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On k-rails in graphs
JOURNAL
OF COMBINATORIAL
THEORY
@)
17, 143-159 (1974)
On k-Rails in Graphs Bo AAGAARD Matematisk
SBRENSEN AND CARSTEN THOMA~~EN
Xnstitut, Aarhus Uninersitet, Aarhus, Denmark Communicated by Frank Harary Received July 25, 1973
A k-rail is the union of k paths each pair of which has exactly the endvertices in common. We first give a sufficient degree-condition for the existence of k-rails in graphs. Next we determine the number of edges required in 3connected graphs to guarantee the existence of Srails, and finally we solve the corresponding problem for graphs with no prescribed connectivity.
1. INTRODUCTION A k-rail is the union of k paths each pair of which has exactly the end vertices in common. For 123 k + 1 > 3, we define f&) as the least integer r so that every graph with n vertices and r or more edges contains a k-rail. It was conjectured by Bollobas and Erdijs [l] that every graph with p(k - 1) + 1 vertices and g(k - 1) kp + 1 or more edges contains a k-rail. If true, this would be best possible and consequently limn+, (l/n)f,(n) = Sk. For k = 4, the conjecture was proved by Bollobis [2] and in [lo] this was obtained from a more general result. Leonard [6] disproved the conjecture for k = 5, and Mader [9] showed that for all integers k, m, where k > 5, m > O,f&z) > &kn + m for some n. The conjecture becomes true of the term “k-rail” is replaced by “k mutually edge-disjoint paths connecting two vertices”. For k = 5, this was proved by Leonard [7], and Mader [9] proved it in the general case. In Section 3 we prove that a graph G has a k-rail for k > 3 if every vertex of G has degree 3 k - 1 and if furthermore every circuit of G has at least two vertices of degree > k. This is based on a result of Mader [8]. In Section 4 we show that the conjecture of Bolloblis and Erdijs becomes true for k = 5 provided the graphs considered are 3-connected. Finally, we show in Section 5 that for n > 6, n # 7, IZ # 12, &(n) = [$z] - 3. 143 Copyright All rights
@?I1974 by Academic Press, Inc. of reproduction in any form reserved.
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We also here show that for k fixed, j&z) > “‘“,,“,
2 (n - k)
for infinitely many n. This disproves the conjecture of Bollobais and Erdiis for all k 3 5.
2. TERMINOLOGY
AND PRELIMINARIES
We consider finite, undirected graphs without loops and without multiple edges. V(G) and E(G) denote the set of vertices and the set of edges, respectively, of the graph G and n(G) = ( V(G)1 and e(G) = I E(G)/. (x, y) denotes the edge joining the vertices x and y. The terms degree, path, connected graph, k-connected graph, connected component, cutvertex, block and complete graph (K,,) mean the same as in Harary [5]. The term circuit means the same as cycle in [5]. The set of vertices joined to the vertex x in the graph G is denoted by N(x, G), and the degree of x in G is denoted by d(x, G), i.e., d(x, G) = ) N(x, G)/. If A, B _CV(G), A n B = o, then an A - B path is a path P joining a vertex x E A and a vertex y E B so that (A u B) n P = {x, y}. An {x} - A path and an (x} - ( y} path will be called an x - A path and an x - ypath, respectively. A k-rail between (or connecting) the vertices x and y is the union of k x - y paths each pair of which has exactly x and y in common. If S C V(G) u E(G), then G - S denotes the graph obtained by deleting from G all vertices of S n V(G) (if any) together with all edges of G incident with them and then deleting all edges of S n E(G) still present. G - {x} and G - {(x, y)} will be denoted by G - x and G - (x, y), respectively. If A C V(G), we define G(A) = G - (V(G) - A). G(A) is said to be the subgraph of G spanned by A. A spanned&graph is a subgraph of the form G(A), where A C V(G). If G is a connected graph, S C V(G), and G - S is disconnected, then S is a separating set of vertices of G. Suppose G is a k-connected but not (k + I)-connected graph and G f &+I . Then G has a separating set S of k vertices and G has two spanned subgraphs G, and G, (which are not necessarily unique) so that G = G, u G, , V(G,) n V(G,) = S, V(G,) - V(G,) # o and V(G,) - V(G,) # o. G, (and also G,) is said to be a k-fragment with attachvertices S. Every k-fragment can be obtained in this way. We need the following versions of Menger’s Theorem: 1. Let x, y E V(G), (x, y) 4 E(G). fl fir all S C V(G) - (x, y} with 1S 1 < k - 1 G - S has an x - ypath, then G has a k-rail between x andy.
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145
2. Let x E V(G) and A C V(G) - {x} so that 1A 1 > k. If for all ScV(G)-{x}withISI
results are special cases of a theorem of Dirac [3, Theorem B].
3. DEGREE-CONDITIONSFOR THE EXISTENCEOF k-RAILS We need the following result of Mader [8, Lemma I]: LEMMA 1. Let G be a graph and H a complete subgraph with vertices vl, va ,..., v, (m 3 1). Let A C E(G) so that (i) (ii)
every edge of A joins a vertex of H and a vertex of G - V(H); if(x, vi) E A, then for all j < i, (x, vi) E E(G) - A.
Let PI, Pz,..., Pn be a k-rail between x1 and x2 where x1, x2 E V(G) - V(H), and assume that at least one of the paths PI , P, ,..., P, contains some edge of A. Then G - A contains a (k - I)-rail PI’, PSI,..., PL-, between x1 and x2 and two paths Pl, P2 so that for i = 1,2, Pi is a path between xi and a vertex which in G is incident with an edge of A, P1 n P2 = 0, and Pi n Pj’ = (xi> for i = 1, 2 and j = 1, 2 ,..., k - 1.
LEMMA 2. Let G be a graph containing a complete subgraph H and a circuit C with no diagonals so that (i) (ii) (iii) (iv)
G - V(H) contains no circuit; H and C have exactly one vertex, x1 say, in common. VXEV(G)-V(H):d(x,G)>k-1111; C contains a vertex x2 so that x2 # x1 and d(x, , G) > k.
Then G contains a k-rail between x1 and x2. Proof. Induction over k. The statement is clearly true for k = 2. Assume that the statement holds for k < k, - 1, and let G be a graph satisfying (i), (ii), (iii), and (iv) with k = k,, b 3. We shall show that G contains a k,-rail between x1 and x2 . Suppose first that x2 is joined to some vertex x, E V(H) - {x1}. G - x3 satisfies (i), (ii), (iii), and (iv) with H replaced by H - x, and k = k, - 1. By the induction hypothesis, G - x, contains a (k, - I)-rail between x1 and x, . If we add the path x1 , x, , x2 , we obtain a k,-rail between x1 and x2 . We may therefore assume that x2 is not joined to any vertex of H - x,. x2 is joined to some vertex x, not contained in C, because d(x, , G) > k,, 3 3. F = G - (V(H) u {x2}) contains no circuit. Let T denote the
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connected component of F containing xQ , and let P be a path of T connecting xQwith a vertex xp so that d(xp , G - V(H)) = 1. Then P n C = o and xp is joined to at least k, - 2 vertices of H. It is easy to see that G contains a 3-rail between x1 and x2, so we may assume that k, > 4. Then X, is joined to some vertex x5 E V(H) - (xl>. Let G’ = G - (V(T) U {x5}). G’ satisfies the conditions (i)-(iv) with H replaced by H - x5 and k = k, - 1. By the induction hypothesis, G’ contains a (k, - 1)-rail between x1 and x2. Adding the path (x1, x5, xp) u P u (x, , x2) we obtain a k,-rail. This concludes the proof of the lemma. THEOREM 1. Let k be a fixed integer > 2. Let G be a graph with j V(G)1 > k + 1, and let H be a complete subgraph with 1 < 1 V(H)1 < k - 1. If the following conditions are satisjed
(1)
Vx E V(G) - V(H): d(x, G) 2 k - 1,
(2) G - V(H) contains a circuit, and every circuit contains at least two vertices of degree k or more (in G), then G contains a k-rail between two vertices of G -
of G - V(H)
V(H).
Proof. For k = 2, there is nothing to prove, so assume that k > 3. We prove the theorem by induction over the number of vertices of G. If 1 V(G)/ = k + 1, then G - V(H) contains two vertices x1, x2 so that d(x, , G) 3 k, d(x, , G) > k. Clearly, G contains a k-rail between x1 and x2 . Assume then that the theorem is true for graphs with fewer than n vertices (n > k + 2), and let G be a graph with n vertices and a complete subgraph H so that (1) and (2) holds. We shall show that G contains a k-rail between two vertices of G - V(H). Let M be a complete subgraph of G so that H C M, and A4 is not a proper subgraph of any complete subgraph of G. V(H) = {vl, v2 ,..,, vls} and V(M) = {vl , v2 ,..., v,}. If m = h = 1, then d(v,, G) = 0. In this case, let u be a vertex of G - v1 so that G - (u, ul> contains a circuit (clearly such a vertex exists). Then (1) and (2) hold with G and H replaced by G - vi and {u}, respectively. By the induction hypothesis, G - v1 contains a k-rail. If m >, k + 1, then G contains a k-rail between v, and v,,-~ . So we may assume that 2 < m < k. We shall consider the following two cases: Case 1. G - V(M) contains no circuit. Lets = min{i / G - {a1 ,..., vi} contains no circuit}. Then s >, h + 1 because G - V(H) contains a circuit. Let C be a shortest circuit of G - {vl , v2 ,..., u,-~). Then C has no diagonals and C contains v, . C is contained in G - V(H) so C contains a vertex z so that z # Y, and d(z, G) > k. By Lemma 2, G contains a k-rail between z and v, .
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Case 2. G - V(W) contains a circuit. Because of the maximality property of M, no vertex of G - V(M) is joined to every vertex of M. For x ES = N(u, , G) - (vl , a2 ,..., urn}, we define
r(x) = min(i I (x, ui) # Z(G)} and we define G’ as the graph obtained from G - v, by adding the edges (x, a++), x E S. Put A = {(x, u,(,)) / x E S} (S and A may be empty). Then G’ satisfies conditions (1) and (2) with H replaced by A4 - urn . By the induction hypothesis, G’ contains a k-rail between two vertices x1, x2 E V(G’) - (V(M) - {u,}) = V(G) -
V(M).
If this k-rail is contained in G, we have finished. In the alternative case, we apply Lemma 1 and conclude that G’ - A = G - v, contains a (k - 1)-rail PI , P, ,..., P,-, between two vertices x1, x2 E V(G) - V(M) and two paths Pl, P2 where for i = 1,2, Pi connects xi with a vertex zi E s u (V(M) - {u,}) = N(v, ) G). Furthermore, P1 n P2 = o, and Pi n Pi = {xi} for i = 1, 2 andj = 1, 2 ,..., k - 1. Let P, denote the path which is the union of the paths PI, P2 and z1 , vln , z2 . Then PI , Pz ,..., PI, is a k-rail of G. This concludes the proof of Theorem 1. THEOREM2. Let G be a graph so that every vertex of G has degree > k - 1 3 2 and so that every circuit contains at least two vertices of degree k or more. Then G contains a k-rail. Proof.
Follows easily from Theorem 1.
Remark. The property required by the graph G in Theorem 2 may be expressed as follows:
(a) Every vertex of G has degree 3 k - 1. (b) If A denotes the set of vertices having degree precisely k - 1, then G(A) contains no circuit. (c) For every x E V(G) - A and for every connected component T of G(A), x is joined to at most one vertex of T. COROLLARY 1. Let G be a graph so that every vertex of G has degree >, k - 1 3 2 and so that no two vertices of degree precisely k - 1 are adjacent. Then G contains a k-rail.
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4. THE NUMBER OF EDGES REQUIRED TO GUARANTEE ~-RAILS
IN j-CONNECTED
THE EXISTENCE OF
GRAPHS
We shall first give some elementary results on 3-fragments. LEMMA 3. Let G be a 3-fragment H = G - E(G({u, v, w})).
with attachvertices
u, v, w. Let
(a) (b)
H + (u, v) + (u, w) + (v, w) is 3-connected. If H - w has a circuit containing u and v, then H+(u, w)+(v, w) is 3-connected. (c) If H has at least jive vertices and x is a cutvertex of H, then H - x has exactly two connected components. One of these has exactly one vertex and this vertex is contained in {u, v, w}. The statement remains true if H is replaced by G.
(d) If d(u, H) >, 2, then H contains two u - {v, w} paths having exactly u in common. (e) If d(u, H) > 2 and d(v, H) 3 2, then H has a circuit which contains u and v. Furthermore, if H - w has no such circuit, then H - u has a circuit containing v, w, and H - v has a circuit containing u, w.
(f ) If G’ is a 3-fragment with attachvertices u’, v’, w’ and G’ n H = .@, then the graph G” obtained from G’ and H by identtyying u, u’ and v, v’ and w, w’ is a 3-connected graph provided every vertex of this graph has degree 3 or more. In particular, if we choose G’ so that d(u’, G’) > 2, d(v’, G’) 2 2, d(w’, G’) > 2, and u’, v’, w’ are independent in G’, we see that His a 3-fragment with attachvertices u, v, w. Proof. For every z E V(H) - {u, v, w}, H contains three z - {u, v, w} paths each pair of which has exactly z in common. From this it immediately follows that if a proper subset of {u, v, w} is deleted from H, then the resulting graph is connected. Also, if x, y are any two vertices of H, then every connected component of H - {x, y> contains at least one of the vertices u, v, w. (a) easily follows from this. Proof of(b). Let x, y E V(H). We shall show that H’ = H + (u, v) + (v, w) - {x, y} is connected. Since every connected component of H - {x, y} contains at least one of the vertices u, v, w, it is sufficient to show that the vertices of {u, v, w} - {x, y} belong to the same connected component of H’. If w $ (x, y} or if (x, y} C {u, v, w}, this is obvious. If, on the other hand, x = w and y 4 {u, v}, then H’ = H - {w, y}, and H’ therefore contains a u - v path.
ON ~-RAILS
IN GRAPHS
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Proofof( Suppose x is a cutvertex of H. Let y E V(H) - (u, u, w, x}. H contains three y - {u, ZJ,w} paths each pair of which has exactly y in common. We may therefore w.1.o.g. assume that H - x contains a y - u path and a y - u path. So u and v belong to the same connected component, M say, of H - X. By a similar argument, we see that for every y’ E V(H) - {u, 0, w, x>, H - x contains either a y’ - u path or a y’ - u path, which implies that y’ E V(M). So H - x has two connected components, namely M and {w}. It is easy to see that (c) still holds if H is replaced by G. Proof@(d). For every x E V(H) - {u}, H - x has either a u - u path or a u - w path by (c). Combining this with Menger’s theorem we obtain (d). Proof of(e). For every x E V(H) - {u, a}, u and u belong to the same connected component of H - x by (c). This implies that H has a circuit containing u and v. Suppose now that every such circuit also contains w. Let C be a circuit containing U, U, w. C - {u, w} = P, U Pz , where PI and Pz are disjoint paths. H - {u, w} is connected and therefore contains a V(P,) - V(P,) path, P3 say. It is easy to see that C U P, contains a circuit C’ which contains u and exactly one of the vertices U, w. By the assumption that H - w has no circuit containing u and 0, C’ contains v and w but not U. Similarly, we see that H - z, has a circuit containing u and w. Proof of(f). We consider G’ and H as subgraphs of G”. Assume that every vertex of G” has degree 3 3 in G”. Let x, y be any vertices. We shall show that G” - {x, y} is connected, From the first paragraph of the proof of the lemma it follows that it is sufficient to show that the vertices of {u, ZJ,w> - {x, y} belong to the same connected component of G” - {x, y}. Also, it follows that this is the case if {x, y} C V(C) or {x, y> C V(H). So we may assume that x E V(H) - V(C) and y E V(G’) - V(H) and that the graphs H - x and G’ - y are both disconnected. Suppose first that V(H) = {x, U, v, w>. Then d(z, G’) > 2 Vz E V(G’). By (c) G’ - y is connected, which is a contradiction. Suppose next that 1 V(H)1 > 5. By (c) H - x has precisely two connected components one of which consists of, say, U. In particular, d(u, H) = 1 and u and w belong to the same connected component of H - X. Then d(u’, G’) > 2, and from (c) it easily follows that G’ - y contains either a u’ - w’ path or a u’ - v’ path. So U, u and w belong to the same connected component of G” - {x, y}. This concludes the proof of Lemma 3. THEOREM 3. Let G be a 3-connected graph which contains no 5-rail. Then e(G) < #(n(G) - 1). Furthermore, ifG has a vertex of degree 3, then e(G) < Q(n(G) - 1).
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Proof. Induction over the number of vertices of G. For 4 < n(G) < 6, it is easy to verify the statement. So we assume that the theorem holds for graphs with fewer than n, vertices (n, > 7) and let G be a 3-connected graph with no 5-rail and with n(G) = n, . We shall then show that e(G) < #(n,, - 1) and that strict inequality holds if G has a vertex of degree 3. We shall consider the following cases separately. Case 1. G has three edges e, , e, , e3 so that G - {e, , e2 , es} is disconnected and each connected component of G - {e, , e2, es>contains more than one vertex.
In this case G - {e, , e2 , es} = G, u G2 , where G, n G, = 0. For i = 1, 2, 3, let xi and yi be the endvertices of ei in G, and G, , respectively. The vertices x1 , x2 , xg , y1 , y2 , y, are distinct because G is 3-connected. For i = 1, 2, Gi is either a K3 or a 3-fragment. For i = 1, 2, let Gi’ denote the graph obtained from Gi by adding a new vertex zi and joining zi to Xl, x2 , xJ( y1 , yz , y3) if i = l/(i = 2). By Lemma 3(f), G,’ and G,’ are 3-connected, and clearly they contain no 5-rail. By the induction hypothesis, e(G,‘) < #(n(Gi’) - 1) for i = 1,2, so e(G) = e(G,‘) + e(G,‘) - 3 < &(n(G1’) + n(Gz’) - 2) - 3 = +n, - 3.
In the next five cases, G has a separating set consisting of three vertices x, y, 2, i.e., G is the union of two spanned subgraphs G1 and G, so that W-2 n W,) = k Y, 4, I/(G) - W,) f 0, W,) - W,) Z 0. G, and G, are 3-fragments, and for i = 1,2, we define Hi = Gi - E(G({xv Y, Z>)>* Case 2. d(x, Hi) 2 2, d(y, Hi) >, 2 and d(z, Hi) 3 2 for Every vertex of G has degree > 4.
i = 1, 2.
Assume w.1.o.g. (Lemma 3(e)) that H1 - z has a circuit containing x and y and that H, - x has a circuit containing y and z. Also, Hz has a circuit containing x and y. G has no 5-rail so (x, y) 4 E(G). By symmetry, (y, z) $ E(G). Let F1 denote the graph obtained from HI by adding the edges (x, y), (y, z), and (x, z). By Lemma 3(a), Fl is 3-connected. We shall show that Fl has no kail. Suppose (reductio ad absurdum) that Fl has a 5-rail. If this 5-rail contains at most two of the edges (x, y), (y, z), and (x, z), then we obtain a 5-rail in G by replacing these edges by appropriate paths in H, (Lemma 3(d)). If the 5-rail of Fl contains all three edges (x, y), (y, z), and (x, z), then the 5-rail connects two vertices vl, DoE {x, y, z}. Let bd = b, Y, z> - h , v2}. Then HI - vg contains a 3-rail between v1 and v2 . Ha has a circuit containing v1 and v2 (Lemma (3(e)) so G contains a 5-r&1. This contradiction shows that Fl has no 5-rail. By the induction hypothesis, e(H,) = e(F,) - 3 < s(n(Hl) - 1) - 3. By symmetry, e(H,) <
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*(n(HJ - 1) - 3. So e(G) < e(H,) + e(H,) + 1 < $(~(Hi) + n(H,) - 2) - 5 = 8@(G) - 0 Case 3. d(x, Hi) 2 2, d(y, Hi) 2 2 for i = 1, 2, d(z, H1) >, 2, d(z, H,) = 1. HI has a circuit containing x and y and H, - z has a circuit containing x and y (Lemma 3(e)), so (x, y) $ E(G). Suppose first that (x, z) E E(G) and (y, z) E E(G). If HI - z has a circuit containing x and y, then this circuit together with a circuit of H, - z and the path x, z, y constitute a 5-rail of G. If, on the other hand, HI - z has not a circuit containing x and y, then HI - x has a circuit containing y and z by Lemma 3(e). By Lemma 3(d), H, contains two y - (x, z} paths having precisely y in common. So H, + (x, z) + (y, z) = G, has a 3-rail between y and z. This shows that G has a 5-rail, a contradiction. We may therefore assume that at most one of the edges (z, y), (z, X) is present in G. Suppose w.1.o.g. that (z, y) 4 E(G). If (x, z) E E(G), we conclude as in Case 2 that for i = 1, 2, Hi + (x, y) + (x, z) + (y, z) is 3-connected and contains no 5-rail, so e(H,) + 3 < $(n(H,) - 1) and e(H,) + 3 < Q(n(H2) - 1) - 4 (because z has degree 3 in H2 + (x, y) + (x, z) + ( y, z)), which implies e(G) = e(H,) + e(H2) + I < &z(G) - 1) - $. If, on the other hand, (x, z) $ E(G), then again H, + (x, y) + (x, z) + (y, z) is 3-connected and contains no 5-rail. By Lemma 3(e), either HI - x has a circuit containing y and z or HI - y has a circuit containing x and z. By symmetry, we may assume that the latter holds. Then by Lemma 3(b), HI + (x, y) + (y, z) is 3-connected. Furthermore, it contains no 5-rail. By the induction hypothesis, e(H,) + 3 < $(n(H,) - 1) - 4 and e(H,) + 2 < Q(n(HJ - I), which implies e(G) = e(K) + e(H,) ,< %(n(G) - 1) - 4. Case 4. 4x, HA 2 2, 4x, HJ 2 2, d(y, HI) > 2, d(z, HI) 3 2, d( y, H,) = d(z, HJ = 1, and every vertex of G has degree 2 4 in G.
Let y1 , z, denote the vertices of H, joined to y and z, respectively. {x, y, , zl} is a separating set of G so H, - {y, z} is a 3-fragment with attachvertices x, yl, z1 . Let Fl denote the graph obtained from G, by adding a new vertex w and joining w to x, y and z, and let F2 denote the graph obtained from H, - {y, z> by adding a new vertex w’ and joining w’ to x, y1 and z1 . None of Fl , F2 contains a 5-rail and both of Fl , F2 are 3-connected by Lemma 3(f). Furthermore, both of Fl , FZ have a vertex of degree 3 so for i = 1, 2, e(Fi) < +(n(F,) - 1) - 4, which implies e(G) = (e(4) - 3) + MFz) - 3) + 2 < 8(44)
+ W2) - 2) - 5 = $(n(G) - 1).
Case 5. 4x, HII 3 2, 4x, Hz) 2 2, d(y, HI) > 2, d( y, HZ) = d(z, Hz) = 1, and d(z, G) = 3.
d(z, HI) > 2,
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We define the vertices y1 , z1 and the graph F, as in Case 4. If n(H,) >, 6, we deduce as in Case 4 that e(F,) < +(n(Fz) - 1) - $. If, on the other hand, n(H,) = 5, clearly Fz is a K4 so in this case e(FJ = 6= $(n(Fz) - 1) - 8. d(z, HI) = d(z, G,) = 2 so G contains none of the edges (x, z) and ( y, z). Suppose first that (x, y) E E(G). G, + (x, z) + ( y, z) is 3-connected (by Lemma 3(a)) and has no 5-rail. For suppose (reductio ad absurdurn) that G, + (x, z) + (y, z) has a 5-rail. If this 5-rail contains at most one of the edges (x, z), (y, z), then this edge may be replaced by a path in Hz and we obtain a 5-rail in G, which is a contradiction. If the 5-rail of G, + (x, z) + (y, z) contains both edges (x, z), (y, z), then one of the five paths contains the path x, z, y (we here use that z has degree < 5 in G, + (x, z) + (y, z)). The path x, z, y may be replaced by a x - y path of Hz and we obtain a 5-rail in G, a contradiction. By the induction hypothesis, e(H,) + 3 < s(n(Hl) - I), so e(G) = e(H,) + e(H,) + 1 = e(H,) + e(F,) < +(n(H,) + n(F,) - 2) - 4 = +z(G) - 4. Suppose next (x, y) 4 E(G). As above, we see that HI + (x, z) + (y, z) has no 5-rail. By Lemma 3(d), HI + (x, y) + (x, z) has no 5-rail. Combining Lemma 3(b), (e), we see that at least one of the graphs HI + (x, y) + (x, z) and HI + (x, z) + (y, z) is 3-connected. So by the induction hypothesis, e(H,) + 2 < $(n(H,) - l), which implies e(G) = e(H,) + e(H,) = e(H,) + e(FJ - 1 < g(n(H,) + n(F,) - 2) - + = &z(G) - 8. Case 6.
d(x, G) = 3, d(x, HI) = d(x, H,) = 1.
Let y, x1 , x2 denote the vertices joined to x. xi E V(HJ for i = 1,2. Suppose w.1.o.g. that n(H,) > n(H,). G1 - x is a 3-fragment with attachvertices x1 , y, z. Let G’ be the graph obtained from G - (x, x1) - (x, y) by identifying x and x1 . By Lemma 3(f), G’ is 3-connected provided every vertex has degree > 3 which is true iff d(y, G) > 4. G’ can also be described as the graph G - x + (x1+ , x2), so obviously G’ contains no 5-rail. If d( y, G) > 4, then e(G’) < Q(n(G’) - I), i.e., e(G) < $(n(G) - 1) - &. So we may assume that d(y, G) = 3. For i = 1,2, let yi be the vertex of V(Hi) - {x, y, z} joined to y. If n(H,) = 5, then n(G) = 7 and e(G) < 11 < $(n(G) - l), so we may assume that n(H,) > 6. Then HI - {x, y} is a 3-fragment with attachvertices x1 , y1 , z. The graph G” obtained from G - (x1 , x) - (x, y) - (y, yJ by identifying the vertices x, x1 and the vertices y, y1 is 3-connected by Lemma 3(f). G” can also be described as the graph G - (x, y} + (x1 , xJ + (y, , y.J, so clearly G contains no 5-rail. By the induction hypothesis, e(G”) < g(n(G”) - l), i.e., e(G) < $(n(G) - 1) - 2. Case 7.
G has a vertex x of degree 3 so that G - x is 3-connected.
ON ~-RAILS
IN GRAPHS
153
x3 denote the vertices joined to x. Suppose first that Let Xl, x2, (x1 , x2) $ E(G). Then G - x + (x1 , x2) is 3-connected and has no 5-rail. By the induction hypothesis, e(G) - 2 < Q@(G) - 1 - I), so e(G) < g@(G) - 1) - 3. Suppose next that (x1 , x2), (x1 , x3), (x2, x3) E E(G). Let H denote the graph G - x - (x1 , x.J - (x1, x3) - (x2, x3). If at least two of the vertices x1 , x2, x, have degree 2 or more in H, then by Lemma 3(e) H has a circuit containing precisely two of the vertices Xl, X2,% * It is then easy to see that G contains a 5-rail. We may therefore assume that d(x, , H) = d(x, , H) = 1. If also d(xQ , H) = 1, Case 7 is reduced to Case 1. So assume that d(x, , H) 3 2. Let x1’, x2’ denote the vertices of H joined to x1 , x2 . {x1’, x2’, x3} is a separating set of G so G - {x, x1, x2} = H - {x1, x2} is a 3-fragment with attachvertices Xl', x2', x3 . Let G’ denote the graph obtained from G - {x, x, , x2} by adding a new vertex y and joining y to x1’, x,’ and xg . Clearly, G’ has no 5-rail, and by Lemma 3(f), G’ is 3-connected. So by the induction hypothesis, e(G’) < @z(G’) - I), i.e., e(G) < $(n(G) - 1). Case 8. G has two vertices x, y so that d(x, G) = d(y, G) = 3, (x, y) E E(G), and N(x, G) n lV( y, G) = o .
N(x, G) = (y, x1 , x2}, N(y, G) = {x, y, , y2}. Let G’ denote the graph obtained from G by contracting the edge (x, y), i.e., G’ is obtained from G - {x, y) by adding a new vertex z and joining z to x1 , x2 , y1 and y2 . It is easy to see that G’ contains no 5-rail. G’ is clearly 2-connected. If G is not 3-connected, then G’ has a separating set {z’, z”}. One of the vertices z’, z” (say z”) must be z. Then {x, y, z’> is a separating set of G. We then have Case 6. If, on the other hand, G’ is 3-connected, then by the induction hypothesis, e(G’) < Q(n(G’) - l), which implies e(G) < $(n(G) - 1) - 8. Case 9.
G is bconnected.
G has no 5-rail so by Corollary d(v, , G) = 4v2 , G) = 4 and
1 G has two vertices v1 , v2 so that
(vl , v2) E E(G).
NV, , G) = {x, Y, z, v,>,
N(v, , G) = {x’, y’, z’, vl>. If (x, y) $ E(G), then G - v1 + (x, y) is 3-connected, has no 5-rail, and has a vertex of degree 3 so by the induction hypothesis e(G) - 3 < Q(n(G) - 2) - *, i.e., e(G) < $(n(G) - 1). So we may assume that G({x, y, z}) is a K3 and also that G((x’, y’, z’}) is a & . If one of the vertices x, y, and z (say x) has degree 4 in G, then as above we may assume that G({v, , y, z>) is a K3 . Then the notation can be chosen so that y = y’ and z = z’. x’ # x because G is assumed to be 4-connected. Each of the vertices x, x’, v1 , v2 is joined to both y and z so G has a 5-rail between y and z, which is a contradiction. We may therefore assume that each of the vertices x, y, and z has degree > 5 in G. G - vr is clearly a 3-fragment with attachvertices x, y, z. By Lemma 3(e), G - vr - (x, y) -
154
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(x, z) - ( y, z) has a circuit containing precisely two of the vertices x, y, z. Then G has a 5-rail between these two vertices, which is a contradiction. We shall now show that at least one of Cases 1-9 occurs. If G is 4-connetted, then Case 9 occurs. If G is not Cconnected but every vertex of G has degree > 4, then we consider a separating set {x, y, z} so that we obtain a 3-fragment, G, say, with the least possible number of vertices. It is then clear that each of x, y, z is joined to at least two vertices of G, - (x, y, z}. So one of Cases l-4 occurs. So assume in what follows that G has a vertex x,, of degree 3. If G - x,, is 3-connected, Case 7 occurs. If G has a separating set {x0, y, z> so that (x,, ,y) EE(G) or (x,, , z) E E(G), then Case 6 occurs. If none of the above-mentioned cases occurs, then G has a separating set consisting of three vertices one of. which is x0 , and if {x,, , y, z} is a separating set and G = G, u G, , where V(G,) n V(G,) = {x0, y, z} and V(G,) - V(G,-,) # o for i = 1, 2, then the notation can be chosen so that x,, is joined to two vertices of G, - {x, , y, z}. We select y and z so that n(G,) is least possible. Then both y and z are joined to at least two vertices of G, - {x, , y, z}, so either Case 1, Case 3, Case 5, or Case 8 occurs (Case 1 occurs if each of y and z is joined to precisely one vertex of G, - (x, , y, z} and n(G,) > 4. Case 8 occurs if n(G,) = 4). This concludes the proof of Theorem 3. Remark. By Theorem 3, every 3-connected graph G with e(G) > *(n(G) - 1) contains a j-rail. If H is a 2-connected graph each vertex of which has degree 3 (except possibly one which has degree 2) and G is obtained from H by adding a new vertex and joining it to every vertex of H, then G has no 5-rail, G is 3-connected, and e(G) = [s@(G) - l)]. So Theorem 3 is best posssible.
5. DETERMINATION LEMMA
4. f&z)
OF&(~)
- 1) + 3 for all n 3 7.
Proof. Let G be a graph containing no 5-rail so that n(G) > 7 and e(G) = f&(G)) - 1. It is easy to see that G is connected. Let G’ be any endblock of G. (An endblock is a block containing at most one cutvertex). Let x denote any vertex of G if G = G’, and let x denote the unique cutvertex of G belonging to G’ if G # G’. By Theorem 1, G’ has a vertex z so that z # x and d(z, G’) = d(z, G) < 4. If d(z, G) < 3, then G - z has no 5-rail and e(G - z) 3 e(G) - 3 so f&(G) - 1)) > f&(G)) - 3 in this case. If d(z, G) = 4 and z is joined to two vertices zl, z, so that (zl , z2) $ E(G), then G - z + (zl , zB) has no 5-rail and again we conclude thatf,(n(G) - 1) > f,@(G)) - 3. So we may assume that d(z, G) = 4 and
0~
/C-RAILS
IN
155
GRAPHS
that N(z, G) = N(z, G’) span a K4 . Any 2-connected graph which contains a K5 as a proper subgraph contains a 5-rail. So we conclude that G’ is a K5 . Let G” be an endblock of G different from G’. We may assume that G” is a K5 . Let G”’ denote the graph obtained from G by deleting the four vertices of G’ which are not cutvertices of G and also the four vertices of G” which are not cutvertices of G. Then n(G”‘) = n(G) - 8 and e(G”‘) = e(G) - 20. Let H be a graph containing no 5-rail so that n(H) = 8 and e(H) = [$(I@) - l)] = 17. By the remark after Theorem 3, such a graph exists. Let G, be the graph obtained from G’” and H by identifying a vertex of G” with a vertex of H. Clearly, G, contains no 5-rail. n(G,) = n(G”‘) +7 = n(G) - 1 and e(G,) = e(G”‘) + 17 = e(G) - 3. So f,(n(G) - 1) > f,(n(GN - 3.
It is easy to see thatf,(6)
= 13, so by Lemma 4f,(n)
< 3n - 5 for all
n 3 6. LEMMA 5. Let G, , G, , and G, be disjoint graphs each of which contains no k-rail. For i = 1, 2, 3, xi , yi E V(G,), (xi , yi) E E(G,), and Gi contains no (k - I)-rail connecting xi and yi . Suppose furthermore that G, contains two vertices x1’, y,’ so that h’, vI’> f ix1 , rI>, (x1’, rI’) E E(GJ, and G contains no (k - I)-rail between x1’ and yl’. Let G denote the graph obtained from G, , Gz and G, by identifving the vertices y1 , xz and y, , x, and y3 , x1 (see Fig. 1). Then G contains no k-rail and G contains no (k - I)rail between x1’ and y,‘.
G@ylxff$y2
_r y3
&
x3
G3
FIG. 1. Construction
of graphs with no k-rails.
156
SBRENSEN
Proof. COROLLARY
AND
THOMASSEN
The proof is not difficult and we leave it to the reader. 2(a). For each k 3 5,
f&) > k(k2k- -*I3- 2 (n -k) for injnitely
(b)
many n. For all m 3 2, m # 4, f,(3m) > 8m - 4.
Proof of (a). Let k be a fixed integer > 5. It is sufficient to prove that for all m 3 0 there exists a graph G” with n(G”) = k + (2k - 3)m and e(G”) = (k(k - 1) - 2)(m + 4) so that G” contains no k-rail and G” contains two vertices zIm, zZmso that (zIm, zz”) E E(G”) and G” contains no (k - 1)-rail between zI” and zZm. We shall prove this by induction over m. Put Go equal to the graph obtained from Kk by deleting an edge. Let zIo denote one of the two vertices of degree k - 2, and let zzo, zzo be two vertices joined to zIo in Go. Having already constructed Gm-l (m > l), we construct G” from G, , G, , and G, as in Lemma 5, where G, = G, = Go and G, = Gm-l and x1 = xz = zIo, y, = y, = zzo, x3 = zy-l, and y, = z?-‘. Then G” contains no k-rail and G” contains no (k - I)-rail between the vertices zIo, zsoE V(G,). n(G”) = n(G”-l) + 2k - 3 = k + (2k - 3)m and e(G”) = e(G”-l) + k(k - 1) - 2 = (k(k - 1) - 2)(m + &)Proof of(b). We shall show that for all m 3 2, m # 4, there exists a graph Hm with e(H”) = Sm - 4 and n(H”) = 3m so that H”’ contains no 5-rail and H” contains two vertices zIm, zzm so that (zIm, zzm) E E(Hn”) and Hm contains no 4-rail between zInz and zzn”. H2 and H3 are shown in Fig. 2. Suppose we have defined H”. We now construct G as in Lemma 5 with G, = H”, G, = H2, G, = H2 or H3, x, = zlnz, y, = zzm, x1 = z12, Yl = z22, xz = z12/(zr3), and y, = zz2/(zz3) if G, = H2/(H3). Then G contains no 5-rail and G contains no 4-rail between the vertices .z12and zs2
FIG.
2.
Two
graphs
with
no S-rails.
0N ~-RAILS
IN GRAPHS
157
(see Fig. 2). n(G) = n(Hm) + n(G,) + n(G,) - 3 = 3(m + 1 + &z(G&) and e(G) = e(H”) + e(GJ + e(G,) = 8(m + Qn(G2) + 1) - 4. So if G, = Hz, then we have constructed Hm+3, and if G, = Ha, we have constructed Hn”+4. Starting with H2 and H3 and using the constructions above, we obtain an Hm for all m 2 5. This completes the proof of Corollary 2. LEMMA 6. Let G be a graph with n(G) > 2. If G contains no 5-rail and e(G) > +n(G) - 4, then either G = KS (in which case e(G) = 10 = &(G) - 4 + 6) or G is a Cconnected graph with n(G) = 7 and e(G) = 15 =Qn(G)-4+&.
Proof. Induction over n(G). For 2 < n(G) < 6, the statement is easily verified. Suppose then that the statement holds for graphs with fewer than n,, vertices (n, 3 7) and let G be a graph containing no 5-rail so that n(G) = n, and e(G) > +n, - 4. We shall show that G is a 4-connected graph with n(G) = 7 and e(G) = 15. (1)
G is 2-connected.
Suppose (reductio ad absurdum) that G is disconnected or G has a cutvertex. Then G has two subgraphs G, and G, so that G = G, u G,, V(G,) - V(G,-,) # lz( for i = 1,2 and 1 V(G,) n V(G,)I < 1. If n(G,) = 1, then by the induction hypothesis e(G) = e(G,)< +n(G,) - 4 + # = #n(G) - 6, which is a contradiction. If n(Gi) > 2 for i = 1,2, then e(Gi) < +n(Gf) - 4 + 6 for i = 1,2, so e(G) = e(G,)+e(G,) < g@(G) + 1) - a$ = &n(G) - 4, which is a contradiction. Proof of (1).
(2)
If x, y E V(G) and (x, y) $ E(G), then G - {x, y} is connected.
Proof of (2). Suppose (reductio ad absurdum) that G - (x, y} is disconnected. Then G has two subgraphs G1 and G, so that G = G, v G, , V(G,) n V(G,) = {x, y} and n(GJ < n,, for i = 1, 2. Clearly, Gi + (x, JJ) contains no 5-rail for i = 1, 2. If both of the graphs Gi + (x, v) are Cconnected, then G contains a 6-rail between x and y, which is a contradiction. So we may assume that G, + (x, JJ) is not 4-connected. By the induction hypothesis, e(G,) + 1 < #n(G,) - 4 and e(G,) + 1 < $n(G,) - 4 + 8 so e(G) = e(G,) + e(G2) < $(n, + 2) - zSB= Qn, - 4,
which is a contradiction. (3)
G is 3-connected.
Proof of (3). Suppose (reductio ad absurdum) that {x, JJ}is a separating set of G. By (2), (x, JJ)E E(G). Let G, , G, be spanned subgraphs of G so that V(G,) n V(G,) = {x, y}, V(G,) - V(G,J # o for i = 1, 2 and
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SBRENSEN AND THOMASSEN
G = G, u Gz . If G, contains a 4-rail between x and y, then clearly G contains a 5-rail between x and y. So G, is not 4-connected and by the induction hypothesis e(G,) < $n(G,) - 4. Similarly, e(G,) < +(G,) - 4. So in, - 4 < e(G) = e(G,) + e(G,) - 1 < $(no + 2) - 9 = +I, - I-$ . From this we deduce e(G,) = Qn(Gi) - 4 for i = 1,2. If both of the graphs Gi - (x, y) (i = 1, 2) have a circuit containing x and y, then G has a 5-rail between x and y, which is a contradiction. So we may assume that G, - (x, y) is not 2-connected. Let G,’ and G,’ be subgraphs of G, so that Gl - (x, y) = G,’ u G2’, V(G,‘) - V(Gi-,) # o for i = 1, 2 and j V(G,‘) n V(G,‘)I = 1. Any 2-connected graph which contains a 4-connetted graph as a proper subgraphs contains a 5-rail. G is 2-connected by (1) so none of G,’ and G,’ are 4-connected. By the induction hypothesis, e(Gi) < &(G;) - 4 for i = 1, 2 so +r(G,) - 5 = e(G,) - 1 = e(G,‘) + e(G,‘) < g(n(G,) + 1) - 8 < &z(G,) - 5, which is a contradiction. G is 3-connected and contains no 5-rail so by Theorem 3 e(G) < [$(no - l)]. So &r, - 4 < [$(q, - l)] and n, >, 7. This implies that n, = 7 and e(G) = 15. It only remains to show that G is 4-connected. Suppose therefore (reductio ad absurdum) that (x, , x2 , xg} is a separating set of G. By Theorem 3, every vertex of G has degree > 4. Therefore every connected component of G - {x1, x2, x,] has at least two vertices. So G - {x, , x2, x3} has two connectem components each of which is a K, . Furthermore every vertex of G - (x1 , x, , x3) is joined to x1 , x, and x3. e(G) > 14 so G({x, , x2, x3}) must contain an edge, say (x, , x2). Then G has a 5-rail between x1 and x2 , which is a contradiction. This concludes the proof of Lemma 6. THEOREM 4.
For n >, 6, n # 7, n # 12,&(n) = [@z] - 3.
Proof. By Lemma 6 and the remark after Theorem 3, we have for n > 6, n # 7, [j(n - I)] + 1
IZ # 12, [%(n - I)] + 1 = [Qn] - 3, so the statement is true for these values of n. For m > 5, we have by Corollary 2(b) and Lemma 6 8m - 3 < fs(3m) < [Q . 3m] - 3 = 8m. - 3. If we combine f,(3m + 3) = 8m + 5, f,(3m + 1) < [$(3m + l)] - 3 = 8m - 1, and Lemma 4, we have f,(3m + 1) = 8m - 1 and f5(3m + 2) = 8m + 2 = [-$(3m + 2)] - 3. So for n 3 15,f,(n) = [@I - 3. In particular,f,(l3) = 31 andf,(l5) = 37. By Lemma 4, f,(14) = 34 = [+ . 141 - 3. This concludes the proof of Theorem 4. Theorem 4 does not cover the cases IZ = 7 and n = 12. Combining Lemma 6 and the remark after Theorem 3 we havef,(7) = 16. We state without proof thatf,(l2) = 28. So for 6 < n < 13,f,(n) = [&z - I)] + 1.
ON k-RAILS
IN GRAPHS
159
REFERENCES I. B. BOLLO& AND P. En&is, On extremal problems in graph theory (in Hungarian), Mat. Lapok. 13 (1962), 143-152. 2. B. BOLLOB~, On graphs with at most three independent paths connecting any two vertices, Studia Sci. Math. Hungar. 1 (1966), 137-140. 3. G. A. DIRAC, Extensions of Menger’s Theorem, J. London Math. Sot. 38 (1963), 148-161. 4. P. Ems, Extremal problems in graph theory, “A Seminar on Graph Theory” (F. Harary, Ed.), pp. 54-59, Holt, Rinehart and Winston, New York, 1967. 5. F. HARARY, “Graph Theory,” Addison-Wesley, Reading, Mass., 1969. 6. J. L. LEONARD, On a conjecture of Bollobas and Erdos, Per. Math. Hungar. 3 (1973), 281-284. 7. J. L. LEONARD, On graphs with at most four line-disjoint paths connecting any two vertices, J. Comb. Theory 13 (1972), 242-250. 8. W. MADER, Existenz gewisser Konfigurationen in n-geslttigten Graphen und in Graphen geniigend grosser Kantendichte, Math. Ann. 194 (1971), 295-312. 9. W. MADER, Ein Extremalproblem des Zusammenhangs von Graphen, Math. Z. 10.
131 (1973), 223-231. C. THOMASSEN,Some
58zb/x7/z-6
homeomorphism
properties of graphs, Math. Nachr. to appear.
OF COMBINATORIAL
THEORY
@)
17, 143-159 (1974)
On k-Rails in Graphs Bo AAGAARD Matematisk
SBRENSEN AND CARSTEN THOMA~~EN
Xnstitut, Aarhus Uninersitet, Aarhus, Denmark Communicated by Frank Harary Received July 25, 1973
A k-rail is the union of k paths each pair of which has exactly the endvertices in common. We first give a sufficient degree-condition for the existence of k-rails in graphs. Next we determine the number of edges required in 3connected graphs to guarantee the existence of Srails, and finally we solve the corresponding problem for graphs with no prescribed connectivity.
1. INTRODUCTION A k-rail is the union of k paths each pair of which has exactly the end vertices in common. For 123 k + 1 > 3, we define f&) as the least integer r so that every graph with n vertices and r or more edges contains a k-rail. It was conjectured by Bollobas and Erdijs [l] that every graph with p(k - 1) + 1 vertices and g(k - 1) kp + 1 or more edges contains a k-rail. If true, this would be best possible and consequently limn+, (l/n)f,(n) = Sk. For k = 4, the conjecture was proved by Bollobis [2] and in [lo] this was obtained from a more general result. Leonard [6] disproved the conjecture for k = 5, and Mader [9] showed that for all integers k, m, where k > 5, m > O,f&z) > &kn + m for some n. The conjecture becomes true of the term “k-rail” is replaced by “k mutually edge-disjoint paths connecting two vertices”. For k = 5, this was proved by Leonard [7], and Mader [9] proved it in the general case. In Section 3 we prove that a graph G has a k-rail for k > 3 if every vertex of G has degree 3 k - 1 and if furthermore every circuit of G has at least two vertices of degree > k. This is based on a result of Mader [8]. In Section 4 we show that the conjecture of Bolloblis and Erdijs becomes true for k = 5 provided the graphs considered are 3-connected. Finally, we show in Section 5 that for n > 6, n # 7, IZ # 12, &(n) = [$z] - 3. 143 Copyright All rights
@?I1974 by Academic Press, Inc. of reproduction in any form reserved.
144
SBRENSEN AND THOMASSEN
We also here show that for k fixed, j&z) > “‘“,,“,
2 (n - k)
for infinitely many n. This disproves the conjecture of Bollobais and Erdiis for all k 3 5.
2. TERMINOLOGY
AND PRELIMINARIES
We consider finite, undirected graphs without loops and without multiple edges. V(G) and E(G) denote the set of vertices and the set of edges, respectively, of the graph G and n(G) = ( V(G)1 and e(G) = I E(G)/. (x, y) denotes the edge joining the vertices x and y. The terms degree, path, connected graph, k-connected graph, connected component, cutvertex, block and complete graph (K,,) mean the same as in Harary [5]. The term circuit means the same as cycle in [5]. The set of vertices joined to the vertex x in the graph G is denoted by N(x, G), and the degree of x in G is denoted by d(x, G), i.e., d(x, G) = ) N(x, G)/. If A, B _CV(G), A n B = o, then an A - B path is a path P joining a vertex x E A and a vertex y E B so that (A u B) n P = {x, y}. An {x} - A path and an (x} - ( y} path will be called an x - A path and an x - ypath, respectively. A k-rail between (or connecting) the vertices x and y is the union of k x - y paths each pair of which has exactly x and y in common. If S C V(G) u E(G), then G - S denotes the graph obtained by deleting from G all vertices of S n V(G) (if any) together with all edges of G incident with them and then deleting all edges of S n E(G) still present. G - {x} and G - {(x, y)} will be denoted by G - x and G - (x, y), respectively. If A C V(G), we define G(A) = G - (V(G) - A). G(A) is said to be the subgraph of G spanned by A. A spanned&graph is a subgraph of the form G(A), where A C V(G). If G is a connected graph, S C V(G), and G - S is disconnected, then S is a separating set of vertices of G. Suppose G is a k-connected but not (k + I)-connected graph and G f &+I . Then G has a separating set S of k vertices and G has two spanned subgraphs G, and G, (which are not necessarily unique) so that G = G, u G, , V(G,) n V(G,) = S, V(G,) - V(G,) # o and V(G,) - V(G,) # o. G, (and also G,) is said to be a k-fragment with attachvertices S. Every k-fragment can be obtained in this way. We need the following versions of Menger’s Theorem: 1. Let x, y E V(G), (x, y) 4 E(G). fl fir all S C V(G) - (x, y} with 1S 1 < k - 1 G - S has an x - ypath, then G has a k-rail between x andy.
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2. Let x E V(G) and A C V(G) - {x} so that 1A 1 > k. If for all ScV(G)-{x}withISI
results are special cases of a theorem of Dirac [3, Theorem B].
3. DEGREE-CONDITIONSFOR THE EXISTENCEOF k-RAILS We need the following result of Mader [8, Lemma I]: LEMMA 1. Let G be a graph and H a complete subgraph with vertices vl, va ,..., v, (m 3 1). Let A C E(G) so that (i) (ii)
every edge of A joins a vertex of H and a vertex of G - V(H); if(x, vi) E A, then for all j < i, (x, vi) E E(G) - A.
Let PI, Pz,..., Pn be a k-rail between x1 and x2 where x1, x2 E V(G) - V(H), and assume that at least one of the paths PI , P, ,..., P, contains some edge of A. Then G - A contains a (k - I)-rail PI’, PSI,..., PL-, between x1 and x2 and two paths Pl, P2 so that for i = 1,2, Pi is a path between xi and a vertex which in G is incident with an edge of A, P1 n P2 = 0, and Pi n Pj’ = (xi> for i = 1, 2 and j = 1, 2 ,..., k - 1.
LEMMA 2. Let G be a graph containing a complete subgraph H and a circuit C with no diagonals so that (i) (ii) (iii) (iv)
G - V(H) contains no circuit; H and C have exactly one vertex, x1 say, in common. VXEV(G)-V(H):d(x,G)>k-1111; C contains a vertex x2 so that x2 # x1 and d(x, , G) > k.
Then G contains a k-rail between x1 and x2. Proof. Induction over k. The statement is clearly true for k = 2. Assume that the statement holds for k < k, - 1, and let G be a graph satisfying (i), (ii), (iii), and (iv) with k = k,, b 3. We shall show that G contains a k,-rail between x1 and x2 . Suppose first that x2 is joined to some vertex x, E V(H) - {x1}. G - x3 satisfies (i), (ii), (iii), and (iv) with H replaced by H - x, and k = k, - 1. By the induction hypothesis, G - x, contains a (k, - I)-rail between x1 and x, . If we add the path x1 , x, , x2 , we obtain a k,-rail between x1 and x2 . We may therefore assume that x2 is not joined to any vertex of H - x,. x2 is joined to some vertex x, not contained in C, because d(x, , G) > k,, 3 3. F = G - (V(H) u {x2}) contains no circuit. Let T denote the
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connected component of F containing xQ , and let P be a path of T connecting xQwith a vertex xp so that d(xp , G - V(H)) = 1. Then P n C = o and xp is joined to at least k, - 2 vertices of H. It is easy to see that G contains a 3-rail between x1 and x2, so we may assume that k, > 4. Then X, is joined to some vertex x5 E V(H) - (xl>. Let G’ = G - (V(T) U {x5}). G’ satisfies the conditions (i)-(iv) with H replaced by H - x5 and k = k, - 1. By the induction hypothesis, G’ contains a (k, - 1)-rail between x1 and x2. Adding the path (x1, x5, xp) u P u (x, , x2) we obtain a k,-rail. This concludes the proof of the lemma. THEOREM 1. Let k be a fixed integer > 2. Let G be a graph with j V(G)1 > k + 1, and let H be a complete subgraph with 1 < 1 V(H)1 < k - 1. If the following conditions are satisjed
(1)
Vx E V(G) - V(H): d(x, G) 2 k - 1,
(2) G - V(H) contains a circuit, and every circuit contains at least two vertices of degree k or more (in G), then G contains a k-rail between two vertices of G -
of G - V(H)
V(H).
Proof. For k = 2, there is nothing to prove, so assume that k > 3. We prove the theorem by induction over the number of vertices of G. If 1 V(G)/ = k + 1, then G - V(H) contains two vertices x1, x2 so that d(x, , G) 3 k, d(x, , G) > k. Clearly, G contains a k-rail between x1 and x2 . Assume then that the theorem is true for graphs with fewer than n vertices (n > k + 2), and let G be a graph with n vertices and a complete subgraph H so that (1) and (2) holds. We shall show that G contains a k-rail between two vertices of G - V(H). Let M be a complete subgraph of G so that H C M, and A4 is not a proper subgraph of any complete subgraph of G. V(H) = {vl, v2 ,..,, vls} and V(M) = {vl , v2 ,..., v,}. If m = h = 1, then d(v,, G) = 0. In this case, let u be a vertex of G - v1 so that G - (u, ul> contains a circuit (clearly such a vertex exists). Then (1) and (2) hold with G and H replaced by G - vi and {u}, respectively. By the induction hypothesis, G - v1 contains a k-rail. If m >, k + 1, then G contains a k-rail between v, and v,,-~ . So we may assume that 2 < m < k. We shall consider the following two cases: Case 1. G - V(M) contains no circuit. Lets = min{i / G - {a1 ,..., vi} contains no circuit}. Then s >, h + 1 because G - V(H) contains a circuit. Let C be a shortest circuit of G - {vl , v2 ,..., u,-~). Then C has no diagonals and C contains v, . C is contained in G - V(H) so C contains a vertex z so that z # Y, and d(z, G) > k. By Lemma 2, G contains a k-rail between z and v, .
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Case 2. G - V(W) contains a circuit. Because of the maximality property of M, no vertex of G - V(M) is joined to every vertex of M. For x ES = N(u, , G) - (vl , a2 ,..., urn}, we define
r(x) = min(i I (x, ui) # Z(G)} and we define G’ as the graph obtained from G - v, by adding the edges (x, a++), x E S. Put A = {(x, u,(,)) / x E S} (S and A may be empty). Then G’ satisfies conditions (1) and (2) with H replaced by A4 - urn . By the induction hypothesis, G’ contains a k-rail between two vertices x1, x2 E V(G’) - (V(M) - {u,}) = V(G) -
V(M).
If this k-rail is contained in G, we have finished. In the alternative case, we apply Lemma 1 and conclude that G’ - A = G - v, contains a (k - 1)-rail PI , P, ,..., P,-, between two vertices x1, x2 E V(G) - V(M) and two paths Pl, P2 where for i = 1,2, Pi connects xi with a vertex zi E s u (V(M) - {u,}) = N(v, ) G). Furthermore, P1 n P2 = o, and Pi n Pi = {xi} for i = 1, 2 andj = 1, 2 ,..., k - 1. Let P, denote the path which is the union of the paths PI, P2 and z1 , vln , z2 . Then PI , Pz ,..., PI, is a k-rail of G. This concludes the proof of Theorem 1. THEOREM2. Let G be a graph so that every vertex of G has degree > k - 1 3 2 and so that every circuit contains at least two vertices of degree k or more. Then G contains a k-rail. Proof.
Follows easily from Theorem 1.
Remark. The property required by the graph G in Theorem 2 may be expressed as follows:
(a) Every vertex of G has degree 3 k - 1. (b) If A denotes the set of vertices having degree precisely k - 1, then G(A) contains no circuit. (c) For every x E V(G) - A and for every connected component T of G(A), x is joined to at most one vertex of T. COROLLARY 1. Let G be a graph so that every vertex of G has degree >, k - 1 3 2 and so that no two vertices of degree precisely k - 1 are adjacent. Then G contains a k-rail.
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4. THE NUMBER OF EDGES REQUIRED TO GUARANTEE ~-RAILS
IN j-CONNECTED
THE EXISTENCE OF
GRAPHS
We shall first give some elementary results on 3-fragments. LEMMA 3. Let G be a 3-fragment H = G - E(G({u, v, w})).
with attachvertices
u, v, w. Let
(a) (b)
H + (u, v) + (u, w) + (v, w) is 3-connected. If H - w has a circuit containing u and v, then H+(u, w)+(v, w) is 3-connected. (c) If H has at least jive vertices and x is a cutvertex of H, then H - x has exactly two connected components. One of these has exactly one vertex and this vertex is contained in {u, v, w}. The statement remains true if H is replaced by G.
(d) If d(u, H) >, 2, then H contains two u - {v, w} paths having exactly u in common. (e) If d(u, H) > 2 and d(v, H) 3 2, then H has a circuit which contains u and v. Furthermore, if H - w has no such circuit, then H - u has a circuit containing v, w, and H - v has a circuit containing u, w.
(f ) If G’ is a 3-fragment with attachvertices u’, v’, w’ and G’ n H = .@, then the graph G” obtained from G’ and H by identtyying u, u’ and v, v’ and w, w’ is a 3-connected graph provided every vertex of this graph has degree 3 or more. In particular, if we choose G’ so that d(u’, G’) > 2, d(v’, G’) 2 2, d(w’, G’) > 2, and u’, v’, w’ are independent in G’, we see that His a 3-fragment with attachvertices u, v, w. Proof. For every z E V(H) - {u, v, w}, H contains three z - {u, v, w} paths each pair of which has exactly z in common. From this it immediately follows that if a proper subset of {u, v, w} is deleted from H, then the resulting graph is connected. Also, if x, y are any two vertices of H, then every connected component of H - {x, y> contains at least one of the vertices u, v, w. (a) easily follows from this. Proof of(b). Let x, y E V(H). We shall show that H’ = H + (u, v) + (v, w) - {x, y} is connected. Since every connected component of H - {x, y} contains at least one of the vertices u, v, w, it is sufficient to show that the vertices of {u, v, w} - {x, y} belong to the same connected component of H’. If w $ (x, y} or if (x, y} C {u, v, w}, this is obvious. If, on the other hand, x = w and y 4 {u, v}, then H’ = H - {w, y}, and H’ therefore contains a u - v path.
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Proofof( Suppose x is a cutvertex of H. Let y E V(H) - (u, u, w, x}. H contains three y - {u, ZJ,w} paths each pair of which has exactly y in common. We may therefore w.1.o.g. assume that H - x contains a y - u path and a y - u path. So u and v belong to the same connected component, M say, of H - X. By a similar argument, we see that for every y’ E V(H) - {u, 0, w, x>, H - x contains either a y’ - u path or a y’ - u path, which implies that y’ E V(M). So H - x has two connected components, namely M and {w}. It is easy to see that (c) still holds if H is replaced by G. Proof@(d). For every x E V(H) - {u}, H - x has either a u - u path or a u - w path by (c). Combining this with Menger’s theorem we obtain (d). Proof of(e). For every x E V(H) - {u, a}, u and u belong to the same connected component of H - x by (c). This implies that H has a circuit containing u and v. Suppose now that every such circuit also contains w. Let C be a circuit containing U, U, w. C - {u, w} = P, U Pz , where PI and Pz are disjoint paths. H - {u, w} is connected and therefore contains a V(P,) - V(P,) path, P3 say. It is easy to see that C U P, contains a circuit C’ which contains u and exactly one of the vertices U, w. By the assumption that H - w has no circuit containing u and 0, C’ contains v and w but not U. Similarly, we see that H - z, has a circuit containing u and w. Proof of(f). We consider G’ and H as subgraphs of G”. Assume that every vertex of G” has degree 3 3 in G”. Let x, y be any vertices. We shall show that G” - {x, y} is connected, From the first paragraph of the proof of the lemma it follows that it is sufficient to show that the vertices of {u, ZJ,w> - {x, y} belong to the same connected component of G” - {x, y}. Also, it follows that this is the case if {x, y} C V(C) or {x, y> C V(H). So we may assume that x E V(H) - V(C) and y E V(G’) - V(H) and that the graphs H - x and G’ - y are both disconnected. Suppose first that V(H) = {x, U, v, w>. Then d(z, G’) > 2 Vz E V(G’). By (c) G’ - y is connected, which is a contradiction. Suppose next that 1 V(H)1 > 5. By (c) H - x has precisely two connected components one of which consists of, say, U. In particular, d(u, H) = 1 and u and w belong to the same connected component of H - X. Then d(u’, G’) > 2, and from (c) it easily follows that G’ - y contains either a u’ - w’ path or a u’ - v’ path. So U, u and w belong to the same connected component of G” - {x, y}. This concludes the proof of Lemma 3. THEOREM 3. Let G be a 3-connected graph which contains no 5-rail. Then e(G) < #(n(G) - 1). Furthermore, ifG has a vertex of degree 3, then e(G) < Q(n(G) - 1).
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Proof. Induction over the number of vertices of G. For 4 < n(G) < 6, it is easy to verify the statement. So we assume that the theorem holds for graphs with fewer than n, vertices (n, > 7) and let G be a 3-connected graph with no 5-rail and with n(G) = n, . We shall then show that e(G) < #(n,, - 1) and that strict inequality holds if G has a vertex of degree 3. We shall consider the following cases separately. Case 1. G has three edges e, , e, , e3 so that G - {e, , e2 , es} is disconnected and each connected component of G - {e, , e2, es>contains more than one vertex.
In this case G - {e, , e2 , es} = G, u G2 , where G, n G, = 0. For i = 1, 2, 3, let xi and yi be the endvertices of ei in G, and G, , respectively. The vertices x1 , x2 , xg , y1 , y2 , y, are distinct because G is 3-connected. For i = 1, 2, Gi is either a K3 or a 3-fragment. For i = 1, 2, let Gi’ denote the graph obtained from Gi by adding a new vertex zi and joining zi to Xl, x2 , xJ( y1 , yz , y3) if i = l/(i = 2). By Lemma 3(f), G,’ and G,’ are 3-connected, and clearly they contain no 5-rail. By the induction hypothesis, e(G,‘) < #(n(Gi’) - 1) for i = 1,2, so e(G) = e(G,‘) + e(G,‘) - 3 < &(n(G1’) + n(Gz’) - 2) - 3 = +n, - 3.
In the next five cases, G has a separating set consisting of three vertices x, y, 2, i.e., G is the union of two spanned subgraphs G1 and G, so that W-2 n W,) = k Y, 4, I/(G) - W,) f 0, W,) - W,) Z 0. G, and G, are 3-fragments, and for i = 1,2, we define Hi = Gi - E(G({xv Y, Z>)>* Case 2. d(x, Hi) 2 2, d(y, Hi) >, 2 and d(z, Hi) 3 2 for Every vertex of G has degree > 4.
i = 1, 2.
Assume w.1.o.g. (Lemma 3(e)) that H1 - z has a circuit containing x and y and that H, - x has a circuit containing y and z. Also, Hz has a circuit containing x and y. G has no 5-rail so (x, y) 4 E(G). By symmetry, (y, z) $ E(G). Let F1 denote the graph obtained from HI by adding the edges (x, y), (y, z), and (x, z). By Lemma 3(a), Fl is 3-connected. We shall show that Fl has no kail. Suppose (reductio ad absurdum) that Fl has a 5-rail. If this 5-rail contains at most two of the edges (x, y), (y, z), and (x, z), then we obtain a 5-rail in G by replacing these edges by appropriate paths in H, (Lemma 3(d)). If the 5-rail of Fl contains all three edges (x, y), (y, z), and (x, z), then the 5-rail connects two vertices vl, DoE {x, y, z}. Let bd = b, Y, z> - h , v2}. Then HI - vg contains a 3-rail between v1 and v2 . Ha has a circuit containing v1 and v2 (Lemma (3(e)) so G contains a 5-r&1. This contradiction shows that Fl has no 5-rail. By the induction hypothesis, e(H,) = e(F,) - 3 < s(n(Hl) - 1) - 3. By symmetry, e(H,) <
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*(n(HJ - 1) - 3. So e(G) < e(H,) + e(H,) + 1 < $(~(Hi) + n(H,) - 2) - 5 = 8@(G) - 0 Case 3. d(x, Hi) 2 2, d(y, Hi) 2 2 for i = 1, 2, d(z, H1) >, 2, d(z, H,) = 1. HI has a circuit containing x and y and H, - z has a circuit containing x and y (Lemma 3(e)), so (x, y) $ E(G). Suppose first that (x, z) E E(G) and (y, z) E E(G). If HI - z has a circuit containing x and y, then this circuit together with a circuit of H, - z and the path x, z, y constitute a 5-rail of G. If, on the other hand, HI - z has not a circuit containing x and y, then HI - x has a circuit containing y and z by Lemma 3(e). By Lemma 3(d), H, contains two y - (x, z} paths having precisely y in common. So H, + (x, z) + (y, z) = G, has a 3-rail between y and z. This shows that G has a 5-rail, a contradiction. We may therefore assume that at most one of the edges (z, y), (z, X) is present in G. Suppose w.1.o.g. that (z, y) 4 E(G). If (x, z) E E(G), we conclude as in Case 2 that for i = 1, 2, Hi + (x, y) + (x, z) + (y, z) is 3-connected and contains no 5-rail, so e(H,) + 3 < $(n(H,) - 1) and e(H,) + 3 < Q(n(H2) - 1) - 4 (because z has degree 3 in H2 + (x, y) + (x, z) + ( y, z)), which implies e(G) = e(H,) + e(H2) + I < &z(G) - 1) - $. If, on the other hand, (x, z) $ E(G), then again H, + (x, y) + (x, z) + (y, z) is 3-connected and contains no 5-rail. By Lemma 3(e), either HI - x has a circuit containing y and z or HI - y has a circuit containing x and z. By symmetry, we may assume that the latter holds. Then by Lemma 3(b), HI + (x, y) + (y, z) is 3-connected. Furthermore, it contains no 5-rail. By the induction hypothesis, e(H,) + 3 < $(n(H,) - 1) - 4 and e(H,) + 2 < Q(n(HJ - I), which implies e(G) = e(K) + e(H,) ,< %(n(G) - 1) - 4. Case 4. 4x, HA 2 2, 4x, HJ 2 2, d(y, HI) > 2, d(z, HI) 3 2, d( y, H,) = d(z, HJ = 1, and every vertex of G has degree 2 4 in G.
Let y1 , z, denote the vertices of H, joined to y and z, respectively. {x, y, , zl} is a separating set of G so H, - {y, z} is a 3-fragment with attachvertices x, yl, z1 . Let Fl denote the graph obtained from G, by adding a new vertex w and joining w to x, y and z, and let F2 denote the graph obtained from H, - {y, z> by adding a new vertex w’ and joining w’ to x, y1 and z1 . None of Fl , F2 contains a 5-rail and both of Fl , F2 are 3-connected by Lemma 3(f). Furthermore, both of Fl , FZ have a vertex of degree 3 so for i = 1, 2, e(Fi) < +(n(F,) - 1) - 4, which implies e(G) = (e(4) - 3) + MFz) - 3) + 2 < 8(44)
+ W2) - 2) - 5 = $(n(G) - 1).
Case 5. 4x, HII 3 2, 4x, Hz) 2 2, d(y, HI) > 2, d( y, HZ) = d(z, Hz) = 1, and d(z, G) = 3.
d(z, HI) > 2,
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We define the vertices y1 , z1 and the graph F, as in Case 4. If n(H,) >, 6, we deduce as in Case 4 that e(F,) < +(n(Fz) - 1) - $. If, on the other hand, n(H,) = 5, clearly Fz is a K4 so in this case e(FJ = 6= $(n(Fz) - 1) - 8. d(z, HI) = d(z, G,) = 2 so G contains none of the edges (x, z) and ( y, z). Suppose first that (x, y) E E(G). G, + (x, z) + ( y, z) is 3-connected (by Lemma 3(a)) and has no 5-rail. For suppose (reductio ad absurdurn) that G, + (x, z) + (y, z) has a 5-rail. If this 5-rail contains at most one of the edges (x, z), (y, z), then this edge may be replaced by a path in Hz and we obtain a 5-rail in G, which is a contradiction. If the 5-rail of G, + (x, z) + (y, z) contains both edges (x, z), (y, z), then one of the five paths contains the path x, z, y (we here use that z has degree < 5 in G, + (x, z) + (y, z)). The path x, z, y may be replaced by a x - y path of Hz and we obtain a 5-rail in G, a contradiction. By the induction hypothesis, e(H,) + 3 < s(n(Hl) - I), so e(G) = e(H,) + e(H,) + 1 = e(H,) + e(F,) < +(n(H,) + n(F,) - 2) - 4 = +z(G) - 4. Suppose next (x, y) 4 E(G). As above, we see that HI + (x, z) + (y, z) has no 5-rail. By Lemma 3(d), HI + (x, y) + (x, z) has no 5-rail. Combining Lemma 3(b), (e), we see that at least one of the graphs HI + (x, y) + (x, z) and HI + (x, z) + (y, z) is 3-connected. So by the induction hypothesis, e(H,) + 2 < $(n(H,) - l), which implies e(G) = e(H,) + e(H,) = e(H,) + e(FJ - 1 < g(n(H,) + n(F,) - 2) - + = &z(G) - 8. Case 6.
d(x, G) = 3, d(x, HI) = d(x, H,) = 1.
Let y, x1 , x2 denote the vertices joined to x. xi E V(HJ for i = 1,2. Suppose w.1.o.g. that n(H,) > n(H,). G1 - x is a 3-fragment with attachvertices x1 , y, z. Let G’ be the graph obtained from G - (x, x1) - (x, y) by identifying x and x1 . By Lemma 3(f), G’ is 3-connected provided every vertex has degree > 3 which is true iff d(y, G) > 4. G’ can also be described as the graph G - x + (x1+ , x2), so obviously G’ contains no 5-rail. If d( y, G) > 4, then e(G’) < Q(n(G’) - I), i.e., e(G) < $(n(G) - 1) - &. So we may assume that d(y, G) = 3. For i = 1,2, let yi be the vertex of V(Hi) - {x, y, z} joined to y. If n(H,) = 5, then n(G) = 7 and e(G) < 11 < $(n(G) - l), so we may assume that n(H,) > 6. Then HI - {x, y} is a 3-fragment with attachvertices x1 , y1 , z. The graph G” obtained from G - (x1 , x) - (x, y) - (y, yJ by identifying the vertices x, x1 and the vertices y, y1 is 3-connected by Lemma 3(f). G” can also be described as the graph G - (x, y} + (x1 , xJ + (y, , y.J, so clearly G contains no 5-rail. By the induction hypothesis, e(G”) < g(n(G”) - l), i.e., e(G) < $(n(G) - 1) - 2. Case 7.
G has a vertex x of degree 3 so that G - x is 3-connected.
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x3 denote the vertices joined to x. Suppose first that Let Xl, x2, (x1 , x2) $ E(G). Then G - x + (x1 , x2) is 3-connected and has no 5-rail. By the induction hypothesis, e(G) - 2 < Q@(G) - 1 - I), so e(G) < g@(G) - 1) - 3. Suppose next that (x1 , x2), (x1 , x3), (x2, x3) E E(G). Let H denote the graph G - x - (x1 , x.J - (x1, x3) - (x2, x3). If at least two of the vertices x1 , x2, x, have degree 2 or more in H, then by Lemma 3(e) H has a circuit containing precisely two of the vertices Xl, X2,% * It is then easy to see that G contains a 5-rail. We may therefore assume that d(x, , H) = d(x, , H) = 1. If also d(xQ , H) = 1, Case 7 is reduced to Case 1. So assume that d(x, , H) 3 2. Let x1’, x2’ denote the vertices of H joined to x1 , x2 . {x1’, x2’, x3} is a separating set of G so G - {x, x1, x2} = H - {x1, x2} is a 3-fragment with attachvertices Xl', x2', x3 . Let G’ denote the graph obtained from G - {x, x, , x2} by adding a new vertex y and joining y to x1’, x,’ and xg . Clearly, G’ has no 5-rail, and by Lemma 3(f), G’ is 3-connected. So by the induction hypothesis, e(G’) < @z(G’) - I), i.e., e(G) < $(n(G) - 1). Case 8. G has two vertices x, y so that d(x, G) = d(y, G) = 3, (x, y) E E(G), and N(x, G) n lV( y, G) = o .
N(x, G) = (y, x1 , x2}, N(y, G) = {x, y, , y2}. Let G’ denote the graph obtained from G by contracting the edge (x, y), i.e., G’ is obtained from G - {x, y) by adding a new vertex z and joining z to x1 , x2 , y1 and y2 . It is easy to see that G’ contains no 5-rail. G’ is clearly 2-connected. If G is not 3-connected, then G’ has a separating set {z’, z”}. One of the vertices z’, z” (say z”) must be z. Then {x, y, z’> is a separating set of G. We then have Case 6. If, on the other hand, G’ is 3-connected, then by the induction hypothesis, e(G’) < Q(n(G’) - l), which implies e(G) < $(n(G) - 1) - 8. Case 9.
G is bconnected.
G has no 5-rail so by Corollary d(v, , G) = 4v2 , G) = 4 and
1 G has two vertices v1 , v2 so that
(vl , v2) E E(G).
NV, , G) = {x, Y, z, v,>,
N(v, , G) = {x’, y’, z’, vl>. If (x, y) $ E(G), then G - v1 + (x, y) is 3-connected, has no 5-rail, and has a vertex of degree 3 so by the induction hypothesis e(G) - 3 < Q(n(G) - 2) - *, i.e., e(G) < $(n(G) - 1). So we may assume that G({x, y, z}) is a K3 and also that G((x’, y’, z’}) is a & . If one of the vertices x, y, and z (say x) has degree 4 in G, then as above we may assume that G({v, , y, z>) is a K3 . Then the notation can be chosen so that y = y’ and z = z’. x’ # x because G is assumed to be 4-connected. Each of the vertices x, x’, v1 , v2 is joined to both y and z so G has a 5-rail between y and z, which is a contradiction. We may therefore assume that each of the vertices x, y, and z has degree > 5 in G. G - vr is clearly a 3-fragment with attachvertices x, y, z. By Lemma 3(e), G - vr - (x, y) -
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(x, z) - ( y, z) has a circuit containing precisely two of the vertices x, y, z. Then G has a 5-rail between these two vertices, which is a contradiction. We shall now show that at least one of Cases 1-9 occurs. If G is 4-connetted, then Case 9 occurs. If G is not Cconnected but every vertex of G has degree > 4, then we consider a separating set {x, y, z} so that we obtain a 3-fragment, G, say, with the least possible number of vertices. It is then clear that each of x, y, z is joined to at least two vertices of G, - (x, y, z}. So one of Cases l-4 occurs. So assume in what follows that G has a vertex x,, of degree 3. If G - x,, is 3-connected, Case 7 occurs. If G has a separating set {x0, y, z> so that (x,, ,y) EE(G) or (x,, , z) E E(G), then Case 6 occurs. If none of the above-mentioned cases occurs, then G has a separating set consisting of three vertices one of. which is x0 , and if {x,, , y, z} is a separating set and G = G, u G, , where V(G,) n V(G,) = {x0, y, z} and V(G,) - V(G,-,) # o for i = 1, 2, then the notation can be chosen so that x,, is joined to two vertices of G, - {x, , y, z}. We select y and z so that n(G,) is least possible. Then both y and z are joined to at least two vertices of G, - {x, , y, z}, so either Case 1, Case 3, Case 5, or Case 8 occurs (Case 1 occurs if each of y and z is joined to precisely one vertex of G, - (x, , y, z} and n(G,) > 4. Case 8 occurs if n(G,) = 4). This concludes the proof of Theorem 3. Remark. By Theorem 3, every 3-connected graph G with e(G) > *(n(G) - 1) contains a j-rail. If H is a 2-connected graph each vertex of which has degree 3 (except possibly one which has degree 2) and G is obtained from H by adding a new vertex and joining it to every vertex of H, then G has no 5-rail, G is 3-connected, and e(G) = [s@(G) - l)]. So Theorem 3 is best posssible.
5. DETERMINATION LEMMA
4. f&z)
OF&(~)
- 1) + 3 for all n 3 7.
Proof. Let G be a graph containing no 5-rail so that n(G) > 7 and e(G) = f&(G)) - 1. It is easy to see that G is connected. Let G’ be any endblock of G. (An endblock is a block containing at most one cutvertex). Let x denote any vertex of G if G = G’, and let x denote the unique cutvertex of G belonging to G’ if G # G’. By Theorem 1, G’ has a vertex z so that z # x and d(z, G’) = d(z, G) < 4. If d(z, G) < 3, then G - z has no 5-rail and e(G - z) 3 e(G) - 3 so f&(G) - 1)) > f&(G)) - 3 in this case. If d(z, G) = 4 and z is joined to two vertices zl, z, so that (zl , z2) $ E(G), then G - z + (zl , zB) has no 5-rail and again we conclude thatf,(n(G) - 1) > f,@(G)) - 3. So we may assume that d(z, G) = 4 and
0~
/C-RAILS
IN
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GRAPHS
that N(z, G) = N(z, G’) span a K4 . Any 2-connected graph which contains a K5 as a proper subgraph contains a 5-rail. So we conclude that G’ is a K5 . Let G” be an endblock of G different from G’. We may assume that G” is a K5 . Let G”’ denote the graph obtained from G by deleting the four vertices of G’ which are not cutvertices of G and also the four vertices of G” which are not cutvertices of G. Then n(G”‘) = n(G) - 8 and e(G”‘) = e(G) - 20. Let H be a graph containing no 5-rail so that n(H) = 8 and e(H) = [$(I@) - l)] = 17. By the remark after Theorem 3, such a graph exists. Let G, be the graph obtained from G’” and H by identifying a vertex of G” with a vertex of H. Clearly, G, contains no 5-rail. n(G,) = n(G”‘) +7 = n(G) - 1 and e(G,) = e(G”‘) + 17 = e(G) - 3. So f,(n(G) - 1) > f,(n(GN - 3.
It is easy to see thatf,(6)
= 13, so by Lemma 4f,(n)
< 3n - 5 for all
n 3 6. LEMMA 5. Let G, , G, , and G, be disjoint graphs each of which contains no k-rail. For i = 1, 2, 3, xi , yi E V(G,), (xi , yi) E E(G,), and Gi contains no (k - I)-rail connecting xi and yi . Suppose furthermore that G, contains two vertices x1’, y,’ so that h’, vI’> f ix1 , rI>, (x1’, rI’) E E(GJ, and G contains no (k - I)-rail between x1’ and yl’. Let G denote the graph obtained from G, , Gz and G, by identifving the vertices y1 , xz and y, , x, and y3 , x1 (see Fig. 1). Then G contains no k-rail and G contains no (k - I)rail between x1’ and y,‘.
G@ylxff$y2
_r y3
&
x3
G3
FIG. 1. Construction
of graphs with no k-rails.
156
SBRENSEN
Proof. COROLLARY
AND
THOMASSEN
The proof is not difficult and we leave it to the reader. 2(a). For each k 3 5,
f&) > k(k2k- -*I3- 2 (n -k) for injnitely
(b)
many n. For all m 3 2, m # 4, f,(3m) > 8m - 4.
Proof of (a). Let k be a fixed integer > 5. It is sufficient to prove that for all m 3 0 there exists a graph G” with n(G”) = k + (2k - 3)m and e(G”) = (k(k - 1) - 2)(m + 4) so that G” contains no k-rail and G” contains two vertices zIm, zZmso that (zIm, zz”) E E(G”) and G” contains no (k - 1)-rail between zI” and zZm. We shall prove this by induction over m. Put Go equal to the graph obtained from Kk by deleting an edge. Let zIo denote one of the two vertices of degree k - 2, and let zzo, zzo be two vertices joined to zIo in Go. Having already constructed Gm-l (m > l), we construct G” from G, , G, , and G, as in Lemma 5, where G, = G, = Go and G, = Gm-l and x1 = xz = zIo, y, = y, = zzo, x3 = zy-l, and y, = z?-‘. Then G” contains no k-rail and G” contains no (k - I)-rail between the vertices zIo, zsoE V(G,). n(G”) = n(G”-l) + 2k - 3 = k + (2k - 3)m and e(G”) = e(G”-l) + k(k - 1) - 2 = (k(k - 1) - 2)(m + &)Proof of(b). We shall show that for all m 3 2, m # 4, there exists a graph Hm with e(H”) = Sm - 4 and n(H”) = 3m so that H”’ contains no 5-rail and H” contains two vertices zIm, zzm so that (zIm, zzm) E E(Hn”) and Hm contains no 4-rail between zInz and zzn”. H2 and H3 are shown in Fig. 2. Suppose we have defined H”. We now construct G as in Lemma 5 with G, = H”, G, = H2, G, = H2 or H3, x, = zlnz, y, = zzm, x1 = z12, Yl = z22, xz = z12/(zr3), and y, = zz2/(zz3) if G, = H2/(H3). Then G contains no 5-rail and G contains no 4-rail between the vertices .z12and zs2
FIG.
2.
Two
graphs
with
no S-rails.
0N ~-RAILS
IN GRAPHS
157
(see Fig. 2). n(G) = n(Hm) + n(G,) + n(G,) - 3 = 3(m + 1 + &z(G&) and e(G) = e(H”) + e(GJ + e(G,) = 8(m + Qn(G2) + 1) - 4. So if G, = Hz, then we have constructed Hm+3, and if G, = Ha, we have constructed Hn”+4. Starting with H2 and H3 and using the constructions above, we obtain an Hm for all m 2 5. This completes the proof of Corollary 2. LEMMA 6. Let G be a graph with n(G) > 2. If G contains no 5-rail and e(G) > +n(G) - 4, then either G = KS (in which case e(G) = 10 = &(G) - 4 + 6) or G is a Cconnected graph with n(G) = 7 and e(G) = 15 =Qn(G)-4+&.
Proof. Induction over n(G). For 2 < n(G) < 6, the statement is easily verified. Suppose then that the statement holds for graphs with fewer than n,, vertices (n, 3 7) and let G be a graph containing no 5-rail so that n(G) = n, and e(G) > +n, - 4. We shall show that G is a 4-connected graph with n(G) = 7 and e(G) = 15. (1)
G is 2-connected.
Suppose (reductio ad absurdum) that G is disconnected or G has a cutvertex. Then G has two subgraphs G, and G, so that G = G, u G,, V(G,) - V(G,-,) # lz( for i = 1,2 and 1 V(G,) n V(G,)I < 1. If n(G,) = 1, then by the induction hypothesis e(G) = e(G,)< +n(G,) - 4 + # = #n(G) - 6, which is a contradiction. If n(Gi) > 2 for i = 1,2, then e(Gi) < +n(Gf) - 4 + 6 for i = 1,2, so e(G) = e(G,)+e(G,) < g@(G) + 1) - a$ = &n(G) - 4, which is a contradiction. Proof of (1).
(2)
If x, y E V(G) and (x, y) $ E(G), then G - {x, y} is connected.
Proof of (2). Suppose (reductio ad absurdum) that G - (x, y} is disconnected. Then G has two subgraphs G1 and G, so that G = G, v G, , V(G,) n V(G,) = {x, y} and n(GJ < n,, for i = 1, 2. Clearly, Gi + (x, JJ) contains no 5-rail for i = 1, 2. If both of the graphs Gi + (x, v) are Cconnected, then G contains a 6-rail between x and y, which is a contradiction. So we may assume that G, + (x, JJ) is not 4-connected. By the induction hypothesis, e(G,) + 1 < #n(G,) - 4 and e(G,) + 1 < $n(G,) - 4 + 8 so e(G) = e(G,) + e(G2) < $(n, + 2) - zSB= Qn, - 4,
which is a contradiction. (3)
G is 3-connected.
Proof of (3). Suppose (reductio ad absurdum) that {x, JJ}is a separating set of G. By (2), (x, JJ)E E(G). Let G, , G, be spanned subgraphs of G so that V(G,) n V(G,) = {x, y}, V(G,) - V(G,J # o for i = 1, 2 and
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SBRENSEN AND THOMASSEN
G = G, u Gz . If G, contains a 4-rail between x and y, then clearly G contains a 5-rail between x and y. So G, is not 4-connected and by the induction hypothesis e(G,) < $n(G,) - 4. Similarly, e(G,) < +(G,) - 4. So in, - 4 < e(G) = e(G,) + e(G,) - 1 < $(no + 2) - 9 = +I, - I-$ . From this we deduce e(G,) = Qn(Gi) - 4 for i = 1,2. If both of the graphs Gi - (x, y) (i = 1, 2) have a circuit containing x and y, then G has a 5-rail between x and y, which is a contradiction. So we may assume that G, - (x, y) is not 2-connected. Let G,’ and G,’ be subgraphs of G, so that Gl - (x, y) = G,’ u G2’, V(G,‘) - V(Gi-,) # o for i = 1, 2 and j V(G,‘) n V(G,‘)I = 1. Any 2-connected graph which contains a 4-connetted graph as a proper subgraphs contains a 5-rail. G is 2-connected by (1) so none of G,’ and G,’ are 4-connected. By the induction hypothesis, e(Gi) < &(G;) - 4 for i = 1, 2 so +r(G,) - 5 = e(G,) - 1 = e(G,‘) + e(G,‘) < g(n(G,) + 1) - 8 < &z(G,) - 5, which is a contradiction. G is 3-connected and contains no 5-rail so by Theorem 3 e(G) < [$(no - l)]. So &r, - 4 < [$(q, - l)] and n, >, 7. This implies that n, = 7 and e(G) = 15. It only remains to show that G is 4-connected. Suppose therefore (reductio ad absurdum) that (x, , x2 , xg} is a separating set of G. By Theorem 3, every vertex of G has degree > 4. Therefore every connected component of G - {x1, x2, x,] has at least two vertices. So G - {x, , x2, x3} has two connectem components each of which is a K, . Furthermore every vertex of G - (x1 , x, , x3) is joined to x1 , x, and x3. e(G) > 14 so G({x, , x2, x3}) must contain an edge, say (x, , x2). Then G has a 5-rail between x1 and x2 , which is a contradiction. This concludes the proof of Lemma 6. THEOREM 4.
For n >, 6, n # 7, n # 12,&(n) = [@z] - 3.
Proof. By Lemma 6 and the remark after Theorem 3, we have for n > 6, n # 7, [j(n - I)] + 1
IZ # 12, [%(n - I)] + 1 = [Qn] - 3, so the statement is true for these values of n. For m > 5, we have by Corollary 2(b) and Lemma 6 8m - 3 < fs(3m) < [Q . 3m] - 3 = 8m. - 3. If we combine f,(3m + 3) = 8m + 5, f,(3m + 1) < [$(3m + l)] - 3 = 8m - 1, and Lemma 4, we have f,(3m + 1) = 8m - 1 and f5(3m + 2) = 8m + 2 = [-$(3m + 2)] - 3. So for n 3 15,f,(n) = [@I - 3. In particular,f,(l3) = 31 andf,(l5) = 37. By Lemma 4, f,(14) = 34 = [+ . 141 - 3. This concludes the proof of Theorem 4. Theorem 4 does not cover the cases IZ = 7 and n = 12. Combining Lemma 6 and the remark after Theorem 3 we havef,(7) = 16. We state without proof thatf,(l2) = 28. So for 6 < n < 13,f,(n) = [&z - I)] + 1.
ON k-RAILS
IN GRAPHS
159
REFERENCES I. B. BOLLO& AND P. En&is, On extremal problems in graph theory (in Hungarian), Mat. Lapok. 13 (1962), 143-152. 2. B. BOLLOB~, On graphs with at most three independent paths connecting any two vertices, Studia Sci. Math. Hungar. 1 (1966), 137-140. 3. G. A. DIRAC, Extensions of Menger’s Theorem, J. London Math. Sot. 38 (1963), 148-161. 4. P. Ems, Extremal problems in graph theory, “A Seminar on Graph Theory” (F. Harary, Ed.), pp. 54-59, Holt, Rinehart and Winston, New York, 1967. 5. F. HARARY, “Graph Theory,” Addison-Wesley, Reading, Mass., 1969. 6. J. L. LEONARD, On a conjecture of Bollobas and Erdos, Per. Math. Hungar. 3 (1973), 281-284. 7. J. L. LEONARD, On graphs with at most four line-disjoint paths connecting any two vertices, J. Comb. Theory 13 (1972), 242-250. 8. W. MADER, Existenz gewisser Konfigurationen in n-geslttigten Graphen und in Graphen geniigend grosser Kantendichte, Math. Ann. 194 (1971), 295-312. 9. W. MADER, Ein Extremalproblem des Zusammenhangs von Graphen, Math. Z. 10.
131 (1973), 223-231. C. THOMASSEN,Some
58zb/x7/z-6
homeomorphism
properties of graphs, Math. Nachr. to appear.