Abstract

The 2-terminal one-to-any problem, which arises in the design of layout systems, is the problem of assigning wach one of n terminals positioned on the upper row of a channel (called entry terminals) to one of m terminals positioned on the lower row (called exit terminals) so that the resulting channel routing problem has minimum density. An optimal solution to this problem is known [1]. In this paper we consider a natural generalization, the 2-color one-to-any problem, in which we have two types of entry terminals, red and blue ones, and exit terminals can be assigned to either type of entry terminal. Red and blue nets created by our algorithm are allowed to run on top of each other in the routing, and the density is defined as the larger of the red density and the blue density. Its minimization is an interesting combinatorial problem. We show how to compute the best achievable density in O( n + m) time, and an assignment achieving this density in O(( n + m)log( n + m)) time.

Full Text
Paper version not known

Talk to us

Join us for a 30 min session where you can share your feedback and ask us any queries you have

Schedule a call