find a simple grammar (a.k.a s-grammar) for the following language:
L={(ab)2mb :m>=0}
[i did this but it is wrong]
S-> aASBB|b
A-> a
B->b
What about this?
S -> aA | T
A -> bB
B -> aC
C -> bS
T -> b
This is a regular grammar - all productions of the form X -> sY or X -> t, and corresponds to a minimal DFA for the language in question via a direct mapping of productions to transactions and nonterminal symbols to states.
Related
I am having trouble understanding what the rules are for adding a lookahead to a core production during the construction of the DFA. To illustrate my confusion, I will be using an online parser generator that exposes all the internal calculations; this_tool. (<- open in a new tab)
(The formating is: NONTERMINAL -> RULE, LOOKAHEADS, where the lookaheads are forward slash sperated)
Using this grammar as an example:
S -> E
E -> ( E )
E -> N O E
E -> N
N -> 1
N -> 2
N -> 3
O -> +
O -> -
Copy and pasting the above grammar into the lalr parser generator will produce a dfa with 12 states (click the >>). My question is finally, why are the goto(0, N) kernel productions ( {[E -> N.O E, $/)]; [E -> N., $/)]} ) initiated with the ) terminal? Where does the ) come from? I would expect the goto(0, N) to be {[E -> N.O E, $]; [E -> N., $]}. Equally the kernel production in the goto(0, ( ) has an 'extra' ).
As the dfa is being constructed, equal cores are merged (the core is the set of productions that introduce a new state by performing closure on that set). State 2 has production [E -> .N, )];, which when merged with [E -> N., $] produces the correct output, but there's no way for state 0 to have known about lookahead of )
Thanks in advance, sorry if this was a confusing and specific question and about using an external website to demonstrate my issue.✌️
The solution is to propagate any newly found lookaheads then 'goto' the states where those lookaheads are cores of.
The method is described in chapter 4 section 7.5 of the Dragon Book 2nd ed.
(here: https://github.com/muthukumarse/books/blob/master/Dragon%20Book%20Compilers%20Principle%20Techniques%20and%20Tools%202nd%20Edtion.pdf)
everybody!
When I learn dragon book, I encountered some trouble. I can't understand the first step in Eaxmple-4.64, which appears in subsection 4.7.5 and page 273.
Problem
At first, Eaxmple-4.61 gives an augmented non-SLR grammar. The original text is as follows:
Example 4.61 : We shall use as an example of the efficient LALR(1) table construction method the non-SLR grammar from Example 4.48, which we reproduce below in its augmented form:
S' -> S
S -> L = R | R
L -> *R | id
R -> L
Then, Eaxmple-4.64 wants to construct the kernels of LALR(1) items for the above grammer. The original text is as follows:
Eaxmple-4.64 : Let us construct the kernels of the LALR(1) items for the grammar of Example 4.61. The kernels of the LR(0) items were shown in Fig. 4.44. When we apply Algorithm 4.62 to the kernel of set of items I0, we first compute CLOSURE({ [S'->.S , #] }), which is
S' -> .S, #
S -> .L = R, #
S -> .R, #
L -> .*R, #/= // why is there a "=".
L -> .id, #/= // why is there a "=".
R -> .L, #
And the pseudo code CLOSUER(I) as follows:
But I think the answer is:
S' -> .S, #
S -> .L = R, #
S -> .R, #
L -> .*R, = // the difference
L -> .id, = // the difference
R -> .L, #
I don't know how the # is derived in L -> .*R, #/= and L -> .id, #/=. Could anybody tell me the reason. Thanks!
Both of them come from the closure of the item R→.L, #, which maps to A→α.Bβ, a with A=R, α=ε, B=L, β=ε, a=# so that FIRST(βa) is {#}, leading to the addition of both productions for L with lookahead #.
How do you identify whether a grammar is LL(1), LR(0), or SLR(1)?
Can anyone please explain it using this example, or any other example?
X → Yz | a
Y → bZ | ε
Z → ε
To check if a grammar is LL(1), one option is to construct the LL(1) parsing table and check for any conflicts. These conflicts can be
FIRST/FIRST conflicts, where two different productions would have to be predicted for a nonterminal/terminal pair.
FIRST/FOLLOW conflicts, where two different productions are predicted, one representing that some production should be taken and expands out to a nonzero number of symbols, and one representing that a production should be used indicating that some nonterminal should be ultimately expanded out to the empty string.
FOLLOW/FOLLOW conflicts, where two productions indicating that a nonterminal should ultimately be expanded to the empty string conflict with one another.
Let's try this on your grammar by building the FIRST and FOLLOW sets for each of the nonterminals. Here, we get that
FIRST(X) = {a, b, z}
FIRST(Y) = {b, epsilon}
FIRST(Z) = {epsilon}
We also have that the FOLLOW sets are
FOLLOW(X) = {$}
FOLLOW(Y) = {z}
FOLLOW(Z) = {z}
From this, we can build the following LL(1) parsing table:
a b z $
X a Yz Yz
Y bZ eps
Z eps
Since we can build this parsing table with no conflicts, the grammar is LL(1).
To check if a grammar is LR(0) or SLR(1), we begin by building up all of the LR(0) configurating sets for the grammar. In this case, assuming that X is your start symbol, we get the following:
(1)
X' -> .X
X -> .Yz
X -> .a
Y -> .
Y -> .bZ
(2)
X' -> X.
(3)
X -> Y.z
(4)
X -> Yz.
(5)
X -> a.
(6)
Y -> b.Z
Z -> .
(7)
Y -> bZ.
From this, we can see that the grammar is not LR(0) because there is a shift/reduce conflicts in state (1). Specifically, because we have the shift item X → .a and Y → ., we can't tell whether to shift the a or reduce the empty string. More generally, no grammar with ε-productions is LR(0).
However, this grammar might be SLR(1). To see this, we augment each reduction with the lookahead set for the particular nonterminals. This gives back this set of SLR(1) configurating sets:
(1)
X' -> .X
X -> .Yz [$]
X -> .a [$]
Y -> . [z]
Y -> .bZ [z]
(2)
X' -> X.
(3)
X -> Y.z [$]
(4)
X -> Yz. [$]
(5)
X -> a. [$]
(6)
Y -> b.Z [z]
Z -> . [z]
(7)
Y -> bZ. [z]
The shift/reduce conflict in state (1) has been eliminated because we only reduce when the lookahead is z, which doesn't conflict with any of the other items.
If you have no FIRST/FIRST conflicts and no FIRST/FOLLOW conflicts, your grammar is LL(1).
An example of a FIRST/FIRST conflict:
S -> Xb | Yc
X -> a
Y -> a
By seeing only the first input symbol "a", you cannot know whether to apply the production S -> Xb or S -> Yc, because "a" is in the FIRST set of both X and Y.
An example of a FIRST/FOLLOW conflict:
S -> AB
A -> fe | ε
B -> fg
By seeing only the first input symbol "f", you cannot decide whether to apply the production A -> fe or A -> ε, because "f" is in both the FIRST set of A and the FOLLOW set of A (A can be parsed as ε/empty and B as f).
Notice that if you have no epsilon-productions you cannot have a FIRST/FOLLOW conflict.
Simple answer:A grammar is said to be an LL(1),if the associated LL(1) parsing table has atmost one production in each table entry.
Take the simple grammar A -->Aa|b.[A is non-terminal & a,b are terminals]
then find the First and follow sets A.
First{A}={b}.
Follow{A}={$,a}.
Parsing table for Our grammar.Terminals as columns and Nonterminal S as a row element.
a b $
--------------------------------------------
S | A-->a |
| A-->Aa. |
--------------------------------------------
As [S,b] contains two Productions there is a confusion as to which rule to choose.So it is not LL(1).
Some simple checks to see whether a grammar is LL(1) or not.
Check 1: The Grammar should not be left Recursive.
Example: E --> E+T. is not LL(1) because it is Left recursive.
Check 2: The Grammar should be Left Factored.
Left factoring is required when two or more grammar rule choices share a common prefix string.
Example: S-->A+int|A.
Check 3:The Grammar should not be ambiguous.
These are some simple checks.
LL(1) grammar is Context free unambiguous grammar which can be parsed by LL(1) parsers.
In LL(1)
First L stands for scanning input from Left to Right. Second L stands
for Left Most Derivation. 1 stands for using one input symbol at each
step.
For Checking grammar is LL(1) you can draw predictive parsing table. And if you find any multiple entries in table then you can say grammar is not LL(1).
Their is also short cut to check if the grammar is LL(1) or not .
Shortcut Technique
With these two steps we can check if it LL(1) or not.
Both of them have to be satisfied.
1.If we have the production:A->a1|a2|a3|a4|.....|an.
Then,First(a(i)) intersection First(a(j)) must be phi(empty set)[a(i)-a subscript i.]
2.For every non terminal 'A',if First(A) contains epsilon
Then First(A) intersection Follow(A) must be phi(empty set).
An ambiguous grammar is given and I am asked to rewrite the grammar to make it unambiguous. In fact, I don't know why the given grammar is ambiguous, let alone rewriting it to an unambiguous one.
The given grammar is S -> SS | a | b , and I have four choices:
A: S -> Sa | Sb | epsilon
B: S -> SS’
S’-> a | b
C: S -> S | S’
S’-> a | b
D: S -> Sa | Sb.
For each choice, I have already know that D is incorrect because it generates no strings at all,C is incorrect because it only matches the strings 'a' and 'b'.
However, I think the answer is A while the correct answer is B.I think B is wrong because it just generates S over and over again, and B can't deal with empty strings.
Why is the given grammar ambiguous？
Why is A incorrect while B is correct？
The original grammar is ambiguous because multiple right-most (or left-most) derivations are possible for any string of at least three letters. For example:
S -> SS -> SSS -> SSa -> Saa -> aaa
S -> SS -> Sa -> SSa -> Saa -> aaa
The first one corresponds, roughly speaking, to the parse a(aa) while the second to the parse (aa)a.
None of the alternatives is correct. A incorrectly matches ε while B does not match anything (like D). If B were, for example,
S -> SS' | S'
S' -> a | b
it would be correct. (This grammar is left-associative.)
I can't seem to figure out the Unrestricted Grammar for
L = (w am bn | w={a,b}* m=number of a's in w n=number of b's in w).
I've constructed the following grammar for it, but it keeps rejecting every string I enter in JFLAP. But manually creating a parse tree for it gives me no problem. Can anyone look at it for me and see what's wrong?
S -> AST | BSU | epsilon
UT -> TU
T -> A
U -> B
A -> a
B -> b
I've downloaded and used JFLAP on your grammar. I think the issue is that you have not used the notation that JFLAP does for grammar entry. It does not used the | symbol, but you have to supply several rules instead. Therefore in JFLAP notation (and still and valid grammar) you would have:
S -> AST
S -> BSU
S -> ε
UT -> TU
T -> A
U -> B
A -> a
B -> b
You would also need to set the empty string as ε in the FLAP preferences. If you can manually create a parse tree you can also do this in JFLAP to show the derivations.