Logical programming. (20 points)

i. Write, execute, and test a Prolog program to simulate a mouse traversing a maze.

Use a predicate d/2 to specify the doors between rooms of the maze, e.g., d(a,b) means that there is a door linking rooms a and b.

Assume that the mouse always starts in the room named enter, and that the room named exit has the only door out of the maze,

although room exit may also have doors leading to other rooms. Show all solutions for the mouse running the maze in each test case.

Use at least the following 5 test cases:

(A) d(enter,b). d(b,c). d(b,e). d(d,e). d(c,d). d(e,f). d(e,exit).

(B) d(enter,e). d(b,c). d(b,e). d(d,e). d(c,d). d(e,f). d(f,exit).

(C)d(enter.e). d(b,c). d(d,e). d(c,d). d(e,f). d(g,e).

(D)d(enter,e). d(b,c). d(d,e). d(c,d). d(e,f). d(c.exit).

(E) d(enter,a). d(a,exit).

ii.Revise your program from part (i) above as follows.

Then run the same 5 test cases from part (i), as well as at least the following additional 5 test cases:

(F) d(enter,b). d(b,e). d(b,c). d(c,d). d(d,e). d(e,exit). cheese(c).

G) d(enter,a). d(a,b). d(b,c). d(c,d). d(d,e). d(e,exit). cheese(a). cheese(d).

(H) d(enter,a). d(a,b). d(b,c). d(b,exit). d(a,exit). d(c,d).

(I) d(enter,e). d(e,c). d(d,e). d(c,d). d(e,f). d(f,exit). cheese(c).

(J) d(enter,a). d(a,b). d(b,c). cheese(b). d(b,exit). d(a,d). d(d,e). d(e,b).

Some mice do not reach the exit because they become weak with hunger after visiting more than 3 rooms.

Add a predicate cheese/1 that specifies which rooms have food (i.e., cheese), e.g., cheese(d) is true if and only if room d contains cheese.

Change your program so that a mouse eats some cheese at least once every 3 rooms traversed.

In other words, the mouse eats a little bit of cheese in any room that contains cheese,

but if the mouse visits more than 3 rooms without eating cheese then that mouse must backtrack and try another path through the maze.