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Feb 22, 2013 at 22:44 answer added Kaz timeline score: 2
Feb 22, 2013 at 19:18 answer added placeholder timeline score: 0
Feb 22, 2013 at 16:44 comment added Dave Tweed In a self-routing network, the individual switches compute their own settings based only on the information available at their two inputs. Global routing uses knowledge of where every input needs to go in order to compute all of the switch settings in parallel. This is in the context of message-passing, in which each message carries destination address information. It's been a long time (30 years) since I last worked with this stuff, but IIRC, a 6-switch network (3 layers of two switches each) is non-blocking (but not self-routing).
Feb 22, 2013 at 16:23 history tweeted twitter.com/#!/StackElectronix/status/304989915354103809
Feb 22, 2013 at 15:35 history edited The Unknown Dev CC BY-SA 3.0
deleted 25 characters in body
Feb 22, 2013 at 15:33 comment added The Unknown Dev I'm not worried about latency at the moment, just functionality. What do you mean by "self-routing?" (Sorry for the stupid question)
Feb 22, 2013 at 15:25 comment added Dave Tweed Here are some search terms that might help: "shuffle-exchange network", of which "Benes" and "Omega" are examples. Does your switch need to be self-routing, or can the settings be computed with global knowledge? How important is latency -- the number of switches in each input-to-output path?
Feb 22, 2013 at 15:17 review First posts
Feb 22, 2013 at 15:18
Feb 22, 2013 at 15:16 history edited The Unknown Dev CC BY-SA 3.0
Improved formatting: made input/output port names italic
Feb 22, 2013 at 15:12 comment added Justin A simple way is to only use 1 output of each of your 2x2 switches (so they are just a 2-to-1 mux). This would require 12 of the 2x2 switches, though.
Feb 22, 2013 at 15:00 history asked The Unknown Dev CC BY-SA 3.0