In article <firstname.lastname@example.org>, Don Deluise <email@example.com> writes:
>This list enumerates all 2 bit binary sequences: > >00 >01 >10 >11 > >This one enumerates all 3 bit binary sequences: > >000 >001 >010 >011 >100 >101 >110 >111 > >Using diagonalization on the first to produce a sequence that is not in the >list fails. It produces '10' which *is* in the list. > >Using diagonalization on the second list also fails. It produces '111' >which *is* in the list. > >Clearly, increasing the length of the sequences does not produce lists in >which diagonalization will achieve its purpose, i.e. to produce binary >sequences of a given length which are not already in the list. > >So if we were to compile an enumeration of infinite length binary >sequences, how do we know that diagonalization produces a sequence not >already in the list?
If you look at what you've done, it only includes changing the first two or three items in the list. Since there are either four or eight items, it's not surprising that you haven't created an item not on the list.
If you examine the case where you change bit/digit n of item n for all items on the list, then you'll see that your resultant has to differ from all of them.
-- Michael F. Stemper #include <Standard_Disclaimer> 2 + 2 = 5, for sufficiently large values of 2