Abstract

Let K c C be a bounded set. In this paper we shall give a simple necessary and sufficient condit ion for K to lie in a rectifiable curve. We say that a set is a rectifiable curve if it is the image of a finite interval under a Lipschitz mapping. Recall that for a connected set F c C, F is a rectifiable curve (not necessarily simple) if and only if l(F) < ~ , where l(-) denotes one dimensional Hausdorff measure. This classical result follows from the fact that on any finite graph there is a tour which covers the entire graph and which crosses each edge (but not necessarily each vertex!) at most twice. If K is a finite set then we are essentially reduced to the classical Traveling Salesman Problem (TSP): Compute the length of the shortest Hami l ton ian cycle which hits all points of K. This is the same, up to a constant multiple, as asking for the inf imum of l(F) where F is a curve, K c F. (Such a F is called a spanning tree in TSP theory.) For infinite sets K, we cannot hope in general to have K be a subset of a Jordan curve. What we should therefore look at is connected sets which conta in K. Let Fmi n be the shortest (minimal) spanning tree. Then we cannot possibly solve our problem for sets K of infinite cardinality if we cannot find F, I(F) < C O/(Fmin) , for any finite set K. (Here and throughout the paper C, Co, C1, c o , etc. denote various universal constants.) While there are several algorithms for computing l(Fml.), these algorithms work for finite graphs satisfying the triangle inequality, and do not use the Euclidean properties of K. (See [13] for an excellent discussion of some of these algorithms.) Therefore these methods cannot solve our problem for general infinite K. We present a method which is a minor modification of a well-known algorithm ("Farthest Insert ion" see [13]) which yields a F with I(F) < C O l(Fmi,). The Farthest Insert ion algorithm has been extensively studied with large numerical calculations on computers, and is experimentally good in the sense that the F produced satisfy I(F) < C O l(F,,,i,) for all examples which have

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