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Input: A binary reference vector $r$, a multi-set of binary vectors $S$, and a distance function d between binary vectors of equal length. Assume that $r$ and all vectors in $S$ have the same length.

Objective: We want to find a binary vector $r'$ that maximizes the number of vectors $s \in S$ for which $d(r', s) \leq k$ for some fixed $k$.

Constraint: We require that the vector we find is within a certain distance $j$ of the reference vector, so $d(r, r') \leq j$.

Formally, we want $$arg \max\limits_{i : d(i,r) \leq j} \sum\limits_{s \in S} [[d(i,s) \leq k]]$$

Where $[[d(i,s) \leq k]]$ is 1 if $d(i,s) \leq k$, else 0.

Details: We specifically want to solve this problem when the distance function $d$ is the Hamming Distance, but it would be nice to be able to substitute any distance function, or at least any distance function built into Mathematica.

Attempts: The following code (adapted from a post here) computes the unconstrained version of the problem that doesn't enforce a maximum distance to a reference vector.

numApprovals[x_?(MatrixQ[#, NumericQ] &), 
   y_?(VectorQ[#, NumericQ] &), k_] := 
  Length[Select[HammingDistance[#, y] & /@ x, # <= k &]];

length = 3;

maximizeApprovals[x_, k_] := 
 Maximize[{numApprovals[x, Array[y, length], k], 
   Thread[0 <= Array[y, length] <= 1]}, 
  Array[y, length] \[Element] Integers]

answerAssoc = 
  maximizeApprovals[{{1, 1, 1}, {0, 1, 0}, {1, 1, 1}}, 
   Ceiling[length/2]];

answer = Values[answerAssoc[[2]]]

However, when I try to adding the additional constraint to the maximizeApprovals function, it doesn't work. The following code evaluates symbolically.

maximizeApprovals[x_, k_, ref_, j_] := 
 Maximize[{numApprovals[x, Array[y, length], k], 
   Thread[0 <= Array[y, length] <= 1], 
   HammingDistance[ref, y] <= j}, 
  Array[y, length] \[Element] Integers]

answerAssoc = 
 maximizeApprovals[{{1, 1, 1}, {0, 1, 0}, {1, 1, 1}}, 
  Ceiling[length/2], {0, 1, 0}, 1]

Adding pattern testing to the inputs doesn't seem to solve the problem.

However, if I have a fixed length, say 3, I can hard-code the constraint corresponding to the HammingDistance by taking the sum of the absolute difference of the corresponding elements and it works fine.

maximizeApprovals[x_?(MatrixQ[#, NumericQ] &), k_?(IntegerQ[#] &), 
  ref_?(VectorQ[#, NumericQ] &), j_?(IntegerQ[#] &)] := 
 Maximize[{numApprovals[x, Array[y, length], k], 
   Thread[0 <= Array[y, length] <= 1], 
   Abs[y[1] - ref[[1]]] + Abs[y[2] - ref[[2]]] + 
     Abs[y[3] - ref[[3]]] <= j}, {Array[y, length] \[Element] 
    Integers}]

But if I use the built-in HammingDistance function, or try to use BitXor, to get it to work for any length, neither works.

maximizeApprovals[x_?(MatrixQ[#, NumericQ] &), k_?(IntegerQ[#] &), 
  ref_?(VectorQ[#, NumericQ] &), j_?(IntegerQ[#] &)] := 
 Maximize[{numApprovals[x, Array[y, length], k], 
   Total[BitXor[ref, y]] <= j, 
   Thread[
    0 <= Array[y, length] <= 1]}, {Array[y, length] \[Element] 
    Integers}]

maximizeApprovals[x_?(MatrixQ[#, NumericQ] &), k_?(IntegerQ[#] &), 
  ref_?(VectorQ[#, NumericQ] &), j_?(IntegerQ[#] &)] := 
 Maximize[{numApprovals[x, Array[y, length], k], 
   HammingDistance[ref, y] <= j, 
   Thread[
    0 <= Array[y, length] <= 1]}, {Array[y, length] \[Element] 
    Integers}]

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  • $\begingroup$ Nitpick: You say S is a set but your example is a list {{1,1,1},{0,1,0},{1,1,1}} where {1,1,1} appears twice. In TeX you write $\leq k$ but in your code it is <k. Your numIssues is not defined. Please always run code in your post starting from a fresh kernel. Potential problems: Something like HammingDistance[{a,b},{c,d}] evaluates to 2 right away, as it should according to the documentation, but this may be a problem if a, b, c, d are symbols to be replaced by 0|1. $\endgroup$
    – user293787
    Commented Jul 27, 2022 at 3:50
  • $\begingroup$ Edited the post to correct for the errors @user293787. Thanks for pointing them out. $\endgroup$
    – B A
    Commented Jul 27, 2022 at 16:37

1 Answer 1

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The following is an adaptation of your code. I just tried to get it to work, still using Maximize, ignoring efficiency. Please Quit[] before trying this:

(* auxiliaries
   did not try distance functions other than Hamming *)
binVec={(0|1)..};
distanceFunction[a:binVec][b:binVec]:=HammingDistance[a,b];

(* now using <= k *)
numApprovals[x:{binVec..},ys:binVec,k_]:=Length[
   Select[Map[distanceFunction[ys],x],(#<=k)&]];

(* version without constraint, but with optional penalty *)
maximizeApprovals[x:{binVec..},k_,penalty_:(0&)]:=Module[{y},
   With[{ys=Array[y,Length[First[x]]]},
     Maximize[{numApprovals[x,ys,k]+penalty[ys],
        Thread[0<=ys<=1]},ys\[Element]Integers]//{First[#],ys/.Last[#]}&]];

(* version with constraint, via suitably large penalty *)
maximizeApprovals[x_,k_,ref_,j_]:= With[{p=-Length[x]-1},
     maximizeApprovals[x,k,If[distanceFunction[ref][#]<=j,0,p]&]];

Comment. I switched to using a penalty since a constraint such as distanceFunction[ref][ys]<=j in Maximize seemed not to work. It makes sense that there are limits to what constraints are allowed, even if in this particular case Maximize will end up brute-forcing anyway I think.

Examples. One without, one with constraint:

maximizeApprovals[{{1,1,1},{0,1,0},{1,0,0}},1]
(* {3,{1,1,0}} *)

maximizeApprovals[{{1,1,1},{0,1,0},{1,0,0}},1,{0,0,0},1]
(* {2,{0,0,0}} *)

Warning. Did not test beyond these examples.

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  • $\begingroup$ The penalty is a nice workaround, but I need the solution to be guaranteed optimal according to the original constraint. I'll try your solution soon and spend some time trying to solve for penalty functions that guarantee optimality in the original problem. Thanks @user293787. $\endgroup$
    – B A
    Commented Jul 27, 2022 at 16:46
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    $\begingroup$ @BA It will be exactly optimal if you choose the penalty negative enough. The number of approvals is at most equal to Length[x] (the number of elements in the set $S$ which you call x in the code). Therefore with a penalty of -Length[x]-1 you are fine and will get the solution you want. See the edit. $\endgroup$
    – user293787
    Commented Jul 27, 2022 at 16:50

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