# Counting number of possible alignments between two DNA sequences using python

I was looking BioPython to calculate the number of possible alignments between two sequences. Let's say there are two input sequences with the length of m and n. The program will count the possible number of alignments based on the length of two sequences.

Is it possible to do that with BioPython? If not, how can I do that?

• Which kind of sequences DNA, RNA or proteins? How have you tried to do so? What do you expect to learn from it? – llrs Oct 23 '17 at 16:43
• And the third question? Why do you need to know? (By the way, if you don't try why do you expect random people on the internet to help you?, Not that someone can't but it is harder to provide help if one doesn't see research effort) – llrs Oct 23 '17 at 16:53
• Seems like a homework assignment where they are asking for the theoretical number of alignments. even down to the typical n and m nomenclature. – Bioathlete Oct 23 '17 at 17:58
• You should define what you mean by "possible alignment". Biopython will only help you use tools that chose among the possible alignments, based on certain criteria. – bli Oct 24 '17 at 11:05

A closed-form solution is offered in An exact formula for the number of alignments between two DNA sequences by Torres, Cabada, and Nieto:

$$f(m,n)=\sum_{k=0}^{min(m,n)}2^{k}\binom{m}{k}\binom{n}{k}$$

If this solution seems reasonable, you could calculate this without BioPython, but with a simple xrange loop, the power operator, and scipy.special.binom:

#!/usr/bin/env python

import sys
import scipy.special

def alignments(m, n):
if m < 0 or n < 0:
raise ValueError('m and n should be non-negative')
s = 0
b = min(m, n) + 1
for k in xrange(0, b):
s += (2 ** k) * scipy.special.binom(m, k) * scipy.special.binom(n, k)
return s

def main():
m = 4
n = 2
a = alignments(m, n)
assert (m == 4 and n == 2 and a == 41), "something went wrong!"
sys.stdout.write("alignments(%d, %d) = %d\n" % (m, n, a))

if __name__ == "__main__":
main()


I don't know what your m and n are, but keep in mind that this solution grows quickly. Also, I did some digging and scipy approximates the calculation of the binomial coefficient, which is probably going to be an issue for large m and n.

If you're working with large values, you may want to look into using the Python long type and consider the accuracy of the answer. The Stack Overflow link offers alternatives to scipy that generate an exact calculation of the binomial coefficient, at the cost of a longer runtime.