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Nov 2, 2017 at 16:53 comment added chi @chelsea the intuition is that at most $poly(m')$ of these tests might have the wrong result by hypothesis. This means that at least $poly(m')+1$ must have the correct result. Hence the majority of the results are correct, so we only need to count the result and follow the majority. The trick is: we have a broken machine that might be wrong on at most $N$ cases, but we know how to perform $2N+1$ queries that would have the same result on another "perfect" machine. Still, we can retrieve what the right result is by a majority argument.
Nov 2, 2017 at 16:34 vote accept chelsea
Nov 2, 2017 at 16:33 comment added chelsea @chi.. Thanks. Can you please explain why $2poly(m')+1$ tests suffice?
Nov 2, 2017 at 14:56 comment added chi @chelsea I'm only appending $\land p$ to the original formula, leaving it as it is. No CNF involved, I'm just adding one clause at the end. I just choose $p$ to be an unused variable, in many (polynomally bounded) ways.
Nov 2, 2017 at 14:26 comment added chelsea @chi can you shed some light on how to rewrite any $\phi$ in an equi-satisfiable way as you have shown? Are you trying to convert it into CNF-form with exponential number of clauses?
Nov 1, 2017 at 17:52 vote accept chelsea
Nov 2, 2017 at 14:26
Nov 1, 2017 at 14:01 history edited chi CC BY-SA 3.0
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Nov 1, 2017 at 13:39 history edited chi CC BY-SA 3.0
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Nov 1, 2017 at 13:34 comment added chi @Ariel Thank you! I indeed missed that point. However, it is enough to test for $2 poly(m')+1$ instances: on these only $poly(m')$ can be wrong, hence the majority decides the outcome. So, at least there's a fix.
Nov 1, 2017 at 13:32 history undeleted chi
Nov 1, 2017 at 13:32 history edited chi CC BY-SA 3.0
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Nov 1, 2017 at 13:27 history deleted chi via Vote
Nov 1, 2017 at 9:53 comment added Ariel What do you do when you see both positive and negative answers?
Nov 1, 2017 at 9:40 history answered chi CC BY-SA 3.0