Tomita算法示例

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GLR分析过程‐7 (0) S Action[10, $]=r6 → S (2) VP → V (4) VP → V NP NP (6) NP → Det N (8) NP → NP PP (1) S → (3) VP → V (7) NP → (9) PP → Pron V Det N Prep Det N $ ' NPVP (5) VP → VP PP 将N, 10压入栈中,N用序号 9代表,10表示状态序号 Pron Prep 42 NP NP
GLR分析过程‐8 Action[13, $]=r9 (0) S → S (2) VP → V (4) VP → V NP NP (6) NP → Det N (8) NP → NP PP (1) S → (3) VP → V (7) NP → (9) PP → Pron V Det N Prep Det N $ ' NPVP (5) VP → VP PP 将栈顶8,2,9,10弹出,将NP, 13压入,NP 用数字序号10代表。13是Go to[8, NP]的值,即8号状态遇到NP转入13号状态 Pron Prep 43 NP NP
Page 1 and 2: GLR 分析算法詹卫东 http://c
Page 3 and 4: LR分析算法的基本思想和基
Page 5 and 6: LR分析算法之 [ 状态构造
Page 7 and 8: 状态构造算法示例‐2 4 {0
Page 9 and 10: 状态构造算法示例‐4 0:
Page 11 and 12: 状态1中有,对应第2条规则,
Page 13 and 14: 分析栈: LR分析算法过程描
Page 15 and 16: LR分析算法过程示例移进
Page 17 and 18: 分析树形成过程示意图‐2
Page 31 and 32: (1)不允许回溯 LR分析算法
Page 33 and 34: (0) (6) (8) ' NP NP GLR分析示例
Page 35 and 36: s: shift r: reduce GLR分析表 35
Page 37 and 38: GLR分析过程‐2 (0) S → S (2)
Page 39 and 40: GLR分析过程‐4 (0) S → S (2)
Page 41: GLR分析过程‐6 Action[2, N]=s1
Page 45 and 46: GLR分析过程‐10 (0) S → S (2
Page 47 and 48: GLR分析过程‐12 满足局部
Page 49 and 50: GLR分析过程‐14 (0) S → S (2

Tomita算法示例

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