Algorithm 2: Get the distance to walk with BFS

Input: $room$, $width$, $height$, $x_{person}$, $y_{person}$  Output: Exit’s x-position ($row$), exit’s y-position ($col$), distance the person must walk - # of        boxes ($distance$), An array with each of the positions (x, y) where the person walks ($path$)  $source$ = QItem($y_{p}erson$ -1, $x_{p}erson$ -1,0);    $visited$ = np.zeros((int($height$),int($width$)), dtype=int)   repeat    repeat    if $room$[i][j] == b’0’ then      $visited$[i][j] = 1    else      $visited$[i][j] = 0    end    until range(0, int($width$); until range(0, int($height$);  q = queue.Queue(); q.put($source$); $visited$[$source$.row][$source$.col] = 1   repeat   p = q.get()    if room[$p.row$][$p.col$] == b’e’ then     return $p.row$ +1, $p.col$ +1, $dist$, bfsPathExit($room$, $width$, $height$, ($x_{person}$ -1,$y_{person}$ -1))   end   The function bfsPathExit is calculated with the Algorithm 2.    if p.row - 1 >= 0 and $visited$[p.row - 1][p.col] == 0 then     q.put(QItem(p.row - 1, p.col, p.dist + 1))       $visited$[p.row - 1][p.col] = 1    end    if p.row + 1 < int($height$) and $visited$[p.row + 1][p.col] == 0 then$\phantom{(}$     q.put(QItem(p.row + 1, p.col, p.dist + 1))       $visited$[p.row + 1][p.col] = 1   end   if p.col - 1 >= 0 and $visited$[p.row][p.col - 1] == 0 then     q.put(QItem(p.row, p.col - 1, p.dist + 1))       $visited$[p.row][p.col - 1] = 1   end   if p.col + 1 < int($width$) and $visited$[p.row][p.col + 1] == 0 then     q.put(QItem(p.row,p.col + 1, p.dist + 1))       $visited$[p.row][p.col + 1] = 1   end  until q.empty() == False;  return -1