Question

I used this hideous and inefficient implementation to find the word that can have the most consecutive last letters removed and still be a word.

Rodeo, for example, is a well-known one: Rodeo, Rode, Rod, Ro.
The program found ‘composers’: Composers, Composer, Compose, Compos, Comp

I was wondering how I should go about creating a program that finds the longest word that can have ANY of its letters (not just the last ones) removed and it still be considered a word:

For example: beast, best, bet, be — would be a valid possibility

Here was my program to find the one that removes consecutive letters (I’m also interested in hearing how this can be improved and optimized):

#Recursive function that finds how many letters can be removed from a word and
#it still be valid.  
def wordCheck(word, wordList, counter):

    if len(word)>=1:
        if word in wordList:
            return (wordCheck(word[0:counter-1], wordList, counter-1))
        else:
            return counter
    return counter


def main():
    a = open('C:\\Python32\\megalist2.txt', 'r+')
    wordList = set([line.strip() for line in a])
    #megaList contains a sorted list of tuple of 
    #(the word, how many letters can be removed  consecutively)
    megaList = sorted([(i, len(i)-1- wordCheck(i, wordList, len(i))) for i in wordList], key= lambda megaList: megaList[1])


    for i in megaList:
        if i[1] > 3:
            print (i)

if __name__ == '__main__':
    main()

Answer 1

Here’s an implementation I just wrote up. It runs in about five seconds with my ~235k word list. The output doesn’t show the whole chain, but you can easily reassemble it from the output.

# Load the words into a dictionary
words = dict((x.strip(), set()) for x in open("/usr/share/dict/words"))

# For each word, remove each letter and see if the remaining word is still
# in the dictionary. If so, add it to the set of shorter words associated with
# that word in the dictionary.
# For example, bear -> {ear, bar, ber}
for w in words:
    for i in range(len(w)):
        shorter = w[:i] + w[i+1:]
        if shorter in words:
            words[w].add(shorter)

# Sort the words by length so we process the shortest ones first
sortedwords = sorted(words, key=len)

# For each word, the maximum chain length is:
#  - the maximum of the chain lengths of each shorter word, if any
#  - or 0 if there are no shorter words for this word
# Note that because sortedwords is sorted by length, we will always
# have maxlength[x] already available for each shorter word x
maxlength = {}
for w in sortedwords:
    if words[w]:
        maxlength[w] = 1 + max(maxlength[x] for x in words[w])
    else:
        maxlength[w] = 0

# Print the words in all chains for each of the top 10 words
toshow = sorted(words, key=lambda x: maxlength[x], reverse=True)[:10]
while toshow:
    w = toshow[0]
    print(w, [(x, maxlength[x]) for x in words[w]])
    toshow = toshow[1:] + list(x for x in words[w] if x not in toshow)

The longest word chain in my dictionary is:

• abranchiate
• branchiate
• branchiae
• branchia
• branchi
• branch
• ranch
• rach
• ach
• ah
• a