Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

What is the most simple, elegant way of implementing a rewrite-system defined as:

$$ \begin{aligned} \Sigma &= \{a_1, a_2, a_3, ...\} \\ N &= \{A_1, A_2, A_3, ...\} \\ \{\alpha_1 , \alpha_2, \alpha_3, ...\} &= \Sigma \bigcup N \\ F &= \{f_1, f_2, f_3, ... \} \\ P &= \left\{ \begin{aligned} A &\rightarrow a \\ A &\rightarrow \alpha \\ A &\rightarrow \alpha_1 | \alpha_2| \alpha_3 | ... \\ A &\rightarrow f(\alpha_1, \alpha_2, \alpha_3, ...) \\ \end{aligned} \right\} \end{aligned} $$

That is: given a set of terminal symbols $\Sigma$, a set of nonterminal symbols $N$, a set of functions $F$ and a set of rewrite rules $P$, starting from an arbitrary but well-formed sentence $S$ of the grammar $G = (\Sigma, N, F, P, S)$, the system should apply the rewrite rules until all non-terminals are resolved, producing sentences like:

$$ f_1(a_3, a_1, f_2(a_4, a_1), a_1, f_3(a_2, f_1(a_1), a_4)) $$

UPDATE: Specification

  • terminal and nonterminal symbols can be e.g. strings;
  • only symbols of $\Sigma$ and $N$ are replaced, functions of $F$ are left intact, though symbols inside them are to be replaced. Therefore g["A"] in the initial sentence below is to be replaced, e.g.:

    g["A"] --applying-7th-rule--> g["B" ~~ "b"];

  • some functions should evaluate immediately to result in a (non)terminal symbol (| = random choice, .. = repeate 1 or more times, ... = repeate 0 or more times);
  • all $f_i$ should be held until termination;
  • rules should be applied randomly: any that fits could be chosen for a certain replacement.

At present, I am struggling with recursively applying structural replacements in held rules, and it's becoming more and more complex and convoluted. There must be a simpler way, concerning Mathematica is by design a very effective rewrite system. Though I have no idea how to start to exploit e.g. regular expressions.

UPDATE: Example

An example follows (note that I used StringExpression, but it can be replaced with List, or similar):

terminals = {"a", "b", "c", "d", "e", "f", "g", "h"};
nonTerminals = {"A", "B", "C", "D"};
functions = {f, g, h, i};

initSentence = g["A"] ... ~~ "B"; (* ... indicates zero or more g["A"] *)
rules = {
   "A" :> "a",
   "B" :> "b" | "c" | "d", (* immediately choose any terminal of the rhs *)
   "C" :> "e" | "f",
   "D" :> "g" | "h",
   "A" :> "a" ~~ "b",
   "A" :> "A" ~~ "a",
   "A" :> "B" ~~ "b",
   "A" :> f["e" | "f"] ~~ "a", (* do not evaluate f but choose "e" or "f" *)
   "A" :> g@"h" ~~ "C",
   "B" :> g@"C", (* do not evaluate g but "C" must be replaced later *)
   "B" :> "B" ~~ h@"e",
   "C" :> f@"f",
   "C" :> "C" ~~ g@"d"

Starting from initSentence one possible process with a valid outcome would be:

g["A"] ... ~~ "B"                 (* initial sentence *)
g["A"] g["A"] ~~ "B"              (* g["A"] is repeated randomly *)
g["A" ~~ "a"] g["A"] ~~ "B"       (* rule A ":>"A "~~"a " is applied *)
g["A" ~~ "a"] g[g@"C"] ~~ "B"     (* rule "B" :> g@"C" is applied *)
g["a" ~~ "a"] g[g@"C"] ~~ "B"     (* rule "A" :> "a" is applied *)
g["a" ~~ "a"] g[g@f@"f"] ~~ "B"   (* rule "C" :> f@"f" is applied *)
g["a" ~~ "a"] g[g@f@"f"] ~~ "b" | "c" | "d"  (* rule "B":>"b"|"c"|"d" is applied *)
g["a" ~~ "a"] g[g@f@"f"] ~~ "d"   (* one terminal is randomly chosen *)
(* terminate and evaluate g, f *)

The point is that the grammar is represented as a set of rules that are both probabilistic (e.g. random choices) and are applied in a random way, i.e. for each symbol (at each step), a random fitting rule should be applied.

share|improve this question
It is not quite clear what you want to do. Grammars usually define a language, which is a set of all possible phrases which can be derived by using the productions of the grammar, and is usually infinite, for interesting cases. To get a specific sentence, you have to provide a sequence of productions to be used, in addition to the grammar, or, equivalently, a parse tree. A more specific example grammar and a specific example of such a parse tree / production set would be useful here. –  Leonid Shifrin Mar 7 '12 at 14:31
For a real-world example of the kind you seem to look for, look at the code of the package SymbolicC` in the Mathematica distribution, particularly function GenerateCode, which generates a string of C code (more or less a sequence of terminal tokens, although they are not separately generated and the joined, but rather joined on the fly), from the Symbolic C representation, which is, in fact, a form of a parse tree for C code. Of course, C grammar is more complex than regular grammars, so this may be an overkill for your purposes. –  Leonid Shifrin Mar 7 '12 at 14:39
I assume that you did try using /. and //. with some Rules, but that does not solve the problem completely. Can you give a simple example where /. doesn't do what you need? I guess the key is in "What is of importance here is that the some functions (e.g. |, cf. Alternatives) should evaluate immediately to result in a (non)terminal symbol, while some fi should be held until termination." but I do not completely understand what you mean by that, and I also don't understand where held expressions come in when you use Replace(All) (as HoldAll doesn't affect replacements) –  Archimedes of Syracuse Mar 7 '12 at 14:41
If the problem is only that you need to prevent some f from evaluating until you reach the terminal point, then Block them during the //. phase: Block[{f1, f2}, expr //. repl]. –  Archimedes of Syracuse Mar 7 '12 at 14:47
"a" ~~ "b" seems to auto-evaluate to "ab", so be careful with ~~ –  Archimedes of Syracuse Mar 7 '12 at 18:34
show 5 more comments

3 Answers

up vote 7 down vote accepted

The following code defines the function randomSentence. It uses recursive descent to generate a sentence from a grammar expresssed as rules:

ClearAll[randomSentence, randomCount]
randomCount[] := RandomVariate[GeometricDistribution[0.5]]

SetAttributes[randomSentence, HoldAll]
randomSentence[rules_, expr_] :=
  , SetAttributes[generate, HoldAll]
  ; Replace[#[[1, 1]] :> #& /@ GatherBy[rules, First], (_ :> r_) :> r, {3}] /.
      (a_ :> {b___}) :> (generate[a] := generate[Alternatives[b]])
  ; generate[a_Alternatives] :=
      generate @@ RandomChoice[List @@ Hold /@ Unevaluated @ a]
  ; generate[(r:(Repeated|RepeatedNull))[e_]] :=
      Hold @@ Evaluate @ ConstantArray[0, randomCount[] + Boole[r===Repeated]] /.
        0 :> generate[e] /.
        Hold -> Composition[generate, List]
  ; generate[l_List] := StringJoin @@ generate /@ Unevaluated @ l
  ; generate[x:_[___]] := generate /@ Unevaluated @ x
  ; generate[x_] := x
  ; generate[expr]

Here is the species grammar, adapted (sic) to the form required by randomSentence:

ClearAll[f, g, h, i]
f[x___] := x <> "er";
g[x___] := x <> "ed";
h[x___] := x <> "ing";
i[x___] := x <> "y";

$rules =
  { "Species" :> "Animal" | "Plant"
  , "Species" :> f@"Action"
  , "Species" :> {"Color" | "Type", "-", "Species"}

  , "Attribute" :> "Type" | "Color"
  , "Attribute" :> {"Animal", "-", f@"Action"}
  , "Attribute" :> {"Color" | "Type", "-", g@"Part"}
  , "Attribute" :> i@"Plant"
  , "Attribute" :> {"Plant" | "Animal", "-", h@"Action"}

  , "Animal" :> "warble"|"shrew"|"whale"|"caiman"|"babuin"|"bat"|"bug"
  , "Plant" :> "bush"|"moss"|"fern"|"grass"|"squash"|"seed"
  , "Part" :> "back"|"head"|"finger"|"tail"|"ear"|"wing"|"thorn"
  , "Color" :> "black"|"red"|"white"|"blue"|"silver"|"crimson"|"dark"
  , "Type" :>  "long"|"cross"|"sharp"|"thick"|"heavy"|"fluffy"|"big"|"wild"
  , "Action" :> "jump"|"kill"|"stalk"|"sting"|"climb"|"crawl"|"eat"

Here is a sample use of the function (note that there is no need to explicitly specify the terminals, non-terminals and functions):

Table[randomSentence[$rules, {{"Attribute", " "}..., "Species"}], {10}] // Column

bat-climbing moss
squashy heavy-killer
squashy white-fingered heavy-red-whale
squashy thick-winged stinger

How It Works

The trickiest part of this problem is to make sure that none of the function expressions in the grammar are evaluated until they are needed -- and even then they may require some preprocessing of their arguments. To make this possible, randomSentence must not evaluate the rules or sentence form passed to it:

SetAttributes[randomSentence, HoldAll]
randomSentence[rules_, expr_] :=

We are going to use a helper function called generate. Again, it must not evaluate any expressions prematurely:

  , SetAttributes[generate, HoldAll]

generate will be able to convert any grammar expression into a string. There are many types of grammar expressions. First, we will teach generate how to deal with each of the non-terminal symbols. We group all of the rules into lists, one for each nonterminal. Then we convert each of those groups into definitions for generate as if they had been specified using Alternatives in the original grammar:

  ; Replace[#[[1, 1]] :> #& /@ GatherBy[rules, First], (_ :> r_) :> r, {3}] /.
      (a_ :> {b___}) :> (generate[a] := generate[Alternatives[b]])

Alternatives expressions are processed by selecting a random alternative from the list and then applying generate to that choice:

  ; generate[a_Alternatives] :=
      generate @@ RandomChoice[List @@ Hold /@ Unevaluated @ a]

Repeated expressions are processed by generating the repeated expression a random number of times. RepeatedNull allows zero occurrences whereas Repeated will have at least one. The tortured logic in this operation is due to the need to make sure that each repetition is not evaluated before it has been duly interpreted as a grammar expression. Also, it is important to ensure that each repetition is generated independently:

  ; generate[(r:(Repeated|RepeatedNull))[e_]] :=
      Hold @@ Evaluate @ ConstantArray[0, randomCount[] + Boole[r===Repeated]] /.
        0 :> generate[e] /.
        Hold -> Composition[generate, List]

Each grammar expression in a list is evaluated independently and then the results are joined together:

  ; generate[l_List] := StringJoin @@ generate /@ Unevaluated @ l

Any function call must be made after the arguments have been individually generated:

  ; generate[x:_[___]] := generate /@ Unevaluated @ x

Anything else is passed unchanged (presumably strings):

  ; generate[x_] := x

Now that generate is defined, all that remains is to use it:

  ; generate[expr]

randomSentence uses the helper function randomCount to generate a random repetition counts. For this example, we are using a geometric distribution where the probability of each successive count is half that of its predecessor. Adjust this distribution to suit your taste.

randomCount[] := RandomVariate[GeometricDistribution[0.5]]
share|improve this answer
add comment

Since no full solution has arrived, I've created the code based on the different sources you have provided.

The function applyRulesRepeated scans through the initial sentence and replaces the first nonterminal symbol it founds while holding all external functions (f, g, ..., for suffixing or alike). Here I used the holding part of Archimedes' answer. This replacement is done one-by-one for each found nonterminal. At the end, when there are only terminals, I substitute these terminals with random choices from the appropriate lexicon variables. I refer to the variables by their symbolname (in $Replacements, solution given by WReach). In the full version, I use a specific function to choose the words by removing them from the original list to prevent repetition (a specific RandomChoice with HoldAll attribute), thus it was necessary to rely on such complicated methods of word-choice (I actually used the Trott-Strzebonski method in the more complicated version). Also I reverted rules to use List instead of StringExpression due to the fact that "a"~~"b" evaluates to "ab" which is unwanted (thanks Archimedes). I was also inspired by Daniel Lichtblau's parser and its more evolved version in Sal Mangano's cookbook.

Here it is, presented as a toy experiment with species names, I hope you will enjoy playing with/extending it:

Regular grammar producing hypothetic animal names

(* Lexicon of different types of words *)
animal = {"warble", "shrew", "whale", "caiman", "babuin", "bat", "bug"};
plant  = {"bush", "moss", "fern", "grass", "squash", "seed"};
part   = {"back", "head", "finger", "tail", "ear", "wing", "thorn"};
color  = {"black", "red", "white", "blue", "silver", "crimson", "dark"};
type   = {"long", "cross", "sharp", "thick", "heavy", "fluffy", "big", "wild"};
action = {"jump", "kill", "stalk", "sting", "climb", "crawl", "eat"};

(* symbols of the grammar *)
$Terminals = {"animal", "plant", "color", "part", "type", "action"};
$NonTerminals = {"Species", "Attribute"};

(* functions not to be evaluated during sentence-generation *)
f[x___] := x <> "er";
g[x___] := x <> "ed";
h[x___] := x <> "ing";
i[x___] := x <> "y";
$Functions = {f, g, h, i};

(* initial sentence specification *)
$Sentence = "S" :> {{"Attribute", " "} .., "Species"};

(* allowed replacements in the grammar *)
$Rules = {
   "Species" :> "animal" | "plant",
   "Species" :> f@"action",
   "Species" :> {"color" | "type", "Species"},

   "Attribute" :> "type" | "color",
   "Attribute" :> {"animal", f@"action"},
   "Attribute" :> {"color" | "type", g@"part"},
   "Attribute" :> i@"plant",
   "Attribute" :> {"plant" | "animal", h@"action"}

(* define replacements for injecting actual words into sentence *)
$Replacements = # -> (ToExpression[#, InputForm, Hold] /. Hold[v_] :> Hold@RandomChoice@v) & /@ $Terminals;

(* parameters *)
$MaxRecursion = 100; (* max number of repeats when resolving Repeated(Null) *)
$MaxIteration = 2; (* max number of recursions when resolving nonterminals *)

(* Aux. function to replace only the first n occasion for each rule, successively *)
replaceEach[expr_, rep_List, args___] := Fold[replaceEach[#1, #2, args] &, expr, rep];
replaceEach[expr_, rule_[lhs_, rhs_], level_: {0, \[Infinity]}, n_: 1,
     opts : OptionsPattern[]] /; MemberQ[{Rule, RuleDelayed}, rule] :=
   ReplacePart[expr, rule[#, rhs] & /@ Position[expr, lhs, level, n, opts]];

(* applyRulesRepeated replaces nonterminals one-by-one in a recursive manner *)
applyRulesRepeated[pattern_] := applyRulesRepeated[pattern, 1];
applyRulesRepeated[pattern_, level_] := 
  Module[{patt, terminals, terminalReplace, nonTerminals, 
    nonTerminalReplace, temp},

   (* resolve : |, .., ... *)
   patt = pattern /. {
      (Verbatim@Alternatives)[x__] :> RandomChoice@List@x,
      (Verbatim@Repeated)[x_] :> Table[x, {RandomInteger@{1, $MaxIteration}}],
      (Verbatim@RepeatedNull)[x_] :> Table[x, {RandomInteger@{0, $MaxIteration}}]

   (* find FIRST nonterminal symbol in sentence *)
   nonTerminals = 
    Cases[patt, _?(MemberQ[$NonTerminals, #] &), \[Infinity], 1];

   (* if no nonterminal was found (or max recursion level is reached) sentence is cosidered ready *)
   If[nonTerminals === {} \[Or] level === $MaxRecursion,
    (* if sentence is ready, replace terminals with actual words and return sentence AND chosen words *)
    terminals = Cases[patt, _?(MemberQ[$Terminals, #] &), \[Infinity]];
    terminalReplace = 
     Thread[terminals -> ReleaseHold[terminals /. $Replacements]];
    ReleaseHold@replaceEach[patt, terminalReplace]
    (* if there is still any unresolved nonterminal, use a rule on it *)
    nonTerminalReplace = 
     Thread[nonTerminals -> (nonTerminals /. 
          Cases[$Rules, _?(MatchQ[First@nonTerminals, First@#] &)])];
     replaceEach[patt, nonTerminalReplace, {0, \[Infinity]}, 1], 
     level + 1]

(* Apply rules repeatedly while the necessary functions are held by Block *)
generateSentence[init_, term_, nonTerm_, rules_, {held___}] := 
  StringJoin@Block[{held}, applyRulesRepeated@Last@init];

Let's generate some random sentences:

    generateSentence[$Sentence, $Terminals, $NonTerminals, $Rules, $Functions],
crimson warble
fluffywinged ferny caiman
bugkiller shrew
whiteeared darkbat
grassclimbing seedy bug
thick whalekilling crimsonmoss
redwinged babuinstinger caiman
whaleeater whale
fluffy warblejumper babuin
caimankiller squash

As you can see, the output is sometimes nonsensical, sometimes quite realistic, but there was no semantic consideration behind the project. Each sentence is well-formed and complies to the (implicit) rules of the grammar. It is nice to realize though that (whale-)cannibalism is simply nature's way of recursion.

share|improve this answer
That's what killer whales will do, given the opportunity. youtube.com/watch?v=K0YpLG88SFc news.nationalgeographic.com/news/2003/03/… –  Daniel Lichtblau Mar 8 '12 at 19:28
add comment

Does this help?

First, let's get rid of StringExpression because it acts as StringJoin and messes things up.

fix[expr_] := expr /. StringExpression -> List

initSentence = fix[initSentence]
rules = fix[rules]

I used initSentence = {g["A"], "B"} here.

Then let's gather the rules into Alternatives expressions:

gatherRules[rules_] := #[[1, 1]] -> Alternatives @@ #[[All, 2]] & /@ 
  GatherBy[rules, First]

define Alternatives temporarily to be RandomChoice (which I don't like, but | is indeed a convenient notation), and do a replacement:

repl[initSentence_, rules_, {heldFunctions___}] :=
 Block[{Alternatives = RandomChoice[{##}] &, heldFunctions},
  initSentence /. gatherRules[rules]

repl[initSentence, rules, {f,g}]

This will prevent f and g from evaluating util the replacement is done.

We can also apply the replacements several times:

Nest[repl[#, rules, {}] &, initSentence, 10]

The reason I didn't use //. is that once the expression gets big, it's very likely it'll never reach a terminal state.


If you want a function to evaluate in some rules, and not evaluate in other rules, then instead of this solution you can wrap them in some head to prevent evaluation, i.e. write H[f]["a","b"] instead of f["a", "b"]. Then when you are ready to evaluate them, just /. H -> Identity.

share|improve this answer
I was just about to write up something like this. Termination is a bit tricky, maybe have repl throw something when initSentence contains no non-terminal symbols using MemberQ? –  Timo Mar 7 '12 at 21:53
Thanks Archimedes, I made good use of your repl function. –  István Zachar Mar 8 '12 at 18:55
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.