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I have been asked to convert:

S → Sa | bSb | bc

to LL(1) so far I have:

S → bY
Y → SbF | cF
F → aF | ε

Is this LL(1)? If not would this be LL(1):

S → bY
Y → bYbF | cF
F → aF | ε

if neither of these would somebody please give me the correct answer and why thanks in advance!

kennytm
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vezbea
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  • What have you done so far to check whether your grammar is LL(1)? Ultimately, that's probably more valuable than just getting a binary yes/no here. – templatetypedef May 21 '16 at 04:48
  • Well it is for a past paper which had no answers available. I've derived some words from it and it seems ok and I'm pretty sure it is ll(1) I'm just unsure about the S in the Y variable and whether I would need to substitute it in or not. – vezbea May 21 '16 at 16:18

1 Answers1

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This is what I would do: 

S → Sa | bSb | bc

  1. Remove left recursion: 

    F -> aF | EPSILON```

  2. Now left factor:

    F -> aF | EPSILON
X -> SbF | cF```

  3. Check the First and Follows:

    S: b
X: b, c
F: a, EPSILON```

```Follows():
S: $, b
X: $, b
F: $, b```

Everything checks out so it is LL(1) parsable.