This chapter presents a 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. 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 this book’s GitHub repository.
This file is in plain text, so you can open it with a text editor, but you can also read it from Julia. The built-in function open
takes a name of the file as a parameter and returns a file stream you can use to read the file:
julia>
fin
=
open
(
"words.txt"
)
IOStream(<file words.txt>)
fin
is a file stream used for input. When it is no longer needed, it has to be closed with close(fin)
.
Julia provides several function for reading, including readline
, which reads characters from the file until it gets to a NEWLINE
and returns the result as a string:
julia>
readline
(
fin
)
"aa"
The first word in this particular list is “aa,” which is a kind of lava.
The file stream keeps track of where it is in the file, so if you call readline
again, you get the next word:
julia>
readline
(
fin
)
"aah"
The next word is “aah,” which is a perfectly legitimate word, so stop looking at me like that.
You can also use a file as part of a for
loop. This program reads words.txt and prints each word, one per line:
for
line
in
eachline
(
"words.txt"
)
println
(
line
)
end
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 (Wetzel Publishing) 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 hasno_e
that returns true
if the given word doesn’t have the letter e in it.
Modify your program from the previous exercise 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 five forbidden letters that excludes the smallest number of words?
Write a function named usesonly
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 usesall
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 isabecedarian
that returns true
if the letters in a word appear in alphabetical order (double letters are okay). 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. The simplest example is:
function
hasno_e
(
word
)
for
letter
in
word
if
letter
==
'e'
return
false
end
end
true
end
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 ∉
(
otin TAB
) operator, but I started with this version because it demonstrates the logic of the search pattern.
avoids
is a more general version of hasno_e
, but it has the same structure:
function
avoids
(
word
,
forbidden
)
for
letter
in
word
if
letter
∈
forbidden
return
false
end
end
true
end
We can return false
as soon as we find a forbidden letter; if we get to the end of the loop, we return true
.
usesonly
is similar except that the sense of the condition is reversed:
function
usesonly
(
word
,
available
)
for
letter
in
word
if
letter
∉
available
return
false
end
end
true
end
Instead of an array of forbidden letters, we have an array of available letters. If we find a letter in word
that is not in available
, we can return false
.
usesall
is similar except that we reverse the role of the word and the string of letters:
function
usesall
(
word
,
required
)
for
letter
in
required
if
letter
∉
word
return
false
end
end
true
end
Instead of traversing the letters in word
, the loop traverses the required letters. If any of the required letters do not appear in word
, we can return false
.
If you were really thinking like a computer scientist, you would have recognized that usesall
was an instance of a previously solved problem, and you would have written:
function
usesall
(
word
,
required
)
usesonly
(
required
,
word
)
end
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.
For isabecedarian
we have to compare adjacent letters, which is a little tricky with a for
loop:
function
isabecedarian
(
word
)
i
=
firstindex
(
word
)
previous
=
word
[
i
]
j
=
nextind
(
word
,
i
)
for
c
in
word
[
j
:
end
]
if
c
<
previous
return
false
end
previous
=
c
end
true
end
An alternative is to use recursion:
function
isabecedarian
(
word
)
if
length
(
word
)
<=
1
return
true
end
i
=
firstindex
(
word
)
j
=
nextind
(
word
,
i
)
if
word
[
i
]
>
word
[
j
]
return
false
end
isabecedarian
(
word
[
j
:
end
])
end
Another option is to use a while
loop:
function
isabecedarian
(
word
)
i
=
firstindex
(
word
)
j
=
nextind
(
word
,
1
)
while
j
<=
sizeof
(
word
)
if
word
[
j
]
<
word
[
i
]
return
false
end
i
=
j
j
=
nextind
(
word
,
i
)
end
true
end
The loop starts at i=1
and j=nextind(word, 1)
and ends when j>sizeof(word)
. Each time through the loop, it compares the i
th character (which you can think of as the current character) to the j
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"
.
Here is a version of ispalindrome
that uses two indices; one starts at the beginning and goes up, and the other starts at the end and goes down:
function
ispalindrome
(
word
)
i
=
firstindex
(
word
)
j
=
lastindex
(
word
)
while
i
<
j
if
word
[
i
]
!=
word
[
j
]
return
false
end
i
=
nextind
(
word
,
i
)
j
=
prevind
(
word
,
j
)
end
true
end
Or we could reduce to a previously solved problem and write
function
ispalindrome
(
word
)
isreverse
(
word
,
word
)
end
using isreverse
from “Debugging”.
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 tests for all possible errors.
Taking hasno_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 nonobvious 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:
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.
Here’s another Car Talk Puzzler:
I was driving on the highway the other day recently and I happened to notice my odometer. Like most odometers nowadays, it shows six digits, in whole miles only—no tenths of a mile. 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 forwards as backwards. 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 did [I] see on the odometer when [I] first looked?
Write a Julia program that tests all the six-digit numbers and prints any numbers that satisfy these requirements.
Here’s a third Car Talk Puzzler that you can solve with a search:
Recently I had a visit with my mom and we realized that the two digits that make up my age when reversed result 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 Julia program that searches for solutions to this Puzzler.
You might find the function lpad
useful.