Graph Descriptions

A Graph Theoretic Point of View

5.1 Introduction

In the preceding chapter, we proved the existence of semigroup solutions to the flow problem on the graph with no sinks. In this chapter, we extend the results of Chapter 4 to graphs with non trivial acyclic part. We will show that asymptotically, the flow will remain in certain parts of the graph with cycles and that these subgraphs where the flow remains asymptotically are those cycles with no outgoing flow. We also show that the flow on the edges in the acyclic part of the graph will be depleted in finite time while the flow in the cyclic parts of the graph with both incoming and outgoing flow will be depleted asymptotically. We start with certain useful graph descriptions which will enable asymptotic description of the flow to be easier.

e^′_i= (v^′_iv^′_j) is an edge ofQ if there exists a vertexv∈V(G)such that→^ei v→^ej. Let us put this in a formal definition:

Definition 5.2.1. A vertex v ∈ G is said to generate an edge eij in Q if there exist edges ei

ande_j in Gsuch that

ei

−→v−→^ej .

We introduce the map

Φ_G :E(G)→V(Q) (5.1)

by definingΦ_G(e) = ¯e, where e¯is the vertex inQcorresponding to the edgeein G. Therefore, it is a one-to-one function. We also defineΨ_G to be the multifunction

Ψ_G :V(G)→2^E(Q). (5.2)

This function assigns to a vertex v ∈V(G) a set of edges generated by v. If u ∈ V_o, then it does not generate any edges in Q. Every vertex inV(G)\V_o generates at least one edge since there are no sinks.

More precisely, ifvis a vertex inV(G)\V_o, then it generates d⁻(v)d⁺(v)edges. To verify this, notice that if v has out degree d⁻(v), then every edge that is incoming at v is adjacent to all thed⁻(v)edges going out ofv, hence there ared⁻(v)edges generated inQfor every incoming edge at v, and since there ared⁺(v) incoming edges atv, there is a total ofd⁻(v)d⁺(v) edges in Qgenerated by v.

On the other hand, the inverse of Ψ_G is a map, that is, if Ψ_G(v)∩Ψ_G(v^′) 6=∅, then v =v^′. Indeed ifǫis generated by a vertexv∈V(G), then there is an incoming edgeuand an outgoing edge w such that the vertices u¯ and w¯ are incident to ǫ. Since an edge has only one tail and one head, ifǫwas generated by another vertexv^′ ∈G, thenv^′ would be the head of uand the tail of w, which is impossible.

Since the graphGhas no sinks, it cannot be acyclic (see Proposition 1.4.2, [6]), hence the line graphQis not acyclic as well. However, ifGhas sources, then there exists an acyclic subgraph of G(and hence, Qhas an acyclic subgraph as well). We shall see how to obtain the maximal acyclic subgraph of Q below. Let V_o(Q) be the set of sources in Q. Then V_o(Q) = Φ_G(e_j) whereej are edges inGwith their tails in Vo(G). LetQ2 be the subgraph ofQ which consists of all cycles and all paths joining the cycles and Q₁ be the graph obtained by deleting Q₂

from Q. Then Vo(Q) ⊂ V(Q1). Let CQ be the cut set separating Q1 and Q2. Then the set E(Q) =C_Q∪E(Q₁)∪E(Q₂).

We can easily determine all the vertices and edges ofQ1by using a topological sorting algorithm, [12]:

1. Pick a vertex v∈V_o(Q) and label it v₁.

2. Delete v₁ from Qand call the resulting graphQ.˜ 3. Pick another vertex v∈V_o( ˜Q) and label it v₂.

4. Delete v₂ from Q˜ and call the resulting graphQ˜ (again).

5. Repeat the above procedure until we get a graphQ˜ =Q₂ such thatV_o( ˜Q₂) =∅or until all the vertices in the graph are labelled.

Lemma 5.2.2. IfQ^′ is the subgraph ofQwhich was left after the execution of the topological sorting algorithm, then it must contain a cycle.

Proof. Ifv∈V(Q^′), then v is the head of some edge inQ^′. Since by assumption every vertex has an outgoing edge, it follows that every vertex in Q^′ has an incoming and outgoing edge, there are no sources and no sinks, so Q^′ must have a cycle, by Proposition 1.4.2, [6].

LetQ1 be the graph whose vertex set contains the vertices{v1, v2, . . .} which were labelled by the topological sorting algorithm. That is, V(Q₁) =V(Q)\V(Q₂). Then we note that Q˜ is a subgraph of Q andVo(Q) ⊆V(Q1). Clearly, if v ∈V(Q2), then v /∈V(Q1). This is because for v ∈ V(Q₂), it must either be on a cycle or an a path between two cycles. In either case v∈V(Q1) would mean that there is a cycle contained inQ1, a contradiction.

Remark 5.2.3. A graph Q(orG) may have more than one cut set, but the cut set separating the acyclic part of Q (that is Q₁) from the cyclic part Q₂ is unique. To verify this, Let C₁ and C₂ be two cuts separatingQ₁ andQ₂. Then note that E(Q) = E(Q₁)∪E(Q₂)∪C₁ = E(Q₁)∪E(Q₂)∪C₂. SinceE(Q₁)∩E(Q₂) =∅, it follows thatC₁ =C₂.

We can also use the topological sorting algorithm to find the maximal acyclic subgraph of G, and we shall call this graphG1 (G2 will then be the graph obtained by deleting G1 fromG).

Lemma 5.2.4. The function ΦG maps the edges of the graph G which form cycles to the vertices of the graphQ which are on cycles or between cycles; i.e

Φ_G(E(G₂)) =V(Q₂); Φ_G(E(G₁))∪C_G =V(Q₁) (5.3) and

Ψ_G(V(G₂)) =E(Q₂)∪C_Q; Ψ_G(V(G₁)) =E(Q₁). (5.4) Proof. Letui1, ei1, ui2, . . . , ei_k−1, ui_k be a cycle inG2. Sinceej−1 andei_j connect throughui_j, there is an edge inΨ_G(u_ij)with head atΦ_G(e_ij)and an edge inΨ_G(u_j+1)with tail atΦ_G(e_ij)so thatΨ_G(ui1)contains an edge connecting Φ_G(ei_k−1)andΦ_G(ei1). Therefore, if there is a cycle in G, there is a corresponding cycle inQ. On the other hand, ifv_i1, e_i1, v_i2, e_i2, . . . , e_il−1, v_il= vi1 is a cycle inQ2, then

Φ⁻¹_G (v_i1),Ψ⁻¹_G (e_i1), . . . ,Ψ⁻¹_G (e_il−1),Φ⁻¹_G (v_il)

is a cycle inG. Hence there is a cycle in G₂ if and only if there is a corresponding cycle inQ₂. Let e ∈ E(G₂), then Φ_G(e) is a vertex in Q. If e is on a cycle, then v is also on a cycle, hence v ∈ V(Q₂). If e is on a path joining two cycles, then there is a v_i1 and v_ik both on cycles such that v_i1, e_i1, . . . , e_ij =e, v_ij+1, e_ij+1, . . . , e_ik−1, v_ik. Let v^′ be a vertex on a cycle, adjacent to the vertex v_i1 (that is; v^′v_i1 is an edge on the cycle). Then Φ_G(v^′v_i1) is a vertex in a corresponding cycle inQ₂. Now sincev_i1 has at least two outgoing edges (one edge is on the cycle and the other is on the path leading to Φ_G(e), which is not on the cycle), Ψ_G(v_i1) consists of at least two edges. Also, v_ik has at least two incoming edges, one edge is on the cycle and the other is on the path which containse. Sincev_ik is on a cycle, there is at least one outgoing edge on the cycle. HenceΦ_G(e) is on a path joining two cycles and therefore belongs to Q₂. This shows thatΦ_G(E(G₂))⊆V(Q₂).

Let u be on a path joining two cycles in Q₂. Then there are vertices u_i1 and u_ik, both on cycles, such thatu_i1, e_i1, . . . , e_j−1, u_ij =u, e_ij, . . . , u_ik is a path between cycles. u_i1 must have at least two outgoing edges; e_i1 (leading to vertex u) and the other (call it e) on a cycle, and at least one incoming edge. So there is some vertexv∈G₂ which generates at least two edges.

In particular, e, e_i1 ∈Ψ_G2(v). Indeed if this was not the case, then we see that ifeis generated byu ande_i1 byv (u6=v), then there would be edges e¯_i,¯e_j,e¯_l ande¯_m such that

¯ ei

−→u−→^e¯j ; −→^e¯l v−→^e¯m,

wheree= ¯eij andei1 = ¯elm. Sinceeandei1 are incident from the same vertex inQ, it follows that e¯_i = ¯e_l, which gives the implication that e¯_i has two different heads, which is impossible.

So indeed e and e_i1 are generated by the same vertex in G₂. On the other hand, u_ik has at least two incoming edges; one on the cycle (call it e^′) and the other is the edges e_ik−1 and at least one outgoing edge (on the cycle). Sou_ik = Φ⁻¹_G (e^′) with e^′ on a cycle inG and the two incoming edges of u_ik are generated by the same vertex v^′ in G₂, with v^′ on a cycle. Indeed to see that e_ik−1, e^′ ∈Ψ_G(v^′), suppose first that they are generated by two different verticesv₁ andv^′. Then there are edgese¯_l1,¯e_l2,e¯_k1,e¯_k2 inG such thate^′ = ¯e_k1k2, e_ik−1 = ¯e_l1l2 and

¯ el1

−→v₁−→^e¯l2 ; −→^e¯k1 v^′ −→^¯ek2 .

Since e^′ ande_ik−1 are incident to the same vertex (u_ik) inQ, it follows that e¯_l2 = ¯e_k2, which gives the implication that e¯_k2 has two different tails, which is impossible. Thus the path starting atΦ⁻¹_G (u_i1)and ending atΦ⁻¹_G (u_ik)is a path joining two different cycles in G₂. Hence Φ_G(E(G₂))⊇V(Q₂) and this gives the equality in (5.3).

From the equationsE(G) =E(G₁)∪E(G₂)∪C_G and V(Q) =V(Q₁)∪V(Q₂), we have V(Q₁) =V(Q)\V(Q₂)

= Φ_G(E(G))\Φ_G(E(G₂))

= Φ_G(E(G₁)∪E(G₂)∪C_G)\Φ_G(E(G₂))

= Φ_G(E(G₁)∪C_G).

To prove that Ψ_G(V(G₁)) = E(Q₁), suppose that v ∈ V(G₁) and let e_ij ∈ Ψ_G(v). Then there exist edges e_i, e_j in E(G) such that −→^ei v −→^ej . We want to show thate_ij ∈ E(Q₁). If v∈V(G₁), then there is no closed path involving it or involving the edgese_iande_j. In particular, Φ_G(e_i)is a source in Q(Φ_G(e_i)∈V(Q₁)) as indeed is any of the d⁺(v) edges incoming atv.

That is, Φ_G(e_k)∈V(Q₁) whenever e_k is an incoming edge of v, for all v ∈V(G₁). Similarly, Φ_G(e_j) ∈ V(Q₁) as well (see from proof of (5.3)) and since e_ij is an edge from Φ_G(e_i) to Φ_G(e_j), it follows that e_ij ∈E(Q₁). Hence, e_ij ∈Ψ_G(V(G₁))⇒ e_ij ∈E(Q₁), implying that Ψ_G(V(G₁))⊆E(Q₁).

Now suppose that e=e_ij ∈E(Q₁). We want to show that this edge is generated by a vertex v in G₁. e_ij ∈E(Q₁) implies thatΦ_G(e_i),Φ_G(e_j)∈ V(Q₁), with Φ_G(e_i) −→^eij Φ_G(e_j). Using the equation in (5.3), we have

Φ⁻¹_G (e_i),Φ⁻¹_G (e_j)∈E(G₁)∪C_G.

This implies that either Φ⁻¹_G (ei) ∈ E(G1) and Φ⁻¹_G (ej) ∈ CG or Φ⁻¹_G (ei),Φ⁻¹_G (ej) ∈ E(G1).

Since

Φ⁻¹_G (ei)

−→ v^Φ

−1 G (ej)

−→ ,

both cannot be in the cut set ofGbecause theneij would not be inE(Q1). Φ⁻¹_G (ei)cannot be in the cut set ofGeither because then Φ⁻¹_G (e_j)∈E(G₁) and this would contradict the second equation of (5.3). In either case (Φ⁻¹_G (ei) ∈E(G1) and Φ⁻¹_G (ej) ∈C_G or Φ⁻¹_G (ei),Φ⁻¹_G (ej)∈ E(G₁)),v∈V(Q₁), henceE(Q₁)⊆Ψ_G(V(G₁)). Thus E(Q₁) = Ψ_G(V(G₁)).

Using this result, we also have that

Ψ_G(V(G)) =E(Q) =C_Q∪E(Q₁)∪E(Q₂)

=C_Q∪E(Q₂)∪Ψ_G(V(G₁))

⇒Ψ_G(V(G)\V(G₁)) =C_Q∪E(Q₂),

henceΨG(V(G2)) =CQ∪E(Q2) which is the first equality in (5.4).

Remark 5.2.5. In most texts on graph theory, the topological sorting algorithm is performed on a directed acyclic graph. We have adapted this algorithm on a graph which may contain cycles to obtain two subgraphs, one of which is acyclic, and the other containing cycles. Most of the results obtained in this section can not be found easily in literature. However, some partial results do exist and have been expanded in this thesis. For example, Theorem 14.4 of [12] states that the vertices of a directed graph can be arranged in a topological order if and only if the graph is acyclic. This can be related to Lemma 5.2.2 in this section in the sense that the vertices of Q₂, which contains cycles, can not be arranged by a topological sorting algorithm.