mjhjmtu

I have a linear problem that I want to solve but the method is quite different from normal, the problem is still Ax=b. However,
in this instant I have A being unknown, apart from the fact that each entry in A can only be zero or 1. Further I have x being known and fixed, the same holds for b.

My question, is it a easy method for getting mathematica to spit out all possible A such that Ax=b given conditions on A.

I tried doing a number of for loops etc, however, I have completely given up after number of hours and the expectation that my effort is incorrect.

Here's a way to code up your problem with Solve automatically:

x = 1, 2, 3;
b = Reverse[x];
sol[x_List, b_List] /; Length[x] === Length[b] := Module[mat, vars,
 mat = Array[[FormalM], Length[x]*1, 1]; (* construct a matrix of variables*)
 vars = Flatten[mat];
 mat /. 
 Solve[And @@ Thread[mat.x == b] (* Construct the equations *) 
 && vars ∈ Integers && And @@ Thread[0 <= vars <= 1] (* constraints *),
 vars
 ]
 ];
matrixSolutions = sol[x, b]

0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0

As you can see, 2 solutions were found. Check that the residuals of the solutions are zero vectors:

#.x - b & /@ matrixSolutions

0, 0, 0, 0, 0, 0

It works, but I don't know how scalable this approach is. Instead of Solve, you can also use NSolve and/or FindInstance (you can just replace Solve with any of those two functions in the code above).

Pretty much the same questions as

FrobeniusSolve with solutions only being 0 or 1 being acceptable

Using @Sjoerd's problem

x = 1, 2, 3;
b = Reverse[x];

Can write solution as

res = Tuples@(Select[FrobeniusSolve[x, #], Max@# <= 1 &] & /@ b)

(* 0, 0, 1, 0, 1, 0, 1, 0, 0, 
 1, 1, 0, 0, 1, 0, 1, 0, 0 *)