Introduction to Set Operations

In this problem we are introduced to set operations. After finishing a course in algebra and geometry almost six years ago, I thought that I'd never have to work with set theory again. Apparently I was wrong. Fortunately, at least in this problem, it turned out to be much easier than I remembered.

For a set A and B, that are both subsets of U (where U is the set {1, 2, ..., n} for a given n), we are asked to find the following six sets: A∪B, A∩B, A−B, B−A, the set complement of A and the set complement of B (with respect to U).

Sample Dataset
10
{1, 2, 3, 4, 5}
{2, 8, 5, 10}

Expected Output
{1, 2, 3, 4, 5, 8, 10}
{2, 5}
{1, 3, 4}
{8, 10}
{8, 9, 10, 6, 7}
{1, 3, 4, 6, 7, 9}

As it turns out, Python has built in set operators, which made this problem a lot easier to solve. The following code is what I came up with. You can also find it on my GitHub.

#read file and format data into sets of integers
data = []
with open('rosalind_seto.txt','r') as f:
    for line in f:
        data.append(line.strip('\n'))
U = set(range(1, int(data[0])+1))
A = set([int(x) for x in data[1].strip('{}').split(',')])
B = set([int(x) for x in data[2].strip('{}').split(',')])

#perform set operations
union = A.union(B)
intersection = set(A).intersection(B)
diff_AB = A-B
diff_BA = B-A
A_comp = U-A
B_comp = U-B

#print answers
print(union)
print(intersection)
print(diff_AB)
print(diff_BA)
print(A_comp)
print(B_comp)

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)