The solution of 4.6.2 that uneliminated left-recursion 4.4.1

[OneirdyniaYi]

The augmented grammar is 1)S'->S 2)S->SS+ 3)S->SS 4)S->a and the closure collection is i0: S'->.S S->.SS+ S->.SS S->.a i1: S'->S. S->S.S+ S->S.S S->.a S->.SS+ THIS 'S' IS THE SECOND 'S' FOLLOWING '.' OF S->S.S+ PRODUCTION S->.SS DITTO i2: S->a. i3: S->SS.+ S->SS. S->.a S->S.S+ THIS 'S' IS S->.SS+ PRODUCTION FROM i1 S->S.S DITTO S->.SS+ THIS 'S' IS THE SECOND 'S' FOLLOWING '.' OF S->S.S+ PRODUCTION S->.SS DITTO i4: S->SS+. i5: S->SS. FIRST(S) = {a} FOLLOW(S) = {a,+,,$} Sum up,the Fig is | a | + | | $ | S |
0 | s2 | | | | 1 | 1 | s2 | | | acc| 3 | 2 | r4 | r4 | r4 | r4 | | 3 | s2 | s4 | s5 | | 3 | 4 | r2 | r2 | r2 | r2 | | 5 | r3 | r3 | r3 | r3 | |