X-Authentication-Warning: delorie.com: mail set sender to geda-user-bounces using -f X-Recipient: geda-user AT delorie DOT com Date: Wed, 12 Dec 2012 06:45:53 +0100 (CET) X-X-Sender: igor2 AT igor2priv To: geda-user AT delorie DOT com X-Debug: to=geda-user AT delorie DOT com from="gedau AT igor2 DOT repo DOT hu" From: gedau AT igor2 DOT repo DOT hu Subject: Re: [geda-user] Find rat lines In-Reply-To: <50C788E9.4070604@laserlinc.com> Message-ID: References: <20121204183305 DOT 6b04c0dc AT jive DOT levalinux DOT org> <20121208112649 DOT 388a9d22 AT jive DOT levalinux DOT org> <1355011808 DOT 19390 DOT 8 DOT camel AT localhost> <1355188647 DOT 12937 DOT 14 DOT camel AT localhost> <201212111911 DOT qBBJBUG3024801 AT envy DOT delorie DOT com> <50C788E9 DOT 4070604 AT laserlinc DOT com> User-Agent: Alpine 2.00 (DEB 1167 2008-08-23) MIME-Version: 1.0 Content-Type: MULTIPART/MIXED; BOUNDARY="0-1060512762-1355291153=:26605" Reply-To: geda-user AT delorie DOT com Errors-To: nobody AT delorie DOT com X-Mailing-List: geda-user AT delorie DOT com X-Unsubscribes-To: listserv AT delorie DOT com Precedence: bulk This message is in MIME format. The first part should be readable text, while the remaining parts are likely unreadable without MIME-aware tools. --0-1060512762-1355291153=:26605 Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed On Tue, 11 Dec 2012, Joshua Lansford wrote: > He he, what if you considered each pin which belonged to net A be a source, > each pin which belongs to net B be a sink, pass a virtual current from each > source to each sink and then colored things by how 'hot' they become. :-P > Might generally help identify the short location. This may work with some modifications (or maybe I am just thinking over the implied details). First, what do we do in case more than 2 nets participate in a short? We could run this for each pair once. I supsect this would cause more shorts to be highlighted than the minimum number of cuts needed for separating the networks, but this may be a minor issue. Second, how do we do the current calculation exactly? Input is one of the nodes in net1 (source) and one of the nodes in net2 (sink); we need to: - identify _all_ paths between them - connect them in parallel and clauclate a resistor network for dissipation - once we could do that reliably, I assume we'd sum dissipations got from different paths for the same edge/node. The simple approach is take each path and increase a 'current counter' for each edge the path includes. A more complicated way is a real series/parallel resistor network of 1 ohm resistors between every source-sink combo. So the method is: 1. select a source and a sink 2. replace all other nodes with simple junctions 3a. place a counter on each edge; or 3b. replace all edges with a 1 ohm resistor and calculate dissipation on each resistor assuming a voltage source (or current source) between source and sink 4. repeat on all source-sink combinations and sum the counter or dissipation per resistor to find the hot spots Using the usual net1=a,b net2=C,D examples, edges (aka resistors) numbered now, it would work like this: 1 2 a--+--b |3 C--+--D 4 5 Paths: a-D: 1,3,5 a-C: 1,3,4 b-C: 2,3,4 b-D: 2,3,5 Dissipation per edge with the simple counter method (3a.): edge 3: 4 edge 1, 2, 4 and 5: 2 So it indeed solves the simple case. Let's find something trickier: 1 a---b |2 |3 C---D 4 We expect to find 2 and 3 be hotter than 1 and 4. a-D: 1,3 and 2,4 a-C: 2 and 1,3,4 b-C: 3,4 and 1,2 b-D: 3 and 1,2,4 With the simple counter all edges are referenced 4 times so no specific hot points. With resistor networks (sch attached with my calculations, please veryify), R2 and R3 are highlighted, so the more complicated method works on this case. However, it's not very hard to construct an example where it would fail: if there are 10 parallel instances of edge 2, power dissipation will drop to 1/10 on them, cooling them down so much that even 1 and 4 seem to be hotter. An obvious fix is merging parallel resistors "where it does not matter", in other words find all connections which are not contributing in the short, and deal with the rest (either find the best cut or just highlight the whole set and let the user decide). In the first example this would result in 3, on the last example in 2,3. But as far as I can tell this was the original problem we tried to solve. So my conclusion: if we can identify which set of edges are contributing in the short, this method could help highlighting the best segment of each set, but alone does not solve the problem. Best regards, Tibor Palinkas --0-1060512762-1355291153=:26605 Content-Type: TEXT/PLAIN; charset=US-ASCII; name=short.sch Content-Transfer-Encoding: BASE64 Content-ID: Content-Description: Content-Disposition: attachment; filename=short.sch diAyMDExMDExNSAyDQpUIDUxODAwIDU0NjAwIDkgMTAgMSAwIDAgMCAxDQph DQpUIDUzNTAwIDU0NjAwIDkgMTAgMSAwIDAgMCAxDQpiDQpUIDUxODAwIDUy NTAwIDkgMTAgMSAwIDAgMCAxDQpDDQpUIDUzNTAwIDUyNTAwIDkgMTAgMSAw IDAgMCAxDQpEDQpDIDUyMzAwIDU0NTAwIDEgMCAwIHJlc2lzdG9yLTIuc3lt DQp7DQpUIDUyNzAwIDU0ODUwIDUgMTAgMCAwIDAgMCAxDQpkZXZpY2U9UkVT SVNUT1INClQgNTI1MDAgNTQ4MDAgNSAxMCAxIDEgMCAwIDENCnJlZmRlcz1S MQ0KVCA1MjcwMCA1NDUwMCA1IDEwIDEgMSAwIDAgMQ0KdmFsdWU9MQ0KfQ0K QyA1MjAwMCA1MzIwMCAxIDkwIDAgcmVzaXN0b3ItMi5zeW0NCnsNClQgNTE2 NTAgNTM2MDAgNSAxMCAwIDAgOTAgMCAxDQpkZXZpY2U9UkVTSVNUT1INClQg NTE3MDAgNTM0MDAgNSAxMCAxIDEgOTAgMCAxDQpyZWZkZXM9UjINClQgNTIw MDAgNTM2MDAgNSAxMCAxIDEgOTAgMCAxDQp2YWx1ZT0xDQp9DQpDIDUzNjAw IDUzMjAwIDEgOTAgMCByZXNpc3Rvci0yLnN5bQ0Kew0KVCA1MzI1MCA1MzYw 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