cycle

问题 I lost a little bit of time in this Python for statement: class MyListContainer: def __init__(self): self.list = [] def purge(self): for object in self.list: if (object.my_cond()): self.list.remove(object) return self.list container = MyListContainer() # now suppose both obj.my_cond() return True obj1 = MyCustomObject(par) obj2 = MyCustomObject(other_par) container.list = [obj1, obj2] # returning not an empty list but [obj2] container.purge() It doesn't work as I expected because when the

问题 I found the jQuery Cycle plugin which is very interesting. So, there is the following script that should return the images: $(function() { // retrieve list of slides from server $.getJSON('slidelist.php', startSlideshow); function startSlideshow(slides) { /* server returns an array of slides which looks like this: [ 'images/beach2.jpg', 'images/beach3.jpg', 'images/beach4.jpg', 'images/beach5.jpg', 'images/beach6.jpg', 'images/beach7.jpg', 'images/beach8.jpg' ] */ var totalSlideCount = 1 +

问题 Given graph adjacency matrix (for ex. g[][]), graph is directed. Needs find count of all graph cycles (if exists) and print them. I tried to wrote this algorithm in Java, sometimes it works correctly. If graph has complex cycles, algorithm return crazy cycles. Please, look at my code and help to resolve this problem public static final int k = 6; public static int g[][] = { { 0, 1, 0, 0, 0, 0 }, { 1, 0, 1, 0, 0, 0 }, { 0, 0, 0, 1, 0, 0 }, { 0, 0, 0, 0, 1, 0 }, { 0, 0, 1, 0, 0, 0 }, { 0, 0, 0,

问题 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. 回答1: This is equivalent

问题 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 回答1: This algorithm will find the length of the shortest cycle: - set `girth` to infinity - for each edge --

问题 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

问题 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

问题 Given two arrays, how do you check if one is a cyclic permutation of the other? For example, given a = [1, 2, 3, 1, 5] , b = [3, 1, 5, 1, 2] , and c = [2, 1, 3, 1, 5] we have that a and b are cyclic permutations but c is not a cyclic permutation of either. Note: the arrays may have duplicate elements. 回答1: The standard trick here is to concatenate one of the arrays with itself, and then try to find the 2nd array in the concatenated array. For example, 'a' concatenated with itself is: [1, 2, 3

问题 I am working with complex networks. I want to find group of nodes which forms a cycle of 3 nodes (or triangles) in a given graph. As my graph contains about million edges, using a simple iterative solution (multiple "for" loop) is not very efficient. I am using python for my programming, if these is some inbuilt modules for handling these problems, please let me know. If someone knows any algorithm which can be used for finding triangles in graphs, kindly reply back. 回答1: A million edges is

问题 Can someone explain how to find the number of Hamiltonian cycles in a complete undirected graph? Wikipedia says that the formula is (n-1)!/2 , but when I calculated using this formula, K3 has only one cycle and K4 has 5. Was my calculation incorrect? 回答1: Since the graph is complete, any permutation starting with a fixed vertex gives an (almost) unique cycle (the last vertex in the permutation will have an edge back to the first, fixed vertex. Except for one thing: if you visit the vertices