#!/usr/bin/env ruby
# Copyright, 2013, by Samuel G. D. Williams. <http://www.codeotaku.com>
#
# 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.

module Build
	module Text
		module Merge
			Difference = Struct.new(:type, :value)
			
			def self.combine(old_text, new_text)
				lcs = lcs(old_text, new_text)
				changes = []
				
				n = 0; o = 0; l = 0
				while o < old_text.size and n < new_text.size and l < lcs.size
					if !similar(old_text[o], lcs[l])
						changes << Difference.new(:old, old_text[o])
						o += 1
					elsif !similar(new_text[n], lcs[l])
						changes << Difference.new(:new, new_text[n])
						n += 1
					else
						changes << Difference.new(:both, lcs[l])
						o += 1; n += 1; l += 1
					end
				end
				
				while o < old_text.size
					changes << Difference.new(:old, old_text[o])
					o += 1
				end
				
				while n < new_text.size
					changes << Difference.new(:old, new_text[n])
					n += 1
				end
				
				changes.map do |change|
					change.value
				end
			end

			# This code is based directly on the Text gem implementation
			# Returns a value representing the "cost" of transforming str1 into str2
			def self.levenshtein_distance(s, t)
				n = s.length
				m = t.length

				return m if n == 0
				return n if m == 0

				d = (0..m).to_a
				x = nil

				n.times do |i|
					e = i+1

					m.times do |j|
						cost = (s[i] == t[j]) ? 0 : 1
						x = [
							d[j+1] + 1, # insertion
							e + 1,      # deletion
							d[j] + cost # substitution
						].min
						d[j] = e
						e = x
					end

					d[m] = x
				end

				return x
			end

			# Calculate the similarity of two sequences, return true if they are with factor% similarity.
			def self.similar(s, t, factor = 0.15)
				return true if s == t
				
				distance = levenshtein_distance(s, t)
				average_length = (s.length + t.length) / 2.0

				proximity = (distance.to_f / average_length)
				
				return proximity <= factor
			end

			LCSNode = Struct.new(:value, :previous)
			
			# Find the Longest Common Subsequence in the given sequences x, y.
			def self.lcs(x, y)
				# Create the lcs matrix:
				m = Array.new(x.length + 1) do
					Array.new(y.length + 1) do
						LCSNode.new(nil, nil)
					end
				end

				# LCS(i, 0) and LCS(0, j) are always 0:
				for i in 0..x.length do m[i][0].value = 0 end
				for j in 0..y.length do m[0][j].value = 0 end

				# Main algorithm, solve row by row:
				for i in 1..x.length do
					for j in 1..y.length do
						if similar(x[i-1], y[j-1])
							# Value is based on maximizing the length of the matched strings:
							m[i][j].value = m[i-1][j-1].value + (x[i-1].chomp.length + y[j-1].chomp.length) / 2.0
							m[i][j].previous = [-1, -1]
						else
							if m[i-1][j].value >= m[i][j-1].value
								m[i][j].value = m[i-1][j].value
								m[i][j].previous = [-1, 0]
							else
								m[i][j].value = m[i][j-1].value
								m[i][j].previous = [0, -1]
							end
						end
					end
				end

				# Get the solution by following the path backwards from m[x.length][y.length]
				lcs = []
				
				i = x.length; j = y.length
				until m[i][j].previous == nil do
					if m[i][j].previous == [-1, -1]
						lcs << x[i-1]
					end
					
					i, j = i + m[i][j].previous[0], j + m[i][j].previous[1]
				end

				return lcs.reverse!
			end
		end
	end
end
