How does $LL(n)$ languages compare with $LR(0)$, for $n>0$?

Question

In the context of languages (not grammars), I know following:

$LL(0) \subset LL(1) \subset LL(2) \subset \cdots \subset LL(k)$

$LR(0) \subset SLR(1) = LALR(1) = LR(1) = SLR(k) = LALR(k) = LR(k)$

$LL(0)\subset LR(0)$

$LL(n)\subset LR(1)$ for all $n$'s

How does $LL(n)$ languages compare with $LR(0)$, for $n>1$?

score 1 · Accepted Answer · answered Dec 14 '19 at 00:18

1

They are incomparable.

Every $LR(0)$ language has the prefix property (no sentence is a prefix of another sentence). But many $LL(k)$ languages don't have this property ($a^*$, for example).
The language $\{a^nb^mc\mid n\ge m\ge 0\}$ is $LR(0)$ but is not $LL(k)$ for any $k$.

answered Dec 14 '19 at 00:18

rici

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Any $LL(k)$ language which is not $LR(0)$? – RajS Dec 14 '19 at 05:46
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@anir: see point 1: $a^*$ is not $LR(0)$ (because it does not have the prefix property). – rici Dec 14 '19 at 06:01

How does $LL(n)$ languages compare with $LR(0)$, for $n>0$?

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