Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 exptools 1/6/84; site ihnet.UUCP Path: utzoo!watmath!clyde!burl!ulysses!mhuxl!ihnp4!ihnet!eklhad From: eklhad@ihnet.UUCP (K. A. Dahlke) Newsgroups: net.puzzle Subject: Re: Don't cross my path ! Message-ID: <150@ihnet.UUCP> Date: Fri, 17-Aug-84 15:06:07 EDT Article-I.D.: ihnet.150 Posted: Fri Aug 17 15:06:07 1984 Date-Received: Sun, 19-Aug-84 01:18:03 EDT References: <684@sbcs.UUCP> <785@abnjh.UUCP> Organization: AT&T Bell Labs, Naperville, IL Lines: 22 The three houses / three utilities problem has been around for a while. The three angry workers going to work is isomorphic, and more interesting. Planar graph theory is sufficient. You are looking for Kerotowski's theorem (name probably badly miss-spelled). It states: a graph can be placed in a plane with no crossings if and only if the graph does not contain a K3,3 or a K5. A K3,3 is the houses/utilities construct. A K5 is five points all interconnected. Another impossible puzzle: five homes have telephones with wires directly connecting each house with the other four houses (no crossings of ccourse). This (a K5) can also never be planar. I find the theorem absolutely amazing. A program could take an arbitrary scrawly graph and tell you if it could be put in a plane, just by checking for K3,3 and K5. I don't know if the program could actually rearrange the graph to make it planar. That would be interesting. Someone in the CAD world has probably solved this problem. -- Karl Dahlke ihnp4!ihnet!eklhad