at our lecture from Theory of Computation, we've been told that
- the set M is recursive, if its indicator function is total recursive function (or partial recursive function, it doesn't make a difference here, since indicator function is always EVERYWHERE defined).
Ok, I thought I understood this. But today, I found this quotation from some book on Wikipedia:
"The characteristic function of a k-place relation is the k-argument function that takes the value 1 for a k-tuple if the relation holds of the k-tuple, and the value 0 if it does not; and a relation is effectively decidable if its characteristic function is effectively computable, and is (primitive) recursive if its characteristic function is (primitive) recursive."
What I don't get here is the difference between "effectively decidable relation" and "recursive relation". In the first case, they say the characteristic function has to be effectively computable (ie. partial recursive, if I'm not mistaken). In the second case the characteristic function has to be recursive (it's what we call total recursive function, am I right?). But on the lecture we've been told there is no difference between these two cases...I'm little confused.
Could someone please enlighten this to me, please?
In article <b8922c23-8bee-469b-a1e3-7cf9aa934...@i29g2000prf.googlegroups.com>,
Twoflower <standa.ku...@gmail.com> wrote: >"The characteristic function of a k-place relation is the k-argument >function that takes the value 1 for a k-tuple if the relation holds of >the k-tuple, and the value 0 if it does not; and a relation is >effectively decidable if its characteristic function is effectively >computable, and is (primitive) recursive if its characteristic >function is (primitive) recursive."
>What I don't get here is the difference between "effectively decidable >relation" and "recursive relation". In the first case, they say the >characteristic function has to be effectively computable (ie. partial >recursive, if I'm not mistaken). In the second case the characteristic >function has to be recursive (it's what we call total recursive >function, am I right?). But on the lecture we've been told there is no >difference between these two cases...I'm little confused.
I think you're seeing a distinction here that was not intended. The statements that
a relation is effectively decidable if its c.f. is effectively computable
and
a relation is recursive if its c.f. is recursive
do not logically imply, and are not intended to suggest, that
there exist relations that are effectively decidable but not recursive, and there exist relations that are recursive but not effectively decidable.
All that they're doing is giving some examples of the principle that when one says that a relation is X, one means that its c.f. is X. That does not mean that two apparently different properties X and Y cannot coincide. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences
> I think you're seeing a distinction here that was not intended. The > statements that
> a relation is effectively decidable if its c.f. is effectively computable
> and
> a relation is recursive if its c.f. is recursive
> do not logically imply, and are not intended to suggest, that
> there exist relations that are effectively decidable but not recursive, > and there exist relations that are recursive but not effectively decidable.
> All that they're doing is giving some examples of the principle that when one > says that a relation is X, one means that its c.f. is X. That does not mean > that two apparently different properties X and Y cannot coincide.
Thank you for your reply Tim. So do you agree with my opinion that effectively decidable relation is the same thing as recursive relation?
> do not logically imply, and are not intended to suggest, that
> there exist relations that are effectively decidable but not recursive, > and there exist relations that are recursive but not effectively decidable.
That's not what I thought, I just see only a terminological difference between what they call effectively decidable relation and recursive relation, because I think that relation is effectively decidable iff it is recursive (ie. set's indicator function is effectively computable iff it is recursive). And so mentioning these two terms in that sentence seems little confusing to me.
In article <b71070ea-7025-468b-a4b2-3ed83e762...@y5g2000hsf.googlegroups.com>,
Twoflower <standa.ku...@gmail.com> wrote: >Thank you for your reply Tim. So do you agree with my opinion that >effectively decidable relation is the same thing as recursive relation?
Yes, I believe that is correct, because as you mentioned, indicator functions are always total.
>That's not what I thought, I just see only a terminological difference >between what they call effectively decidable relation and recursive >relation, because I think that relation is effectively decidable iff >it is recursive (ie. set's indicator function is effectively >computable iff it is recursive). And so mentioning these two terms in >that sentence seems little confusing to me.
Perhaps the parentheses around "primitive" weren't there originally and were added as an afterthought. -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences
> In article <b71070ea-7025-468b-a4b2-3ed83e762...@y5g2000hsf.googlegroups.com>,
> Twoflower <standa.ku...@gmail.com> wrote: > >Thank you for your reply Tim. So do you agree with my opinion that > >effectively decidable relation is the same thing as recursive relation?
> Yes, I believe that is correct, because as you mentioned, indicator functions > are always total.
> >That's not what I thought, I just see only a terminological difference > >between what they call effectively decidable relation and recursive > >relation, because I think that relation is effectively decidable iff > >it is recursive (ie. set's indicator function is effectively > >computable iff it is recursive). And so mentioning these two terms in > >that sentence seems little confusing to me.
> Perhaps the parentheses around "primitive" weren't there originally and were > added as an afterthought.
Yes, that would make sense to me :) Thank you again Tim.