julia 中 se 的所有子集

Question

你能帮我吗？我怎样才能做一个可以找到一个集合的所有子集的代码

例如

我想在 Julia 中编写此约束。这是一个子游览约束。但我不知道如何找到 S 集的所有子集。

@constraint(ILRP,
            c7[k in totalK, t in totalH],
            sum(x[i,j,k,t] for i=1:totalS, j=1:totalS)<=size(S)-1);

非常感谢

Answer 1

您可以使用powersetCombinatorics.jl 包中的函数获取它，例如：

julia> using Combinatorics

julia> x = [1:5;]
5-element Array{Int64,1}:
 1
 2
 3
 4
 5

julia> powerset(x)
Base.Iterators.Flatten{Array{Combinatorics.Combinations{Array{Int64,1}},1}}(Combinatorics.Combinations{Array{Int64,1}}[Combinations{Array{Int64,1}}([1, 2, 3, 4, 5], 0), Combinations{Array{Int64,1}}([1, 2, 3, 4, 5], 1), Combinations{Array{Int64,1}}([1, 2, 3, 4, 5], 2), Combinations{Array{Int64,1}}([1, 2, 3, 4, 5], 3), Combinations{Array{Int64,1}}([1, 2, 3, 4, 5], 4), Combinations{Array{Int64,1}}([1, 2, 3, 4, 5], 5)])

julia> collect(powerset(x))
32-element Array{Array{Int64,1},1}:
 []
 [1]
 [2]
 [3]
 [4]
 [5]
 [1, 2]
 [1, 3]
 [1, 4]
 [1, 5]
 [2, 3]
 [2, 4]
 [2, 5]
 [3, 4]
 [3, 5]
 [4, 5]
 [1, 2, 3]
 [1, 2, 4]
 [1, 2, 5]
 [1, 3, 4]
 [1, 3, 5]
 [1, 4, 5]
 [2, 3, 4]
 [2, 3, 5]
 [2, 4, 5]
 [3, 4, 5]
 [1, 2, 3, 4]
 [1, 2, 3, 5]
 [1, 2, 4, 5]
 [1, 3, 4, 5]
 [2, 3, 4, 5]
 [1, 2, 3, 4, 5]

请注意，默认情况下powerset返回一个迭代器以避免分配所有子集。您还可以传递第二个和第三个位置参数powerset来限制返回子集的最小和最大大小，例如：

julia> collect(powerset(x, 2, 3))
20-element Array{Array{Int64,1},1}:
 [1, 2]
 [1, 3]
 [1, 4]
 [1, 5]
 [2, 3]
 [2, 4]
 [2, 5]
 [3, 4]
 [3, 5]
 [4, 5]
 [1, 2, 3]
 [1, 2, 4]
 [1, 2, 5]
 [1, 3, 4]
 [1, 3, 5]
 [1, 4, 5]
 [2, 3, 4]
 [2, 3, 5]
 [2, 4, 5]
 [3, 4, 5]