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I have a feeling that the following question might have been studied: Suppose I have a finite alphabet $A,$ with $|A| = n,$ and a string $S$ of length $N.$ A string can be said to contain a $k$-th power, if $$S = S_1 S_2^k S_3,$$ where exponentiation means that $S_2$ is repeated $k$ times, and $S_1$ and $S_3$ might be empty. Let $P(n, N)$ be the function defined by: every string of length $N$ over an alphabet of length $n$ contains a $P(n, N)$-th power. Now, the question(s):

  1. What is the (asymptotic) behavior of $P(n, N)?$
  2. Given a string $S$ as above, how hard is it to compute whether $S$ contains a $k$-th power?
  3. What can one say about random strings?
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  • $\begingroup$ 1. Would $P(n,N)=1$ for all $n>1$, if you say every string. 2. Can do it through a linear scan, so $O(N)$. $\endgroup$
    – Fei Gao
    Commented Mar 13, 2014 at 1:50
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    $\begingroup$ @FeiGao 1. No. Suppose $n=2,$ for simplicity, with $A=\{a, b\}.$ Then if the string contains no square, then it contains no $aa$ or $bb,$ so must have the form (wlog) $S=ababab\dots,$ which means that it contains a very high power of $ab.$ $\endgroup$
    – Igor Rivin
    Commented Mar 13, 2014 at 1:58
  • $\begingroup$ @FeiGao Linear Scan? How do you detect powers, and where do you store them? $\endgroup$
    – Igor Rivin
    Commented Mar 13, 2014 at 1:59
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    $\begingroup$ If I understand your definition of $P(n,N)$ then since there exist infinite cubefree words over a binary alphabet (see for example the Thue-Morse sequence), one should have $P(2,N) = 2$ when $N\geq 4$. And since the Thue-Morse sequence could be viewed as a word over any $n$-ary alphabet but only using 2 symbols, shouldn't this mean $P(n,N)\leq 2 \forall n\geq 2,N$? Or did I misread what you are asking? $\endgroup$
    – ARupinski
    Commented Mar 13, 2014 at 2:03
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    $\begingroup$ continuing ARupinski's comment: Since there exist arbitrarily long square-free words over ternary alphabet (en.wikipedia.org/wiki/Square-free_word), then $P(n,N)=1$ for every $n\ge 3$. $\endgroup$
    – Jan Kyncl
    Commented Mar 13, 2014 at 3:06

3 Answers 3

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regarding Question 2:

By Theorem 8.4.3 in M. Lothaire, Applied Combinatorics on Words and subsequent discussion, the number of $k$th powers in $S$ can be determined in time $O(N)$.

An algorithm for finding squares in DNA sequences has been implemented: http://bioinfo.lifl.fr/mreps//

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  • $\begingroup$ Interesting! I wonder why this is a useful thing to do in the DNA setting... $\endgroup$
    – Igor Rivin
    Commented Mar 13, 2014 at 13:57
  • $\begingroup$ The most common terms for squares or powers in DNA seem to be tandem repeats, microsatellites, or minisatellites, according to the length of the period, and they appear to be useful for many things, like determining parentage or diagnosing Huntington's disease... $\endgroup$
    – Jan Kyncl
    Commented Mar 14, 2014 at 0:02
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Regarding the 3rd question, I will show this:

Theorem. For a random binary word of length $n$, the expected number of $h$th powers is $$ \sim \frac{n}{2^{h-1}-1}. $$ Proof. A basic event about occurrences of powers of a word in a binary word is $$ S_{i,j,h} = \{w\in \{0,1\}^n \mid \text{$w$ has an $h$th power of length $h\cdot i$ ending in position $j$}\} $$ $$ = \{w \mid w = av^hb, |av^h|=j, |v|=i\} $$ We let $p$ denote the probability of 1 as opposed to 0; namely $p=1/2$. We may assume $w$ has odd length $n = 2k-1$. Then $\mathbb P(S_{i,j,h}) = p^{(h-1)i}$, and the ranges for the variables are $$ h\cdot i\le j\le 2k-1 = n $$ Let $W_{n}$ be a uniformly distributed random word of length $n$, and let $s^{(c)}(w)$ be the number of $c$th powers in the word $w$. So $$ \mathbb E \sum_{j=hi}^{2k-1} \mathbf{1}_{S_{i,j,h}} = (2k-hi)p^{(h-1)i} $$ is the expected number of $h$th powers of length $i$, for $hi\le 2k-1$. Then $$ \mathbb E s^{(h)}(W_{2k-1}) = \sum_{i=1}^{\lfloor(2k-1)/h\rfloor}(2k-hi)p^{(h-1)i} $$ is the expected number of $h$th powers in a word of length $n=2k-1$. Let us take $n\rightarrow\infty$ and divide by $n$; we get $$ \sum_{i=1}^\infty p^{(h-1)i} = \frac{p^{h-1}}{1-p^{h-1}} = \frac1{2^{h-1}-1} $$

Corollary. The expected number of squares, cubes, and 4th powers is $\sim n$, $n/3$, $n/7$, respectively.

The total number of nontrivial powers (squares and higher) overall will be something like $$ n\cdot \sum_{h=2}^\infty \frac1{2^{h-1}-1} $$ which is a Lambert series with value $\approx 1.606695n$.

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  • $\begingroup$ In the interest of clarity, I recommend you use "random binary word" in your theorem. $\endgroup$
    – The Masked Avenger
    Commented Mar 13, 2014 at 14:04
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The comments have covered the bulk of the behaviour of P(n,N) for nontrivial values of n, showing that only when n=1 should P have values exceeding 2 (or n at most 2 to get a value more than 1). A sorted suffix tree might help in improving the time bound suggested for searching for k powers, but the problem is clearly solvable in low degree polynomial time.

Regarding random strings, it is easiest to analyze powers of single letters as words of length L can be considered as letters in an alphabet of size n^L. The proportion of words of length N having a kth power in the first k letters is seen to be n^{1-k}, and although the probabilities of having a kth power of length k starting at the cth character of a string of length N are not independent, the actual probability of having a kth power in a string of length N greater than k is close to (N +1-k)n^{1-k} for small values of N, say N at most n^{k/2}.

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