السلام عليكم ورحمة الله وبركاته..
أرجو مساعدتي عندي برنامج بالذكاء الإصناعي يبحث مشكلة حل 8-puzzle
باستخدام طريقة بحث بـA*
ولكن عندي إشكالية في فهم بعض التعليمات ''' فهل أجد عندكم استعداد لمساعدتي
/*
Solution of the 8 puzzle.
Uses A-Star algorithm.
*/
#include <stdio.h>
#include <string.h>
#include <iostream.h>
#include <math.h>
//#include <stdlib.h>
#define DOWN 0
#define UP 1
#define LEFT 2
#define RIGHT 3
#define H2
struct elementstruct
{
int block[9];
char* str;
int pathcost;
int valid;
int totalcost;
elementstruct* next;
};
int heur(int block[]);
void prepend(elementstruct* newnode, elementstruct* oldnode, int operator1);
int goal(int* block);
int notonqueue(int block[]);
elementstruct* bestnodefromqueue();
void print_block(int* block);
int apply (int* newstate, int* oldstate, int op);
elementstruct* newelement();
int op(char);
char to_char(int i);
char rep[] = "dulr";
int notvalid1[4] = { 6, 0, 0, 2 };
int notvalid2[4] = { 7, 1, 3, 5 };
int notvalid3[4] = { 8, 2, 6, 8 };
int applyparam[4] = { +3, -3, -1, +1 };
int goal_block[9] = { 0, 1, 2, 3, 4, 5, 6, 7, 8}; //8 indicates no tile
int maxdepth;
elementstruct* top;
int main()
{
int block[9];
printf("\nThe Eight Puzzle!\n");
printf("\nPlease Enter the initial state of the game \n"
" [Represent tiles with numbers 1 to 8, and the blank space as 'x'.\n"
" Start writing them from left to right for each row. Start from the topmost row to the bottommost row.\n"
" Your final string will look similar to this '1 4 2 3 x 6 7 8 5'.\n"
" Do not forget the spaces in between the characters]\n");
int i = 0;
while(i<9)
{
char chr;
chr = fgetc(stdin);
if (chr==32) continue;
if (chr=='x') block = 8;
else if (chr >= '1' && chr <= '9') block = chr - '1';
else { printf("Invalid Input. Example of valid input...2 1 3 4 7 5 6 8 x.", chr); return 1; }
i++;
}
fgetc(stdin); //flush out the end of line character
printf("\n Now Enter the Goal State in a similar way. (Typical. 1 2 3 4 5 6 7 8 x): ");
i = 0;
while(i<9)
{
char chr;
chr = fgetc(stdin);
if (chr==32) continue;
if (chr=='x') goal_block = 8;
else if (chr >= '1' && chr <= '9') goal_block = chr - '1';
else { printf("chr=%d. Invalid Input. Example of valid input...2 1 3 4 7 5 6 8 x.",(int) chr); return 1; }
i++;
}
printf("Enter the maximum depth you want to search (<25 is solved quickly): ");
scanf("%d", &maxdepth);
printf("\nWorking...");
top = newelement();
for(i=0; i<9; i++)
top->block = block;
top->totalcost = heur(block);
elementstruct* newnode = newelement();
while (1)
{
elementstruct* node = bestnodefromqueue();
if (node == NULL) {
printf("done!\n");
printf("There is no solution to this of less than %d depth.\n", maxdepth);
printf("Try increasing the depth by 5.\n");
printf("If there is no solution within 35-40 depth, the pattern is usually unsolvable.\n\n");
break;
}
else if (goal(node->block)) {
char chr[15];
printf("done. \nFound the solution of least number of steps (%d).", node->pathcost);
printf("\nWant a graphical display of each step? (Y/N)?");
scanf("%s", chr);
if(chr[0] =='n' || chr[0]=='N') {
printf("\n (Move Blank u=up, d=down, l=left, r=right)\n");
printf(node->str);
printf("\n");
break;
}
int block2[9];
for (i=0; i<node->pathcost; i++)
{
print_block(block);
apply(block2, block, op(node->str));
for(int j=0; j<=8; j++)
block[j] = block2[j];
}
print_block(block);
printf("\nGraphical Display Complete.\nThe steps taken were: (Move blank u=up, d=down, l-left, r=right)\n");
printf(node->str);
printf("\n");
break;
}
if (node->totalcost > maxdepth) continue;
for(i=0; i<=3; i++) {
if (apply(newnode->block, node->block, i) == -1)
continue;
if (notonqueue(newnode->block)) {
prepend(newnode, node, i);
newnode = newelement();
if (newnode==NULL) { printf ("ERROR!! insufficient memory!! Try decreasing depth!"); return 1; }
}
}
}
return 0;
}
int heur(int* block)
{
#ifdef H2
int to_return = 0;
for(int i=0; i<9; i++)
{
to_return += abs((i/3) - (block/3));
to_return += abs((i%3) - (block%3));
}
return to_return;
#else
int to_return = 0;
for(int i=0; i<9; i++)
{
if (block != i) to_return++;
}
return to_return;
#endif
}
void prepend(elementstruct* newnode, elementstruct* oldnode, int op)
{
newnode->next = top;
top = newnode;
strcpy(newnode->str, oldnode->str);
newnode->str[oldnode->pathcost] = rep[op];
newnode->str[oldnode->pathcost+1] = 0;
newnode->pathcost = oldnode->pathcost+1;
newnode->totalcost = newnode->pathcost + heur(newnode->block);
if (newnode->totalcost < oldnode->totalcost) newnode->totalcost = oldnode->totalcost;
}
int goal(int* block)
{
int* g_block = goal_block;
for(int i=0; i<9; i++)
if ((*(block++))!=(*(g_block++)))
return 0;
return 1;
}
int notonqueue(int* block)
{
int i;
elementstruct* t = top;
while (t!=NULL)
{
for(i=0; i<9; i++)
if (t->block != block) break;
if (i==9) return 0;
t = t->next;
}
return 1;
}
elementstruct* bestnodefromqueue()
{
elementstruct* t = top;
int min_totalpathcost = 1000;
int totalpathcost;
elementstruct* to_return = NULL;
while (t != NULL)
{
if (t->valid==1 && t->totalcost < min_totalpathcost)
{
min_totalpathcost = t->totalcost;
to_return = t;
}
t = t->next;
}
if (to_return != NULL) to_return->valid = 0;
return to_return;
}
int apply (int* newstate, int* oldstate, int op)
{
int j;
int blank;
for (j=0; j<9; j++)
if (oldstate[j]==8) { blank=j; break; }
if (blank==notvalid1[op] || blank==notvalid2[op] || blank==notvalid3[op])
return -1;
for (j=0; j<9; j++)
newstate[j] = oldstate[j];
newstate[blank] = newstate[blank+applyparam[op]];
newstate[blank+applyparam[op]] = 8;
return 1;
}
elementstruct* newelement()
{
elementstruct* t = new elementstruct;
if (t==NULL) return NULL;
t->valid = 1;
t->str = new char[maxdepth+1];
if (t->str ==NULL) return NULL;
t->str[0] = 0;
t->pathcost = t->totalcost = 0;
t->next = NULL;
return t;
}
void print_block(int* block)
{
printf("\n");
printf ("\n-------");
printf ("\n|%c|%c|%c|", to_char(block[0]), to_char(block[1]), to_char(block[2]));
printf ("\n-------");
printf ("\n|%c|%c|%c|", to_char(block[3]), to_char(block[4]), to_char(block[5]));
printf ("\n-------");
printf ("\n|%c|%c|%c|", to_char(block[6]), to_char(block[7]), to_char(block[8]));
printf ("\n-------");
}
char to_char(int i)
{
if (i>=0 &&i<=7) return i+'1';
else if (i==8) return 'x';
else { printf("ERROR in Program!"); return -1; }
}
int op(char i)
{
switch (i)
{
case 'd': return 0;
case 'u': return 1;
case 'l': return 2;
case 'r': return 3;
default: printf("ERROR!"); return -1;
}
}
لاتخيبون ظني تكفووووووووووووون أبي أحد يشرح
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هذي أول مشاركاتي
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