GO-Compiler1: Parsing-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

Sir, cannot we have a reduce move and a GOTO action in same canonical item ?

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Not for LR(0) parser

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I just realized how dumb that question was. Sorry sir

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If there is a reduce move we place it only in the action part right???

So what happens when we have a reduce item and Goto item in the same state??

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When we reduce using a production say $A \to Bcd$, we pop 3 items from stack - since production used for REDUCE has 3 symbols, and move to a state given by GOTO(S, A) where S is the stack top after the pop moves.

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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"

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

@Arjun sir, isn't B false?

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Yes. Just corrected the option

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Sir, can we say that since a given regular grammar can be ambiguous and LR(0) parsers can't parse any ambiguous grammar, so option (d) is false ?

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Can you give an ambiguous regular grammar?

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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

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.