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From: Wolfgang Lux <wlux_at_uni-muenster.de>

Date: Thu, 08 Nov 2007 11:55:09 +0100

Sergio Antoy wrote:

*> Not exactly. The semantics of non-determinism we intend in Curry
*

*> is such that computations are confluent (except for the order and
*

*> timing in which non-deterministic alternatives are produced), but
*

*> here I see the problem as an interaction between order of
*

*> evaluation, sharing and non-deterministic choices. The meaning of
*

*>
*

*> (f x ? g x where x free)
*

*> as
*

*> (f x where x free) ? (g x where x free)
*

*>
*

*> makes sense. Unfortunately, the language syntactically allows the
*

*> first form and I am afraid one cannot catch all the variations of
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*> this problem at compile time. Giving the intended semantics
*

*> statically might be quite difficult.
*

I'm sorry Sergio, but I do not understand you problem here. IIUC,

you are saying that (I'm using let notation here to make things

syntactically sound)

let x free in (f x ? g x)

does not make sense because you cannot evaluate f x and g x in

parallel for, say

f 0 = success

g 1 = success

since given either instantiation of x the other argument fails.

But then this is not a particular property of (?). For instance,

given

let x free in const (f x) (g x)

you get the same failure when you evaluate both (f x) and (g x).

This shows that you should not naively evaluate arguments that

are not demanded by the program. And note that (?) does not

demand either of its arguments, since they are not used in *every*

possible reduction. (BTW, note that two branches of (?) are equal

to const and (flip const), respectively.)

What you can do, however, is using the fact that each of (?)'s

arguments is used in *one* non-deterministic branch of the

computation. So may perform a parallel reduction of the arguments,

but this *must* be done in an or-parallel manner, i.e., using

independent instances for the yet free and unevaluated variables

of the program. Hence, there is no difference between

let x free in (f x ? g x)

and

(let x free in f x) ? (let x free in g x)

Regards

Wolfgang

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curry_at_lists.RWTH-Aachen.DE

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Received on Do Nov 08 2007 - 12:01:37 CET

Date: Thu, 08 Nov 2007 11:55:09 +0100

Sergio Antoy wrote:

I'm sorry Sergio, but I do not understand you problem here. IIUC,

you are saying that (I'm using let notation here to make things

syntactically sound)

let x free in (f x ? g x)

does not make sense because you cannot evaluate f x and g x in

parallel for, say

f 0 = success

g 1 = success

since given either instantiation of x the other argument fails.

But then this is not a particular property of (?). For instance,

given

let x free in const (f x) (g x)

you get the same failure when you evaluate both (f x) and (g x).

This shows that you should not naively evaluate arguments that

are not demanded by the program. And note that (?) does not

demand either of its arguments, since they are not used in *every*

possible reduction. (BTW, note that two branches of (?) are equal

to const and (flip const), respectively.)

What you can do, however, is using the fact that each of (?)'s

arguments is used in *one* non-deterministic branch of the

computation. So may perform a parallel reduction of the arguments,

but this *must* be done in an or-parallel manner, i.e., using

independent instances for the yet free and unevaluated variables

of the program. Hence, there is no difference between

let x free in (f x ? g x)

and

(let x free in f x) ? (let x free in g x)

Regards

Wolfgang

_______________________________________________

curry mailing list

curry_at_lists.RWTH-Aachen.DE

http://MailMan.RWTH-Aachen.DE/mailman/listinfo/curry

Received on Do Nov 08 2007 - 12:01:37 CET

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