# !/usr/bin/env python 
# -*- coding:utf-8 -*- 
# Author: wsw 
# 简单实现SGA算法 
import numpy as np 
from scipy.optimize import fsolve, basinhopping 
import random 
import timeit 
 
 
# 根据解的精度确定染色体(chromosome)的长度 
# 需要根据决策变量的上下边界来确定 
def getEncodedLength(delta=0.0001, boundarylist=[]): 
 # 每个变量的编码长度 
 lengths = [] 
 for i in boundarylist: 
  lower = i[0] 
  upper = i[1] 
  # lamnda 代表匿名函数f(x)=0,50代表搜索的初始解 
  res = fsolve(lambda x: ((upper - lower) * 1 / delta) - 2 ** x - 1, 50) 
  length = int(np.floor(res[0])) 
  lengths.append(length) 
 return lengths 
 pass 
 
 
# 随机生成初始编码种群 
def getIntialPopulation(encodelength, populationSize): 
 # 随机化初始种群为0 
 chromosomes = np.zeros((populationSize, sum(encodelength)), dtype=np.uint8) 
 for i in range(populationSize): 
  chromosomes[i, :] = np.random.randint(0, 2, sum(encodelength)) 
 # print('chromosomes shape:', chromosomes.shape) 
 return chromosomes 
 
 
# 染色体解码得到表现型的解 
def decodedChromosome(encodelength, chromosomes, boundarylist, delta=0.0001): 
 populations = chromosomes.shape[0] 
 variables = len(encodelength) 
 decodedvalues = np.zeros((populations, variables)) 
 for k, chromosome in enumerate(chromosomes): 
  chromosome = chromosome.tolist() 
  start = 0 
  for index, length in enumerate(encodelength): 
   # 将一个染色体进行拆分，得到染色体片段 
   power = length - 1 
   # 解码得到的10进制数字 
   demical = 0 
   for i in range(start, length + start): 
    demical += chromosome[i] * (2 ** power) 
    power -= 1 
   lower = boundarylist[index][0] 
   upper = boundarylist[index][1] 
   decodedvalue = lower + demical * (upper - lower) / (2 ** length - 1) 
   decodedvalues[k, index] = decodedvalue 
   # 开始去下一段染色体的编码 
   start = length 
 return decodedvalues 
 
 
# 得到个体的适应度值及每个个体被选择的累积概率 
def getFitnessValue(func, chromosomesdecoded): 
 # 得到种群规模和决策变量的个数 
 population, nums = chromosomesdecoded.shape 
 # 初始化种群的适应度值为0 
 fitnessvalues = np.zeros((population, 1)) 
 # 计算适应度值 
 for i in range(population): 
  fitnessvalues[i, 0] = func(chromosomesdecoded[i, :]) 
 # 计算每个染色体被选择的概率 
 probability = fitnessvalues / np.sum(fitnessvalues) 
 # 得到每个染色体被选中的累积概率 
 cum_probability = np.cumsum(probability) 
 return fitnessvalues, cum_probability 
 
 
# 新种群选择 
def selectNewPopulation(chromosomes, cum_probability): 
 m, n = chromosomes.shape 
 newpopulation = np.zeros((m, n), dtype=np.uint8) 
 # 随机产生M个概率值 
 randoms = np.random.rand(m) 
 for i, randoma in enumerate(randoms): 
  logical = cum_probability >= randoma 
  index = np.where(logical == 1) 
  # index是tuple,tuple中元素是ndarray 
  newpopulation[i, :] = chromosomes[index[0][0], :] 
 return newpopulation 
 pass 
 
 
# 新种群交叉 
def crossover(population, Pc=0.8): 
 """ 
 :param population: 新种群 
 :param Pc: 交叉概率默认是0.8 
 :return: 交叉后得到的新种群 
 """ 
 # 根据交叉概率计算需要进行交叉的个体个数 
 m, n = population.shape 
 numbers = np.uint8(m * Pc) 
 # 确保进行交叉的染色体个数是偶数个 
 if numbers % 2 != 0: 
  numbers += 1 
 # 交叉后得到的新种群 
 updatepopulation = np.zeros((m, n), dtype=np.uint8) 
 # 产生随机索引 
 index = random.sample(range(m), numbers) 
 # 不进行交叉的染色体进行复制 
 for i in range(m): 
  if not index.__contains__(i): 
   updatepopulation[i, :] = population[i, :] 
 # crossover 
 while len(index) > 0: 
  a = index.pop() 
  b = index.pop() 
  # 随机产生一个交叉点 
  crossoverPoint = random.sample(range(1, n), 1) 
  crossoverPoint = crossoverPoint[0] 
  # one-single-point crossover 
  updatepopulation[a, 0:crossoverPoint] = population[a, 0:crossoverPoint] 
  updatepopulation[a, crossoverPoint:] = population[b, crossoverPoint:] 
  updatepopulation[b, 0:crossoverPoint] = population[b, 0:crossoverPoint] 
  updatepopulation[b, crossoverPoint:] = population[a, crossoverPoint:] 
 return updatepopulation 
 pass 
 
 
# 染色体变异 
def mutation(population, Pm=0.01): 
 """ 
 
 :param population: 经交叉后得到的种群 
 :param Pm: 变异概率默认是0.01 
 :return: 经变异操作后的新种群 
 """ 
 updatepopulation = np.copy(population) 
 m, n = population.shape 
 # 计算需要变异的基因个数 
 gene_num = np.uint8(m * n * Pm) 
 # 将所有的基因按照序号进行10进制编码，则共有m*n个基因 
 # 随机抽取gene_num个基因进行基本位变异 
 mutationGeneIndex = random.sample(range(0, m * n), gene_num) 
 # 确定每个将要变异的基因在整个染色体中的基因座(即基因的具体位置) 
 for gene in mutationGeneIndex: 
  # 确定变异基因位于第几个染色体 
  chromosomeIndex = gene // n 
  # 确定变异基因位于当前染色体的第几个基因位 
  geneIndex = gene % n 
  # mutation 
  if updatepopulation[chromosomeIndex, geneIndex] == 0: 
   updatepopulation[chromosomeIndex, geneIndex] = 1 
  else: 
   updatepopulation[chromosomeIndex, geneIndex] = 0 
 return updatepopulation 
 pass 
 
 
# 定义适应度函数 
def fitnessFunction(): 
 return lambda x: 21.5 + x[0] * np.sin(4 * np.pi * x[0]) + x[1] * np.sin(20 * np.pi * x[1]) 
 pass 
 
 
def main(max_iter=500): 
 # 每次迭代得到的最优解 
 optimalSolutions = [] 
 optimalValues = [] 
 # 决策变量的取值范围 
 decisionVariables = [[-3.0, 12.1], [4.1, 5.8]] 
 # 得到染色体编码长度 
 lengthEncode = getEncodedLength(boundarylist=decisionVariables) 
 for iteration in range(max_iter): 
  # 得到初始种群编码 
  chromosomesEncoded = getIntialPopulation(lengthEncode, 10) 
  # 种群解码 
  decoded = decodedChromosome(lengthEncode, chromosomesEncoded, decisionVariables) 
  # 得到个体适应度值和个体的累积概率 
  evalvalues, cum_proba = getFitnessValue(fitnessFunction(), decoded) 
  # 选择新的种群 
  newpopulations = selectNewPopulation(chromosomesEncoded, cum_proba) 
  # 进行交叉操作 
  crossoverpopulation = crossover(newpopulations) 
  # mutation 
  mutationpopulation = mutation(crossoverpopulation) 
  # 将变异后的种群解码，得到每轮迭代最终的种群 
  final_decoded = decodedChromosome(lengthEncode, mutationpopulation, decisionVariables) 
  # 适应度评价 
  fitnessvalues, cum_individual_proba = getFitnessValue(fitnessFunction(), final_decoded) 
  # 搜索每次迭代的最优解，以及最优解对应的目标函数的取值 
  optimalValues.append(np.max(list(fitnessvalues))) 
  index = np.where(fitnessvalues == max(list(fitnessvalues))) 
  optimalSolutions.append(final_decoded[index[0][0], :]) 
 # 搜索最优解 
 optimalValue = np.max(optimalValues) 
 optimalIndex = np.where(optimalValues == optimalValue) 
 optimalSolution = optimalSolutions[optimalIndex[0][0]] 
 return optimalSolution, optimalValue 
 
 
solution, value = main() 
print('最优解: x1, x2') 
print(solution[0], solution[1]) 
print('最优目标函数值:', value) 
# 测量运行时间 
elapsedtime = timeit.timeit(stmt=main, number=1) 
print('Searching Time Elapsed:(S)', elapsedtime)