At many places (for example in this answer here), I have seen it is written that an LR(0) grammar cannot contain ε productions.
Also in Wikipedia I have seen statements like: An ε free LL(1) grammar is also SLR(1).
Now the problem which I am facing is that I cannot reason out the logic behind these statements.
Well, I know that LR(0) grammars accept the languages accepted by a DPDA by empty stack, i.e. the language they accept must have prefix property. [This prefix property can, however, be dealt with if we assume end markers and as such given any language the prefix property shall always be satisfied. Many texts like Theory of Computation by Sipser assume this end marker to simply their argument]. That being said, we can say (informally?) that a grammar is LR(0) if there is no state in the canonical collection of LR(0) items that have a shift-reduce conflict or reduce-reduce conflict.
With this background, I tried to consider the following grammar:
S -> Aa
A -> ε
canonical collection of LR(0) items
In the above DFA, I find that there is no state which has a shift-reduce conflict or reduce-reduce conflict.
So this grammar should be LR(0) as per my analysis. But it also has ε production.
Isn't this example contradicting the statement:
"no grammar with ε productions can be LR(0)"
I guess if I know the logic behind the above quoted statement then I can understand the concept better.
Actually my main problem arose with the statement :
An ε free LL(1) grammar is also SLR(1).
When I asked one of my friends, he gave the argument that as the LL(1) grammar is ε free hence it is LR(0) and hence it is SLR(1).
But I could not understand his logic either. When I asked him about reasoning, he started sharing post regarding "grammar with ε productions can never be LR(0)"...
But personally I could not think of any logic as to how "ε free LL(1) grammar is SLR(1)". Is it really related to the above property of "grammar with ε productions cannot be LR(0)"? If so, please do help me out.. If not, then should I consider asking a separate question for the second confusion?
I have got my concepts of compiler design from the dragon book by Ullman only. Also the knowledge of TOC from Ullman and from few other texts like Sipser, Linz.
A notable feature of your grammar is that A could just be eliminated. It serves absolutely no purpose. (By "eliminated", I mean simply removing all references to it; leaving productions otherwise intact.)
It is true that it's existence doesn't preclude the grammar from being LR(0). Similarly, a grammar with an unreachable non-terminal and an ε-production for that non-terminal could also be LR(0).
So it would be more accurate to say that a grammar cannot be LR(0) if it has a productive non-terminal with both an ε-production and some other productive production. But since we usually only consider reduced grammars without pointless non-terminals, I'm not sure that this additional pedantry serves much purpose.
As for your question about ε-free LL(1) grammars, here's a rough outline:
If an ε-free grammar is not LR(0), then there is some state with both a shift and a reduce action. Since the grammar is ε-free, that state was reached by way of a shift or a goto. The previous state must then have had two different productions with the same FIRST set, contradicting the LL(1) condition.
Is there an easy way to tell whether a simple grammar is suitable for recursive descent? Is eliminating left recursion and left factoring the grammar enough to achieve this ?
Not necessarily.
To build a recursive descent parser (without backtracking), you need to eliminate or resolve all predict conflicts. So one definitive test is to see if the grammar is LL(1); LL(1) grammars have no predict conflicts, by definition. Left-factoring and left-recursion elimination are necessary for this task, but they might not be sufficient, since a predict conflict might be hiding behind two competing non-terminals:
list ::= item list'
list' ::= ε
| ';' item list'
item ::= expr1
| expr2
expr1 ::= ID '+' ID
expr2 ::= ID '(' list ')
The problem with the above (or, at least, one problem) is that when the parser expects an item and sees an ID, it can't know which of expr1 and expr2 to try. (That's a predict conflict: Both non-terminals could be predicted.) In this particular case, it's pretty easy to see how to eliminate that conflict, but it's not really left-factoring since it starts by combining two non-terminals. (And in the full grammar this might be excerpted from, combining the two non-terminals might be much more difficult.)
In the general case, there is no algorithm which can turn an arbitrary grammar into an LL(1) grammar, or even to be able to say whether the language recognised by that grammar has an LL(1) grammar as well. (However, it's easy to tell whether the grammar itself is LL(1).) So there's always going to be some art and/or experimentation involved.
I think it's worth adding that you don't really need to eliminate left-recursion in a practical recursive descent parser, since you can usually turn it into a while-loop instead of recursion. For example, leaving aside the question of the two expr types above, the original grammar in an extended BNF with repetition operators might be something like
list ::= item (';' item)*
Which translates into something like:
def parse_list():
parse_item()
while peek(';'):
match(';')
parse_item()
(Error checking and AST building omitted.)
As far as I know, Left recursion is not a problem for LR parser.And I know that an ambiguous grammar can't be parsed by any kind of parser.So if I have an ambiguous grammar as follows,how can I remove ambiguity so that I can check if this grammar is SLR(1) or not?
E->E+E|E-E|(E)|id
And one more question,is left factoring needed for a grammar to check if the grammar is LL(1) or SLR(1)?
Any help will be appreciated.
Any parser generator you are likely to encounter will be able to handle the ambiguities in your grammar simply.
Your grammar produces shift/reduce conflicts. These are not necessarily a problem (as are reduce/reduce conflicts). The default action on a shift/reduce conflict in every parser generator is to shift, which solves your problem. There are usually mechanisms (as in YACC or Bison) to ignore this as a warning.
You can remove the conflicts in your grammar by setting up multiple levels of expressions so that you force the precedence of the operators.
According to "Recursive descent parser" on Wikipedia, recursive descent without backtracking (a.k.a. predictive parsing) is only possible for LL(k) grammars.
Elsewhere, I have read that the implementation of Lua uses such a parser. However, the language is not LL(k). In fact, Lua is inherently ambiguous: does a = f(g)(h)[i] = 1 mean a = f(g); (h)[i] = 1 or a = f; (g)(h)[i] = 1? This ambiguity is resolved by greediness in the parser (so the above is parsed as the erroneous a = f(g)(h)[i]; = 1).
This example seems to show that predictive parsers can handle grammars which are not LL(k). Is it true they can, in fact, handle a superset of LL(k)? If so, is there a way to find out whether a given grammar is in this superset?
In other words, if I am designing a language which I would like to parse using a predictive parser, do I need to restrict the language to LL(k)? Or is there a looser restriction I can apply?
TL;DR
For a suitable definition of a recursive descent parser, it is absolutely correct that only LL(k) languages can be parsed by recursive descent.
Lua can be parsed with a recursive descent parser precisely because the language is LL(k); that is, an LL(k) grammar exists for Lua. [Note 1]
1. An LL(k) language may have non-LL(k) grammars.
A language is LL(k) if there is an LL(k) grammar which recognizes the language. That doesn't mean that every grammar which recognizes the language is LL(k); there might be any number of non-LL(k) grammars which recognize the language. So the fact that some grammar is not LL(k) says absolutely nothing about the language itself.
2. Many practical programming languages are described with an ambiguous grammar.
In formal language theory, a language is inherently ambiguous only if every grammar for the language is ambiguous. It is probably safe to say that no practical programming language is inherently ambiguous, since practical programming languages are deterministically parsed (somehow). [Note 2].
Because writing a strictly non-ambiguous grammar can be tedious, it is pretty common for the language documentation to provide an ambiguous grammar, along with textual material which indicates how the ambiguities are to be resolved.
For example, many languages (including Lua) are documented with a grammar which does not explicitly include operator precedence, allowing a simple rule for expressions:
exp ::= exp Binop exp | Unop exp | term
That rule is clearly ambiguous, but given a list of operators, their relative precedences and an indication of whether each operator is left- or right-associative, the rule can be mechanically expanded into an unambiguous expression grammar. Indeed, many parser generators allow the user to provide the precedence declarations separately, and perform the mechanical expansion in the course of producing the parser. The resulting parser, it should be noted, is a parser for the disambiguated grammar so the ambiguity of the original grammar does not imply that the parsing algorithm is capable of dealing with ambiguous grammars.
Another common example of ambiguous reference grammars which can be mechanically disambiguated is the "dangling else" ambiguity found in languages like C (but not in Lua). The grammar:
if-statement ::= "if" '(' exp ')' stmt
| "if" '(' exp ')' stmt "else" stmt
is certainly ambiguous; the intention is that the parse be "greedy". Again, the ambiguity is not inherent. There is a mechanical transformation which produces an unambiguous grammar, something like the following:
matched-statement ::= matched-if-stmt | other-statement
statement ::= matched-if-stmt | unmatched-if-stmt
matched-if-stmt ::= "if" '(' exp ')' matched-statement "else" matched-statement
unmatched-if-stmt ::= "if" '(' exp ')' statement
| "if" '(' exp ')' matched-statement "else" unmatched-if-stmt
It is quite common for parser generators to implicitly perform this transformation. (For an LR parser generator, the transformation is actually implemented by deleting reduce actions if they conflict with a shift action. This is simpler than transforming the grammar, but it has exactly the same effect.)
So Lua (and other programming languages) are not inherently ambiguous; and therefore they can be parsed with parsing algorithms which require unambiguous deterministic parsers. Indeed, it might even be a little surprising that there are languages for which every possible grammar is ambiguous. As is pointed out in the Wikipedia article cited above, the existence of such languages was proven by Rohit Parikh in 1961; a simple example of an inherently-ambiguous context-free language is
{anbmcmdn|n,m≥0} ∪ {anbncmdm|n,m≥0}.
3. Greedy LL(1) parsing of Lua assignment and function call statements
As with the dangling else construction above, the disambiguation of Lua statement sequences is performed by only allowing the greedy parse. Intuitively, the procedure is straight-forward; it is based on forbidding two consecutive statements (without intervening semicolon) where the second one starts with a token which might continue the first one.
In practice, it is not really necessary to perform this transformation; it can be done implicitly during the construction of the parser. So I'm not going to bother to generate a complete Lua grammar here. But I trust that the small subset of the Lua grammar here is sufficient to illustrate how the transformation can work.
The following subset (largely based on the reference grammar) exhibits precisely the ambiguity indicated in the OP:
program ::= statement-list
statement-list ::= Ø
| statement-list statement
statement ::= assignment | function-call | block | ';'
block ::= "do" statement-list "end"
assignment ::= var '=' exp
exp ::= prefixexp [Note 3]
prefixexp ::= var | '(' exp ')' | function-call
var ::= Name | prefixexp '[' exp ']'
function-call ::= prefixexp '(' exp ')'
(Note: (I'm using Ø to represent the empty string, rather ε, λ, or %empty.)
The Lua grammar as is left-recursive, so it is clearly not LL(k) (independent of the ambiguity). Removing the left-recursion can be done mechanically; I've done enough of it here in order to demonstrate that the subset is LL(1). Unfortunately, the transformed grammar does not preserve the structure of the parse tree, which is a classic problem with LL(k) grammars. It is usually simple to reconstruct the correct parse tree during a recursive descent parse and I'm not going to go into the details.
It is simple to provide an LL(1) version of exp, but the result eliminates the distinction between var (which can be assigned to) and function-call (which cannot):
exp ::= term exp-postfix
exp-postfix ::= Ø
| '[' exp ']' exp-postfix
| '(' exp ')' exp-postfix
term ::= Name | '(' exp ')'
But now we need to recreate the distinction in order to be able to parse both assignment statements and function calls. That's straight-forward (but does not promote understanding of the syntax, IMHO):
a-or-fc-statement ::= term a-postfix
a-postfix ::= '=' exp
| ac-postfix
c-postfix ::= Ø
| ac-postfix
ac-postfix ::= '(' exp ')' c-postfix
| '[' exp ']' a-postfix
In order to make the greedy parse unambiguous, we need to ban (from the grammar) any occurrence of S1 S2 where S1 ends with an exp and S2 starts with a '('. In effect, we need to distinguish different types of statement, depending on whether or not the statement starts with a (, and independently, whether or not the statement ends with an exp. (In practice, there are only three types because there are no statements which start with a ( and do not end with an exp. [Note 4])
statement-list ::= Ø
| s1 statement-list
| s2 s2-postfix
| s3 s2-postfix
s2-postfix ::= Ø
| s1 statement-list
| s2 s2-postfix
s1 ::= block | ';'
s2 ::= Name a-postfix
s3 ::= '(' exp ')' a-postfix
4. What is recursive descent parsing, and how can it be modified to incorporate disambiguation?
In the most common usage, a predictive recursive descent parser is an implementation of the LL(k) algorithm in which each non-terminal is mapped to a procedure. Each non-terminal procedure starts by using a table of possible lookahead sequences of length k to decide which alternative production for that non-terminal to use, and then simply "executes" the production symbol by symbol: terminal symbols cause the next input symbol to be discarded if it matches or an error to be reported if it doesn't match; non-terminal symbols cause the non-terminal procedure to be called.
The tables of lookahead sequences can be constructed using FIRSTk and FOLLOWk sets. (A production A→ω is mapped to a sequence α of terminals if α ∈ FIRSTk(ω FOLLOWk(A)).) [Note 5]
With this definition of recursive descent parsing, a recursive descent parser can handle precisely and solely LL(k) languages. [Note 6]
However, the alignment of LL(k) and recursive descent parsers ignores an important aspect of a recursive descent parser, which is that it is, first and foremost, a program normally written in some Turing-complete programming language. If that program is allowed to deviate slightly from the rigid structure described above, it could parse a much larger set of languages, even languages which are not context-free. (See, for example, the C context-sensitivity referenced in Note 2.)
In particular, it is very easy to add a "default" rule to a table mapping lookaheads to productions. This is a very tempting optimization because it considerably reduces the size of the lookahead table. Commonly, the default rule is used for non-terminals whose alternatives include an empty right-hand side, which in the case of an LL(1) grammar would be mapped to any symbol in the FOLLOW set for the non-terminal. In that implementation, the lookahead table only includes lookaheads from the FIRST set, and the parser automatically produces an empty right-hand side, corresponding to an immediate return, for any other symbol. (As with the similar optimisation in LR(k) parsers, this optimization can delay recognition of errors but they are still recognized before an additional token is read.)
An LL(1) parser cannot include a nullable non-terminal whose FIRST and FOLLOW sets contain a common element. However, if the recursive descent parser uses the "default rule" optimization, that conflict will never be noticed during the construction of the parser. In effect, ignoring the conflict allows the construction of a "greedy" parser from (certain) non-deterministic grammars.
That's enormously convenient, because as we have seen above producing unambiguous greedy grammars is a lot of work and does not lead to anything even vaguely resembling a clear exposition of the language. But the modified recursive parsing algorithm is not more powerful; it simply parses an equivalent SLL(k) grammar (without actually constructing that grammar).
I do not intend to provide a complete proof of the above assertion, but a first step is to observe that any non-terminal can be rewritten as a disjunction of new non-terminals, each with a single distinct FIRST token, and possibly a new non-terminal with an empty right-hand side. It is then "only" necessary to remove non-terminals from the FOLLOW set of nullable non-terminals by creating new disjunctions.
Notes
Here, I'm talking about the grammar which operates on a tokenized stream, in which comments have been removed and other constructs (such as strings delimited by "long brackets") reduced to a single token. Without this transformation, the language would not be LL(k) (since comments -- which can be arbitrarily long -- interfere with visibility of the lookahead token). This allows me to also sidestep the question of how long brackets can be recognised with an LL(k) grammar, which is not particularly relevant to this question.
There are programming languages which cannot be deterministically parsed by a context-free grammar. The most notorious example is probably Perl, but there is also the well-known C construct (x)*y which can only be parsed deterministically using information about the symbol x -- whether it is a variable name or a type alias -- and the difficulties of correctly parsing C++ expressions involving templates. (See, for example, the questions Why can't C++ be parsed with a LR(1) parser? and Is C++ context-free or context-sensitive?)
For simplicity, I've removed the various literal constants (strings, numbers, booleans, etc.) as well as table constructors and function definitions. These tokens cannot be the target of a function-call, which means that an expression ending with one of these tokens cannot be extended with a parenthesized expression. Removing them simplifies the illustration of disambiguation; the procedure is still possible with the full grammar, but it is even more tedious.
With the full grammar, we will need to also consider expressions which cannot be extended with a (, so there will be four distinct options.
There are deterministic LL(k) grammars which fail to produce unambiguous parsing tables using this algorithm, which Sippu & Soisalon-Soininen call the Strong LL(k) algorithm. It is possible to augment the algorithm using an additional parsing state, similar to the state in an LR(k) parser. This might be convenient for particular grammars but it does not change the definition of LL(k) languages. As Sippu & Soisalon-Soininen demonstrate, it is possible to mechanically derive from any LL(k) grammar an SLL(k) grammar which produces exactly the same language. (See Theorem 8.47 in Volume 2).
The recursive definition algorithm is a precise implementation of the canonical stack-based LL(k) parser, where the parser stack is implicitly constructed during the execution of the parser using the combination of the current continuation and the stack of activation records.
Is it possible to construct an LR(0) parser that could parse a language with both prefix and postfix operators? For example, if I had a grammar with the + (addition) and ! (factorial) operators with the usual precedence then 1+3! should be 1 + 3! = 1 + 6 = 7, but surely if the parser were LR(0) then when it had 1+3 on the stack it would reduce rather than shift?
Also, do right associative operators pose a problem? For example, 2^3^4 should be 2^(3^4) but again, when the parser have 2^3 on the stack how would it know to reduce or shift?
If this isn't possible is there still a way to use an LR(0) parser, possibly by altering the grammar to add brackets in the appropriate places?
LR(0) parsers have a weakness in that they can only parse prefix-free languages, languages where no string in the language is a prefix of any other. This generally makes it a bit tricky to parse expressions like these, since something like 5 is a prefix of 5!. This also explains why it's hard to get right-associative operators - given a production like
S → F | F ^ S
the parser will have a shift/reduce conflict after seeing an F because it can't tell whether to extend it or to reduce again. This is related to the prefix-free property mentioned earlier.
This weakness of LR(0) is one of the reasons why people don't use it much in practice. SLR(1) and LALR(1) parsers can usually parse these grammars because they have a token of lookahead that lets them decide whether to shift or reduce. In the above case, the parsers wouldn't encounter shift/reduce conflicts because when deciding whether to reduce an F or shift a ^, they can see to shift the ^ because there's no correct string where a ^ should appear after an S.