Re: [Scheme-reports] Cycle detection problems: #442, #338, "equal?"

[Ray Dillinger <bear@x>]

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On 08/30/2012 11:57 AM, Ray Dillinger wrote:

> The behavior I would prefer, however, is that outlined by Per
> Bothner.
> 
> To state it precisely in terms clear enough for the standard and
> also calculated to tell people exactly how to implement it:
> 
> "If one operand has a cycle and the other does not, then the 
> predicate returns #f.  If all have cycles, then iff the printed 
> representations of the structures are identical up to the point
> when all have entered their cycles at least twice, then the
> predicate returns #t."

Actually, as I consider the above, there is one case where I'm
not comfortable with this definition. Vectors can have elements
whose printed representation is not part of the "infinite
printed representation" if there is a cycle that starts before
those elements.  Consider two arrays, each of which has itself
as its own second element:

V1 ==>   #1=#(1 #1 6)
V2 ==>   #2=#(1 #2 9)

(equal? V1 v2) ==> ?

The infinite printed representation is identical, well past the
point where both have entered their cycle at least twice. Both
would show "#(1 #(1 #(1 #(1 #(1 #(1 ......"  But they can't be
considered equal? because

(equal? (vector-ref V1 2) (vector-ref V2 2)) => #f

It requires cycle-aware print to distinguish the written
representation.

So, given the above, I think it's necessary that equal? should
detect structural isomorphism (written representation under
cycle-aware printing is equal) rather than generation of
identical infinite sequences ("infinite" written representation
up to second entry of cycle is equal).

					Bear
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