单纯性算法是解决线性规划的经典方法,上世纪50年代就提出了,其基本思想是在可行域内沿着边遍历所有的顶点,找出最优值,即为算法的最优值。
算法的执行过程如下:
__author__ = 'linfuyuan' class Problem: def __init__(self): self.obj = 0 self.coMatrix = [] self.b = [] self.c = [] def pivot(self, basic, l, e): # the l-th line self.b[l] /= self.coMatrix[e][l] origin = self.coMatrix[e][l] for i in range(len(self.coMatrix)): self.coMatrix[i][l] /= origin # the other lines for i in range(len(self.b)): if i != l: self.b[i] = self.b[i] - self.b[l] * self.coMatrix[e][i] for i in range(len(self.b)): if i != l: origin = self.coMatrix[e][i] for j in range(len(self.coMatrix)): self.coMatrix[j][i] = self.coMatrix[j][i] - origin * self.coMatrix[j][l] origin = self.c[e] for i in range(len(self.coMatrix)): self.c[i] = self.c[i] - self.coMatrix[i][l] * origin self.obj = self.obj - self.b[l] * origin basic[l] = e def clone(self, another): self.obj = another.obj for i in another.b: self.b.append(i) for i in another.c: self.c.append(i) for v in another.coMatrix: newv = [] for i in v: newv.append(i) self.coMatrix.append(newv) basic = [] problem = Problem() def readProblem(filename): with open(filename)as f: var_num = int(f.readline()) constrait_num = int(f.readline()) matrix = [([0] * var_num) for i in range(constrait_num)] for i in range(constrait_num): line = f.readline() tokens = line.split(' ') for j in range(var_num): matrix[i][j] = float(tokens[j]) problem.b.append(float(tokens[-1])) for i in range(var_num): var = [] for j in range(constrait_num): var.append(matrix[j][i]) problem.coMatrix.append(var) line = f.readline() tokens = line.split(' ') for i in range(var_num): problem.c.append(float(tokens[i])) def getMinCIndex(c): min, minIndex = 1, 0 for i in range(len(c)): if c[i] < 0 and c[i] < min: min = c[i] minIndex = i if min > 0: return -1 else: return minIndex def getLamdaVector(evector): ld = [] for i in range(len(evector)): ld.append(evector[i]) return ld def simplex(basic, problem): minCIndex = getMinCIndex(problem.c) while minCIndex != -1: ld = getLamdaVector(problem.coMatrix[minCIndex]) # find the l line l, min = -1, 10000000000 for i in range(len(problem.b)): if ld[i] > 0: if problem.b[i] / ld[i] < min: l = i min = problem.b[i] / ld[i] if l == -1: return False problem.pivot(basic, l, minCIndex) minCIndex = getMinCIndex(problem.c) return True def initialSimplex(basic, problem): min, minIndex = 1000000000, -1 for i in range(len(problem.b)): if problem.b[i] < min: min = problem.b[i] minIndex = i for i in range(len(problem.b)): basic.append(i+len(problem.b)) if min >= 0: return True else: originC = problem.c newC = [] for i in range(len(problem.c)): newC.append(0) newC.append(1) problem.c = newC x0 = [] for i in range(len(problem.b)): x0.append(-1) problem.coMatrix.append(x0) problem.pivot(basic, minIndex, -1) res = simplex(basic, problem) if res == False or problem.obj != 0: return False else: problem.c = originC problem.coMatrix.pop() # Gaussian row operation counter = 0 for i in basic: if problem.c[i] != 0: origin = problem.c[i] for j in range(len(problem.c)): problem.c[j] -= problem.coMatrix[j][counter] * origin problem.obj -= problem.b[counter] * origin counter += 1 return True filename = raw_input('please input the problem description file: ') readProblem(filename) if initialSimplex(basic, problem): res = simplex(basic, problem) if res: print 'the optimal obj is %.2f' % problem.obj index = ['0.00'] * len(problem.coMatrix) counter = 0 for i in basic: index[i] = '%.2f' % problem.b[counter] counter += 1 print 'the solution is {%s}' % ' '.join(index) else: print 'no feasible solution' else: print 'no feasible solution' raw_input('please input any key to quit.')希望对大家有所帮助。
原文地址:http://blog.csdn.net/xanxus46/article/details/42032721