Graphs and Minimum Cuts(Karger's Min-Cut Algorithm)

Posted on 2013-08-03 21:23  冰天雪域  阅读(337)  评论(0编辑  收藏  举报

 

Graphs 

    Two ingredients

     1. vertices (nodes) v

     2. edges(undirected or directed)

Examples: road networks, the web, social networks


The minimum Cut problem

    Input: undirected graph G = (V, E)   (parallel edges allowed)

    Goal: compute a cut with fewest number of Crossing edges (a min cut)


 

Sparse vs. Dense Graphs

    let n = # of vertices, m = # of edges

    In most applications, m is Omega(n) and O(n^2)

        In a "sparse graph", m is O(n) or close to it

        In a "dense graph",  m is closer to Theta(n^2)


Two ways to represent a Graph

    1. The Adjacency Matrix 

    2. The Adjacency List

    Which one is better?  Depends on graph density and operation needed.



Random Contraction Algorithm

    while there are more than 2 vertices:

        -pick a remaining edge(u, v) uniformly at random

        -merge(or "contract") u and v into a single vertex

        -remove self-loops

    return cut represented by final 2 vertices




Karger's Min-Cut Algorithm -------Random Contraction Algorithm(Python code):

 

import random
import copy
import time

def contract(ver, e): 
    while len(ver) > 2: #create a new graph every time (not efficient)
        ind = random.randrange(0, len(e))      
        [u, v] = e.pop(ind)   #pick a edge randomly
        ver.remove(v)    #remove v from vertices
        newEdge = list()
        for i in range(len(e)):
            if e[i][0] == v: e[i][0] = u
            elif e[i][1] == v: e[i][1] = u
            if e[i][0] != e[i][1]: newEdge.append(e[i])   # remove self-loops
        e = newEdge
    return(len(e))  #return the number of the remained edges 

if __name__ == '__main__':
    f = open('kargerMinCut.txt')
    _f = list(f) 
    edges = list()          #initialize vertices and edges
    vertices = list()
    for i in range(len(_f)):     #got 2517 different edges
        s = _f[i].split()
        vertices.append(int(s[0]))
        for j in range(1, len(s)):
            if [int(s[j]), int(s[0])] not in edges:
                edges.append([int(s[0]), int(s[j])])  

    result = list()
    starttime = time.clock() 
    for i in range(2000):  #we take n^2logn times so that the Pr(allfail) <= 1/n where n is the number of vertics
        v = copy.deepcopy(vertices)   #notice: deepcopy
        e = copy.deepcopy(edges)
        r = contract(v, e)
        result.append(r)
    endtime = time.clock()
    #print(result)
    print(min(result))
    print(endtime - starttime)


 


 



    

 

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