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Play on a classic: computers
Posted 14 March 2014 - 05:42 PM
Posted 15 March 2014 - 03:35 AM
i personally see a clear way to stratigize about all n cases.
Posted 15 March 2014 - 04:58 AM
i personally see a clear way to stratigize about all n cases.Spoiler for
Edited by Rainman, 15 March 2014 - 04:59 AM.
Posted 15 March 2014 - 10:13 PM
hmm interesting. ill have to think about it some more.
Posted 17 March 2014 - 01:26 AM
After watching "I Robot," I want these to be good and evil robots.
Vidi vici veni.
Posted 18 March 2014 - 10:54 AM
N-1 (except for N=3 or 4 where you can do better).
Let's put M=(N-1)/2.
If we ask all i to test all i+1, the most unlucky case occurs when all answers are GOOD. Consider that all bad computers are followed by all good ones. We only can be sure the last one is good - there is no bad computer after a good one.
If the answer is BAD:
- discard both (you never discard two good computers)
- substract 1 from i (if i=0, take anyone)
Now, each computer said the next one is GOOD, so we are in the case (B)(B)(B)..GGGG.
In the worst case:
a) we always discarded a good computer and a bad one
b) both were tested
c) there were M bad computers and therefore M discardments
d) b+c imply 2*M tests.
If a bad computer says BAD, the above still applies.
I just hope I did not forget something...
The results of the tests can be GOOD and BAD in each case. 3 and 4 are special cases, so you cannot use them for induction.
Edited by harey, 18 March 2014 - 11:03 AM.
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