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
解决方法
伪代码
For each production RA → RB RC:
if P[j,A] = true
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
For each production RA → RB RC:
if P[j,B] == true and P[j+k,C] == true,then set P[j,A] = true
希望这可以帮助！
