我在使用模拟退火算法来解决 n 个皇后问题时遇到了一些麻烦。基本上，我让它寻找更好的更多，它工作得很好，但是我运行一个公式来检查它是否应该采取“坏”的举动。据我了解，公式是 e^（板状态计算变化）/CurrentTemperature。该数字应与随机双精度或浮点数进行比较，如果随机数大于等式，则算法应采取“坏”举动。我遇到的问题是公式总是非常接近 1 或超过 1。这是我的一些代码（让我知道是否应该提供更多）：
temperature = n*100; //initializes starting temperature
currentTemp = n*100;
int cooldown = n*2; //initializes cool down temperature
float examine = 0; //this is the change in board calculation
int cost = 1;
boolean betterMove = false;
queen = new int[n];
int[][] board = graph; // generates a board of n size
float ran = 0; //random float to compare to
double compareAgainst = 0; //formula variable
cost = calculate(board, n); //calculates the cost
while (cost > 0 && currentTemp > 0)
{
// chooses a random queen to move that has a heuristic higher than zero
int Q = rand.nextInt(n);
while (queen[Q] == 0)
Q = rand.nextInt(n);
betterMove = false;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
if (board[i][j] == 1 && j == Q)
{
while (!betterMove)
{
int move = i;
while (move == i)
move = rand.nextInt(n); //pick a random move
tempBoard[i][j] = 0; //erase old position
tempBoard[move][j] = 1; //set new position
examine = calculate(tempBoard, n) - calculate(board, n); //calculates the difference between the change in boards
ran = rand.nextFloat(); //generates random number to compare against
compareAgainst = Math.pow(Math.E, (-examine / currentTemp)); //formula out of the book, basically is e^(change in board state divided by currentTemp)
if (calculate(tempBoard, n) < calculate(board, n)) //if this is a better move
{
for (int a = 0; a < n; a++)
for (int b = 0; b < n; b++)
board[a][b] = tempBoard[a][b]; //set it to the board
cost = calculate(board, n);
currentTemp -= cooldown; //cool down the temperature
betterMove = true;
}
else if(calculate(tempBoard,n) >= calculate(board,n)) //if this is a worse move
{
if(verbose == 1) //outputs whether or not this is a bad move and outputs function value and random float for simulated annealing purposes
{
System.out.println("This is a worse move");
System.out.println("The numbers for Simulated Annealing:");
System.out.println("Random number = " + ran);
System.out.println("Formula = " + compareAgainst);
System.out.println("Examine = " + examine);
}
if(ran > compareAgainst) //if the random float is greater than compare against, take the bad move
{
for (int a = 0; a < n; a++)
for (int b = 0; b < n; b++)
board[a][b] = tempBoard[a][b]; //take the move
cost = calculate(board, n);
currentTemp-= cooldown;
betterMove = true;
}
else //if not, do not take the move
{
for (int a = 0; a < n; a++)
for (int b = 0; b < n; b++)
tempBoard[a][b] = board[a][b];
}
currentTemp-= cooldown;
betterMove = true;
}
}
}
i = n;
j = n;
}
}
}
}
我尝试了很多方法，例如将检查变量设为负数或取棋盘状态之间差异的绝对值。此外，被调用的计算函数基本上是扫描棋盘并返回有多少皇后发生冲突，这是一个 int。让我知道是否需要更多说明。谢谢
