cycle

Suppose we have a weighted undirected graph, and the number of edge |E| is about k times of node number |V|, i.e, |E| ~= k * |V|. Now, one node, say v, is choosen, and we want to find the cycle containing v with the minimum average cost. (i.e., average cost means average edge weight along a cycle.) Is there any efficient algorithms? Sorry, I missed one point-the cycle do not need to contain all nodes in this Graph. This makes it different from Hamilton cycle problem. This is equivalent to the Hamiltonian cycle problem , which is NP-complete. There is no worst-case-polynomial algorithm able to

After much searching I managed to find these two Cycle plugin options to allow for transparency in Internet Explorer cleartype: true, cleartypeNoBg: true, The problem I'm getting now is a black block thing around the image that appears during the bounce transition than gets removed once the transition is complete and the image is sitting in the frame. It's only happening on IE. Any ideas on how to fix this? Reading the documentation it seems that clearTypeNoBG does the opposite of what you think: it tells the plugin to NOT correct for transparency under IE. Leave it on false. cleartype is

Hey guys i need to output the girth of a given graph represented as an adjacency matrix.Could anyone give me some hints how i can use an adjacency matrix or an adjacency list to obtain the girth of a graph ? Thanks Example: graph one: 0 1 0 0 1 0 0 1 0 0 0 0 0 1 0 0 graph two: 0 1 0 0 1 1 0 1 0 0 0 1 0 1 0 0 0 1 0 1 1 0 0 1 0 The result: Girth of graph 1: infinity Girth of graph 2: 5 This algorithm will find the length of the shortest cycle: - set `girth` to infinity - for each edge -- remove the edge from the graph -- measure the distance between the edge endpoints -- if `girth` is longer

I have a directed graph and my problem is to enumerate all the minimal (cycles that cannot be constructed as the union of other cycles) directed cycles of this graph. This is different from what the Tarjan algorithm outputs. For instance, for the directed graph at this wikipedia page , I would like to get c <-> d and d <-> h as two separated directed cycles. I don't if this problem is polynomial or not. I ran over a paper "Enumerating Minimal Dicuts and Strongly Connected Subgraphs" that seems to conclude that this problem is incremental polynomial (I don't know what it means), but I am unable

I'm currently trying to make a cheap, simple slide show navigation box for a client (something along the lines of what, say, the Red Sox or Gamespot use on their sites, but far, far simpler). So far, it's actually coming along nicely, with one problem - the images don't appear upon a first visit. They only appear after the page is reloaded. I think it may be some sort of runtime issue, or perhaps a cache issue, but I'm not sure how to fix it. My code: PHP: if (isset($_GET['start']) && "true" === $_GET['start']) { $images = array(); if ($dir = dir('images')) { //$count = 0; while(false !== (

For a small project I'm working on I need to cycle through a list. For each element of this cycle I have to start another cycle through the same list, with the former element as first element of the new cycle. For example I'd like to be able to produce something like this: 1, 2, 3, 4, 1, 2, 3, 4, 1, ... 2, 3, 4, 1, 2, 3, 4, 1, 2, ... 3, 4, 1, 2, 3, 4, 1, 2, 3, ... 4, 1, 2, 3, 4, 1, 2, 3, 4, ... 1, 2, 3, 4, 1, 2, 3, 4, 1, ... ... I thought that copying a itertools.cycle after each .next() would conserve the current state, so that I can begin the new cycle with the element from the "outer" cycle

I am trying to build a slideshow using the jQuery Cycle plugin that contains another level of nested slideshows within some of the top-level slides. The main "container" slides slide horizontally, and then—for "container" slides that contain more than one image on the left-hand side—those images slide up vertically. Sample here, because I imagine it's hard to visualize: http://dl.dropbox.com/u/2890872/js-test/index.html You'll only get to the second top-level slide before it stops, even thought there are six top-level slides (which is my issue—the second slide only contains one left-hand inner