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    <p>Here's a solution in Python for an arbitrary number of sequences. You could use it to test your solution for 2D, 3D cases. It closely follows <a href="http://en.wikipedia.org/wiki/Longest_common_subsequence_problem#LCS_function_defined" rel="nofollow">Wikipedia's algorithm</a>:</p> <pre><code>#!/usr/bin/env python import functools from itertools import starmap @memoize def lcs(*seqs): """Find longest common subsequence of `seqs` sequences. Complexity: O(len(seqs)*min(seqs, key=len)*reduce(mul,map(len,seqs))) """ if not all(seqs): return () # at least one sequence is empty heads, tails = zip(*[(seq[0], seq[1:]) for seq in seqs]) if all(heads[0] == h for h in heads): # all seqs start with the same element return (heads[0],) + lcs(*tails) return max(starmap(lcs, (seqs[:i]+(tails[i],)+seqs[i+1:] for i in xrange(len(seqs)))), key=len) def memoize(func): cache = {} @functools.wraps(func) def wrapper(*args): try: return cache[args] except KeyError: r = cache[args] = func(*args) return r return wrapper </code></pre> <p>Note: without memoization it is an exponential algorithm (<a href="http://www.wolframalpha.com/input/?i=RSolve%5B%7Ba%5Bn%5D+%3D%3D+K+a%5Bn-1%5D+%2B+K%2C+a%5B0%5D+%3D+K%7D%2C+a%5Bn%5D%2C+n%5D" rel="nofollow">wolfram alpha</a>):</p> <pre><code>$ RSolve[{a[n] == K a[n-1] + K, a[0] = K}, a[n], n] a(n) = (K^(n + 1) - 1) K/(K - 1) </code></pre> <p>where <code>K</code> == <code>len(seqs)</code> and <code>n</code> == <code>max(map(len, seqs))</code></p> <h3>Examples</h3> <pre><code>&gt;&gt;&gt; lcs("agcat", "gac") ('g', 'a') &gt;&gt;&gt; lcs("banana", "atana") ('a', 'a', 'n', 'a') &gt;&gt;&gt; lcs("abc", "acb") ('a', 'c') &gt;&gt;&gt; lcs("XMJYAUZ", "MZJAWXU") ('M', 'J', 'A', 'U') &gt;&gt;&gt; lcs("XMJYAUZ") ('X', 'M', 'J', 'Y', 'A', 'U', 'Z') &gt;&gt;&gt; lcs("XMJYAUZ", "MZJAWXU", "AMBCJDEFAGHI") ('M', 'J', 'A') &gt;&gt;&gt; lcs("XMJYAUZ", "MZJAWXU", "AMBCJDEFAGUHI", "ZYXJAQRU") ('J', 'A', 'U') &gt;&gt;&gt; lcs() #doctest: +IGNORE_EXCEPTION_DETAIL Traceback (most recent call last): ... ValueError: &gt;&gt;&gt; lcs(*"abecd acbed".split()) ('a', 'b', 'e', 'd') &gt;&gt;&gt; lcs("acd", lcs("abecd", "acbed")) ('a', 'd') &gt;&gt;&gt; lcs(*"abecd acbed acd".split()) ('a', 'c', 'd') </code></pre>
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