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We can generate a Boolean function of $n$ variables (take $n=4$ as an example) as follows:

f = BooleanFunction[10, 4]

I want to express $f$ as a polynomial. This is almost achieved by converting the function to the "ANF" form. All that's left is to get Mathematica to replace the And operator by multiplication, and the Xor operator by addition. Any idea how I can do that?

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    $\begingroup$ Why not make the replacements directly? expr /. {And -> Times, Xor -> Plus} $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Mar 9, 2016 at 17:16
  • $\begingroup$ @J.M. Close, but not enough for the general case BooleanConvert[BooleanFunction[1, 4], "ANF"] /. {And -> Times, Xor -> Plus} $\endgroup$
    – Dr. belisarius
    Commented Mar 10, 2016 at 2:53
  • $\begingroup$ @Dr. bel, the Not[] is indeed a problem. OP did not give instructions on how it should be dealt with, however. BTW: no need for BooleanConvert[]: BooleanFunction[1, 4, "ANF"]. $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Mar 10, 2016 at 3:01

1 Answer 1

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I believe this is what you want:

bf[n_, m_] := BooleanFunction[n, m, "DNF"] /. {Not[a_] :> (a - 1)^2, And -> Times,
              Xor -> Plus, Or -> Plus} /. (Plus[a__] &) :> (Unitize[Plus[a]] &)

bf[10, 4]
(* (#1 - 1)^2 (#2 - 1)^2 #4 & *)

Checking:

bt[n_, m_] := Boole /@ BooleanTable[BooleanFunction[n, m]]

And @@ ((bt[#, 6] == bf[#, 6] @@@ Tuples[{1, 0}, 6]) & /@ Range[2^6])

( * True *)
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  • $\begingroup$ It works, somehow. But not always return a proper polynomial. Try for example bf[300, 10]. The Unitize[ ] stays in there. Looking for a better option. $\endgroup$
    – Dr. belisarius
    Commented Mar 10, 2016 at 16:17

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