I have two vectors p={p1,p2,...,pK} and q={q1,q2,...,qK} of size K. I would like to create a function which is dynamic in the sense that its definition depends on K because it uses all cross-terms (keeping one term constant) which can possibly be created with these two vectors.

As an example, for K=2 and keeping the term p1 constant, and assuming some other function g is already apropriately defined, I wish to create the following function fp1:


Edit thanks to the comment by @march: I forgot to mention that each variable can have two values 0 and 1. In this example, since p1 is the "reference" (or "constant"), we get the 8 following cross products for the two values of p2, q1 and q2: {p2,1-p2}X{q1,1-q1}X{q2,1-q2}. For K=3, I would thus expect there to be the sum of 2^5 terms for the 5 variables {p2,p3,q1,q2,q3}.

How can I achieve these combinations for any K? How about if I want to create equivalent combinations but using three vectors p, q and r instead of the two above p and q?

I have not been able to get very far, but the following code defines my vectors in preparation of the cross products:

pk=Table[p[k], {k, 2, K}];
qk=Table[q[k], {k, K}];

It also seems to me that Tuples might help, but I have not figured out how...

I am a Mathematica novice so do please do not hesitate to provide other approaches if mine seems incorrect or poor practice.

  • $\begingroup$ It's not clear to me what the general structure of the expression is supposed to be. One example is good, but maybe a second example or a more complete description of what you're looking for is necessary to understand what you're trying to do. For instance, it's not clear what "p1 is a constant" really means. Please edit your post to include this extra information. $\endgroup$
    – march
    Commented Jun 29, 2023 at 21:53
  • $\begingroup$ That said, does this work? Clear[p, q]; n = 2; ps = Array[p, n - 1, 2]; qs = Array[q, n]; elems = Subsets[Join[ps, qs, 1 - qs, 1 - ps], {2 n - 1}]; Times @@ # g @@ # & /@ elems // Total $\endgroup$
    – march
    Commented Jun 29, 2023 at 21:58
  • $\begingroup$ @march Thanks, I have edited the post to attempt to be more specific. I will try your suggestion as soon as I get back to my computer. $\endgroup$
    – tyogi
    Commented Jun 29, 2023 at 22:24
  • $\begingroup$ @march your suggested code does not work for my purposes: right off the bat, it seems to create more combinations, 20 in all, instead of the expected 8. Two problems I notice: 1. among those combinations are mixes of p2 and (1-p2), whereas it should be one or the other, and 2. the placement of arguments in g should be for (p2,q1,q2) in that order, whereas the order seems to have permutations as well. $\endgroup$
    – tyogi
    Commented Jun 30, 2023 at 0:46
  • 1
    $\begingroup$ What if you replaced the Subsets line with Tuples@Transpose[{Join[qs, ps], 1 - Join[qs, ps]}]? $\endgroup$
    – march
    Commented Jun 30, 2023 at 3:36

2 Answers 2


Try this:

fp1[p_, q_] := Module[{el = Join[p, q], t},
  t = Function[x, el /. Thread[x -> 1 - x]] /@ Subsets[el];
  (Times @@ # g @@ #) & /@ t

E.g. with K=2:

K = 2;
pk = Table[p[k], {k, 2, K}];
qk = Table[q[k], {k, K}];
fp1[pk, qk] // TableForm

enter image description here

Or a bit more complicated with K=3:

enter image description here

  • $\begingroup$ This works as specified (provided we use Total at the end), thanks! The line Function[x, el /. Thread[x -> 1 - x]] /@ Subsets[el] was a bit tough for me to understand but I think I got it :) $\endgroup$
    – tyogi
    Commented Jun 30, 2023 at 17:18
  • $\begingroup$ Thread[x -> 1 - x] creates a replacement rule that replaces all elements x in the subset by x and 1-x. This rule is then applied to el. $\endgroup$ Commented Jun 30, 2023 at 17:26

Here's my version from the comments, wrapped up in a function.

Clear[p, q, g]
fp1[n_] := Module[{ps = Array[p, n - 1, 2], qs = Array[q, n]},
             Times @@ # g @@ # & /@ Tuples@Transpose[{Join[qs, ps], 1 - Join[qs, ps]}] // Total

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