I always forget how much I love the simple problems. Everyone plays
when we
have them and the solutions tend to be quite creative. This time was no
different. Everyone did see MenTaLguY’s state machine parser, right?
Solving the problem, even with the suggested extra credit, is easy stuff
though.
Let’s have a look at Bob S.'s solution to see just how easy.
Here’s the
start of the code:
set to dictionary file to load
DICT = ‘/usr/share/dict/words’
class Morse
@@words = nil
LETTER = Hash[*%w/
A .- N -.
B -… O —
C -.-. P .–.
D -… Q --.-
E . R .-.
F …-. S …
G --. T -
H … U …-
I … V …-
J .— W .–
K -.- X -…-
L .-… Y -.–
M – Z --…
/]
…
Here we see some setup work. Bob defines a constant to point at the
dictionary
on his system. The comment encourages you to tweak this for your
system.
Next we have a the start of a class definition. This class is really
just used
as a namespace. No objects are ever constructed from it. Given that, a
module
definition might have better represented it’s purpose.
The class variable will hold the actual dictionary words, assuming the
user
requests that we load it. More on that in a bit.
Finally, we have a wonderful little trick that many solvers picked up
on. It’s
possible to write some Ruby that will allow you to paste the translation
chart
from the quiz directly into your code and actually have that be a
meaningful
data structure. Here we see the whole thing converted into a whitespace
delimited Array, which is then splatted into Hash form. Every other
element
becomes a key or value, so we end up with Morse code values keyed by the
English
letter it translates to. Very clever stuff.
Here’s the dictionary loading code:
…
loads dictionary file to limit the words output
def self.load_dictionary(path)
@@words = {}
File.open(path, ‘r’) do |f|
while word = f.gets
@@words[word.chomp.upcase] = true
end
end
end
…
This method creates a Hash to hold the words, reads the indicated file
line by
line, peels off line endings, and normalizes word case, filling the Hash
as it
reads. Most solutions that included dictionary handling had a chunk of
code
very similar to this. This particular version could be simplified a
touch by
using File.foreach().
The last method of Bob’s class does the heavy lifting:
…
returns list of words starting with prefix that can be made from
code
def self.words(code = @code, prefix = ‘’)
results = []
if code == ‘’
results << prefix if @@words.nil? || @@words.include?(prefix)
else
LETTER.sort.each do |l, p|
if code[0, p.length] == p
results += words(code[p.length,code.length], prefix + l)
end
end
end
results
end
end
…
This is a very straightforward recursive decoder. We get two parameters
here:
the remaining code (ignore that unused default) and any prefix we should
apply
to words generated from that code. You can see that the method defines,
fills,
and returns an Array of results. How those results are generated is the
interesting bit.
The if statement branch comes into play when we run out of code to
translate.
In that case, the word is added to the results, as long as no dictionary
was
loaded or the word is in the loaded dictionary. This means that you can
run
this code without using a dictionary to see all options or with a
dictionary to
see only likely matches.
The else branch handles all cases where we have some code left. When
that
happens, each letter is tried at the start of the code. If it matches,
the
method recurses using the code after the matched letter and the existing
prefix
plus the new letter. All words generated from the recursive calls are
added to
the current result set. This generates all possible combinations.
I should note here that there was some discussion over similar code from
Donald
A. Ball Jr. that yielded the words to a block instead of collecting them
in an
Array. This is probably a good move for this particular decoder, since
there
are a surprising 5,104 possible translations of the simple code provided
in the
quiz. Collecting larger translations in an Array might get a bit
resource
intensive. Bob’s method is easily converted:
changed to use a block by JEG2
def self.words(code, prefix = ‘’, &block)
if code == ‘’
block[prefix] if @@words.nil? || @@words.include?(prefix)
else
LETTER.sort.each do |l, p|
if code[0, p.length] == p
results += words(code[p.length,code.length], prefix + l,
&block)
end
end
end
end
Note the three simple changes: I added the block as a parameter, I pass
finished words to the block instead of placing them in an Array, and I
hand the
block down the stack when we recurse. This removes the need for the
results
Array, so I have also trimmed that code.
Getting back to Bob’s solution, here is the application code that makes
it run:
…
Morse.load_dictionary(DICT) if ARGV.delete(’-d’)
abort “Usage: #{$0} [-d] code […]” if ARGV.empty?
ARGV.each do |code|
puts “#{code}:”
puts Morse.words(code) # Morse.words(code) { |w| puts w } with the
block
end
First we see the optional dictionary flag interpretation. We’ve already
talked
about how the code changes depending on whether or not a word list is
loaded.
The next line is a usage statement printed by abort(). We don’t see
that method
too often in quiz solutions, but it prints a message to STDERR, when
provided,
and then terminates the program. Note that the usage statement tells us
this
code differs slightly from the quiz interface, taking the code as a
command-line
argument instead of reading it from STDIN.
That last iterator just walks the provided codes printing the
translations
returned from a call to Morse.words().
—.-- -…–.-.-… ---- .-.-…-… .–…— -.-…–. .-…–…
-----.-… -.-.----… (.-.-.-.----…-… -…-.-- --.-.-.-.-
-…–.--…
…—.-…–…----.)
Tomorrow we have another easy, though more practical, challenge…
