Question

I have a set of variables S, and a boolean function f defined on S as follows:

f(x₁, x₂, … x_n) = True iff f(x_i, x_j) = True ∀ 1 ≤ i ≤ n ∀ 1 ≤ j ≤ n, n > 1, else False.

f(a, b) is known and f(a, a) is True ∀ a, b in S.

I would appreciate some help in designing a fast algorithm that can return all subsets of S upon which f returns True.

As an example, let S = [a, b, c] and f(a, b) = f(b, c) = f(a, c) = True. The algorithm should then return [[a, b], [a, c], [b, c], [a, b, c]].

I have thought of four strategies to improve on brute force search:

1) The order of parameters of f doesn’t matter.

2) Use the fact that f(a, a) is True and f(x_i, x_j) = f(x_j, x_i) so only i < j needs checking.

2) Use the fact that f(x₁, x₂, … x_n+1) = f(x₁, x₂, … x_n) ∧ (f(x_i, x_n+1) ∀ 1 ≤ i ≤ n) where ∀ denotes iterated conjunction.

3) note that 2) implies that if f(x₁, x₂, … x_n) returns False, then f(x₁, x₂, … x_n+Δ) also does, potentially reducing the solution space.

4) Returning False as soon as soon as f(x_i, x_j) is false for some i, j.

If you want to write some code, I would appreciate it if you could give it in python.

Many thanks.

Answer 1

The two-argument function f(a, b) can be seen as a symmetric, reflexive relation on S, which can be seen as an undirected graph.

Viewed that way, f(x1, ..., xn) is true iff {x1, ..., xn} forms a complete subgraph.

From there, you end up at the clique problem which, unfortunately, turns out to be NP-complete. In other words, a fast algorithm is unlikely to exist.