# Re: Shirt Reader (#140)

**URL:** <https://www.ruby-forum.com/t/re-shirt-reader-140/115643>\
**Category:** Ruby\
**Created:** [September 27, 2007, 5:14am UTC](https://www.ruby-forum.com/t/re-shirt-reader-140/115643 "2007-09-27T05:14:16Z")\
**Posts on this page:** 1\
**Page:** 1

<div class="post-metadata">

**Author:** ![jameskelly](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/j/3ab097/32.png) [@jameskelly](https://www.ruby-forum.com/u/jameskelly)\
**Post date:** [September 27, 2007, 5:14am UTC](https://www.ruby-forum.com/t/re-shirt-reader-140/115643/1 "2007-09-27T05:14:16Z")

</div>

My solution’s algorithm is fairly simple: Take the words, take their  
pronunciations in IPA or a suitable equivalent, and then find the word  
with the pronunciation that requires the least amount of changes to turn  
it into the concatenated pronunciations of the terms (I am indebted to  
Steve D for the knowledge that this is called the “Levenshtein  
Distance”).

> ruby Rebus…rb (B+YOU+TEA+FULL)  
> BEAUTIFUL

I had plans for performing this on every combination of the synonyms of  
the given terms, and acquired both a thesaurus and pronunciation  
dictionary from [http://www.dcs.shef.ac.uk/research/ilash/Moby/](http://www.dcs.shef.ac.uk/research/ilash/Moby/), but,  
unfortunately, they both had quite a few more entries than they should  
have. Apparently, “skim milk” is a synonym for “cheese.” Additionally,  
according to the Moby Dictionary, “contretemps” is pronounced the way  
it’s spelled, and “ember” is not a word, but “embers” is. To use the  
thesaurus would be to make a program that already takes a minute or so  
to run take a day instead. For those reasons, as well as the fact that  
the dictionary contains a lot that it shouldn’t have (what’s a  
“Fayme?”), and due to some unintuitive pronunciation differences, my  
program doesn’t use a thesaurus, and gets a lot of things wrong.

> ruby Rebus.rb (E+SCENT+SHELLS)  
> CENTRAL’S

(Yes, the unweighted Levensgtein distance really is lower than to  
“ESSENTIALS” when using Moby Pronunciation Dictionary.)

> ruby Rebus.rb (FAN+TASK+TICK)  
> FANTASKTIK

Note on my program’s input: only expressions within parentheses are  
evaluated (allowing for multiple evaluation and leaving known terms  
unchanged in a multi-rebus phrase that must be evaluated), everything  
must be cnverted to all caps, and operators must be shown. The latter is  
because I’ve implemented a crude interpreation of subtraction in  
rebuses.

> ruby Rebus.rb (LOVE-L)  
> OF

As can be seen, my program’s output is regretfully only mediocre.  
Changes that would greatly improve output include some mechanism to  
differentiate between using a letter’s name and phoneme (i.e.: “aitch”  
versus the h-sound) (at present the dictionary just uses the name),  
using a more restrictive dictionary and thesaurus, and modifying the  
levenshtein\_distance method to make changes between similar vowels and  
consonants have a lower weight than changes between dissimilar vowels  
and consonants.

But alas, the period for this quiz is ending.

Anyway, here’s my code:

#$thesaurus = File.open(“mobythes.aur”) do |f|

# h = Hash.new

# f.readlines.each do |line|

# words=line.chomp.split(‘,’).map{|s|s.upcase}

# h[words[0]] = words[1…-1]

# end

# h

#end

#This only works if the beginning comments are manually removed  
$pronunciations = File.open(“cmudict0.3”) do |f|  
NUMBER\_OF\_SYMBOLS = 39

h = Hash.new

lines = f.readlines  
lines[0…NUMBER\_OF\_SYMBOLS].each do |line|  
h[lines[0…(line =~ /[A-Z]/)]] = line.split[1…-1]  
end  
lines[NUMBER\_OF\_SYMBOLS…-1].each do |line|  
words = line.split(/\s+/)  
h[words.first] = words[1…-1]  
end  
h  
end

$words = $pronunciations.keys

#See [Levenshtein distance - Wikipedia](http://en.wikipedia.org/wiki/Levenshtein_distance#The_algorithm)  
def levenshtein\_distance(a, b)  
prev\_row = (0…b…length).to\_a  
cur\_row = [0] \* b.length  
1.upto(a.length) do |i|  
cur\_row[0] = i  
1.upto(b.length) do |j|  
cost = (a[i-1] == b[j-1]) ? 0 : 1  
cur\_row[j] = [prev\_row[j]+1,  
cur\_row[j-1]+1,  
prev\_row[j-1]+cost].min  
end  
prev\_row = cur\_row  
cur\_row = cur\_row.dup  
end  
prev\_row.last  
end

#def all\_synonym\_combinations(word\_arr)

# if word\_arr.size == 1

# ($thesaurus[word\_arr.first] || [word\_arr.first]).map{|w|[w]}

# else

# next\_combs = all\_synonym\_combinations(word\_arr[1…-1])

# next\_combs\_length = next\_combs.flatten.length

# synonyms = ($thesaurus[word\_arr[0]] || []) \<\< word\_arr[0]

# synonyms = synonyms.inject([]){|a,w| a+=[[w]]\*next\_combs\_length}

# synonyms.flatten.zip(next\_combs\*synonyms.length

# ).map{|a|a.flatten}

# end

#end

rebus = ARGV[0].chomp

expressions = rebus.scan(/(?:().+?(?:))/)

expressions.each do |expression|

ops = expression.scan(/[±]/)  
terms = expression.gsub(/[()]/,“”).scan(/[^±]+/)  
#all\_synonym\_combinations(terms).each do |terms|  
pronunciation = $pronunciations[terms[0]]  
terms[1…-1].each\_with\_index do |term,idx|  
if ops[idx] == “+”  
pronunciation += $pronunciations[term]  
else  
#Set difference is a very crude interpretation of subtraction in  
rebuses  
#(especially considering, in cmudict, a letter’s pronunciation  
is that  
# of its name, not its main phoneme)  
pronunciation -= $pronunciations[term]  
end  
end  
reps = $words.sort\_by{|word|  
levenshtein\_distance($pronunciations[word],  
pronunciation)}  
rep = terms.index(reps[0]) ? reps[1] : reps[0]  
rebus[expression] = rep  
#end  
end

puts rebus.gsub(“+”," ")

```
  ____________________________________________________________________________________

```

Luggage? GPS? Comic books?  
Check out fitting gifts for grads at Yahoo! Search  
[http://search.yahoo.com/search?fr=oni\_on\_mail&p=graduation+gifts&cs=bz](http://search.yahoo.com/search?fr=oni_on_mail&p=graduation+gifts&cs=bz)