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I recently met the concept of LR, LL etc. Which category is Lua? Are there, or, can there be implementations that differ from the official code in this aspect?
LR, LL and so on are algorithms which attempt to find a parser for a given grammar. This is not possible for every grammar, and you can categorize on the basis of that possibility. But you have to be aware of the difference between a language and a grammar.
It might be possible to create an LR(k) parser for a given grammar, for some specific value of k. If so, the grammar is LR(k). Note that an LR(k) grammar is also an LR(k+1) grammar, and that an LL(k) grammar is also LR(k). So these are not categories in the sense that every grammar is in exactly one category.
Any language can be recognised by many different grammars. (In fact, an unlimited number). These grammars can be arbitrarily complex. You can always write a grammar for a given language which is not even context-free. We say that a language is <X> if there exists a grammar for that language which is <X>. But the fact that a specific grammar for that language is not <X> says nothing.
One interesting theorem demonstrates that if there is an LR(k) grammar for any language, then it is possible to derive an LR(1) grammar for that language. So while the k parameter is useful for describing grammars, languages can only be LR(0) or LR(1). This is not true of LL(k) languages, though.
Lua as a language is basically LR(1) and LL(2). The grammar is part of the reference manual, except that the published grammar doesn't specify operator precedences or a few rules having to do with newlines. The actual parser is a hand-written recursive-descent parser (at least, the last time I looked) with a couple of small deviations in order to handle operator precedence and the minor deviations from LL(1). However, there exist LALR(1) parsers for Lua as well.
I confused Exactly !!!!!!
I read following example in one of my professor note.
1) we have a SLR(1) Grammar G as following. we use SLR(1) parser generator and generate a parse table S for G. we use LALR(1) parser generator and generate a parse table L for G.
S->AB
A->dAa
A-> lambda (lambda is a string with length=0)
B->aAb
Solution: the number of elements with R (reduce) in S is more than L.
but in one site i read:
2) Suppose T1, T2 is created with SLR(1) and LALR(1) for Grammar G. if G be a SLR(1) Grammar which of the following is TRUE?
a) T1 and T2 has not any difference.
b) total Number of non-error entries in T1 is lower than T2
c) total Number of error entries in T1 is lower than T2
Solution:
The LALR(1) algorithm generates exactly the same states as the SLR(1) algorithm, but it can generate different actions; it is capable of resolving more conflicts than the SLR(1) algorithm. However, if the grammar is SLR(1), both algorithms will produce exactly the same machine (a is right).
any one could describe for me which of them is true?
EDIT: infact my question is why for a given SLR(1) Grammar, the parse table of LALAR(1) and SLR(1) is exactly the same, (error and non-error entries are equal and number of reduce is equal) but for the above grammar, the number of Reduced in S is more than L.
I see in another book that in general we have:
Summary:
1) for the above grammar that i wrote in question 1, why number of reduced is different?
2) if we have a SLR(1) Grammar, why the table is exactly the same? (number of reduced and error entries become the same)
Both of these statements are true!
One of your questions was why SLR(1) and LALR(1) parsers have the same states as one another. SLR(1) parsers are formed by starting with an LR(0) automaton, then augmenting each production with lookahead information from FOLLOW sets. In an LALR(1) parser, we begin with an LR(1) parser (where each production has very precise lookahead information), then combine any two states that have the same underlying LR(0) state. This results in an LR(0) automaton with additional information because each LR(0) state corresponds to at least one LR(1) state and each LR(1) state corresponds to some underlying LR(0) state.
SLR(1) and LALR(1) parsers both have the same set of states, which are the same states as in an LR(0) parser. The parsers differ only in what actions they perform in each state.
In both SLR(1) and LALR(1) parsers, each item has an associated set of lookahead tokens. Whenever the parser enters a state with a reduce item in it, the parser will perform that reduction if the next token of input is in the lookahead set. In an SLR(1) parser, the lookahead set is the FOLLOW set for the nonterminal on the left-hand side of the production. In an LALR(1) parser, the lookahead set is, appropriately, called the LA set for the combination of the nonterminal in the production and the automaton state.
You can prove that the LA sets used in an LALR(1) parser are subsets of the FOLLOW sets used in SLR(1) parsers. This means that LALR(1) parsers will never have more reduce actions than SLR(1) parsers, and in some cases the LALR(1) parsers will choose to shift when an SLR(1) parser would have a shift/reduce conflict.
Hope this helps!
Answer to Q1:
First of all you need to create DFA for SLR(1) and LALR(1) parsers. I created DFA for both of them.
Link to images of DFAs SLR(1) and LALR(1) DFAs
For SLR(1) I got 10 states and 10 reduce entries whereas for LALR(1) I created DFA for CLR(1) with 13 states which got minimized to 10 states with 7 reduce entries. Thats answers your first question.
Answer to Q2:
G is SLR(1) grammar, then surely there are no conflicts (or error) S-R or R-R in the SLR(1) table. LALR(1) has more power than SLR(1),therefore there is also no conflict in LALR(1) table for given grammar G. Lets see option by option
(c) : there no error in T1 and T2 (wrong option)
(b) : Non-error entries means shift entries and reduce entries. It should be clearly noted that in bottom up parsers from parser to parser only rules for reduce entries changes while for that of shift entries remain same. For e.g in LR(0) reduce entries are made in each column, for SLR(1) it is done in FOLLOW of left hand side variable, while in CLR(1) and LALR(1) reduce entries are made in lookahead symbols. Thus reduce entries changes from parser to parser but shift entries are same.
We have also already proved in Q1 where reduce entries of SLR(1) parsing table are more than that of LALR(1). Therefore proving (b) option to be incorrect.
(a) T1 and T2 may come out to be same but not always. And other important thing is that multiple choice questions sometimes wants you to choose most appropriate option. Thus for me (a) is the answer
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What is the difference between LR, SLR, and LALR parsers?
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I understand both LR and LALR are bottom-up parsing algorithms, but what's the difference between the two?
What's the difference between LR(0), LALR(1), and LR(1) parsing? How can I tell if a grammar is LR(0), LALR(1), or LR(1)?
At a high level, the difference between LR(0), LALR(1), and LR(1) is the following:
An LALR(1) parser is an "upgraded" version of an LR(0) parser that keeps track of more precise information to disambiguate the grammar. An LR(1) parser is a significantly more powerful parser that keeps track of even more precise information than an LALR(1) parser.
LALR(1) parsers are a constant factor larger than LR(0) parsers, and LR(1) parsers are usually exponentially larger than LALR(1) parsers.
Any grammar that can be parsed with an LR(0) parser can be parsed with an LALR(1) parser and any grammar that can be parsed with an LALR(1) parser can be parsed with an LR(1) parser. There are grammars that are LALR(1) but not LR(0) and LR(1) but not LALR(1).
More formally, an LR(k) parser is a bottom-up parser that works by maintaining a stack of terminals and nonterminals. The parser is controlled by a finite automaton that determines, based on the current state of the parser and the next k tokens of input, whether to shift a new token onto the stack or reduce the top symbols of the stack by applying a production in reverse.
In order to keep track of enough information to make a determination about whether to shift or reduce, LR(k) parsers have each state correspond to a "configurating set," a set of productions annotated with the following information:
How much of the production has been seen so far, and
What tokens to expect after the production has been completed (the lookahead)
The first of these pieces of information is used to determine whether the parser may need to do a reduction - if none of the productions in a current state have been completed, there's no reason to do a reduction. The second of these pieces of information is used when doing a reduction to determine whether the reduction should be performed. When deciding whether to reduce, an LR(k) parser looks at the next k tokens of the input stream. If they match the lookahead tokens, the parser will reduce, and otherwise the parser does nothing.
Problems arise in an LR(k) parser when there are conflicts about what the parser should do in a given state. One type of conflict, a shift/reduce conflict, comes up when the parser is in a state where a production has been completed, but the lookahead symbols for that production conflict are also used by another uncompleted production in the state. This means that the parser can't tell whether to perform the reduction or not. A second type of conflict is a reduce/reduce conflict, where the parser knows it has to do a reduction, but two or more reductions are possible and it can't tell which to do.
Intuitively, as k gets larger and larger, the parser has more and more precise information available to it to determine when to shift and when to reduce. If a grammar is not LR(0), for example, the parser might have a state where given no lookahead at all it can't determine whether to shift or to reduce. However, that grammar might still be LR(1) because given an extra token of lookahead, it may be able to recognize that it should definitely shift and not reduce or definitely reduce and not shift.
The problem with LR(k) parsers is that as k gets larger, the number of states can increase exponentially. Lookahead in LR(k) parsers is handled by building more and more states in the parser to correspond to different combinations of productions and lookaheads, so as the number of possible lookaheads increases so does the number of states. Consequently, LR(1) parsers are commonly too large to be practical, and LR(2) or greater is almost unheard of in practice.
LALR(1) was invented as a compromise between the space efficiency of LR(0) parsers and the expressive power of LR(1) parsers. There are several ways to think about what an LALR(1) parser is. Originally, LALR(1) parsers were specified as a transformation that converts LR(1) automata into smaller automata. Although an LR(1) parser may have many more states than an LR(0) automaton, the only difference is that an LR(1) parser may have multiple copies of any particular state in an LR(0) automaton, each annotated with different lookahead information. An LALR(1) parser can be formed by starting with an LR(1) parser, then combining together all states that have the same "core" (the set of productions and their positions), then aggregating all the lookahead information together. This results in a parser that has the same number of states as an LR(0) parser but retains some amount of information about lookaheads to help avoid LR conflicts.
Another view of LALR(1) grammars uses the "LALR-by-SLR" method. LALR(1) parsers can be constructed by starting with an LR(0) parser for a grammar, then creating a new grammar for the language that annotates nonterminals with information about what states in the LR(0) parser they correspond to. The information about the FOLLOW sets of the nonterminals in that grammar can then be used to compute the lookaheads in the LR(0) parser.
The net result is that
LR(0) parsers are small, but not very expressive.
LALR(1) parsers are slightly larger due to the lookahead information, but very expressive.
LR(1) parsers are huge, but extremely expressive.
As for your second question - how do you determine whether a grammar is LR(1) or LALR(1) - the standard approach is to try to build the parsing automata for the LR(1) parser and LALR(1) parser and checking for conflicts. To build the LR(1) parser, you build up the LR(1) configurating sets, then check to see if any of those configurating sets have a shift/reduce conflict or reduce/reduce conflict. To construct an LALR(1) parser, you can either build the LR(1) parser and then condense configurating sets with the same core or can use the LALR-by-SLR method based on the LR(0) parser for the language. More details about how to construct these configurating sets are available in most compilers textbooks. You can also check out the lecture notes from a compilers course I taught in Summer 2012, which cover all of the above parsing methods and a few others.
Hope this helps!
LR(0), SLR(1), LALR(1) parsers all have the same number of states. Minimal LR(1) parsers will have a few more states if the grammar requires it, to avoid reduce-reduce conflicts.
Canonical LR(1) parsers will have many more states, too many for medium or large computer languages.
SLR(1) parser generators build an LR(0) state machine and determine the k=1 lookaheads by examining the grammar (which may report erroneous conflicts).
LALR(1) parser generators build an LR(0) state machine and determine the k=1 lookaheads by examining the LR(0) state machine (which is very complicated).
Canonical LR(1) parser generators build an LR(1) state machine.
Minimal LR(1) parser generators build an LR(1) state machine and merge compatible states during the build process.
The parsing algorithm for a good LALR(1) parser is different in two ways: (1) It should have shift-reduce actions, which reduces the number of states by about 30% and makes the parser faster, and (2) it must do one or more reductions when detecting a syntax error, which makes error recovery more complicated.
The parsing algorithm for a canonical LR(1) parser (1) does not have shift-reduce actions and (2) does not make any reductions when detecting a syntax error, which makes error recovery simpler.
There is another case, called minimal LR(1), which uses the same parsing algorithm and error recovery algorithm as LALR(1). Minimal LR(1) parsers offer the power of LR(1) and their size is almost as small as LALR(1). The LRSTAR Parser Generator creates minimal LR(1) parsers for C++ programmers.
First off, I have checked similar questions and none has quite the information I seek.
I want to know if, given a context-free grammar, it is possible to:
Know if there exists or not an equivalent LL(1) grammar. Equivalent in the sense that it should generate the same language.
Convert the context-free grammar to the equivalent LL(1) grammar, given it exists. The conversion should succeed if an equivalent LL(1) grammar exists. It is OK if it does not terminate if an equivalent does not exists.
If the answer to those questions are yes, are such algorithms or tools available somewhere ? My own researches have been fruitless.
Another answer mentions that the Dragon Book has an algorithms to eliminate left recursion and to left factor a context-free grammar. I have access to the book and checked it but it is unclear to me if the grammar is guaranteed to be LL(1). The restriction imposed on the context-free grammar (no null production and no cycle) are agreeable to me.
From university compiler courses I took I remember that LL(1) grammar is context free grammar, but context free grammar is much bigger than LL(1). There is no general algorithm (meaning not NP hard) to check and convert from context-free (that can be transformed to LL(1)) to LL(1).
Applying the bag of tricks like removing left recursion, removing first-follow conflict, left-factoring, etc. are similar to mathematical transformation when you want to integrate a function... You need experience that is sometimes very close to an art. The transformations are often inverse of each other.
One reason why LR type grammars are being used now a lot for generated parsers is that they cover much wider spectrum of context free grammars than LL(1).
Btw. e.g. C grammar can be expressed as LL(1), but C# cannot (e.g. lambda function x -> x + 1 comes to mind, where you cannot decide if you are seeing a parameter of lambda or a known variable).
Is there a simple way to determine whether a grammar is LL(1), LR(0), SLR(1)... just from looking on the grammar without doing any complex analysis?
For instance: To decide whether a BNF Grammar is LL(1) you have to calculate First and Follow sets - which can be time consuming in some cases.
Has anybody got an idea how to do this faster?
Any help would really be appreciated!
First off, a bit of pedantry. You cannot determine whether a language is LL(1) from inspecting a grammar for it, you can only make statements about the grammar itself. It is perfectly possible to write non-LL(1) grammars for languages for which an LL(1) grammar exists.
With that out of the way:
You could write a parser for the grammar and have a program calculate first and follow sets and other properties for you. After all, that's the big advantage of BNF grammars, they are machine comprehensible.
Inspect the grammar and look for violations of the constraints of various grammar types. For instance: LL(1) allows for right but not left recursion, thus, a grammar that contains left recursion is not LL(1). (For other grammar properties you're going to have to spend some quality time with the definitions, because I can't remember anything else off the top of my head right now :).
In answer to your main question: For a very simple grammar, it may be possible to determine whether it is LL(1) without constructing FIRST and FOLLOW sets, e.g.
A → A + A | a
is not LL(1), while
A → a | b
is.
But when you get more complex than that, you'll need to do some analysis.
A → B | a
B → A + A
This is not LL(1), but it may not be immediately obvious
The grammar rules for arithmetic quickly get very complex:
expr → term { '+' term }
term → factor { '*' factor }
factor → number | '(' expr ')'
This grammar handles only multiplication and addition, and already it's not immediately clear whether the grammar is LL(1). It's still possible to evaluate it by looking through the grammar, but as the grammar grows it becomes less feasable. If we're defining a grammar for an entire programming language, it's almost certainly going to take some complex analysis.
That said, there are a few obvious telltale signs that the grammar is not LL(1) — like the A → A + A above — and if you can find any of these in your grammar, you'll know it needs to be rewritten if you're writing a recursive descent parser. But there's no shortcut to verify that the grammar is LL(1).
One aspect, "is the language/grammar ambiguous", is a known undecidable question like the Post correspondence and halting problems.
Straight from the book "Compilers: Principles, Techniques, & Tools" by Aho, et. al.
Page 223:
A grammar G is LL(1) if and only if whenever A -> alpha | beta are two distinct productions of G, the following conditions hold:
For no terminal "a" do both alpha and beta derive strings beginning with "a"
At most one of alpha and beta can derive the empty string
If beta can reach the empty transition via zero or more transitions, then alpha does not derive any string beginning with a terminal in FOLLOW(A). Likewise, if alpha can reach the empty transition via zero or more transitions, then beta does not derive any string beginning with a terminal in FOLLOW(A)
Essentially this is a matter of verifying the grammar passes the Pairwise Disjointness Test and also does not involve Left Recursion. Or more succinctly a grammar G that is left-recursive or ambiguous cannot be LL(1).
Check whether the grammar is ambiguous or not. If it is, then the grammar is not LL(1) because no ambiguous grammar is LL(1).
ya there are shortcuts for ll(1) grammar
1) if A->B1|B2|.......|Bn
then first(B1)intersection first(B2)intersection .first(Bn)=empty set then it is ll(1) grammar
2) if A->B1|epsilon
then B1 intersection follow(A)is empty set
3) if G is any grammar such that every non terminal derives only one production then the grammar is LL(1)
p0 S' → E
p1 E → id
p2 E → id ( E )
p3 E → E + id
Construct the LR(0) DFA, the FOLLOW set for E and the SLR action/goto tables.
Is this an LR(0) grammar? Prove your answer.
Using the SLR tables, show the steps (shifts, reductions, accept) of an LR parser parsing: id ( id + id )