具有类和for循环的MPI

问题描述

我想使用MPI并行处理图形中的哈密顿路径的计算。 因此，我实现了这一点：

from mpi4py import MPI
import random,time

comm = MPI.COMM_WORLD
my_rank = comm.Get_rank()
p = comm.Get_size()
numOfNodes = 10
numOfProblems = 11
class Graph:
    def __init__(self,numOfNodes):
        if numOfNodes > 0:
            self.numOfNodes = numOfNodes
        else:
            print("Error")

    def calculateMaxPairs(self):
        self.maxPairs = self.numOfNodes*(self.numOfNodes - 1)//2

    def generatePairs(self):
        self.calculateMaxPairs()

        self.pairs = []
        startRange = self.numOfNodes
        endRange = (self.numOfNodes - 10)*3 + 18
        numOfPairs = random.randint(startRange,endRange)

        while len(self.pairs) != numOfPairs:
            try:
                startNode = random.randint(1,self.numOfNodes)
                endNode = random.randint(1,self.numOfNodes)
                if startNode == endNode:
                    raise ValueError
            except ValueError:
                pass
            else:
                pair = (startNode,endNode)
                invertedPair = (endNode,startNode)

                if pair not in self.pairs and invertedPair not in self.pairs:
                    self.pairs.append(pair)
        self.hamiltonianPath = []

    def generatePathLink(self):
        self.graphLink = {}
        for x in self.pairs:
            x = str(x)
            splitNode = x.split(',')
            a = int(splitNode[0][1:])
            b = int(splitNode[1][:-1])
            try:
                if b not in self.graphLink[a]:
                    self.graphLink[a].append(b)
            except KeyError:
                self.graphLink[a] = []
                self.graphLink[a].append(b)
            finally:
                try:
                    if a not in self.graphLink[b]:
                        self.graphLink[b].append(a)
                except KeyError:
                    self.graphLink[b] = []
                    self.graphLink[b].append(a)
                finally:
                    pass

    def findpaths(self,start,end,path = []):
        path = path + [start]
        if start == end:
            return [path]
        if start not in self.graphLink:
            return []
        paths = []
        for node in self.graphLink[start]:
            if node not in path:
                newpaths = self.findpaths(node,path)
                for newpath in newpaths:
                    paths.append(newpath)
                    if (len(newpath) == self.numOfNodes):
                        self.hamiltonianPath = newpath
                        raise OverflowError
        return paths

    def exhaustiveSearch(self):
        try:
            allPaths = []
            for startNode in self.graphLink:
                for endNode in self.graphLink:
                    newPaths = self.findpaths(startNode,endNode)
                    for path in newPaths:
                        if (len(path) == self.numOfNodes):
                            allPaths.append(path)
            return allPaths
        except OverflowError:
            return self.hamiltonianPath
        else:
            pass

    def isHamiltonianPathExist(self):
        time_start = time.clock()
        self.generatePathLink()

        if len(self.graphLink) != self.numOfNodes:
            time_elapsed = (time.clock() - time_start)
            return [[],time_elapsed]
        else:
            result = self.exhaustiveSearch()
            time_elapsed = (time.clock() - time_start)
            if len(result) == 0:
                print("There isn't any Hamiltonian Path.")
            else:
                print("Computing time:",round(time_elapsed,2),"seconds\n")
            return [result,time_elapsed]
            comm.send(result,dest=0)
            

yes = 0
no = 0
total_computing_time = 0
for x in range(1,numOfProblems + 1):
    if my_rank !=0:
        graph = Graph(numOfNodes)
        graph.generatePairs()
        output = graph.isHamiltonianPathExist()
    else:
        for procid in range(1,p):
            result = comm.recv(source=procid)
            time_elapsed = comm.recv(source=procid,tag=12)
            total_computing_time += time_elapsed
            if len(result) == 0:
                no += 1
            else:
                yes += 1

print("Have Hamiltonian Path:",yes)
print("Don't Have Hamiltonian Path:",no)
print("Total computing time for %s problems in %s processes: %s s"%(numOfProblems,p,str(round(total_computing_time,2))))

您可以在此脚本中看到两个部分。
第一个是我们将生成图并计算哈密尔顿路径的地方。
第二个是我们告诉脚本在多个处理器中并行运行此图生成脚本的地方。
这里的问题是它生成图形并计算每个处理器中的路径，而不是在它们之间划分工作。
我在哪里做错了？

解决方法

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