From: Mat Schaffer <schapht@...>
Date: 2006-07-10T03:43:53+09:00
Subject: [SOLUTION] [QUIZ] Panagrams (#86)

This isn't much to look at, but it appears to be > 48 hours since the  
posting.  So I thought I'd post it anyway.  I still can't generate  
the example pangrams given at http://www.cs.indiana.edu/~tanaka/GEB/ 
pangram.txt but it does generate this pangram in 1-5 minutes:

This terribly inefficient pangram contains five a's, two b's, three  
c's, two d's, thirty-one e's, six f's, four g's, ten h's, sixteen  
i's, one j, one k, three l's, two m's, twenty n's, thirteen o's, two  
p's, one q, twelve r's, twenty-eight s's, twenty-eight t's, three  
u's, three v's, nine w's, four x's, six y's and one z.

It's a basic randomized robinsoniziz..ing.  With a rather absurd  
amount of test code.
-Mat

-- pangram.rb
#!/usr/bin/ruby

# Ruby Quiz 86, Pangrams: Solution by Mat Schaffer <schapht@gmail.com>
# uses numeric_spell library available from http://tanjero.com/svn/ 
plugins/numeric_spell/

require 'numeric_spell'

class SelfDocumentingPangram
   LETTERS = (?a..?z).to_a

   def initialize(starter_string = "This test starter contains ")
     @start = starter_string
   end

   def to_s
     current = count(@start)
     actual = count(add_count(current))
     while current != actual
       LETTERS.each do |letter|
         current[letter] = rand_between(current[letter], actual[letter])
       end
       actual = count(add_count(current))
     end
     add_count(current)
   end

   def rand_between a,b
     range = (a - b).abs + 1
     rand(range) + [a,b].min
   end

   def add_count(counts)
     @start + counts_to_s(counts)
   end

   def count_to_s(char, count)
     if count != 1
       count.spell + " " + char.chr + "'s"
     else
       count.spell + " " + char.chr
     end
   end

   def counts_to_s(count)
     string_counts = []
     LETTERS.each do |letter|
       string_counts << count_to_s(letter, count[letter])
     end
     last = string_counts.pop
     string_counts.join(", ") + " and " + last + "."
   end

   def count(string)
     count = Hash.new(0)
     string.downcase.each_byte do |letter|
       if LETTERS.include? letter
         count[letter] += 1
       end
     end
     count
   end
end

if ARGV[0] =~ /test/i
   require 'test/unit'

   class TestSelfDocumentingPangram < Test::Unit::TestCase
     # checks that count will yield accurate counts for only letters,  
ignoring case
     def test_count
       # check basic case containing only a..z
       string = ('a'..'z').to_a.to_s
       count = SelfDocumentingPangram.new.count(string)
       assert_equal(26, count.length)
       count.each do |key, value|
         assert_equal(1, value)
       end

       # check case for a..z, A..Z, and some punctiation that we're  
likely to use
       string = (('a'..'z').to_a + ('A'..'Z').to_a + ['\'', ',', '.',  
'-']).to_s
       count = SelfDocumentingPangram.new.count(string)
       assert_equal(26, count.length)
       count.each do |key, value|
         assert_equal(2, value)
       end
     end

    def test_count_to_s
       assert_equal("one a", SelfDocumentingPangram.new.count_to_s(? 
a, 1))
       assert_equal("fifteen z's",  
SelfDocumentingPangram.new.count_to_s(?z, 15))
       assert_equal("forty-two d's",  
SelfDocumentingPangram.new.count_to_s(?d, 42))
     end

     def test_counts_to_s
       start = "The last of these contained "
       expected = "two a's, zero b's, one c, one d, four e's, one f,  
zero g's, two h's, one i, zero j's, zero k's, one l, zero m's, two  
n's, two o's, zero p's, zero q's, zero r's, two s's, four t's, zero  
u's, zero v's, zero w's, zero x's, zero y's and zero z's."
       pangram = SelfDocumentingPangram.new
       result = pangram.counts_to_s(pangram.count(start))
       assert_equal(expected, result)
     end

     def test_rand_between
       100.times do
         a = rand(100)
         b = [a, rand(100)].max
         c = SelfDocumentingPangram.new.rand_between(a,b)
         assert (a..b) === c, "#{c} is not between #{a} and #{b}"
       end
     end

     def test_add_count
       pangram = SelfDocumentingPangram.new("hi ")
       count = Hash.new(0)
       expected = "hi " + pangram.counts_to_s(Hash.new(0))
       assert_equal(expected, pangram.add_count(Hash.new(0)))
     end

     # runs the SelfDocumentingPangram class to verify that it can  
produce the pangrams found at
     # http://www.cs.indiana.edu/~tanaka/GEB/pangram.txt
     def test_to_s
       pangram1 = "This pangram tallies five a's, one b, one c, two  
d's, twenty-eight e's, eight f's, six g's, eight h's, thirteen i's,  
one j, one k, three l's, two m's, eighteen n's, fifteen o's, two p's,  
one q, seven r's, twenty-five s's, twenty-two t's, four u's, four  
v's, nine w's, two x's, four y's and one z."
       assert_equal(pangram1, SelfDocumentingPangram.new("This  
pangram tallies ").to_s)

       #pangram2 = "This computer-generated pangram contains six a's,  
one b, three c's, three d's, thirty-seven e's, six f's, three g's,  
nine h's, twelve i's, one j, one k, two l's, three m's, twenty-two  
n's, thirteen o's, three p's, one q, fourteen r's, twenty-nine s's,  
twenty-four t's, five u's, six v's, seven w's, four x's, five y's and  
one z."
       #assert_equal(pantram2, SelfDocumentingPangram.new("This  
computer-generated pangram contains ").to_s)
     end

     # This is mainly a sanity check to see that a pangram will  
evaluate to itself when counted and regenerated
     def test_approach
       prefix = "This pangram tallies "
       solution = "This pangram tallies five a's, one b, one c, two  
d's, twenty-eight e's, eight f's, six g's, eight h's, thirteen i's,  
one j, one k, three l's, two m's, eighteen n's, fifteen o's, two p's,  
one q, seven r's, twenty-five s's, twenty-two t's, four u's, four  
v's, nine w's, two x's, four y's and one z."
       pangram = SelfDocumentingPangram.new(prefix)
       assert_equal(solution, pangram.add_count(pangram.count 
(solution)))

       prefix = "This terribly inefficient pangram contains "
       solution = "This terribly inefficient pangram contains five  
a's, two b's, three c's, two d's, thirty-one e's, six f's, four g's,  
ten h's, sixteen i's, one j, one k, three l's, two m's, twenty n's,  
thirteen o's, two p's, one q, twelve r's, twenty-eight s's, twenty- 
eight t's, three u's, three v's, nine w's, four x's, six y's and one z."
       pangram = SelfDocumentingPangram.new(prefix)
       assert_equal(solution, pangram.add_count(pangram.count 
(solution)))
     end
   end
else
   puts SelfDocumentingPangram.new("This terribly inefficient pangram  
contains ").to_s
end



On Jul 7, 2006, at 11:01 AM, Ruby Quiz wrote:

> The three rules of Ruby Quiz:
>
> 1.  Please do not post any solutions or spoiler discussion for this  
> quiz until
> 48 hours have passed from the time on this message.
>
> 2.  Support Ruby Quiz by submitting ideas as often as you can:
>
> http://www.rubyquiz.com/
>
> 3.  Enjoy!
>
> Suggestion:  A [QUIZ] in the subject of emails about the problem  
> helps everyone
> on Ruby Talk follow the discussion.  Please reply to the original  
> quiz message,
> if you can.
>
> -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- 
> =-=-=-=-=-=-=
>
> by Darren Kirby
>
> One thing that interests me are word puzzles and language oddities.  
> One such
> example is the self-documenting panagram. If a panagram is a  
> sentence that uses
> every letter in the alphabet, then a self-documenting panagram is a  
> sentence
> that enumerates its own letter count. Simple enough, but what if we  
> state that
> the letter count must be spelled ie: 'twenty-seven' instead of  
> '27'.  Now we
> have a challenge.
>
> A while back I wrote a script in Python that finds these sentences.  
> Today I
> rewrote it in Ruby and it found me this sentence:
>
> 	Darren's ruby panagram program found this sentence which contains  
> exactly
> 	nine 'a's, two 'b's, five 'c's, four 'd's, thirty-five 'e's, nine  
> 'f's,
> 	three 'g's,  nine 'h's, sixteen 'i's, one 'j', one 'k', two 'l's,  
> three 'm's,
> 	twenty-seven 'n's, fourteen 'o's,  three 'p's, one 'q', fifteen 'r's,
> 	thirty-four 's's, twenty-two 't's, six 'u's, six 'v's, seven 'w's,  
> six 'x's,
> 	seven 'y's, and one 'z'.
>
> My script does have its problems, and I would love to see what kind  
> of code the
> Ruby experts could come up with to find self-documenting panagrams.
>
> There is a lot more info on self-documenting panagrams at this  
> address:
>
> 	http://www.cs.indiana.edu/~tanaka/GEB/pangram.txt
>