如何将此语法转换为LL（1）？

Question

S → Y | 1X

X → 1X | 0

Y → Y 0 | 1X1 | 2X2

I do not understand how to do factoring or substitution when there are more than one of the same symbol. Thanking you.

Answer 1

X → 1 X | 0

refers to a 0-or-more 1 followed by 0 , so it's equivalent to

X → 1* 0

The same approach can be used to remove your left-recursion. You can rewrite

S → Y | 1 X
X → 1 X | 0
Y → Y 0 | 1 X 1 | 2 X 2

as

S → Y | 1 X
X → 1* 0
Y → ( 1 X 1 | 2 X 2 )* 0

In EBNF:

S → Y | 1 X
X → 1* 0
Y → A* 0
A → 1 X 1 | 2 X 2

In BNF:

S → Y | 1 X
X → 1* 0
Y → A* 0
A → 1 X 1 | 2 X 2

1* → 1 1* | Ɛ
A* → A A* | Ɛ

If all you wanted to do was to eliminated left-recursion, you're done.

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If you want to eliminate common prefixes too, you're not done because both sub-rules of S can start with 1 X . To fix this, inline and distribute to obtain the following:

S → 0
  | 1 X 1 Y
  | 2 X 2 Y
  | 1 X
X → 1* 0
Y → A* 0
A → 1 X 1 | 2 X 2

Now, we're finally in a position to factor out the common 1 X .

S → 0
  | 1 X ( 1 Y )?
  | 2 X 2 Y
X → 1* 0
Y → A* 0
A → 1 X 1 | 2 X 2

In EBNF:

S → 0 | 1 X B? | 2 X 2 Y
X → 1* 0
Y → A* 0
A → 1 X 1 | 2 X 2
B → 1 Y

In BNF:

S → 0 | 1 X B? | 2 X 2 Y
X → 1* 0
Y → A* 0
A → 1 X 1 | 2 X 2
B → 1 Y

B? → B    | Ɛ
1* → 1 1* | Ɛ
A* → A A* | Ɛ