Ex4 Q1
http://tau-cm2014.wikidot.com/forum/t-857896/ex4-q1
Posts in the discussion thread "Ex4 Q1"Mon, 23 Sep 2019 12:12:33 +0000http://tau-cm2014.wikidot.com/forum/t-857896#post-2025903(no title)
http://tau-cm2014.wikidot.com/forum/t-857896/ex4-q1#post-2025903
Mon, 05 May 2014 14:51:42 +0000(not) (a)
Well, unfortunately the example appears to be trivial if we know that irrationalNumber representation is infinite in digits, so let's assume computer doesn't know that.
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http://tau-cm2014.wikidot.com/forum/t-857896#post-2025896Ex4 Q1
http://tau-cm2014.wikidot.com/forum/t-857896/ex4-q1#post-2025896
Mon, 05 May 2014 14:40:00 +0000(not) (a)
In order to show that some L not in RE, is it enough to show that Universal machine never accepts (because it never stops) any word that supposed to be in that language ? For example L={irrationalNumber | #digits in irrationallNumber representation = 999}. L is not in RE, because in order to accept some irrational number Universal machine has to check every digit, and # of digits in irrationalNumber representation is infinite. So, there's no decider and no acceptor for the language => language not in RE. Is it enough for a proof ?
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