Mathematica Generate Binary Numbers with Locked Bits

Question

I have a very specific Mathematica question. I am trying to generate all the binary numbers around certain 'locked' bits. I am using a list of string values to denote which bits are locked e.g. {"U","U,"L","U"}, where U is an "unlocked" mutable bit and L is a "locked" immutable bit. I start with a temporary list of random binary numbers that have been formatted to the previous list e.g. {0, 1, 1, 0}, where the 1 is the locked bit. I need to find all the remaining binary numbers where the 1 bit is constant. I've approached this problem recursively, iteratively, and with a combination of both with no results. This is for research I am doing at my university.

I am building a list of base 10 forms of the binary numbers. I realize that this code is completely wrong. This is just one attempt.

    Do[
      If[bits[[pos]] == "U", 
        AppendTo[returnList, myFunction[bits, temp, pos, returnList]]; ],
    {pos, 8, 1}]

    myFunction[bits_, bin_, pos_, rList_] :=
      Module[{binary = bin, current = Length[bin], returnList = rList},
        If[pos == current,
          Return[returnList],
          If[bits[[current]] == "U",
          (*If true*)
            If[! MemberQ[returnList, FromDigits[binary, 2]],
            (*If true*)
              AppendTo[returnList, FromDigits[binary, 2]];
              binary[[current]] = Abs[binary[[current]] - 1],
            (*If false*)
              binary[[current]] = 0;
              current = current - 1]; ,
          (*If false*)
            current = current - 1];
          returnList = myFunction[bits, binary, pos, returnList];  
        Return[returnList]]]

Answer 1

You can use Tuples and Fold to generate only bit sets that you are interested in.

bits = {"U", "U", "L", "U"};

Fold[
  Function[{running, next}, 
    Insert[running, 1, next]], #, Position[bits, "L"]] & /@ Tuples[{0, 1}, Count["U"]@bits]

(*
{{0, 0, 1, 0}, {0, 0, 1, 1}, {0, 1, 1, 0}, {0, 1, 1, 1}, 
 {1, 0, 1, 0}, {1, 0, 1, 1}, {1, 1, 1, 0}, {1, 1, 1, 1}}
*)

Hope this helps.