ON INDUCED SUBGRAPHS WITH ALL DEGREES ODD A.D. SCOTT Abstract. Gallai proved that the vertex set of any graph can be partitioned into two sets, each inducing a subgraph with all degrees even. We prove that every connected graph of even order has a vertex partition into sets inducing subgraphs with all degrees odd, and give bounds for the number of sets of this type required for vertex partitions and vertex covers. We also give results on the partitioning and covering problems for random graphs. Note: For the final version of this paper, see the journal publication. 1. Introduction Gallai (see [10], Exercise 5.19) proved that every graph G has a ver- tex partition into two sets, each of which induces a subgraph with all degrees even. Thus every graph G contains an induced subgraph with all degrees even on at least djGj=2e vertices. For induced subgraphs with all degrees odd, however, the situation is less clear. Clearly no subgraph with odd degrees can have an isolated vertex, so we must restrict our attention to graphs without isolated vertices. Let us write f(G) for the maximum order of an induced subgraph of G with all degrees odd and f(n) for the minimum of f(G) over the set of graphsp of order n without isolated vertices. Caro [5] proved that f(n) ≥ c n for n ≥ 2. It was proved in [12] that (1) f(n) ≥ cn= log n: Best possible bounds for f(T ) where T is a tree have been proved by Radcliffe and Scott [11], while Berman, Wang and Wargo [3] have proved optimal bounds for graphs with maximum degree 3. It has been conjectured (see [5]) that there is c > 0 such that f(n) ≥ cn for n ≥ 2. In this paper we prove results about partitioning and covering by induced subgraphs with all degrees odd. We begin by asking under what conditions a graph G has a vertex partition into sets that induce subgraphs with all degrees odd. A graph with all degrees odd must have even order, so clearly every component of a graph with such a 1 2 A.D. SCOTT vertex partition must have even order. Surprisingly, this trivial neces- sary condition is in fact sufficient, and we prove this in x2. It would be interesting to determine the number of sets required for a vertex partition of a graph of order n into sets inducing subgraphs with all degrees odd. Itp is known that o(n) sets suffice and we give an example showing that c n sets may be required. In x3, we look at vertex coverings by sets that induce subgraphs with all degrees odd. It turns out that we can in general achieve this with far fewer sets than for a vertex partition: we prove that every graph has a vertex covering of this type with c log2 n sets and give an example showing that we may need c log n sets. We would expect that for most graphs we could do rather better than this, and in x4 we prove that this is the case. In particular, we 1 prove that almost every graph G 2 G(n; 2 ) has a vertex partition (and hence a vertex covering) with three sets inducing subgraphs with all degrees odd. In x5, we turn to residue classes modulo k. In particular, following Caro [5] and Caro, Krasikov and Roditty [6], for a graph G without isolated vertices, we define fk(G) to be the maximum order of an in- duced subgraph of G with all degrees congruent to 1 modulo k. We give a lower bound on fk(G) in terms of the chromatic number of G, and prove that every connected claw-free graph G of order n satis- p fies fk(G) ≥ (1 + o(1)) n=12. This improves upon a result of Caro [5], who gave a lower bound of c(n log n)1=3. Best possible bounds for fk(T ), where T is a tree, are given by Berman, Radcliffe, Scott, Wang and Wargo [2]. Finally, in x6, we consider some open questions. 2. Partitions into induced subgraphs Under what conditions does a graph G have a vertex partition into sets inducing subgraphs with all degrees odd? Given a graph G, we say that a partition V1;:::;Vk of V (G) is a good partition of G if G[Vi] has all degrees odd for i = 1; : : : ; k. Any subgraph with all degrees odd must have an even number of vertices, so if G has a good parti- tion then it must have an even number of vertices in each component. Surprisingly, this trivial condition is also sufficient. Theorem 1. Let G be a graph. Then G has a good partition if and only if every component of G has even order. Proof. If a graph G has a good partition then it is clearly necessary that every component of G have even order. We prove by induction ON INDUCED SUBGRAPHS WITH ALL DEGREES ODD 3 on jGj that if every component of G has even order then G has a good partition. If jGj = 2 then the statement is trivial. We suppose that jGj > 2, that every component of G has even order, and that the inductive statement is true for graphs of smaller order. We may assume that G is connected (and thus has even order), or else we could partition every component of G separately. If, for every edge xy in G, we have d(x) = 1 or d(y) = 1, then G is a star: since jGj is even, G must itself have all degrees odd, so we can take the trivial partition. Otherwise, we may assume that we can find an edge xy with d(x) > 1 and d(y) > 1. Let G0 = G n fx; yg. If G0 is connected, then by the inductive hypothesis we can find some 0 good partition V1;:::;Vk of G , in which case V1;:::;Vk; fx; yg is a good partition of G. Also, if any component S of G0 has even order then we can find a good partition of S, and since GnS is connected we can find a good partition of G n S and combine these partitions to give a good partition of G. Thus we may assume that every component of G0 has odd order. 0 Let X1;:::;Xi be the components of G adjacent to x but not to y; 0 let Y1;:::;Yj be the components of G adjacent to y but not to x; and 0 let Z1;:::;Zk be the components of G adjacent to both x and y. Since G has even order, i + j + k must be even. If k > 0 then let i k [ [ X = fxg [ V (Xm) [ V (Zm); m=1 m=2 and j [ Y = fyg [ V (Ym): m=1 Exactly one of jXj and jY j is odd: adding V (Z1) to that set gives a partition of V (G) into two sets inducing non-empty connected graphs with even order. Both graphs have a good partition by the inductive hypothesis, and these can be combined to obtain a good partition for G. We may therefore assume that k = 0, so that i and j are both even or both odd. Now if i and j are both odd, then let i [ X = fxg [ V (Xm); m=1 4 A.D. SCOTT and j [ Y = fyg [ V (Ym): m=1 Both G[X] and G[Y ] are connected graphs with even order smaller than jGj, so both have good partitions, and hence G does as well. Otherwise, both i and j are even (and i + j ≥ 2). Let i [ X = fx; yg [ V (Xm); m=1 and i [ Y = fx; yg [ V (Ym): m=1 Let V1;:::;Vp be a good partition of G[X] and W1;:::;Wq be a good partition of G[Y ]. We may assume that x 2 V1 and y 2 W1, and so y 2 V1 and x 2 W1. Then V1 [ W1;V2;:::;Vp;W2;:::;Wq is a good partition of G. Can we say anything further about the subgraphs induced by the vertex sets in a good partition? We know from Theorem 1 that if G is a connected graph of even order, then G has a good partition. We say that G is a basic odd graph if G is minimal with respect to these partitions: in other words, G has all degrees odd, and the only good partition of G is the trivial partition fV (G)g. It is easy to characterize the basic odd graphs. Lemma 2. A graph G is a basic odd graph iff it is a tree with all degrees odd. Proof. Suppose first that G is a tree with all degrees odd and it has a good partition into sets V1;:::;Vk, with k ≥ 2. Let xy be any edge in G between V1 and V (G) n V1, and let Gx and Gy be the components of G with the edge xy removed, where x 2 V (Gx) and y 2 V (Gy). Since every vertex of Gx has odd degree (in G and hence in Gx), except for x, which has even degree in Gx, we must have jGxj odd, and similarly jGyj odd. However, there are clearly no edges other than xy between Vi \ V (Gx) and Vi \ V (Gy) , for i = 1; : : : ; k, so V1 \ V (Gx);:::;Vk \ V (Gx) must be a good partition of Gx, which is impossible, since jGxj is odd.