CS606 Assignment 3 Solution Spring 2021 | CS606 Assignment No. 3 100% Correct Solution by Mustakbil Corner

CS606 Assignment 3 Solution Spring 2021 | CS606 Assignment No. 3 100% Correct Solution by Mustakbil Corner 

Following is the Question 

Question:

 

Consider the following CFG. 

 

S ®tTd | fFd | tFe | fTe

T ®a

F ®  a

 

 

For above grammar construct the canonical collection for LR(1) items.

 

 

 

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Cs606 Assignment 3 Solution BY Mustakbil Corner

 

 

 

 

Question:

 

Consider the following CFG. 

 

S ®tTd | fFd | tFe | fTe

T ®a

F ®  a

 

 

For above grammar construct the canonical collection for LR (1) items.

 

 

 

Solution

 

Add augment production, insert  ^. (Dot) symbol at the first position for  every Production of  “G”. Also add the look ahead.

 

 

S` à.S , $

 

I0 State :

 

Add augment production to the I0 State and compute the closure

 

I0 = closure (S` à.S )

 

Add all production starting with “S” in  I0 State because “.” Is followed by non-terminal, So the I0 State becomes  

 

 

S à.S , $

 

S à.tTd , $

 

S à.fFd , $

 

S à.tFe , $

 

S à.fTe , $

 

I1 = Go to ( I0 , S ) = S` àS. , $ (Accepting State)

I2 = Go to ( I0 , t ) = S àt.Td , $

Add all production starting with small “t” in I2 State because “.” Is followed by non-terminal.

So, the I2 State becomes

 

I2     =  S à.tTd , $

 

 S à.tTd , $

 S à  t.Fe , $

 T à.a , d

  F à.b , e

 

 

I3 = Go to ( I0 , f ) = S àf .Fd , $

 

Add all production starting with small “f” in I3 State because “.” Is followed by non-terminal.

So, the I3 State becomes

 

I3 = S à f .Fd , $

 

            S àf .Te , $

 

T à.a , e

 

 F à.b , d

 

 

 

I4 = Go to ( I2 , T ) = Closure ( S àtT . d ,$ ) = S àtT . d ,$

 

I5 = Go to ( I2 , F ) = Closure ( S àtF . e ,$ ) = S àtF . e ,$

 

I6 = Go to ( I2 , a ) = Closure ( T àa ., d )  = T àa ., d

 

I7 = Go to ( I3 , F ) = Closure ( F àb ., e  ) = F àb ., e

 

I8 = Go to ( I2 , b ) = Closure ( S àfF .d , $  ) = S àfF .d , $ 

 

I9 = Go to ( I3 , T ) = Closure ( S àfT . e ,$  ) = S àfT . e ,$ 

 

I10 = Go to ( I3 , a ) = Closure ( T àa ., e  ) = T àa ., e  (*)

 

I11 = Go to ( I3 , b ) = Closure ( F àb ., d  ) = F àb ., d (*)

 

I12 = Go to ( I4 , d ) = Closure ( S àtTd ., $  ) = S àtTd ., $  (*)

 

I13 = Go to ( I5 , e ) = Closure ( S àtFe ., $  ) = S àtFe ., $  (*)

 

I14 = Go to ( I8 , F ) = Closure ( S àtFd ., $  ) = S àtFd ., $  (*)

 

I15 = Go to ( I9 , e ) = Closure ( S àfTe ., $  ) = S àfTe ., $  (*)

 

 

 

 

This is 100% Correct Solution

 

NOTE: Do not Copy Make your Own

 

 

 

 

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