c# - 如何在 C# 中解决应用程序响应问题
问题描述
我正在尝试在 C# 中实现遗传算法。遗传有交叉法和变异法。种群大小为 5 条染色体，其中每条染色体是整数的二维数组（矩阵 10x10）遗传应循环 50 次，并在每个循环中调用交叉和变异，如下所示：
DateTime startTiming = DateTime.Now;
TimeSpan startGenetic;
// Make a population for each layer - as we have 4 layers, thus 4 population arrays after the intial eavaluation for the intial population and have differernt values in the aterations > 1
static List<int[,]> populationLayer1 = new List<int[,]>();
static List<int[,]> populationLayer2 = new List<int[,]>();
static List<int[,]> populationLayer3 = new List<int[,]>();
static List<int[,]> populationLayer4 = new List<int[,]>();
// 4 layers - we need 4 arrays
double[,] FitnessValLayer1 = new double[5, 2]; // for all "5" chromosome we store "2" Values : (Fitness Value - Fitness Ratio) in each layer (as we have 4 layers)
double[,] FitnessValLayer2 = new double[5, 2]; // for all "5" chromosome we store "2" Values : (Fitness Value - Fitness Ratio) in each layer (as we have 4 layers)
double[,] FitnessValLayer3 = new double[5, 2]; // for all "5" chromosome we store "2" Values : (Fitness Value - Fitness Ratio) in each layer (as we have 4 layers)
double[,] FitnessValLayer4 = new double[5, 2]; // for all "5" chromosome we store "2" Values : (Fitness Value - Fitness Ratio) in each layer (as we have 4 layers)
// 4 RouletteWeel values because of the 4 layers
int[] RouletteWheelLayer1 = new int[10];
int[] RouletteWheelLayer2 = new int[10];
int[] RouletteWheelLayer3 = new int[10];
int[] RouletteWheelLayer4 = new int[10];
public async Task Genetic_Algorithm(List<int[,]> population)
{
cancelSource = new CancellationTokenSource();
//In this step just duplicate the initial population
populationLayer1 = population.ToList();
populationLayer2 = population.ToList();
populationLayer3 = population.ToList();
populationLayer4 = population.ToList();
int round = 0;
for (geneticIteration = 0; geneticIteration < minIteration; geneticIteration++)
{
round = geneticIteration;
//----------------------//
//Calculate Fitness
try
{
// Calculate the fitness Function and the Fitness Ratio
await FitnessFunctionAsync(populationLayer1, populationLayer2, populationLayer3, populationLayer4, cancelSource.Token); // Fitness Function
}
catch (Exception ex)
{
// Write output to the file.
Trace.Write("when calling (FitnessFunctionAsync)..." + ex.Message);
Trace.Flush();
}
//----------------------//
// To Do :
//get 4 arrays for the fitness for each layer
//----------------------//
// Note : Each layer has different values for fitness so the changes on the population will not be the same in the 4 layers
//---------------------//
//RouletteWeel
RouletteWheelLayer1 = RouletteWheel_Selection(FitnessValLayer1);
RouletteWheelLayer2 = RouletteWheel_Selection(FitnessValLayer2);
RouletteWheelLayer3 = RouletteWheel_Selection(FitnessValLayer3);
RouletteWheelLayer4 = RouletteWheel_Selection(FitnessValLayer4);
//Crossover
// populationLayer1 = CrosssOver(RouletteWheelLayer1, populationLayer1);
//populationLayer2 = CrosssOver(RouletteWheelLayer2, populationLayer2);
// populationLayer3 = CrosssOver(RouletteWheelLayer3, populationLayer3);
//populationLayer4 = CrosssOver(RouletteWheelLayer4, populationLayer4);
//Mutation
//populationLayer1 = Mutation(RouletteWheelLayer1, populationLayer1);
// populationLayer2 = Mutation(RouletteWheelLayer2, populationLayer2);
// populationLayer3 = Mutation(RouletteWheelLayer3, populationLayer3);
//populationLayer4 = Mutation(RouletteWheelLayer4, populationLayer4);
// 4 layers - re-intialize
FitnessValLayer1 = new double[5, 2];
FitnessValLayer2 = new double[5, 2];
FitnessValLayer3 = new double[5, 2];
FitnessValLayer4 = new double[5, 2];
// 4 RouletteWeel - re-intialize
RouletteWheelLayer1 = new int[10];
RouletteWheelLayer2 = new int[10];
RouletteWheelLayer3 = new int[10];
RouletteWheelLayer4 = new int[10];
}
InvokeUpdateControls();
}
不幸的是，当实现交叉或突变时，即满足概率时，我的应用程序没有响应。以下是两种方法：
public List<int[,]> CrosssOver(int[] RouletteWheel, List<int[,]> population)
{
double rndNumber1 = 0.0;
int rndNumber2 = 0;
Random rnd = new Random();
int chrom1 = 0;
int chrom2 = 0;
for (int i = 0; i < population.Count; i++) // For all Chromosomes
{
rndNumber1 = rnd.Next(0, 11) / 10.00; // generate a random number to check the probability for crossover
chrom1 = RouletteWheel[rnd.Next(0, 10)];
chrom2 = RouletteWheel[rnd.Next(0, 10)];
if (rndNumber1 <= Pc) /* check if we will do Crossover */
{
rndNumber2 = rnd.Next(0, rows - 1); // determine the crossover point randomly by generating number between 0 and rows-1
for (int j = 0; j < rows; j++)
{
for (int v = 0; v < columns; v++)
{
if (j == rndNumber2) /* copy from same chromosome */
{
try
{
population[chrom1][j, v] = population[chrom2][j, v];
}
catch (Exception ex)
{
// Write output to the file.
Trace.Write("crossover..." + ex.Message);
Trace.Flush();
return population;
}
}
}
}
}
}
return population;
} // end-cross-over
变异方法：
public List<int[,]> Mutation(int[] RouletteWheel, List<int[,]> population)
{
double rndNumber1 = 0.0;
int chrom1 = 0;
int rndNumber2 = 0;
Random rnd = new Random();
for (int i = 0; i < population.Count; i++) // For all Chromosomes
{
rndNumber1 = rnd.Next(0, 11) / 100.00; // generate a random number between 0 and 10 and divide result by 100
chrom1 = RouletteWheel[rnd.Next(0, 10)];
if (rndNumber1 <= Pm) /* check if we will do Crossover */
{
rndNumber2 = rnd.Next(0, rows); // determine the crossover point randomly by generating number between 0 and rows -1
for (int j = 0; j < rows; j++)
{
for (int v = 0; v < columns; v++)
{
if (j == rndNumber2) /* Mutate the cell that is equal to 1 */
{
try
{
if (population[chrom1][j, v] == 0)
{
population[chrom1][j, v] = 1;
}
}
catch (Exception ex)
{
// Write output to the file.
Trace.Write("mutation..." + ex.Message);
Trace.Flush();
return population;
}
}
}
}
}
}
return population;
}
由于这些方法，我的应用程序被卡住了......如果突变和交叉成功完成，我需要进入下一个迭代。
解决方案
不幸的是，在这种情况下，唯一真正的答案是将有问题的代码移动到一个单独的线程中，这样主线程仍然可以用来抽取消息队列。
确保它不能被触发两次，确保在它运行时没有其他东西可以写入数据，并且即使在任何其他线程中读取任何数据（例如，显示中间结果）也要非常小心。
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