Colourings
4.1 Edge colourings
Colourings of edges and vertices of a graph G are useful, when one is interested in classifying relations between objects.
There are two sides of colourings. In the general case, a graphGwith a colouring αis given, and we study the properties of this pairGα = (G,α). This is the situation, e.g., in transportation networks with bus and train links, where the colour (buss, train) of an edge tells the nature of a link.
In the chromatic theory,Gis first given and then we search for a colouring that the satisfies required properties. One of the important properties of colourings is ‘proper-ness’. In a proper colouring adjacent edges or vertices are coloured differently.
Edge chromatic number
DEFINITION. Ak-edge colouringα: EG → [1,k]of a graphGis an assignment of k colours to its edges. We writeGαto indicate thatGhas the edge colouringα.
A vertexv∈Gand a colouri∈[1,k]areincidentwith each other, ifα(vu) =ifor somevu∈ G. Ifv∈Gis not incident with a colouri, theniisavailableforv.
The colouringαisproper, if no two adjacent edges obtain the same colour:α(e1)6=
α(e2)for adjacente1ande2.
Theedge chromatic numberχ′(G)ofGis defined as
χ′(G) =min{k| there exists a properk-edge colouring ofG}.
Ak-edge colouringαcan be thought of as a partition{E1,E2, . . . ,Ek}ofEG, where Ei = {e | α(e) = i}. Note that it is possible that Ei = ∅ for some i. We adopt a simplified notation
Gα[i1,i2, . . . ,it] = G[Ei1∪Ei2∪ · · · ∪Eit]
for the subgraph ofGconsisting of those edges that have a colouri1,i2, . . . , orit. That is, the edges having other colours are removed.
Lemma 4.1.Each colour set Eiin a proper k-edge colouring is a matching. Moreover, for each graph G,∆(G)≤χ′(G)≤εG.
Proof. This is clear. ⊓⊔
Example 4.1.The three numbers in Lemma 4.1 can be equal. This happens, for in-stance, whenG=K1,nis a star. But often the inequalities are strict.
A star, and a graph withχ′(G) =4.
Optimal colourings
We show that for bipartite graphs the lower bound is always optimal:χ′(G) =∆(G). Lemma 4.2.Let G be a connected graph that is not an odd cycle. Then there exists a 2-edge colouring (that need not be proper), in which both colours are incident with each vertex v with dG(v)≥2.
Proof. Assume thatGis nontrivial; otherwise, the claim is trivial.
(1) Suppose first thatGis eulerian. IfGis an even cycle, then a 2-edge colouring exists as required. Otherwise, since nowdG(v)is even for allv,Ghas a vertexv1with dG(v1) ≥ 4. Lete1e2. . .et be an Euler tour of G, where ei = vivi+1(and vt+1 = v1).
Define
α(ei) =
(1, ifiis odd , 2, ifiis even .
Hence the ends of the edges ei fori ∈ [2,t−1] are incident with both colours. All vertices are among these ends. The conditiondG(v1)≥4 guarantees this forv1. Hence the claim holds in the eulerian case.
(2) Suppose then thatG is not eulerian. We define a new graphG0 by adding a vertexv0toGand connectingv0to eachv∈Gof odd degree.
In G0 every vertex has even degree including v0 (by the handshaking lemma), and henceG0is eulerian. Let e0e1. . .et be an eulerian tour ofG0, whereei = vivi+1. By the previous case, there is a required colouringαof G0as above. Now,αrestricted toEGis a colouring ofG as required by the claim, since each vertexviwith odd degreedG(vi)≥3 is entered and departed at least once in the tour by an edge of the original graphG:ei−1ei.
v0 1
2 1
2 2 1
⊓
⊔ DEFINITION. For ak-edge colouringαofG, let
cα(v) =|{i|vis incident withi∈[1,k]}|.
4.1 Edge colourings 45 Ak-edge colouringβis animprovementofα, if
∑
v∈G
cβ(v)>
∑
v∈G
cα(v). Also,αisoptimal, if it cannot be improved.
Notice that we always havecα(v)≤dG(v), and ifαis proper, thencα(v) =dG(v), and in this caseαis optimal. Thus an improvement of a colouring is a change towards a proper colouring. Note also that a graphGalways has an optimalk-edge colouring, but it need not have any properk-edge colourings.
The next lemma is obvious.
Lemma 4.3.An edge colouringαof G is proper if and only if cα(v) = dG(v)for all vertices v∈G.
Lemma 4.4.Let α be an optimal k-edge colouring of G, and let v ∈ G. Suppose that the colour i is available for v, and the colour j is incident with v at least twice. Then the connected component H of Gα[i,j]that contains v, is an odd cycle.
Proof. Suppose the connected componentH is not an odd cycle. By Lemma 4.2, H has a 2-edge colouringγ: EH → {i,j}, in which bothiandjare incident with each vertexxwithdH(x)≥2. (We have renamed the colours 1 and 2 toiandj.) We obtain a recolouringβofGas follows:
β(e) =
(γ(e), ife∈ H, α(e), ife∈/ H.
Since dH(v) ≥ 2 (by the assumption on the colour j) and in β both coloursiand j are now incident with v, cβ(v) = cα(v) +1. Furthermore, by the construction of β, we have cβ(u) ≥ cα(u) for all u 6= v. Therefore ∑u∈Gcβ(u) > ∑
u∈Gcα(u), which contradicts the optimality ofα. HenceHis an odd cycle. ⊓⊔ Theorem 4.1 (KÖNIG(1916)).If G is bipartite, thenχ′(G) =∆(G).
Proof. Let αbe an optimal ∆-edge colouring of a bipartiteG, where∆ = ∆(G). If there were av∈Gwithcα(v)< dG(v), then by Lemma 4.4,Gwould contain an odd cycle. But a bipartite graph does not contain such cycles. Therefore, for all verticesv, cα(v) =dG(v). By Lemma 4.3,αis a proper colouring, and∆=χ′(G)as required. ⊓⊔ Vizing’s theorem
In general we can haveχ′(G)> ∆(G)as one of our examples did show. The following important theorem, due to VIZING, shows that the edge chromatic number of a graph Gmisses∆(G)by at most one colour.
Theorem 4.2 (VIZING(1964)).For any graph G, ∆(G)≤χ′(G)≤∆(G) +1.
Proof. Let ∆ = ∆(G). We need only to show thatχ′(G) ≤ ∆+1. Suppose on the contrary thatχ′(G)> ∆+1, and letαbe an optimal(∆+1)-edge colouring ofG.
We have (trivially)dG(u)<∆+1<χ′(G)for allu∈G, and so Claim 1.For each u∈G, there exists an available colour b(u)for u.
Moreover, by the counter hypothesis,αis not a proper colouring, and hence there exists av∈ Gwithcα(v)< dG(v), and hence a colouri1that is incident withvat least with j≥1, such that the claim holds for these. Suppose, contrary to the claim, thatv is not incident withb(uj) =ij+1.
We can recolour the edgesvuℓ byiℓ+1forℓ∈[1,j], and obtain in this way an improvement ofα. Herevgains a new colour ij+1. Also, each uℓ gains a new colouriℓ+1
(and may loose the colouriℓ). Therefore, for eachuℓ ei-ther its number of colours remains the same or it in-creases by one. This contradicts the optimality ofα, and proves Claim 2.
Let t be the smallest index such that for some r < t, it+1 =ir. Such an indextexists, becausedG(v)is finite.
Let then the colouring γ be obtained from β by re-colouring the edges vuj by ij+1 for r ≤ j ≤ t. Now,
4.2 Ramsey Theory 47