拆分一个字符串的有效字使用动态规划字符串 [英] Split a string to a string of valid words using Dynamic Programming
问题描述
我需要找到一个动态规划算法来解决这个问题。我试过,但不能看着办吧。这里的问题是:
您给出的n个字符S [1 ... N],您认为是其所有的标点已经消失已损坏的文本文档的字符串(这样它看起来像itwasthebestoftimes ......)。希望使用字典,它可在一个布尔函数字典的形式来重建文件(*),使得对于任何字符串瓦特,字典(w)的具有值1,如果w是一个有效的字,并具有值0并非如此。
- 在给一个动态编程算法来决定是否字符串s [*]可改组为有效字序列。运行时间应该是最多为O(n ^ 2),假设每次调用词典需要单位时间。
- 在事件的字符串是有效的,让你的算法输出字的对应序列。
让您压实文档的长度为N。
让B(n)是一个布尔值:如果此文件可以被分成从位置N文件中启动的话
。B(N)为真(因为空字符串可分成0字)。 鉴于B(N),B(N - 1)... B(N - K)(1 N - - K)通过考虑开始字符n的所有单词 - 如果有1 - K,你可以构造b任何这样的话,瓦,B(N - K - 1 + LEN(W))集,然后设置B(N - K - 1)为true。如果没有这样的话,那么B组(N - K - 1)假
最后,你计算B(0),它会告诉你,如果整个文档可以分割成词。
在伪code:
高清try_to_split(DOC):
N = LEN(DOC)
B = [假] *(N + 1)
B〔N] = TRUE
对于i在范围(N - 1,-1,-1):
字起始位置i:
如果B〔1 + LEN(字)]:
B〔I] = TRUE
打破
回复B
有一些技巧,你可以做的就是'字起始位置i'高效,但你要求为O(n ^ 2)算法,因此你可以查找每一个在字典中开始我的字符串。
要生成的话,你可以修改上面的算法来存储好话,或只是产生这样的:
高清generate_words(DOC,B,IDX = 0):
长度= 1
而真正的:
断言B(IDX)
如果IDX == LEN(DOC):返回
字= DOC [IDX:IDX +长]
如果字在字典和B(IDX +长):
输出(字)
IDX + =长度
长度= 1
下面b是该算法的第一部分产生的布尔数组
I need to find a dynamic programming algorithm to solve this problem. I tried but couldn't figure it out. Here is the problem:
You are given a string of n characters s[1...n], which you believe to be a corrupted text document in which all punctuation has vanished (so that it looks something like "itwasthebestoftimes..."). You wish to reconstruct the document using a dictionary, which is available in the form of a Boolean function dict(*) such that, for any string w, dict(w) has value 1 if w is a valid word, and has value 0 otherwise.
- Give a dynamic programming algorithm that determines whether the string s[*] can be reconstituted as a sequence of valid words. The running time should be at most O(n^2), assuming that each call to dict takes unit time.
- In the event that the string is valid, make your algorithm output the corresponding sequence of words.
Let the length of your compacted document be N.
Let b(n) be a boolean: true if the document can be split into words starting from position n in the document.
b(N) is true (since the empty string can be split into 0 words). Given b(N), b(N - 1), ... b(N - k), you can construct b(N - k - 1) by considering all words that start at character N - k - 1. If there's any such word, w, with b(N - k - 1 + len(w)) set, then set b(N - k - 1) to true. If there's no such word, then set b(N - k - 1) to false.
Eventually, you compute b(0) which tells you if the entire document can be split into words.
In pseudo-code:
def try_to_split(doc):
N = len(doc)
b = [False] * (N + 1)
b[N] = True
for i in range(N - 1, -1, -1):
for word starting at position i:
if b[i + len(word)]:
b[i] = True
break
return b
There's some tricks you can do to get 'word starting at position i' efficient, but you're asked for an O(N^2) algorithm, so you can just look up every string starting at i in the dictionary.
To generate the words, you can either modify the above algorithm to store the good words, or just generate it like this:
def generate_words(doc, b, idx=0):
length = 1
while true:
assert b(idx)
if idx == len(doc): return
word = doc[idx: idx + length]
if word in dictionary and b(idx + length):
output(word)
idx += length
length = 1
Here b is the boolean array generated from the first part of the algorithm.
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