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Re: [tlaplus] n-ary Cartesian product



Hello,

your problem is not well specified because {S1, S2} = {S2, S1} but S1 \X S2 is different from S2 \X S1. Also, I don't understand your remark about the output: the cartesian product is a set, but its elements are sequences.

Assuming that your operator takes a *sequence* of sets, i.e. Cartesian(<<S1, S2, ..., Sn>>), you can write the following in TLA+.

Range(S) == { S[i] : i \in 1 .. Len(S) }
Cartesian(S) == 
  LET U == UNION Range(S)
  IN  {s \in Seq(U) : /\ Len(s) = Len(S)
                      /\ \A i \in 1 .. Len(s) : s[i] \in S[i]}

However, TLC will not be available to interpret this because of the quantification over the infinite set Set(U). The following should work in principle (I haven't actually tried):

Cartesian(S) == 
  LET U == UNION Range(S)
      FSeq == [ (1 .. Len(s)) -> U ]
  IN  {s \in FSeq : \A i \in 1 .. Len(s) : s[i] \in S[i]}

However, you probably want to override this operator definition by a Java method.

Stephan

On 20 Oct 2020, at 13:27, Mariusz Ryndzionek <mryndzionek@xxxxxxxxx> wrote:

Hello,
I need n-ary Cartesian product operator. Something that would do:
Cartesian({S1, S2, .., Sn}) = S1 \X S2 \X .. Sn

The output shouldn't necessarily be sequences. Sets will do.
Is there already something like this in TLA+?

Regards,
Mariusz

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