Question

What does "eliminate left recursion" mean in parser grammar?

Asked by: arantius

I've been working on a project that would involve parsing arithmetic-like expressions.

I found this page:

http://compilers.iecc.com/comparch/article/06-01-092

And I mostly understand it, but I could really use some more info about what the phrase "eliminate left-recursion" means there, and some detail on how that gets from the first to the second grammar listed.  Also, what are the epsilons, where do they come from, why?

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Asked On
2007-06-07 at 07:50:43ID22619240
Tags

left

,

recursion

,

eliminate

,

grammar

Topics

Parsers

,

Miscellaneous Programming

Participating Experts
1
Points
125
Comments
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Answers

 

by: Infinity08Posted on 2007-06-07 at 08:12:21ID: 19234274

This grammar :

      E -> E+T | E-T | T
      T -> T*F | T/F | F
      F -> +F | -F | number | (E)

uses left recursion. Let's look at the second rule specifically :

      T -> T*F | T/F | F

This allows strings like :

      F
      F*F
      F/F
      F*F*F
      F/F*F
      etc.

or any other combinations of multiplying and dividing factors (F).

The T is on the left side of the rule results :

      T*F
      T/F

which is why it's called left recursion ... ie. new content is added on the left when constructing the string :

      T
      T*F                               <--- because of first option T -> T*F
      T/F*F                            <--- because of second option T -> T/F
      F/F*F                            <--- because of third option T -> F

This poses a problem for recursive descent parsers, because they can't parse something like that. An LALR parser by contrast prefers left recursion.

So, if we make use of a recursive descent parser, we'll have to eliminate the left recursion, and form a new grammar that can be used by a recursive descent parser.

One option is to do this (from the page you linked to) :

      T -> FT'
      T' -> *FT' | /FT' | \epsilon

ie. now right recursion is used (the T is on the right side of the options), and since \epsilon stands for the empty string, these rules describe the same set of possible strings. For example, we can form :

      T                                  
      FT'                                <--- because of the first rule T -> FT'
      F/FT'                             <--- because of second option T' -> /FT'
      F/F*FT'                         <--- because of first option T' -> *FT'
      F/F*F                            <--- because of third option T' -> \epsilon

 

by: Infinity08Posted on 2007-06-07 at 08:13:27ID: 19234287

You can also check out this wiki page for more info on left recursion :

http://en.wikipedia.org/wiki/Left_recursion

 

by: arantiusPosted on 2007-06-07 at 08:25:01ID: 19234387

You mentioned the second rule uses left recursion.  Does the first, also?  Because it says "E+T" in it?  If so, would it still use left recursion if it was instead:

E -> T+E | T-E | T

 

by: Infinity08Posted on 2007-06-07 at 08:58:54ID: 19234695

>> You mentioned the second rule uses left recursion.  Does the first, also?

Yes ... I just picked one of them as example - the other can be dealt with in exactly the same way.


>> If so, would it still use left recursion if it was instead:
>>
>> E -> T+E | T-E | T

No, that's right recursive.


Note that there is an extra problem with changing rules like this : the associativity can change.

Take for example the string "T + T + T".

parsing it with the original rule :

      E -> E+T | E-T | T

we get :

      T + T + T
      E + T
      (E + T) + T
      (T + T) + T

while with your rule :

      E -> T+E | T-E | T

we get :

      T + T + T
      T + E
      T + (T + E)
      T + (T + T)

and with the "fixed" rule from the link you posted :

      E -> TE'
      E' -> +TE' | -TE' | \epsilon

we get :

      T + T + T
      TE'
      T + TE'
      T + (T + TE')
      T + (T + T)

Note that the associativity changed from left ((T + T) + T) to right (T + (T + T)). In the case of addition, this is not a problem, but it can be a problem with other operations, like subtraction, ie. : ((T - T) - T)  is not the same as (T - (T - T)).

20120131-EE-VQP-002

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