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I'm exploring the use of SAT solvers to find a permutation of the numbers 1, 2, and 3, subject to specific constraints. Initially, I considered employing nine variables: $x_{1,1}, x_{1,2}, \dots, x_{3,3}$, where $x_{i,j}$ is true if the number $i$ is placed in position $j$ in the permutation.

However, since there are only $3! = 6$ possible permutations, it seems excessive to use nine variables. I'm contemplating using just three boolean variables, $y_1, y_2, y_3$. My idea is to map each permutation to a unique combination of true or false values of these $y_i$'s. But I'm finding it challenging to formulate the constraints under this scheme.

Could there be a more efficient encoding strategy that strikes a balance between the number of variables and their expressive power? I'm seeking a method that simplifies the representation of permutations and constraints in a SAT solver context.

NonalcoholicBeer
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    If you're looking for non-repeating permutations, your three numbers will only ever have two degrees of freedom - Once you've picked your first and second numbers, your third is chosen for you - it has to be the number you haven't picked yet. So, you only really need to encode which number was chosen first, and which was chosen second.

    If you can choose with repetition, obviously this won't work - you really will need to encode all three numbers for each variable, because you'll always have three choices at each selection.

     – Lee Davis-Thalbourne Jan 11 '24 at 03:36

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In isolation, the obvious answer is to enumerate all 6 permutations and check which ones satisfy the conditions.

Without more context, this is not answerable. There are many ways to represent a permutation of 3 elements. We can't tell you which is best, because that depends on the rest of your model.

Here are some candidate options to consider:

  • A one-hot encoding of which of the 6 permutations has been chosen. This requires 6 boolean variables.

  • A binary encoding of which of the 6 permutations has been chosen. This requires $\lceil \lg 6 \rceil = 3$ boolean variables.

  • For each pair $i,j$, a boolean variable indicating whether $\pi(i)<\pi(j)$. This requires 6 boolean variables.

D.W.
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