Consider the following state diagram which accepts the alphabet {0,1} and accepts if the input string has two consecutive 0's or 1's:
01001 --> Accept
101 --> Reject
How would I write the production rules to show this? Is it just:
D -> C0 | B1 | D0 | D1
C -> A0 | B0
B -> A1 | C1
And if so, how would the terminals (0,1) be differentiated from the states (A,B,C) ? And should the state go before or after the input? That is, should it be A1 or 1A for example?
The grammar you suggest has no A: it's not a non-terminal because it has no production rules, and it's not a terminal because it's not present in the input. You could make that work by writing, for example, C → 0 | B 0, but a more general solution is to make A into a non-terminal using an ε-rule: A → ε and then
C → A 0 | B 0.
B0 is misleading, because it looks like a single thing. But it's two grammatical symbols, a non-terminal (B) and a terminal 0.
With those modifications, your grammar is fine. It's a left linear grammar; a right linear grammar can also be constructed from the FSA by considering in-transitions rather than out-transitions. In this version, the epsilon production corresponds to final states rather than initial states.
A → 1 B | 0 C
B → 0 C | 1 D
C → 1 B | 0 D
D → 0 D | 1 D | ε
If it's not obvious why the FSM corresponds to these two grammars, it's probably worth grabbing a pad of paper and constructing a derivation with each grammar for a few sample sentences. Compare the derivations you produce with the progress through the FSM for the same input.
Related
P → PL | L
L → N; | M; | C
N → print E
M → print "W"
W → TW | ε
C → if E {P} | if E {P} else {P}
E → (EOE) | V (note: this has a variable O)
O → + | - | *
V → 0 | 1 | 2 | 3 (note: this has a terminal 0 (zero))
T → a | b | c | d
For the above grammar G, is it not LL(1) because it evokes FIRST/FIRST conflict when trying to predict the production of P?
I am really struggling on how to prove that it is not LL(1)...
Any help or advice would be very thankful!
A left-recursive grammar cannot be LL(k) for any k. P → P L is left-recursive.
In addition, L has two productions starting with the same terminal, so it is impossible to choose between them with only one symbol of lookahead.
I have been asked to convert:
S → Sa | bSb | bc
to LL(1) so far I have:
S → bY
Y → SbF | cF
F → aF | ε
Is this LL(1)? If not would this be LL(1):
S → bY
Y → bYbF | cF
F → aF | ε
if neither of these would somebody please give me the correct answer and why thanks in advance!
This is what I would do:
S → Sa | bSb | bc
Remove left recursion:
F -> aF | EPSILON```
Now left factor:
F -> aF | EPSILON
X -> SbF | cF```
Check the First and Follows:
S: b
X: b, c
F: a, EPSILON```
```Follows():
S: $, b
X: $, b
F: $, b```
Everything checks out so it is LL(1) parsable.
I am new to this topic of left recursion and left factoring, please help me in determining whether this grammar is left recursive or left factored, if it is then why ?
S-> aAd | bBd | aBe | bAe | cA | cB
Is it Left recursive? Answer: No.
By definition, "A grammar is left-recursive if we can find some non-terminal A which will eventually derive a sentential form with itself as the left-symbol."
example:
Immediate left recursion occurs in rules of the form
A -> Aa | b
where a and b are sequences of nonterminals and terminals, and b doesn't start with A.
Clearly not the case in:
S-> aAd | bBd | aBe | bAe | cA | cB
Is it Left factored? Answer: Yes.
By definition, in left factoring, it is not clear which two alternative production to choose, to expand a non-terminal.
This ambiguity occurs, when you have two alternative production that starts with same terminal/non-terminal.
In your case, I can see that thrice, two alternative paths:
S-> aAd | aBe
S-> bBd | bAe
S-> cA | cB
If I remove the left factoring then the grammar becomes:
S-> aA'
A'-> Ad | Be
S-> bB'
B'-> Bd | Ae
S-> cC'
C'-> A | B
This slide explains the same in simpler words
How can i implement an eliminator for this?
A := AB |
AC |
D |
E ;
This is an example of so called immediate left recursion, and is removed like this:
A := DA' |
EA' ;
A' := ε |
BA' |
CA' ;
The basic idea is to first note that when parsing an A you will necessarily start with a D or an E. After the D or an E you will either end (tail is ε) or continue (if we're in a AB or AC construction).
The actual algorithm works like this:
For any left-recursive production like this: A -> A a1 | ... | A ak | b1 | b2 | ... | bm replace the production with A -> b1 A' | b2 A' | ... | bm A' and add the production A' -> ε | a1 A' | ... | ak A'.
See Wikipedia: Left Recursion for more information on the elimination algorithm (including elimination of indirect left recursion).
Another form available is:
A := (D | E) (B | C)*
The mechanics of doing it are about the same but some parsers might handle that form better. Also consider what it will take to munge the action rules along with the grammar its self; the other form requires the factoring tool to generate a new type for the A' rule to return where as this form doesn't.
I guess the same logic is applied in both of them, i.e replacing the matched strings with the corresponding non-terminal elements as provided in the production rules.
Why do they categorize LL as top down and LR as bottom-up?
Bottom up parsing:
Bottom-up parsing (also known as
shift-reduce parsing) is a strategy
for analyzing unknown data
relationships that attempts to
identify the most fundamental units
first, and then to infer higher-order
structures from them. It attempts to
build trees upward toward the start
symbol.
Top-down parsing:
Top-down parsing is a strategy of
analyzing unknown data relationships
by hypothesizing general parse tree
structures and then considering
whether the known fundamental
structures are compatible with the
hypothesis.
Top down parsing
involves to generating the string from first non-terminal.
Example: recursive descent parsing,non-recursive descent parsing, LL parsing, etc.
The grammars with left recursive and left factoring do not work.
Might occur backtracking.
Use of left most derivation
Things Of Interest Blog
The difference between top-down parsing and bottom-up parsing
Given a formal grammar and a string produced by that grammar, parsing is figuring out the production process for that string.
In the case of the context-free grammars, the production process takes the form of a parse tree. Before we begin, we always know two things about the parse tree: the root node, which is the initial symbol from which the string was originally derived, and the leaf nodes, which are all the characters of the string in order. What we don't know is the layout of nodes and branches between them.
For example, if the string is acddf, we know this much already:
S
/|\
???
| | | | |
a c d d f
Example grammar for use in this article
S → xyz | aBC
B → c | cd
C → eg | df
Bottom-up parsing
This approach is not unlike solving a jigsaw puzzle. We start at the bottom of the parse tree with individual characters. We then use the rules to connect the characters together into larger tokens as we go. At the end of the string, everything should have been combined into a single big S, and S should be the only thing we have left. If not, it's necessary to backtrack and try combining tokens in different ways.
With bottom-up parsing, we typically maintain a stack, which is the list of characters and tokens we've seen so far. At each step, we shift a new character onto the stack, and then reduce as far as possible by combining characters into larger tokens.
Example
String is acddf.
Steps
ε can't be reduced
a can't be reduced
ac can be reduced, as follows:
reduce ac to aB
aB can't be reduced
aBd can't be reduced
aBdd can't be reduced
aBddf can be reduced, as follows:
reduce aBddf to aBdC
aBdC can't be reduced
End of string. Stack is aBdC, not S. Failure! Must backtrack.
aBddf can't be reduced
ac can't be reduced
acd can be reduced, as follows:
reduce acd to aB
aB can't be reduced
aBd can't be reduced
aBdf can be reduced, as follows:
reduce aBdf to aBC
aBC can be reduced, as follows:
reduce aBC to S
End of string. Stack is S. Success!
Parse trees
|
a
| |
a c
B
| |
a c
B
| | |
a c d
B
| | | |
a c d d
B
| | | | |
a c d d f
B C
| | | |\
a c d d f
| |
a c
| | |
a c d
B
| /|
a c d
B
| /| |
a c d d
B
| /| | |
a c d d f
B C
| /| |\
a c d d f
S
/|\
/ | |
/ B C
| /| |\
a c d d f
Example 2
If all combinations fail, then the string cannot be parsed.
String is acdg.
Steps
ε can't be reduced
a can't be reduced
ac can be reduced, as follows:
reduce ac to aB
aB can't be reduced
aBd can't be reduced
aBdg can't be reduced
End of string. Stack is aBdg, not S. Failure! Must backtrack.
ac can't be reduced
acd can be reduced, as follows:
reduce acd to aB
aB can't be reduced
aBg can't be reduced
End of string. stack is aBg, not S. Failure! Must backtrack.
acd can't be reduced
acdg can't be reduced
End of string. Stack is is acdg, not S. No backtracking is possible. Failure!
Parse trees
|
a
| |
a c
B
| |
a c
B
| | |
a c d
B
| | | |
a c d g
| |
a c
| | |
a c d
B
| /|
a c d
B
| /| |
a c d g
| | |
a c d
| | | |
a c d g
Top-down parsing
For this approach we assume that the string matches S and look at the internal logical implications of this assumption. For example, the fact that the string matches S logically implies that either (1) the string matches xyz or (2) the string matches aBC. If we know that (1) is not true, then (2) must be true. But (2) has its own further logical implications. These must be examined as far as necessary to prove the base assertion.
Example
String is acddf.
Steps
Assertion 1: acddf matches S
Assertion 2: acddf matches xyz:
Assertion is false. Try another.
Assertion 2: acddf matches aBC i.e. cddf matches BC:
Assertion 3: cddf matches cC i.e. ddf matches C:
Assertion 4: ddf matches eg:
False.
Assertion 4: ddf matches df:
False.
Assertion 3 is false. Try another.
Assertion 3: cddf matches cdC i.e. df matches C:
Assertion 4: df matches eg:
False.
Assertion 4: df matches df:
Assertion 4 is true.
Assertion 3 is true.
Assertion 2 is true.
Assertion 1 is true. Success!
Parse trees
S
|
S
/|\
a B C
| |
S
/|\
a B C
| |
c
S
/|\
a B C
/| |
c d
S
/|\
a B C
/| |\
c d d f
Example 2
If, after following every logical lead, we can't prove the basic hypothesis ("The string matches S") then the string cannot be parsed.
String is acdg.
Steps
Assertion 1: acdg matches S:
Assertion 2: acdg matches xyz:
False.
Assertion 2: acdg matches aBC i.e. cdg matches BC:
Assertion 3: cdg matches cC i.e. dg matches C:
Assertion 4: dg matches eg:
False.
Assertion 4: dg matches df:
False.
False.
Assertion 3: cdg matches cdC i.e. g matches C:
Assertion 4: g matches eg:
False.
Assertion 4: g matches df:
False.
False.
False.
Assertion 1 is false. Failure!
Parse trees
S
|
S
/|\
a B C
| |
S
/|\
a B C
| |
c
S
/|\
a B C
/| |
c d
Why left-recursion is a problem for top-down parsers
If our rules were left-recursive, for example something like this:
S → Sb
Then notice how our algorithm behaves:
Steps
Assertion 1: acddf matches S:
Assertion 2: acddf matches Sb:
Assertion 3: acddf matches Sbb:
Assertion 4: acddf matches Sbbb:
...and so on forever
Parse trees
S
|
S
|\
S b
|
S
|\
S b
|\
S b
|
S
|\
S b
|\
S b
|\
S b
|
...