For a connected graph G let L(G) denote the maximum number of leaves in any spanning tree of G. We give a simple construction and a complete proof of a result of Storer that if G is a connected cubic graph on n vertices, then L(G) L r(n/4) + 21, and this is best possible for all (even) n.The main idea is to count the number of "dead leaves" as the tree is being constructed. This method of amortized analysis is used to prove the new result that if G is also 3-connected, then L(G) 2 r(n/3) + (4/3)1, which is best possible for many n. This bound holds more generally for any connected cubic graph that contains no subgraph K4 -e. The proof is rather elaborate since several reducible configurations need to be eliminated before proceeding with the many tricky cases in the construction. denote the maximum number of leaves in any spanning tree of G. We are interested here in L ( G ) for cubic (3-regular) graphs G.Suppose T is a spanning tree for a connected cubic graph G on n vertices. Necessarily, n is even. Let di denote the number of vertices of degree i in T, i = 1,2,3. Then the number of vertices n = d , + d , + d,, while the sum of the degrees 2n -2 = d , + 2d2 + 3d3. It follows that L(T) = d , = d, + 2.
Consequently, L(G)is maximized over such graphs G when it contains T with dz as small as possible, that is, d , = 0. Hence, L(G) I (n/2) + 1. This bound is attained for all (even) n by taking the caterpillar in which (n/2) -1 vertices form a path, and a leg (leaf vertex) is joined to each interior vertex of the path, while two legs are joined to each end of the path. This is the desired tree T, which can be embedded in a suitable graph G by adding a cycle through the leaves of T (see Figure 1).The more interesting question then is to minimize L(G), i.e., to obtain a lower bound on L ( G ) over all such graphs G in terms of n. This problem was proposed and solved in 1981 by Storer. \ Theorem 1 [ 5 ] . If G is a connected cubic graph on n vertices, then L(G) 2 w 4 ) + 21.This bound is best possible for all (even) n. For example, if n ES 0 (mod 4), then take G to be a circular "necklace" of n/4 "beads," where each bead is K4e (meaning K4 with one edge deleted), as shown in Figure 2. If n = 2 (mod 4), we may take G to be a necklace in which there are (n -6)/4 beads C( n 12 + 1 FIGURE 1. A Cubic Graph G with L(G) = (n/2) + 1 FIGURE 2. Extremal Graph for Theorem 1