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Coding Challenge

Summary

The prompt stated that the collection of subsequences could be combined to form exactly one supersequence. For this exercise, I assumed that the latter portion of each subsequence overlapped the front of exactly one other subsequence, aside from the end fragment. This assumption works if the subsequences are random and long enough. With real genomic data, the presence of long repeated sequences might break this assumption.

Algorithm

The data structure I used was a linked list. For each pair, I tried to overlay a "start" sequence on an "end" sequence. If I found a match, I recorded the index of overlap, and removed the "end" sequence from a list of unmatched subsequences. The last sequence left from this list was the root of the linked list.

Then, starting from the root, I concatenated each string (truncated at the point of overlap) to form the supersequence.

To test whether one "start" sequence overlapped with an "end" sequence, I took the first half of the "end" sequence and tested whether it was a substring of the "start" sequence. If it overlapped at character N, I compared the last N characters of the "start" sequence with the first N characters of the "end" sequence. If this was a match, I knew that the sequences overlapped.

I estimated the run time of this approach to be O(n*n*l), where "n" is the number of sequences, and "l" is the approximate length of each sequences. The reasoning is that I compare each string to every other string (n*n) and each string comparison is about O(l) in the worst case, when only the last character differs between the two strings.

Alternative Approach

I looked into other data structures that could improve on this, and implemented a suffix tree for string comparison. For this, I first stored the sort order of each possible suffix of the "start" sequence. Using this array, a search for the "end" fragment would take O(log l) time with binary search. Since the "end" fragment wouldn't usually be an exact suffix, when doing a binary search I truncated the search string to match the length of the "end" fragment.

This approach ended up being much slower than using Python's native string comparison operator, even without the cost of building the suffix tree. I'm not sure why.

Usage

The script uses BioPython to parse FASTA. To install:

easy_install -f http://biopython.org/DIST/ biopython

The supersequence is returned by:

SequenceAssembler('./FASTA_FILE_NAME.txt').super_sequence()

For testing, I used the data from the prompt, and used the script generate_test_data.py to generate a few more test scripts.

Files

My solution is in sequence_assembler.py. The file can run the larger data set with: python run.py

Tests can be run with: python test.py

The file alternative_sequence_assembler_with_suffix_array.py has the alternative implementation with suffix arrays.

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