A considerable repeated substring $t \leqslant s \in \Sigma^*$ is a string of length $2$ that occurs at least 3 times disjointly within $s$ or a string of length $3$ or more that occurs at least $2$ times disjointly within $s$.

So for $|\Sigma| = 1$, $s = a^n$ it's $|s| = 6$, since we have $(aa)(aa)(aa)$ or $(aaa)(aaa)$ but not less than 6, since you can't do the same for $n = 5, 4, 3, 2, 1$.

Thus it's going to be a function of alphabet size.

For $|\Sigma| = 2$. I'll work out the possibilities:

$$ aaaaabbbbbaba $$

So it's $|s| = 13$. Not too shabby. My strategy was to make the string as long as possible before switching chars.

What's the general estimate?

One more example: $|\Sigma| = 3$ $$ s = aaaaabbbbbcccccacbacabcba $$ which is probably not quite exact, since I'm not sure how I'm actually deciding, but $f(3) = |s| = 25$.


I'm fairly sure that $f(n) = 2n^2 + 2n + 1$, where $f(n)$ is the maximal length of a string in a language of $n$ symbols that does not have considerable repeated substrings. Here's why:

Let $|\Sigma| = n$. In a string of length $n^2 + 1$, there are $n^2$ two-character substrings. If this has minimal repetition of two-character strings, then that means every two-character string is visited -- except that you've said that interlocking repetitions don't count, so in the case of the strings $aa$, $bb$, $cc$ and so on the substring that begins at the second character might be the same without counting as a new repetition (e.g., "$aaa$" has two "$aa$" substrings, but they only count as one disjoint instance). There are $n$ of those. So the maximum-length string with no repetitions of a two-character substring has length $n^2 + n + 1$ (the $+1$ at the end is just to give the last two-character substring its second character).

Then we can get another $n^2 + n$ by repeating the sequence, burning up our second repetition of each two-character substring; the $+1$ at the end of the previous string gets subsumed, but we get a new $+1$ at the end of this one. Now we have a string of length $2n^2 + 2n + 1$, but every two-character substring has two disjoint instances, so we can't extend it any further. Note that this matches the examples you gave.

Notice that I haven't dealt with the three-character sequences; that might restrict the length further when $n$ is larger, but it doesn't seem to have had any effect for low $n$.

  • $\begingroup$ So it's $O(|\Sigma|^2)$ nice! $\endgroup$ – BananaCats Author Aug 3 '19 at 5:22

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