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*Subject*: Graph theory question*From*: jamesd at echeque.com (James A. Donald)*Date*: Fri, 27 Oct 2017 12:43:38 +1000*In-reply-to*: <[email protected]>*References*: <[email protected]> <CALmccy42Ehg8mRTh76QfnG4Kxz01ZBi8+FG42d3gbJDVuDfZtw@mail.gmail.com> <[email protected]> <[email protected]> <[email protected]> <[email protected]> <[email protected]>

In order to solve a sybil problem, want to show that each peer is accepted as a peer by three peers, and we don't want these peers to be sybils. Related directed graph problem: Google trying to distinguish real links from networks of fake websites giving links to each other. Suppose you have an undirected graph in which every vertex is connected to at least three other vertexes (because we collapse vertices with only one connection or two connections) Now suppose we want to make sure that the total graph is well connected - that every subgraph is connected to at least three vertexes of the rest of the graph. We want to collapse subgraphs that are only connected to the rest of the graph by one connection into the vertex by which they are connected, and collapse subgraphs that are only connected to rest of the graph by two connections into a single edge. And suppose our graph is very large, thousands, perhaps millions, of vertices Purported peers that have only one connection are clients of the entity to which they are connected, and he is responsible for their good behavior, similarly those that have only two connections. Three good peer connections make you a peer of all the other peers, two good connections do not make you a peer - don't get equal treatment to the vertices by which you are connected, get graylist treatment. --- This email has been checked for viruses by Avast antivirus software. https://www.avast.com/antivirus

**Follow-Ups**:**Graph theory question***From:*violarisgeorge at gmail.com (George Violaris)

**References**:**Primary Sources for "The Woman Who Smashed Codes" on Elizabeth Friedman***From:*jya at pipeline.com (John Young)

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