all subset of a se in julia

Question

would you please help me? How can I do a code which can find all subset of a set

for example

I want to code this constraint in Julia. It is a subtour constraint. But I don't know how I can find all subsets of S set.

@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);

Thanks very much

score 7 · Answer 1 · answered Jan 05 '19 at 14:56

7

You can get it using powerset function from the Combinatorics.jl package, e.g.:

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]

Note that by default powerset returns an iterator to avoid allocating all subsets. Also you can pass a second and third positional argument to powerset to limit minimum and maximum size of the returned subset, e.g.:

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]

answered Jan 05 '19 at 14:56

Bogumił Kamiński

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Thank you very much. I used your valuable helps. but when I use in constraints it gets me error. – Soma Jan 06 '19 at 08:29
Thank you very much. I used your valuable helps. but when I use in constraints it gets me error. "totalI=3; s=[1:totalI;] S=collect(powerset(s)) for i=2:size(S,1) @constraint(ILRP,c7[ k in totalK,t in totalH, s in size(S[i],1)],sum(x[i,j,k,t] for i=1:size(S[i],1), j=1:size(S[i],1))<=size(S[i],1)-1) end" this error ERROR: UndefVarError: i not defined if you don't mind, please help me gain why the code don't identify i – Soma Jan 06 '19 at 08:43
1

You are reusing `i` variable twice. Once in the outer loop and once inside the constraint. You should use unique variable names, e.g. change `for i=2:size(S,1)` to `for outer_i=2:size(S,1)` and correct the code inside to use `outer_i` appropriately. – Bogumił Kamiński Jan 06 '19 at 08:50
if you don't mind, please help me again. can I use powerset(x, 2:5)? that it gives me only subsets with 2,3,4 and 5 elements. – Soma Jan 06 '19 at 10:18
1

`powerset(x, 2, 5)` – Bogumił Kamiński Jan 06 '19 at 13:02
yes it is very useful. thanks veryyyyyyyyyyyyyy muchhhhhhhhhh. :) Im sorry for asking again what you said before. – Soma Jan 06 '19 at 15:27

score 1 · Accepted Answer · answered Jan 05 '19 at 14:59

Don't know whether this is what you seek:

using Combinatorics

   function subsets(A::AbstractArray,r::Union{AbstractArray,Integer})
  o= Array{Array{eltype(A),1},1}(undef,0)         
  if typeof(r)<:Integer
      r>length(A) && (r=[length(A)])
      r=[r...]
  elseif typeof(r)<:UnitRange
      r[end]>length(A) && (r=1:r[length(A)])
  else
      !issubset(r,1:length(A)) && (r=intersect(r,1:length(A)))
  end
  for n = r
      a=combinations(A,n)
      for i in a     
          push!(o,i)
      end
  end
  return o
end

subsets(A::AbstractArray) = subsets(A,1:length(A))

It can List all Subsets or Subsets up to a certain limit (length): eg:

 julia> subsets(1:3)
 7-element Array{Array{Int64,1},1}:
  [1]
  [2]
  [3]
  [1, 2]
  [1, 3]
  [2, 3]
  [1, 2, 3]
  julia> subsets(1:3,2)
  3-element Array{Array{Int64,1},1}:
  [1, 2]
  [1, 3]
  [2, 3]

all subset of a se in julia

2 Answers2