Ruby 从较大的句子中找出重叠最少的两个短语

我有：

我想从数组中找出两个在6-10个单词范围内且在单词方面重叠最少的短语

比如：

phrase = "will have to buy online pass from EA to play online but its in perfect condition" 

phrases = ["its",
"perfect condition",
"but its",
"in perfect condition",
"from EA",
"buy online pass from EA",
"to play online but its in perfect condition",
"online",
"online pass",
"play online but its in perfect condition",
"online but its",
"EA",
"will have to buy online pass from EA to play online but its in perfect condition",
"have to buy online pass from EA to play online but its in perfect condition",
"u",
"pass",
"to buy online pass from EA"]

太好了。。最好的方法是什么？这个怎么样

result = ["to buy online pass from EA", "play online but its in perfect condition"]

---

这是相似的。。。你是说这个？有两个不重叠的吗？在任何顺序中？任何顺序在6之内是的，事实上如果保持顺序就好了。这找到了一个很好的答案，但遗漏了短语在头部和尾部重叠但彼此不完全包含在一起的例外情况。

result = Array.new
phrases.each do |p|
  result.push(p) if(phrase.include?(p) && (6..10).include?(p.split.size))
end 
#remove entries that are substr of others   
result.each do |r|
  result.delete(r) if (t = result.clone ; t.delete(r) ; t.any? {|v| v.include?(r)})
end   
print result.inspect 
#["to play online but its in perfect condition", "to buy online pass from EA"]

split_phrases = phrases.map {|phrase| phrase.split }

# find number of words of overlap between two word vectors
def overlap(p1,p2)
  s1 = p1.size
  s2 = p2.size

  # make p1 the longer phrase
  if s2 > s1
    s1,s2 = s2,s1
    p1,p2 = p2,p1
  end

  # check if p2 is entirely contained in p1
  return s2 if p1.each_cons(s2).any? {|p| p == p2}

  longest_prefix = (s2-1).downto(0).find { |len| p1.first(len) == p2.last(len) }
  longest_suffix = (s2-1).downto(0).find { |len| p2.first(len) == p1.last(len) }

  [longest_prefix, longest_suffix].max
end

def best_two_phrases_with_minimal_overlap(wphrases, minlen=6, maxlen=10)
  # reject too small or large phrases, evaluate every combination, order by word overlap
  scored_pairs = wphrases.
    select {|p| (minlen..maxlen).include? p.size}.
    combination(2).
    map { |pair| [ overlap(*pair), pair ] }.
    sort_by { |tuple| tuple.first }

  # consider all pairs with least word overlap
  least_overlap = scored_pairs.first.first
  least_overlap_pairs = scored_pairs.
    take_while {|tuple| tuple.first == least_overlap }.
    map {|tuple| tuple.last }

  # return longest pair with minimal overlap
  least_overlap_pairs.sort_by {|pair| pair.first.size + pair.last.size }.last
end

puts best_two_phrases_with_minimal_overlap(split_phrases).map{|p| p.join ' '}

# to play online but its in perfect condition
# to buy online pass from EA