Re: LR(1) paser generator based on Pager's algorithm

[Joel E. Denny]

On Mon, 2 Apr 2007, Joel E. Denny wrote:

> > In my impression, this can intuitively 
> > cover all such cases, and this problem is already solved. A strict proof 
> > may be needed. Then as Knuth said, be alerted, since "I have proved it, 
> > but did not try it"..
> 
> Here's a counter-example:
> 
>   S: 'a' A 'a'
>    | 'b' A 'b'
>    ;
>   A: 'a' 'a' 'a'
>    | 'a' 'a'
>    ;
> 
> Here's the additional itemset that must be split:
> 
>   A: 'a' . 'a' 'a' ['a' 'b']
>    | 'a' . 'a'     ['a' 'b']

But hmm... I easily foresee the upcoming S/R conflict by inspection.  
This looks promising, but I need to think more about whether that works in 
general.  I have to run now.