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Microlibrary finding all common subsequences between two sequences in polynomial time.

Project description

A re-usable Python micro-library that finds all of the subsequences shared between two sequences (like strings or lists) in polynomial time.

Author, License, and Conditions

(c) A. Samuel Pottinger (http://gleap.org), 2013 Released under the [MIT license](http://opensource.org/licenses/MIT). Don’t forget to be awesome.

Installation (pip)

pip install py_common_subseq

Installation (manual / single file)

This mico-library is a single file and engineers that may prefer to include the file directly instead of using pip can simply copy py_common_subseq/py_common_subseq.py into an accessible location. This micro-library does not have any additional dependencies beyond the Python standard library.

Quickstart

` >>> import py_common_subseq >>> test_seq_1 = 'qwer' >>> test_seq_2 = 'qewr' >>> py_common_subseq.count_common_subsequences(test_seq_1, test_seq_2) 12 >>> py_common_subseq.find_common_subsequences(test_seq_1, test_seq_2) set(['', 'qer', 'wr', 'qwr', 'er', 'qr', 'e', 'qw', 'q', 'r', 'qe', 'w']) >>> py_common_subseq.find_common_subsequences(test_seq_1, test_seq_2, sep=',') set(['', ',q,w,r', ',e,r', ',e', ',w,r', ',q,w', ',q,r', ',w', ',r', ',q', ',q,e', ',q,e,r']) `

Full API

`count_common_subsequences(seq_1, seq_2)` Find the number of common subsequences between two collections.

This function finds the number of common subsequences between two collections but does not return an actual listing of those subsequences. This is more space efficient O(len(seq_1)) than find_common_subsequences.

  • seq_1: Any integer indexable collection (list, tuple, etc.). The first collection to find subsequences in.

  • seq_2: Any integer indexable collection (list, tuple, etc.). The second collection to find subsequences in.

  • return: Integer. The number of common subsequences between seq_1 and seq_2.

`find_common_subsequences(seq_1, seq_2)` Find the number of common subsequences between two collections.

This function finds the common subsequences between two collections and returns an actual listing of those subsequences. This is less space efficient (O(len(seq_1)^2)) than count_common_subsequences.

  • seq_1: Any integer indexable collection (list, tuple, etc.). The first collection to find subsequences in.

  • seq_2: Any integer indexable collection (list, tuple, etc.). The second collection to find subsequences in.

  • sep: Seperator to put between elements when constructing a subsequence. Keyword argument defaulting to ‘’.

  • empty_val: The value to use to represent the empty set. Keyword argument defaulting to ‘’.

  • return: Python standard lib set. Set of subsequences in common between seq_1 and seq_2.

Motivation / Background

While the longest common subsequence allows for the comparison of sequences, some problem domains also benefit from the additional information hiding in the second, third, fourth, etc. largest common subsequences ignored by typical LCS. This micro-library provides an implementation of the dynamic programming solution for finding all common subsequences as presented in All Common Subsequences (http://dl.acm.org/citation.cfm?id=1625377 - calACS-DP) by Hui Wang (http://www.ulster.ac.uk/staff/h.wang.html). This micro-library also adds some space efficiency improvements and functionality to list common subsequences (semi-formal proof below).

Testing

Within the py_common_subseq folder, run:

python test_py_common_subseq.py

Unit tests do not have any dependencies beyond the Python standard library.

Overview of Space and Time complexity

The algorithm runs in O(|A|x|B|)` time where |A| is the length of the first sequence provided and |B| is the length of the second sequence. Space complexity is as follows:

count_common_subsequences: 2 * min(len(seq_1), len(seq_2)) or O(min(|A|, |B|)) find_common_subsequences: 2 * min(len(seq_1), len(seq_2))^2 or O(min(|A|, |B|))^2

See the discussion below for additional detail.

Overview of Deviations and Optimizations

Similar to the well-documented space optimization for the dynamic programming solution to the Longest Common Subsequence problem, both count_common_subsequences and find_common_subsequences only maintains the “current” and “previous” rows of the table that Hui Wang’s algorithm requires. As proven below, this reduces the space complexity to the following:

count_common_subsequences: 2 * min(len(seq_1), len(seq_2)) or O(min(|A|, |B|)) find_common_subsequences: 2 * min(len(seq_1), len(seq_2))^2 or O(min(|A|, |B|))^2

Additionally, unlike Professor Wang’s original paper, find_common_subsequences modifies the algorithm’s table to contain the set of subsequences achieved at the point of the algorithm’s execution as opposed to the cardinality of that set.

Discussion / proof of correctness

See README.md

Discussion of time complexity

See README.md

Discussion of space optimization

See README.md

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