Chapter 9  Case study: word play 第 9 章 案例研究：文字游戏

本页译自 Think Python 2e（Allen B. Downey）· Chapter 9 Case study: word play。代码块保留英文原文不翻译；正文段段对照，中文块可用右下角按钮隐藏。

9.1 Reading word lists 9.1 读词表

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://thinkpython.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 is a common name for a file object used for input. Mode fin indicates that this file is open for reading (as opposed to fin for writing).

The file object provides several methods for reading, including open0002, which reads characters from the file until it gets to a newline and returns the result as a string:

The first word in this particular list is “aa,” which is a kind of lava. The sequence open 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 open0002 again, you get the next word:

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 open0:

You can also use a file object as part of a fin loop. This program reads open00033 and prints each word, one per line:

Exercise 1

Write a program that reads open00037 and prints only the words with more than 20 characters (not counting whitespace).

9.2 Exercises 9.2 习题

There are solutions to these exercises in the next section. You should at least attempt each one before you read the solutions.

Exercise 2

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 open0003 that returns open 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 have no “e.”

Exercise 3

Write a function named open00 that takes a word and a string of forbidden letters, and that returns open 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?

Exercise 4

Write a function named open00047 that takes a word and a string of letters, and that returns open if the word contains only letters in the list. Can you make a sentence using only the letters open000? Other than “Hoe alfalfa?”

Exercise 5

Write a function named open0005 that takes a word and a string of required letters, and that returns open if the word uses all the required letters at least once. How many words are there that use all the vowels open0? How about open00?

Exercise 6

Write a function called open00061



that returns open if the letters in a word appear in alphabetical order (double letters are ok). How many abecedarian words are there?

9.3 Search 9.3 搜索

All of the exercises in the previous section have something in common; they can be solved with the search pattern we saw in Section 8.6. The simplest example is:

The fin loop traverses the characters in open. If we find the letter “e”, we can immediately return open0; 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 open.

open00 is a more general version of open0007 but it has the same structure:

We can return open0 as soon as we find a forbidden letter; if we get to the end of the loop, we return open.

open00083 is similar except that the sense of the condition is reversed:

Instead of a list of forbidden letters, we have a list of available letters. If we find a letter in open that is not in open00087, we can return open0.

open0009 is similar except that we reverse the role of the word and the string of letters:

Instead of traversing the letters in open, the loop traverses the required letters. If any of the required letters do not appear in the word, we can return open0.

If you were really thinking like a computer scientist, you would have recognized that open0009 was an instance of a previously-solved problem, and you would have written:

This is an example of a program development method called problem recognition, which means that you recognize the problem you are working on as an instance of a previously-solved problem, and apply a previously-developed solution.

9.4 Looping with indices 9.4 带索引的循环

I wrote the functions in the previous section with fin loops because I only needed the characters in the strings; I didn’t have to do anything with the indices.

For open00104



we have to compare adjacent letters, which is a little tricky with a fin loop:

An alternative is to use recursion:

Another option is to use a open0 loop:

The loop starts at fin and ends when open00114


. Each time through the loop, it compares the ith character (which you can think of as the current character) to the i+1th 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 open0.

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 open0011. 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 open00123


(see Exercise 6) that uses two indices; one starts at the beginning and goes up; the other starts at the end and goes down.

Or, if you noticed that this is an instance of a previously-solved problem, you might have written:

Assuming you did Exercise 9.

9.5 Debugging 9.5 调试

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.

Taking open0012 as an example, there are two obvious cases to check: words that have an ’e’ should return open0; words that don’t should return open. 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 open00133. 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!

程序测试能用来证明缺陷的存在，却永远无法证明缺陷不存在！

— Edsger W. Dijkstra

—— Edsger W. Dijkstra

9.6 Glossary 9.6 术语表

file object:

A value that represents an open file.

problem recognition:

A way of solving a problem by expressing it as an instance of a previously-solved problem.

special case:

A test case that is atypical or non-obvious (and less likely to be handled correctly).

file object 文件对象：

代表一个已打开文件的值。

problem recognition 问题识别：

一种解题方式：把当前问题表述为某个已解决问题的实例，即归约到已解决的问题。

special case 特例：

不典型或不显眼的测试用例（也因此更容易被处理错）。

9.7 Exercises 9.7 习题

Exercise 7

This question is based on a Puzzler that was broadcast on the radio program Car Talk (open00135




























):

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?

给我一个含有三组连续双写字母的单词。我先举两个差一点就合格、但其实不合格的例子。比如 committee，c-o-m-m-i-t-t-e-e，要不是中间偷偷冒出个「i」，它就完美了。再比如 Mississippi：M-i-s-s-i-s-s-i-p-p-i，把那几个「i」抽掉就行了。但确实有个单词含有三组紧挨着的双写字母，据我所知它可能是唯一的一个。当然也许还有五百个，只不过我只想得出一个。那个词是什么？

Write a program to find it. Solution: open00137




























.

Exercise 8 Here’s another Car Talk Puzzler (open00139




























):

“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.

「前几天我在高速上开车，无意间瞥了一眼里程表。和大多数里程表一样，它显示六位数字，只记整英里。所以，假如我的车跑了 300,000 英里，我看到的就是 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.

「那天我看到的可有意思了。我注意到最后 4 位是回文的，也就是正着读和倒着读一样。比如 5-4-4-5 就是个回文，所以我的里程表可能显示 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!

「又开了一英里，最后 5 位成了回文。比如它可能显示 3-6-5-4-5-6。再过一英里，6 位中间的那 4 位是回文。准备好了吗？再过一英里，6 位全都是回文！

“The question is, what was on the odometer when I first looked?”

「问题是：我第一眼看到时，里程表上是多少？」

Write a Python program that tests all the six-digit numbers and prints any numbers that satisfy these requirements. Solution: open00141




























.

Exercise 9 Here’s another Car Talk Puzzler you can solve with a search (open00143




























):

“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.

「最近我去看我妈，我们发现我年龄的两位数字倒过来正好是她的年龄。比如她 73 岁，我就 37 岁。我们琢磨这些年来这种情况出现过几次，可聊着聊着就跑题了，一直没算出答案。

“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?”

「回到家我算出来，到目前为止我们俩年龄的数字互为倒序已经出现过六次。我还算出，运气好的话过几年还会再来一次；运气特别好的话，那之后还能再来一次。也就是说，总共会出现 8 次。那么问题是：我现在多大？」

Write a Python program that searches for solutions to this Puzzler. Hint: you might find the string method open0 useful.

Solution: open00147




























.