Question

我正在研究Ruby中的字符串重建算法（动态编程示例中的经典，将空间少的文本转换为正常间隔文本）。 以下代码是纯红宝石，您可以复制粘贴并立即开始测试，它在80％的时间内工作并且往往会中断，字典变得越大。我用超过80k字的词典对它进行了测试，它的效果不太好，大约70％的时间都是如此。

如果有一种方法可以让它在字典中出现100％的效果，请告诉我。

以下是代码:(它间距很大，应该非常易读）

# Partially working string reconstruction algo in pure Ruby

# the dictionary
def dict(someWord)
  myArray = [" ", "best", "domain", "my", "successes", "image", "resizer", "high", "tech", "crime", "unit", "name", "edge", "times", "find", "a", "bargain", "free", "spirited", "style", "i", "command", "go", "direct", "to", "harness", "the", "force"]
  return !!(myArray.index(someWord))
end


# inspired by http://cseweb.ucsd.edu/classes/wi12/cse202-a/lecture6-final.pdf

## Please uncomment the one you wanna use
#
# (all the words used are present in the dictionary above)
#
# working sentences
  x = ' ' + "harnesstheforce"
# x = ' ' + "hightechcrimeunit"
#
# non working sentences
# x = ' ' + "findabargain"
# x = ' ' + "icommand"

puts "Trying to reconstruct #{x}"

# useful variables we're going to use in our algo
n = x.length
k = Array.new(n)
s = Array.new(n)
breakpoints = Hash.new
validBreakpoints = Hash.new

begin

  # let's fill k
  for i in 0..n-1
    k[i] = i
  end

  # the core algo starts here
  s[0] = true
  for k in 1..n-1

    s[k] = false

    for j in 1..k
      if s[j-1] && dict(x[j..k])
        s[k] = true

        # using a hash is just a trick to not have duplicates
        breakpoints.store(k, true)
      end
    end

  end

  # debug
  puts "breakpoints: #{breakpoints.inspect} for #{x}"

  # let's create a valid break point vector

  i=1
  while i <= n-1 do

    # we choose the longest valid word
    breakpoints.keys.sort.each do |k|

      if i >= k
        next
      end

      # debug: when the algo breaks, it does so here and goes into an infinite loop
      #puts "x[#{i}..#{k}]: #{x[i..k]}"
      if dict(x[i..k])
        validBreakpoints[i] = k
      end
    end

    if validBreakpoints[i]
      i = validBreakpoints[i] + 1
    end

  end

  # debug
  puts "validBreakpoints: #{validBreakpoints.inspect} for #{x}"

  # we insert the spaces at the places defined by the valid breakpoints
  x = x.strip
  i = 0
  validBreakpoints.each_key do |key|
    validBreakpoints[key] = validBreakpoints[key] + i
    i += 1
  end

  validBreakpoints.each_value do |value|
    x.insert(value, ' ')
  end

  puts "Debug: x: #{x}"


  # we capture ctrl-c
rescue SignalException
  abort

# end of rescue
end

Answer 1

请注意，对于包含单字符字符的字符串，您的算法会失败。这是一个一个错误的错误。在这些单词之后你忽略了断点，因此你最终得到了一个单词（"abargain"），而你的词典中没有这个单词。

更改

   if i >= k
     next
   end

到

   if i > k
     next
   end

或更多类似Ruby的

   next if i > k

另请注意，只要字符串包含的内容不是单词，您就会遇到无限循环：

if validBreakpoints[i]        # will be false 
  i = validBreakpoints[i] + 1 # i not incremented, so start over at the same position
end

您最好将此视为错误

return '<no parse>' unless validBreakpoints[i] # or throw if you are not in a function
i = validBreakpoints[i] + 1

"inotifier"的问题是您的算法存在缺陷。总是选择最长的单词并不好。在这种情况下，检测到的第一个“有效”断点位于"in"之后，为您留下非单词"otifier"。

Ruby：字符串重建算法，只能部分工作

1 个答案: