一个数学游戏

一个古老的数学游戏，很早就知道了，但一直编程没编正确。今天终于正确了。

有1、2、3、4、5、6这6种牌，每种牌有3张。开始总点和记为24，然后两个人轮流选一张牌，然后从总点和里减去牌面值，直到总点和非正。谁取完牌后，总点和为0，则胜；若为负，则败。

 

例如：

24

A：3

21

B：6

15

A：1

14

B：2

12

A：5

7

B：3

4

A：4

A胜

 

问先取是否有必胜策略，或后取是否有？

 

 

下面是一段人和电脑玩的代码

 

#include <cstdio>#include <cstdlib>#include <ctime>const int N = 6;const int initV = 3;int A[N];int maxDepth=0;int lastChoice= -1;void status() {printf("Remains : ");for (int i=0;i<N;++i)printf("%3d %2d | ", i+1, A[i]);printf("/n");}// play 返回和为 sum 时的最佳策略和最大概率double play(int sum, int depth) {if (depth>maxDepth) {maxDepth = depth;//status();}if (sum==0) return 0.0;if (sum<0) return 1.0;double minHope=1;for (int i=0;i<N;++i)if (A[i]>0) {--A[i];double hope = play(sum-1-i, depth+1);++A[i];if (hope<minHope) {minHope = hope;if (depth==0) lastChoice = i;if (minHope< 1E-6) break;}}return 1-minHope;}int main() {srand((unsigned)time(0));for (int i=0;i<N;++i) A[i] = initV;int humFirst = 0;printf("Hum first(0,1) ? ");scanf("%d", &humFirst);humFirst %= 2;int sum = 24;while(sum>0) {printf("sum=%d /n", sum);int c=0;if (humFirst>0) {status();while(!(c>=1 && c<=N && A[c-1]>0)) {if (c>=1 && c<=N) printf("%d leaves nothing!/n", c);printf("choose (1..%d) : ", N);scanf("%d", &c);}--c;} else {play(sum, 0);c = lastChoice;while(!(c>=0 && c<N && A[c]>0)) {c = rand()%N;}printf("Computer choose %d/n", c+1);}sum -= (c+1);--A[c];humFirst = (humFirst+1)%2;}if ((sum==0) ^ (humFirst==0)) printf("YOU LOSE!/n");elseprintf("YOU WIN!/n");return 0;}

 

下面是加强的版本，表征胜败不是用1,0简单表达，而是用了-1.0~1.0之间的实数，符号表示胜败，绝对值大小表示胜或败的强度，这样可以让电脑选择败强度低的局面（复杂，人能获胜的支路相对整个博弈树来要小，需要仔细选择），这样人可能容易出错。

另外在选择方案时，加入了一点随机性，以免每次博弈都是同样选择顺序（当，然胜败应该相同）。

 

#include <cstdio>#include <cstdlib>#include <ctime>#include <cmath>const int N = 6;const int initV = 3;int A[N];int maxDepth=0;int lastChoice= -1;void status() {printf("Remains : ");for (int i=0;i<N;++i)printf("%3d %2d | ", i+1, A[i]);printf("/n");}// play 返回和为 sum 时的最佳策略和最大概率double play(int sum, int depth) {if (depth>maxDepth) {maxDepth = depth;//status();}if (sum==0) return -1.0;if (sum<0) return 1.0;const double T = 0.001;double minHope=1.0;double negHope=0.0;double posHope=0.0;for (int i=0;i<N;++i)if (A[i]>0) {--A[i];double hope = play(sum-1-i, depth+1);++A[i];if (hope>0) posHope += hope;else negHope += hope;if (depth==0) {//printf("%d hope = %lf/n", i+1, hope);double chance=0.0;if (minHope>0 && hope<0) {minHope = hope;lastChoice = i;} else if (hope-minHope<-0.01 || ( fabs(hope-minHope)<0.01 && (chance=static_cast<double>(rand()%1024)/1024) > 0.6) ) {minHope = hope;lastChoice = i;} //printf("minhope %lf dhope %lf chance %lf/n", minHope, hope-minHope,chance);}}if (negHope < 0.0 && posHope> 0.0) {return -negHope/(posHope-negHope+1E-6);} else{return -posHope/(posHope-negHope+1E-6);}}int main(int argc, char* argv[]) {srand((unsigned)time(0));for(int i=rand()%13 + 3;i>0; --i) rand();for (int i=0;i<N;++i) A[i] = initV;int humFirst = 0;printf("Hum first(0,1) ? ");scanf("%d", &humFirst);humFirst %= 2;int sum = 24;if (argc==2) {sscanf(argv[1],"%d", &sum);}while(sum>0) {printf("sum=%d /n", sum);int c=0;if (humFirst>0) {status();while(!(c>=1 && c<=N && A[c-1]>0)) {if (c>=1 && c<=N) printf("%d leaves nothing!/n", c);printf("choose (1..%d) : ", N);scanf("%d", &c);}--c;} else {play(sum, 0);c = lastChoice;while(!(c>=0 && c<N && A[c]>0)) {c = rand()%N;}printf("Computer choose %d/n", c+1);}sum -= (c+1);--A[c];humFirst = (humFirst+1)%2;}if ((sum==0) ^ (humFirst==0)) printf("YOU LOSE!/n");elseprintf("YOU WIN!/n");return 0;}