4 Concluding Remarks

We do not know the integrality ratio for the relaxation we propose. It is possible that a better rounding procedure can be discovered. Here are the worst examples of which we are aware. For k = 3, the following example (which also appeared in 7]) satisses all the new constraints and shows that the integrality ratio is at least 16 15. Consider the graph G = (V; E). V = fS f1; 2; 3gj 1 jSj 2g, where f1g; f2g; f3g are the terminals. E = ffS; Tgj S 6 = T; jS \ Tj = 1g. The edges fS; Tg with S or T of size 1 (between a terminal and a nonterminal) have cost 2, and the edges with jSj = jTj = 2 (between two nonterminals) have cost 1. It is not hard to see that the optimum multiway cut has cost 8 (by enumerating over all distinct assignments of the nonterminals to terminals). On the other hand, assigning length 1 2 to all the edges is a feasible (optimal) solution to the relaxation (i.e., place nonterminal fi; jg midway between terminals fig and fjg in the 2-simplex). This solution has value 7:5. Thus, the ratio of the integral optimum to the fractional optimum for this example is at least 16 15. The example can be generalized to k = 4. The graph has vertex set V = fS f1; 2; 3; 4gj 1 jSj 2g, where f1g; f2g; f3g; f4g are the terminals. The edge set is deened as in the previous example: E = ffS; Tgj S 6 = T; jS \ Tj = 1g. The 12 edges fS; Tg such that jSj = 1 or jTj = 1 (between a terminal and a nonterminal) have cost 3, and the 12 edges fS; Tg such that jSj = jTj = 2 (between two nonterminals) have cost 1. By exhaustive search, one can verify that the optimum multiway cut has cost 26. There is a feasible fractional solution in which all edges have length 1 2. Its cost is 24, so the integrality ratio in this case is at least 13 12. 8 Now suppose that i j and that by the time i is processed, neither u nor v has been assigned to a vertex. For uv to be cut, we must have 2 I L (I R n I L). The crux of the …