$7\mathrm{.}33($Maze Traversal$)$ The grid of hashes $($#$)$ and dots $(.)$ in Fig. $7\mathrm{.}27$ is a two$-$dimensional built$-$

in array representation of a maze. In the two$-$dimensional built$-$in array, the hashes represent the

walls of the maze and the dots represent squares in the possible paths through the maze. Moves can

be made only to a location in the built$-$in array that contains a dot.

There is a simple algorithm for walking through a maze that guarantees finding the exit

$($assuming that there is an exit$).$ If there is not an exit, you'll arrive at the starting location again,

Place your right hand on the wall to your right and begin walking forward. Never remove your

hand from the wall. If the maze turns to the right, you follow the wall to the right. As long as you

do not remove your hand from the wall, eventually you'll arrive at the exit of the maze. There may

be a shorter path than the one you've taken, but you are guaranteed to get out of the maze if you

follow the algorithm.

Do not solve this project with recursion. Instead, follow the right$-$hand rule as described in the text and below.

Program a rat $($your maze runner$)$ to have a direction: up$,$ right, down, and left $($can be coded $0,1,2,3),$ as its forward direction. Based on which of the four directions, it should then try to go right relative to its forward direction. E$.$g$.$ if facing up then step right, if facing right then step down, facing down then left, etc. But in the case of facing up$,$ if you can't go right, then try to go forward. If not, try going left, otherwise step backward $($which you can always do$).$ If you successfully take a step in that direction, change the direction for the next iteration. You may backtrack, of course, as you never become blocked. As you step forward, change the. to $*$; if you backtrack, then change the $*$ to $-.$ Of course, don't go through a wall #$.$ Print the matrix$/$maze at each step. Stop when the rat finds the exit. There is a lot of similar code $(16$ cases$)$ but it is not easy to abstract to a function. so you don't have to try. code in C$++$ please