module Lexorank

Inspired by github.com/DevStarSJ/LexoRank/blob/master/lexo_rank.rb licensed under MIT License

Permission is hereby granted, free of charge, to any person obtaining a copy of this software and associated documentation files (the “Software”), to deal in the Software without restriction, including without limitation the rights to use, copy, modify, merge, publish, distribute, sublicense, and/or sell copies of the Software, and to permit persons to whom the Software is furnished to do so, subject to the following conditions:

The above copyright notice and this permission notice shall be included in all copies or substantial portions of the Software.

THE SOFTWARE IS PROVIDED “AS IS”, WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.

Constants

MAX_CHAR
MIN_CHAR
VERSION

Public Instance Methods

get_char(str, i, default_char) click to toggle source

# File lib/lexorank.rb, line 78
def get_char(str, i, default_char)
  i >= str.length ? default_char : str[i]
end

mid(prev, after) click to toggle source

# File lib/lexorank.rb, line 73
def mid(prev, after)
  middle_ascii = ((prev.ord + after.ord) / 2).round
  middle_ascii.chr
end

value_between(_before_, _after_) click to toggle source

# File lib/lexorank.rb, line 31
def value_between(_before_, _after_)
  before = _before_ || MIN_CHAR
  after = _after_ || MAX_CHAR

  rank = ''

  (before.length + after.length).times do |i|
    prev_char = get_char(before, i, MIN_CHAR)
    after_char = get_char(after, i, MAX_CHAR)

    if prev_char == after_char
      rank += prev_char
      next
    end

    mid_char = mid(prev_char, after_char)
    if mid_char == prev_char || mid_char == after_char
      rank += prev_char
      next
    end

    rank += mid_char
    break
  end

  # Problem: If we try to get a rank before the character '0' or after 'z' the algorithm would return the same char
  # This first of all breaks a possible unique constraint and of course makes no sense when ordering the items.
  #
  # Thoughts: I think this issue will never happen with the Lexorank::Rankable module
  # Why? Let's look at '0' as a rank:
  # Because the algorithm always chooses the char in between two other chars, '0' can only happen when before is nil and after is '1'
  # In this case the algorithm will return '0U' though. This means there will never be an item with rank '0' which is why this condition
  # should never equal to true.
  #
  # Please report if you have another opinion about that or if you reached the exception! (of course you can force it by using `value_between(nil, '0')`)
  if rank >= after
    raise InvalidRankError, "This rank should not be achievable using the Lexorank::Rankable module! Please report to https://github.com/richardboehme/lexorank/issues! " +
      "The supplied ranks were #{_before_.inspect} and #{_after_.inspect}. Please include those in the issue description."
  end
  rank
end