I'm currently working on an answer set program to create a timetable for a school.
The rule base I use looks similar to this:
teacher(a). teacher(b). teacher(c). teacher(d). teacher(e). teacher(f).teacher(g).teacher(h).teacher(i).teacher(j).teacher(k).teacher().teacher(m).teacher(n).teacher(o).teacher(p).teacher(q).teacher(r).teache(s).teacher(t).teacher(u).
teaches(a,info). teaches(a,math). teaches(b,bio). teaches(b,nawi). teaches(c,ge). teaches(c,gewi). teaches(d,ge). teaches(d,grw). teaches(e,de). teaches(e,mu). teaches(f,de). teaches(f,ku). teaches(g,geo). teaches(g,eth). teaches(h,reli). teaches(h,spo). teaches(i,reli). teaches(i,ku). teaches(j,math). teaces(j,chem). teaches(k,math). teaches(k,chem). teaches(l,deu). teaches(l,grw). teaches(m,eng). teaches(m,mu). teachs(n,math). teaches(n,geo). teaches(o,spo). teaches(o,fremd). teaches(p,eng). teaches(p,fremd). teaches(q,deu). teaches(q,fremd). teaches(r,deu). teaches(r,eng). teaches(s,eng). teaches(s,spo). teaches(t,te). teaches(t,eng). teaches(u,bio). teaches(u,phy).
subject(X) :- teaches(_,X).
class(5,a). class(5,b). class(6,a). class(6,b). class(7,a). class(7,b). class(8,a). class(8,b). class(9,a). class(9,b). class(10,a). class(10,b).
%classes per week (for class 5 only at the moment)
classperweek(5,de,5). classperweek(5,info,0). classperweek(5,eng,5). classpereek(5,fremd,0). classperweek(5,math,4). classperweek(5,bio,2). classperweek(5,chem,0). classperweek(5,phy,0). classperweek(5,ge,1). classperweek(5,grw,0). cassperweek(5,geo,2). classperweek(5,spo,3). classperweek(5,eth,2). classperwek(5,ku,2). classperweek(5,mu,2). classperweek(5,tec,0). classperweek(5,nawi,0) .classperweek(5,gewi,0). classperweek(5,reli,2).
room(1..21).
%for monday to friday
weekday(1..5).
%for lesson 1 to 9
slot(1..9).
In order to creat a timetable I wanted to create every possible combination of all predicats I'm using and then filter all wrong answers.
This is how I created a timetable:
{timetable(W,S,T,A,B,J,R):class(A,B),teacher(T),subject(J),room(R)} :- weekday(W), slot(S).
Up to this point everything works, except that this solution is probably relatively inefficient.
To filter that no class uses the same room at the same time I formulated the following constraint.
:- timetable(A,B,C,D,E,F,G), timetable(H,I,J,K,L,M,N), A=H, B=I, G=N, class(D,E)!=class(K,L).
It looks like this makes to problem so big that the grounding fails, because I get the following error message
clingo version 5.4.0
Reading from timetable.asp
Killed
Therefore, I was looking for a way to create different instances of timetable without getting too many "meaningless" answers created by the choiserule.
One possibility I thought of is to use a negation cycle. So you could replace the choiserule
{a;b} with a :- not b. b :- not a. and exclude all cases where rooms are occupied twice.
Unfortunately I do not understand this kind of approach enough to apply it to my problem.
After a lot of trial and error (and online search), I have not found a solution to eliminate the choicerule and at the same time eliminate the duplication of rooms and teachers at the same time.
Therefore I wonder if I can use this approach for my problem or if there is another way to not create many pointless answersets at all.
edit: rule base will work now and updated the hours per lesson for class 5
I think you're looking for something like:
% For each teacher and each timeslot, pick at most one subject which they'll teach and a class and room for them.
{timetable(W,S,T,A,B,J,R):class(A,B),room(R),teaches(T,J)} <= 1 :- weekday(W);slot(S);teacher(T).
% Cardinality constraint enforcing that no room is occupied more than once in the same timeslot on the timetable.
:- #count{uses(T,A,B,J):timetable(W,S,T,A,B,J,R)} > 1; weekday(W); slot(S); room(R).
to replace your two rules.
Note that this way clingo won't generate spurious ground terms for teachers teaching a subject they don't know. Additionally by using a cardinality constraint as opposed to a binary clause, you get a big-O reduction in the grounded size (from O(n^2) in the number of rooms to O(n)).
Btw, you may be missing answers because of typos in the input. I would suggest phrasing it as:
teacher(a;b;c;d;e;f;g;h;i;j;k;l;m;n;o;p;q;r;s;t;u).
teaches(
a,info;
a,math;
b,bio;
b,nawi;
c,ge;
c,gewi;
d,ge;
d,grw;
e,de;
e,mu;
f,de;
f,ku;
g,geo;
g,eth;
h,reli;
h,spo;
i,reli;
i,ku;
j,math;
j,chem;
k,math;
k,chem;
l,deu;
l,grw;
m,eng;
m,mu;
n,math;
n,geo;
o,spo;
o,fremd;
p,eng;
p,fremd;
q,deu;
q,fremd;
r,deu;
r,eng;
s,eng;
s,spo;
t,te;
t,eng;
u,bio;
u,phy
).
subject(X) :- teaches(_,X).
class(
5..10,a;
5..10,b
).
%classes per week (for class 5 only at the moment)
classperweek(
5,de,5;
5,info,0;
5,eng,5;
5,fremd,0;
5,math,4;
5,bio,2;
5,chem,0;
5,phy,0;
5,ge,1;
5,grw,0;
5,geo,2;
5,spo,3;
5,eth,2;
5,ku,2;
5,mu,2;
5,tec,0;
5,nawi,0;
5,gewi,0;
5,reli,2
).
room(1..21).
%for monday to friday
weekday(1..5).
%for lesson 1 to 9
slot(1..9).
From the below code, I was attempting to retrieve 250 observations rather than 177. The gap is due to the fact that the call only considers trading days which is fine to me.
s='SX5E INDEX';
f='LAST_PRICE'
t= datestr(today()-250,'mm/dd/yy');
T= datestr(today(),'mm/dd/yy');
[dt,~]=history(con,s,f,t,T)
However, is there a way of retrieving the last 250 observations from today(), whatever the starting date t is ?
Best
EDIT
#Daniel : Based on your suggestion, and Going forward with while loop, I've ended up with the below way around which is free from any Matlab default calendar setting. Thanks
while l~=p
n=p-l;
t=t-n;
[dt,~]=history(con,s,f,t,T);
l=length(dt);
end
Maybe my idea to use isbusday wasn't well explained in the comments. Here is what I would try:
n=250;
m=n;
while(sum(isbusday(today()-n:today()))<m)
missing=m-sum(isbusday(today()-n:today()));
n=n+missing;
end
Count the number of missing days, add the missing days and check again (in case you added a holiday)
You should end up with n the total number of days you have to query.
(Lacking the toolbox, I was unable to test the code)