Time 5, 3, 8, 2, 7, 3, 9, 3, 3, 5, 7
You have to schedule players (which uses Time) to three different showers. Get the best solution. So far, my solution is:
use_module(library(clpfd)).
shower(S, E, D, Done) :-
D = [5, 3, 8, 2, 7, 3, 9, 3, 3, 5, 7],
R = [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
length(D, N),
length(S, N),
length(E, N),
domain(S, 0, 100),
domain(E, 0, 100),
Done in 0..100,
ready(E, Done),
tasks(S, D, E, R, Tasks),
cumulative(Tasks, [limit(3)]),
labeling([minimize(Done)], [Done|S]).
tasks([], [], [], [], []).
tasks([S|Ss], [D|Ds], [E|Es], [R|Rs], [T|Ts]) :-
T = task(S, D, E, R, 0),
tasks(Ss, Ds, Es, Rs, Ts).
ready([], _).
ready([E|Es], Done) :-
Done #>= E,
ready(Es, Done).
If I run the program:
?- shower(S,D).
it will print:
D = 19,
S = [0,0,0,3,5,5,8,8,11,14,12]
Done is the total time (maximum time), S is the time of start for each of the tasks, E is the ending time for each of the tasks while D is the duration of the tasks.
So far, so good.
But now, I am struggling with the other things. Precisely:
1) How can I also print which player is using which shower?
2) How can I limit one shower to be able to run a maximum of four players?
I am a total newbie to Prolog, and it is quite likely that this might be the last program I am writing on it. I managed so far to do this, but I need to finish this (you guess it, it is an assignment). It is the very last thing I have to do on this course, and I haven't ever done any constraint logic programming before.
Thanks for reading and eventually replying on this!
If you use the cumulatives/[2,3] constraint instead of the cumulative/1 constraint, then you will get the assigned machine for 'free'.
By use of cumulatives each single machine can be given individual resource capacity.
This shows your problem solved by use of cumulatives:
:- use_module(library(clpfd)).
:- use_module(library(lists)).
go( Ss, Es, Ms) :-
Ss = [S1, S2, S3, S4,S5,S6,S7], %Starttimes
Es = [E1, E2, E3, E4,E5,E6,E7], %Endtimeds
Ms = [M1, M2, M3, M4,M5,M6,M7], %MachineIds
domain(Ss, 0, 10),
domain(Es, 0, 10),
domain(Ms, 1, 4),
Tasks = [
task( S1, 3, E1, 1, M1 ),
task( S2, 4, E2, 1, M2 ),
task( S3, 1, E3, 1, M3 ),
task( S4, 1, E4, 1, M4 ),
task( S5, 1, E5, 1, M5 ),
task( S6, 1, E6, 1, M6 ),
task( S7, 4, E7, 1, M7 )
],
%All machines has resource capacity = 1
Machines = [
machine( 1, 1 ),
machine( 2, 1 ),
machine( 3, 1 ),
machine( 4, 1 )
],
cumulatives(Tasks, Machines, [bound(upper)] ),
maximum( MaxEndTime, Es ),
%The variables to lable:
append([Ms, Ss ], Vars),
labeling( [minimize(MaxEndTime)], Vars).
And when launched you get:
| ?- go( Ss, Es, Ms) .
Ss = [0,0,3,0,1,2,0],
Es = [3,4,4,1,2,3,4],
Ms = [1,2,1,3,3,3,4] ?
As you can see: task 1 is assigned to machine 1 with start time 0 task 2 is assigned to machine 2 with start time 0 task 3 is assigned to machine 1 with start time 3 . .