有限自动机的构造与识别
一、实验目标
1、掌握有穷状态自动机的概念;
2、掌握有穷状态自动机的存储及表示方法;
3、掌握有穷状态自动机与正则式之间的关系。
二、实验要求
1、输入正规式;
2、构造该正规式的有穷状态自动机;
3. 以五元组形式输出。
三、算法
参见教材的转换规则。
练习:
² (a|b)*abb
² l(l|d)*
² 1(1010*|1(010)*1)*0
四、完成算法设计、编码和调试工作,完成实验报告。
#include <string.h> #include <stdio.h> #include <stdlib.h> int main() { char p[30][30]; //存放文法 char q[30][30]; int line=0; int n; int i,j; int count=0; int k,t=0; int flag=0; int l,m=0; char VN[30] = { '\0' }; //存放非终结符号 char VT[30] = { '\0' }; //存放终结符号 printf( "\t请输入规则个数:" ); scanf( "%d" ,&n); line = n; for (i = 0; i < 30; i++) //给字符串数组p、q全部赋值为'\0' for (j=0;j<30;j++){p[i][j]= '\0' ;q[i][j]= '\0' ;} printf( "\t请输入文法:\n" ); for (i = 0; i < line; i++) {printf( "\t" );scanf( "\t%s" ,p[i]);} //把字符分为终结符号合非终结符号 l=0;m=0; for (i = 0;i < line; i++) { for (j = 0;j < 30&&(p[i][j] != '\0' );j++) { // 非终结符号放入数组VN中 if ((p[i][j]<= 'z' && p[i][j]>= 'a' )||(p[i][j]<= '9' && p[i][j]>= '0' )) { flag = 0; for (t=0; VN[t] != '\0' ;t++){ if (VN[t] == p[i][j]){flag = 1; break ;}} if (flag == 0){VN[l] = p[i][j];l++;} } // 终结符号放入数组VT中 if (p[i][j]<= 'Z' && p[i][j]>= 'A' ) { flag = 0; for (t = 0; t<30&&(VT[t] != '\0' ); t++){ if (VT[t] == p[i][j]){flag = 1; break ;}} if (flag==0){VT[m] = p[i][j];m++;} } } //把规则右部分分离放入数组q中 count = 0; k =0; for (i = 0;i < line;i++) { for (j = 4;j < 30 && (p[i][j] != '\0' );j++) { if ((p[i][j]<= 'z' && p[i][j]>= 'a' )||(p[i][j]<= 'Z' && p[i][j]>= 'A' )||(p[i][j]<= '9' && p[i][j]>= '0' )){q[count][k] = p[i][j];k++;} else {count++;k =0;} } count++;k=0; } //判断是确定的还是非确定的有穷状态自动并进行前半部分打印 //判断依据:q数组中每一行字符串是否相同 flag = 0; for (i=0;i<count;i++) { for (j=i+1;j<count;j++) { if (strcmp(q[i],q[j])==0){flag=1; break ;} if (flag==0){VT[m]=p[i][j];m++;} } } } // 把规则右部分分离放入数组 q 中 count=0; k=0; for (i=0;i<line;i++) { for (j=4;j<30&&(p[i][j]!= '\0' );j++) { if ((p[i][j]<= 'z' &&p[i][j]>= 'a' )||(p[i][j]<= 'Z' &&p[i][j]>= 'A' )||(p[i][j]<= '9' &&p[i][j]>= '0' )){q[count][k]=p[i][j];k++;} else {count++;k=0;} } count++;k=0; } // 判断是确定的还是非确定的有穷状态自动机并进行前半部分打印 // 判断依据q 数组中每一行字符串是否相同 flag=0; for (i=0;i<count;i++) { for (j=i+1;j<count;j++) { if (strcmp(q[i],q[j])==0) {flag=1; break ;} } } if (flag==1){printf( "\t是非确定的有穷状态自动机,即 NFA\n\n" ); printf( "\t构造的有穷状态自动机为: \n" ); printf( "\tNFA N= ( K ,∑, M , {S} , {Z} ) \n" );} else {printf( "\t是确定的有穷状态自动机,即 DFA\n\n\n" ); printf( "\t构造的有穷状态自动机为: \n" ); printf( "\tDFA D= ( K ,∑, M , {S} , {Z} ) \n" );} printf( "\t其中:\n\tK={S" ); for (i=0;i<30&&(VT[i]!= '\0' );i++){printf( " , %c" ,VT[i]);} printf( "}\n" );printf( "\t∑ ={" ); for (i=0;i<30&&(VN[i]!= '\0' );i++){printf( "%c " ,VN[i]);} printf( "}\n" );k=0;count=0; for (i=0;i<line;i++) { j=4; while (p[i][j]!= '\0' ) { if (k<4){q[count][k]=p[i][k];k++;} else { if ((p[i][j]<= 'z' && p[i][j]>= 'a' )||(p[i][j]<= 'Z' && p[i][j]>= 'A' )||(p[i][j]<= '9' &&p[i][j]>= '0' )){q[count][k]=p[i][j];k++;j++;} if (p[i][j]== '|' ){count++;k=0;j++;} } } count++;k=0; } printf( "\n" );printf( "\tM:\n" );l=0; while (VN[l]!= '\0' ) { printf( "\tM(S,%c)={" ,VN[l]); for (i=0;i<30;i++) { for (j=4;j<30&&(q[i][j]!= '\0' );j++) { if (VN[l]==q[i][j]&&(q[i][j+1]== '\0' )&&(q[i][j-1]== '=' )) printf( "%c" ,q[i][0]); } } printf( "}\t" );l++; } printf( "\n" ); l=0; k=0; while (VT[k]!= '\0' ) { l=0; while (VN[l]!= '\0' ) { printf( "\tM(%c,%c)={" ,VT[k],VN[l]); for (i=0;i<30;i++) { for (j=4;j<30&&(q[i][j]!= '\0' );j++) { if (VT[k]==q[i][j]&&VN[l]==q[i][j+1]) printf( "%c" ,q[i][0]); } } printf( "}\t" );l++; } k++;printf( "\n" ); } system( "pause" ); } |