GGALLAI MULTIGRAPHS
KYLE PULA
Abstract.
A complete edge-colored graph or multigraph is called Gallai if itlacks rainbow triangles. We give a construction of all finite Gallai multigraphs. Background
A complete, edge-colored graph without loops lacking rainbow triangles is called
Gallai after Tibor Gallai, who gave an iterative construction of all finite graphsof this sort [3]. Some work progress has been made on the more general problemof understanding edge-colored graphs lacking rainbow n -cycles for a fixed n . Inparticular, Ball, Pultr, and Vojt˘echovsk´y give algebraic results about the sequence( n : G lacks rainbow n -cycles) as a monoid [2], and Vojt˘echovsk´y extends the workof Alexeev [1] to find the densest arithmetic progression contained in this sequence[5].While searching for a general construction of graphs lacking rainbow n -cycles,we were confronted with the task of understanding a different generalization. Wecall a complete, simple, edge-colored multigraph Gallai if it lacks rainbow triangles.(By complete here we mean only that each pair of vertices is connected by at leastone edge.) Mubayi and Diwan make a conjecture about the possible color densitiesin Gallai multigraphs having at most three colors [4].The main result of this paper is a construction of all finite Gallai multigraphs.1.1.
Basic Notation.
We denote vertices using lowercase letters such as u, v, and w ,sets of vertices using uppercase letters such as U, V, and W , and colors using up-percase letters such as A, B, and C . Given two sets of vertices, U and V , we write U V for the set of edges connecting vertices of U to vertices of V . This notationwill also be used with singletons, u and v , to refer to the edges connecting u and v . To denote the set of colors present in a set of edges, say U V , we write
U V whenthere is no risk of ambiguity. Otherwise we refer explicitly to the coloring at hand,i.e. ρ [ U V ]. If
U V = { A } , we will often shorten notation by writing U V = A . Ifthe edges of an n -cycle contain no repeated colors, we say it is rainbow .Many of our results will be stated in terms of mixed graphs. A mixed graph is atriple M = ( V, E, A ) with vertices V , undirected edges E , and directed edges A . Wesay M is complete if every pair of distinct vertices is connected by a single directedor undirected edge. The weak components of a directed graph are the componentsof the graph that results from replacing each directed edge with an undirected edge.For our purposes, the weak components of a mixed graph M = ( V, E, A ) will be theweak components of the directed graph (
V, A ). Note that this notion of componentdisregards undirected edges.We use the term rooted tree to refer to a directed graph that is transitive andwhose transitive reduction forms a tree in the usual sense. If (
V, A ) is a rooted tree, a r X i v : . [ m a t h . C O ] J un K. PULA then its root , written 1 V , is the unique vertex having the property that there is adirected edge from 1 V to every other vertex in V .1.2. Construction of Gallai Graphs.
It is easy to see that the following con-struction yields Gallai graphs.Let ( G = ( V, E ) , ρ ) be a complete, edge-colored graph such that | ρ ( V V ) | ≤ v ∈ V , let ( G v = ( V v , E v ) , ρ v ) be a Gallai graph. Construct a newcomplete graph on (cid:83) v ∈ V V v with edge-coloring ρ (cid:48) defined by ρ (cid:48) ( xy ) := (cid:26) ρ v ( xy ) if x, y ∈ V v ρ ( vw ) if x ∈ V v , y ∈ V w , and v (cid:54) = w . Gallai showed [3] that every finite Gallai graph can be built iteratively by theabove construction, i.e. for every Gallai graph, G = ( V, E ), there exists a nontrivialpartition V = ∪ V i such that | V i V j | = 1 for i (cid:54) = j and | ∪ i (cid:54) = j V i V j | ≤ Decomposition of Gallai Multigraphs
While our purpose is to give a construction of Gallai multigraphs, the mostsubstantive step is in developing the appropriate decomposition. Before statingthis result, we describe our basic techniques and introduce a few definitions.2.1.
Basic Techniques: Maximality and Dominance.
Let ( G = ( V, E ) , ρ ) bea Gallai multigraph. We will in all cases assume that distinct edges connecting thesame vertices are colored distinctly. We also think of V ⊆ N and thus having anatural ordering. We say that ( G = ( V, E ) , ρ ) is uniformly colored if ρ ( e ) = ρ ( e )for all e i ∈ E .We call uv isolated if for every w (cid:54)∈ { u, v } , uw = vw and | uw | = 1. Noticethat if uv is isolated we can reduce the multigraph by collapsing the edge(s) uv .Likewise, given any multigraph, we can arbitrarily introduce new isolated edgeswithout introducing rainbow triangles. We therefore call a multigraph reduced if itcontains no isolated edges.We call uv maximal if no new color can be added to uv without introducing arainbow triangle. Here we allow the possibility that uv has “all possible colors”and thus is maximal. Likewise, ( G = ( V, E ) , ρ ) is maximal if uv is maximal for all u, v ∈ V .Let ( G = ( V, E ) , ρ ) be a maximal Gallai multigraph. For u, v ∈ V , notice that | uv | ≥ uv is isolated. Therefore, if G is reduced, | uv | = 1 or 2 forall u, v ∈ V . Furthermore, if G is not reduced, we can reach a reduced Gallaimultigraph by successively collapsing isolated edges of G .To construct all Gallai multigraphs, it therefore suffices to construct the reducedmaximal ones. In Section 2.2, we develop a basic decomposition of any maximalreduced Gallai multigraph and then in Section 3 reverse this decomposition toconstruct all finite reduced Gallai multigraphs. Lemma 2.1.
Suppose ( G = ( V, E ) , ρ ) is a maximal Gallai multigraph. If u, v ∈ V and A ∈ uv , then for all B (cid:54)∈ U V , there is w ∈ V \ { u, v } and C (cid:54)∈ { A, B } suchthat either A ∈ uw and C ∈ wv or C ∈ uw and A ∈ wv .Proof. Since G is maximal and B (cid:54)∈ uv , we can find w (cid:54) = u, v such that u, v, w would form a rainbow triangle if B were to be added to uv . Thus we may find X ∈ uw and Y ∈ vw such that X, Y, and B are distinct. However, since A ∈ uv , |{ X, Y, A }| ≤ A = X or A = Y . Let C be the other color. (cid:3) ALLAI MULTIGRAPHS 3
While Theorem 2.2 will follow from Theorem 2.3, our general decompositionresult, we present it separately here because of its importance in understanding themost basic structure of a maximal reduced Gallai multigraph.
Theorem 2.2.
The vertices of a reduced maximal Gallai multigraph that are con-nected by two edges form uniformly colored cliques.Proof.
Let ( G = ( V, E ) , ρ ) be a reduced maximal Gallai multigraph. Let u, v, w ∈ V . Suppose uv = { A, B } and vw = { C, D } . If { A, B } (cid:54) = { C, D } , then we find arainbow triangle no matter the colors of uw . Suppose then that uv = vw = { A, B } .Certainly uw ⊆ { A, B } . Suppose uw = A . Then by Lemma 2.1, we may find x ∈ V \ { u, w } such that, without loss of generality, A ∈ ux and C ∈ wx . Then v, x, w contains a rainbow triangle. (cid:3) Theorem 2.3 is primarily an explanation of how each of these uniformly coloredcliques are related to each other, and the following relation on sets of vertices playsa central role in this analysis. Let ( G = ( V , E ) , ρ ) be a Gallai multigraph. For U, V ⊆ V disjoint, we say that U dominates V and write U → V iff | U V | > U = { u } , V = { v } and u < v or(2) | U | > | V | > u ∈ U and v ∈ V , uv = uV .Given U, V ⊆ V , we write Σ(
U, V ) for the map from U to the powerset of U V defined by u (cid:55)→ uV . When U → V , Σ( U, V ) completely describes the relationshipbetween U and V and we call it the signature of U → V .Given a reduced maximal Gallai multigraph ( G = ( V , E ) , ρ ), we will describe itsstructure through a sequence of edge-colored mixed graphs M n ( G ) = ( V n , E n , A n )defined as follows:(1) V := V ,(2) A := { ( u, v ) ∈ V : u → v } , and(3) E := {{ u, v } ∈ [ V ] : | ρ [ uv ] | = 1 } ,and for n ≥ (cid:48) ) V n is the partition of V induced by the weak components of M n − ( G ),(2 (cid:48) ) A n := { ( U, V ) ∈ V n : U → V } , and(3 (cid:48) ) E n := {{ U, V } ∈ [ V n ] : | ρ [ U V ] | = 1 } .For each n , ρ induces a list-coloring, ρ (cid:48) , of E n ∪ A n by ρ (cid:48) ( e ) = ρ [ U V ] where e = ( U, V ) or e = { U, V } . Likewise, Σ induces a partition of A n by ( U , V ) ∼ Σ ( U , V ) if and only if Σ( U , V ) = Σ( U , V ). Note that ( U , V ) ∼ Σ ( U , V ) if and only if U = U and ρ [ uV ] = ρ [ uV ] for all u ∈ U .Figure (1) shows an example of this sequence for a particular Gallai multigraph.For readability, we show only those edges in M n ( G ) that contribute to the formationof directed edges in M n +1 ( G ). The hash marks on the directed edges in M ( G )indicate whether the signatures agree or disagree.2.2. Decomposition of Maximal Gallai Multigraphs.
We may now state ourmain result.
Theorem 2.3.
Let G be a reduced maximal Gallai multigraph, H an induced sub-graph of G , and M n ( H ) = ( V n , E n , A n ) the sequence described above. Then (1) M n ( H ) is complete, (2) | ρ (cid:48) ( e ) | = (cid:26) if e ∈ E n if e ∈ A n , K. PULA M ( G ) M ( G )
46 7 8 0 9E F1 2 3 5A B B B A B C DB A B B A B M ( G ) M ( G ) G Figure 1.
Sequence of M n ( G ) for a Gallai multigraph.(3) the weak components of M n ( H ) are rooted trees, and (4) if ( U, V ) , ( V, W ) ∈ A n , then ( U, V ) ∼ Σ ( U, W ) for all n ≥ . For convenience, if M k ( H ) has properties (1)-(4) for all k ≤ n , we will say that H has the tree property for n .As we are primarily interested in decomposing reduced maximal Gallai multi-graphs, our most important application of Theorem 2.3 is when H = G . We will,however, need the result in this greater generality in a key technical step in Section3.2.2.3. Proof of Theorem 2.3.
Throughout this section, we assume ( G = ( V , E ) , ρ )is a reduced maximal Gallai multigraph and H is an induced subgraph of G . Lemma 2.4.
Suppose
U, V, W ⊆ V disjoint, U → V , and { A, B } = U V . (1) If U W = C / ∈ {
A, B } , then V W = C . (2) If V W = C / ∈ {
A, B } , then either C ∈ U W or U → W and Σ( U, V ) =Σ(
U, W ) . If we also know that either U → W, W → U, or | U W | = 1 andthat U always dominates with the same colors (i.e., whenever U → U (cid:48) , then U U (cid:48) = { A, B } ), then either U W = C or U → W and Σ( U, V ) = Σ(
U, W ) . (2 (cid:48) ) If W is a single vertex, we need only require C ∈ V W in (2) .Proof. (1) Fix v ∈ V and w ∈ W . Since U → V , we may select u A , u B ∈ U suchthat A ∈ u A v and B ∈ u B v . Observe that the triangle w, u A , v forces vw ⊆ { A, C } while w, u B , v forces vw ⊆ { B, C } . Thus vw = C . Since v and w were arbitrary, V W = C . ALLAI MULTIGRAPHS 5 (2) Fix u ∈ U, w ∈ W, v ∈ V . Since U V = { A, B } . We are in one of thefollowing cases: uv = A , uv = B , or uv = { A, B } . If uv = { A, B } , then vw = C forces uw = C . Suppose then that C (cid:54)∈ U W . If uv = A , then uw ⊆ { A, C } and thus uw = A . Likewise, if uv = B , then uw = B . We thus have either C ∈ U W or U → W and Σ( U, V ) = Σ(
U, W ).Suppose we also know that we are in one of the following cases:(i) U → W and U W = { A, B } ,(ii) W → U , or(iii) | U W | = 1.Again, if C (cid:54)∈ U W , then we must be in case (i). We would like to concludethat if C ∈ U W , then we are in case (iii) and thus
U W = C . Suppose W → U .To avoid a rainbow triangle, U W ⊆ {
A, B, C } . In particular, since | U W | ≥ A ∈ U W or B ∈ U W . Assume A ∈ U W and fix w ∈ W such that A ∈ wU . Wemay then choose u ∈ U and v ∈ V such that B ∈ uv but now u, v, w is a rainbowtriangle. Thus W (cid:54)→ U and U W = C .(2 (cid:48) ) Let W = { w } and V = { v } and repeat the proof of (2). (cid:3) Lemma 2.5. H has the tree property for .Proof. It is clear that M ( H ) is complete. The rest of the claim is essentially arestatement of Theorem 2.2. By the definition of dominance between single vertices,each complete, uniformly colored clique from Theorem 2.2 becomes a linear orderedset of vertices and thus a rooted tree. In this context, property (4) of Theorem 2.3is simply the observation that these cliques are uniformly colored. (cid:3) Before proceeding, we introduce some convenient notation. Elements of V n areby definition subsets of V ( H ). We will however at times want to speak of theirstructure as rooted trees. For U ∈ V n , we write Υ( U ) to refer to the set of elementsof V n − contained in U and 1 U to refer to the root of Υ( U ). Notice that 1 U ∈ V n − has its own tree structure and thus we may refer to 1 U , 1 U , etc. We may continuethis recursion until we reach a single vertex. We write U to refer to this singlevertex. Similarly, for u ∈ V ( H ), we write [ u ] n to refer to the unique U ∈ V n containing u . Lastly, we point out how this notation fits together. For U ∈ V n ,[ U ] n = U , [ U ] n − = 1 U , [ U ] n − = 1 U , ..., and [ U ] = U .We also associate a set of colors with each member of V n as follows. For u ∈ V , (cid:98) u := ∪ u → v uv and for U ∈ V n +1 , (cid:98) U := (cid:99) U for n ≥
0. Lemma 2.6 demonstrates theimportance of this notation.
Lemma 2.6.
Suppose H has the tree property for n and ( U, V ) ∈ A n +1 . Then U always dominates with the same two colors (cid:98) U , i.e. (1) U V = (cid:98) U and (2) | (cid:98) U | = 2 .Proof. It is clear that | (cid:98) U | = 2 since (cid:98) U is defined inductively and dominance betweentwo vertices must be with exactly two colors. Likewise, (1) certainly holds for U, V ∈ V .Since H has the tree property for n , | U (cid:48) V | = 1 for all U (cid:48) ∈ Υ( U ), and since U → V , we know | U V | ≥
2. Suppose we find C ∈ U V \ (cid:98) U . Then fix U C ∈ Υ( U )such that U C V = C . If U C = 1 U , then for all U (cid:48) ∈ Υ( U ) \ { U } , 1 U → U (cid:48) and K. PULA we may apply part (1) of Lemma 2.4 to get that U (cid:48) V = C and thus U V = C , acontradiction.Suppose then that 1 U → U C . By induction, 1 U U C = (cid:98) U . We may now applypart (2) of Lemma 2.4 to get that either C ∈ U V or 1 U → V . The former hasalready been ruled out while the latter contradicts the assumption that 1 U and 1 V were in different components of V n . (cid:3) The following lemma is useful because it allows us to locate a vertex in U thatis connected to the rest of U by only the colors contained in (cid:98) U . Lemma 2.7. If H has the tree property for n and U ∈ V n +1 , then U U = (cid:98) U .Proof. By Lemma 2.6, | (cid:98) U | = 0 or 2. If (cid:98) U = ∅ , then U is a single vertex, i.e. U = { U } , and thus U U = ∅ .For n = 0, U is either a single vertex, in which case (cid:98) U = ∅ , or U is a nontrivialuniformly colored clique, in which case (cid:98) U is by definition U U .For n ≥
1, if U is a single vertex, again (cid:98) U = ∅ . Otherwise, by induction1 U U = (cid:99) U = (cid:98) U . But since U = U , we have 1 U U = (cid:98) U , and by Lemma 2.6,1 U U (cid:48) = (cid:98) U for every U (cid:48) ∈ Υ( U ) \ { U } . Finally, given that U ∈ U , U U (cid:48) ⊆ U U (cid:48) and thus U U = (cid:98) U . (cid:3) Lemmas 2.8 and 2.9 will be used in situations where a tree is connected toanother tree or vertex by a color not present in the dominating colors of the firsttree.
Lemma 2.8.
Suppose H has the tree property for n , U, V ∈ V n +1 distinct, and V (cid:48) ∈ Υ( V ) such that C ∈ U V (cid:48) \ (cid:98) U . Then U V (cid:48) = C .Proof. If 1 U V (cid:48) = C , then for every U (cid:48) ∈ Υ( U ) \ { U } we may apply part (1) ofLemma 2.4 with 1 U → U (cid:48) and V (cid:48) to get that U (cid:48) V (cid:48) = C and thus U V (cid:48) = C . Supposethen that U (cid:48) ∈ Υ( U ) \ { U } such that U (cid:48) V (cid:48) = C . Applying part (2) of Lemma 2.4with 1 U → U (cid:48) and V (cid:48) , we get that 1 U V (cid:48) = C or 1 U → V (cid:48) . Since 1 U and V (cid:48) are indifferent components of V n +1 , we must be in the former case, and by the previouscase we are done. (cid:3) Lemma 2.9.
Suppose H has the tree property for n , U ∈ V n +1 , and v ∈ V suchthat U v = C (cid:54)∈ (cid:98) U . Then U v = C .Proof. First observe that v (cid:54)∈ U since by Lemma 2.7, U U = (cid:98) U . Next let k bemaximal such that [ U ] k v = C. If k = n + 1, we are done. Suppose k < n + 1 andselect u ∈ [ U ] k +1 \ [ U ] k . We may apply (1) of Lemma 2.4 with [ U ] k → u and v to get that uv = C and thus violating the maximality of k . (cid:3) Note that in Lemma 2.9 we do not require that v be in V ( H ) but rather in thelarger set V . Lemma 2.10.
For n ≥ , suppose H has the tree property for n and U, V ∈ V n +1 such that U V ⊆ (cid:98) U = (cid:98) V = { A, B } Then the following statements are equivalent: (1) U → V , (2) there is x ∈ V \ ( U ∪ V ) such that U → x and V x = C (cid:54)∈ { A, B } , and (3) V (cid:54)→ U and U V = { A, B } .If the statements are true, then Σ( U, V ) = Σ(
U, x ) . ALLAI MULTIGRAPHS 7
Proof. (1 ⇒
2) We may assume 1 U V = A . Since G is maximal and B (cid:54)∈ U V ,there is x ∈ V such that A ∈ U x and C ∈ x V or C ∈ U x and A ∈ x V for C (cid:54)∈ { A, B } . Notice that if C ∈ U x , since C (cid:54)∈ (cid:98) U , U x cannot contain multipleedges and thus C = U x . Likewise, if C ∈ V x . Furthermore, in either case, since C (cid:54)∈ U U = V V , we must conclude that x ∈ V \ ( U ∪ V ). Suppose we are inthe latter case, i.e. C = U x and A ∈ x V . By Lemma 2.9, U x = C . We maythen apply part (1) of Lemma 2.4 with U → V and x to get that V x = C , whichcontradicts our assumption that A ∈ V x .We must then be in the former case, i.e. A ∈ U x and C = x V and, againby Lemma 2.9, V x = C . Note that if we can show that C (cid:54)∈ U x , we may thenapply part (2) of Lemma 2.4 with U → V and x to get that U → x and thatΣ( U, V ) = Σ(
U, x ), which is exactly what we would like to prove.To this end, suppose C ∈ U x and let k be minimal such that C ∈ [ U ] k x .If k = 0, we have that C ∈ U x and it again follows that U x = C , which is acontradiction. Thus k >
0. Fix u ∈ [ U ] k such that C ∈ ux . Since k is minimal, u ∈ [ U ] k \ [ U ] k − and thus [ U ] k − → u . Recall that (cid:91) [ U ] i = (cid:98) U for all i ≤ n andthus [ U ] k − u = { A, B } .We may now apply part (2 (cid:48) ) of Lemma 2.4 with [ U ] k − → u and x to getthat either C ∈ [ U ] k − x or [ U ] k − → x and Σ([ U ] k − , x ) = Σ([ U ] k − , u ). Bythe minimality of k , we must be in the latter case. Then we may apply part(2) of Lemma 2.4 with [ U ] k − → x and V to get that either C ∈ [ U ] k − V or[ U ] k − → V . Both of theses cases contradict the assumption that 1 U V = A . Thus C (cid:54)∈ U x .(2 ⇒
3) We may apply part (2) of Lemma 2.4 with U → x and V to get thateither C ∈ U V or U → V and Σ( U, x ) = Σ(
U, V ). Since C (cid:54)∈ U V ⊆ {
A, B } , we areleft with the latter case; U → V and thus U V = (cid:98) U = { A, B } and V (cid:54)→ U .(3 ⇒ U V = A . As argued in (1 ⇒ x ∈ V \ ( U ∪ V ) such that either A ∈ U x and V x = C or U x = C and A ∈ V x for some C (cid:54)∈ { A, B } . Suppose we are in the latter case. Since V (cid:54)→ U and U V = { A, B } , there must be U A , U B ∈ Υ( U ) and V (cid:48) ∈ Υ( V ) such that U A V (cid:48) = A and U B V (cid:48) = B . Then x, U A , V (cid:48) forces xV (cid:48) ⊆ { A, C } while xU B V (cid:48) forces xV (cid:48) ⊆ { B, C } . Thus xV (cid:48) = C and C ∈ V x .We may then let k be minimal such that C ∈ [ V ] k x . If k = 0, then C ∈ V x and V x = C , which contradicts our assumption that A ∈ V x . Therefore k > v ∈ [ V ] k \ [ V ] k − such that C ∈ vx . We now apply part(2 (cid:48) ) of Lemma 2.4 with [ V ] k − → v and x to get that either C ∈ [ V ] k − x or[ V ] k − → x . By the minimality of k , we must be in the latter case and we mayapply part (2) of Lemma 2.4 with [ V ] k − → x and 1 U (recall our assumption that U x = C ) to get that either C ∈ U [ V ] k − or [ V ] k − → U . Both of these casescontradict the assumption that 1 U V = A .We therefore may assume that A ∈ U x and V x = C . It is either the case that U → V or U (cid:54)→ V . If we suppose that U (cid:54)→ V , then we are in the case just handledwith the roles of U and V reversed. Since that assumption leads to a contradiction,we have that U → V . (cid:3) Lemma 2.11. If H has the tree property for n , then M n +1 ( H ) is complete. K. PULA
Proof.
Let
U, V ∈ V n +1 . If | U V | = 1, then { U, V } ∈ E n +1 . Suppose then that | U V | >
1. Notice that if (cid:98) U = ∅ , then U is a single vertex and thus V → U and( V, U ) ∈ A n +1 .We may therefore assume | (cid:98) U | = | (cid:98) V | = 2 and consider the following cases:Case 1: (cid:98) U (cid:54) = (cid:98) V and | U V | >
2. We may then select C U , C V ∈ U V distinctsuch that C U (cid:54)∈ (cid:98) U and C V (cid:54)∈ (cid:98) V and U (cid:48) ∈ Υ( U ) , V (cid:48) ∈ Υ( V ) such that C V ∈ U (cid:48) V and C U ∈ U V (cid:48) . By Lemma 2.8, U (cid:48) V = C V and U V (cid:48) = C U . This implies that C V = U (cid:48) V (cid:48) = C U , a contradiction.Case 2: (cid:98) U (cid:54) = (cid:98) V and | U V | = 2. Then we may assume there is C ∈ U V such that C (cid:54)∈ (cid:98) U . Let U V = { C, D } . Select V C ∈ Υ( V ) such that C ∈ U V C . By Lemma 2.8, U V C = C . Now select V D ∈ Υ( V ) such that D ∈ U V D . Observe that C (cid:54)∈ U V D since that would imply that D (cid:54)∈ U V D = C . Thus U V D = D . Since U V = { C, D } ,we have accounted for every element of Υ( V ) and V → U , i.e. ( V, U ) ∈ A n +1 .Case 3: (cid:98) U = (cid:98) V = { A, B } and C ∈ U V \ {
A, B } . Fix U C ∈ Υ( U ) such that C ∈ U C V . By Lemma 2.8, U C V = C and thus for every V (cid:48) ∈ Υ( V ) , C ∈ U V (cid:48) . ApplyingLemma 2.8 again, gives us that
U V (cid:48) = C and thus U V = C . Thus { U, V } ∈ E n +1 .Case 4: U V = (cid:98) U = (cid:98) V . If V (cid:54)→ U , apply (3 ⇒
1) from Lemma 2.10 to get that U → V . (cid:3) Lemma 2.12.
Suppose H has the tree property for n . The weak components of M n +1 ( H ) are transitive, and if ( U, V ) , ( V, W ) ∈ A n +1 , then ( U, V ) ∼ Σ ( U, W ) .Proof. Let
U, V, W ∈ V n +1 such that U → V and V → W . By Lemma 2.6, | (cid:98) U | = | (cid:98) V | = 2. We consider two cases: (cid:98) U (cid:54) = (cid:98) V and (cid:98) U = (cid:98) V .Suppose (cid:98) U (cid:54) = (cid:98) V and let A ∈ (cid:98) U \ (cid:98) V . Fix U A ∈ Υ( U ) such that U A V = A and V , V ∈ Υ( V ) such that V W (cid:54) = V W . Fix W (cid:48) ∈ Υ( W ). We have that U A , V , W (cid:48) forces U A W (cid:48) ⊆ { A, V W (cid:48) } while U A , V , W (cid:48) forces U A W (cid:48) ⊆ { A, V W (cid:48) } and thus U A W (cid:48) = A . Since W (cid:48) was arbitrary, U A W = A . Note that since | U A W | = 1, wehave ruled out the possibility that W → U . By Lemma 2.11, we will be done ifwe can show that | U W | >
1. Observe that we could also choose U B ∈ Υ( U ) suchthat U B V = B (cid:54) = A . If it happens that B (cid:54)∈ (cid:98) V , by the same reasoning as above U B W = B so that { A, B } ⊆
U W and thus U → W and Σ( U, V ) = Σ(
U, W ).Suppose then that (cid:98) U = { A, B } and (cid:98) V = { B, C } . We can now find U A , U B ∈ Υ( U ) such that U A W = A and U B W ⊆ { B, C } . Therefore | U W | > U → W . By Lemma 2.6, U W = (cid:98) U = { A, B } and thus U B W = B . Therefore, Σ( U, V ) = Σ(
U, W ).We now consider the case (cid:98) U = (cid:98) V = { A, B } . Note that U W ⊆ {
A, B } since wemay otherwise easily form a rainbow triangle. We now have the setup for (1 ⇒ U → V and have x ∈ V \ ( U ∪ V ) such that U → x , Σ( U, V ) =Σ(
U, x ), and xV = C (cid:54)∈ { A, B } . Applying part (1) of Lemma 2.4 to V → W and x , we have that xW = C . Now apply part (2) of Lemma 2.4 with U → x and W toget that either C ∈ U W or U → W and Σ( U, W ) = Σ(
U, x ) = Σ(
U, V ). We havealready ruled out the former while the latter is what we sought to prove. (cid:3)
Lemma 2.13.
Suppose H has the tree property for n . The weak components of M n +1 ( H ) form rooted trees. ALLAI MULTIGRAPHS 9
Proof.
After Lemma 2.12, we need only show that for U , U , V ∈ V n +1 distinct, if U → V and U → V , then either U → U or U → U . By Lemma 2.11, it sufficesto show | U U | > | U U | = 1. Observe that if | U V ∪ U V ∪ U U | >
2, then we must find arainbow triangle in U , U , V . Thus we may assume U U = A ∈ (cid:99) U = (cid:99) U = { A, B } .As in the proof of Lemma 2.10, since (cid:99) U = (cid:99) U , we may select x ∈ V \ ( U ∪ U ) suchthat U x = C (cid:54)∈ { A, B } and A ∈ U x (or with the roles of U and U reversed).We may then apply part (1) of Lemma 2.4 with U → V and x to get that V x = C and apply part (2) of Lemma 2.4 with U → V and x to get that either C ∈ U x or U → x .First we consider the case C ∈ U x . Let k be minimal such that C ∈ [ U ] k x . If k = 0, by Lemma 2.9, U x = C , which contradicts our assumption that A ∈ U x .Thus k > u ∈ [ U ] k \ [ U ] k − such that C ∈ ux . We may applypart (2 (cid:48) ) of Lemma 2.4 with [ U ] k − → u and x to get that either C ∈ [ U ] k − x or [ U ] k − → x . By the minimality of k , we must be in the latter case. However,by assumptions that U U = A and U x = C and thus we may locate a rainbowtriangle in U , [ U ] k − , x .We turn now to the second case, U → x . We may apply part (2) of Lemma2.4 with U → x and U to get that either C ∈ U U or U → U , which bothcontradict our assumption that U U = A . Thus U → U or U → U . (cid:3) Taking Lemmas 2.11, 2.12, and 2.13 together we have proved Theorem 2.3.3.
Construction of Finite Gallai Multigraphs
In Section 2.2, we found that any reduced maximal Gallai multigraph (
G, ρ ) canbe decomposed into a sequence of mixed graphs, ( M k ( G ) , ρ (cid:48) , ∼ Σ ) , having certainproperties. We now reverse this process to construct all finite Gallai multigraphs.An example of this construction is presented in Figure 2. Construction 1 (Edge-colored Multigraph Construction Γ) . Given a triple ( M =( V , E , A ) , ρ (cid:48) , ∼ Σ ) with M a complete mixed graph, ρ (cid:48) a list coloring of E ∪A such that | ρ (cid:48) ( e ) | = 1 for e ∈ E and | ρ (cid:48) ( e ) | = 2 for e ∈ A , and ∼ Σ an equivalence on membersof A sharing an initial vertex, construct a complete, edge-colored multigraph ( G =( V, E ) , ρ ) as follows: (1) Replace each u ∈ V with ( G u = ( E u , V u ) , ρ u ) , a uniformly colored Gallaimultigraph such that if ( u, v ) ∈ A for some v ∈ V , then | V u | ≥ and ρ u [ V u V u ] = ρ (cid:48) (( u, v )) . (2) For each u ∈ V , ρ (cid:22) E u := ρ u . (3) For u, v ∈ V distinct, connect V u and V v as follows: (a) if { u, v } ∈ E , then ρ ( { w , w } ) = ρ (cid:48) ( { u, v } ) for w ∈ V u and w ∈ V v ; (b) if ( u, v ) ∈ A , then define ρ such that (i) V u → V v , (ii) ρ [ V u V v ] = ρ (cid:48) (( u, v )) , and (iii) Σ( V u , V v ) = Σ( V u , V w ) whenever ( u, v ) ∼ Σ ( u, w ) .It is easy to see that ρ can be defined in this way whenever | V u | ≥ asrequired above. We write Γ((
M, ρ (cid:48) , ∼ Σ )) for the family of edge-colored multigraphs resulting fromall possible choices of { ( G u , ρ u ) } in step (1) and permissible definitions of ρ in step(3.b).We use the following notation in the construction of Gallai multigraphs. • G = { ( G i , ρ i ) } is the family of all finite Gallai multigraphs. • G r = { ( G i , ρ i ) } is the family of all finite reduced Gallai multigraphs. • G + r is the family of all induced subgraphs of maximal members of G r . • M = { ( M n ( H ) , ρ, ∼ Σ ) : H ∈ G + r and n ∈ N } . • T is the family of rooted trees in M . • M ∗ is the family of triples ( M, ρ, ∼ Σ ) such that Γ(( M, ρ, ∼ Σ )) ⊆ G . • T ∗ is the family of rooted trees in M ∗ .Lastly, we let G ( n ) = { ( G = ( V, E ) , ρ ) : | V | ≤ n } and likewise for each of thefamilies defined above. We will construct a family M (cid:48) and show that • M ⊆ M (cid:48) ⊆ M ∗ and • G + r ⊆ Γ[ M ] ⊆ Γ[ M (cid:48) ] ⊆ Γ[ M ∗ ] ⊆ G .In particular, we will construct a family of Gallai multigraphs,Γ[ M (cid:48) ], containingthe reduced maximal ones. AAAC AB CDC DAB AABAA AB0 21 3AB AB ABCDAA
Figure 2. ( M, ρ, ∼ Σ ) (left) and a member of Γ(( M, ρ, ∼ Σ )) (right) Lemma 3.1.
Suppose ( M n ( G ) = ( V , A , E ) , ρ, ∼ Σ ) ∈ M . If ρ (cid:48) is a coloring of A∪E such that ρ (cid:48) ( e ) ∈ ρ ( e ) for all e ∈ A ∪ E and ρ (cid:48) ( e ) = ρ (cid:48) ( e ) whenever e ∼ Σ e , then ( M n ( G ) , ρ (cid:48) ) lacks rainbow triangles.Proof. Fix
U, V, W ∈ V distinct. Recall that
U, V, and W are disjoint subsets ofvertices of G and that ( U, V ) ∈ A if and only if U → V in G . We now considereach of the general cases presented in Figure 3. Fix v ∈ V and w ∈ W .Case (A): Select u ∈ U and note that ρ (cid:48) colors the triangle U, V, W just as u, v, w is colored in G .Case (B): Select u ∈ U such that uV = ρ (cid:48) (( U, V )). Again,
U, V, W is colored inthe same way as u, v, w .Cases (C) and (E): Since (
U, V ) ∼ Σ ( U, W ), ρ (cid:48) (( U, V )) = ρ (cid:48) (( U, W )).Case (D): Since (
U, V ) (cid:54)∼ Σ ( U, W ), we know Σ(
U, V ) (cid:54) = Σ( U, W ) and we maythus select u ∈ U such that uV (cid:54) = uW . It might happen that uV = ρ (cid:48) (( U, V ))and uW = ρ (cid:48) (( U, W )). In this case, we again note that ρ (cid:48) colors U, V, W in thesame way that u, v, w is colored in G . Suppose then that we are in the alternate ALLAI MULTIGRAPHS 11
UV W UV W UV W UV W UV W ( A ) ( B ) ( C ) ( D ) ( E ) Figure 3.
General possible relations between
U, V, and W .case: uV = ρ (cid:48) (( U, W )) and uW = ρ (cid:48) (( U, V )). This however also forces
V W = ρ (cid:48) ( { V, W } ) ∈ { ρ (cid:48) (( U, V )) , ρ (cid:48) (( U, W )) } . (cid:3) Lemma 3.2. G + r ⊆ Γ[ M ] ⊆ G and thus M ⊆ M ∗ .Proof. To see that Γ[ M ] ⊆ G , let ( M = ( V , A , E ) , ρ, ∼ Σ ) ∈ M and fix ( G =( V, E ) , ρ (cid:48) ) ∈ Γ((
M, ρ, ∼ Σ )). Let { ( G u = ( E u , V u ) , ρ u ) } be the family of Gallaimultigraphs used in step (1) of the construction of ( G, ρ (cid:48) ). We now show that(
G, ρ (cid:48) ) lacks rainbow triangles.First note that for a given triangle, u, v, w , if any two of the vertices correspondto the same vertex of M , i.e. fall in the same V i , then u, v, w must have a repeatedcolor.Therefore we only need to consider the triangles formed by vertices from different G i . In particular, we may form V (cid:48) ⊆ V by selecting a single vertex from each of V i and consider ( V (cid:48) , E (cid:48) , ρ (cid:48) ), the induced graph of G by V (cid:48) .We will be done if we show that ( V (cid:48) , E (cid:48) , ρ (cid:48) ) lacks rainbow triangles. Noticethat this triple is equivalent to an edge-coloring of M n ( G ) of the form described inLemma 3.1 and thus ( V (cid:48) , E (cid:48) , ρ (cid:48) ) lacks rainbow triangles.Finally, note that H ∈ Γ(( M ( G ) , ρ , ∼ Σ )) for each H ∈ G + r and thus G + r ⊆ Γ[ M ]. (cid:3) We now construct a family M (cid:48) such that M ⊆ M (cid:48) ⊆ M ∗ . Construction 2 (Forest Construction ∆ F ) . Suppose we have at our disposal
N ⊆M ∗ . Set T N = N ∩ T and form a triple ( M (cid:48) = ( V (cid:48) , E (cid:48) , A (cid:48) ) , ρ (cid:48) , ∼ Σ (cid:48) ) as follows: (1) Fix ( M = ( V , E , A ) , ρ, ∼ Σ ) ∈ N . (2) For each u ∈ V , fix ( T u = ( V u , E u , A u ) , ρ u , ∼ Σ u ) ∈ T N such that if ( u, v ) ∈A for some v ∈ V , then (a) (cid:99) T u = ρ (( u, v )) (this is just an issue of labeling) and (b) | V u | ≥ . (3) Set V (cid:48) := ∪ u ∈V V u . (4) Set A (cid:48) := ∪ u ∈V A u . (5) Set E (cid:48) := ( ∪ u ∈V E u ) (cid:83) {{ w , w } : w ∈ V u , w ∈ V v , and u (cid:54) = v } . (6) Define ρ (cid:48) as follows: (a) ρ (cid:48) (cid:22) A u ∪ E u := ρ u for all u ∈ V ; (b) for { w , w } ∈ E (cid:48) where w ∈ V u , w ∈ V v , and u (cid:54) = v , (i) if { u, v } ∈ E , then ρ (cid:48) ( { w , w } ) := ρ ( { u, v } ) ; (ii) if ( u, v ) ∈ A , then define ρ (cid:48) on V u V v so that (A) V u → V v , (B) ρ (cid:48) [ V u V v ] = ρ (( u, v )) , and (C) Σ( V u , V v ) = Σ( V u , V w ) whenever ( u, v ) ∼ Σ ( u, w )Key to this construction is understanding when and how step (6.b.ii) can beaccomplished. We call this the signature configuration problem and resolve it inSection 3.1We write ∆ F ( N ) for the family of all mixed graphs resulting from one iterationof this construction beginning with N . Notice that ∆ F ( N ) constructs no rootedtrees that were not already present in N , and we will therefore need a separateconstruction for trees. We present this tree construction in Section 3.2 and write∆ T ( N ) for the resulting family of trees. Finally, we set ∆( N ) := ∆ F ( N ) ∪ ∆ T ( N ).We let M be the family of elements in M having no directed edges (and are thusessentially Gallai graphs) and for n ≥
0, set M n +1 := ∆( M n ) and M (cid:48) := ∪ ∞ n =0 M n . Theorem 3.3.
M ⊆ M (cid:48) ⊆ M ∗ and thus G + r ⊆ Γ( M (cid:48) ) ⊆ G . We prove Theorem 3.3 after addressing the signature configuration problem inSection 3.1 and describing the tree construction ∆ T in Section 3.2.3.1. Signature Configuration Problem.
Here we address the following basicproblem arising in step (6.b.ii) of the ∆ F construction: given T ∈ T and a setof vertices U , describe all possible ways to join T and U with two colors so that T → U and T ∪ U lacks rainbow triangles.First observe that, since T → U , the set of vertices U is irrelevant and that wemay just as well assume there is only a single vertex u . Now suppose we would liketo define dominance between T = ( V, E, A ) and u using the colors A and B . Up torelabeling, we may assume (cid:98) T = { A, B } . The signature of T → u is determined bythe map Σ : V → { A, B } given by v (cid:55)→ vu .For v , v ∈ V , if v v (cid:54)⊆ { A, B } , then to avoid a rainbow triangle it must be thecase that Σ( v ) = Σ( v ), i.e. v u = v u . We therefore partition V by first definingthe relation v ∼ v if v v (cid:54)⊆ { A, B } and then extending ∼ to an equivalence. Lemma 3.4. T ∪ u lacks rainbow triangles if and only if Σ is constant on ∼ classes.Proof. The statement follows directly from the definition of ∼ . (cid:3) Theorem 3.5.
For T = ( V, E, A ) ∈ T and M ∈ M , the ways of joining T and M so that T → M and T ∪ M lacks rainbow triangles correspond exactly to choicefunctions Σ : (
V / ∼ ) (cid:16) (cid:98) T .Proof. The only difference between this statement and Lemma 3.4 is that T is nowdominating a set of vertices rather than a single vertex, but since T → M , whatevercolors may be present in M cannot form rainbow triangles with vertices from T .The requirement that Σ be onto corresponds to the fact that dominance requiresmultiple colors and | (cid:98) T | = 2. (cid:3) Tree Construction ∆ T . We now describe how to construct rooted trees ofsize n + 1 out of a mixed graph of size n . Here we use the fact that Theorem 2.3holds for all of G + r , rather than just the maximal members.We would like to construct T = ( V, E, A ) ∈ T using elements of M having fewerthan | V | vertices. Perhaps the most natural approach would be to consider thecollection of subtrees { T i } in V \ T as in the left-most image of Figure 4 and then ALLAI MULTIGRAPHS 13 T T T T M M ! T Figure 4.
Constructing a tree from a smaller forest.list all possible ways to assign signatures to 1 T → T i . One quickly realizes thatwith this approach you must also describe how each pair of { T i } is related.Notice, however, that since T ∈ T ⊆ M , V corresponds to the induced subgraphof some reduced maximal Gallai multigraph, and therefore V \ V does as well. Wemay thus use Theorem 2.3 to show that V \ V decomposes into a mixed graph in M . In particular, Lemma 3.6 shows that if we remove the root from T , we are leftwith M ∈ M whose weak components are { T i } , as in the center image of Figure 4.We may now collapse M into M (cid:48) , as in the right-most image of Figure 4. Lemma 3.6.
Let ( G = ( V, E ) , ρ ) ∈ G + r and M k ( G ) = ( V k , E k , A k ) for all k ≤ n .Let V T ∈ V n and let H be the induced subgraph of G by V \ V T . Let M k ( H ) =( V (cid:48) k , E (cid:48) k , A (cid:48) k ) for all k ≤ n . Set V T ( k ) = { W ∈ V k : W ⊆ V T } . Then V k = V (cid:48) k ∪ V T ( k ) for all k ≤ n .Proof. The claim holds for k = 0 since V = V , V (cid:48) = V \ V T , and V T (0) = V T .Let k + 1 ≤ n be minimal such that V k +1 (cid:54) = V (cid:48) k +1 ∪ V T ( k + 1). By definition, V T ( k + 1) agrees with V k +1 on the V T portion of V . It must be the case then that V k +1 \ V T ( k + 1) (cid:54) = V (cid:48) k +1 . That is, V k +1 \ V T ( k + 1) and V (cid:48) k +1 represent two differentpartitions of V \ V T . However, by the minimality of k + 1, V k \ V T ( k ) = V (cid:48) k . Notethat V (cid:48) k +1 is completely determined by the dominance relations between membersof V (cid:48) k . Likewise, since V T ∈ V n , none of the members of V (cid:48) k is in a dominancerelation with any member of V T ( k ) and thus V (cid:48) k +1 \ V T ( k + 1) is also determined bythe dominance relations between members of V (cid:48) ( k ). Thus V k +1 \ V T ( k + 1) = V (cid:48) k +1 ,a contradiction. (cid:3) We will be done once we specify all ways in which we can assign signaturesto the directed edges from 1 T to the vertices of M (cid:48) . In particular, a choice ofsignatures corresponds exactly to a partition of the vertices of M (cid:48) . What then arethe necessary and sufficient conditions on this partition to produce a valid choiceof signatures? Lemma 3.7 answers this question in much the same way as Lemma3.4 did for the signature configuration problem. Lemma 3.7.
For ( T = ( V, E, A ) , ρ, ∼ ) ∈ T and { u, v } ∈ E , if ρ ( { u, v } ) (cid:54)⊆ (cid:98) T , then Σ(1 T , u ) = Σ(1 T , v ) .Proof. Suppose (cid:98) T = { A, B } and C ∈ ρ ( { u, v } ) \ { A, B } . If Σ(1 T , u ) (cid:54) = Σ(1 T , v ),then we may form a rainbow triangle by selecting A for (1 T , u ), B for (1 T , v ), and C for { u, v } . (cid:3) Construction 3 (Tree Construction: ∆ T ) . For
N ⊆ M ∗ , fix ( M = ( V, E, A ) , ρ, ∼ Σ ) ∈ N and colors { A, B } . Let V = ∪ V i be the vertex partition of M into weak com-ponents. Let T be a new vertex and form ( T := ( V ∪ T , E (cid:48) , A (cid:48) ) , ρ (cid:48) , ∼ Σ (cid:48) ) as follows: (1) E (cid:48) := E ; (2) A (cid:48) := A ∪ ( { T } × { V } ) ; (3) ρ (cid:48) ( e ) := { A, B } for all e ∈ { T } × { V } and ρ (cid:48) (cid:22) E ∪ A := ρ ; (4) Define ∼ Σ (cid:48) on A (cid:48) by (a) (1 T , u ) ∼ Σ (cid:48) (1 T , v ) whenever ( u, v ) ∈ A , (b) if V i V j (cid:54)⊆ { A, B } , then (1 T , u ) ∼ Σ (cid:48) (1 T , v ) for all u ∈ V i , v ∈ V j , (c) if ( u, v ) ∼ Σ ( u, v ) , then ( u, v ) ∼ Σ (cid:48) ( u, v ) for all u, v i ∈ V . We write ∆ T ( N ) for the family of mixed graphs resulting from all possible choicesof ( M, ρ, ∼ Σ ) ∈ N and possible definitions of ∼ Σ in step (4). Theorem 3.8. T ( n + 1) ⊆ ∆ T ( M ( n )) ⊆ ∆ T ( M ∗ ) ⊆ T ∗ .Proof. Since M ( n ) ⊆ M ∗ , we have ∆ T ( M ( n )) ⊆ ∆ T ( M ∗ ). We now show that∆ T ( M ∗ ) ⊆ T ∗ . Fix M ∈ M ∗ , T ∈ ∆ T ( M ), and G ∈ Γ( T ).By construction, it is clear that T is a tree with root 1 T so we only need to verify T ∈ T ∗ , i.e. G ∈ G . Since M ∈ M ∗ , the portion of G corresponding to M , thatis G with 1 T removed, will certainly lack rainbow triangles. Likewise, if a trianglefalls entirely in the portion of G corresponding to 1 T , then it will have at most twocolors. We thus only need to consider the general types of triangles presented inFigure 5. T T T T T U U U U UV V V ( A ) ( B ) ( C ) ( D ) ( E ) Figure 5.
Case (A) cannot lead to a rainbow triangle since 1 T T ⊆ (cid:98) T = 1 T U . Likewise, incase (B), since 1 T → U , the vertex in 1 T must be connected to U by a single color.Cases (C) and (D) cannot contain a rainbow triangle since Σ(1 T , U ) = Σ(1 T , V ).Finally, by construction, if ρ ( U, V ) (cid:54)⊆ (cid:98) T , then Σ(1 T , U ) = Σ(1 T , V ). Since incase (E), Σ(1 T , U ) (cid:54) = Σ(1 T , V ) it must be the case that ρ ( U, V ) ⊆ (cid:98) T and thus thefigure cannot contain a rainbow triangle. We have thus shown that ∆ T ( M ∗ ) ⊆ T ∗ .We now show T ( n + 1) ⊆ ∆ T ( M ( n )). Fix T ∈ T ( n + 1). Since T ( n + 1) ⊆ M , T = M k ( H ) for some H ∈ G + r . Let H (cid:48) be the subgraph of H induced by removingthe vertices corresponding to 1 T . Again, H (cid:48) ∈ G + r and thus M k ( H (cid:48) ) ∈ M . ByLemma 3.6 however, M k ( H (cid:48) ) will have one less vertex than T and thus M k ( H (cid:48) ) ∈M ( n ). Since T ∈ ∆ T ( M k ( H (cid:48) )) ⊆ ∆ T ( M ( n )), we have T ( n + 1) ⊆ ∆ T ( M ( n )). (cid:3) Proof of Theorem 3.3.
We may now show that
M ⊆ M (cid:48) ⊆ M ∗ . ALLAI MULTIGRAPHS 15
Proof.
It is clear that
M ⊆ M (cid:48) .Certainly M ⊆ M ∗ since M is just the family of Gallai graphs. Now con-sider M n +1 = ∆ T ( M n ) ∪ ∆ F ( M n ) where M n ⊆ M ∗ . Theorem 3.8 shows that∆ T ( M n ) ⊆ M ∗ .Fix ( M = ( V, E, A ) , ρ, ∼ Σ ) ∈ ∆ F ( M n ) and ( G, ρ ) ∈ Γ( M ). Recall that M was constructed by replacing vertices of M (cid:48) ∈ M n with trees { T i } ⊆ M n andjoining these trees subject to certain constraints. Recall further than G was formedby replacing the vertices of M , which as just mentioned can be thought of asthe vertices of the collection { T i } , with complete edge-colored multigraphs, say { G i } ∈ G .Now consider a triangle u, v, w in G . Let u fall in the portion of G correspondingto T u , v in T v , and w in T w . Up to relabeling, we may then assume we are in oneof the following cases:(1) T u = T v = T w ,(2) T u , T v , and T w are distinct,(3) T v = T w and T u → T v , or(4) T u = T v and T u → T w .Cases (1) and (2) are resolved by the facts that T u ∈ M ∗ and M (cid:48) ∈ M ∗ ,respectively. Case (3) is straightforward since Σ( T u , T v ) = Σ( T u , v ) = Σ( T u , w ) andthus uv = uw .Case (4) requires some consideration of Σ( T u , T w ). If it happens that Σ( T u , T w )( u ) =Σ( T u , T w )( v ), then uw = vw and we are done. Suppose then that Σ( T u , T w )( u ) (cid:54) =Σ( T u , T w )( v ). By construction, this implies that u and v are not connected by apath in T u whose colors fall outside (cid:99) T u . In particular, uv ⊆ (cid:99) T u . Since we also knowthat uw, vw ⊆ (cid:99) T u and | (cid:99) T u | = 2, we are done. (cid:3) Open Problems
We approached the topic of Gallai multigraphs in an attempt to generalize theconstruction of Gallai graphs for graphs lacking rainbow 4-cycles. We remain in-terested in this problem and in the following somewhat more general questionssuggested by the multigraph perspective:(1) Can ∆ be generalized to construct finite multigraphs lacking rainbow n -cycles for a fixed n ?(2) Is there a construction of all finite (not necessarily complete) graphs ormultigraphs lacking rainbow triangles?5. Acknowledgments
I would like to thank Rick Ball and Petr Vojt˘echovsk´y for suggesting to me thetopic of Gallai graphs, and Petr for his many helpful comments. I would also liketo thank Peter Keevash for informing me of [4].
References [1] B. Alexeev,
On lengths of rainbow cycles , to appear in Electronic J. Combinatorics.[2] R. N. Ball, A. Pultr and P. Vojt˘echovs´ky,
Colored graphs without colorful cycles , toappear in Combinatorica. [3] T. Gallai,
Transitiv oreintierbare Graphen , Acta Math. Acad. Sci. Hungar (1967),2566. English translation by F. Maffray and M. Preissmann, in Perfect Graphs, editedby J. L. Ramirez-Alfonsin and B. A. Reed, John Wiley and Sons.[4] D. Mugayi and A. Diwan, Tur´an’s Theorem with Colors , submitted.[5] P. Vojt˘echovsk´y,
Periods in Missing Lengths of Rainbow Cycles , submitted to Journalof Graph Theory.
E-mail address , Pula: [email protected]@math.du.edu