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Word Ladders - Medium


Word ladder is a word game invented by Lewis Carroll. A word ladder puzzle begins with two words, and to solve the puzzle one must find a chain of other words to link the two, in which two adjacent words (that is, words in successive steps) differ by one letter. Eg. SPORT <-> SHORT
Source: Wikipedia

If you've followed my previous Dijkstra's Shortest Path Algorithm Tutorial, you'll be able to solve this challenge with minimum effort.

Essentially, we create a graph where each word is a vertex and there exists an edge if two words differ by a single character. To find the word ladder is basically running any shortest path algorithm. You could also use a Breadth First Search.

Some important observations

  • The length of the source and destination words must be the same for a path to exist.
  • The path from source to destination  can only contain words of the same length. So we can prune off the rest of the dictionary.
  • The graph is undirected.
Here's my implementation, it uses the modules from the tutorial.

import sys,graphs,itertools
def prepare_graph():
diff_by_1 = lambda (x,y): len(x) == len(y) and sum(1 for a,b in zip(x,y) if a != b) == 1
graph = {}
for x,y in filter(diff_by_1,itertools.product(words,repeat=2)):
if x not in graph: graph[x] = []
if y not in graph: graph[y] = []
graph[x].append( ( y , 1) )
graph[y].append( ( x , 1) )
return graph
if __name__ == '__main__':
if len(sys.argv) >= 3 and len(sys.argv[1]) == len(sys.argv[2]):
src,dst = sys.argv[1:]
words = filter(lambda w : len(w) == len(src), sys.stdin.read().split()) #only consider those words with same length as parameter
if src in words and dst in words:
print graphs.dijkstra(prepare_graph(),src,dst)
else:
print "Both words should be present in the dictionary."
else:
print "Missing command line parameters."
view raw Word Ladder hosted with ❤ by GitHub


To run the program, "wordladder.py SOURCE DESTINATION < dict.txt"
For eg.

C:\>wordladder.py BOX TEN < words.txt
(4, ['BOX', 'BOD', 'BED', 'BEN', 'TEN'])

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