a note on mixed graphs and directed splitting off

A Note on Mixed Graphsand Directed Splitting Off

Steffen EnniDEPARTMENT OF MATHEMATICS AND COMPUTER SCIENCE,

ODENSE UNIVERSITY,DENMARK

E-mail: [email protected]

Received December 12, 1994; revised October 24, 1997

Abstract: We give counterexamples to two conjectures of Bill Jackson in Someremarks on arc-connectivity, vertex splitting, and orientation in graphs and digraphs(Journal of Graph Theory 12(3):429–436, 1988) concerning orientations of mixedgraphs and splitting off in digraphs, and prove the first conjecture in the (di-) Eu-lerian case(s). Beside that we solve a degree constrained non-uniform directedaugmentation problem for di-Eulerian mixed graphs. c© 1998 John Wiley & Sons, Inc. J

Graph Theory 27: 213–221, 1998

Keywords: mixed graphs, splitting theorems, augmentation, orientations, edge-connectivity

1. INTRODUCTION

A mixed graph G = (V,A ∪ E) has vertex set V undirected edge set E and arcset A. We allow multiple edges but not loops. We write an arc directed from u tov as u → v and an edge with end vertices u and v as uv. For a set X ⊆ V we letd(X)(dG(X) if we want to emphasize the graph we study) denote the number ofedges betweenX and V −X , for a vertex v we call d({v}) = d(v) the degree of v.The number of arcs directed from V −X to X is denoted by ρ(X) and we defineδ(X) = ρ(V −X). An orientation of G is an orientation of the edge set E and welet ~d(X) denote the number of edges which in this orientation are directed from Xto V −X . If d(v) is even for all v ∈ V we sayG is Eulerian and if ρ(v) = δ(v) for

c© 1998 John Wiley & Sons, Inc. CCC 0364-9024/98/040213-09

214 JOURNAL OF GRAPH THEORY

all v ∈ V we say G is di-Eulerian. Let e = s → u, f = v → s be two oppositelydirected arcs. Splitting off {e, f}means that we delete e and f from the graph andadd a new directed arc v → u, the result is denoted by Gef . For a mixed graphG = (V,A∪E) and a set of (possibly directed) edgesM we letG+M denote themixed graph G = (V,A ∪ E ∪M).

We let λ(u, v;G) denote the maximum number of pairwise edge-disjoint pathsfrom u to v where arcs may only be used in the direction agreeing with the arc. Bya variation of Menger's theorem we have

λ(u, v;G) = min{d(X) + δ(X) : u ∈ X ⊆ V − v}.If λ(u, v;G) ≥ k for every pair u /= v ∈ V we say that G is k-edge-connected.

In [4] the following two conjectures appear.

Conjecture 1 (Jackson). Let M = (V,A ∪ E) be a mixed graph with directedarc set A and undirected edge set E. There exists an orientation ~E of E such thatthe resulting digraph G = (V,A ∪ ~E) satisfies

λ(u, v;G) ≥ min{bdM (X)/2c+ δM (X) : u ∈ X ⊆ V − v} (1)

for all u /= v ∈ V.Conjecture 2 (Jackson). Let G = (V + s,A) be a digraph for which ρ(s) =δ(s). There is a pair of arcs {e = s → u, f = v → s} incident with s so thatsplitting off e and f the result Gef satisfies

λ(u, v;Gef ) ≥ min{λ(u, v;G), λ(v, u;G)}. (2)

The purpose of this article is to give counterexamples to both conjectures and toprove Conjecture 1 in the special cases where the mixed graph is either Eulerian ordi-Eulerian. Beside this we apply the proof techniques developed by Frank [2] tosolve the degree constrained non-uniform directed augmentation problem on mixeddi-Eulerian graphs. A variation of this problem is solved for general starting mixedgraphs without degree constraints in [1].

2. ORIENTATIONS OF MIXED GRAPHS

First for any integer k > 0 we construct a k-edge-connected mixed graph withoutany orientation satisfying (1). We call an orientation satisfying (1) good.

Let G be a mixed graph with vertex set V := {ui, vi}6i=1 ∪ {ai, bi}3i=1 ∪{x1, x2, y1, y2}, undirected edge setE := {aibi}3i=1 and directed arc set as follows.For each of the three different 2-tuples (i, j) of {1, 2, 3} add k multiple arcs ofthe following types {vl → ai, vl → aj , ai → vl+3, aj → vl+3, bi → ul, bj →ul, ul+3 → bi, ul+3 → bj} using different values of l for different tuples. Further-more add k multiple arcs of each of the following types; {ui → x1, x2 → vi}3i=1,{vi → y2, y1 → ui}6i=4 and x1 → x2, y2 → y1. The construction is shown inFigure 1 for k = 1.

MIXED GRAPHS AND DIRECTED SPLITTING OFF 215

FIGURE 1. A counterexample to Conjecture 1 for k = 1.

A good orientation ~G of G must satisfy λ(vi, ui; ~G) ≥ k + 1 for i ∈ {1, 2, 3}and λ(ui, vi; ~G) ≥ k + 1 for i ∈ {4, 5, 6}. In the directed part of G we haveλ(vi, ui) = λ(ui+3, vi+3) = k for i ∈ {1, 2, 3}, which mean that we have to orientthe edges so that each of the six pairs is connected through the orientation. Bysymmetry we can orient the edge a1b1 as a1 → b1. This implies that in order tosatisfy the pair u4, v4 we have to orient the edge a2b2 as b2 → a2. Now we bothhave to orient the edge a3b3 as a3 → b3 in order to satisfy v3, u3 and as b3 → a3 inorder to satisfy u5, v5, which is impossible.

Let us define the following notation similar to the one used in [3] before weproceed.

µ(u, v) := min{d(X) + 2δ(X) : u ∈ X ⊆ V − v}R(X) := max{µ(u, v) : u ∈ X, v /∈ X}γ(X) := d(X) + 2δ(X)

(Note that if a mixed graph is di-Eulerian then µ, R and γ are symmetric.)For an orientation ~G of a mixed graph G satisfying (1) we have λ(u, v; ~G) ≥

bµ(u, v)/2cwhich by the directed edge version of Menger's Theorem is equivalentto

~d(X) + δ(X) ≥ R̂(X)/2 (3)


for all non-empty subsets X ⊂ V . Where f̂ := 2bf/2c and ~d is the number ofedges which are oriented from X to V −X .

Similar to the approach in [5] and [3] we will define a perfect matching M onthe set of vertices with odd degree inG as a good odd-vertex pairing ifM satisfies

dM (X) ≤ γ(X)− R̂(X) = d(X) + 2δ(X)− R̂(X) (4)

for all X ⊂ V .

Theorem 1. If a mixed graph G has a good odd-vertex pairing M then it hasan orientation satisfying (1).

Proof. InG+M the undirected edges induce an Eulerian subgraph. It is well-known that an Eulerian graph can be decomposed into edge-disjoint cycles. If weorient the edges in the same direction along each cycle then for a subset of verticesX we have ~dG+M (X) = dG+M (X)/2. Using thatM is a good odd-vertex pairinggives for X ⊂ V

~d(X) + δ(X) ≥ (d(X) + dM (X))/2 + δ(X)− dM (X)= (d(X)− dM (X))/2 + δ(X)

≥ R̂(X)/2.

Observe dG(X) ≡ dM (X) modulo 2 so (dG(X) − dM (X))/2 is always aninteger.

We will now show that every di-Eulerian mixed graph has a good odd-vertexpairing.

Theorem 2. Every mixed di-Eulerian graph has a good odd-vertex pairing.

Proof. By induction on the number of arcs. If A = ∅ then µ is equivalent tothe normal edge-connectivity λ in undirected graphs and in this case the statementis precisely Nash-Williams' Pairing Theorem as stated in [3].

Let now A /= ∅. There is a cycle ~C ⊆ A, because of the di-Eulerian arcs.Consider G′ := G − ~C + C where C is an undirected copy of ~C. The graph G′is di-Eulerian and by induction there is a good odd-vertex pairing of G′. Note thatd′(v) ≡ d(v) modulo 2 and γ′(X) = γ(X) for all subsets X ⊂ V . From this weget dM (X) ≤ γ′(X)− R̂′(X) = γ(X)− R̂(X) for all subsets X of V , that is Mis a good odd-vertex pairing of G as well.

We are now in a position to state and prove the main theorem of this section.

Theorem 3. Let G = (V,A ∪ E) be a mixed Eulerian or di-Eulerian graph.Then G has an orientation ~G satisfying

λ(u, v; ~G) ≥ min{bdG(X)/2c+ δG(X) : u ∈ X ⊆ V − v} (5)

for all u /= v ∈ V. In addition this orientation may be chosen so that

|ρ(v)− δ(v)| ≤ 1 for all v ∈ V.


Proof. IfG is Eulerian then there are no odd vertices and in this case the theoremis easily seen to be true using the same arguments as in the proof of Theorem 1, fromwhich it follows that this orientation is di-Eulerian and thus satisfies the secondpart of the theorem.

If G is di-Eulerian then by Theorem 2 there is a good odd-vertex pairing whichby Theorem 1 implies that there is a good orientation satisfying (5). A vertex vis incident with at most one edge from the odd-vertex pairing and when we orientalong cycles in a consistent way as in the proof of Theorem 1 and remove the edgesfrom the pairing, we get an orientation satisfying the second part of the theorem.

3. DIRECTED SPLITTING

In this section we give a k-edge-connected counterexample to Conjecture 2 for allnatural numbers k. Consider a directed graphGwith vertex set V := {s, 1, 2, 3, 4}and arc set E := {s → 1, s → 3, 1 → 2, 1 → 2, 2 → s, 2 → 4, 3 → 2, 3 →2, 4 → s, 4 → 1, 4 → 3}. The construction is shown in Figure 2. In G wehave λ(1, 2) = λ(2, 1) = λ(3, 2) = λ(2, 3) = 2 and it is easy to check that nosplitting off of edges {e, f} incident with s preserves both λ(2, 3;Gef ) ≥ 2 andλ(2, 1;Gef ) ≥ 2.

The reason to this is that the four possible pairs of splittable arcs are covered byfour tight sets (a set X is tight if either ρ(X) = max{λ(u, v) : v ∈ X,u /∈ X} orδ(X) = max{λ(u, v) : u ∈ X, v /∈ X}) which all separates 2 from either 1 or 3.The four sets are {1, 2}, {2, 3}, {1, 4} and {3, 4}.

By giving all arcs inG except 3→ 2 and 1→ 2 multiplicity k and 3→ 2, 1→ 2multiplicity 2k the example becomesk-edge-connected andλ(2, v) = λ(v, 2) = 2kfor v = 1, 3. And all the 4k2 possible pairs of splittable arcs are covered by one ofthe all ready mentioned tight sets.

If one does not allow multiple arcs then the following construction can replace anarc with multiplicityk. Letu→ v have multiplicityk and replaceu→ vwithk newverticesv1, v2, . . . , vk and arcsvi → vj (i /= j) andu→ vi, vi → v (i = 1, . . . , k).By replacing all arcs in the digraph G above with such a construction the result isk-edge-connected and still no splitting at s is possible.

In [1] it is shown that one can not hope to preserve the same kind of minimumlocal connectivity when we split off undirected edges in mixed graphs. This seemsto indicate that there is not much room left for further splitting results in graphspreserving edge-connectivity in some sense. So far the extensions of Mader’ssplitting off theorems either consider undirected splitting or directed splitting at avertex incident with only one type of edges. Therefore it is tempting to considerdirected or undirected splitting at a vertex incident with both edges and arcs, butthe example in Figure 3 shows that even if bothE andA are (di-) Eulerian then theminimum local connectivity between two vertices can not be preserved when wesplit a pair of edges or a pair of arcs. In the example λ(u, v) = λ(v, u) = 2 andnone of the two types of splitting off at s preserves this number.


FIGURE 2. A 1-edge-connected counterexample to Conjecture 2.

In the remaining part of this article we will need the following splitting offresult from [1, Theorem 2.3]. For sake of simplicity we only formulate the di-Eulerian case.

Theorem 4 (Bang-Jensen, Frank, Jackson). Let M = (V + s,A ∪ E) be amixed di-Eulerian graph. Assume that s is incident only with directed arcs andρM (s) = δM (s). Then for every arc f = s→ t there is an arc e = u→ s so that

λ(x, y;M ef ) ≥ λ(x, y;M) for every x, y ∈ V.If we allow splitting of a mixed pair of edges, that is replace the arc s→ u and

the edge sv by two arcs {v → u, s→ v} then we have the following.

Theorem 5. Let G = (V + s,A ∪E) be a mixed di-Eulerian graph, let f = stbe an arc incident with s. There is either an arc e = v → s or an undirected edgee = sv so that when we split off {e, f} the result Gef satisfies

λ(u, v;Gef ) ≥ λ(u, v;G) for all u, v ∈ V.

Proof. Replace every undirected edge incident with s by a pair of oppositelydirected arcs. From Theorem 4 it follows that there is an arc e splittable with f .Performing this splitting and replace the remaining of the newly created oppositedirected arcs with edges the theorem follows.

FIGURE 3. There is no splitting at s.


4. DIRECTED AUGMENTATION OF MIXED DI-EULERIAN GRAPHS

LetG = (V,A∪E) be a mixed di-Eulerian graph and let f : V → Z+, g :→ Z+∪{∞}, f ≤ g be two non-negative integer-valued functions and let r : V ×V → Z+

be a symmetric function which we shall call a demand. For a set X ⊆ V we leth(X) =

∑(h(x) : x ∈ X) for h = f, g. Further let R(X) := max(r(u, v) : u ∈

X, v /∈ X) for all X ⊂ V .Call an augmentation of G with a set F of new directed arcs good if G+ :=

(V,A ∪ E ∪ F ) satisfies λ(u, v;G+) ≥ r(u, v) for all u, v ∈ V . By a version ofMenger's Theorem this is equivalent to requiring

βG+(X) ≥ R(X) for all X ⊂ V, (6)

where we define β as β(X) = d(X) + ρ(X).We want to determine a necessary and sufficient condition for the existence of a

good augmentation using γ edges satisfying upper and lower bounds on the numberof new arcs entering and leaving each vertex. In [1] the authors solve this problemfor any starting mixed graph (with certain restrictions on r) without upper and lowerbounds on the number of new arcs entering and leaving each vertex. It does notseem possible to apply the techniques we use here to solve the degree constrainedaugmentation problem for general mixed graphs.

Theorem 6. A mixed di-Eulerian graph G = (V,A ∪ E) has a good augmen-tation with a set F of γ directed arcs satisfying

f(v) ≤ ρF (v) = δF (v) ≤ g(v) (7)

if and only if

γ ≤ g(V ) (8)

R(X)− β(X) ≤ g(X) (9)

for all X ⊂ V and

k∑i=1

(R(Xi)− β(Xi)) + f(X0) ≤ γ (10)

holds for all partitions {X0, X1, . . . , Xk} of V where only X0 may be empty.

Proof of necessity. IfG+F is a good augmentation with γ arcs then by (7) weget (8) immediately and forX ⊂ V we haveR(X)−β(X) ≤ ρF (X) ≤∑(ρF (v) :v ∈ X) ≤ g(X) which shows that (9) must hold. Using the left inequality in (7)we get

k∑i=1

(R(Xi)− β(Xi)) + f(X0) ≤k∑i=1

ρF (Xi) + f(X0) ≤ |F | = γ

for any partition {X0, X1, . . . , Xk} of V , which shows that (10) holds.


Proof of sufficiency. Add a new vertex s to G and g(v) parallel undirectededges between s and v for all v ∈ V . From the construction and by (9) we have

f(v) ≤ d′(v) ≤ g(v), v ∈ V (11)

and d′(s) = g(V ) ≥ γ, where d′ is the degree function with respect to thenew edges.

Define S(X) := βG(X) + d′(X)−R(X). The condition in (6) is equivalent to

S(X) ≥ 0 for all X ⊂ V. (12)

Call a set X ⊂ V critical if S(X) = 0.Delete the edges incident with s one by one as long as the result G+ satisfies

(11) and (12). We claim that this process ends with d′(s) ≤ γ. At this point wereplace each undirected edge incident to s with a pair of oppositely directed arcs.The result is a mixed di-Eulerian graph for which ρ(s) = δ(s) = d′(s) and thetheorem follows from d′(s) applications of Theorem 4 when we split off the edgesincident to s and finally discard s. From (11) it now follows that the set of edgesconstructed by the splitting off satisfies (7).

It can only fail if d′(s) > γ and no edge can be deleted. That happens when everyedge sv either enters a critical set or d′(v) = f(v). The setM := {v ∈ V : d′(v) >f(v)} can be covered by a family of critical sets. LetF := {X1, X2, . . . , Xk} be afamily of critical sets covering M , chosen such that

∑(|X| : X ∈ F) is minimum

among the families where k is minimum. There are two possible cases.Case 1. F consists of disjoint sets. Let X0 := V − ∪ki=1Xi, then F ∪ {X0} is

a partition of V and we have

k∑i=1

(R(Xi)− β(Xi)) + f(X0) =k∑i=1

d′(Xi) + f(X0) = d′(s) > γ,

a violation of (10).Case 2. There are two intersecting sets X,Y ∈ F . The following inequalities

are well-known and can be proved by counting the contributions from the arcs andedges on both sides, we refer to [1, Proposition 1.1 and 1.3] for a proof.

In a mixed graph G = (V,E ∪ A), if X and Y are subsets of V such thatρ(X ∩ Y ) = δ(X ∩ Y ) then:

β(X) + β(Y ) ≥ max{β(X ∪ Y ) + β(X ∩ Y ), β(X − Y ) + β(Y −X)}.Further if (X ∪ Y ) /= V then

d(X − Y ) + d(Y −X) + 2d̄(X,Y ) = d(X) + d(Y ) ≥ d(X ∪ Y ) + d(X ∩ Y ).

Where d̄(X,Y ) denotes the number of undirected edges between X ∩ Y and V −(X ∪ Y ).

Finally R(X) is a skew-supermodular function, that is, R satisfies at least oneof the following inequalities for arbitrary X,Y ⊆ V (G) [1, Lemma 1.1]:

R(X) +R(Y ) ≤ R(X ∪ Y ) +R(X ∩ Y ) (13)


R(X) +R(Y ) ≤ R(X − Y ) +R(Y −X) (14)

If (13) holds for X and Y then 0 + 0 = S(X) + S(Y ) = βG(X) + d′(X) −R(X) + βG(Y ) + d′(Y ) − R(Y ) ≥ βG(X ∪ Y ) + d′(X ∪ Y ) − R(X ∪ Y ) +βG(X ∩ Y ) + d′(X ∩ Y )−R(X ∩ Y ) = S(X ∪ Y ) + S(X ∩ Y ) ≥ 0 + 0. Thatis, X ∪ Y is critical, a contradiction to the minimality of k in the choice of F .Therefore (14) must hold, but then

0 + 0 = S(X) + S(Y ) = β(X) + d′(X)−R(X) + β(Y ) + d′(Y )−R(Y )≥ β(X − Y ) + d′(X − Y )−R(X − Y )

+ β(Y −X) + d′(Y −X)−R(Y −X) + 2d̄(X,Y )= S(X − Y ) + S(Y −X) + 2d̄(X,Y ) ≥ 0 + 0 + 2d̄(X,Y ).

That is X − Y and Y − X are both critical and d̄(X,Y ) = 0. But then therecannot exist any edges between s and X ∩ Y and we can exchange X,Y withX − Y, Y −X in F and still cover M . This is a contradiction to the minimality of∑

(|X| : X ∈ F). �

References

[1] J. Bang-Jensen, A. Frank, and B. Jackson, Preserving and increasing localedge-connectivity in mixed graphs, SIAM J. Discrete Math. 8 (1995), 155–178.

[2] A. Frank, Augmenting graphs to meet edge-connectivity requirements, SIAMJ. Discrete Math. 5 (1992), 159–170.

[3] A. Frank, Applications of submodular functions, in London Math. Soc. Lec-ture Note Ser., vol. 187 (1993) pp. 85–136.

[4] B. Jackson, Some remarks on arc-connectivity, vertex splitting, and orienta-tion in graphs and digraphs, J. Graph Theory 12 (1988), 429–436.

[5] C. S. J. A. Nash-Williams, On orientations, connectivity and odd-vertex-pairings in finite graphs, Canad. J. Math. 12 (1960), 555–567.

[6] C. S. J. A. Nash-Williams, Well-balanced orientations of finite graphs andunobtrusive odd-vertex-pairings, in Recent Progress in Combinatorics, W. T.Tutte (ed.), (1969), pp. 133–149.

a note on mixed graphs and directed splitting off

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