Parsing Expression Grammar (PEG) is a way to define a recursive-descent parser with limited backtracking. Its properties are useful in many applications. In spite of its apparent similarity to Extended Backus-Naur Form (EBNF), PEG often defines quite a different language, sometimes difficult to describe exactly. However, a recent result by Medeiros shows that an EBNF grammar having the LL(1) property can be transcribed verbatim into a PEG that not only defines the same language, but also parses the input in exactly the same way. We show that such transcription is possible for a wider class of EBNF grammars, which is interesting because the backtracking of PEG is often a convenient way to circumvent just the LL(1) restriction.
alternatives of Literal by looking at any predefined number of characters ahead. But, treated as a PEG parser, this grammar recognizes exactly the language defined by its EBNF interpretation.
To see how it happens, suppose the PEG parser is presented with the input
”10B”. It performs a series of calls Literal → DecimalLiteral → [
In the following, we try to answer this question: given an EBNF grammar, when can I transcribe it directly into PEG and obtain a correct parser? 2
We consider a finite alphabet of letters. A finite string of letters is a word. The string of 0 letters is called the empty word and is denoted by ". The set of all words is ∗. A subset of ∗ is a language.
As usual, we write XY to mean the concatenation of languages X and Y , that is, the set of all words xy where x ∈ X and y ∈ Y .
For x; y ∈ ∗, x is a pre x of y if y = xu for some u ∈ ∗. We write x ≤ y to mean that x is a prefix of y. For X ⊆ ∗, the set of all prefixes of x ∈ X is denoted by Pref(X).
A relation R on E is a subset of E × E. As usual, we write R(e) to mean the set of all e′ such that (e; e′) ∈ R, and R(E) to mean the union of R(e) for all e ∈ E ⊆ E. The transitive and reflexive closure of R is denoted by R∗, and the product of relations R and S by R × S. 3
We consider a grammar G over the alphabet that will be interpreted as either PEG or EBNF. The grammar is a set of rules of the form A 7→ e, where e is a syntax expression and A is the name given to it. The expression e has one of these forms: { e1 e2 (Sequence), { e1|e2 (Choice), where each of e1; e2 is an expression name, a letter from , or ". The set of all names appearing in the rules is denoted by N . For A ∈ N , e(A) is the expression e in A 7→ e. We define E = N ∪ ∪ ".
When G is interpreted as PEG, the rule A 7→ e is a definition A ← e of parsing expression e, and e1|e2 is the ordered choice. When G is interpreted as EBNF, the rule is a production A → e, and e1|e2 is the unordered choice. For obvious reasons, we assume G to be free from left recursion.
This grammar is reduced to bare bones in order to simplify the reasoning. Any full EBNF grammar or PEG without predicates can be reduced to such form by steps like these: { Replacing [a1-an] by a1|a2| : : : |an; { Rewriting e1e2e3 and e1|e2|e3 as e1(e2e3) respectively e1|(e2|e3) and introducing new names for expressions in parentheses; { Replacing e+ by ee∗; { Replacing e∗ by A 7→ eA | ". 3.1
The PEG Interpretation When G is interpreted as PEG, the elements of E are parsing procedures that can call each other recursively. In general, parsing procedure is applied to a string from ∗ at a position indicated by some ”cursor”. It tries to recognize a portion of the string ahead of the cursor. If it succeeds, it ”consumes” the recognized portion by advancing the cursor and returns ”success”; otherwise, it returns ”failure” and does not consume anything (does not advance the cursor). The action of different procedures is as follows: { ": Indicate success without consuming any input. { a ∈ : If the text ahead starts with a, consume it and return success.
Otherwise return failure. { A 7→ e1 e2: Call e1. If it succeeded, call e2 and return success if e2 succeeded.
If e1 or e2 failed, reset cursor as it was before the invocation of e1 and return failure. { A 7→ e1|e2: Call e1. Return success if it succeeded. Otherwise call expression e2 and return success if e2 succeeded or failure if it failed.
Backtracking occurs in the Sequence expression. If e1 succeeds and consumes some input, and then e2 fails, the cursor is moved back to where it was before trying e1. If this Sequence was called as the first alternative in a Choice, Choice has an opportunity to try another alternative on the same input. However, the backtracking is limited: once e1 in the Choice e1|e2 succeeded, e2 will never be tried on the same input, even if the parse fails later on.
Following [
The relation itself is defined using the method of ”natural semantics”: [e] x PEG X
holds if and only if it can be proved using the inference rules in Figure 1. The
rules come from [
The proof tree of [e] x PEG X, when followed from bottom up, mimics procedure calls in the process of parsing the string x. Rule seq.1p with Z = fail corresponds to backtracking in Sequence, and rule choice.2p to calling e2 after e1 failed.
According to Lemma 2.3.1 in [
(empty.1p) (letter.2p) (seq.1p) (seq.2p) (choice.1p) (choice.2p) ["] x PEG x
b ̸= a [b] ax PEG fail (letter.1p) (letter.3p) [a] ax PEG x [a] " PEG fail [e1] xyz PEG yz [e2] yz PEG Z e(A) = e1e2
[A] xyz PEG Z [e1] x PEG fail e(A) = e1e2
[A] x PEG fail [e1] xy PEG y e(A) = e1|e2
[A] xy PEG y [e1] x PEG fail [e2] xy PEG Y e(A) = e1|e2
[A] xy PEG Y where Y denotes y or fail and Z denotes z or fail. The grammar G with ”7→” interpreted as ”→” corresponds quite exactly to the traditional Context-Free Grammar (CFG), so we shall speak of its interpretation as CFG. Traditionally, CFG is a mechanism for generating words: { " generates empty word. { a ∈ generates itself. { A 7→ e1e2 generates any word generated by e1 followed by any word generated by e2.
{ A 7→ e1|e2 generates any word generated by e1 or e2.
The language L(e) of e ∈ E is the set of all words that can be generated by e. (empty.1c) (seq.1c) (choice.1c) (choice.2c) ["] x CFG x [e1] xyz CFG yz [e2] yz CFG z e(A) = e1e2 [A] xyz CFG z (letter.1c)
[a] ax CFG x [e1] xy CFG y e(A) = e1|e2
[A] xy CFG y [e2] xy CFG y e(A) = e1|e2
[A] xy CFG y
Following [
The proof tree of [e] x CFG y, when followed from bottom up, mimics procedure calls in an imagined recursive-descent parser processing the string x. The procedures correspond to elements of E, and the procedure for Choice always chooses the correct alternative, either by full backtracking or by an oracle.
It can be verified that L(e) = {x ∈ ∗ : [e] xy CFG y for some y ∈ ∗}. It can also be verified that if x ∈ L(e), [e] xy CFG y holds for every y ∈ ∗. 4
The following has been proved as Lemma 4.3.1 in [
Proof. We spell out the proof sketched in [
By induction hypothesis, [e1] xy CFG y, so [A] xy CFG x follows from choice.1c. { [A] xy PEG x derived from [e1] xy PEG fail, [e2] xy PEG y, and e(A) = e1e2 using choice.2p. By induction hypothesis, [e2] xy CFG y, so [A] xy CFG x follows from choice.2c. ⊓⊔
Known examples show that the reverse of Proposition 1 is, in general, not true. We intend to formulate some conditions under which the reverse does hold. The problem is complicated by another known fact, namely that [e] xy PEG y does not imply [e] xy PEG y′ for every y′ ∈ ∗: the action of e depends on the string ahead. We have to consider the action of e in relation to the entire input string.
Let $, the end-of-text marker, be a letter that does not appear in any rules. Define some S ∈ N as the starting rule of the grammar. As L(e) does not depend on the input ahead, we have [S] w$ CFG $ if and only if w ∈ L(S). We note that every partial result in the proof tree of [S] w$ CFG $ has the form [e] xy$ CFG y$ for some e ∈ E and x; y ∈ ∗.
For A ∈ N , define Tail(A) to be any string that appears ahead of the ”CFG parser” just after it completed the call to A while processing some w ∈ L(S). Formally, y$ ∈ Tail(A) if there exists a proof tree for [S] w$ CFG $ that contains [A] xy$ CFG y$ as a partial result.
L(e1) ∩ Pref(L(e2) Tail(A)) = ∅; (1) there exists a proof of [S] w$ PEG $ for each w ∈ L(S). Moreover, for every partial result [e] xy$ CFG y$ in the proof tree of [S] w$ CFG $ there exists a proof of [e] xy$ PEG y$.
Proof. Take any w ∈ L(S). It means there exists proof tree for [S] w$ CFG $. We are going to show that for each partial result [e] xy$ CFG y$ in that tree there exists a proof of [e] xy$ PEG $. We show it using induction on the height of the proof tree. (Induction base) Suppose the proof of the partial result [e] xy$ CFG y$ consists of one step. Then it has to be the proof of ["] x$ CFG x$ or [a] ax$ CFG x$ using empty.1c or letter.1c, respectively. But then, ["] x$ PEG x$ respectively [a] ax$ PEG x$ by empty.1p or letter.1p. (Induction step) Assume that for every partial result [e] xy$ CFG y$ that has ppraortoifaltrreeesuolft h[ee]igxhyt$ nCF≥G y1$ twheitrhe pexroisotfs tareperoofofheoifgh[et] nx+y$1P.EIGts yl$a.stCsotnepsidmeursat be one of these: { [A] xyz$ CFG z$ derived from [e1] xyz$ CFG yz$, [e2] yz$ CFG z$, and e(A) = e1e2 using seq.1c. By induction hypothesis, [e1] xyz$ PEG yz$ and [e2] yz$ PEG z$, so [A] xyz$ PEG z$ follows from seq.1p. { [A] xy$ CFG y$ derived from [e1] xy$ CFG y$ and e(A) = e1|e2 using choice.1c. By induction hypothesis, [e1] xy$ PEG y$, so [A] xy$ PEG y$ follows from choice.1p. { [A] xy$ CFG y$, derived from [e2] xy$ CFG y$ and e(A) = e1|e2 using choice.2c. By induction hypothesis, [e2] xy$ PEG y$. But, to use choice.2p we also need to verify that [e1] xy$ PEG fail.
Assume that (1) holds for A and suppose that there is no proof of [e1] xy$ PEG fail. According to Lemma 1, there exists a proof of [e1] uv$ PEG v$ where uv$ = xy$. According to Proposition 1, there exists a proof of [e1] uv$ CFG v$, so u ∈ L(e1). From [e2] xy$ CFG y$ follows x ∈ L(e2). From [A] xy$ CFG y$ follows y$ ∈ Tail(A). From uv$ = xy$ follows u ≤ xy, so u ∈ Pref(L(e2) Tail(A)), which contradicts (1). We must thus conclude that there exists a proof of [e1] xy$ PEG fail, so there exists a proof of [A] xy$ PEG y$ using choice.2p.
The proof of [S] w$ PEG $ exists as a special case of the above.
⊓⊔ Proposition 3. The two interpretations of G are equivalent if every Choice A 7→ e1|e2 ∈ G satis es (1). They are equivalent not only by accepting the same language, but also by parsing the input in the same way.
Note that the condition is sufficient, but not necessary. Unfortunately, (1) is
not easy to check. The languages appearing in it are context-free languages. The
problems of inclusion and emptiness of intersection are in general undecidable
for these languages, so one has to consider special cases, or use a conservative
approximation. In fact, condition (1) is identical to the ”general semi disjointness”
considered by Schmitz in [
As it will be seen, the LL(1) property implies (1), which accounts for the
result obtained in [
nonempty word in X has a nonempty prefix in Y . Lemma 2. For any X; Y; Z ⊆ ∗: (a) If X ⊑ Y and Y ⊑ Z then X ⊑ Z. (b) If PX ⊑ X and PY ⊑ Y then PX ∪ PY ⊑ X ∪ Y . (c) If " ∈= X and PX ⊑ X then PX ⊑ XY . (d) If " ∈ X, PX ⊑ X, and PY ⊑ Y then PX ∪ PY ⊑ XY . ∗, and write Y ⊑ X if each
(b) If x ̸= " is in X, it has a nonempty prefix in PX . If it is in Y , it has a nonempty prefix in PY . (c) If " ∈= X, each x ∈ XY has a nonempty prefix in X This, in turn, has a nonempty prefix in PX . (d) If " ∈ X, we have XY = (X − ")Y ∪ Y . From (c) we have PX ⊑ (X − ")Y . The stated result follows from (b). ⊓⊔ 6
We say that X ⊆ ∗ and Y ⊆ ∗ are pre x-disjoint, and write X ≍ Y to mean that for all nonempty x ∈ X and y ∈ Y neither x ≤ y nor y ≤ x.
One can easily see that prefix-disjoint languages are disjoint, but the reverse is in general not true.
Lemma 3. Let X ⊆ + and Y ⊆ implies X ∩ Pref(Y ) = ∅.
∗. For any PX ⊑ X and PY ⊑ Y , PX ≍ PY Proof. Let X; Y; PX ; PY be as stated. Suppose that X ∩ Pref(Y ) ̸= ∅, so there exists x such that x ∈ X, x ∈ Pref(Y ). Since x ̸= ", we have x = uv for some nonempty u ∈ PX and v ∈ ∗. As x ∈ Pref(Y ), we have xt ∈ Y for some t ∈ ∗. As xt ̸= ", we have xt = pq for some nonempty p ∈ PY ; q ∈ ∗. This gives uvt = pq; this means either u ≤ p or p ≤ u, which contradicts PX ≍ PY .
⊓⊔ Proposition 4. The two interpretations of G are equivalent if for every Choice A 7→ e1|e2 ∈ G: { " ∈= L(e1) { There exist P1 ⊑ L(e1) and P2 ⊑ L(e2) Tail(A) such that P1 ≍ P2. Proof. Consider any Choice A 7→ e1|e2 ∈ G. Assume " ∈= L(e1) and P1, P2 as stated. By Lemma 3, we have L(e1) ∩ Pref(L(e2) Tail(A)) = ∅, and the stated result follows from Proposition 3. ⊓⊔ Proposition 5. The two interpretations of G are equivalent if for every Choice A 7→ e1|e2 ∈ G: { " ∈= L(e1). { If " ∈= L(e2), there exist P1 ⊑ L(e1) and P2 ⊑ L(e2) such that P1 ≍ P2. { If " ∈ L(e2), there exist P1 ⊑ L(e1), P2 ⊑ L(e2), and PT ⊑ Tail(A) such that P1 ≍ (P2 ∪ PT ).
Proof. Consider any Choice A 7→ e1 | e2 ∈ G. Assume " ∈= L(e1), P1 ⊑ L(e1) and P2 ⊑ L(e2). Suppose " ∈= L(e2). By Lemma 2(c), P2 ⊑ L(e2) Tail(A), so P1 ≍ P2 satisfies Proposition 4. Suppose now that " ∈ L(e2) and PT ⊑ Tail(A). By Lemma 2(d), P2∪PT ⊑ L(e2) Tail(A), so P1 ≍ P2∪PT satisfies Proposition 4. ⊓⊔ 7
One can easily see that the classical First(e) and Follow(A) are prefix covers of e and Tail(A) that consist of one-letter words. For such sets, X ≍ Y is the same as X ∩Y = ∅, so First(e1) ∩ First(e2) = ∅ and First(e1) ∩ (First(e2)∪ Follow(A)) = ∅ are special cases of the last two conditions in Proposition 5. They are exactly the LL(1) conditions.
We are now going to look at other sets that may be used instead of First(e) and Follow(A). For A ∈ N define: { First(A) = {e1; e2} if e(A) = e1| e2, { First(A) = {e1} if e(A) = e1 e2 and " ∈= L(e1), { First(A) = {e1; e2} if e(A) = e1 e2 and " ∈ L(e1).
For E ⊆ E, let F IRST (E) be the family of subsets of E defined as follows: { E ∈ F IRST (E). { If F belongs to F IRST (E), the result of replacing its member A ∈ N by all elements of First(A) also belongs to F IRST (E). { nothing else belongs to F IRST (E) unless its being so follows from the above.
These definitions are illustrated in Figure 3. An arrow from e1 to e2 means e2 ∈ First(e1). One can easily see that F IRST (E) contains the classical First(E).
A 7→ C | D B 7→ E A C 7→ a | b D 7→ b | X E 7→ d | Y X 7→ c D Y 7→ c E FIRST (A) = {{A}; {C; D}; {a; b; D}; {C; b; X}; {a; b; X}; {a; b; c}} FIRST (B) = {{B}; {E}; {d; Y }; {d; c}}
Lemma 4. For each A ∈ N holds L(First(A)) ⊑ L(A).
Proof. In case e(A) = e1| e2 we have L(A) = L(e1) ∪ L(e2) = L(e1| e2). Each language is its own prefix cover, so L{e1; e2} ⊑ L(e1| e2).
In case e(A) = e1 e2 we have L(A) = L(e1)L(e2). Again, L(e1) ⊑ L(e1) and L(e2) ⊑ L(e2).
If " ∈= e1, we have, by Lemma 2(c), L{e1} ⊑ L(e1)L(e2).
If " ∈ e1, we have, by Lemma 2(d), L{e1; e2} = L(e1) ∪ L(e2) ⊑ L(e1)L(e2). ⊓⊔ Lemma 5. For each F ∈ F IRST (E) holds L(F ) ⊑ L(E).
Proof. We prove the Proposition by induction on the number of replacements. (Induction base) As each language is its own prefix cover, L(E) ⊑ L(E). (Induction step) Assume L{e1; e2; : : : ; en} ⊑ L(E). As sets are not ordered, we can always consider replacing e1. Let First(e1) = {f1; f2}. (The proof will be similar if First(e1) has only one element.) According to Lemma 4, L{f1; f2} ⊑ L(e1). We also have L{e2; : : : ; en} ⊑ L{e2; : : : ; en}. Using Lemma 2(b), we obtain:
L{f1; f2; e2; : : : ; en} = L{f1; f2} ∪ L{e2; : : : ; en}
⊑ L(e1) ∪ L{e2; : : : ; en} = L{e1; e2; : : : ; en}: From Lemma 2(a) follows L{f1; f2; e2; : : : ; en} ⊑ L(E).
We can apply this result directly to "-free grammars: Proposition 6. If none of the rules contains ", the two interpretations of G are equivalent if for every Choice A 7→ e1|e2 ∈ G there exist First1 ∈ F IRST (e1) and First2 ∈ F IRST (e2) such that L(First1) ≍ L(First2). ⊓⊔ Proof. Suppose the required First1 and First2 exist. If G is "-free, we have " ∈= L(e1) and " ∈= L(e2). From Lemma 5 we have L(First1) ⊑ L(e1) and L(First2) ⊑ L(e2). The stated result follows from Proposition 5. ⊓⊔ 8
To handle the case when G is not "-free, we need to find suitable prefix covers of
Tail(A). For this purpose, we borrow from [
For e ∈ E, define Next(e) = {e1 ∈ E : exists A 7→ ee1 ∈ G}.
Finally, define Follow = Last∗ × Next.
Lemma 6. For each A ∈ N , L(Follow(A)) ⊑ Tail(A).
Proof. Consider some A ∈ N and y$ ∈ Tail(A). By definition, there is a proof of [S] w$ CFG $ that contains [A] xy$ CFG y$ as one of the partial results. This partial result must be used in a subsequent derivation. This derivation can only result in one of the following: (a) [A1] xy$ CFG y$ where e(A1) = A | e from [e] xy$ CFG y$ using choice.1c. (b) [A1] xy$ CFG y$ where e(A1) = e |A from [e] xy$ CFG y$ using choice.2c. (c) [A1] zxy$ CFG y$ where e(A1) = e A from [e] zxy$ CFG xy$ using seq.1c. (d) [A1] xy$ CFG y$ where e(A1) = A e from [e] "y$ CFG y$ using seq.1c. (e) [A1] xy′y′′$ CFG y′′$ where e(A1) = A B, y′y′′ = y and y′ ̸= " from [B] y′y′′$ CFG y′′$ using seq.1c.
In each of the cases (a)-(d), the result is similar to the original one, and the alternative derivations (a)-(e) apply again. We may have a chain of derivations (a)-(d), but it must end with (e) as y must eventually be reduced. We have thus in general a sequence of steps of this form: [A0] xy$ CFG y$ : : : as in (a)-(d) [A1] x1y$ CFG y$ : : : as in (a)-(d) [A2] x2y$ CFG y$ : : : as in (a)-(d)
: : : [An−1] xn−1y$ CFG y$ : : : as in (a)-(d) [An] xny$ CFG y$ [B] y′y′′$ CFG y′′$ e(An+1) = AnB
[An+1] xny′y′′$ CFG y′′$ where A0 = A and n ≥ 0. We have A1 ∈ Last(A0), A2 ∈ Last(A1), etc., An ∈ Last(An−1), and B ∈ Next(An), which means B ∈ Follow(A). From [B] y′y′′$ CFG y′′$ we have y′ ∈ L(B); since y = y′y′′ and y′ ̸= ", y$ has a nonempty prefix in L(B) ⊆ L(Follow(A)). ⊓⊔ Proposition 7. The two interpretations of G are equivalent if for every Choice A 7→ e1|e2 ∈ G: { " ∈= L(e1). { If " ∈= L(e2), there exist First1 ∈ F IRST (e1) and First2 ∈ F IRST (e2) such that L(First1) ≍ L(First2). { If " ∈ L(e2), there exist First1 ∈ F IRST (e1), First2 ∈ F IRST (e2), and Follow ∈ F IRST (Follow(A)) such that L(First1) ≍ L(First2 ∪ Follow).
Proof. Suppose the required First1, First2, and Follow exist. From Lemma 5 we have L(First1) ⊑ L(e1) and L(First2) ⊑ L(e2). From Lemmas 5 and 6 we have L(Follow) ⊑ Tail(A). The stated result follows from Proposition 5. ⊓⊔ 9
We have shown that a direct transcription of an EBNF grammar to equivalent PEG is possible for grammars outside the LL(1) class. These are the grammars where the choice of the way to proceed is made by examining the input within the reach of one parsing procedure instead of examining a single letter. One can call them ”LL(1P)” grammars, the ”1P” meaning ”one procedure”. However, checking the conditions stated by Proposition 7 is not as simple as verifying the LL(1) property. The families F IRST can be constructed in a mechanical way, and are presumably not very large. But, checking the disjointness of their members may be difficult. It becomes more difficult as one moves up the graph in Figure 3; at the top of that graph we come close to the condition (1).
If the sets First1, First2, and Follow satisfying Proposition 7 exist, the
information can be used for the improvement suggested by Mizushima et al.
in [
The class of EBNF grammars that can be directly transcribed to PEG includes cases where the choice of the way to proceed is made by looking at the input within the reach of more than one parsing procedure. The following is an example of such grammar:
InputLine → Assignment | Expression Assignment → Name "=" Expression Expression → Primary ([+-] Primary)∗ Primary → Name | Number Name → [a-z]+
Number → [
Last but not least: it appears that natural semantics, as used by Medeiros to define the meaning of a grammar, is a very convenient tool to investigate the grammar’s properties.