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.
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\n'
The first word in this particular list is “aa”, which is a kind of lava.
\n represents the newline character that separates 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\n'
The next word is “aah”, which is a perfectly legitimate word, so stop looking at me like that. Or, if it’s the newline character 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)
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.
Write a program that reads words.txt and prints only the words that have no “e”. Compute the percentage of 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.
Write a program that prompts the user to enter a string of forbidden letters and then prints 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?
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
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
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.
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.
is_abecedarian we have to compare adjacent letters, which is a
little tricky with a for loop:
def is_abecedarian(word): previous = word 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 > word: 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 $i$th character (which you can think of as the current character) to the $i+1$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<j: if word[i] != word[j]: return False i = i+1 j = j-1 return True
Or we could reduce to a previously solved problem and write:
def is_palindrome(word): return is_reverse(word, word)
is_reverse from Section [isreverse].
Testing programs is hard. The functions in this chapter are relatively easy to test because you can check the results by hand. Even so, it is somewhere between difficult and impossible to choose a set of words that test for all possible errors.
has_no_e as an example, there are two obvious cases to check:
words that have an ‘e’ should return False, and words that
don’t should return True. You should have no trouble coming
up with one of each.
Within each case, there are some less obvious subcases. Among the words that have an “e”, you should test words with an “e” at the beginning, the end, and somewhere in the middle. You should test long words, short words, and very short words, like the empty string. The empty string is an example of a special case, which is one of the non-obvious cases where errors often lurk.
In addition to the test cases you generate, you can also test your program with a word list like words.txt. By scanning the output, you might be able to catch errors, but be careful: you might catch one kind of error (words that should not be included, but are) and not another (words that should be included, but aren’t).
In general, testing can help you find bugs, but it is not easy to generate a good set of test cases, and even if you do, you can’t be sure your program is correct. According to a legendary computer scientist:
Program testing can be used to show the presence of bugs, but never to show their absence!
— Edsger W. Dijkstra
This question is based on a Puzzler that was broadcast on the radio program Car Talk (http://www.cartalk.com/content/puzzlers):
Give me a word with three consecutive double letters. I’ll give you a couple of words that almost qualify, but don’t. For example, the word committee, c-o-m-m-i-t-t-e-e. It would be great except for the ‘i’ that sneaks in there. Or Mississippi: M-i-s-s-i-s-s-i-p-p-i. If you could take out those i’s it would work. But there is a word that has three consecutive pairs of letters and to the best of my knowledge this may be the only word. Of course there are probably 500 more but I can only think of one. What is the word?
Write a program to find it. Solution: http://thinkpython2.com/code/cartalk1.py.
Here’s another Car Talk Puzzler (http://www.cartalk.com/content/puzzlers):
``I was driving on the highway the other day and I happened to notice my odometer. Like most odometers, it shows six digits, in whole miles only. So, if my car had 300,000 miles, for example, I’d see 3-0-0-0-0-0.
``Now, what I saw that day was very interesting. I noticed that the last 4 digits were palindromic; that is, they read the same forward as backward. For example, 5-4-4-5 is a palindrome, so my odometer could have read 3-1-5-4-4-5.
``One mile later, the last 5 numbers were palindromic. For example, it could have read 3-6-5-4-5-6. One mile after that, the middle 4 out of 6 numbers were palindromic. And you ready for this? One mile later, all 6 were palindromic!
“The question is, what was on the odometer when I first looked?”
Write a Python program that tests all the six-digit numbers and prints any numbers that satisfy these requirements. Solution: http://thinkpython2.com/code/cartalk2.py.
Here’s another Car Talk Puzzler you can solve with a search (http://www.cartalk.com/content/puzzlers):
``Recently I had a visit with my mom and we realized that the two digits that make up my age when reversed resulted in her age. For example, if she’s 73, I’m 37. We wondered how often this has happened over the years but we got sidetracked with other topics and we never came up with an answer.
“When I got home I figured out that the digits of our ages have been reversible six times so far. I also figured out that if we’re lucky it would happen again in a few years, and if we’re really lucky it would happen one more time after that. In other words, it would have happened 8 times over all. So the question is, how old am I now?”
Write a Python program that searches for solutions to this Puzzler. Hint: you might find the string method zfill useful.
This chapter presents one of Python’s most useful built-in types, lists. You will also learn more about objects and what can happen when you have more than one name for the same object.
Like a string, a list is a sequence of values. In a string, the values are characters; in a list, they can be any type. The values in a list are called elements or sometimes items.
There are several ways to create a new list; the simplest is to enclose
the elements in square brackets (
[10, 20, 30, 40] ['crunchy frog', 'ram bladder', 'lark vomit']
The first example is a list of four integers. The second is a list of three strings. The elements of a list don’t have to be the same type. The following list contains a string, a float, an integer, and (lo!) another list:
['spam', 2.0, 5, [10, 20]]
A list within another list is nested.
A list that contains no elements is called an empty list; you can create
one with empty brackets,
As you might expect, you can assign list values to variables:
>>> cheeses = ['Cheddar', 'Edam', 'Gouda'] >>> numbers = [42, 123] >>> empty =  >>> print(cheeses, numbers, empty) ['Cheddar', 'Edam', 'Gouda'] [42, 123] 
The syntax for accessing the elements of a list is the same as for accessing the characters of a string—the bracket operator. The expression inside the brackets specifies the index. Remember that the indices start at 0:
>>> cheeses 'Cheddar'
Unlike strings, lists are mutable. When the bracket operator appears on the left side of an assignment, it identifies the element of the list that will be assigned.
>>> numbers = [42, 123] >>> numbers = 5 >>> numbers [42, 5]
The one-eth element of numbers, which used to be 123, is now 5.
Figure [fig.liststate] shows the state diagram for cheeses, numbers and empty.
Lists are represented by boxes with the word “list” outside and the elements of the list inside. cheeses refers to a list with three elements indexed 0, 1 and 2. numbers contains two elements; the diagram shows that the value of the second element has been reassigned from 123 to 5. empty refers to a list with no elements.
List indices work the same way as string indices:
Any integer expression can be used as an index.
If you try to read or write an element that does not exist, you get an IndexError.
If an index has a negative value, it counts backward from the end of the list.
The in operator also works on lists.
>>> cheeses = ['Cheddar', 'Edam', 'Gouda'] >>> 'Edam' in cheeses True >>> 'Brie' in cheeses False
The most common way to traverse the elements of a list is with a for loop. The syntax is the same as for strings:
for cheese in cheeses: print(cheese)
This works well if you only need to read the elements of the list. But if you want to write or update the elements, you need the indices. A common way to do that is to combine the built-in functions range and len:
for i in range(len(numbers)): numbers[i] = numbers[i] * 2
This loop traverses the list and updates each element. len returns the number of elements in the list. range returns a list of indices from 0 to $n-1$, where $n$ is the length of the list. Each time through the loop i gets the index of the next element. The assignment statement in the body uses i to read the old value of the element and to assign the new value.
A for loop over an empty list never runs the body:
for x in : print('This never happens.')
Although a list can contain another list, the nested list still counts as a single element. The length of this list is four:
['spam', 1, ['Brie', 'Roquefort', 'Pol le Veq'], [1, 2, 3]]
The + operator concatenates lists:
>>> a = [1, 2, 3] >>> b = [4, 5, 6] >>> c = a + b >>> c [1, 2, 3, 4, 5, 6]
The operator repeats a list a given number of times:
>>>  * 4 [0, 0, 0, 0] >>> [1, 2, 3] * 3 [1, 2, 3, 1, 2, 3, 1, 2, 3]
The first example repeats four times. The second example repeats the list three times.
The slice operator also works on lists:
>>> t = ['a', 'b', 'c', 'd', 'e', 'f'] >>> t[1:3] ['b', 'c'] >>> t[:4] ['a', 'b', 'c', 'd'] >>> t[3:] ['d', 'e', 'f']
If you omit the first index, the slice starts at the beginning. If you omit the second, the slice goes to the end. So if you omit both, the slice is a copy of the whole list.
>>> t[:] ['a', 'b', 'c', 'd', 'e', 'f']
Since lists are mutable, it is often useful to make a copy before performing operations that modify lists.
A slice operator on the left side of an assignment can update multiple elements:
>>> t = ['a', 'b', 'c', 'd', 'e', 'f'] >>> t[1:3] = ['x', 'y'] >>> t ['a', 'x', 'y', 'd', 'e', 'f']
Python provides methods that operate on lists. For example, append adds a new element to the end of a list:
>>> t = ['a', 'b', 'c'] >>> t.append('d') >>> t ['a', 'b', 'c', 'd']
extend takes a list as an argument and appends all of the elements:
>>> t1 = ['a', 'b', 'c'] >>> t2 = ['d', 'e'] >>> t1.extend(t2) >>> t1 ['a', 'b', 'c', 'd', 'e']
This example leaves t2 unmodified.
sort arranges the elements of the list from low to high:
>>> t = ['d', 'c', 'e', 'b', 'a'] >>> t.sort() >>> t ['a', 'b', 'c', 'd', 'e']
Most list methods are void; they modify the list and return None. If you accidentally write t = t.sort(), you will be disappointed with the result.
To add up all the numbers in a list, you can use a loop like this:
def add_all(t): total = 0 for x in t: total += x return total
total is initialized to 0. Each time through the loop, x gets one element from the list. The += operator provides a short way to update a variable. This augmented assignment statement,
total += x
is equivalent to
total = total + x
As the loop runs, total accumulates the sum of the elements; a variable used this way is sometimes called an accumulator.
Adding up the elements of a list is such a common operation that Python provides it as a built-in function, sum:
>>> t = [1, 2, 3] >>> sum(t) 6
An operation like this that combines a sequence of elements into a single value is sometimes called reduce.
Sometimes you want to traverse one list while building another. For example, the following function takes a list of strings and returns a new list that contains capitalized strings:
def capitalize_all(t): res =  for s in t: res.append(s.capitalize()) return res
res is initialized with an empty list; each time through the loop, we append the next element. So res is another kind of accumulator.
An operation like
capitalize_all is sometimes called a
map because it “maps” a function (in this case the
method capitalize) onto each of the elements in a sequence.
Another common operation is to select some of the elements from a list and return a sublist. For example, the following function takes a list of strings and returns a list that contains only the uppercase strings:
def only_upper(t): res =  for s in t: if s.isupper(): res.append(s) return res
isupper is a string method that returns True if the string contains only upper case letters.
An operation like
only_upper is called a filter
because it selects some of the elements and filters out the others.
Most common list operations can be expressed as a combination of map, filter and reduce.
There are several ways to delete elements from a list. If you know the index of the element you want, you can use pop:
>>> t = ['a', 'b', 'c'] >>> x = t.pop(1) >>> t ['a', 'c'] >>> x 'b'
pop modifies the list and returns the element that was removed. If you don’t provide an index, it deletes and returns the last element.
If you don’t need the removed value, you can use the del operator:
>>> t = ['a', 'b', 'c'] >>> del t >>> t ['a', 'c']
If you know the element you want to remove (but not the index), you can use remove:
>>> t = ['a', 'b', 'c'] >>> t.remove('b') >>> t ['a', 'c']
The return value from remove is None.
To remove more than one element, you can use del with a slice index:
>>> t = ['a', 'b', 'c', 'd', 'e', 'f'] >>> del t[1:5] >>> t ['a', 'f']
As usual, the slice selects all the elements up to but not including the second index.
A string is a sequence of characters and a list is a sequence of values, but a list of characters is not the same as a string. To convert from a string to a list of characters, you can use list:
>>> s = 'spam' >>> t = list(s) >>> t ['s', 'p', 'a', 'm']
Because list is the name of a built-in function, you should avoid using it as a variable name. I also avoid l because it looks too much like 1. So that’s why I use t.
The list function breaks a string into individual letters. If you want to break a string into words, you can use the split method:
>>> s = 'pining for the fjords' >>> t = s.split() >>> t ['pining', 'for', 'the', 'fjords']
An optional argument called a delimiter specifies which characters to use as word boundaries. The following example uses a hyphen as a delimiter:
>>> s = 'spam-spam-spam' >>> delimiter = '-' >>> t = s.split(delimiter) >>> t ['spam', 'spam', 'spam']
join is the inverse of split. It takes a list of strings and concatenates the elements. join is a string method, so you have to invoke it on the delimiter and pass the list as a parameter:
>>> t = ['pining', 'for', 'the', 'fjords'] >>> delimiter = ' ' >>> s = delimiter.join(t) >>> s 'pining for the fjords'
In this case the delimiter is a space character, so join
puts a space between words. To concatenate strings without spaces, you
can use the empty string,
'', as a delimiter.
If we run these assignment statements:
a = 'banana' b = 'banana'
We know that a and b both refer to a string, but we don’t know whether they refer to the same string. There are two possible states, shown in Figure [fig.list1].
In one case, a and b refer to two different objects that have the same value. In the second case, they refer to the same object.
To check whether two variables refer to the same object, you can use the is operator.
>>> a = 'banana' >>> b = 'banana' >>> a is b True
In this example, Python only created one string object, and both a and b refer to it. But when you create two lists, you get two objects:
>>> a = [1, 2, 3] >>> b = [1, 2, 3] >>> a is b False
So the state diagram looks like Figure [fig.list2].
In this case we would say that the two lists are equivalent, because they have the same elements, but not identical, because they are not the same object. If two objects are identical, they are also equivalent, but if they are equivalent, they are not necessarily identical.
Until now, we have been using “object” and “value” interchangeably, but it is more precise to say that an object has a value. If you evaluate , you get a list object whose value is a sequence of integers. If another list has the same elements, we say it has the same value, but it is not the same object.
If a refers to an object and you assign b = a, then both variables refer to the same object:
>>> a = [1, 2, 3] >>> b = a >>> b is a True
The state diagram looks like Figure [fig.list3].
The association of a variable with an object is called a reference. In this example, there are two references to the same object.
An object with more than one reference has more than one name, so we say that the object is aliased.
If the aliased object is mutable, changes made with one alias affect the other:
>>> b = 42 >>> a [42, 2, 3]
Although this behavior can be useful, it is error-prone. In general, it is safer to avoid aliasing when you are working with mutable objects.
For immutable objects like strings, aliasing is not as much of a problem. In this example:
a = 'banana' b = 'banana'
It almost never makes a difference whether a and b refer to the same string or not.
When you pass a list to a function, the function gets a reference to the
list. If the function modifies the list, the caller sees the change. For
delete_head removes the first element from a list:
def delete_head(t): del t
Here’s how it is used:
>>> letters = ['a', 'b', 'c'] >>> delete_head(letters) >>> letters ['b', 'c']
The parameter t and the variable letters are aliases for the same object. The stack diagram looks like Figure [fig.stack5].
Since the list is shared by two frames, I drew it between them.
It is important to distinguish between operations that modify lists and operations that create new lists. For example, the append method modifies a list, but the + operator creates a new list.
Here’s an example using append:
>>> t1 = [1, 2] >>> t2 = t1.append(3) >>> t1 [1, 2, 3] >>> t2 None
The return value from append is None.
Here’s an example using the + operator:
>>> t3 = t1 +  >>> t1 [1, 2, 3] >>> t3 [1, 2, 3, 4]
The result of the operator is a new list, and the original list is unchanged.
This difference is important when you write functions that are supposed to modify lists. For example, this function does not delete the head of a list:
def bad_delete_head(t): t = t[1:] # WRONG!
The slice operator creates a new list and the assignment makes t refer to it, but that doesn’t affect the caller.
>>> t4 = [1, 2, 3] >>> bad_delete_head(t4) >>> t4 [1, 2, 3]
At the beginning of
bad_delete_head, t and
t4 refer to the same list. At the end, t
refers to a new list, but t4 still refers to the original,
An alternative is to write a function that creates and returns a new list. For example, tail returns all but the first element of a list:
def tail(t): return t[1:]
This function leaves the original list unmodified. Here’s how it is used:
>>> letters = ['a', 'b', 'c'] >>> rest = tail(letters) >>> rest ['b', 'c']
Careless use of lists (and other mutable objects) can lead to long hours of debugging. Here are some common pitfalls and ways to avoid them:
Most list methods modify the argument and return None. This is the opposite of the string methods, which return a new string and leave the original alone.
If you are used to writing string code like this:
word = word.strip()
It is tempting to write list code like this:
t = t.sort() # WRONG!
Because sort returns None, the next operation you perform with t is likely to fail.
Before using list methods and operators, you should read the documentation carefully and then test them in interactive mode.
Pick an idiom and stick with it.
Part of the problem with lists is that there are too many ways to do things. For example, to remove an element from a list, you can use pop, remove, del, or even a slice assignment.
To add an element, you can use the append method or the + operator. Assuming that t is a list and x is a list element, these are correct:
t.append(x) t = t + [x] t += [x]
And these are wrong:
t.append([x]) # WRONG! t = t.append(x) # WRONG! t + [x] # WRONG! t = t + x # WRONG!
Try out each of these examples in interactive mode to make sure you understand what they do. Notice that only the last one causes a runtime error; the other three are legal, but they do the wrong thing.
Make copies to avoid aliasing.
If you want to use a method like sort that modifies the argument, but you need to keep the original list as well, you can make a copy.
>>> t = [3, 1, 2] >>> t2 = t[:] >>> t2.sort() >>> t [3, 1, 2] >>> t2 [1, 2, 3]
In this example you could also use the built-in function sorted, which returns a new, sorted list and leaves the original alone.
>>> t2 = sorted(t) >>> t [3, 1, 2] >>> t2 [1, 2, 3]
You can download solutions to these exercises from http://thinkpython2.com/code/list_exercises.py.
Write a function called
nested_sum that takes a list of lists of
integers and adds up the elements from all of the nested lists. For
>>> t = [[1, 2], , [4, 5, 6]] >>> nested_sum(t) 21
Write a function called cumsum that takes a list of numbers and returns the cumulative sum; that is, a new list where the $i$th element is the sum of the first $i+1$ elements from the original list. For example:
>>> t = [1, 2, 3] >>> cumsum(t) [1, 3, 6]
Write a function called
middle that takes a list and returns a new
list that contains all but the first and last elements. For example:
>>> t = [1, 2, 3, 4] >>> middle(t) [2, 3]
Write a function called
chop that takes a list, modifies it by
removing the first and last elements, and returns None. For
>>> t = [1, 2, 3, 4] >>> chop(t) >>> t [2, 3]
Write a function called
is_sorted that takes a list as a parameter and
returns True if the list is sorted in ascending order and
False otherwise. For example:
>>> is_sorted([1, 2, 2]) True >>> is_sorted(['b', 'a']) False
Two words are anagrams if you can rearrange the letters from one to
spell the other. Write a function called
is_anagram that takes two
strings and returns True if they are anagrams.
Write a function called
has_duplicates that takes a list and returns
True if there is any element that appears more than once.
It should not modify the original list.
This exercise pertains to the so-called Birthday Paradox, which you can read about at http://en.wikipedia.org/wiki/Birthday_paradox.
If there are 23 students in your class, what are the chances that two of you have the same birthday? You can estimate this probability by generating random samples of 23 birthdays and checking for matches. Hint: you can generate random birthdays with the randint function in the random module.
You can download my solution from http://thinkpython2.com/code/birthday.py.
Write a function that reads the file words.txt and builds a list with one element per word. Write two versions of this function, one using the append method and the other using the idiom t = t + [x]. Which one takes longer to run? Why?
To check whether a word is in the word list, you could use the in operator, but it would be slow because it searches through the words in order.
Because the words are in alphabetical order, we can speed things up with a bisection search (also known as binary search), which is similar to what you do when you look a word up in the dictionary (the book, not the data structure). You start in the middle and check to see whether the word you are looking for comes before the word in the middle of the list. If so, you search the first half of the list the same way. Otherwise you search the second half.
Either way, you cut the remaining search space in half. If the word list has 113,809 words, it will take about 17 steps to find the word or conclude that it’s not there.
Write a function called
in_bisect that takes a sorted list and a
target value and returns True if the word is in the list
and False if it’s not.
Or you could read the documentation of the bisect module and use that! Solution: http://thinkpython2.com/code/inlist.py.
Two words are a “reverse pair” if each is the reverse of the other. Write a program that finds all the reverse pairs in the word list. Solution: http://thinkpython2.com/code/reverse_pair.py.
Two words “interlock” if taking alternating letters from each forms a new word. For example, “shoe” and “cold” interlock to form “schooled”. Solution: http://thinkpython2.com/code/interlock.py. Credit: This exercise is inspired by an example at http://puzzlers.org.
Write a program that finds all pairs of words that interlock. Hint: don’t enumerate all pairs!
Can you find any words that are three-way interlocked; that is, every third letter forms a word, starting from the first, second or third?