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How to find equal elements in two lists – Artificial Intelligence

I have the following set of facts:

lc('Dr. Smith', 'Algorithms').
lc('Dr. Jones', 'Models & Analysis').
lc('Dr. Smith', 'Operating Systems').
lc('Dr. Jones', 'Artificial Intelligence').
lc('Dr. Smith', 'Models of Computation').
lc('Dr. Smith', 'Discrete Math').
lc('Dr. Jones', 'Information Retrieval').
lc('Dr. Jones', 'Computer Vision').

ctl('Algorithms', 'MWF, 9:00 - 11:00 a.m.', 'McB 209').
ctl('Models & Analysis', 'MWF, 9:00 - 11:00 a.m.', 'McB 211').
ctl('Operating Systems', 'TTH, 9:00 - 11:00 a.m.', 'McB 306').
ctl('Artificial Intelligence', 'TTH, 3:00 - 5:00 p.m.', 'McB 311').
ctl('Models of Computation', 'TTH, 11:00 - 1:00 p.m.', 'McB 204').
ctl('Discrete Math', 'TTH, 3:00 - 5:00 p.m.', 'McB 204').
ctl('Information Retrieval', 'MWF, 3:00 - 5:00 p.m.', 'McB 205').
ctl('Computer Vision', 'MWF, 1:00 - 3:00 p.m.', 'NEB 2182').

And I’ve been able to write the following, that gives me the schedule for a lecturer:

schedule(Lecturer, X, Y, Z) :- ctl(X, Y, Z), lc(Lecturer, X).

But now I’m trying to write a rule to find when do Dr. Jones and Dr. Smith teach at the same time. Any ideas?

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2

Answers


  1. 0

    _something like that:

    samehours(Prof1,Prof2,H) :-
   Prof1 = Prof2,
   lc(Prof1, Sub1), lc(Prof2,Sub2),
   ctl(Sub1,H,_), ctl(Sub2, H, _).

    result: 

    H = 'MWF, 9:00 - 11:00 a.m.' ;
H = 'TTH, 3:00 - 5:00 p.m.' ;
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  2. 0

    You can use your own function to extend this utility, by doing:

    same_time(X, Y, R) :-
    schedule(X, _, B, _),
    schedule(Y, _, B, _),
    X = Y,
    R = B.

    which is simply telling you what teachers X and Y teaches both at the same time B, as returned in the list R.

    And, if you only want the times when more than one teacher co-occur in the teaching schedule, you can simply remove the initial vars X and Y, and even replace R with the final result itself, just as follows:

    same_time(B) :-
    schedule(X, _, B, _),
    schedule(Y, _, B, _),
    X = Y.

    The return will still be the times B when more than one teacher is giving classes.

    For the conflicts, you may keep the part of your statement that is working, and simply extend it to:

    same_teachertime(X) :-
    schedule(X, V1, B, _),
    schedule(X, V2, B, _),
    V1 = V2.

schedulingconflict(X, Y, X):- ctl(X, A, B), ctl(Y, A, B), X = Y.
schedulingconflict(_, _, X):- same_teachertime(X).

    This considers that being the same teacher, in the same time, teaching different subjects, is a conflict, since the room conflicts are being solved with the statement you wrote.

    Regards!

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