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Nov 30, 2017 at 11:50 vote accept Karen
Nov 15, 2018 at 9:32
Nov 18, 2017 at 22:10 comment added poncho To emphesize the point that SEJPM was making, you could compute $\operatorname{Enc}(\frac{a+b}2)$, however, the $/2$ part doesn't mean what you expect; if $a=0$ and $b=1$, then $\operatorname{Enc}(\frac{a+b}2) = \operatorname{Enc}((n+1)/2)$, which is probably not what you're looking for...
Nov 18, 2017 at 21:22 comment added SEJPM I hope this clears the main confusion, if anything is left unclear, ask!
Nov 18, 2017 at 21:22 history answered SEJPM CC BY-SA 3.0