Abstract problem description:
The way I see it, unparsing means to create a token stream from an AST, which when parsed again produces an equal AST.
So parse(unparse(AST)) = AST holds.
This is the equal to finding a valid parse tree which would produce the same AST.
The language is described by a context free S-attributed grammar using a eBNF variant.
So the unparser has to find a valid 'path' through the traversed nodes in which all grammar constraints hold. This bascially means to find a valid allocation of AST nodes to grammar production rules. This is a constraint satisfaction problem (CSP) in general and could be solved, like parsing, by backtracking in O(exp(n)).
Fortunately for parsing, this can be done in O(n³) using GLR (or better restricting the grammar). Because the AST structure is so close to the grammar production rule structure, I was really surprised seeing an implementation where the runtime is worse than parsing: XText uses ANTLR for parsing and backtracking for unparsing.
Questions
Is a context free S-attribute grammar everything a parser and unparser need to share or are there further constraints, e.g. on the parsing technique / parser implementation?
I've got the feeling this problem isn't O(exp(n)) in general - could some genius help me with this?
Is this basically a context-sensitive grammar?
Example1:
Area returns AnyObject -> Pedestrian | Highway
Highway returns AnyObject -> "Foo" Car
Pedestrian returns AnyObject -> "Bar" Bike
Car returns Vehicle -> anyObjectInstance.name="Car"
Bike returns Vehicle -> anyObjectInstance.name="Bike"
So if I have an AST containing
AnyObject -> AnyObject -> Vehicle [name="Car"] and I know the root can be Area, I could resolve it to
Area -> Highway -> Car
So the (Highway | Pedestrian) decision depends on the subtree decisions. The problem get's worse when a leaf might be, at first sight, one of several types, but has to be a specific one to form a valid path later on.
Example2:
So if I have S-attribute rules returning untyped objects, just assigning some attributes, e.g.
A -> B C {Obj, Obj}
X -> Y Z {Obj, Obj}
B -> "somekeyword" {0}
Y -> "otherkeyword" {0}
C -> "C" {C}
Z -> "Z" {Z}
So if an AST contains
Obj
/ \
"0" "C"
I can unparse it to
A
/ \
B C
just after I could resolve "C" to C.
While traversing the AST, all constraints I can generate from the grammar are satisfied for both rules, A and X, until I hit "C". This means that for
A -> B | C
B -> "map" {MagicNumber_42}
C -> "foreach" {MagicNumber_42}
both solutions for the tree
Obj
|
MagicNumber_42
are valid and it is considered that they have equal semantics ,e.g. syntactic sugar.
Further Information:
unparsing in XText
grammar constraints for unparsing, see Serializer: Concrete Syntax Validation
Question 1: no, the grammar itself may not be enough. Take the example of an ambiguous grammar. If you ended up with a unique leftmost (rightmost) derivation (the AST) for a given string, you would somehow have to know how the parser eliminated the ambiguity. Just think of the string 'a+b*c' with the naive grammar for expressions 'E:=E+E|E*E|...'.
Question 3: none of the grammar examples you give is context sensitive. The lefthand-side of the productions are a single non-terminal, there is no context.
Related
I wanted to write a parser based on John Hughes' paper Generalizing Monads to Arrows. When reading through and trying to reimplement his code I realized there were some things that didn't quite make sense. In one section he lays out a parser implementation based on Swierstra and Duponchel's paper Deterministic, error-correcting combinator parsers using Arrows. The parser type he describes looks like this:
data StaticParser ch = SP Bool [ch]
data DynamicParser ch a b = DP (a, [ch]) -> (b, [ch])
data Parser ch a b = P (StaticParser ch) (DynamicParser ch a b)
with the composition operator looking something like this:
(.) :: Parser ch b c -> Parser ch a b -> Parser ch a c
P (SP e2 st2) (DP f2) . P (SP e1 st1) (DP f1) =
P (SP (e1 && e2) (st1 `union` if e1 then st2 else []))
(DP $ f2 . f1)
The issue is that the composition of parsers q . p 'forgets' q's starting symbols. One possible interpretation I thought of is that Hughes' expects all our DynamicParsers to be total such that a symbol parser's type signature would be symbol :: ch -> Parser ch a (Maybe ch) instead of symbol :: ch -> Parser ch a ch. This still seems awkward though since we have to duplicate information putting starting symbol information in both the StaticParser and DynamicParser. Another issue is that almost all parsers will have the potential to throw which means we will have to spend a lot of time inside Maybe or Either creating what is essentially the "monads do not compose problem." This could be remedied by rewriting DynamicParser itself to handle failure or as an Arrow transformer, but this is straying quite a bit from the paper. None of these issues are addressed in the paper, and the Parser is presented as if it obviously works, so I feel like I must me missing something basic. If someone can catch what I missed that would be super helpful.
I think the deterministic parsers described by Swierstra and Duponcheel are a bit different from traditional parsers: they do not handle failure at all, only choice.
See also the invokeDet function in the S&D paper:
invokeDet :: Symbol s => DetPar s a -> Input s -> a
invokeDet (_, p) inp = case p inp [] of (a, _) -> a
This function clearly assumes it will always be able to find a valid parse.
With the arrow version of the parsers described by Hughes you can write a examples like this:
main = do
let p = symbol 'a' >>> (symbol 'b' <+> symbol 'c')
print $ invokeDet p "ab"
print $ invokeDet p "ac"
Which will print the expected:
'b'
'c'
However, if you write a "failing" parse:
main = do
let p = symbol 'a' >>> (symbol 'b' <+> symbol 'c')
print $ invokeDet p "ad"
It will still print:
'c'
To make this behavior a bit more sensible, Swierstra and Duponcheel also introduce error-correction. The output 'c' is expected if we assume the erroneous character d has been corrected to be a c in the input. This requires an extra mechanism which presumably was too complicated to include in Hughes' paper.
I have uploaded the implementation I used to get these results here: https://gist.github.com/noughtmare/eced4441332784cc8212e9c0adb68b35
For more information about a more practical parser in the same style (but no longer deterministic and no longer limited to LL(1)) I really like the "Combinator Parsing: A Short Tutorial" by Swierstra. An interesting excerpt from section 9.3:
A subtle point here is the question how to deal with monadic parsers. As we described in [13] the static analysis does not go well with monadic computations, since in that case we dynamically build new parses based on the input produced thus far: the whole idea of a static analysis is that it is static. This observation has lead John Hughes to propose arrows for dealing with such situations [7]. It is only recently that we realised that, although our arguments still hold in general, they do not apply to the case of the LL(1) analysis. If we want to compute the symbols which can be recognised as the first symbol by a parser of the form p >>= q then we are only interested in the starting symbols of the right hand side if the left hand side can recognise the empty string; the good news is that in that case we statically know what value will be returned as a witness, and can pass this value on to q, and analyse the result of this call statically too. Unfortunately we will have to take special precautions in case the left hand side operator contains a call to pErrors in one of the empty derivations, since then it is no longer true that the witness of this alternative can be determined statically.
The full parser implementation by Swierstra can be found in the uu-parsinglib package, although I do not know how many of the extensions are implemented there.
Given a parser
newtype Parser a = Parser { parse :: String -> [(a,String)] }
(>>=) :: Parser a -> (a -> Parser b) -> Parser b
p >>= f = Parser $ \s -> concat [ parse (f a) s' | (a, s') <- parse p s ]
return :: a -> Parser a
return a = Parser (\s -> [(a,s)])
item :: Parser Char
item = Parser $ \s -> case cs of
"" -> []
(c:cs) -> [(c,cs)]
We can see that item consumes part of the input string given to it ("abc" -> [('a', "bc")]). Is there ever a case where a parser would produce additional string output or replace/modify it (e.g. Parser $ \s -> [((), 'a':s)])? I suspect that this might be the case with context-sensitive grammars but have trouble coming up with a sensible example.
Is there a reason why it would make sense to do this for a real-world problem?
References
Monadic Parsing in Haskell
Here are a couple of cases where it is convenient to inject tokens into the input stream. (How this is actually integrated into the parsing pipeline is another question.)
Macro expansion, in the style of the C/C++ preprocessing phase. This is arguably not the best model for macro expansion; hygienic macros would more likely be expanded using a tree transformation, as with C++ template resolution. But the token-oriented preprocessor is not going away soon. Since it is not tightly coupled with the language syntax, the easiest implementation is to substitute the macro (and arguments if applicable) with the tokens from its expansion.
Ecmascript-style automatic semi-colon insertion (ASI). The language syntax requires a semi-colon to be inserted into the token stream under certain precisely-defined circumstances, which are difficult (at least) to incorporate in a CFG. Since ASI is only possible if the next token in the input stream cannot be shifted (and done other conditions), it can certainly be integrated into the parser loop.
Similarly, indentation-aware block syntax (as in Haskell and Python, for example) can certainly be implemented by replacing leading whitespace with an injected INDENT token or some number of injected DEDENTs. Since this substitution is dependent on parse context (it isn't done inside parentheses, for example), injection inside the parser could be a reasonable approach.
That's not an exhaustive list, but it might be at least indicative. Not all of those cases necessarily involve context-sensitivity (I believe ASI could, in theory, be handled with a context-free grammar although I have no intention of trying) and not all instances of context-sensitivity necessarily require token injection (the ambiguity in C between type and variable names only requires selecting the correct token).
Showing that the reverse of a word for a regular language L is also regular
I am confused as to how I am to approach this question, i've been stuck for hours: For a word x, we use x^r to denote its reverse. For a language L, we use L^r to denote {x^r where x is in the set of L}. Show that if L is regular then so is L^r
If L is regular, then there exists some regular grammar which generates it. It can be always represented as either a left-regular grammar, or a right-regular grammar. Let's assume that it's left-regular grammar G_l(the proof for right-regular grammar is analogous).
This grammar has productions of two types; the terminating-type:
A -> a, where A is non-terminal and a is either a terminal or empty string (epsilon)
or the chaining type:
B -> Ca, where B, C are non-terminals and a is a terminal
When we apply reverse to a regular language, we basically also apply it to the tails of productions (since heads are just single non-terminals). It's going to be proved later on. So we get a new grammar G_r, with productions:
A -> a, where A is non-terminal and a is either a terminal or empty string (epsilon)
B -> aC, where B, C are non-terminals and a is a terminal
But hey, it's a right-regular grammar! So the language it accepts is also regular.
There is one thing to do - to show that reversing tails actually does the thing it's supposed to. We're going to prove it very simply:
If L contains \epsilon, then there is production 'S -> \epsilon' in G_l. Since we don't touch productions like that, it's also present in G_r.
If L contains a, a word composed of a single terminal, then it's similar to the above
If L contains aZ, where a is a terminal and Z is a word from the language constructed from chopping off the first terminals out of words in L, then L^r contains (because of changes to the chaining productions) (Z^r)a. Z is also a regular language, since it can be constructed by dropping the first "level" of left-productions from G_l, which leaves us with a regular grammar.
I hope it helped. There's also an arguably easier way of doing that by reversing edges of the relevant finite automata and changing accepting and entry states a bit.
Goal: find a way to formally define a grammar that recognizes elements from a set 0 or 1 times in any order. Subsequently, I want to parse it and generate an AST as well.
For example: Say the set of valid strings in my language is {A, B, C}. I want to define a grammar that recognizes all valid permutations of any number of those elements.
Syntactically valid strings would include:
(the empty string)
A,
B A, and
C A B
Syntactically invalid strings would include:
A A, and
B A C B
To be clear, defining all possible permutations explicitly in a CFG is unacceptable for my purposes, since larger sets would be impossible to maintain.
From what I understand, such a language fails the pumping lemma for context free languages, so the solution will not be context free or regular.
Update
What I'm after is called a "permutation language", which Benedek Nagy has done some theoretical work on as an extension to context free languages.
Regarding a parser generator, I've only found talk of implementing parsers with a permutation phase (link). Parsers evidently have an exponential lower bound on the size of resulting CFG, and I haven't found any parser generators that support it anyhow.
A sort-of solution to this problem was written in ANTLR. It uses semantic predicates to 'code around' the issue.
Assuming that the set of alternative strings is fixed and known in advance, say of size n, one can come up with a (non context-free) grammar of size O(n!). This is not asymptotically smaller than enumerating all permutations, so I suppose it cannot be considered a good solution. I believe that this grammar can be reformulated as a context-sensitive grammar (although in the form I'm suggesting below it is not).
For the example {a, b, c} mentioned in the question, one such grammar is the following. I'm using lower case letters for terminal symbols and upper case letters for non-terminals, as is customary. S is the initial non-terminal symbol.
S ::= XabcY
XabcY ::= aXbcY | bXacY | cXabY
XabY ::= ab | ba
XacY ::= ac | ca
XbcY ::= bc | cb
Non-terminals X and Y enclose the substring in the production which has not been finalized yet; this substring will eventually be replaced by a permutation of the terminals that are given between X and Y (in some arbitrary order).
Is there a (simple) way, within a parsing expression grammar (PEG), to express an "unordered sequence"? A rule such as
Rule <- A B C
requires A, B and C to match in order. A rule such as
Rule <- (A B C) / (B C A) / (C A B) / (A C B) / (C B A) / (B A C)
allows them to match in any order (which is what we want) but it is cumbersome and inapplicable in practice with more terms in the sequence.
Is the only solution to use a syntactically looser rule such as
Rule <- (A / B / C){3}
and semantically check that each rule matches only once?
The fact that, e.g., Relax NG Compact Syntax has an "unordered list" operator to parse XML make me hint that there is no obvious solution.
Last question: do you think the addition of such an operator would bring ambiguity to PEG?
Grammar rules express precisely the sequence of forms that you want, regardless of parsing engine (e.g., PEG, LALR, LL(k), ...) that you choose.
The only way to express that you want all possible sequences of just of something using BNF rules is the big ugly rule you proposed.
The standard solution is to simply define:
rule <- (A | B | C)*
(or whatever syntax your parser generator accepts for lists) and semantically count that only 3 forms are provided and they are unique.
Often people building parser generators add special "extended BNF" notations to let them describe special circumstances; you gave an example use {3} as special syntax implying that you only wanted "3 of" under the assumption the parser generator accepts this notation and does the appropriate enforcement. One can imagine an extension notation {unique} to let you describe your situation. I've never seen a parser generator that implemented that idea.