Bug fix in LL1 table generator

[cmaglie]

In the example exposed in #151:

%%

S -> A;
A -> "a" | ;

Parsing mode: LL(1).

Grammar:

    1. S -> A
    2. A -> 'a'
    3.    | ε

we get the wrong LL1 Parsing Table:

LL(1) parsing table:

┌───┬─────┬───┐
│   │ "a" │ $ │
├───┼─────┼───┤
│ S │ 1   │   │
├───┼─────┼───┤
│ A │ 2   │ 3 │
└───┴─────┴───┘

The problem is in the condition that checks the existence of ε, because a production may have other terminals besides ε.
In our case, the First(A) set equals to "a", ε. When we fill the table we should consider "a" and replace ε with the Follow(S) set.

After this patch we get the expected result:

LL(1) parsing table:

┌───┬─────┬───┐
│   │ 'a' │ $ │
├───┼─────┼───┤
│ S │ 1   │ 1 │
├───┼─────┼───┤
│ A │ 2   │ 3 │
└───┴─────┴───┘

Should fix #151

[DmitrySoshnikov]

Looks good, thanks for fixing it! Added one comment on the code.

[cmaglie]

Looks good, thanks for fixing it!

Thank you for making such good courses!