Given a word $w$ over $\{ [, ] \}$, the alphabet of the two square brackets, the Dyck correction problem is to find the shortest sequence of edit operations that would make $w$ a Dyck word, i.e., a word in which the squared brackets are balanced.

I can imagine a few practical applications, but my bibliography search yielded nothing. It looks like dynamic programming though. Any ideas?

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    $\begingroup$ Check: arxiv.org/pdf/1904.08402.pdf $\endgroup$ Mar 6, 2021 at 19:15
  • $\begingroup$ That's an interesting problem! what about the immediate generalization from Dyck languages to context-free languages, have you looked for literature on that question? $\endgroup$
    – a3nm
    Mar 12, 2021 at 11:38
  • $\begingroup$ Or maybe even something easier: what about visibly pushdown languages? $\endgroup$ Mar 12, 2021 at 22:03
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    $\begingroup$ Btw, might be related: arxiv.org/pdf/1504.08259 $\endgroup$ Mar 12, 2021 at 22:07

2 Answers 2


The generalized problem, concerning Dyck($s$), for $s$ distinct pairs of parenthesis, was studied by Barna Saha in a paper entitled The Dyck Language Edit Distance Problem in Near-linear Time

B. Saha, "The Dyck Language Edit Distance Problem in Near-Linear Time," 2014 IEEE 55th Annual Symposium on Foundations of Computer Science, Philadelphia, PA, USA, 2014, pp. 611-620, doi: 10.1109/FOCS.2014.71.

Indeed, the problem is easy for $s=1$. For $s\ge2$ the exact complexity is not known. The classical techniques for correcting context-free languages are $O(sn^3)$ (Valiant 1975). On the other hand, the usual string editing problem which can be solved in quadratic time can be reduced to Dyck($s$).

Saha's result gives a nearly linear time algorithm that returns an approximation of the true edit distance. I am not sure whether this algorithm can produce the edit sequence.


If by "edit operations" you mean single character insertions, deletions, and substitutions then I believe a greedy algorithm works.

Let $x_i$ be the number of ] minus the number of [ in the first $i$ characters. Every edit operation changes each $x_i$ by at most 2. Hence you will need at least $\lceil \frac {\max x_i} 2 \rceil$ operations to correct the word. One way to achieve this is to substitute all the earliest ] for [ and delete one ] if the number is odd.

Finally, you need $x_n$ to be 0. The above operations will reduce $x_n$ by $\max x_i$ and this is unavoidable, since an operation at position $i$ affects exactly those positions after $i$ and by an equal amount. Hence you will need a final $\lceil \frac {\max x_i - x_n} 2 \rceil$ operations to finish correcting the word. One way to achieve this is to substitute the latest [ for ] and delete one [ if the number is odd.

Since the solution is so simple, I doubt there's any literature on it.

  • $\begingroup$ I am not sure I understand what you mean by saying: "One way to achieve this is to substitute all the earliest ] for [ and delete one ] if the number is odd." Do you mean this is what you do when xi is positive? $\endgroup$
    – Yossi Gil
    Mar 10, 2021 at 16:58
  • $\begingroup$ $\max x_i$ and $\max x_i - x_n$ are always nonnegative so yes. $\endgroup$ Mar 11, 2021 at 10:39

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