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Consider the following CFG:

$S\rightarrow Aa\mid ca$

$A\rightarrow c\mid d$

How many conflict occur in $CLR\left ( 1 \right )$ Parsing construction ?

I think $LR\left ( 0 \right )$ there is $1$ conflict, but in $SLR\left ( 1 \right )orCLR\left ( 1 \right )$ there won’t be any conflict. Someone verify it.

in Compiler Design by Veteran (117k points)
edited by | 66 views
There will be a SR conflict in both LR(0) and SLR(1)
S -> .ca and A -> .c    -------------->(on c)    S->c.a and A->c.
So, answer of this question will be 0

No, the grammar is neither LR(0), SLR(1) nor CLR(1).

Even incase of CLR(1) there will be a SR conflict b/w

S -> c.a , $
S -> c. , a
how CLR(1) has conflict ?

Is their look aheads same?
S -> c.a , $  ( it is shift in this case, so this will reside in a's column)

S -> c. , a ( it is reduce in this case, this goes in the same column as well)

we will compare lookaheads only for reduce not for shift.

yes, I got it now.

Actually lookaheads are not different.

It is same for the above two cases.

$S\rightarrow c.a,$$

$S\rightarrow c.,$$

by the way can u chk it too

pls check it again, lookaheads must be different.

S -> c.a , $
S -> c. , a

@balchandar reddy san

If different lookahead, then there won't be any SR conflict in CLR(1). Because, GOTO works only those rows, which have lookaheads

S -> c.a , $ ( this is shift conflict so, we have to take 'a'),

we have to take lookahead only in case of reduce conflict

1 Answer

0 votes

Grammer is not CLR(1) because it have Shift-Reduce conflict. 

S->c.a $   (Shift production)

A->c.   a  (Reduce production)

The shift reduce condition in CLR(1) is that Shifting symbol should not be appearing look ahead list of reducing production.

Here 'a' is shift in 1st production and  it also appear look ahead list of 2nd production.


And another alternative grammer is ambiguous it  have 2 parse tree for the string 'ca' if the grammer is ambigious it can not be CLR(1) 

by (403 points)

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