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Test by Bikram | Compiler Design | Test 1 | Question: 12

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@bikram sir ,

The first closure itself has a reduce move ( epsilon to X) and a shift move ( a or b) and as reduce moves are placed in the entire row for LR(0) there is a shift reduce conflict. 

Is my reasoning correct ?

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Closure [$S' \rightarrow .S$] = 

$S' \rightarrow .S$

$S \rightarrow X$

$X \rightarrow .YX$

$X \rightarrow .$

$Y \rightarrow .aY$

$Y \rightarrow .b$

There are two SR conflicts(shift on $a, b$ & reduce on $X \rightarrow .$ in this $LR(0) $ item.

Hence it is not LR(0) grammar

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