Prolog: Finding all solutions

Question

The title may look like it\'s a dime a dozen, but it\'s not. The object of this program is to take these classes (needs)

needs([[ece2090,1,m,13,16],
[ece3520,1,t         


        

Answer 1

This is probably an assignment, so I'll stop short of giving away a complete solution. But I would recommend structuring your information differently, to be more typical of how it's done in Prolog:

need(ece2090, 1, [m, 13, 16]).
need(ece3520, 1, [tu, 11, 14]).
need(ece4420, 1, [w, 13, 16]).

qualified_for(joel, ece2090).
qualified_for(joel, ece2010).
qualified_for(joel, ece3520).
qualified_for(joel, ece4420).
qualified_for(sam, ece2090).
qualified_for(sam, ece4420).
qualified_for(pete, ece3520).

busy(joel, [m, 13, 16]).
% busy(sam, []).  % absence of this fact means sam isn't busy
busy(pete, [w, 13, 16]).
busy(pete, [f, 9, 12]).

Here, rather than having embedded lists in facts, each fact stands on its own. Each distinct functor is like one table in a database that you can query, and the different functors are different tables that are interrelated based upon some argument.

So if you want to know what joel is qualified for:

| ?- qualified_for(joel, C).

C = ece2090 ? ;
C = ece2010 ? ;
C = ece3520 ? ;
C = ece4420

yes
| ?-

If you need a list of all the qualified TAs:

| ?- setof(TA, C^qualified_for(TA, C), AllTAs).

AllTAs = [joel,pete,sam]

yes
| ?-

If you want to see what needed classes joel is available for and qualified to teach:

| ?-  need(Class, I, Schedule), qualified_for(joel, Class), \+ busy(joel, Schedule).

Class = ece3520
I = 1
Schedule = [tu,11,14] ? a

Class = ece4420
I = 1
Schedule = [w,13,16]

(1 ms) yes
| ?-

The assumption above is that "schedule" blocks are all described in the same way, [day, start, end] as you show, so they can simply be matched as entire 3-element lists. With a little extra effort, this can be more generalized to check days and time intervals in more detail. But that doesn't appear to be a requirement for you. The above are good building blocks to help solve your problem. It only requires a few lines of code (I used only 12 lines, plus the above facts). :)