The Levenshtein or edit distance between two strings is the minimum number of edits (adding a letter, removing a letter or changing a letter) required to transform one into the other.

Assume that we have two strings, each of size $n$ and that consist of letters drawn uniformly IID from an alphabet of $k$ letters. What is the (possibly approximate) distribution of their Levenshtein distance?

This can be seen as an extreme value problem since the Levenshtein distance is the minimum number of mismatches over all the possible ways to align the two sequences. For an alignment chosen randomly among those with $m$ insertions and $m$ deletions, the number of mismatches has a Binomial distribution with parameters $n-m$ and $1-1/k$, so it is tempting to compute the distribution of the minimum of all those Binomial variables (as in this question), but in this case, the Binomial variables are not IID.

A solution to this problem would be useful in biology (where $k=4$ in the case of DNA sequences) because it would give the statistical properties of the alignment between unrelated sequences.


2 Answers 2


I guess it is not a direct answer, but you can try to simulate the scenario and check the empirical distribution to have a rough idea (R code below).

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So it seems that for strings longer than ~ 100 the distribution is symmetric and quite narrow around .53 times the length of the string.

# dependencies

library(ggplot2); theme_set(theme_classic())

# settings

alphabet <- c("A", "C", "G", "T")
Nsim <- 1e3
read_lengths <- seq(60, 500, 20)

# function to create a random string of length "n" using letters of the alphabet "alph"

random_read <- function(n, alph=alphabet) paste(sample(alph, size=n, replace=T), collapse="")

# simulate

res <- mclapply(read_lengths,
                function(N) replicate(Nsim, levenshteinDist(random_read(N), random_read(N))),

# arrange results as data.frame

res_df <- data.frame(dist=unlist(res),
                     length=rep(read_lengths, sapply(res, length)))

# plot densities

       aes(x=dist / length, col=length, group=length)) +
  geom_density() +
  ggtitle("Distribution of Levenshtein distance / length")

       aes(x=length, y=dist / length, col=length)) +
  geom_violin(aes(group=length)) +
  geom_smooth(col="black", lwd=1) +
  ggtitle("Distribution of Levenshtein distance / length")
  • $\begingroup$ In case you want so simulate from an organism that do not have the letters evenly distributed, you can include sampling probabilities via the "prob" argument in calling of "sample" of the "random_read" function definition. $\endgroup$
    – qenvio
    Commented Feb 6, 2015 at 13:00
  • 2
    $\begingroup$ Thanks for this analysis! I had done a similar one up to length 10,000 (the average keeps decreasing it seems). +1 for the good work, but I cannot check "accept" because I am looking for an analytical solution. $\endgroup$
    – gui11aume
    Commented Feb 6, 2015 at 18:32

This looks a bit like the Tracy-Widom distribution perhaps? For longest common subsequences there is a "famous" result relating the distribution of LCS between random strings (under a bunch of limiting assumptions) to the Tracy-Widom distribution: https://arxiv.org/abs/q-bio/0410012. https://www.quantamagazine.org/beyond-the-bell-curve-a-new-universal-law-20141015/

  • $\begingroup$ It is a very interesting connection! I was not aware of this "famous" result, but it is spectacular! $\endgroup$
    – gui11aume
    Commented Oct 24, 2020 at 1:25

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