Question

请帮助优化this正在使用的MiniZinc代码：

任务：会议有6个时段。参加会议的有3位发言者，每位发言者都可以在某些位置使用。每个发言者将呈现预定数量的时段。

目标：制作最早完成发言人的时间表。

示例：发言人A，B＆amp; C.谈话持续时间= [1,2,1]

演讲者可用性：

+---+------+------+------+
|   | Sp.A | Sp.B | Sp.C |
+---+------+------+------+
| 1 |      | Busy |      |
| 2 | Busy | Busy | Busy |
| 3 | Busy | Busy |      |
| 4 |      |      |      |
| 5 |      |      | Busy |
| 6 | Busy | Busy |      |
+---+------+------+------+

链接到工作的MiniZinc代码：http://pastebin.com/raw.php?i=jUTaEDv0

我希望优化的内容：

% ensure allocated slots don't overlap and the allocated slot is free for the speaker
constraint 
    forall(i in 1..num_speakers) (
        ending_slot[i] = starting_slot[i] + app_durations[i] - 1
    ) /\
    forall(i,j in 1..num_speakers where i < j) (
        no_overlap(starting_slot[i], app_durations[i], starting_slot[j], app_durations[j])
    ) /\
    forall(i in 1..num_speakers) (
        forall(j in 1..app_durations[i]) (
            starting_slot[i]+j-1 in speaker_availability[i]
        )
    ) 
;

预期解决方案：

+---+----------+----------+----------+
|   |   Sp.A   |   Sp.B   |   Sp.C   |
+---+----------+----------+----------+
| 1 | SELECTED | Busy     |          |
| 2 | Busy     | Busy     | Busy     |
| 3 | Busy     | Busy     | SELECTED |
| 4 |          | SELECTED |          |
| 5 |          | SELECTED | Busy     |
| 6 | Busy     | Busy     |          |
+---+----------+----------+----------+

Answer 1

我是hakank（原始模型的作者）。如果我理解正确，你现在的问题是如何提供表格以获得最佳解决方案，而不是真正找到解决方案本身（我测试的所有FlatZinc求解器都能立即解决）。

创建表格的一种方法是有一个帮助矩阵（“m”），如果选择了一个扬声器（1），忙（-1）或不可用（0），它包含信息：

array[1..num_slots, 1..num_speakers] of var -1..1: m;

然后必须在矩阵和其他决策变量（“starting_slot”和“ending_slot”）中连接信息：

% connect to matrix m
constraint
   forall(t in 1..num_slots) (
      forall(s in 1..num_speakers) (
         (not(t in speaker_availability[s]) <-> m[t,s] = -1) 
          /\
          ((t >= starting_slot[s] /\ t <= ending_slot[s]) <-> m[t,s] = 1)
     )
 )

然后矩阵“m”可以这样打印：

% ...
++
[ 
   if s = 1 then "\n" else " " endif ++
   if fix(m[t,s]) = -1 then 
      "Busy    " 
   elseif fix(m[t,s]) =  1 then 
      "SELECTED" 
   else
      "        "
   endif
 | t in 1..num_slots, s in 1..num_speakers

] ;

与往常一样，有不止一种方法可以做到这一点，但我对此已经解决了，因为它非常直接。

这是完整的模型： http://www.hakank.org/minizinc/scheduling_speakers_optimize.mzn

更新：添加模型的输出：

Starting:  [1, 4, 3]
Durations: [1, 2, 1]
Ends:      [1, 5, 3]
z:         5

SELECTED Busy             
Busy     Busy     Busy    
Busy     Busy     SELECTED
         SELECTED         
         SELECTED Busy    
Busy     Busy             
----------
==========

更新2：另一种方法是使用累积/ 4而不是no_overlap / 4，这应该更有效，即

constraint 
    forall(i in 1..num_speakers) (
    ending_slot[i] = starting_slot[i] + app_durations[i] - 1
    ) 
    % /\ % use cumulative instead (see below)
    % forall(i,j in 1..num_speakers where i < j) (
    %   no_overlap(starting_slot[i], app_durations[i], starting_slot[j], app_durations[j])
    % ) 
    /\
    forall(i in 1..num_speakers) (
    forall(j in 1..app_durations[i]) (
        starting_slot[i]+j-1 in speaker_availability[i]
           )
    ) 

    /\ cumulative(starting_slot, app_durations, [1 | i in 1..num_speakers], 1)
;

这是更改后的版本（给出相同的结果） http://www.hakank.org/minizinc/scheduling_speakers_optimize2.mzn （我也跳过了表示矩阵“m”并在输出部分做了所有演示。）

对于这个简单的问题实例，没有明显的区别，但对于较大的实例，这应该更快。（对于较大的实例，可能需要测试不同的搜索启发式而不是“求解最小化z”。）

Answer 2

正如我对您之前的问题Constraint Programming: Scheduling speakers in shortest time所评论的那样，累积约束适用于此。我没有Minizinc代码方便，但ECLiPSe（http://eclipseclp.org）中有模型：

:- lib(ic).
:- lib(ic_edge_finder).
:- lib(branch_and_bound).

solve(JobStarts, Cost) :-
    AllUnavStarts = [[2,6],[1,6],[2,5]],
    AllUnavDurs   = [[2,1],[3,1],[1,1]],
    AllUnavRess   = [[1,1],[1,1],[1,1]],
    JobDurs = [1,2,1],
    Ressources = [1,1,1],
    length(JobStarts, 3),
    JobStarts :: 1..9,

    % the jobs must not overlap with each other
    cumulative(JobStarts, JobDurs, Ressources, 1),

    % for each speaker, no overlap of job and unavailable periods
    (
        foreach(JobStart,JobStarts),
        foreach(JobDur,JobDurs),
        foreach(UnavStarts,AllUnavStarts),
        foreach(UnavDurs,AllUnavDurs),
        foreach(UnavRess,AllUnavRess)
    do
        cumulative([JobStart|UnavStarts], [JobDur|UnavDurs], [1|UnavRess], 1)
    ),

    % Cost is the maximum end date
    ( foreach(S,JobStarts), foreach(D,JobDurs), foreach(S+D,JobEnds) do true ),
    Cost #= max(JobEnds),

    minimize(search(JobStarts,0,smallest,indomain,complete,[]), Cost).

优化工作调度MiniZinc代码 - 约束编程

2 个答案: