parsing - LR(1) 解析器中的左递归

Question

LR(1) 解析器可以解析这种类型的文法吗？

S -> SA  | A
A -> aSb | ab

我正在尝试编写一个实现这种类型解析器的 Java 程序，但我只能在没有左递归的语法上得到正确的结果。

Answer 1

LR(1) 解析器可以处理某些类型的左递归，但并非所有左递归语法都是 LR(1)。

让我们看看你的特定语法是否是 LR(1)。增强语法给出

S' → S

S → SA | 一个

A → aSb | 抗体

因此，我们的配置集是

 (1)
 S' -> .S    [$]     (Go to 2)
 S  -> .SA   [$a]    (Go to 2)
 S  -> .A    [$a]    (Go to 3)
 A  -> .aSb  [$a]    (Shift on a and go to 4)
 A  -> .ab   [$a]    (Shift on a and go to 4)

 (2)
 S' -> S.    [$]     (Accept on $)
 S  -> S.A   [$a]    (Go to 3)
 A  -> .aSb  [$a]    (Shift on a and go to 4)
 A  -> .ab   [$a]    (Shift on a and go to 4)

 (3)
 S  -> A.    [$a]    (reduce on $ or a)

 (4)
 A  -> a.Sb  [$a]    (Go to 6)
 A  -> a.b   [$a]    (Shift on b and go to 10)
 S  -> .SA   [ab]    (Go to 11)
 S  -> .A    [ab]    (Go to 12)
 A  -> .aSb  [ab]    (Shift on a and go to 8)
 A  -> .ab   [ab]    (Shift on a and go to 8)

 (5)
 A  -> ab.   [$a]    (Reduce on a or $)

 (6)
 A  -> aS.b  [$a]    (Shift on b and go to 7)
 S  -> S.A   [ab]    (Go to 13)
 A  -> .aSb  [ab]    (Shift on a and go to 8)
 A  -> .ab   [ab]    (Shift on a and go to 8)

 (7)
 A  -> aSb.  [$a]    (Reduce on a or $)

 (8)
 A  -> a.Sb  [ab]    (Go to 14)
 A  -> a.b   [ab]    (Shift on b and go to 16)
 S  -> .SA   [ab]    (Go to 11)
 S  -> .A    [ab]    (Go to 12)
 A  -> .aSb  [ab]    (Shift on a and go to 8)
 A  -> .ab   [ab]    (Shift on a and go to 8)

 (9)
 S  -> SA.   [$a]    (Reduce on a or $)

 (10)
 A  -> ab.   [$a]    (Reduce on a or b)

 (11)
 S  -> S.A   [ab]    (Go to 13)
 A  -> .aSb  [ab]    (Shift on a and go to 8)
 A  -> .ab   [ab]    (Shift on a and go to 8)

 (12)
 S  -> A.    [ab]    (Reduce on a or b)

 (13)
 S  -> SA.   [ab]    (Reduce on a or b)

 (14)
 A  -> aS.b  [ab]    (Shift on b and go to 15)
 S  -> S.A   [ab]    (Go to 13)
 A  -> .aSb  [ab]    (Shift on a and go to 8)
 A  -> .ab   [ab]    (Shift on a and go to 8)   

 (15)
 A  -> aSb.  [ab]    (Reduce on a or b)

 (16)
 A  -> ab.   [ab]    (Reduce on a or b)

该语法中没有移位/归约或归约/归约冲突，因此应该是 LR(1)（除非我在某处犯了错误！）

希望这可以帮助！

Answer 2

原来它也是一个 SLR 语法，所以 LR 是矫枉过正。LR 不需要 16 个状态，而 SLR 只需要 10 个状态。

(1)
 S' -> .S    S => (Go to 2)
 S  -> .SA   A => (Go to 3)
 S  -> .A    a => (Shift and go to 4)
 A  -> .aSb     
 A  -> .ab   

 (2)
 S' -> S.    $ => (Accept on $)
 S  -> S.A   A => (Go to 5)
 A  -> .aSb  a => (Shift and go to 6)
 A  -> .ab   

 (3)
 S  -> A.    [$b] => (reduce on $ or b)

 (4)
 A  -> a.Sb  S => (Go to 7)
 A  -> a.b   A => (Go to 9)
 S  -> .SA   a => (Shift and go to 6)
 S  -> .A    b => (Shift and go to 8)
 A  -> .aSb  
 A  -> .ab   

 (5)
 S  -> SA.   [$b] => (reduce on $ or b)

 (6)
 A  -> a.Sb  S => (Go to 7)
 A  -> a.b   A => (Go to 3)
 S  -> .SA   a => (Shift and go to 4)
 S  -> .A    b => (Shift and go to 8)
 A  -> .aSb  
 A  -> .ab   

 (7)
 A  -> aS.b  A => (Go to 5)
 S  -> S.A   a => (Shift and go to 6)
 A  -> .aSb  b => (Shift and go to 10)
 A  -> .ab   

 (8)
 A  -> ab.   [$b] => (reduce on $ or b)

 (9)
 S  -> A.    [$b] => (reduce on $ or b)

 (10)
 A  -> aSb.  [$b] => (reduce on $ or b)