八数码问题的过程表示及其实现

过程式知识表示是将有关某一问题领域的知识， 连同如何使用这些知识的方法，均隐式的表达为 一个求解问题的过程，每个过程是一段程序，完 成对具体情况的处理。过程式不像陈述式那样具有固定的形式，如何描述知识完 全取决于具体问题。

例：八数码问题   人工智能及其应用

c语言实现：

#include<stdio.h>//空格按箭头方向移动，回到起始位置void zero_back(int start[], int array[], int *zero_place, int length){int i;int temp;for (i = 0; i < length-1; i++){temp = start[*zero_place];start[*zero_place] = start[array[i + 1]];start[array[i + 1]] = temp;*zero_place = array[i + 1];}temp = start[*zero_place];start[*zero_place] = start[array[0]];start[array[0]] = temp;*zero_place = array[0];return;}//_flag 为1则按abcdefgh方式移动；_flag不为1则使1和0不在位置上void zero_move(int start[], int array[], int *zero_place, int length, int _flag){int i;int j = 0;//int length;int temp;//length = sizeof(array)/sizeof(array[0]);for (i = 0; i < length; i++){if (array[i] == *zero_place){j = i;break;}}if (j < length - 1){temp = start[*zero_place];start[*zero_place] = start[array[j + 1]];start[array[j + 1]] = temp;*zero_place = array[j + 1];}else{if (_flag == 1){temp = start[*zero_place];start[*zero_place] = start[array[0]];start[array[0]] = temp;*zero_place = array[0];}else{temp = start[*zero_place];start[*zero_place] = start[array[5]];start[array[5]] = temp;*zero_place = array[5];}}return;}void _input(int start[]){int i;printf("请给定初始状态：\n");printf("3*3矩阵形式输入,空格用0代替\n");for (i = 0; i < 9; i++){scanf("%d", &start[i]);if ((i + 1) % 3 == 0){printf("\n");}}}void _print(int start[]){int i;for (i = 0; i < 9; i++){                                     printf("%d\t", start[i]);if ((i + 1) % 3 == 0){printf("\n");}}}void main(){int i;int switch_case;int start[9];int zero_place;int array_m[9] = { 8,7,6,3,0,1,2,5,4 };int array_a[8] = { 0,3,6,7,8,5,4,1 };int array_b[8] = { 3,6,7,8,5,2,1,4 };int array_c[6] = { 3,6,7,8,5,4 };int array_d[8] = { 3,0,1,4,5,2,1,0 };int array_e[6] = { 3,6,7,8,5,4 };int array_f[4] = { 3,6,7,4 };int array_g[13] = { 3,0,1,2,5,4,7,8,5,2,1,0,3 };int array_h[4] = { 3,6,7,4 };_input(start);for (i = 0; i < 9; i++){if (start[i] == 0){zero_place = i;break;}}switch_case = 1;while (switch_case < 10){switch (switch_case){case 1:for (i = 0; i < 50; i++){if (start[2] != 0 && start[2] != 1){switch_case = 2;break;}else{zero_move(start, array_m, &zero_place, 9, 0);}}printf("step 1\n");_print(start);break;case 2:for (i = 0; i < 50; i++){if (start[0] == 1){switch_case = 3;break;}else{zero_move(start, array_a, &zero_place, 8, 1);}}printf("step 2\n");_print(start);break;case 3:for (i = 0; i < 50; i++){if (start[1] == 2){if (start[2] == 3){switch_case = 6;}else{switch_case = 4;}break;}else{zero_move(start, array_b, &zero_place, 8, 1);}}printf("step 3\n");_print(start);break;case 4:for (i = 0; i < 50; i++){if (start[4] == 3){switch_case = 5;break;}else{zero_move(start, array_c, &zero_place, 6, 1);}}printf("step 4\n");_print(start);break;case 5:zero_back(start, array_d, &zero_place, 8);switch_case = 6;printf("step 5\n");_print(start);break;case 6:for (i = 0; i < 50; i++){if (start[5] == 4){if (start[8] == 5){switch_case = 9;}else{switch_case = 7;}break;}else{zero_move(start, array_e, &zero_place, 6, 1);}}printf("step 6\n");_print(start);break;case 7:for (i = 0; i < 100; i++){if (start[4] == 5){switch_case = 8;break;}else{zero_move(start, array_f, &zero_place, 4, 1);}}printf("step 7\n");_print(start);break;case 8:zero_back(start, array_g, &zero_place, 13);switch_case = 9;printf("step 8\n");_print(start);break;case 9:for (i = 0; i < 50; i++){if (start[7] == 6&&start[4]==0){if (start[3] == 8 && start[6] == 7){printf("问题有解，达到目标状态\n");}else{printf("问题无解，达不到目标状态\n");}switch_case = 10;break;}else{zero_move(start, array_h, &zero_place, 4, 1);}}printf("step 9\n");_print(start);break;default:break;}}}/**************************测试情况：201 465 378 有解；          **************************/

运行结果如下：

请给定初始状态：
3*3矩阵形式输入,空格用0代替
2 0 1

4 6 5

3 7 8

step 1
2       1       5
4       6       0
3       7       8
step 2
1       6       5
0       8       7
2       4       3
step 3
1       2       8
4       0       6
3       7       5
step 4
1       2       8
0       3       4
7       5       6
step 5
1       2       3
0       4       8
7       5       6
step 6
1       2       3
7       0       4
5       6       8
step 7
1       2       3
0       5       4
6       7       8
step 8
1       2       3
0       7       4
6       8       5
问题有解，达到目标状态
step 9
1       2       3
8       0       4
7       6       5
请按任意键继续. . .