21点的dp解

   一个21点的dp解,https://github.com/openai/gym/blob/master/gym/envs/toy_text/blackjack.py,实验环境,和真实的21点规则有点小区别,主要是Blackjack的大小计算有点区别.

#include 
#pragma warning(disable:4996)
using namespace std;

double dpp[11][11][2][2];
int maction[11][11][2];
double dp[22][6][2];
void dpdealer()
{
	for (int j = 1; j >= 0; j--)
	{
		for (int i = 21; i >= 1; i--)
		{
			if (i >= 17)
			{
				dp[i][i - 17][j] = 1;
			}
			else
			{
				if (i >= 7 && j == 1 && i <= 11)
				{
					dp[i][i - 7][j] = 1;
					continue;
				}
				int nxtj = j;
				for (int k = 1; k <= 10; k++)
				{
					double prob = 1.0 / 13;
					if (k == 10)
					{
						prob = 4.0 / 13;
					}
					if (k == 1)
						nxtj = 1;
					else
						nxtj = j;
					int nxtstat = i + k;
					if (i + k > 21)
					{
						dp[i][5][j] += prob;
					}
					else
					{
						for (int f = 0; f <= 5; f++)
						{
							dp[i][f][j] += prob * dp[nxtstat][f][nxtj];
						}
					}
				}
			}
		}
	}
}

void dpplayer()
{
	for (int k = 0; k < 2; k++)
	{
		for (int i = 9; i >= 0; i--)//corresponse to 12-21
		{
			/*
			if (i == 9)
			{
				for (int j = 0; j < 10; j++)
				{
					dpp[i][j][k][0] = 1;
					maction[i][j][k] = 0;
				}
				continue;
			}*/
			for (int j = 1; j <= 10; j++)
			{
				double winp = 0;
				for (int mk = 0; mk < i - 5; mk++)
				{
					winp += dp[j][mk][j == 1];
				}
				winp += dp[j][5][j == 1];
				double losep = 0;
				for (int mk = max(i - 4, 0); mk < 5; mk++)
				{
					losep += dp[j][mk][j == 1];
				}
				dpp[i][j - 1][k][0] = winp - losep;
			}
			for (int j = 1; j <= 10; j++)
			{
				for (int mk = 1; mk <= 10; mk++)
				{
					double prob = 1.0 / 13;
					if (mk == 10)
						prob = 4.0 / 13;
					int nxtstat = i + mk;

					if (k == 0)
					{
						if (nxtstat <= 9)
						{
							dpp[i][j - 1][k][1] += prob * max(dpp[nxtstat][j - 1][k][0], dpp[nxtstat][j - 1][k][1]);
						}
						else
						{
							dpp[i][j - 1][k][1] -= prob;
						}
					}
					else
					{
						if (nxtstat > 9)
						{
							nxtstat -= 10;
							dpp[i][j - 1][k][1] += prob * max(dpp[nxtstat][j - 1][k - 1][0], dpp[nxtstat][j - 1][k - 1][1]);
						}
						else
						{
							dpp[i][j - 1][k][1] += prob * max(dpp[nxtstat][j - 1][k][0], dpp[nxtstat][j - 1][k][1]);
						}
					}
				}
				maction[i][j - 1][k] = (dpp[i][j - 1][k][0] >= dpp[i][j - 1][k][1]) ? 0 : 1;
			}
		}
	}
}

int main()
{
	dpdealer();
	printf("without ace prob\n");
	for (int i = 1; i <= 10; i++)
	{
		for (int j = 0; j <= 5; j++)
			printf("%.2f ", dp[i][j][0]);
		printf("\n");
	}
	printf("with ace prob\n");
	for (int i = 1; i <= 10; i++)
	{
		for (int j = 0; j <= 5; j++)
			printf("%.2f ", dp[i][j][1]);
		printf("\n");
	}
	dpplayer();
	printf("\n\n");


	printf("without ace policy\n");
	for (int i = 9; i >= 0; i--)
	{
		for (int j = 0; j <= 9; j++)
		{
			printf("%d ", maction[i][j][0]);
		}
		printf("\n");
	}

	printf("with ace policy\n");
	for (int i = 9; i >= 0; i--)
	{
		for (int j = 0; j <= 9; j++)
		{
			printf("%d ", maction[i][j][1]);
		}
		printf("\n");
	}
}

运行结果如下: 

 和标准的解相同:

这个项目教会了大家怎么和机器人玩21点