GATE Overflow | Compiler Design | Test 1 | Parsing | Question: 5

Question

Which of the following statements regarding $LR(0)$ parser is FALSE?

• A. A $LR(0)$ configurating set cannot have multiple reduce items
• B. A $LR(0)$ configurating set cannot have  both shift as well as reduce items
• C. If a reduce item is present in a $LR(0)$ configurating set it cannot have any other item
• D. A $LR(0)$ parser can parse any regular grammar

Comments

LR(0) is a top-down parser, it can not parse ambiguous grammar, even all parser can't parse ambiguous grammar and every regular grammar need't be unambiguous  

example: A-->Ba/aa

                  B-->a

for above regular grammar has two parse tree for string "aa"

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@Arjun @GO Classes Deepak Poonia

Hi sir, If I take a grammar:

S -> M A
M ->
A -> a

Then, this grammer is LR(0) and also have GOTO action with REDUCED action. So, can I say that Op C is also false?

Please correct me sir If I am wrong.

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I am not getting how 1,2,3 are true,as LR(0) parser can have both r-r and s-r conflict possible,

hence a LR(0) configurating set can have multple reduce item,etc

Answer 1

No grammar with empty production can be LR(0). But empty rules are allowed in regular grammar as the only condition for a grammar to be regular is to be either left linear or right linear.

Comments

Ambiguous Regular:

S→A | B
A→a
B→a

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Yes. 
https://math.stackexchange.com/questions/1773779/can-a-regular-grammar-be-ambiguous

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I have doubt in option C.. it can be the case that A-> a. is a reduce move and along with it we can have other items as B->a.C and C->.d (in the same state of LR0 items)

Answer 2

Since LR(0) parser places reduce-moves in the entire row of "Action", having anything more than just the reduce-move in the state having final-item would lead to SR or RR conflict.

So, Options A, B and C are true.

 

So Option D must be false. But why is that?

Intuitively, there can be a Regular Language that has its grammar such that the state having a final-item would have some additional item.

But to give a counter example:

$S\rightarrow aA$,

$A\rightarrow b|\epsilon$

It has an $\epsilon$ production. Any grammar with $\epsilon$ production can't be LR(0), because in the parsing table, we'd put reduce-

moves in the corresponding row, on which a shift-move can result in an SR conflict.

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In contrast, an SLR(1) parser can have multiple reduce-moves, or both shift and reduce moves in the same state.

Answer 3

 

1. A LR(0) configurating set cannot have multiple reduce items => This is true bcoz multiple reduce items will create a RR conflict 
2. A LR(0) configurating set cannot have  both shift as well as reduce items => This is true bcoz multiple reduce items will create a SR conflict
3. If a reduce item is present in a LR(0) configurating set it cannot have any other item => True bcoz in LR(0) the entire row is filled with reduce part. So having anything else will 100% be a conflict.
4. A LR(0) parser can parse any regular grammar . False => Bcoz LR(0) cant parse Ambiguos Grammars and Regular G can be ambigious too.

So Ans is D