问题
I want to build a graph from a list of words with Hamming distance of (say) 1, or to put it differently, two words are connected if they only differ from one letter (lol -> lot).
so that given
words = [ lol, lot, bot ]
the graph would be
{
'lol' : [ 'lot' ],
'lot' : [ 'lol', 'bot' ],
'bot' : [ 'lot' ]
}
The easy way is to compare every word in the list with every other and count the different chars; sadly, this is a O(N^2) algorithm.
Which algo/ds/strategy can I use to to achieve better performance?
Also, let's assume only latin chars, and all the words have the same length.
回答1:
Assuming you store your dictionary in a set(), so that lookup is O(1) in the average (worst case O(n)).
You can generate all the valid words at hamming distance 1 from a word:
>>> def neighbours(word):
... for j in range(len(word)):
... for d in string.ascii_lowercase:
... word1 = ''.join(d if i==j else c for i,c in enumerate(word))
... if word1 != word and word1 in words: yield word1
...
>>> {word: list(neighbours(word)) for word in words}
{'bot': ['lot'], 'lol': ['lot'], 'lot': ['bot', 'lol']}
If M is the length of a word, L the length of the alphabet (i.e. 26), the worst case time complexity of finding neighbouring words with this approach is O(L*M*N).
The time complexity of the "easy way" approach is O(N^2).
When this approach is better? When L*M < N, i.e. if considering only lowercase letters, when M < N/26. (I considered only worst case here)
Note: the average length of an english word is 5.1 letters. Thus, you should consider this approach if your dictionary size is bigger than 132 words.
Probably it is possible to achieve better performance than this. However this was really simple to implement.
Experimental benchmark:
The "easy way" algorithm (A1):
from itertools import zip_longest
def hammingdist(w1,w2): return sum(1 if c1!=c2 else 0 for c1,c2 in zip_longest(w1,w2))
def graph1(words): return {word: [n for n in words if hammingdist(word,n) == 1] for word in words}
This algorithm (A2):
def graph2(words): return {word: list(neighbours(word)) for word in words}
Benchmarking code:
for dict_size in range(100,6000,100):
words = set([''.join(random.choice(string.ascii_lowercase) for x in range(3)) for _ in range(dict_size)])
t1 = Timer(lambda: graph1()).timeit(10)
t2 = Timer(lambda: graph2()).timeit(10)
print('%d,%f,%f' % (dict_size,t1,t2))
Output:
100,0.119276,0.136940
200,0.459325,0.233766
300,0.958735,0.325848
400,1.706914,0.446965
500,2.744136,0.545569
600,3.748029,0.682245
700,5.443656,0.773449
800,6.773326,0.874296
900,8.535195,0.996929
1000,10.445875,1.126241
1100,12.510936,1.179570
...
I ran another benchmark with smaller steps of N to see it closer:
10,0.002243,0.026343
20,0.010982,0.070572
30,0.023949,0.073169
40,0.035697,0.090908
50,0.057658,0.114725
60,0.079863,0.135462
70,0.107428,0.159410
80,0.142211,0.176512
90,0.182526,0.210243
100,0.217721,0.218544
110,0.268710,0.256711
120,0.334201,0.268040
130,0.383052,0.291999
140,0.427078,0.312975
150,0.501833,0.338531
160,0.637434,0.355136
170,0.635296,0.369626
180,0.698631,0.400146
190,0.904568,0.444710
200,1.024610,0.486549
210,1.008412,0.459280
220,1.056356,0.501408
...
You see the tradeoff is very low (100 for dictionaries of words with length=3). For small dictionaries the O(N^2) algorithm perform slightly better, but that is easily beat by the O(LMN) algorithm as N grows.
For dictionaries with longer words, the O(LMN) algorithm remains linear in N, it just has a different slope, so the tradeoff moves slightly to the right (130 for length=5).
回答2:
There's no need to take a dependency on the alphabet size. Given a word bot, for example, insert it into a dictionary of word lists under the keys ?ot, b?t, bo?. Then, for each word list, connect all pairs.
import collections
d = collections.defaultdict(list)
with open('/usr/share/dict/words') as f:
for line in f:
for word in line.split():
if len(word) == 6:
for i in range(len(word)):
d[word[:i] + ' ' + word[i + 1:]].append(word)
pairs = [(word1, word2) for s in d.values() for word1 in s for word2 in s if word1 < word2]
print(len(pairs))
回答3:
Ternary Search Trie supports Near-Neighbor Searching pretty well.
If your dictionary is stored as TST then, I believe, average complexity of lookups while building your graph would be close to O(N*log(N)) on real world word dictionaries.
And check Efficient auto-complete with a ternary search tree article.
回答4:
Here is linear O(N) algorithm, but with big constant factor (R * L * 2). R is radix (for latin alphabet it is 26). L is a medium length of word. 2 is a factor of adding/replacing wildcard character. So abc and aac and abca are two ops wich leads to hamming distance of 1.
It is written in Ruby. And for 240k words it takes ~250Mb RAM and 136 seconds on average hardware
Blueprint of graph implementation
class Node
attr_reader :val, :edges
def initialize(val)
@val = val
@edges = {}
end
def <<(node)
@edges[node.val] ||= true
end
def connected?(node)
@edges[node.val]
end
def inspect
"Val: #{@val}, edges: #{@edges.keys * ', '}"
end
end
class Graph
attr_reader :vertices
def initialize
@vertices = {}
end
def <<(val)
@vertices[val] = Node.new(val)
end
def connect(node1, node2)
# print "connecting #{size} #{node1.val}, #{node2.val}\r"
node1 << node2
node2 << node1
end
def each
@vertices.each do |val, node|
yield [val, node]
end
end
def get(val)
@vertices[val]
end
end
The algorithm itself
CHARACTERS = ('a'..'z').to_a
graph = Graph.new
# ~ 240 000 words
File.read("/usr/share/dict/words").each_line.each do |word|
word = word.chomp
graph << word.downcase
end
graph.each do |val, node|
CHARACTERS.each do |char|
i = 0
while i <= val.size
node2 = graph.get(val[0, i] + char + val[i..-1])
graph.connect(node, node2) if node2
if i < val.size
node2 = graph.get(val[0, i] + char + val[i+1..-1])
graph.connect(node, node2) if node2
end
i += 1
end
end
end
来源：https://stackoverflow.com/questions/31100623/efficiently-build-a-graph-of-words-with-given-hamming-distance