为什么这个语法在 LR(0) 中有 Reduce/Reduce 冲突？

Question

I have the following grammar:

S -> a b D E
S -> A B E F
D -> M x
E -> N y
F -> z
M -> epsilon
N -> epsilon

My textbook says there is a Reduce/Reduce conflict in LR(0). I built a diagram and found out that there is a state:

S -> a b . D E
S -> A B . E F
D -> . M x
E -> . N y
M -> .
N -> .

The textbook says that it's a Reduce/Reduce conflict. I'm trying to figure out why. If I build the SLR table I get the following row (3 is the state above):

That's because:

• Follow(M)={x} so we can do reduce to rule 6 from state 3.
• Follow(N)={y} so we can do reduce to rule 7 from state 3.

I was taught that there is a conflict S/R if there is a cell with S/R and conflict R/R if there is a cell with R/R. But I don't see two Rs in the same cell in the table. So why is it a reduce/reduce conflict?

Answer 1

You show an SLR(1) parsing table, in which the columns correspond to a lookahead of length 1. It's correct, and there is no conflict.

But here we're talking about an LR(0) machine, in which there is no lookahead. (That's the 0 in LR(0).) The only decision the machine can make is to shift or reduce, and since it cannot use lookahead, it can only use the state itself. A given state must be either a shift state or a reduce state (and, if a reduce state, which production is being reduced).

(In case it's confusing, and it often is, the concept of lookahead does not refer to the use of the shifted symbol to decide which state to transition to. The transition is taken based on the shifted symbol, which is at that point no longer part of the lookahead.)

So in that state, there is no possible shift action; in all items in the itemset, either the dot is at the end or the next symbol is a non-terminal (implying a GOTO action after returning from a reduce).

But the state does not have a unique reduction. Depending on the lookahead, the parsers needs to choose to reduce M or to reduce N. And since there is no lookahead, the decision cannot be made and hence it's a conflict.