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 | |