Tom Caylor wrote:
> An even simpler case is the following:
>
>        inputs
>        1 2 3 4 5 6 7 8 9
> f1:    1 2 3 4 5 6 7 8 9 ...   (identity)
> f2:    2 1 3 4 5 6 7 8 9 ...   (switch 1 and 2)
> f3:    3 2 1 4 5 6 7 8 9 ...   (switch 1 and 3)
> f4:    4 2 3 1 5 6 7 8 9 ...   (switch 1 and 4)
> f5:    5 2 3 4 1 6 7 8 9 ...   (switch 1 and 5)
> ...
> fn:    fn(n) 2 3 4 5 6 7 8 9 ... fn(n-2) fn(n-1) 1 fn(n+1) fn(n+2) ...
> (switch 1 and fn(n))
> ...
> So let's do the diagonalization move.  Let g(n) = fn(n)+1.
> But from inspection of the table, we see that fn(n) = 1.  So g(n) = 1+1
> = 2. Putting this in a table:
>        inputs
>        1 2 3 4 5 6 7 8 9 ...
> g:     2 2 2 2 2 2 2 2 2 ...
>
> But from inspection, we see that g does not fit anywhere in the table
> of f1,f2,f3,... because it does not follow the rule of "switch 1 and
> something else" [and also it doesn't follow fn(i)=i for all i not equal
>
> to 1 or fn(n+1) ].
>
> So therefore g is not part of the list.  I.e. g is not a total
> computable function.
>
> However, as is plainer to see with this example, how can g(n)=2 (for
> all n) not be computable and total?  It is not one-to-one, but does
> that make it not computable or not total?
>
> Tom

In fact, my first example, where g(n) = n+1, is in fact one-to-one.

Tom


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