We show several upper and lower bounds on vsrc(G), both for general graphs and for graphs G that belong to a specific graph class. Crucial in our analysis are the connections between very strong rainbow colorings and decompositions of the input graph into cliques. The graph ˆG used in the following lemma (defined in the preliminaries in Section 5.1.5)
is important for our hardness reductions.
Lemma 5.47. Let G be any graph. Then (1) src(G) ≤ vsrc(G) ≤ cp(G)(cp(G) + 1)/2. (2) src( ˆG) ≤ vsrc( ˆG) ≤ cp(G)(cp(G) + 1)/2.
Proof. Let C = C1, . . . , Cr be the set of cliques in an optimal clique partition of G; that
is, r = cp(G). For a vertex v, let c(v) denote the clique in C that contains v. We define the set of colors as C1
∪ C2
this set is at most cp(G)(cp(G) + 1)/2. We color an edge uv ∈ E(G) by {c(u), c(v)}. It only remains to show that this edge coloring is indeed a very strong rainbow coloring of
G.
For the sake of contradiction, suppose that the edge coloring is not a very strong rainbow coloring of G. Then there exist two vertices s, t ∈ V (G), a shortest path P between s and t, and two distinct edges uv, wx ∈ E(P ) that received the same color. If
c(u) = c(v), then the color of uv is {c(u)}. Since wx has the same color, we get that c(w) = c(x) = c(u). This implies that P uses two edges within the same clique. Then P can be shortcut, contradicting that P is a shortest path between s and t. Hence, c(u) 6= c(v), implying also that c(w) 6= c(x). Since colors of uv and wx are the same, we
can without loss of generality assume now that c(u) = c(w) and c(v) = c(x). (Note that it is possible that either u = w or v = x but not both at the same time). Then either the edge uw or the edge vx will shortcut P , a contradiction.
To see the second part of the lemma, color each edge ˆuv incident on the universal
vertex ˆu in ˆG by c(v), in addition to the above coloring of the edges of G. Suppose this
was not a very strong rainbow coloring of ˆG. Then there exists vertices u, v such that uˆuv is a shortest path and uˆu and v ˆu are colored the same. But then u and v are in the
same clique Ci in C. But then uv can shortcut uˆuv, a contradiction.
The following lemma is more consequential for our upper bounds.
Lemma 5.48. Let G be any graph. Then vsrc(G) ≤ is(G) = ecc(G).
Proof. Let U = {x1, x2, . . . , xn} be a universe and F = {S1, S2, . . . , Sm} be a family
of subsets of U such that, G is the intersection graph of F and |U | = is(G). Let vi be
the vertex of G corresponding to the set Si. We take U = {x1, x2, . . . , xn} as the set of
colors, and color an edge between vertices vi and vj with any x ∈ Si∩ Sj. (Note that this intersection is non-empty because of the presence of the edge vivj). Suppose for the
sake of contradiction that this is not a very strong rainbow coloring of G. Then there exist two vertices s, t ∈ V (G), a shortest path P between s and t, and two edges vivj
and vavb in P that received the same color. Let x be this color. By the construction of
the coloring, we have that x ∈ Si∩ Sj∩ Sa∩ Sb. Hence, {vi, vj, va, vb} induce a clique in
G. But then the path P can be shortcut, a contradiction.
We remark that a similar result to the above lemma for the case of src(G), was proved independently by Lauri [102, Prop. 5.3].
Corollary 5.49. Let G be any graph. Then vsrc(G) ≤ min{b|V (G)|2/4c, |E(G)|}. Proof. Directly from ecc(G) ≤ min{b|V (G)|2/4c, |E(G)|} for any graph [61]. Corollary 5.50. Let G be any graph.
(1) If G is chordal, then src(G) ≤ vsrc(G) ≤ |V (G)| − χ(G) + 1. (2) If G is circular-arc, then src(G) ≤ vsrc(G) ≤ |V (G)|.
(3) src(L(G)) ≤ vsrc(L(G)) ≤ |V (G)|, where L(G) is the line graph of G. These bounds are (almost) tight in general.
Proof. In each of the three cases, we express the graph as an intersection graph over a
suitable universe. Then, Lemma 5.48 implies that the size of the universe is an upper bound on the vsrc of the graph. The required result directly follows in each case.
Every chordal graph is the intersection graph of subtrees of a tree [72]. It is also known that the number of vertices of this tree only needs to be at most |V (G)| − ω(G) + 1. (For completeness, we provide a proof of this in Lemma 5.51 below). Since ω(G) = χ(G)
for chordal graphs, the required statement follows.
For a circular arc graph G, consider any set of arcs whose intersection graph is G. We now construct a different intersection representation. Take the set of second (considering a clockwise ordering of points) endpoints of all arcs as the universe U . Take Si ⊆ U as the elements of U contained in the i-th arc. It is easy to see that G is the intersection graph of F =nS1, S2, . . . , S|E(G)|
o
.
Finally, consider L(G). We construct an intersection representation with universe
V (G). For each uv ∈ E(G), let Suv = {u, v}. Then L(G) is the intersection graph of
F = {Se: e ∈ E(G)}.
The (almost) tightness for all the three cases follow by taking G as a path. For a path G, we have vsrc(G) = |V (G)| − 1 and vsrc(L(G)) = |V (G)| − 2. Paths are both chordal and circular-arc.
We now prove Lemma 5.51 used in the proof of Corollary 5.50. We remark that this lemma is a known fact and the proof is only given for the sake of completeness.
Lemma 5.51. Let G be any chordal graph. Then G can be represented as the intersection graph of the subtrees of a tree on at most n − ω(G) + 1 vertices.
Proof. Let n = |V (G)|. Since G is a chordal graph, it has a perfect elimination order [75].
In fact, the lexicographic search algorithm that construct a perfect elimination order implies the existence of such an order v1, v2, . . . , vn such that vn, vn−1. . . , vn−ω(G)+1
forms a maximum clique. For any i = 1, . . . , n, let Gi = G[vn, vn−1, . . . , vi].
We claim that for n − ω(G) + 1 ≥ i ≥ 1, Gi can be represented as the intersection
graph of subtrees Sn, Sn−1, . . . , Si of a tree Ti on at most n − i − ω(G) + 2 vertices, and
there exists a bijection fi from vertices of Ti to maximal cliques of Gi such that for each
n ≥ k ≥ i, Sk = Ti[u ∈ V (Ti) : vk∈ fi(u)].
We prove the claim by downwards induction on i. The base case is when i =
n − ω(G) + 1. In this case, the statement follows by taking Ti as a single vertex u and
fi(u) as the clique {vn, vn−1, . . . , vi}.
Now, assuming the statement is true for i, we prove it for i−1. Since vi−1is simplicial in Gi−1, NGi−1(vi−1) is a subset of some maximal clique C of Gj. Let t be the vertex in
Tj such that fj(t) = C. If all vertices in C are adjacent to vi−1 in Gi−1, then we take
Ti−1 = Ti, fi−1(t) = fi(t) ∪ {vi−1}, and fi−1(t0) = fi(t0) for all t0 ∈ V (Tj+1) such that
t0 6= t. It is easy to see that the statement follows in this case. Now suppose that not all vertices in C are adjacent to vi−1in Gi−1. Then we take V (Ti−1) = V (Tj) ∪ {u} and
E(Ti−1) = E(Ti) ∪ {u, t} where u is a new vertex introduced with fi−1(u) = NGi−1[vi−1]. For all u0 ∈ V (Ti−1) such that u0 6= u, we take fi−1(u0) = fi(u). The statement follows from this construction.
By taking i = 1 in the statement of the claim, it follows that G can be represented as the intersection graph of subtrees of a tree with at most n − ω(G) + 1 vertices.
Corollary 5.52. Let G be any k-perfectly orientable graph. Then, src(G) ≤ vsrc(G) ≤ k|V (G)|.
Proof. Consider any orientation of the edges of G such that the outgoing neighbors of
each vertex v can be partitioned into k or fewer cliques. Let C(v) denote these set of cliques for each vertex v. Let C0(v) := {S ∪ {v} : S ∈ C(v)}. Note that C0(v) is also a set of cliques. Observe that S
v∈V (G)C0(v) is an edge clique cover of G because, every
edge is outgoing from some vertex v, and will thus be covered by a clique in C0(v). Hence, vsrc(G) ≤ ecc(G) ≤ k|V (G)|.
Since any k-perfectly groupable graph is also k-perfectly orientable, the above bound also applies to k-perfectly groupable graphs. In this context, we prove an interesting converse of the above bound.
Lemma 5.53. Let G be any graph. If vsrc(G) ≤ k, then G is k-perfectly groupable. Proof. Consider an optimal very strong rainbow coloring µ of G. Consider an arbitrary
vertex v of G and let c be any color used in µ. Define the set Q(c) = {u ∈ N (v) :
µ(vu) = c}. Suppose there exist two non-adjacent vertices u, w in Q(c). Then uvw is
a shortest path between u and w, and thus uv and vw cannot have the same color, a contradiction to the definition of Q. Hence, for each color c used in µ, Q(c) is a clique. Since the number of colors is at most k, the edges incident on v can be covered with at most k cliques. Hence, G is k-perfectly groupable.