# Better version of anagram solver.
#    This version prunes the recursion if prefix is impossible
from BinarySearch import *

def anagrams(lexicon, charsToUse, prefix=''):
  """
  Return a list of anagrams, formed with prefix followed by charsToUse.

  lexicon          a list of words (presumed to be alphabetized)
  charsToUse       a string which represents the characters to be arranged
  prefix           a prefix which is presumed to come before the arrangement
                   of the charsToUse (default empty string)
  """
  global numCalls
  numCalls += 1

  # you can uncomment the following line for convenient debugging info
  # print '  '*len(prefix) + 'Call with "%s(%s)"' % (prefix, charsToUse)
  
  solutions = []
  if len(charsToUse) > 1:
    for i in range(len(charsToUse)):         # pick charsToUse[i] next
      newPrefix = prefix + charsToUse[i]
      if prefixSearch(lexicon, newPrefix):   # worth exploring
        newCharsToUse = charsToUse[ : i] + charsToUse[i+1 : ]
        solutions.extend(anagrams(lexicon, newCharsToUse, newPrefix))
  else:     # check to see if we have a good solution
    candidate = prefix + charsToUse
    if search(lexicon, candidate):           # use binary search
      solutions.append(candidate)
  return solutions

if __name__ == '__main__':
    from FileUtilities import *
    print 'Will begin by reading the file of words.'
    words = readWordFile()
    words.sort()

    puzzle = raw_input('enter a string of characters to use: ')
    while puzzle:
        numCalls = 0
        solutions = anagrams(words, puzzle.replace(' ',''))    # ignore spaces from input
        print '(total number of calls was %d)'%numCalls
        print 'There were %d solutions reported' % len(solutions)
        verbose = raw_input('Would you like to see them? [y/n] ')
        if verbose and verbose.strip()[0].lower() == 'y':
            print '\n'.join(solutions)

        puzzle = raw_input('enter a string of characters to use: ')