + 1 nodes and consider the problem of finding a simple
digraph H = (V, F ) fitting a degree specification (mo , mi ) for which D + H is k-node-connected. Our
present purpose is to show how to embed this problem into the framework of [8]. To this end, define a
bi-set function p as follows. We say that a bi-set X = (XO , XI ) is non-trivial, if XI 6= ; and XK 6= V .
Type 1: For a non-trivial bi-set X = (XO , XI ) with XI ✓ XO ⇢ V , let p(X) = k |XO XI | %D (X).
Type 2: For u 2 V , let p(V
u, V
u) = mo (u).
Type 3: For v 2 X ⇢ V , let p(X, v) = mi (v)
(|X|
264
1).
Claim 58 The bi-set function p is positively crossing supermodular.
By applying the main min-max result of [8] to this p, one obtains the following characterization.
e i (V ) =
Theorem 59 Given a starting digraph D = V, A) and a degree-specification (mo , mi ) on V with m
=m
e o (V ), there is a simple digraph H fitting (mo , mi ) for which D + H is k-node-connected if and only
if the maximum total p-value of an independent family of bi-sets is at most .
In the special case when A = ; the theorem provides a characterization of the existence of a simple
k-connected digraph fitting (mo , mi ). This characterization can be simplified in a way analogous to the
approach applied in Theorem 45. The details will be worked out in the journal-version of the present
work.
7
Investigations in progress
Theorem 7 can be generalized to the case when a matroid M = (S, r) is given and a simple graph
G = (S, T ; E) satisfying (3) and r( G (Y )) p(Y ) is required.
By using g-polymatroids, we were able to generalize Theorem 12 to the case when instead of a strict
degree prescription only lower and upper bound functions are given on S. This allows us to generalize all
results of Sections 4 and 5 accordingly. For example, given a digraph D = (V, A) and non-negative integers
f1 , . . . , fk , g1 , . . . , gk , the existence of k branchings B1 , . . . , Bk with fi  |Bi |  gi can be characterized.
These extensions and algorithmic aspects of the results will be discussed in the journal version of the
present work.
Acknowledgement
We are grateful to Nóra Bödei, for her indispensable help in exploring and understanding the paper [11].
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