Linguistic Complexity of a Genome solved by 147

Aug. 7, 2012, midnight by Rosalind Team

Topics: String Algorithms

Getting Repetitive

We have seen that every genome contains a large number of repeats and noted that the Alu repeat recurs around a million times on the human genome. Yet exactly how repetitive is the human genome?

To frame such a vague question mathematically, we first need to make the observation that if the genome were formed by adding nucleobases randomly, with each base having a 1/4 probability of being added at each nucleotide position, then we should expect to see a huge number of different substrings in the genome. Yet (to take a simple case) the genome containing only adenosine and forming the DNA string "AAAAAA...AAA" has relatively very few distinct substrings.

Now, real genomes are formed by a process that chooses nucleotides somewhere in between these two extreme cases, and so to quantify just how random this process is, we need to take the percentage of distinct substrings appearing in a genome with respect to the maximum possible number of distinct substrings that could appear in a genome of the same length.


Given a length $n$ string $s$ formed over an alphabet $\mathscr{A}$ of size $a$, let the "substring count" $\textrm{sub}(s)$ denote the total number of distinct substrings of $s$. Furthermore, let the "maximum substring count" $m(a, n)$ denote the maximum number of distinct substrings that could appear in a string of length $n$ formed over $\mathscr{A}$.

The linguistic complexity of $s$ (written $\textrm{lc}(s)$) is equal to $\frac{\textrm{sub}(s)}{m(a, n)}$; in other words, $\textrm{lc}(s)$ represents the percentage of observed substrings of $s$ to the total number that are theoretically possible. Note that $0 < \textrm{lc}(s) < 1$, with smaller values of $\textrm{lc}(s)$ indicating that $s$ is more repetitive.

As an example, consider the DNA string ($a = 4$) $s = \textrm{ATTTGGATT}$. In the following table, we demonstrate that $\textrm{lc}(s) = \frac{35}{40} = 0.875$ by considering the number of observed and possible length $k$ substrings of $s$, which are denoted by $\textrm{sub}_{k}(s)$ and $m(a, k, n)$, respectively. (Observe that $m(a, n) = \sum_{k=1}^{n}{m(a,k,n)} = 40$ and $\textrm{sub}(s)= \sum_{k=1}^{n}\textrm{sub}_{k}(s) = 35$.)

$k$ $\textrm{sub}_{k}(s)$ $m(a, k, n)$
1 3 4
2 5 8
3 6 7
4 6 6
5 5 5
6 4 4
7 3 3
8 2 2
9 1 1
Total 35 40

Given: A DNA string $s$ of length at most 100 kbp.

Return: The linguistic complexity $\textrm{lc}(s)$.

Sample Dataset


Sample Output



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