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I have a linear grammar $G$ and a string $s$. $G$ is is not limited to right or left linear only but rather has a mix of rules of both types.

Is there an algorithm to determine whether $s \in L(G)$ in less than cubic time?

Converting to CNF and applying the CKY algorithm takes cubic time, and I am wondering if there is a more efficient algorithm.

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  • $\begingroup$ You can apply CKY directly, without transforming (it's essentially a dynamic programming algorithm; transforming to CNF is needed only for guaranteed running time). It'll take quadratic time. $\endgroup$
    – user114966
    Oct 5 '20 at 12:11
  • $\begingroup$ @Dmitry Why would the CKY take quadratic time here? $\endgroup$
    – Eve George
    Oct 5 '20 at 14:36
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    $\begingroup$ CYK is a dynamic programming algorithm with $n^2$ states. For each state, computation takes $n^{k-1}$ time, where $k$ is the maximum number of nonterminals on the RHS of a rule. For CNF, $k=2$, for linear grammar, $k=1$. $\endgroup$
    – user114966
    Oct 5 '20 at 16:46
  • $\begingroup$ @Dmitry According to en.wikipedia.org/wiki/CYK_algorithm it loops over the input 3 times and thus the cubic time. Would this not be needed to be adjusted in some way to get to $O(n^{2})$? $\endgroup$
    – Eve George
    Oct 5 '20 at 20:52
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    – Glorfindel
    Oct 8 '20 at 5:18
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Yes, you can. As Dmitry explains, CYK parsing can be used to parse a linear grammar in $O(n^2)$ time, where $n$ is the length of the input word.

CYK parsing is a dynamic programming algorithm that sets $P[l,s,R]$ to be true if $a_{s .. s+l-1}$ can be generated from the non-terminal $R$, where $a_{1..n}$ is the input word.

Note that there is a recurrence relation that can be used to compute $P[l,s,R]$ from values of the form $P[l',s',R']$ with $l'<l$, in $O(1)$ time. In particular, if we have the rule $R \to b_1\cdots b_k Q c_1 \cdots c_m$, then we should set $P[l,s,R]$ to true if $a_{s .. s+k} = b_{1 .. k}$ and $P[l-k-m,s+k,Q]$ is true and $a_{s+l-m .. s+l-1} = c_{1 .. m}$. Do this for each rule with $R$ on the left-hand side. If none of them sets $P[l,s,R]$ to true, set $P[l,s,R]$ to false. This takes $O(1)$ time (treating the size of the grammar as a constant).

We repeat this for each of the entries $P[l,s,R]$. There are $O(n^2)$ such entries (treating the size of the grammar as a constant), and each entry takes $O(1)$ time to compute, so the entire algorithm runs in $O(n^2)$ time.

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