Talk:Computable function

message: Should this be merged with Recursive function? --Saforrest 00:27, 27 January 2006 (UTC)

message: Isn't this definition of a computable function: 
A partial function f:\subseteq \mathbb{N} \to \mathbb{N} 
is called computable if the graph of f is a recursively enumerable set.
circular with the definition of a recursively enumerable set:
A countable set S is called recursively enumerable if there exists a partial computable function f : \mathbb{N} \to S such that S is the range of f?
--Michael Stone 00:24, 11 March 2006 (UTC)
thread_title:  Circularity of Definitions 

message: I have problems finding any info (on Wikipedia) about functions which are not computable. Now, it could be there exists an article about such a topic and that I should just look harder, but if it is so, the I think it should be linked from "See also" section of this article. If not, what a shame, lets write one :-) --Dijxtra 09:18, 7 April 2006 (UTC)
thread_title:  Functions which are not computable 

message: From the description of Church's Thesis.
:“There must be exact instructions (i.e. a program), finite in length, for the procedure.”
:“If the procedure is given a k-tuple x in the domain of f, then after a finite number of discrete steps the procedure must terminate and produce f(x).”
:“If the procedure is given a k-tuple x which is not in the domain of f, then the procedure might go on forever, never halting. Or it might get stuck at some point, but it must not pretend to produce a value for f at x.”
Later on, an example of a computable function ...
:The function f such that f(n) = 1 if there is a sequence of n consecutive fives in the decimal expansion of π, and f(n) = 0 otherwise, is computable.
If I am to take the rendition of Church's thesis on face value, this example seems wrong.  So, if for a certain n, call it m, there is no sequence of m 5s in the expansion of pi, then (1) the function is defined at m and (2) the procedure will continue ad infinitum without producing a result.  This contradicts the second statement, above.  Church's thesis, as described here, says nothing about the procedure going on forever when given an input which _is_ in the domain of f.  It only speaks, in the third part, of going on forever if given an input which is _not_ in the domain of f.  So, I think that the example should be:
The function f whose domain is {n in the Natural numbers | there is a sequence of n consecutive fives in the decimal expansion of pi} and which is defined to be 1 on that domain, is computable.
Could someone please either correct this function of correct Church's thesis?  —Preceding unsigned comment added by 72.74.113.222 (talk) 05:50, 7 September 2007 (UTC)
replies: text: I'm afraid it's clearly computable in classical logic; it's just that we don't know what the computation is.  Let N be the largest number of consecutive 5's in the decimal expansion of π, (with ∞ being a possible value for N).  Then f(n) is 1 if n ≤ N, and 0 otherwise.  — Arthur Rubin |  (talk) 08:19, 7 September 2007 (UTC)

text: I see this was already in the article, explained better than I did.  I don't know what the anon is talking about.  — Arthur Rubin |  (talk) 12:25, 7 September 2007 (UTC)

text: This is a standard example in courses on computability theory, and the anon has made the mental mistake that many people do when learning this.  Most people think of a computable function intensionally - it is linked to a particular algorithm to compute it. The anon is saying that since the obvious attempt at an algorithm to compute this function doesn't work, it is therefore not computable. But the actual definition of a computable function is extensional - a function is computable if there is any algorithm that computes it, even if that algorithm is "trying  to compute a different function".   In this situation, the algorithm that computes this function is just a short program that makes no mention of π. That's because, extensionally, this function is only barely related to π. — Carl (CBM · talk) 13:00, 7 September 2007 (UTC)

text: Someone asked for a citation. This example is an exercise in Fundamentals of Mathematical Logic, Hinman, 2005 - see 4.2.48 on p. 340. This example is actually extremely common in texts, as it is immediately accessible to students and the problem of whether there are arbitrarily long runs of a single digit in π hasn't been solved yet. — Carl (CBM · talk) 13:26, 7 September 2007 (UTC)
thread_title: An Error Either in an Example or in Church's thesis

message: I think the claim about the busy beaver function: "every finite sequence of Σ values, such as Σ(0), Σ(1), Σ(2), ..., Σ(n) for any given n, is (trivially) computable" is false. 
At the very least, it's contradictory to the claim on the busy beaver page that "there is a true-but-unprovable sentence of the form 'Σ(10↑↑10) = n' "
As, if the fist statement were true, one should "trivially" be able to compute the finite sequence Σ(0), Σ(1), Σ(2), ..., Σ(10↑↑10) thus "trivially" computing Σ(10↑↑10).
Also, as a general note, I think articles should avoid claims that something is "(trivially) computable" unless a simple proof is given or referenced. — Preceding unsigned comment added by 18.111.124.28 (talk • contribs) 18:25, 4 October 2011
:Every finite sequence of natural numbers is indeed trivially computable.  For example, the program can just have a list of the numbers, and if you ask it if n is the mth number in the list, it just looks it up and says yes or no. --Trovatore (talk) 19:42, 4 October 2011 (UTC)
replies: text: There is no problem defining the sequence, and ZFC is irrelevant here.  It's true that there is no particular number that ZFC proves equal to BB(7918) (assuming you've reported this correctly; I haven't checked that point, but if it's not 7918 it's something else).  But that has nothing at all to do with being able to define the sequence.  A deep-dive into this is kind of beyond the scope of this page; if you'd care to ask a question at WP:RD/Math or WP:RD/Computing I'd be happy to say more. --Trovatore (talk) 18:59, 21 August 2017 (UTC)
replies: text: This is all correct. For each n, in ZFC we can define "the sequence of the first n Busy Beaver numbers and we can prove that finite sequence is computable. It is true that ZFC cannot prove the exact values of the sequence but it can prove the sequence to be computable. By analogy, ZFC proves that the infinite sequence that is all 1s if the Continuum Hypothesis holds and all 0s otherwise is computable, even though ZFC cannot prove the exact values of the sequence. — Carl (CBM · talk) 23:05, 21 August 2017 (UTC)
thread_title:  Correctness of Busy Beaver Example 

message: Please note that most standard texts refer to these functions as UNcomputable and not INcomputable.  — Preceding unsigned comment added by 132.205.45.13 (talk) 17:08, 23 August 2012 (UTC)
replies: text: Good catch.  I hadn't noticed that it said incomputable in two spots. --Trovatore (talk) 21:47, 23 August 2012 (UTC)
thread_title:  Uncomputable and not incomputable