This is a question from the Dragon Book. This is the grammar:

$S \to AaAb \mid BbBa $
$A \to \varepsilon$
$B \to \varepsilon$

The question asks how to show that it is LL(1) but not SLR(1).

To prove that it is LL(1), I tried constructing its parsing table, but I am getting multiple productions in a cell, which is contradiction.

Please tell how is this LL(1), and how to prove it?

  • 6
    $\begingroup$ I am not very fammiliar with grammars, but it seems that language of this grammar is finite. $L=\{ab,ba\}$ $\endgroup$
    – Nejc
    Nov 19, 2012 at 14:44
  • $\begingroup$ @Nejc: Yes it does seems like that! $\endgroup$ Nov 19, 2012 at 14:45

3 Answers 3


First, let's give your productions a number.

1 $S \to AaAb$
2 $S \to BbBa$
3 $A \to \varepsilon$
4 $B \to \varepsilon$

Let's compute the first and follow sets first. For small examples such as these, using intuition about these sets is enough.

$$\mathsf{FIRST}(S) = \{a, b\}\\ \mathsf{FIRST}(A) = \{\}\\ \mathsf{FIRST}(B) = \{\}\\ \mathsf{FOLLOW}(A) = \{a, b\}\\ \mathsf{FOLLOW}(B) = \{a, b\}$$

Now let's compute the $LL(1)$ table. By definition, if we don't get conflicts, the grammar is $LL(1)$.

    a | b |
S | 1 | 2 |
A | 3 | 3 |
B | 4 | 4 |

As there are no conflicts, the grammar is $LL(1)$.

Now for the $SLR(1)$ table. First, the $LR(0)$ automaton.

$$\mbox{state 0}\\ S \to \bullet AaAb\\ S \to \bullet BbBa\\ A \to \bullet\\ B \to \bullet\\ A \implies 1\\ B \implies 5\\ $$$$\mbox{state 1}\\ S \to A \bullet aAb\\ a \implies 2\\ $$$$\mbox{state 2}\\ S \to Aa \bullet Ab\\ A \to \bullet\\ A \implies 3\\ $$$$\mbox{state 3}\\ S \to AaA \bullet b\\ b \implies 4\\ $$$$\mbox{state 4}\\ S \to AaAb \bullet b\\ $$$$\mbox{state 5}\\ S \to B \bullet bBa\\ b \implies 6\\ $$$$\mbox{state 6}\\ S \to Bb \bullet Ba\\ B \to \bullet\\ B \implies 7\\ $$$$\mbox{state 7}\\ S \to BbB \bullet a \\ a \implies 8\\ $$$$\mbox{state 8}\\ S \to BbBa \bullet \\ $$

And then the $SLR(1)$ table (I assume $S$ can be followed by anything).

    a     | b     | A | B |
0 | R3/R4 | R3/R4 | 1 | 5 |
1 | S2    |       |   |   |
2 | R3    | R3    | 3 |   |
3 |       | S4    |   |   |
4 | R1    | R1    |   |   |
5 |       | S4    |   |   |
6 | R4    | R4    |   | 7 |
7 | S8    |       |   |   |
8 | R2    | R2    |   |   |

There are conflicts in state 0, so the grammar is not $SLR(1)$. Note that if $LALR(1)$ was used instead, then both conflicts would be resolved correctly: in state 0 on lookahead $a$ $LALR(1)$ would take R3 and on lookahead $b$ it would take R4.

This gives rise to the interesting question whether there is a grammar that is $LL(1)$ but not $LALR(1)$, which is the case but not easy to find an example of.

  • $\begingroup$ Thanks! I had constructed the First & Follow correctly, but I made a mistake in constructing the table. $\endgroup$ Nov 20, 2012 at 5:44

If you are not asked, you don't have to construct the LL(1) table to prove that it is an LL(1) grammar. You just compute the FIRST/FOLLOW sets as Alex did:

$\qquad \begin{align} \operatorname{FIRST}(S)&={a,b} \\ \operatorname{FIRST}(A)&={ε} \\ \operatorname{FIRST}(B)&={ε} \\ \operatorname{FOLLOW}(A)&={a,b} \\ \operatorname{FOLLOW}(B)&={a,b} \end{align}$

And then, by definition an LL(1) grammar has to:

  1. If $A \Rightarrow a$ and $A \Rightarrow b$ are two different rules of the grammar, then it should be that $\operatorname{FIRST}(a) \cap \operatorname{FIRST}(b) = \emptyset$. Hence, the two sets haven't any common element.
  2. If for any non-terminal symbol $A$ you have $Α \Rightarrow^* ε$, then it should be that $\operatorname{FIRST}(A) \cap \operatorname{FOLLOW}(A) = \emptyset$. Hence, if there is a zero production for a non-terminal symbol, then the FIRST and FOLLOW sets can't have any common element.

So, for the given grammar:

  1. We have $\operatorname{FIRST}(AaAb) \cap \operatorname{FIRST}(BbBa) = \emptyset$ since $\operatorname{FIRST}(AaAb) = \{a\}$ while $\operatorname{FIRST}(BbBa) = \{b\}$ and they don't have any common elements.
  2. $\operatorname{FIRST}(A) \cap \operatorname{FOLLOW}A) = \emptyset$ since $\operatorname{FIRST}(A) = \{a,b\}$ while $\operatorname{FOLLOW}(A) = \emptyset$, and now $\operatorname{FIRST}(B) \cap \operatorname{FOLLOW}(B) = \emptyset$ since $\operatorname{FIRST}(B) = \{ε\}$ while $\operatorname{FOLLOW}(B) = \{a,b\}$.

As for the SLR(1) analysis I think it is flawless!

  • $\begingroup$ Welcome! In order to improve this answer, why don't you apply what you state to the grammar at hand? $\endgroup$
    – Raphael
    Dec 3, 2012 at 21:19
  • $\begingroup$ Happy to be here!! Answered your request and I think I gave a thorough explanation! $\endgroup$
    – Ethan
    Dec 3, 2012 at 22:04
  • $\begingroup$ Thanks! Note that we can use LaTeX here, as I did edit in for your maths. $\endgroup$
    – Raphael
    Dec 3, 2012 at 22:23
  • 1
    $\begingroup$ Wow Thanks! this is a great explanation. But I think there is some mistake in the application. Isn't First(A) = {epsilon}? I think you swapped the FIRST and FOLLOW. $\endgroup$ Dec 4, 2012 at 5:59
  • 1
    $\begingroup$ FIRST(A) is indeed epsilon but since you are looking to calculate the whole right member's FIRST set, A -> ε just shows that we have an empty production and the first terminal symbol you see (and therefore the FIRST set of it) is terminal symbol a. Hope this helped! $\endgroup$
    – Ethan
    Dec 5, 2012 at 10:33

Search for a sufficient condition which makes a grammar LL(1) (hint: look at the FIRST sets).

Search for a needed condition which all SLR(1) grammars must meet (hint: look at the FOLLOW sets).


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