I am defining a linear associative multiplication operation Mult[]:
ClearAll[Mult]
SetAttributes[Mult, {Flat, OneIdentity}];
Mult[A___, a_, B___] := a Mult[A, B] /; NumberQ[a]
Then I try to evaluate a simple expression, which (I believe) does not match the rule written above:
Mult[X, Y]
I got the error Recursion depth of 1024 exceeded. It is definitely related to the attribute Flat, but I do not see how. What am I doing wrong?
I find Flat hard to work with. Despite seeing it in use numerous times I'm never quite sure what it will do until I try it, so maybe I'm not the person to answer your question. Nevertheless I shall try.
Consider this related, reduced example:
Attributes[foo] = {Flat, OneIdentity};
foo[a_] := "inert" /; Print @ HoldForm[a]
foo[X, Y];
foo[X,Y]X
Y
Critically the entire expression foo[X, Y] is substituted for the pattern a_ in one of the match attempts. This leads to infinite recursion.
One way to get around this is to prevent evaluation of that expression, which is what I used HoldForm for above. Applied to your original function:
ClearAll[Mult]
Attributes[Mult] = {Flat, OneIdentity};
Mult[A___, a_, B___] := a Mult[A, B] /; NumberQ[Unevaluated @ a]
Mult[X, Y]
Mult[3, x, 7, y, z]
Mult[X, Y]21 Mult[x, y, z]
This is basically a duplicate of question (5067). You can use my answer to that question to resolve your issue:
ClearAll[Mult]
SetAttributes[Mult, {Flat, OneIdentity}];
Verbatim[Mult][A___, a_, B___] := a Mult[A,B] /; NumberQ[a]
Then:
Mult[X, Y]
Mult[3, X, 7, Y, Z]
Mult[X, Y]
21 Mult[X, Y, Z]
No recursion errors.
Comparison
Using Verbatim should be much quicker than the accepted answer, because those patterns where a gets matched with a Mult object should never happen. The accepted answer (using Mult2):
ClearAll[Mult2]
Attributes[Mult2] = {Flat, OneIdentity};
Mult2[A___, a_, B___] := a Mult2[A,B] /; NumberQ[Unevaluated@a]
And a comparison:
SeedRandom[1]
r = RandomChoice[Join[Alphabet[], Range[10]], 30];
r1 = Mult @@ r; //AbsoluteTiming
r2 = Mult2 @@ r; //AbsoluteTiming
ReplaceAll[Mult->List] @ r1 === ReplaceAll[Mult2->List] @ r2
{0.000206, Null}
{0.025536, Null}
True
Flatto begin with? Doesn't an actual solution need to preserve that pattern-matching behavior? (I am assuming that this is a toy example and not the actual, complete problem.) – Mr.Wizard Sep 24 '18 at 14:53Mult[a, Mult[b, c]]still evaluates toMult[a, b, c]. – Carl Woll Sep 24 '18 at 14:58