Get the algorithm to remove the indirect left recursion from a grammar from Aho et al. (2006). Use this algorithm to remove all left recursion from the following grammar: S→ Aa | Bb A→ Aa | Abc | c I Sb B → bb

Step-by-step explanation:

Here is the gramamr

S -> Aa | Bb

A -> Aa | Abc | c | Sb

B -> bb

Now for first gramamr step introduce new symbol S'

S -> BbS'

S' -> e | S'aA

Now for second gramamr statement...

A -> SbA' | cA'

A' -> e | A'a | A'bc

For third grammar statement it is not required since it doesn't contain any recursion

So final gramamr will be

S -> BbS'

S' -> e | S'aA

A -> SbA' | cA'

A' -> e | A'a | A'bc

B -> bb

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.