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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Received on Thu Jun 15 2006 - 12:03:19 PDT