I have this grammar
E -> E + i
E -> i
The augmented grammar
E' -> E
E -> E + i
E -> i
Now I try to expand the item set 0
I0)
E' -> .E
+E -> .E + i
+E -> .i
Then, since I have .E in I0 I would expand it but then I will get another E rule, and so on, this is my first doubt.
Assuming that this is alright the next item sets are
I0)
E' -> .E
+E -> .E + i
+E -> .i
I1) (I moved the dot from I0, no variables at rhs of dot)
E' -> E.
E -> E. + i
E -> i.
I2) (I moved the dot from I1, no vars at rhs of dot)
E -> E +. i
I3) (I moved the dot from I2, also no vars)
E -> E + i.
Then I will have this DFA
I0 -(E, i)-> I1 -(+)-> I2 -(i)-> I3
| |
+-(∅)-> acpt <-(∅)--+
I'm missing something because E -> E + i must accept i + i + .. but the DFA doesn't goes back to the I0, so it seems wrong to me. My guess is that it should have a I0 to I0 transition, but I then I don't know that to do with the dot.
What you call the "expansion" of the item set is actually a closure; that's how it's described in all the descriptions of the algorithm I've seen (at least in textbooks). Like any closure operation, you just keep on doing the transformation until you reach a fixed-point; once you've included the productions for E, they're included.
But the essential point is that you're not building a DFA. You're building a pushdown automaton, and the DFA is just one part of it. The DFA is used for shift operations; when you shift a new terminal (because the current parse stack is not a handle), you do a state transition according to the DFA. But you also push the current state onto the PDA's stack.
The interesting part is what happens when the automaton decides to perform a reduction, which replaces the right-hand side of a production with its left-hand side non-terminal. (The right-hand side at the top of the stack is called a "handle".) To do the reduction, you unwind the stack, popping each right-hand side symbol (and the corresponding DFA state) until you reach the beginning of the production. What that does is rewind the DFA to the state it was in before it shifted the first symbol from the right-hand side. (Note that it is only at this point that you know for sure which production was used.) With the DFA thus reset, you can now shift the non-terminal which was encountered, do the corresponding DFA transition, and continue with the parse.
The basis for this procedure is the fact that the parser stack is at all times a "viable prefix"; that is, a sequence of symbols which are the prefix of some right sentential form which can be derived from the start symbol. What's interesting about the set of viable prefixes for a context-free grammar is that it is a regular language, and consequently can be recognised by a DFA. The reduction procedure given above precisely represents this recognition procedure when handles are "pruned" (to use Knuth's original vocabulary).
In that sense, it doesn't really matter what procedure is used to determine which handle is to be pruned, as long as it provides a valid answer. You could, for example, fork the parse every time a potential handle is noticed at the top of the stack, and continue in parallel with both forks. With clever stack management, this parallel search can be done in worst-case O(n3) time for any context-free grammar (and this can be reduced if the grammar is not ambiguous). That's a very rough description of Earley parsers.
But in the case of an LR(k) parser, we require that the grammar be unambiguous, and we also require that we can identify a reduction by looking at no more than k more symbols from the input stream, which is an O(1) operation since k is fixed. If at each point in the parse we know whether to reduce or not, and if so which reduction to choose, then the reductions can be implemented as I outlined above. Each reduction can be performed in O(1) time for a fixed grammar (since the maximum size of a right-hand side in a particular grammar is fixed), and since the number of reductions in a parse is linear in the size of the input, the entire parse can be done in linear time.
That was all a bit informal, but I hope it serves as an intuitive explanation. If you're interested in the formal proof, Donald Knuth's original 1965 paper (On the Translation of Languages from Left to Right) is easy to find and highly readable as these things go.
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.
I have the following grammar:
S -> a b D E
S -> A B E F
D -> M x
E -> N y
F -> z
M -> epsilon
N -> epsilon
My textbook says there is a Reduce/Reduce conflict in LR(0). I built a diagram and found out that there is a state:
S -> a b . D E
S -> A B . E F
D -> . M x
E -> . N y
M -> .
N -> .
The textbook says that it's a Reduce/Reduce conflict. I'm trying to figure out why. If I build the SLR table I get the following row (3 is the state above):
That's because:
Follow(M)={x} so we can do reduce to rule 6 from state 3.
Follow(N)={y} so we can do reduce to rule 7 from state 3.
I was taught that there is a conflict S/R if there is a cell with S/R and conflict R/R if there is a cell with R/R. But I don't see two Rs in the same cell in the table. So why is it a reduce/reduce conflict?
You show an SLR(1) parsing table, in which the columns correspond to a lookahead of length 1. It's correct, and there is no conflict.
But here we're talking about an LR(0) machine, in which there is no lookahead. (That's the 0 in LR(0).) The only decision the machine can make is to shift or reduce, and since it cannot use lookahead, it can only use the state itself. A given state must be either a shift state or a reduce state (and, if a reduce state, which production is being reduced).
(In case it's confusing, and it often is, the concept of lookahead does not refer to the use of the shifted symbol to decide which state to transition to. The transition is taken based on the shifted symbol, which is at that point no longer part of the lookahead.)
So in that state, there is no possible shift action; in all items in the itemset, either the dot is at the end or the next symbol is a non-terminal (implying a GOTO action after returning from a reduce).
But the state does not have a unique reduction. Depending on the lookahead, the parsers needs to choose to reduce M or to reduce N. And since there is no lookahead, the decision cannot be made and hence it's a conflict.
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.
LL(1) Grammar:
(1) Var -> ID DimList
(2) DimList -> ε DimList'
(3) DimList' -> Dim DimList'
(4) DimList' -> ε
(5) Dim -> [ CONST ]
And, in the script that I am reading, it says that the function FIRST(ε DimList') gives {#, [}. But, how?
My guess is that since the right side of (2) begins with ε, it skips epsilon and takes FIRST(DimList') which is, considering (3) and (5), equal to {[}, BUT also, because of (4), takes FOLLOW(DimList') which is {#}.
Other way it could be is that, since (2) begins with ε it skips epsilon and takes FIRST(DimList') BUT ALSO takes FOLLOW(DimList) from (2)...
First one makes more sense to me, though I'm still in the process of learning basics of LL(1) grammars so I would appreciate if someone takes the time to make this clear, thank you.
EDIT: And, of course, it could be that neither of these is true.
The usual definition of the FIRST function would result in FIRST(Dimlist) (or, if you like, FIRST(ε Dimlist') being {ε, [}. ε is in FIRST(ε Dimlist') because both ε and Dimlist' are nullable. [ is an element because it could be the first symbol in a derivation of ε Dimlist, which is the same as saying that it could be the first symbol in a derivation of Dimlist'.
Another way of saying this is that:
FIRST(ε Dimlist' #) = {#, [}
We usually then define the function PREDICT:
PREDICT(ω) = FIRST(ω FOLLOW(ω))
and we can see that
PREDICT(Dimlist) = FIRST(Dimlist FOLLOW(Dimlist)) = {#, [}
Here, FIRST(ω) is the set of strings of terminals (of length ≤ 1) which could appear at the beginning of a derivation of ω, while PREDICT(ω) is the set of strings of terminals (of length ≤ 1) which could be present in the input when a derivation of ω is possible.
It's not uncommon to confuse FIRST and PREDICT, but it's better to keep the difference straight.
Note that all of these functions can be generalized to strings of maximum length k, which are usually written FIRSTk, FOLLOWk and PREDICTk, and the definition of PREDICTk is similar to the above:
PREDICTk(ω) = FIRSTk(ω FOLLOWk(ω))
What is the difference between Left Factoring and Left Recursion ? I understand that Left factoring is a predictive top down parsing technique. But I get confused when I hear these two terms.
Left factoring is removing the common left factor that appears in two productions of the same non-terminal. It is done to avoid back-tracing by the parser. Suppose the parser has a look-ahead, consider this example:
A -> qB | qC
where A, B and C are non-terminals and q is a sentence.
In this case, the parser will be confused as to which of the two productions to choose and it might have to back-trace. After left factoring, the grammar is converted to:
A -> qD
D -> B | C
In this case, a parser with a look-ahead will always choose the right production.
Left recursion is a case when the left-most non-terminal in a production of a non-terminal is the non-terminal itself (direct left recursion) or through some other non-terminal definitions, rewrites to the non-terminal again (indirect left recursion).
Consider these examples:
(1) A -> Aq (direct)
(2) A -> Bq
B -> Ar (indirect)
Left recursion has to be removed if the parser performs top-down parsing.
Left Factoring is a grammar transformation technique. It consists in "factoring out" prefixes which are common to two or more productions.
For example, going from:
A → α β | α γ
to:
A → α A'
A' → β | γ
Left Recursion is a property a grammar has whenever you can derive from a given variable (non terminal) a rhs that begins with the same variable, in one or more steps.
For example:
A → A α
or
A → B α
B → A γ
There is a grammar transformation technique called Elimination of left recursion, which provides a method to generate, given a left recursive grammar, another grammar that is equivalent and is not left recursive.
The relationship/confusion between both terms probably derives from the fact that both transformation techniques may need to be applied to a grammar before being able to derive a predictive top down parser for it.
This is the way I've seen the two terms used:
Left recursion: when one or more productions can be reached from themselves with no tokens consumed in-between.
Left factoring: a process of transformation, turning the grammar from a left-recursive form to an equivalent non-left-recursive form.
left factor :
Let the given grammar :
A-->ab1 | ab2 | ab3
1) we can see that, for every production, there is a common prefix & if we choose any production here, it is not confirmed that we will not need to backtrack.
2) it is non deterministic, because we cannot choice any production and be assured that we will reach at our desired string by making the correct parse tree.
but if we rewrite the grammar in a way that is deterministic and also leaves us flexible enough to convert it into any string that is possible without backtracking, it will be:
A --> aA',
A' --> b1 | b2| b3
now if we are asked to make the parse tree for string ab2 and now we don't need back tracking. Because we can always choose the correct production when we get A' thus we will generate the correct parse tree.
Left recursion :
A --> Aa | b
here it is clear that the left child of A will always be A if we choose the first production,this is left recursion .because , A is calling itself over and over again .
the generated string from this grammar is :
ba*
since this cannot be in a grammar ... we eliminate the left recursion by writing :
A --> bA'
A' --> E | aA'
now we will not have left recursion and also we can generate ba* .
Left Recursion:
A grammar is left recursive if it has a nonterminal A such that there is a derivation A -> Aα | β where α and β are sequences of terminals and nonterminals that do not start with A.
While designing a top down-parser, if the left recursion exist in the grammar then the parser falls in an infinite loop, here because A is trying to match A itself, which is not possible.
We can eliminate the above left recursion by rewriting the offending production. As-
A -> βA'
A' -> αA' | epsilon
Left Factoring: Left factoring is required to eliminate non-determinism of a grammar. Suppose a grammar, S -> abS | aSb
Here, S is deriving the same terminal a in the production rule(two alternative choices for S), which follows non-determinism. We can rewrite the production to defer the decision of S as-
S -> aS'
S' -> bS | Sb
Thus, S' can be replaced for bS or Sb
Here is a simple way to differentiate between both terms:
Left Recursion:
When leftmost Element of a production is the Producing element itself (Non Terminal Element).
e.g. A -> Aα / Aβ
Left Factoring:
When leftmost Element of a production (Terminal element) is repeated in the same production.
e.g. A -> αB / αC
Furthermore,
If a Grammar is Left Recursive, it might result into infinite loop hence we need to Eliminate Left Recursion.
If a Grammar is Left Factoring, it confuses the parser hence we need to Remove Left Factoring as well.
left recursion:= when left hand non terminal is same as right hand non terminal.
Example:
A->A&|B where & is alpha.
We can remove left ricursion using rewrite this production as like.
A->BA'
A'->&A'|€
Left factor mean productn should not non deterministic. .
Example:
A->&A|&B|&C