I was wondering if I could say any of the following is true.

Given a grammar $G$,

  1. If the LALR parser has reduce-reduce conflict for $G$, then the SLR parser also has reduce-reduce conflict for $G$.
  2. If the SLR parser has reduce-reduce conflict for $G$, then the LALR parser also has reduce-reduce conflict.
  3. The LALR parser has reduce-reduce conflict for $G$ if and only if SLR also has reduce-reduce conflict for $G$.

Since SLR $\subset$ LALR, I think point 1 is true. Is this wrong?

Further more, based on a few examples that I have come across it seems like points 2 & 3 are also true. Is this correct. If so, could you please point me to the proof. If not could you please give me a counter example?


1 Answer 1


Note: I'm assuming that when you wrote $SLR$ and $LALR$ that you actually meant $SLR(1)$ and $LALR(1)$.

It is certainly the case that if a grammar shows a conflict in an $LALR(1)$ automaton, that conflict will also be present in the $SLR(1)$ automaton, because the two automata have the same states and the $LALR(1)$ lookaheads for any production are a subset of the $SLR(1)$ lookaheads.

The remaining two proposals are false, as can be demonstrated by the same counter-example.

The classic example of an $LALR(1)$ grammar which is not $SLR(1)$ (taken directly from the Dragon book) is the abstraction of a grammar which attempts to discriminate between "lvalue" and "rvalue" syntaxes. (Roughly speaking, an "lvalue" is something which can be assigned to, so-named because it can appear on the left-hand side of an assignment operator; an "rvalue" is a value which cannot be assigned to, which means that it can only appear on the right-hand side.

It's usually convenient to include the production $R\to L$, which says that "lvalues" can also appear on the right-hand side of the assignment operator. But it is precisely this fact which leads to the conflict:

$$\begin{align}S &\to L = R\\ S &\to R\\ L &\to * R\\ L &\to id\\ R &\to L\\ \end{align}$$ As the Dragon book points out, that grammar leads to an $SLR(1)$ conflict in the state reached from $\text{GOTO}(I_0, L)$. That state ($I_2$ in the Dragon book exposition) contains the items $S\to L\;\cdot = R$ and $R\to L\;\cdot$. Since $=$ is in $\text{FOLLOW}(R)$, the state has a shift-reduce conflict. In the $LALR(1)$ automaton, that conflict is resolved.

Since that's a shift-reduce conflict, it doesn't address your question, which concerns reduce-reduce conflicts. But it's easy to modify the grammar slightly in order to turn the shift-reduce conflict into a reduce-reduce conflict. All that's necessary is to introduce a new intermediate non-terminal:

$$\begin{align}S &\to L' = R\\ S &\to R\\ L' &\to L\\ L &\to * R\\ L &\to id\\ R &\to L\\ \end{align}$$

That changes the itemset for state $I_2$ to include the items $\{L'\to L\;\cdot, R\to L\;\cdot\}$, which is now an $SLR(1)$ reduce-reduce conflict, for the same reason that the previous example had a shift-reduce conflict. Also for the same reason, the $LALR(1)$ algorithm resolves that conflict using more precisely-computed lookahead sets.

That disproves your statement 2; since the "if and only if" in statement 3 requires both statement 1 and statement 2 to be true, it also disproves statement 3.


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