Finding the smallest solution set, if one exists (two multipliers)

Note: This is the two-multipliers variation of this problem

Given a set A, consisting of floats between 0.0 and 1.0, find a smallest set B such that for each a in A, there is either a value where a == B[x], or there is a pair of unique values where a == B[x] * B[y].

For example, given

$ A = [0.125, 0.25, 0.5, 0.75, 0.9]

A possible (but probably not smallest) solution for B is

$ B = solve(A)
$ print(B)
[0.25, 0.5, 0.75, 0.9]

This satisfies the initial problem, because A[0] == B[0] * B[1], A[1] == B[1], etc., which allows us to recreate the original set A. The length of B is smaller than that of A, but I’m guessing there are smaller answers as well.

I assume that the solution space for B is large, if not infinite. If a solution exists, how would a smallest set B be found?

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Notes:

• We're not necessarily limited to the items in A. B can consist of any set of values, whether or not they exist in A.
• Since items in A are all 0-1 floats, I'm assuming that B will also be 0-1 floats. Is this the case?
• This may be a constraint satisfaction problem, but I'm not sure how it would be defined?
• Since floating point math is generally problematic, any answer should frame the algorithm around rational numbers.

Sort the array. For each pair of elements Am, An ∈ A, m < n - calculate their ratio.

Check if the ratio is equal to some element in A, which is not equal to Am nor to An.

Example:

A = { 0.125, 0.25, 0.5, 0.75, 0.9 }

(0.125, 0.25): 0.5    <--- bingo
(0.125, 0.5 ): 0.25   <--- bingo
(0.125, 0.75): 0.1(6)
(0.125, 0.9 ): 0.13(8)
(0.25 , 0.5 ): 0.5
(0.25 , 0.75): 0.(3)
(0.25 , 0.9 ): 0.2(7)
(0.5  , 0.75): 0.(6)
(0.5  , 0.9 ): 0.(5) 
(0.75 , 0.9 ): 0.8(3)

The numerator (0.125) is redundant (= 0.25 * 0.5) or (= 0.5 * 0.25)

We can do better by introducing new elements:

Another example:

A = { 0.1, 0.11, 0.12, 0.2, 0.22, 0.24 }

(0.1 , 0.11): 0.(90)        ***
(0.1 , 0.12): 0.8(3)        +++
(0.1 , 0.2 ): 0.5     <--------
(0.1 , 0.22): 0.(45)
(0.1 , 0.24): 0.41(6)
(0.11, 0,12): 0.91(6)       ~~~
(0.11, 0.2 ): 0.55
(0.11, 0.22): 0.5     <--------
(0.11, 0.24): 0.458(3)
(0.12, 0.2 ): 0.6
(0.12, 0.22): 0.(54)
(0.12, 0.24): 0.5     <--------
(0.2 , 0.22): 0.(90)        ***
(0.2 , 0.24): 0.8(3)        +++
(0.22. 0.24): 0.91(6)       ~~~

Any 2 or more pairs (a1,a2), (a3,a4), (... , ...) with a common ratio f can be replaced with { a1, a3, ..., f }.

Hence adding 0.5 to our set makes { 0.1, 0.11, 0.12 } redundant.

B = (0.2, 0.22, 0.24, 0.5}

We are now (i the general case) left with an optimization problem of selecting which of these elements to remove and which of these factors to add in order to minimize the cardinality of B (which I leave as an exercise to the reader).

Note that there is no need to introduce numbers greater than 1. B can also be represented as { 0.1, 0.11, 0.12, 2} but this set has the same cardinality.