Mathl. Comput. Modelling Vol. 17, No. 11, pp. 85-88, 1993 0895-7177193 $6.00 + 0.00

Printed in Great Britain. All rights reserved Copyright@ 1993 Pergamon Press Ltd

SPANNING SUBGRAPHS OF A HYPERCUBE IV: ROOTED TREES

FRANK HARARY

Department of Computer Science

New Mexico State University, Las Cruces, NM 88003, U.S.A.

MARTIN LEWINTER

Department of Mathematics

SUNY Purchase, Purchase, NY 10577, U.S.A.

Abstract-A rooted spanning tree T of a hypercube Qn with root at origin u = (O,O, ,O) has a function g : E(T) -+ {-l,+l} d fi d e ne as follows. For each edge zcy of T with dT(u,z) < dT(u, y), let g(zy) = c(yi - xi). Another function f is defined by f(T,u) = cg(q). We observe that f depends on the embedding of T, is odd-valued, and we obtain sharp bounds for f. We derive an odd-interpolation theorem for the values of f over all spanning trees of Qn.

1. INTRODUCTION

The hypercube Qn is defined recursively by Q1 = K2 while Qn = Qn-l x K2. Its node-set V(Qn) is the set of all ordered binary n-tuples. Let nodes x, y E V(Qn) have binary labels

(x1,... ,x,) and (YI,... , yn). Then xy E E(Q,) if and only if c (xi - yili( = 1. More generally, d(U, V) = c [Ui - Vii.

Hypercubes are important in computer architecture, especially for parallel processing. An open problem is to find a characterization of the spanning trees of Qn. Partial results can be found in [l-4].

Let N, = {1,2,... , n}. An edge xy of Qn is in the set Ei of edges of dimension i if xi # yi. Obviously we have the partition E(Q,) = E = UEi. Moreover, Qn - Ei consists of two copies of Qn_lr denoted Hi and Hi’. For convenience, let Hi be the copy of Qn-l containing the node Ug = (0,. . . , O), i.e., Hi is the induced subgraph on all nodes with 0 in the ith place, while all the nodes of Hi’ have a 1 in the i th place. Let E(zy) be the dimension set of edge xy, and let Hz and H3/ be the copies of Qn_l containing x, y respectively.

Observe that if xy E E(Q,), then d(x,uo) < d(y,uo) if and only if C(yi - xi) = 1. Let T be a rooted spanning tree embedded in Qn with root in the origin ~0 = (0,. . ,O), without loss of generality. We define a function g : E(T) + (-1, +l} a s o f 11 ows. Given e = xy E E(T) such that d~(x,~o) < &-(y, UO), let g(e) = C(yi -xi). In Figure 1, g(xy) = -1 while g(ulz) = 1. The edges of the spanning tree are directed away from the root to facilitate calculating g, since the second node of an arc of the resulting oriented tree is further from the root than the first node of the arc.

The function g may be used to give a sign to each edge of T. Note that g(xy) = 1 if and only

if the node closer to uo in Qn is also closer to uo in T. Alternatively, the arc gy is leaving the copy of Qn-l containing uo if and only if g(xy) = 1.

Finally, given an embedded, rooted spanning tree T of Qn with root ‘1~0, define the function f

by f(T,uo) = C eEECTj g(e). For the tree of Figure 1, f(T, ug) = 5. It should be noted that f depends both on the embedding and on the choice of root.

In the next section, we present various results about this function including bounds. In Sec- tion 3, we discuss interpolating properties.

* Research supported in part by ONR Grant N0014-90-J-1860.

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86 F. HARARY, M. LEWINTER

Figure 1. A rooted tree in Q3.

2. PROPERTIES OF THE FUNCTION f

We begin by establishing bounds for this function.

THEOREM 1. Given a rooted, embedded spanning tree T of Qn with root uc,

Furthermore, both bounds are tight.

PROOF. Since IE(T)I = 2” - 1, the upper bound follows at once. The lower bound is a conse- quence of the fact that each dimension set must contain at least one positive edge of T, since at least one edge of T must be directed from H - i to Hi’ for each i. Now even if all the remaining 2n - 1 - n edges of T are negative, it follows that 2n + 1 - 2n = n - (2n - 1 - n) 5 f(T, UO), thus verifying the lower bound.

Now let T be a spanning tree resulting from a breadth-first search with uc as root. Then given any node z E V(Qn), we have d&,(uo,z) = dT(uO,z), implying that g(e) = 1 for all edges of T.

It follows that f(T,uo) = [IS( = 2” - 1, p roving that the upper bound is tight. We show the lower bound is tight by the following recursively defined construction beginning

with n = 2, as the case n = 1 is trivial. Let T2 be the spanning tree of Q3 depicted in Figure 2, in which tia is the unique antipodal node of uc.

Figure 2. The a-cube as a step in a recursive construction.

Note that all edges on the us - ZLO path of Tz are positive, while the remaining edge of T3 is negative. Now f(T, ZLO) = 1 = 2 ‘2 + 1 - 22, so the lower bound is tight for n = 2. Construct T3

as follows. Take two copies of Q2 and embed T2 in both of them. Add an edge from ii0 in the copy containing uc to its corresponding node ~0’ in the other copy (called Q3’). Clearly, g(&, tic’) = 1. Those edges of T3 in Q3’ which correspond to positive edges in Q3 will be oppositely directed and, therefore, negative, while those edges of T3 in Q2’ which correspond to negative edges in Q3 will be similarly directed and, hence, will also be negative. It follows that f(T3, ug) = -1 = 2 . 3 + 1 - 23. This construction can be extended to construct a sequence

T2, T3, ..’ such that f(T,, 2~0) = 271-t 1 - 2n and T, is a spanning tree of Qn. I

COROLLARY 1A. The function f(T, uo) is odd-valued.

Rooted trees 87

Figure 3. The caterpillar T with code (1, 1, 1, 1) used to illustrate the values of the function.

PROOF. Let T have n positive and m negative edges. Then f(T, ~0) = n - m. Since n + m = IE(T)I which is odd, it follows that n and m have opposite parity, implying that n - m is odd. I

The spanning tree of T of Qs, shown in Figure 3, has each node labelled with the value of f(T, ~0) when that node is taken to be the root us.

Note that when the sum C f(T, ~0) is taken over all nodes of Qs, each regarded as the origin, we get 24 = 3. 23. This suggests that this sum C f(T, ~0) = n2n for every spanning tree T of Qn. The next result asserts that this is indeed the case.

THEOREM 2. Given Qn with node-set wj, j = 1,. . . , 2n, and an embedded spanning tree T, then denoting f(T, vj) by fj(T), we have

5 fj(T) = n2n. j=l

PROOF. Given an embedded spanning tree T of Qn, a division set Ei, and a root vj E V(Q,),

let fj(Ei) = Es(e), h w ere the sum is taken over all edges e E Ei n E(T). It follows that fj (T) = Cz, fj (Ei), from which one obtains

Now each vj contributes at least one positive edge to each Ei, yielding a total of n2n for the right side of (1). Suppose, however, that for some dimension set Ei, IEi n E(T)1 > 1. Then given any root Vj of T, consider its corresponding node Uk such that vj vk E Ei. Let L be the Uj - vk path of T (bear in mind that vj ‘& need not belong to E(T)). Clearly fj(LnEi) = fk(LfIEi) = 1, since an edge of LnEi has the same sign whether Vj or vk is chosen as the root of T, and successive edges of L n Ei alternate in sign, with the first and last edges being positive.

On the other hand, the sign of any edge of (T - L) n Ei with vi as root is opposite to its sign when ?& that the

is the root, thereby, in effect, cancelling these contributions to the right side of (l), so sum remains n2n. I

3. ODD-INTERPOLATION

Given a graph G with spanning tree set S = {Tl, . . . ,T,}, an integer-valued function f inter- polates over S, if given Ti, Tj E S and an integer k such that f (Ti) < k < f (Tj), then there exists a spanning tree Th E S for which f(Th) = k. S ome interpolation results can be found in [5-71. If the above definition holds only when k is odd, f is called odd-interpolating.

Now by Corollary 1A of the preceding section, f (T, UO) does not interpolate. We will show, however, that it is odd-interpolating. We require the following lemma, in which AAB denotes the symmetric difference of sets A and B, i.e., AAB = (A U B) - (A n B).

88 F. HARARY, M. LEWINTER

LEMMA 3A. If Tl and TZ are rooted embedded spanning trees of Q,, with root u. such that

E(Tl)AE(fi) = { el,e2} with el and e2 adjacent edges, then

If(Tl,uo) - f(T2,uo)l = 0 or 2.

PROOF. One may view T2 as Tl - el + e2, i.e., edge el in Tl “pivots” on one of its nodes to become edge e2 in T2. In fact, the remaining edges of Tl are also in T2. Let el = xy while e2 = xz, i.e., x is the pivoting node. Now is xy is directed from x to y, then the pivot would disconnect tree T2 as there would no longer be a path from ug to y in T2. Then edge xy must be directed from y to x. It follows that the pivot has no effect on g(e) for all edges e E E(Tl - xy), in which case f(Tz,~g) = f(T I, ~0) - g(xy) + g(xz). This implies that If(Tl, ~0) - f(T2, uo)l = Ig(xy) - g(xz)I = 0 or 2. I

THEOREM 3. The function f (T, ug) is odd-interpolating on the spanning tree set of Qn.

PROOF. Given spanning trees Ti,Tj and an odd integer k, with f(Ti,uo) < k < f(Tj,~o), let

Tl,Tz,... , T, be a sequence of spanning trees of Qn such that T, = Tl and Tj = T,, and E(Tk)AE(Tk+1) consists of two adjacent. edges for k = 1,2,. . . , m - 1, and apply Lemma 3a. 1

Note that, for an embedded spanning tree T of Qn, f(T, UO) is not odd-interpolating with respect to the choice of root UO, as Figure 3 shows. The values on the nodes in this figure are -1, 3, and 7, while there are no values 1 or 5. From this example it is clear that f(T, UO) does not odd-interpolate with respect to the choice of root.

REFERENCES

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