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
//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
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. . .
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. . .