Expected Number of Restriction Sites

This problem is very similar to "Introduction to Random Strings". As in that problem, we are given a string, s, and an array, A, containing some GC contents. However, in this case we are also given an integer, n, representing the length of a second string, t. t is the random string formed with each given GC content and we are asked to find the probability of finding s as a substring of each t. What we need to realise to solve this problem is that the number of opportunities for finding s in t is equal to n-len(s)+1. To get the overall probability of finding s in t, we simply need to add all the individual probabilities of randomly forming s with the specified GC contents.

Sample Dataset
10
AG
0.25 0.5 0.75

Expected Output
0.422 0.563 0.422

The following is the code I wrote to solve the problem. When I wrote it I became aware that there is a difference in how Python3 and Python2 handles rounding of 0.5. In Python2 it is rounded up, but in Python3 it is rounded down. On an earlier problem I had to run my program using Python2 because the answer I got from Python3 wasn't accepted by Rosalind. In this case however, the sample data set I got yielded the same answer regardless of which version I used (suggesting it no cases of having to round 0.5 occurred in this data set). (Note: if the following code is to be run with Python2 the formating of the output needs to be rewritten for it to work).

data = []
with open('rosalind_eval.txt', 'r') as f:
    for line in f:
        data.append(line.strip('\n'))
n = int(data[0])
s = data[1]
A = [float(x) for x in data[2].split()]

AT, GC = 0, 0
for nt in s:
    if nt == 'A' or nt == 'T':
        AT += 1
    elif nt == 'G' or nt == 'C':
        GC += 1

B = [None]*len(A)
for i, j in enumerate(A):
    P = (((1 - j)/2)**AT)*((j/2)**GC)*(n - len(s)+1)
    B[i] = '%0.3f' % P
print(*B, sep=' ')

Edit Distance

In this problem we are looking for the Levenshtein distance between two protein sequences. The distance is calculated as the number of single character changes (insertion, deletion or substitution) needed to turn one sequence into the other. One way to do this is described in the Wikipedia page and involves building a matrix of distances between the prefixes of the two strings. The last value to be calculated (in the bottom right position of the matrix) then equals the distance between the two strings.

Sample Dataset
>Rosalind_39
PLEASANTLY
>Rosalind_11
MEANLY

Expected Output
5

The following python3 code initiates a matrix M with len(s) rows and len(t) columns. It then fills the matrix based on if and what type of change is needed between the two strings. The last value is then printed as the answer.

from Bio import SeqIO
seqs = []
with open('sampledata.fasta', 'r') as f:
    for record in SeqIO.parse(f, 'fasta'):
        seqs.append(str(record.seq))
s = seqs[0]
t = seqs[1]
M = [[0 for j in range(len(t) + 1)] for i in range(len(s) + 1)]
for i in range(1, len(s) + 1):
    M[i][0] = i
for i in range(1, len(t) + 1):
    M[0][i] = i

for i in range(1, len(s) + 1):
    for j in range(1, len(t) + 1):
        if s[i - 1] == t[j - 1]:
            M[i][j] = M[i - 1][j - 1]
        else:
            M[i][j] = min(M[i - 1][j] + 1, M[i][j - 1] + 1,
                          M[i - 1][j - 1] + 1)
print(M[len(s)][len(t)])

Introduction to Alternative Splicing

In this problem we continue looking at subsets. This time we imagine them as exons and want to know how many different combinations we can form from subsets of the set. We are given two integers n and m, which are the length of the set ([1,2,3,4,5,6] in the sample case) and the minimum length of the subsets to be formed. The order of the exons cannot be altered, so for n=3 and m=2 we would get the four subsets [1,2], [1,3], [2,3] and [1,2,3]. The answer should be given modulo 1000000.

Sample Dataset
6 3

Expected Output
42

The problem can quite easily be solved by counting all the different combinations of each length from m to  n, and then summarising them to form the final answer. The number of combinations can be calculated as n!/(k!(n-k)!), where k is the length of the subset. In python3 code this can look like:

from math import factorial
n, m = 6, 3
subsets = 0
for k in range(m, n + 1):
    subsets += (factorial(n) // (factorial(k) * factorial(n - k)))
print(subsets % 1000000)

At first I used / instead of //, which worked fine for the sample dataset, but when I downloaded another dataset the numbers were bigger and therefore they got too big for float division and I got the error message "OverflowError: integer division result too large for a float". A quick Google search took me to this thread on Stack Overflow that solved my problem.

Counting Subsets

In this problem we are introduced to set theory (causing flashbacks from the linear algebra course...). For a set {1,2,3,...,n}, we are asked to return the number of subsets of the set. What we need to remember is that the empty set {} and the whole set {1,2,3,...,n} are also subsets (aptly explained here). So, to find the total number of subsets, all we need to do is calculate 2^n. Since this number quickly gets very large, the problem states that we should return the answer modulo 1 000 000.

Sample Dataset
3

Expected Output
8

The following program is probably the shortest one yet:

n = 3                      

print(2**n % 1000000)      

Or, if you like, a fancier version where the user can input n in the terminal:

n = int(input('Enter n: '))

print(2**n % 1000000)      

Matching Random Motifs

This problem is very similar to “Introduction to Random Strings”. In this case however, we are given a sting s, and are asked to calculate the probability of randomly forming s when generating N nr of stings of equal length of s and a GC content of x.

Sample Dataset
90000 0.6
ATAGCCGA

Sample Output
0.689

To calculate the probability of randomly getting a string that equals s when constructing it with GC content x, we can use the same equation as in introduction to random strings. To go from this probability (lets call it s_prob) to the probability of getting at least one string that is equal to s we use the following equation:

P(at least 1 match of s) = 1 − P(no matches out of N strings) = 1 − [1 - s_prob]^N

The following program makes the above calculations and outputs the answer with three significant figures:

N = 90000
x = 0.6
s = 'ATAGCCGA'
AT = 0
GC = 0
for nt in s:
    if nt == 'A' or nt == 'T':
        AT += 1
    elif nt == 'G' or nt == 'C':
        GC += 1

s_prob = (((1 - x) / 2)**AT) * (((x) / 2)**GC)
prob = 1 - (1 - s_prob)**N
print('%0.3f' % prob)

Creating a Distance Matrix

This problem was fairly similar to "Counting Point Mutations" and "Error Correction in Reads". We are given a set of DNA strings and are asked to return the distance matrix of the strings. The distance between two given strings can be calculated by dividing the Hamming distance (i.e. the number of nucleotides that differ between the stings) with the length of the sequences (given that all sequences are of equal length). These values should then be printed on matrix form with 5 significant figures.

Sample Dataset

>Rosalind_9499

TTTCCATTTA

>Rosalind_0942

GATTCATTTC

>Rosalind_6568

TTTCCATTTT

>Rosalind_1833

GTTCCATTTA

Expected Output

0.00000 0.40000 0.10000 0.10000

0.40000 0.00000 0.40000 0.30000

0.10000 0.40000 0.00000 0.20000

0.10000 0.30000 0.20000 0.00000

To write this program I reused parts of the code from "Counting Point Mutations" and "Error Correction in Reads" and modified it to suit this problem. The following is the final code, which took me only 20 minutes to write. Hurray!

from Bio import SeqIO

reads = []

with open('sampledata.fasta', 'r') as f:

    for record in SeqIO.parse(f, 'fasta'):

        reads.append(str(record.seq))

read_len = len(reads[0])

for curr_read in reads:

    distance = []

    for comp_read in reads:

        hamming = 0

        for nt1, nt2 in zip(curr_read, comp_read):

            if nt1 != nt2:

                hamming += 1

        distance.append(str.format('{0:.5f}', hamming / read_len))

    print(*distance, sep=' ')

Maximum Matchings and RNA Secondary Structures

This problem is very similar to the problem presented in "Perfect Matchings and RNA Secondary Structures". The difference is that in this problem we don't have the same number of A as U and G as C, so we don't get any perfect matchings. Instead we are given the task to find the total possible number of maximum matchings for a given RNA string.

Sample Dataset
>Rosalind_92
AUGCUUC

Expected Output
6

As in "Perfect Matchings and RNA Secondary Structures" the programming is very simple and it all comes down to the maths. We can simplify the problem by looking at AU and GC separately, and can describe the number of maximum matches for each part as max(A,U)!/(max(A,U)-min(A,U))! and max(G,C)!/(max(G,C)-min(G,C))!, respectively. To get the total number of maximum matches, simply multiply the answers.

For the first dataset I tried, I ran the program using Python 3.5 and got the result 434412400153932848464251158517855537428692992. This was deemed incorrect by Rosalind. After seeing a tip online, I ran the program using Python 2.7, and got 434412400153932864602133769790674698240000000 instead. So I tried my program with a new dataset and ran it with Python 2.7 and this time my answer was accepted.

Here follows the code I wrote to solve this problem (works in both 2.7 and 3.5):

from Bio import SeqIO
from math import factorial
sequence = ''
with open('sampledata.fasta', 'r') as f:
    for record in SeqIO.parse(f, 'fasta'):
        sequence = str(record.seq)

A, U, G, C = 0, 0, 0, 0
for nt in sequence:
    if nt == 'A':
        A += 1
    elif nt == 'U':
        U += 1
    elif nt == 'G':
        G += 1
    elif nt == 'C':
        C += 1

AU = factorial(max(A, U)) / factorial(max(A, U) - min(A, U))
GC = factorial(max(G, C)) / factorial(max(G, C) - min(G, C))
print(int(AU * GC))