[SOLUTION] #83 Short but Unique

Here is my quick solution to quiz #83.
First I find all possible abbreviations, then try to find unique ones
for each string. If this isn’t possible, then it tries for at least a
different abbreviation for each string. Finally, it falls back to the
heighest weighted abbreviation.

Weighting is currently a bit simplistic, and could probably be improved.
Weight is increased for
A) including the first character
B) cutting out ‘word boundary’ characters (’ ', ‘_’, ‘-’, ‘.’)


Brian Mattern

#Code below:

require ‘pp’

module Compressor

gets all compressed versions of str

def self.compress(str, len, ellipses = ‘…’)
len = len.to_i
return case
when str.size <= len
[str]
else
ret = []
weight = {}
cutout = str.size - len + ellipses.size - 1
(0…(len-ellipses.size)).to_a.each do |i|
a = str.dup
a[i…(i + cutout)] = ellipses
w = 0
w += 1 if i > 0
[’ ', ‘_’, ‘-’, ‘.’].each do |c|
w += 1 if str[i…(i+cutout)].include?©
end
ret << a
weight[a] = w

         end

         ret.sort{|s1, s2| weight[s2] <=> weight[s1] }
       end

end

def self.compress_array(arr, len, ellipses = ‘…’)
candidates = {}
arr.each { |s| candidates[s] = self.compress(s, len, ellipses) }

results = {}
candidates.each { |k, v|
  # first try to find a completely unique abbreviation
  results[k] = v.find { |s|
    candidates.all? { |k2, v2| k2 == k or !v2.include?(s)}
  }

  # if none was found, just pick one that's different from the other 

chosen ones
results[k] = v.find { |s| !results.values.include?(s) } if
results[k].nil?

  # if we still don't have one, pick the first one (heighest 

weighted)
results[k] = v.first if results[k].nil?
}
arr.collect{ |s| results[s] }
end
end

class Array
def compress(len = 10)
Compressor.compress_array(self, len)
end
end

pp ARGV[1…-1].compress(ARGV[0])

One of the other solutions preferred cutting out the middle of words
(although it also found ambiguous abbreviations when non-ambiguous ones
exist).

To do something similar with mine, replace this line:

w += 1 if i > 0

with:

#weight by distance of cutout from middle of word (middle being
highest)
ideal = (str.size - cutout) / 2
w += i if i <= ideal
w += (ideal * 2) - i if i > ideal


brian

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