What loop condition is needed to exit the maze?

Question

C++ / C#

Hello, I wrote a program to find the exit from the labyrinth (two-dimensional array), but I just can’t come up with a condition for the program to end when it hits the exit coordinates (y=9, x=6).

Program code

#include "iostream"
#include "windows.h"

using namespace std;

int main()
{
  int labirint[10][10] =   //Сам лабиринт
  { { 1,1,1,1,1,1,1,1,1,1 },
  { 1,0,1,1,0,1,0,0,0,1 },
  { 1,0,0,0,0,1,0,1,0,1 },
  { 1,1,1,1,0,1,0,1,0,1 },
  { 1,0,0,0,0,1,0,1,0,1 },
  { 1,1,0,1,1,1,0,1,0,1 },
  { 1,1,0,0,0,0,0,1,0,1 },
  { 1,1,1,1,1,1,1,1,0,1 },
  { 1,0,0,0,0,0,0,0,0,1 },
  { 1,1,1,1,1,1,0,1,1,1 } };
  int direction = 1; //Направление робота
  int y = 0, x = 1; //Координаты робота

  while (1) {   //Цикл, из которого надо выйти при нахождении выхода

    Sleep(1000);
    system("cls");

    labirint[y][x] = 2;   //Указание робота

    for (int i = 0; i < 10; i++) {   //Вывод лабиринта в консоли
      for (int a = 0; a < 10; a++) {
        if (labirint[i][a] == 1) {
          printf("%c", char(219));
        }
        else if (labirint[i][a] == 2) {
          printf("%c", char(253));
        }
        else {
          cout << " ";
        }	
      }
      cout << endl;
    }

    labirint[y][x] = 0; //Очистка хвоста робота
    cout << y << "  " << x << endl;
    switch (direction) {
    case 1: //Вверх
      while (labirint[y][x - 1] == 1 && labirint[y + 1][x] == 0) {
        y++;
      }
      if (labirint[y][x - 1] == 0) {
        direction = 3;
        x--;
      }
      else {
        direction = 2;
      }
      break;
    case 2: //Налево
      while (labirint[y + 1][x] == 1 && labirint[y][x + 1] == 0) {
        x++;
      }
      if (labirint[y + 1][x] == 0) {
        direction = 1;
        y++;
      }
      else {
        direction = 4;
      }
      break;
    case 3: //Направо
      while (labirint[y - 1][x] == 1 && labirint[y][x - 1] == 0) {
        x--;
      }
      if (labirint[y - 1][x] == 0) {
        direction = 4;
        y--;
      }
      else {
        direction = 2;
      }
      break;
    case 4: //Вниз
      while (labirint[y][x + 1] == 1 && labirint[y - 1][x] == 0) {
        y--;
      }
      if (labirint[y][x + 1] == 0) {
        direction = 2;
        x++;
      }
      else {
        direction = 3;
      }
    }
  } 
  system("pause");
    return 0;
}

The algorithm for finding the exit is built on the basis of the rule of one side.
What is the condition to exit the loop?

Answer 1

You must be at a point on the perimeter - this is a set of points with coordinates {x:any,y:0}, {x:any,y:9},{x:0,y:any}, {x:9,y: any}.