ReplaceRepeated has an option MaxIterations which indicates Mathematica is keeping track of the number of iterations. Is there a way to obtain the actual number of iterations used in an evaluation of ReplaceRepeated?

(The use case is to obtain the additive persistence of a number by counting the number of times we use a rule that sums digits until we end up with a single digit)

number = 11113132434242342342342342535646657758768872132131111119
sumrule = value_ :> Total[IntegerDigits[value]]
number //. sumrule
(*  and ideally a magic function, e.g., count = `InternalValue`Iterations  etc  *)
  • $\begingroup$ possible duplicate of How to visualize pattern matching process? $\endgroup$
    – xzczd
    Commented Feb 12, 2015 at 13:21
  • $\begingroup$ @xzczd I don't see the duplication. The linked Q/answer/comments describe an entirely new function to visualize replace steps. Yes, you can count the iterations it uses. But I'm just seeking a single value query/return using the built-in ReplaceRepeated. $\endgroup$
    – PlaysDice
    Commented Feb 12, 2015 at 13:29
  • $\begingroup$ But you're asking for "a way to obtain the actual number of replacement operation", not a option of ReplaceRepeated that demonstrate the actual number of replacement operation :) $\endgroup$
    – xzczd
    Commented Feb 12, 2015 at 13:39
  • $\begingroup$ Edits made for 'clarity' $\endgroup$
    – PlaysDice
    Commented Feb 12, 2015 at 13:53
  • $\begingroup$ Tangentially related: (20181) $\endgroup$
    – Mr.Wizard
    Commented Feb 12, 2015 at 14:26

3 Answers 3


If we wish to handle an arbitrary list of rules in the second parameter of ReplaceRepeated we can by using a single wrapping rule with a counter and handing off the actual processing to ReplaceAll. This avoids the memory overhead of keeping all intermediate results for FixedPointList.

countReplace[expr_, rules_] :=
  Module[{i = -1},
    {expr //. all_ :> (i++; all /. rules), i}


  {17, 4381, 423},
  {x_ /; x > 200 :> x/2, x_ /; x > 10 :> x - 1, x_?Positive :> x - 0.3}
{{-0.2, -0.29375, -0.15}, 166}

This also works inside held expressions, as mentioned by Jacob:

countReplace[Hold[1728], n_Integer :> RuleCondition[n/2]]
{Hold[27/2], 7}

The example is contrived as I couldn't think of another replacement that would not be infinite. RuleCondition is used to force evaluation of the right-hand-side; that behavior is not characteristic of countReplace itself. Update: With Jacob's own example:

c = 3;

countReplace[HoldComplete @ Hold @ Hold @ c, Hold[c] :> c]
{HoldComplete[c], 2}
  • $\begingroup$ +1 for the embedded counter and starting at -1. $\endgroup$
    – PlaysDice
    Commented Feb 12, 2015 at 18:01
  • $\begingroup$ +1. It took me a while to realise why the examples would cause infinite replacement and it was luck that my own example did not suffer from it. I also like the name countReplace. $\endgroup$ Commented Feb 13, 2015 at 12:38
  • $\begingroup$ there's deep and then there's Mr Wizard DEEP ! Learning so much ! $\endgroup$
    – PlaysDice
    Commented Feb 13, 2015 at 14:02
  • $\begingroup$ @PlaysDice And then there is Leonid deep. If you haven't discovered that layer yet order food in; you'll be reading for a while. ;-) $\endgroup$
    – Mr.Wizard
    Commented Feb 13, 2015 at 18:30
i = 0;
number = 11113132434242342342342342535646657758768872132131111119;
sumrule = value_ :> (i++; Total[IntegerDigits[value]]);
number //. sumrule

This is a quick and dirty way to do this.

  • $\begingroup$ Hah! I didn't know you could just put a counter in a rule. I like quick and dirty. It seems to give an answer that is +1, perhaps due to a final 'ReplaceRepeated' which finds that there is no change. I can easily subtract one $\endgroup$
    – PlaysDice
    Commented Feb 12, 2015 at 14:05
  • 1
    $\begingroup$ ReplaceRepeated needs to try one more time to do a ReplaceAll after value reaches 9 in order to verify that it is done. $\endgroup$
    – bbgodfrey
    Commented Feb 12, 2015 at 15:01

Here is another solution, that works with a list of rules and with held expressions

cRepRep2[expr_, snd_] :=
 {Last@#, Length@# - 2} &@
  FixedPointList[# /. snd &, expr]

Here is a similar solution, that saves on memory by using FixedPoint rather than FixedPointList. It also works with held expressions

cRepRep3[expr_, snd_] :=
  {c = -1, resExpr},
  resExpr = 
   FixedPoint[# /. snd &, expr, SameTest -> ((c++; SameQ[#, #2]) &)];
  {resExpr, c}

Below is the dirty way made into a function, which is less nice than the other solutions.

cRepRep[expr_, (Rule | RuleDelayed)[pat_, rep_]] :=
 Module[{c = 0},
  {ReplaceRepeated[expr, pat :> (c++; rep)], c}

Here is function that tries to let all the replacements be done by ReplaceRepeated, while also working for held expressions. Unfortunately there is a problem with it, which it shares with cRepRep, as we will see further below.

cRepRep4[expr_, (Rule | RuleDelayed)[pat_, rep_]] :=     
 cRepRep4[expr, {pat :> rep}]

cRepRep4[expr_, rules_List] :=
 Module[{c = 0, rLen, cIncrCondArrHeld, repsHeld, newRepsHeld, 
   newRulesHeld, resExpr},
  rLen = Length[rules];
  cIncrCondArrHeld = 
   DeleteCases[Hold@Evaluate@ConstantArray[Hold[c++; True], rLen], 
    Hold, {2, Infinity}, Heads -> True];
  repsHeld = Hold[rules][[All, All, 2]];
  newRepsHeld = 
    Hold@Evaluate@Thread[Join[repsHeld, cIncrCondArrHeld]], {2}];
  newRulesHeld =
    Hold@Evaluate@Thread@Prepend[newRepsHeld, rules[[All, 1]]], {2}];
  resExpr =
    {0 -> ReplaceRepeated, {2, 0} -> Unevaluated}

  {resExpr, c}

Unfortunately ReplaceRepeated can terminate in two ways, which causes trouble. Either it makes a last replacement which results in the same expression, or all the rules fail to match. I think it is not straightforward, if at all possible, to get this information from ReplaceRepeated. Therefore cRepRep, cRepRep4 will fail, like in the following example.

c = 0; (*just to make things harder*)
cRepRep4[HoldComplete@Hold@Hold@c, {Hold[c] :> c}]
cRepRep4[number, {9-> 8, sumrule}]
{HoldComplete[c], 2} (*correct*)
{8, 5} (*count is one too high*)

Mr.Wizard's solution does not suffer from this because the replacement by the rules is really done by ReplaceAll and the rule in ReplaceRepeated always matches, so that ReplaceRepeated always terminates because of "identical expressions".

Good examples

c = 3; (*just to make things harder*)
cRepRep3[HoldComplete@Hold@Hold@c, Hold[c] :> c]
cRepRep3[number, {9 -> 8, sumrule}]
{HoldComplete[c], 2}
{8, 4}
  • $\begingroup$ Nice, thank you. Like Carlo's answer, it gives +1 more iterations than is 'needed', but I'm guessing MMA includes the final iteration where it sees no change from the (pen)ultimate iteration. So the MMA calculation requires #iterations + (1 final step with no change). but 'real world' iterations are c minus 1. $\endgroup$
    – PlaysDice
    Commented Feb 12, 2015 at 14:15
  • $\begingroup$ ReplaceRepeated is able to handle a list of rules; it appears this function is not. Can you fix that? $\endgroup$
    – Mr.Wizard
    Commented Feb 12, 2015 at 14:32
  • $\begingroup$ @Mr.Wizard ah yes I was not satisfied anyway. Let's see $\endgroup$ Commented Feb 12, 2015 at 15:30
  • 1
    $\begingroup$ Cool. cRepRep2 gives the 'correct' answer for iterations = 3. Nice hinting by @Mr.Wizard with FixedPointList and his 20181 $\endgroup$
    – PlaysDice
    Commented Feb 12, 2015 at 15:53
  • 1
    $\begingroup$ @Mr.Wizard I went into a coding frenzy, but failed to find a solution that uses ReplaceRepeated. My mistake was that I thought adding a rule _/;(c++;False) to the beginning of the list of rules would work, but here c++ gets evaluated to often. It is also hard to make things work for held expressions, as the original approach fails here. $\endgroup$ Commented Feb 12, 2015 at 17:10

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