blocks|key|1180148|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1180149|由于我们不能想出两个不同的标签，我们必须找到Q*e乘积最大的标签T，并返回该标签T|blockquote|1180150|1180151|1180152|是啊，听起来不错。q是三元组(转换)概率，e是发射概率。如你所说|offset|length|style|CODE|1180153|📷|atomic|1180154|在每个阶段的不同路径之间是不变的，因此最大值仅依赖于其他两个路径。|1180155|每个标签序列应以位置-2和-1处的两个星号开头。所以第一个假设是正确的：|1180156|​|1180157|1180158|1180159|如果我们假设|1180160|1180161|位置k的最后两个标签是u和v的最大概率，根据我们刚才所说的开始星号，基本情况是|1180162|1180163|1180164|。|1180165|不过，在一般情况下，您有两个错误。发射概率是有条件的。同样在三元组中，|1180166|1180167|重复两次，给出的公式不正确：|1180168|1180169|![](https://latex.codecogs.com/gif.latex?tag(WORD--_1%2529&space;%255Ctimes&space;q(tag_k%7C&space;tag_k-2,&space;tag_k-1%2529&space;%255Ctimes&space;e(&space;word_k&space;%7Ctag_k%2529%2529)|1180170|1180171|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/121c10da60cd31e1a4e1f7cef87d8b0c.gif|imageAlt|1|https://ask.qcloudimg.com/http-save/yehe-900000/3c5459033f359c4767fb43db6df2257d.gif|2|https://ask.qcloudimg.com/http-save/yehe-900000/2edad1d0014d462e260fcf7d8be011ef.gif|3|https://ask.qcloudimg.com/http-save/yehe-900000/8fd2224586d5372bc9fe18b378b6ce6b.gif|4|https://latex.codecogs.com/gif.latex?%255Cinline&space;tag_k_-_1|5|LINK|MUTABLE|url|https://latex.codecogs.com/gif.latex?tag(WORD_k%2529&space;=&space;argmax(%255Cprod&space;(k-1,&space;tag_k_-_2,&space;tag_k_-_1%2529&space;%255Ctimes&space;q(tag_k%7C&space;tag_k-2,&space;tag_k-1%2529&space;%255Ctimes&space;e(&space;word_k&space;%7Ctag_k%2529%2529^0|0|0|0|0|9|1|L|1|0|0|1|0|0|0|A|2|D|2|0|0|0|1|1|0|0|0|0|1|2|0|2|1|B|1|D|1|0|0|0|1|3|0|0|0|0|1|4|0|0|0|4|4Q|5|0|0^^$0|@$1|2|3|-4|4|5|6|20|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|21|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|22|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|23|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|24|7|@$H|25|I|26|J|K]|$H|27|I|28|J|K]]|8|@]|9|$]]|$1|L|3|M|4|N|6|29|7|@]|8|@$H|2A|I|2B|1|2C]]|9|$]]|$1|O|3|P|4|5|6|2D|7|@]|8|@]|9|$]]|$1|Q|3|R|4|5|6|2E|7|@$H|2F|I|2G|J|K]|$H|2H|I|2I|J|K]]|8|@]|9|$]]|$1|S|3|T|4|5|6|2J|7|@]|8|@]|9|$]]|$1|U|3|M|4|N|6|2K|7|@]|8|@$H|2L|I|2M|1|2N]]|9|$]]|$1|V|3|T|4|5|6|2O|7|@]|8|@]|9|$]]|$1|W|3|X|4|5|6|2P|7|@]|8|@]|9|$]]|$1|Y|3|M|4|N|6|2Q|7|@]|8|@$H|2R|I|2S|1|2T]]|9|$]]|$1|Z|3|10|4|5|6|2U|7|@$H|2V|I|2W|J|K]|$H|2X|I|2Y|J|K]|$H|2Z|I|30|J|K]]|8|@]|9|$]]|$1|11|3|T|4|5|6|31|7|@]|8|@]|9|$]]|$1|12|3|M|4|N|6|32|7|@]|8|@$H|33|I|34|1|35]]|9|$]]|$1|13|3|14|4|5|6|36|7|@]|8|@]|9|$]]|$1|15|3|16|4|5|6|37|7|@]|8|@]|9|$]]|$1|17|3|M|4|N|6|38|7|@]|8|@$H|39|I|3A|1|3B]]|9|$]]|$1|18|3|19|4|5|6|3C|7|@]|8|@]|9|$]]|$1|1A|3|T|4|5|6|3D|7|@]|8|@]|9|$]]|$1|1B|3|1C|4|5|6|3E|7|@]|8|@$H|3F|I|3G|1|3H]]|9|$]]|$1|1D|3|T|4|5|6|3I|7|@]|8|@]|9|$]]|$1|1E|3|-4|4|5|6|3J|7|@]|8|@]|9|$]]]|1F|$1G|$4|1H|1I|1J|9|$1K|1L|1M|-4]]|1N|$4|1H|1I|1J|9|$1K|1O|1M|-4]]|1P|$4|1H|1I|1J|9|$1K|1Q|1M|-4]]|1R|$4|1H|1I|1J|9|$1K|1S|1M|-4]]|1T|$4|1H|1I|1J|9|$1K|1U|1M|-4]]|1V|$4|1W|1I|1X|9|$1Y|1Z]]]]

<blockquote>
 since we can't come up with 2 different tags, we basically have to
 find the tag T for which the q * e product is maximal, and return that
</blockquote>

Yeah, sounds about right. <code>q</code> is the trigram (transition) probability and <code>e</code> is named the emission probability. As you said <img src="https://latex.codecogs.com/gif.latex?%5Cinline&space;%5Cprod" title="\prod" /> is unchanged between different paths in each stage, so the max is only dependent on the other two.

Each tag sequence should start with two asterisks at positions <code>-2</code> and <code>-1</code>. So the first assumption is correct:

<img src="https://latex.codecogs.com/gif.latex?%5Cinline&space;tag_k_-1&space;=&space;tag_k_-2&space;=&space;%5Cleft&space;%5C%7B&space;*&space;%5Cright&space;%5C%7D" title="tag_k_-1 = tag_k_-2 = \left \{ * \right \}" />

If we assume <img src="https://latex.codecogs.com/gif.latex?%5Cinline&space;%5Cprod&space;(k,&space;u,&space;v)" title="\prod (k, u, v)" /> to be the maximum probability that the last two tags at position <code>k</code> are <code>u</code> and <code>v</code>, based on what we just said about the beginning asterisks, the base case would be 

<img src="https://latex.codecogs.com/gif.latex?%5Cinline&space;%5Cprod&space;(0,&space;*,&space;*)&space;=&space;1" title="\prod (0, *, *) = 1" /> .

You had two errors in the general case though. The emission probability is a conditional. Also in the trigram, <img src="https://latex.codecogs.com/gif.latex?%5Cinline&space;tag_k_-_1" title="tag_k_-_1" /> is repeated two times and the formula given is incorrect:

<img src="https://latex.codecogs.com/gif.latex?tag(WORD_k)&space;=&space;argmax(%5Cprod&space;(k-1,&space;tag_k_-_2,&space;tag_k_-_1)&space;%5Ctimes&space;q(tag_k|&space;tag_k-2,&space;tag_k-1)&space;%5Ctimes&space;e(&space;word_k&space;|tag_k))" title="tag(WORD_k) = argmax(\prod (k-1, tag_k_-_2, tag_k_-_1) \times q(tag_k| tag_k-2, tag_k-1) \times e( word_k |tag_k))" />

I'm currently trying to implement the viterbi algorithm in python, more specifically the version presented in an online course. 

As it stands, the algorithm is presented that way:
given a sentence with K tokens, we have to generate K tags .

We assume that tag K-1 = tag K-2 = '*', then for k going from 0 to K,
we set the tag for the token as follows :
tag(WORD_k) = argmax(p(k-1, tag_k-2, tag_k-1) * e( word_k, tag_k) * q(tag_k, tag_k-1, tag_k-1))

From my understanding this is straightforward because the p parameters are already calculated on each step (we go from 1 forward, and we already know p0), and max for the e and q params can be calculated by one iteration through the tags (since we can't come up with 2 different tags, we basically have to find the tag T for which the q * e product is maximal, and return that). This saves a lot of time, since we are almost at linear time in terms in big O notation, instead of exponential complexity, which we would get if we iterated through all possible word/tag combinations.

Am I getting the core of the algorithm correctly or am I missing something out? 

Thanks in advance

Trying to understand the VITERBI algorithm a bit better

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我目前正在尝试用python实现维特比算法，更具体地说就是在线课程中介绍的版本。现在，算法是这样给出的:给定一个带有K个标记的句子，我们必须生成K个标记。我们假设标签K-1 =标签K-2 = '*'，那么对于从0到K的K，我们设置令牌的标签如下: tag(WORD_k) = argmax(p(k-1，tag_k-2，tag_k-1) * e( word_k，tag_k) * q(tag_k，tag

问试图更好地理解VITERBI算法
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