java – 无法理解CYK算法的伪代码
发布时间:2020-12-14 19:35:38 所属栏目:Java 来源:网络整理
导读:我正在阅读关于 CYK algorithm的内容,并且有一部分我无法理解的伪代码.整个伪代码是: let the input be a string S consisting of n characters: a1 ... an.let the grammar contain r nonterminal symbols R1 ... Rr.This grammar contains the subset Rs
我正在阅读关于
CYK algorithm的内容,并且有一部分我无法理解的伪代码.整个伪代码是:
let the input be a string S consisting of n characters: a1 ... an. let the grammar contain r nonterminal symbols R1 ... Rr. This grammar contains the subset Rs which is the set of start symbols. let P[n,n,r] be an array of booleans. Initialize all elements of P to false. for each i = 1 to n for each unit production Rj -> ai set P[i,1,j] = true for each i = 2 to n -- Length of span for each j = 1 to n-i+1 -- Start of span for each k = 1 to i-1 -- Partition of span for each production RA -> RB RC if P[j,k,B] and P[j+k,i-k,C] then set P[j,i,A] = true if any of P[1,x] is true (x is iterated over the set s,where s are all the indices for Rs) then S is member of language else S is not member of language 这些部分让我很困惑: for each production RA -> RB RC if P[j,A] = true 有人会给出一些关于这些伪代码的提示吗? 解决方法
伪代码
应该用以下方式解释.假设P [j,B]为真.这意味着从位置j开始的k个字符形成的字符串可以从非终结RB导出.如果P [j k,i – k,C]也为真,那么从位置j k开始的i-k个字符形成的字符串可以从非终结RC导出.因此,由于RA→RB RC是生产,因此从位置j开始的i个字符形成的字符串可以从RA导出. 我认为这可能有助于将伪代码解释为
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