Case study: word play ===================== This chapter presents the second case study, which involves solving word puzzles by searching for words that have certain properties. For example, we’ll find the longest palindromes in English and search for words whose letters appear in alphabetical order. And I will present another program development plan: reduction to a previously solved problem. Reading word lists ------------------ For the exercises in this chapter we need a list of English words. There are lots of word lists available on the Web, but the one most suitable for our purpose is one of the word lists collected and contributed to the public domain by Grady Ward as part of the Moby lexicon project (see http://wikipedia.org/wiki/Moby_Project). It is a list of 113,809 official crosswords; that is, words that are considered valid in crossword puzzles and other word games. In the Moby collection, the filename is 113809of.fic; you can download a copy, with the simpler name words.txt, from http://thinkpython2.com/code/words.txt. This file is in plain text, so you can open it with a text editor, but you can also read it from Python. The built-in function open takes the name of the file as a parameter and returns a **file object** you can use to read the file. :: >>> fin = open('words.txt') fin is a common name for a file object used for input. The file object provides several methods for reading, including readline, which reads characters from the file until it gets to a newline and returns the result as a string: :: >>> fin.readline() 'aa\r\n' The first word in this particular list is “aa”, which is a kind of lava. The sequence ``\r\n`` represents two whitespace characters, a carriage return and a newline, that separate this word from the next. The file object keeps track of where it is in the file, so if you call readline again, you get the next word: :: >>> fin.readline() 'aah\r\n' The next word is “aah”, which is a perfectly legitimate word, so stop looking at me like that. Or, if it’s the whitespace that’s bothering you, we can get rid of it with the string method strip: :: >>> line = fin.readline() >>> word = line.strip() >>> word 'aahed' You can also use a file object as part of a for loop. This program reads words.txt and prints each word, one per line: :: fin = open('words.txt') for line in fin: word = line.strip() print(word) Exercises --------- There are solutions to these exercises in the next section. You should at least attempt each one before you read the solutions. Write a program that reads words.txt and prints only the words with more than 20 characters (not counting whitespace). In 1939 Ernest Vincent Wright published a 50,000 word novel called *Gadsby* that does not contain the letter “e”. Since “e” is the most common letter in English, that’s not easy to do. In fact, it is difficult to construct a solitary thought without using that most common symbol. It is slow going at first, but with caution and hours of training you can gradually gain facility. All right, I’ll stop now. Write a function called ``has_no_e`` that returns True if the given word doesn’t have the letter “e” in it. Modify your program from the previous section to print only the words that have no “e” and compute the percentage of the words in the list that have no “e”. Write a function named avoids that takes a word and a string of forbidden letters, and that returns True if the word doesn’t use any of the forbidden letters. Modify your program to prompt the user to enter a string of forbidden letters and then print the number of words that don’t contain any of them. Can you find a combination of 5 forbidden letters that excludes the smallest number of words? Write a function named ``uses_only`` that takes a word and a string of letters, and that returns True if the word contains only letters in the list. Can you make a sentence using only the letters acefhlo? Other than “Hoe alfalfa?” Write a function named ``uses_all`` that takes a word and a string of required letters, and that returns True if the word uses all the required letters at least once. How many words are there that use all the vowels aeiou? How about aeiouy? Write a function called ``is_abecedarian`` that returns True if the letters in a word appear in alphabetical order (double letters are ok). How many abecedarian words are there? Search ------ All of the exercises in the previous section have something in common; they can be solved with the search pattern we saw in Section [find]. The simplest example is: :: def has_no_e(word): for letter in word: if letter == 'e': return False return True The for loop traverses the characters in word. If we find the letter “e”, we can immediately return False; otherwise we have to go to the next letter. If we exit the loop normally, that means we didn’t find an “e”, so we return True. You could write this function more concisely using the in operator, but I started with this version because it demonstrates the logic of the search pattern. is a more general version of ``has_no_e`` but it has the same structure: :: def avoids(word, forbidden): for letter in word: if letter in forbidden: return False return True We can return False as soon as we find a forbidden letter; if we get to the end of the loop, we return True. ``uses_only`` is similar except that the sense of the condition is reversed: :: def uses_only(word, available): for letter in word: if letter not in available: return False return True Instead of a list of forbidden letters, we have a list of available letters. If we find a letter in word that is not in available, we can return False. ``uses_all`` is similar except that we reverse the role of the word and the string of letters: :: def uses_all(word, required): for letter in required: if letter not in word: return False return True Instead of traversing the letters in word, the loop traverses the required letters. If any of the required letters do not appear in the word, we can return False. If you were really thinking like a computer scientist, you would have recognized that ``uses_all`` was an instance of a previously solved problem, and you would have written: :: def uses_all(word, required): return uses_only(required, word) This is an example of a program development plan called **reduction to a previously solved problem**, which means that you recognize the problem you are working on as an instance of a solved problem and apply an existing solution. Looping with indices -------------------- I wrote the functions in the previous section with for loops because I only needed the characters in the strings; I didn’t have to do anything with the indices. For ``is_abecedarian`` we have to compare adjacent letters, which is a little tricky with a for loop: :: def is_abecedarian(word): previous = word[0] for c in word: if c < previous: return False previous = c return True An alternative is to use recursion: :: def is_abecedarian(word): if len(word) <= 1: return True if word[0] > word[1]: return False return is_abecedarian(word[1:]) Another option is to use a while loop: :: def is_abecedarian(word): i = 0 while i < len(word)-1: if word[i+1] < word[i]: return False i = i+1 return True The loop starts at i=0 and ends when i=len(word)-1. Each time through the loop, it compares the :math:`i`\ th character (which you can think of as the current character) to the :math:`i+1`\ th character (which you can think of as the next). If the next character is less than (alphabetically before) the current one, then we have discovered a break in the abecedarian trend, and we return False. If we get to the end of the loop without finding a fault, then the word passes the test. To convince yourself that the loop ends correctly, consider an example like ``'flossy'``. The length of the word is 6, so the last time the loop runs is when i is 4, which is the index of the second-to-last character. On the last iteration, it compares the second-to-last character to the last, which is what we want. Here is a version of ``is_palindrome`` (see Exercise [palindrome]) that uses two indices; one starts at the beginning and goes up; the other starts at the end and goes down. :: def is_palindrome(word): i = 0 j = len(word)-1 while i