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Apr 23, 2014 at 18:47 vote accept Dan
Apr 2, 2014 at 7:59 history edited Dan CC BY-SA 3.0
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Apr 2, 2014 at 2:05 comment added Denis Hirschfeldt Constructing an infinite computable sequence of halting programs shows that the Halting Problem is not bi-immune, but that argument doesn't work for Chaitin's constant.
Apr 2, 2014 at 1:59 answer added Denis Hirschfeldt timeline score: 6
Apr 2, 2014 at 1:29 comment added 喻 良 Recursion theory people call what you called as bi-immuness. It contains all the weakly-random and weakly-generic reals.
Apr 1, 2014 at 23:47 answer added Joel David Hamkins timeline score: 4
Apr 1, 2014 at 22:41 comment added Gerry Myerson Knuth gets stuck into the question of ways to define randomness in Section 3.5 of Volume 2 (Seminumerical Algorithms) of The Art Of Computer Programming, though maybe you are already beyond anything Knuth covers there.
Apr 1, 2014 at 22:40 comment added Richard Stanley Isn't it the case that if we choose $f$ randomly (each value $f(n)$ independent, with probability 1/2 that $f(n)=0$), then the probability is 1 that $f$ is absolutely random?
Apr 1, 2014 at 21:30 comment added Per Alexandersson Why does not Chaitin's constant work? How do you construct g and h in this case? The n'th bit is given by the parity of the number of programs of size n, that halts, and I see no way to "peek" at a certain subsequence of this... Related: see also en.wikipedia.org/wiki/Chaitin%27s_constant#Super_Omega
Apr 1, 2014 at 21:21 history asked Dan CC BY-SA 3.0