# Posix Pangrams (#97)

**URL:** <https://www.ruby-forum.com/t/posix-pangrams-97/71938>\
**Category:** Ruby\
**Created:** [October 12, 2006, 3:22pm UTC](https://www.ruby-forum.com/t/posix-pangrams-97/71938 "2006-10-12T15:22:02Z")\
**Posts on this page:** 2\
**Page:** 1

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**Author:** ![bbazzarrakk](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/b/47e85d/32.png) [@bbazzarrakk](https://www.ruby-forum.com/u/bbazzarrakk)\
**Post date:** [October 12, 2006, 3:22pm UTC](https://www.ruby-forum.com/t/posix-pangrams-97/71938/1 "2006-10-12T15:22:02Z")

</div>

This problem turns out to be pretty tough. Finding the best pangram for  
a given  
constraint can be NP-Hard, as explained by Ken B.:

```
http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-talk/219003

```

Given that, we’re just interested in code that gets reasonably close to  
ideal  
results in a decent amount of time. All of the submissions did that  
quite well,  
but I want to show off Ezwan Aizat Bin Abdullah F.'s code below. I’ve  
Rubyified it just a bit, but the code still works the same. For  
reference, it  
finds this pangram in under a second on my box:

```
String: zcat jobs newgrp mv qhold tty uux mkfifo
Panagram? true
Words? 8
Length? 33
Dupes? 7

```

Let’s start with the easy stuff:

```
#!/usr/bin/ruby

# removed c99 fort77 m4
words = %w[ admin alias ar asa at awk basename batch bc bg cal cat cd
            cflow chgrp chmod chown cksum cmp comm command compress cp
            crontab csplit ctags cut cxref date dd delta df diff 

```

dirname  
du echo ed env ex expand expr false fc fg file find fold  
fuser gencat get getconf getopts grep hash head iconv id  
ipcrm ipcs jobs join kill lex link ln locale localedef  
logger  
logname lp ls mailx make man mesg mkdir mkfifo more mv  
newgrp  
nice nl nm nohup od paste patch pathchk pax pr printf prs  
ps  
pwd qalter qdel qhold qmove qmsg qrerun qrls qselect qsig  
qstat qsub read renice rm rmdel rmdir sact sccs sed sh  
sleep  
sort split strings strip stty tabs tail talk tee test time  
touch tput tr true tsort tty type ulimit umask unalias  
uname  
uncompress unexpand unget uniq unlink uucp uudecode  
uuencode  
uustat uux val vi wait wc what who write xargs yacc zcat ]  
words\_line = words.join(" ")

```
# ...

```

This code begins by filling an Array with the Posix utilities, except  
for the  
three that contain digits in their names. Everyone dropped these to  
keep things  
simple. Note that the words are also joined to form one long line of  
utility  
names.

Next we need to enhance String a bit with some capabilities we will need  
to do  
our work:

```
# ...

class String
  def letters(&block)
    scan(/[a-z]/, &block)
  end

  def pangram?
    letters.uniq.length == 26
  end

  def duplicated_letters
    seen = Hash.new { |found, char| found[char] = 1; 0 }
    letters.inject(0) { |sum, l| sum + seen[l] }
  end
end

# ...

```

We will need easy access to an Array of letters (without the spaces), so  
a  
simple wrapper is created over scan(). From there we can check pangram  
status  
by ensuring that we have 26 unique letters. Finally, when printing  
statistics  
it would be nice to be able to see how many letters are duplicates, so  
we add a  
method for that too.

Now the code builds up some handy constants in preparation for the  
search:

```
# ...

OCCURRENCE = Hash.new { |counts, char| counts[char] = 0 }
words_line.letters { |char| OCCURRENCE[char] += 1 }

WORDS_LENGTH = words_line.delete(" ").length

# ...

```

Here we see the OCCURRENCE Hash being filled. It will be the key to the  
search  
algorithm. The Hash is simply a count of how many times each letter is  
present  
in all of the names. “e” appears the most at a count of 62 times and  
“z” is  
present only once.

Another constant is filled with the overall character count.

Here’s a method used to print statistics when we have our answer:

```
# ...

def stats(line)
  puts "String: #{line}"
  puts "Pangram? #{line.pangram?}"
  puts "Words? #{line.split.length}"
  puts "Length? #{line.delete(' ').length}"
  puts "Dupes? #{line.duplicated_letters}"
end

# ...

```

I think that one is pretty self documenting.

We’re finally ready for the primary search algorithm and here it comes:

```
# ...

=begin
 Suitability should be determined by
  * least number of letters takes out the length of the computer
  * no used letters
  * no duplicates
=end
def suitability(value, line)
  amount, used = 0, ""

  value.letters do |char|
    amount += if used.include?(char) || line.include?(char)
      -OCCURRENCE[char]
    else
      WORDS_LENGTH / OCCURRENCE[char]
    end
    used << char
  end

  amount
end

# ...

```

This method grades the passed value for inclusion in the line. Each  
letter is  
scored, creating an overall suitability rating for this value.

If the letter is already in the line or the letters we have seen so far  
from  
this value, the OCCURRENCE value for that letter is tacked onto the  
score as a  
penalty. This ensures the duplicates, especially of common letters, are  
costly  
so the code will try to avoid them.

If this is the first time the letter has been encountered, the score is  
the  
OCCURRENCE for that letter divided into the total letter count. This  
makes it  
valuable for the code to place letters like the “z” early on, since we  
don’t  
have any choices on that one and thus will also need to accept the  
letters it  
comes with.

Here’s the final piece of the puzzle, where we see suitability() put to  
work:

```
# ...

line = ""

until line.pangram?
  words.sort! do |x, y|
    suitability(y, line) <=> suitability(x, line)
  end

  line += "#{words.shift} "
end

stats line

```

The process is easy to follow here. First an empty line is created.  
Then we  
loop until the line has become a pangram. At each step, the utility  
list is  
sorted by suitability() and the best word added to the line. Just  
before exit,  
the program dumps the statistics as the result.

My thanks to all you pangram hunters. As always you taught me clever  
new tricks  
in optimization.

Tomorrow we’ll play with the king of pathfinding algorithms…

---

<div class="post-metadata">

**Author:** ![bbazzarrakk](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/b/47e85d/32.png) [@bbazzarrakk](https://www.ruby-forum.com/u/bbazzarrakk)\
**Post date:** [October 14, 2006, 12:36pm UTC](https://www.ruby-forum.com/t/posix-pangrams-97/71938/2 "2006-10-14T12:36:03Z")

</div>

Ruby Q. [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) writes:

> My thanks to all you pangram hunters. As always you taught me  
> clever new tricks in optimization.

For the record, this is how I found the shortest possible solution:

1. Look at the words. Recognize “zcat” needs to be in there, and  
remove it from the list.

2. Download dance.w[1](http://www-cs-faculty.stanford.edu/~knuth/programs/dance.w) from Donald E. Knuth, tangle and compile.

3. Write a small program prep.rb to preprocess the words for the  
expected input format:

4. Run it like this:

5. Look at the shortest solutions (1 and 2), and write them down as  
unix commands.

6. Since dance.w uses DLX to find optimal matches, these are the  
shortest solution (but not with least possible words, as I first  
assumed, too). We needed to remove “zcat”, because there is no way  
with it to have an optimal match (i.e. every letter only once).

Enjoy, “and know thy tools”,