I wrote up some notes at williamumboh.com/comp30026-20...
I wrote up some notes at williamumboh.com/comp30026-20...
The usual diagonalization proof avoids the encoding issue as it constructs a machine D that on input <M> runs H on <M>, <M> and does the opposite.
The usual diagonalization proof avoids the encoding issue as it constructs a machine D that on input <M> runs H on <M>, <M> and does the opposite.
Does this work?
Does this work?
Consider the following machine L:
On input w:
Run H to see if L accepts w
Accept if H rejects
Reject if H accepts
Consider the following machine L:
On input w:
Run H to see if L accepts w
Accept if H rejects
Reject if H accepts