How to find second last value in Answer Set Programming (ASP Clingo) - answer-set-programming

Question

I'm modelling a university curriculum timetable with Answer Set Programming (Clingo). The requirements are that each lesson must be asssigned to a specific week , day and start /end hours until the total hours are reached.
In a day there are max 8 hours and each lesson is 2 hours minimum. Also the scheduled lessons are until the 10th week.
week(1..10). time(9,11;11,13;14,16;16,18).
So first I generated each assigned lesson slot
TotalHours { assigned(Week,Day,Start,End,Course,Teacher) : day(Day), time(Start,End), week(Week) } TotalHours :-
lesson(Course,Teacher,TotalHours).
After that for other requirements and rules I needed to find the last course assigned in the timetable. I don't know if this is a good way but I was able to find a solution
MaxWeek = #max {Week : assigned(Week,_,_,_,Course,_)},
MaxDay = #max {Day : assigned(MaxWeek,Day,_,_,Course,_)},
MaxStart= #max {Start : assigned(MaxWeek,MaxDay,Start,_,Course,_)},
assigned(_,_,_,_,Course,_).
I'm new to ASP and so far I couldn't find a good way to find the second last lesson of the course assigned in the timetable(Week,Day,Start,End) needed for other schedule requirements.
So given for example
assigned(1,Monday,9,11,History,John) , assigned(1,Tuesday,11,13,Math,Smith), assigned(4,Tuesday,16,18,History,John),assigned(5,Monday,11,13,History,John)
I want to find the second last lesson of History which is
assigned(4,Tuesday,16,18)
Any tips or solutions is really appreciated

Answer 1

Consider having a second assigned predicate (with a different arity) to be something more absolute, like:
assigned((Week-1)*7*24+Day+Start,End,Course,Teacher) :- assigned(Week,Day,Start,End,Course,Teacher).
Now you just have one field that is completely ordered over all assignments, which is total hours since start of the year.
To find the last courses, the last aggregate performs very badly, and can be optimized using chaining constraints:
auxLastCourse(C,T) :- assigned(T,_,C,_).
auxLastCourse(C,T-1) :- auxLastCourse(C,T), T > 0.
lastCourse(C,T) :- auxLastCourse(C,T), not auxLastCourse(C,T+1).
Also note that in your example aren't any courses during the night, so this could be optimized further by not beeing so finegranular and explicitly representing every hour, but maybe only hours that actually occur in a possible schedule. THis would mean replacing the T-1 in the aux code with an explicit next relation over weeks and days and hours. Currently I'm just too lazy to add something like this.
For the second last course you can do exactly the same thing:
auxSecondLastCourse(C,T) :- assigned(T,_,C,_), not lastCourse(C,T).
auxSecondLastCourse(C,T-1) :- auxSecondLastCourse(C,T), T > 0.
secondLastCourse(C,T) :- auxSecondLastCourse(C,T), not auxSecondLastCourse(C,T+1).
PS: Note that I assume that your days are given as numbers 0..6, and not as words as in your example, I'm sure you will figure out how to translate them using another predicate.
PPS: I haven't tested the code, as you did'nt provide an MWE.