Re: Count and Say (#138)

This week’s quiz was fairly easy. The hardest part was finding and
choosing a way to turn the numbers into their English equivalents.
Initially I used an array that extended to TWENTY, but that was
insufficient. I had solved Ruby Q. #25 and some of the other old ones
myself a while ago, but lost it due to a harddrive crash and couldn’t
find a copy. I then just picked someone else’s solution to Ruby Q. #25
that added a to_en method to Integer.

Here’s my solution, minus the Integer#to_en method. I commented out the
code to print out the elements of the cycle, as it turns out the cycle
is about 430 elements long!

def count_and_say(str)
(‘A’…‘Z’).map{|l| (str.count(l) > 0) ?
[str.count(l).to_en.upcase, l] : “”}.join(’ ‘).squeeze(’ ')
end

order = ARGV[0].chomp.to_i
prev_results = {}
element = “LOOK AND SAY”
for n in (0…order)
if prev_results[element]
puts “Cycle of length #{n-prev_results[element]} starting” +
" at element #{prev_results[element]}"
#puts “Cycle’s elements are:”
#puts (prev_results[element]…n).to_a.map{|n|
prev_results.invert[n]}
break
else
prev_results[element] = n
end
element = count_and_say(element)
end

----- Original Message ----
From: Ruby Q. [email protected]
To: ruby-talk ML [email protected]
Sent: Thursday, September 6, 2007 7:00:20 AM
Subject: [QUIZ] Count and Say (#138)

The three rules of Ruby Q.:

  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 Q. by submitting ideas as often as you can:

http://www.rubyquiz.com/

  1. Enjoy!

Suggestion: A [QUIZ] in the subject of emails about the problem helps
everyone
on Ruby T. follow the discussion. Please reply to the original quiz
message,
if you can.

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

by Martin DeMello

Conway’s “Look and Say” sequence
(Look-and-say sequence - Wikipedia) is a sequence of
numbers in
which each term “reads aloud” the digits of the previous term. For
instance, the
canonical L&S sequence starts off 1, 11, 21, 1211, 111221, …, because:

* 1 is read off as "one 1" or 11.
* 11 is read off as "two 1's" or 21.
* 21 is read off as "one 2, then one 1" or 1211.
* 1211 is read off as "one 1, then one 2, then two 1's" or 111221.
* 111221 is read off as "three 1, then two 2, then one 1" or 312211.

Over on rec.puzzles, Eric A. proposed a variant in which the letters of
a
sentence are grouped, then “counted aloud”, omitting the "s"s for the
plural
form. Thus, seeding the sequence with “LOOK AND SAY”, we get:

0. LOOK AND SAY
1. TWO A ONE D ONE K ONE L ONE N TWO O ONE S ONE Y
2. ONE A ONE D SIX E ONE K ONE L SEVEN N NINE O ONE S TWO T TWO W 

ONE Y
3. ONE A ONE D TEN E TWO I ONE K ONE L TEN N NINE O THREE S THREE T
ONE V
THREE W ONE X ONE Y

and so on. (Note the difference between this and the L&S sequence–the
letters
are counted rather than read in order). Eric wants to know when the
sequence
enters a cycle, and how long that cycle is. Well?

Here’s my solution. Coincidentally, it uses the same
Integer#to_english method that JEG2 posted from quiz 25 (not
included, as integer_to_english.rb).

–Usage:
$ ./138_count_and_say.rb LOOK AND SAY
Took 179 cycles to enter a cycle of length 429

#!/usr/bin/env ruby -rubygems

integer_to_english.rb → http://blade.nagaokaut.ac.jp/cgi-bin/

scat.rb/ruby/ruby-talk/135449
%w(facet/string/chars facet/enumerable/injecting facet/symbol/to_proc
integer_to_english).each(&method(:require))

class String
def letter_histogram
upcase.gsub(/[^A-Z]/,‘’).chars.injecting(Hash.new(0)){|h, l| h
[l] += 1}
end

def count_and_say
letter_histogram.sort_by{|l,n| l}.map{|(l, n)| “#
{n.to_english.upcase} #{l}”}.join(" ")
end
end

class Object
def detect_cycles
ary = [self]
loop do
val = yield(ary.last)
if ary.include? val
return [ary.index(val)+1, ary.length - ary.index(val)]
end
ary << val
end
end
end

if FILE == $PROGRAM_NAME
tail, cycle_length = ARGV.join.detect_cycles(&:count_and_say)
puts “Took #{tail} cycles to enter a cycle of length #{cycle_length}”
end

Here is my solution for both the number puzzle and the letter one. I
didn’t
realize that the numbers wouldn’t converge until I watched it take a
looong
time and followed the wikepedia link. In any case, the code handles
both
count_and_say and look_and_say. As others have said, the Fixnum.say
method
was more work than the rest of the puzzle.

Expect a block that takes a first argument of ‘count’ or ‘reorder’

depending on what

we need to do with the current string.

In hindsight, this could have taken two Proc objects instead:

count_proc
and reorder_proc
def find_cycle string
puts “Finding a cycle for #{string}”
sequence = {}
until sequence[string]
sequence[string] = sequence.length
string.gsub!(/ /,’’) # we ignore all spaces
#STDOUT.write "#{string.length}:#{string[0…50]} " #progress
feedback
string = yield( ‘reorder’, string )
previous_char, count = string[0…0], 1
counts = string.split(’’)[1…-1].inject([]) do |memo,obj|
if previous_char == obj
count += 1
else
memo << [yield(‘count’,count),previous_char]
previous_char, count = obj, 1
end
memo
end
counts << [yield(‘count’,count),string[-1…-1]]
string = counts.flatten.join(’ ')
end
“Cycle found at position #{sequence.length}, duplicating position
#{sequence[string]}: #{string}”
end

def count_and_say string = “1”
find_cycle string do |operation, value|
# in the numerical mode, each operation is a null operation
case operation
when ‘reorder’
value # no need to re-order the string
when ‘count’
value.to_s # just make sure it is a string
end
end
end

def look_and_say string = “LOOK AND SAY”
find_cycle string do |operation, value|
case operation
when ‘reorder’
value.split(’’).sort.join
when ‘count’
value.to_i.say
end
end
end

class Fixnum
NUMBERS = %w[zero one two three four five six seven eight nine ten
eleven twelve thirteen fourteen fifteen sixteen seventeen
eighteen
nineteen twenty]
TENS = %w[zero ten twenty thirty forty fifty sixty seventy eighty
ninety]
BIG_DIGITS = %w[zero ten hundred thousand]
def ones_digit; (self % 10); end
def tens_digit; ((self / 10) % 10); end
def say
result = []
if NUMBERS[self % 100]
result << NUMBERS[self % 100] if self == 0 or (self % 100) != 0
else
result << NUMBERS[self.ones_digit] if self.ones_digit > 0
result << TENS[self.tens_digit]
end
str = self.to_s
str[0…-3].reverse.split(’’).each_with_index do |char,idx|
result << BIG_DIGITS[idx+2]
result << NUMBERS[char.to_i] if char.to_i > 0
end
result.reverse.collect {|i| i.upcase }.join(’ ')
end
end

I found this to be a lot of fun - just the right amount of work for a
lazy
Saturday.
=> “Cycle found at position 607, duplicating position 178: ONE A ONE D
EIGHTEEN E FIVE F TWO G THREE H EIGHT I ONE K TWO L ELEVEN N TEN O THREE
R
TWO S TEN T TWO U FIVE V SIX W TWO X ONE Y”

(I hope this is actually correct!)
JB