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6.3.4 Retraction of J f Theorem 6.3.4 indicates that the connectivity of space Jf can indeed be captured via a space of lower dimension, Jpf . However, space Jpf cannot be used for motion planning because, by de nition, it may happen that Jpf OJ = ; that is, some portion of Jpf is not obstacle-free. In this section we de ne a 2D space Df Jf that is entirely obstacle-free and, like Jpf , captures the connectivity of Jf . De nition 6.3.10. [57]. A subset A of a topological space X is called a retract of X if there exists a continuous map r : X A, called a retraction, such that r(a) = a for any a A. De nition 6.3.11. [57]. A subset A of space X is a deformation retract of X if there exists a retraction r and a continuous map f :X I X such that f (x, 0) = x f (x, 1) = r(x) f (a, t) = a, x X a A and t I (6.6) (6.5)

These options are almost always presented and controlled independently, which is unfortunate due to the strong relationship they possess In basic training, engineers are told that they can increase the effort of placement and routing to achieve better results We practice increasing the effort levels in lab experiments and see the improvements In practice, we see similar improvements So what s the problem The problem is that if you turn the router effort level up beyond trivial effort before you have an optimal placement, you are wasting time Because of the fact that most projects have schedules with deadlines and the fact that most FPGA designers go through many iterations before achieving timing closure, this is certainly something worth considering The fundamental concept to understand here is that routing is extremely dependent on placement.

In other words, set A X is a deformation retract of X if X can be continuously deformed into A. We show below that Df is a deformation retract of Jf . Let Jp , Jpf , and Jf be as de ned in the previous section; then we have the following lemma. Lemma 6.3.7. Jp is a deformation retract of J, and Jpf is a deformation retract of Jf . De ne r(j1 , j2 , l3 ) = (j1 , j2 , 0). It follows from Lemma 6.3.2 that r is a retrac 1 tion. Since for Type 1 and 2 obstacles Pm (Pm (OJ )) = OJ , then, if J contains

selling off information to the news or to the opposing counsel I didn t stop to wonder at all where Scott was He was old news and I was happy that we had left on reasonably good terms By 7:00 PM, I had to pull myself away from work and nearly forgot to visit the 31st floor As promised, the night security guard had a badge for me when I showed up and it worked perfectly The floor was just as empty as it had been the day before, with probably 15 people that I could see, sitting at their desks A few of them were playing cards and the others were surfing the web None of them were trading.

For a design of any complexity, a good route can only be achieved with a good placement Most FPGA design engineers have not run.

only Type 1 and Type 2 obstacles that is, Jf = J (OJ 1 OJ 2 ) it follows that Jf = J f and Jpf is a deformation retract of Jf In the general case, all obstacle types (including Type 3) can be present, and Jpf is not necessarily a deformation retract of Jf Theorem 635 Let Q1 = OJ 3 Jf , and Q2 = Jp Jf Then, Df = Q1 Q2 is a deformation retract of Jf Q1 and Q2 , are respectively, the obstacle-free portion of OJ 3 and Jp It is easy to see that Df Jpf Since Jf J f (Theorem 633) and Jpf is a deformation = = retract of J f (Lemma 637), then Df is a deformation retract of Jf .

Let D denote the 2D space obtained by patching all the holes in Df so that D Jp It is obvious that D is a deformation retract of J = Theorem 636 Given two points js , jt Df , if there exists a path pJ Jf connecting js and jt , then there exists a path pD Df , such that pD pJ From Theorem 635, Df is a deformation retract of Jf Let r be the retraction as in Ref 57, II4; then p = r(p) must be an equivalent path in Df On the other hand, if js and jt are not connected in Jf , then by de nition js and jt are not connected in Df either.