Algorithms: Design and Analysis, Part 1
Download the text file here . (Right click and save link as)The file contains the adjacency list representation of a simple undirected graph. There are 200 vertices labeled 1 to 200. The first column in the file represents the vertex label, and the particular row (other entries except the first column) tells all the vertices that the vertex is adjacent to. So for example, the<nobr style="border:0px; padding:0px; margin:0px; max-width:none; max-height:none; vertical-align:0px; line-height:normal; text-decoration:none; white-space:nowrap!important"><span class="math" id="MathJax-Span-1" style="display:inline; position:static; border:0px; padding:0px; margin:0px; vertical-align:0px; line-height:normal; text-decoration:none"><span style="display:inline-block; position:relative; border:0px; padding:0px; margin:0px; vertical-align:0px; line-height:normal; text-decoration:none; width:23px; height:0px; font-size:18px"><span style="display:inline; position:absolute; border:0px; padding:0px; margin:0px; vertical-align:0px; line-height:normal; text-decoration:none; top:-41px; left:0px"><span class="mrow" id="MathJax-Span-2" style="display:inline; position:static; border:0px; padding:0px; margin:0px; vertical-align:0px; line-height:normal; text-decoration:none"><span class="msubsup" id="MathJax-Span-3" style="display:inline; position:static; border:0px; padding:0px; margin:0px; vertical-align:0px; line-height:normal; text-decoration:none"><span style="display:inline-block; position:relative; border:0px; padding:0px; margin:0px; vertical-align:0px; line-height:normal; text-decoration:none; width:22.4px; height:0px"><span style="display:inline; position:absolute; border:0px; padding:0px; margin:0px; vertical-align:0px; line-height:normal; text-decoration:none; top:-41px; left:0px"><span class="mn" id="MathJax-Span-4" style="font-family:MathJax_Main; display:inline; position:static; border:0px; padding:0px; margin:0px; vertical-align:0px; line-height:normal; text-decoration:none">6</span></span><span style="display:inline; position:absolute; border:0px; padding:0px; margin:0px; vertical-align:0px; line-height:normal; text-decoration:none; top:-48.3px; left:9px"><span class="texatom" id="MathJax-Span-5" style="display:inline; position:static; border:0px; padding:0px; margin:0px; vertical-align:0px; line-height:normal; text-decoration:none"><span class="mrow" id="MathJax-Span-6" style="display:inline; position:static; border:0px; padding:0px; margin:0px; vertical-align:0px; line-height:normal; text-decoration:none"><span class="mi" id="MathJax-Span-7" style="font-family:MathJax_Math; display:inline; position:static; border:0px; padding:0px; margin:0px; vertical-align:0px; line-height:normal; text-decoration:none; font-size:13px; font-style:italic">t</span><span class="mi" id="MathJax-Span-8" style="font-family:MathJax_Math; display:inline; position:static; border:0px; padding:0px; margin:0px; vertical-align:0px; line-height:normal; text-decoration:none; font-size:13px; font-style:italic">h</span></span></span></span></span></span></span></span></span></span></nobr>row looks like : "6 155 56 52 120 ......". This just means that the vertex with label 6 is adjacent to (i.e., shares an edge with) the vertices with labels 155,56,52,120,......,etc
Your task is to code up and run the randomized contraction algorithm for the min cut problem and use it on the above graph to compute the min cut. (HINT: Note that you'll have to figure out an implementation of edge contractions. Initially, you might want to do this naively, creating a new graph from the old every time there's an edge contraction. But you should also think about more efficient implementations.) (WARNING: As per the video lectures, please make sure to run the algorithm many times with different random seeds, and remember the smallest cut that you ever find.) Write your numeric answer in the space provided. So e.g., if your answer is 5, just type 5 in the space provided.
第三单元课件中的算法python实现代码如下:
import copy
import random
def contraCut(mapD,edgeList):
while len(mapD)>2:
[u,v]=edgeList.pop(random.randrange(0,len(edgeList)-1))
while([v,u] in edgeList):
edgeList.remove([v,u])
while([u,v] in edgeList):
edgeList.remove([u,v])
for ind in range(0,len(edgeList)):
if edgeList[ind][0]==v:edgeList[ind][0]=u
if edgeList[ind][1]==v:edgeList[ind][1]=u
mapD[u]=mapD[u]-{v}
mapD[v]=mapD[v]-{u}
for [x,y] in mapD.items():
if v in y:
mapD[x]=(mapD[x]|{u})-{v}
mapD[u]=mapD[u]|mapD[v]
del mapD[v]
return len(edgeList)/2
if __name__ == '__main__':
f=open('kargerMinCut.txt','r')
mapDict={}
for line in f.readlines():
tmp=[int(x) for x in line.split()]
mapDict[tmp[0]]=set(tmp[1:])
f.close()
edgeList=[]
for [x,y] in mapDict.items():
edgeList.extend([[x,v] for v in y])
numList=[]
for i in range(20):
cpmapDict=copy.deepcopy(mapDict)
cpedgeList=copy.deepcopy(edgeList)
#print cpmapDict
num=contraCut(cpmapDict,cpedgeList)
numList.append(num)
numList.sort()
print num,
i+=1
print numList
读取的txt文件sample如下:
1 3 5
2 4 5
3 1
4 2
5 1 2
每行第一列代表pointer,后面跟的是邻接点。
本文介绍了一种解决图论中最小割问题的随机化收缩算法,并提供了Python代码实现。通过不断收缩边来寻找图中的最小割,算法需多次运行以找到最佳解。

被折叠的 条评论
为什么被折叠?



