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The Shortest Common Supersequence (SCS) problem is an NP-hard problem (https://en.wikipedia.org/wiki/Shortest_common_supersequence), which occurs in problems originating from various domains, e.g. Bio Genetics. Given a set of strings, the common supersequence of minimal length is sought after. Below a set of algorithms is given, which I used in approximating and/or backtracking the optimal solution(s).

Python, 310 lines
 ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310``` ```import random import copy import time def supersequence(x, y): """ True if and only if all elements in y occur in x in order (x is a supersequence of y)""" idx = 0 try: for i in y: idx = x.index(i, idx)+1 except ValueError, e: return False return True def subsequence(x, y): """ x is a subsequence of y <==> y is a supersequence of x """ return supersequence(y, x) def is_supersequence_of_sequences(sequence, sequences): """ True if sequence is a supersequence of every s in sequences """ result = True for s in sequences: if not supersequence(sequence, s): return False return result def remove_redundant_sequences(redundant_sequences, debug=False): """ Removes every doublure sequence and every sequence that is a subsequence of another """ if debug: print "Removing redundant sequences:" sequences = [] for i,s in enumerate(redundant_sequences): remove = False for j in range(i+1, len(redundant_sequences)): if s == redundant_sequences[j]: remove = True if debug: print " Found doublure ", s, "==", redundant_sequences[j] if not remove: sequences.append(s) sequences_p = [] for i,s1 in enumerate(sequences): found = False for j,s2 in enumerate(sequences): if ((i != j) and (supersequence(s2, s1))): found = True if debug: print " Found supersequence", s2, "for", s1 break if not found: sequences_p.append(s1) if debug: print "Removed %s redundant sequence(s). Reduced set:"%(len(redundant_sequences)-len(sequences_p)) for i,s in enumerate(sequences_p): print " s%s:"%i, s return sequences_p def lower_bound(scs): """ Finds a lower bound based on some fast algorithms.""" def scs_length(s1, s2): """ shortest common supersequence length for two single sequences can be computed using dynamic programming. Based on algorithm for computing Longest Common Subsequence (== len(s1)+len(s2)-SCS )""" S = [] for i in range(0, len(s1)+1): S.append([-1]*(len(s2)+1)) for i in range(len(s1)+1): S[i][0] = i for j in range(len(s2)+1): S[0][j] = j for i in range(1, len(s1)+1): for j in range(1, len(s2)+1): if s1[i-1] == s2[j-1]: S[i][j] = min(S[i-1][j]+1, S[i][j-1]+1, S[i-1][j-1]+1) else: S[i][j] = min(S[i-1][j]+1, S[i][j-1]+1) return S[len(s1)][len(s2)] def find_max_scs_length_of_2_combinations(sequences): """ Returns maximum scs_length(s1,s2) for any s1, s2 in sequences. Note that this does not guarantee the length of the actual SCS of the entire set of sequences. """ result = -1 for i in range(len(sequences)): for j in range(i+1, len(sequences)): length = scs_length(sequences[i], sequences[j]) if length > result: result = length return result def max_count_occurrences(sequences): """ for every value v in the alphabet, for all s in sequences, count occurrences of v in s. The length of the SCS has be least the sum of maximum occurrences.""" lower_bound = 0 for v in set([s for seq in sequences for s in seq]): count_v = -1 for s in sequences: count = s.count(v) if count > count_v: count_v = count lower_bound += count_v return lower_bound return max(max_count_occurrences(scs), find_max_scs_length_of_2_combinations(scs)) def upper_bound(sequences, max_run_time_random_seconds = 1): """ Finds an upper bound based on some fast approximation algorithms.""" def alphabet_leftmost(sequences): """ Approximation algorithm by looping through a random permutation on the alphabet """ seqs = copy.deepcopy(sequences) common_supersequence = [] permutation = list(set([x for s in seqs for x in s])) random.shuffle(permutation) i = 0 while any(seqs): found = False for s in seqs: if s and s[0] == permutation[i]: found = True break if found: for s in seqs: if s and s[0] == permutation[i]: s.remove(permutation[i]) common_supersequence.append(permutation[i]) i = (i+1) % len(permutation) return common_supersequence def alphabet_leftmost_rand(sequences): """ Approximation algorithm using random selection on alphabet values. """ seqs = copy.deepcopy(sequences) common_supersequence = [] while any(seqs): firsts = list(set([s[0] for s in seqs if s])) next = firsts[random.randint(0, len(firsts)-1)] for s in seqs: if s and s[0] == next: s.remove(next) common_supersequence.append(next) return common_supersequence def majority_merge(seqs): """ Quite fast approximation algorithm """ sequences = copy.deepcopy(seqs) common_supersequence = [] while any(sequences): firsts = [(s[0], len(s)) for s in sequences if s] counts = {} for x in firsts: if x[0] in counts: counts[x[0]] += x[1] #use +1 if not weighted MM else: counts[x[0]] = x[1] #use 1 if not weighted MM most_common = sorted(counts, key=lambda x: counts[x], reverse=True)[0] common_supersequence.append(most_common) for s in sequences: if s and s[0] == most_common: s.remove(most_common) if s == []: sequences.remove([]) return common_supersequence trivial = sum(len(s) for s in sequences) mm = len(majority_merge(sequences)) alm, almr = 10000000, 10000000 start = time.time() while time.time() < start + max_run_time_random_seconds: alm_test = alphabet_leftmost(sequences) if len(alm_test) < alm: alm = len(alm_test) almr_test = alphabet_leftmost_rand(sequences) if len(almr_test) < almr: almr = len(almr_test) upperbounds = [trivial, mm, alm, almr] solutions = bfs_genetic(sequences, keep_best = 1000, best_so_far = min(upperbounds)) if solutions: upperbounds.append(sorted(solutions)[0]) return min(upperbounds) def backtrack_shortest_common_supersequences(sequences, upperbound=None): """ backtrack with pruning, then filter out the shortest SCSs (they are not unique!)""" if not upperbound: upperbound = 10000000 result = backtrack_scs(sequences, [], upperbound) return [x for x in result if len(x) == min([len(y) for y in result])] def backtrack_all_common_supersequences(sequences): """ backtrack all supersequences (not just shortest) """ upperbound = 10000000 return backtrack_scs(sequences, [], upperbound, False) def backtrack_scs(sequences, choices, best_so_far, prune=True): """ Recursive DFS algorithm, potentially risky with large #sequences or long s in sequences""" solutions = [] if any(sequences): # no need to evaluate subsolution choices if it will become longer than best_so_far. if not prune or len(choices) + max([len(s) for s in sequences]) <= best_so_far: firsts = set([s[0] for s in sequences if s]) for c in firsts: indices = [] for i,s in enumerate(sequences): if s and s[0] == c: s.remove(c) indices.append(i) for solution in backtrack_scs(sequences, choices+[c], best_so_far): solutions.append(solution) for i in indices: sequences[i] = [c]+sequences[i] else: solutions = [choices] if len(choices) < best_so_far: best_so_far = len(choices) return solutions def bfs_genetic(sequences, keep_best = 100000000, best_so_far = 100000000): """ Breath first search algorithm. Prunes away up to keep_best subsolutions according to metric() """ def metric((choices, remaining_sequences)): # TODO: find good metric, given current choices and remaining sequences to solve. # return sum(len(s) for s in remaining_sequences) return len(remove_redundant_sequences(copy.deepcopy(remaining_sequences))) Q = [] solutions = {} firsts = set([s[0] for s in sequences if s]) for f in firsts: seqs = copy.deepcopy(sequences) for s in seqs: if s and s[0] == f: s.remove(f) Q.append(([f], seqs)) depth = 1 while Q: if depth < len(Q[0][0]): if len(Q) > keep_best: # print "Throwing away", len(Q) - keep_best, "out of", len(Q), "at depth", depth Q = sorted(Q, key=metric)[:keep_best] depth += 1 else: choices, seqs = Q.pop(0) if len(choices) + max([len(s) for s in seqs]) <= best_so_far: if any(seqs): firsts = set([s[0] for s in seqs if s]) for f in firsts: sequences = copy.deepcopy(seqs) for s in sequences: if s and s[0] == f: s.remove(f) Q.append((choices[:]+[f], sequences)) else: if depth in solutions: solutions[depth].append(choices[:]) else: solutions[depth] = [choices[:]] return solutions def pretty_print_scs_with_sequences(scs, sequences): print "SCS: " + "".join([str(s) for s in scs]) for i,aseq in enumerate(sequences): seq = [-1]*len(scs) idx = 0 for v in aseq: while scs.index(v, idx) > idx: idx += 1 seq[idx] = v idx += 1 print " s%s: "%i + "".join([str(s) for s in seq]).replace("-1", ".") if __name__=="__main__": def run_SCS_algorithms(sequences): sequences = remove_redundant_sequences(sequences, debug=True) lowerbound, upperbound = lower_bound(sequences), upper_bound(sequences) print "Lowerbound on Shortest Common Supersequence:", lowerbound print "Upperbound on Shortest Common Supersequence:", upperbound solutions = backtrack_shortest_common_supersequences(sequences, upperbound) print "Backtrack found %s SCSs of optimal length %s."%(len(solutions), len(solutions[0])) print "E.g.", solutions[0], "is valid:", is_supersequence_of_sequences(solutions[0], sequences) pretty_print_scs_with_sequences(solutions[0], sequences) sequences_1 = [ [1,2,3], [1,2,5], [3,1,5,4], [1,2,1,5], [1,2,5]] sequences_2 = [ [1,2,1,2,3], # "acacg" [1,4,1,3,1], # "ataga" [2,1,2,3,4], # "cacgt" [3,4,1,1,4]] # "ctaat" sequences_3 = [ [5,2,1,6,3,6,3,1], [1,4,1,3,1,5,1,2], [2,1,6,3,4,4,2,3], [3,4,5,1,4,6,1,2]] sequences_4 = [ [5,1,3,5,2,6], [1,5,2,1,3,2,3,1], [5,1,3,5,2,1,6,2], [3,5,1,6,2,4,6,1,2]] run_SCS_algorithms(sequences_4) ```

Usually SCS is applied to DNA strings, with nucleotides like "ataga" and "cacgt". My code operates on integers, the mapping is left to the user. Running the above script results in the following output:

``````Removing redundant sequences:
Found supersequence [5, 1, 3, 5, 2, 1, 6, 2] for [5, 1, 3, 5, 2, 6]
Removed 1 redundant sequence(s). Reduced set:
s0: [1, 5, 2, 1, 3, 2, 3, 1]
s1: [5, 1, 3, 5, 2, 1, 6, 2]
s2: [3, 5, 1, 6, 2, 4, 6, 1, 2]
Lowerbound on Shortest Common Supersequence: 13
Upperbound on Shortest Common Supersequence: 14
Backtrack found 6 SCSs of optimal length 14.
E.g. [5, 1, 3, 5, 2, 1, 3, 6, 2, 3, 4, 6, 1, 2] is valid: True

SCS: 51352136234612
s0: .1.5213.23..1.
s1: 513521.62.....
s2: ..35.1.62.4612
``````
 Created by Rutger Saalmink on Wed, 2 Oct 2013 (MIT)