We can find first and follow for left recursive grammer
Example:
E$\rightarrow$E+T/T
T$\rightarrow$T*F/F
F$\rightarrow$(E)/id
initialize first(x) to the empty set for every non-terminal x.
for every epsilon production of the form X$\rightarrow$$\epsilon$, add epsilon to first(x)
for every non epsilon productions of the form X$\rightarrow$a$\beta$ where a is a terminal. Add a to the first(x)
algorithm is;
repeat
let x$\rightarrow$y1,y2........yk be a non-terminal production
for i=1 to k
if yi is a terminal then break
add every non terminal from first(yi) to first(x)
if $\epsilon$not belongs to first(yi) then break
end for
until no more non terminals can be added to any set
for above examle productions are:
E$\rightarrow$E+T
E$\rightarrow$T
T$\rightarrow$T*F
T$\rightarrow$F
F$\rightarrow$(E)
F$\rightarrow$id
there are no epsilon productions in this grammer. And there are 2 productions whose R.H.S begins with terminal.
so before loop
FIRST(E) = {}
FIRST(T) = {}
FIRST(F) = { (, id }
T → F implies that we should add FIRST(F) to FIRST(T)
now,
FIRST(E) = {}
FIRST(T) = { (, id }
FIRST(T) = { (, id }
FIRST(F) = { (, id }
going through every production no more non terminals added to any of these sets
FOLLOW(E) = { $ }
FOLLOW(T) = { }
FOLLOW(F) = { }
E→E+T implies that + should be added to FOLLOW(E) and that everything in FOLLOW(E) should be added to FOLLOW(T):
FOLLOW(E) = { $, + }
FOLLOW(T) = { $, + }
FOLLOW(F) = { }
T→T*F implies that * should be added to FOLLOW(T) and that everything in
FOLLOW(T) should be added to FOLLOW(F):
FOLLOW(E) = { $, + }
FOLLOW(T) = { $, +, * }
FOLLOW(F) = { $, +, * }
F→(E) implies that ) should be added to FOLLOW(E):
FOLLOW(E) = { $, +, ) }
FOLLOW(T) = { $, +, * }
FOLLOW(F) = { $, +, * }
E→T implies that everything in FOLLOW(E) should be added to FOLLOW(T):
FOLLOW(E) = { $, +, ) }
FOLLOW(T) = { $, +, *, ) }
FOLLOW(F) = { $, +, * }
T→F implies that everything in FOLLOW(T) should be added to FOLLOW(F):
FOLLOW(E) = { $, +, ) }
FOLLOW(T) = { $, +, *, ) }
FOLLOW(F) = { $, +, *, ) }
