I'm preparing midterm exam, so I found some questions and tried to solve it.
But my answer is not as same as given answer
Question:
Consider The Following Grammar, Which Generates Email Addresses:
Addr → Name#Name.id
Name → id | id.Name
This grammar, as written, is not LL(1). Rewrite the grammar to eliminate all LL(1) conflicts.
The question asked us to eliminate LL(1) conflicts
According to my understood, it can be eliminated by left factoring.
Therefore, my answer is
Addr → Name#Name.id
Name → idName'
Name' → epsilon | .Name
However, the given answer is
Addr → Name#Name.id
Name → idName'
Name' → epsilon | .idName'
Is there any wrong of my understood?
Any advise in this post is appreciated.
Thanks
As far as I can see, neither your solution nor the one you show as the given answer is LL(1). However, the given answer is a more accurate reflection of the left-factoring procedure. Your solution leaves the artefact Name -> id Name', which now serves no purpose; expanding Name in the grammar to its only production eliminates an unnecessary derivation step. (Indeed, you could eliminate both instances of Name, the same way.)
But if the question is asking you to come up with an LL(1) grammar, then the given answer is not going to work.
Take a simple address, me#example.com, and imagine that the parser runs until just after example, which is an id. At this point the parser is expecting a Name', which is either empty or something which starts with .. (In your grammar, the something is '.' Name and in the solution you quote, it's '.' id Name, but either way the expected next token is ..) So the parser needs to decide between the empty production and the production starting with a .. Since the lookahead is ., the parser could certainly choose the production starting with .. But what about choosing the empty production? In that case, the next token will match whatever follows the second Name in Addr -> Name # Name . id, which is also a .. In other words, looking at the ., the parser has two apparently valid options: the empty production or the production starting with .. And that's true whether it's using your grammar or the solution's grammar.
The simplest solution (which is not very general) is to rearrange the Addr rule. Addr wants there to be at least two ids after the #, which could just as well be achieved with:
Addr -> Name # id . Name
You still need to left-factor Name to make that work, but now there is no problem with Name's FOLLOW set; it can no longer be followed by a ., and the decision after example is forced.
Related
So i have this grammar :
S -> (D)
D -> EF
E -> a|b|S
F -> *D | +D | ε
First of all, books solution uses the P -> pBq , First(q) - {ε} is subset of FOLLOW(B) for the rule D -> EF but that rule has only 2 symbols do we assume ε infront of E (ε being the p in pBq)?
And secondly i can't understand how to calculate Follow(E).
FOLLOW(E) consists of every terminal symbol which can immediately follow E in some derivation step. That's the precise definition; it's not very complicated.
For a simple grammar, you should be able to figure out all the FOLLOW sets just be looking at the grammar and applying a little bit of common sense. It would probably be a good idea to do that, since it will give you a better idea of how the algorithm works.
As a side note, it's maybe worth mentioning that ε is not a thing. Or at least, it's not a grammar symbol. It's one of several conventions used to make the empty sequence visible, just like 0 is a way to make nothing visible. Sometimes that's useful, but it's important to not let it confuse you. (Abuse of notation is endemic in mathematics, which can be frustrating.)
So, what can follow E? E only appears in one place on the right-hand sdie of that grammar, in the production D → E F. So clearly any symbol which be the first symbol of F must be in FOLLOW(E). The symbols which could be at the start of F are + and *, since as mentioned, ε is not a grammar symbol. (Many definitions of FIRST allow ε to be a member of that set, along with any actual terminal symbol. That's an example of the abuse of notation I was talking about in the previous paragraph, since it makes it look like ε is a terminal symbol. But it isn't. It's nothing.)
F is what we call a "nullable" non-terminal, because it can derive the empty sequence (which was written as ε so that you can see it). In other words, it's possible for F to disappear completely in a derivation step. And if it does disappear, then E might be at the end of the production D → E F. If E is at the end of D, then it can be followed by anything which could follow D, which includes ). D can also appear at the end of a derivation of F, which means that F could be followed by anything which could follow F, a tautology which adds no information whatsoever.
So it's easy to see that FOLLOW(F) = {*, +, )}, and you can use that to check your understanding of any algorithm to compute follow sets.
Now, I don't know what book you are referring to (and it would have been courteous to mention that in your original question; sources should always be correctly cited). But the book I happen to have in front of me --the Dragon Book-- has a pretty similar algorithm. The Dragon book uses a simple convention for writing statements like that. Probably your book does, too, but it might not be the same convention. You should check what it says and make sure that you typed the copied statement correctly, respecting whatever formatting used to indicate what the symbols stand for.
In the Dragon book, some of the conventions include:
Lower case characters at the start of the alphabet. –a, b, c,…– are terminals (as well as actual symbols like * and +).
Upper case characters at the start of the alphabet. –A, B, C,…– are non-terminals.
S is the start symbol.
Upper case characters at the end of the alphabet. –X, Y, Z– stand for arbitrary grammar symbols (either terminals or non-terminals).
$ is the marker used to indicate the end of the input.
Lower-case Greek letters –α, β, γ,…– are possibly-empty strings of grammar symbols.
The phrase "possibly empty" is very important, so I'm repeating it.
With that convention, they write the rules for computing the FOLLOW set:
Place $ in FOLLOW(S).
For every production A → αBβ, copy everything from FIRST(&beta) except ε into FOLLOW(B).
If there is a production A → αB or a production A → αBβ where FIRST(β) contains ε, place everything in FOLLOW(A) into FOLLOW(B).
As mentioned above, α is a possibly-empty string of grammar symbols. So it might not be visible.
Keep doing steps 2 and 3 until no new symbols are added to any follow set.
I'm pretty sure that the algorithm in your book differs only in notation conventions.
to determine if my parser is working correctly i need to find a lr(2+) grammar. After a quick research i have found this grammar and i believe that it is lr(2). However, i am not sure how to determine this.
Terminals: b, e, o, r, s
NonTerminals: A, B, E, Q, SL
Start: P
Productions:
P -> A
A -> E B SL E | b e
B -> b | o r
E -> e | Ɛ
SL -> s SL | s
I would be glad, if someone is able to confirm or deny that this grammar is lr(2) and at best give me a brief explanation on how to determine it by myself.
Thank you very much!
I'm pretty sure it's LR(2), but I don't have an LR(2) parser generator handy to test it, which would be the definitive way to do the test. Of course, you could generate the parser tables by hand. It's not that complicated a grammar, so it shouldn't take you too long.
It's certainly not LR(1), as can be seen from the pair of inputs:
b e
b s e
The left-most derivations are:
P->A->b e
P->E B SL E->B SL E->b SL E->b s E->b s e
So at the beginning of the parse, the parser can either shift a b in order to follow the first derivation chain or reduce an empty sequence to E in order to proceed with the second derivation chain. The second token is needed to choose between these two options, hence a lookahead of at least 2 is required.
As a side note, it should be pretty simple to mine StackOverflow for LR(2) grammars; they come up from time to time in questions. Here's a few I found by searching for LALR(2): (I used a Google search with site:stackoverflow.com because SO's own search engine doesn't do well with search patterns which aren't words. Not that Google does it well, but it does do it better.)
Solving bison conflict over 2nd lookahead
Solving small shift reduce conflict
Persistent Shift - Reduce Conflict in Goldparser
How to reduce parser stack or 'unshift' the current token depending on what follows?
I didn't verify the claims in those questions and answers, and there are other questions which didn't seem to have as clear a result.
The most classic LALR(2) grammar is the grammar for Yacc itself, which is pretty ironic. Here's a simplified version:
grammar: %empty | grammar production
production: ID ':' symbols
symbols: %empty | symbols symbol
symbol: ID | QUOTED_LITERAL
That simple grammar leaves out actions and the optional semicolon. But it captures the essence of the LALR(2)-ness of the grammar, which is precisely the result of the semicolon being optional. That's not a complaint; the grammar is unambiguous so the semicolon really is redundant and no-one should be forced to type a redundant 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).
What are FIRST and FOLLOW sets? What are they used for in parsing?
Are they used for top-down or bottom-up parsers?
Can anyone explain me FIRST and FOLLOW SETS for the following set of grammar rules:
E := E+T | T
T := T*V | T
V := <id>
They are typically used in LL (top-down) parsers to check if the running parser would encounter any situation where there is more than one way to continue parsing.
If you have the alternative A | B and also have FIRST(A) = {"a"} and FIRST(B) = {"b", "a"} then you would have a FIRST/FIRST conflict because when "a" comes next in the input you wouldn't know whether to expand A or B. (Assuming lookahead is 1).
On the other side if you have a Nonterminal that is nullable like AOpt: ("a")? then you have to make sure that FOLLOW(AOpt) doesn't contain "a" because otherwise you wouldn't know if to expand AOpt or not like here: S: AOpt "a" Either S or AOpt could consume "a" which gives us a FIRST/FOLLOW conflict.
FIRST sets can also be used during the parsing process for performance reasons. If you have a nullable nonterminal NullableNt you can expand it in order to see if it can consume anything, or it may be faster to check if FIRST(NullableNt) contains the next token and if not simply ignore it (backtracking vs predictive parsing). Another performance improvement would be to additionally provide the lexical scanner with the current FIRST set, so the scanner does not try all possible terminals but only those that are currently allowed by the context. This conflicts with reserved terminals but those are not always needed.
Bottom up parsers have different kinds of conflicts namely Reduce/Reduce and Shift/Reduce. They also use item sets to detect conflicts and not FIRST,FOLLOW.
Your grammar would't work with LL-parsers because it contains left recursion. But the FIRST sets for E, T and V would be {id} (assuming your T := T*V | T is meant to be T := T*V | V).
Answer :
E->E+T|T
left recursion
E->TE'
E'->+TE'|eipsilon
T->T*V|T
left recursion
T->VT'
T'->*VT'|epsilon
no left recursion in
V->(id)
Therefore the grammar is:
E->TE'
E'->+TE'|epsilon
T->VT'
T'->*VT'|epsilon
V-> (id)
FIRST(E)={(}
FIRST(E')={+,epsilon}
FIRST(T)={(}
FIRST(T')={*,epsilon}
FIRST(V)={(}
Starting Symbol=FOLLOW(E)={$}
E->TE',E'->TE'|epsilon:FOLLOW(E')=FOLLOW(E)={$}
E->TE',E'->+TE'|epsilon:FOLLOW(T)=FIRST(E')={+,$}
T->VT',T'->*VT'|epsilon:FOLLOW(T')=FOLLOW(T)={+,$}
T->VT',T->*VT'|epsilon:FOLLOW(V)=FIRST(T)={ *,epsilon}
Rules for First Sets
If X is a terminal then First(X) is just X!
If there is a Production X → ε then add ε to first(X)
If there is a Production X → Y1Y2..Yk then add first(Y1Y2..Yk) to first(X)
First(Y1Y2..Yk) is either
First(Y1) (if First(Y1) doesn't contain ε)
OR (if First(Y1) does contain ε) then First (Y1Y2..Yk) is everything in First(Y1) except for ε as well as everything in First(Y2..Yk)
If First(Y1) First(Y2)..First(Yk) all contain ε then add ε to First(Y1Y2..Yk) as well.
Rules for Follow Sets
First put $ (the end of input marker) in Follow(S) (S is the start symbol)
If there is a production A → aBb, (where a can be a whole string) then everything in FIRST(b) except for ε is placed in FOLLOW(B).
If there is a production A → aB, then everything in FOLLOW(A) is in FOLLOW(B)
If there is a production A → aBb, where FIRST(b) contains ε, then everything in FOLLOW(A) is in FOLLOW(B)
Wikipedia is your friend. See discussion of LL parsers and first/follow sets.
Fundamentally they are used as the basic for parser construction, e.g., as part of parser generators. You can also use them to reason about properties of grammars, but most people don't have much of a need to do this.