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Tomita算法示例

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GLR分析过程‐2<br />

(0) S<br />

→ S<br />

(2) VP → V<br />

(4) VP → V NP NP<br />

(6) NP → Det N<br />

(8) NP → NP PP<br />

将栈顶0,3弹出,将NP, 1压入,NP用数字序号1代表。<br />

后一个1是Go to[0, NP]的值,即0遇到NP转入1号状态<br />

Action[1, V]=s6<br />

将V, 6压入栈中,V用序号2代表,6表示状态序号<br />

Action[6, Det]=s2<br />

(1) S →<br />

(3) VP → V<br />

(7) NP →<br />

(9) PP →<br />

Pron V Det N Prep Det N $<br />

'<br />

NPVP<br />

(5) VP → VP PP<br />

将Det, 2压入栈中,Det用序号3代表,2表示状态序号<br />

Action[2, N]=s10<br />

Pron<br />

Prep<br />

37<br />

NP<br />

NP

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