Rule for amount of facts - prolog

Question

For a school project I want to make a planning maker in Prolog. I want Prolog to make a planning for every day. On each day a certain amount of employees are needed and employees can only work on certain days. I want Prolog to make a planning that plans the right amount of people on each day. For that I wrote the following code:
workingday_employeesneeded(monday, 2).
workingday_employeesneeded(tuesday, 1).
workingday_employeesneeded(wednesday, 2).
employee_availability(tom, monday).
employee_availability(thomas, monday).
employee_availability(timme, monday).
employee_availability(timo, monday).
employee_availability(tom, tuesday).
planning(Employee, Day) :-
workingday_employeesneeded(Day, Amount),
employee_availability(Employee, Day).
planning(Employee, Day) :-
aggregate_all(count, planning(Employee, Day), Count),
workingday_employeesneeded(Day, Amount),
Count <= Amount.
However, I can't get Prolog to give me the right result, as I query the following Prolog gives me all the options, not regarding the amount of employees needed.
?- planning(X, Y).
X = tom,
Y = monday ;
X = thomas,
Y = monday ;
X = timme,
Y = monday ;
X = timo,
Y = monday ;
X = tom,
Y = tuesday ;
false.
Can you guy's see what I'm doing wrong? Thanks in advance!
EDIT:
I thought it might be handy to make a list of employees for each day in the planning. So I edited the code to the following (also fixing some syntax errors pointed out in the comments);
planning_on_day(Day, Employees) :-
workingday_employeesneeded(Day, Amount),
findall(E, employee_availability(E, Day), Employees),
length(Employees, Amount).
The following problem still exists; if there are more employees available than needed the program does not print the planning for that day instead of only picking the first N employees.
Do you guys have suggestions to fix that problem?

Answer 1

Simply your predicate fails because first you use findall/3 then you constraint the length of the list. For instance, for monday there are 4 employees available, you find all of them with findall/3 and store into Employees. Then you check the lenght of the list and it fails. To solve it you need find all the available employees and then find a subset of the list with desired length. So your code will be:
subset([], []).
subset([E|Tail], [E|NTail]):-
subset(Tail, NTail).
subset([_|Tail], NTail):-
subset(Tail, NTail).
planning_on_day(Day, Employees) :-
workingday_employeesneeded(Day, Amount),
findall(E, employee_availability(E, Day), E),
length(Employees,Amount),
subset(E,Employees).
?- planning_on_day(monday,P).
P = [tom, thomas]
P = [tom, timme]
P = [tom, timo]
P = [thomas, timme]
P = [thomas, timo]
P = [timme, timo]
false
?- planning_on_day(tuesday,P).
P = [tom]
false
?- planning_on_day(wednesday,P).
false
Then, if you want find the plan of the week you can add:
isDifferent(_, []).
isDifferent(X, [H | T]) :-
X \= H,
isDifferent(X, T).
allDifferent([]).
allDifferent([H | T]) :-
isDifferent(H, T),
allDifferent(T).
solve([],Plan,Plan):-
flatten(Plan,P),
allDifferent(P).
solve([Day|T],LT,Plan):-
workingday_employeesneeded(Day, Amount),
planning_on_day(Day,PlanD),
length(A,Amount),
subset(PlanD,A),
append(LT,[PlanD],LT1),
solve(T,LT1,Plan).
?- solve([monday,tuesday],[],L).
L = [[thomas, timme], [tom]]
L = [[thomas, timo], [tom]]
L = [[timme, timo], [tom]]