## algorithms – Search in array with blackbox

Given array $$A(1..n)$$, and black box, and a number $$x$$. we want to
find out a $$i$$ such that $$A(i)=x$$. Our black box have three inputs,
array $$A$$ and at most $$k$$ numbers from set $$S={1,2,…,n}$$, and $$x$$.
Black box return true if there is $$iin k$$ that $$A(i)=x$$. How we can
show that the number of times we use black box is $$O(frac{n}{k}+log k)$$?

I think as follow: divide set $$S$$ to $$k$$ sets with size $$frac{n}{k}$$ and then we use block at most $$k$$ times. Now if after $$k^{th}$$ the black box return true the on array $$A$$ we search an interval with size $$k$$. So the number of calls are $$O(frac{n}{k}+k)$$. But the term $$k$$ contradict with $$log k$$. Any help be appreciated.

## pseudocode – Use magic-pivot as a black-box to obtain a deterministic quick-sort algorithm with worst-case runningtime of O(nlogn)

Suppose we have an array A(1 :n) of n distinct numbers. For any element A(i), we define the rank of A(i), denoted by rank (A(i)), as the number of elements in A that are strictly smaller than A(i) plus one; so rank (A(i)) is also the correct position of A(i) in the sorted order of A. Suppose we have an algorithm magic-pivot that given any array B (1 :m) (for anym >0), returns an element B(i) such that m/3≤rank(B(i))≤2m/3 and has worst-case runtime O(n)1. Example:if B= (1,7,6,2,13,3,5,11,8), then magic-pivot(B) will return one arbitrary number among {3,5,6,7}(since sorted order of B is (1,2,3,5,6,7,8,11,13))

## Can not connect with power cord in blackbox os [on hold]

I am updating my new black box and can not connect to a LAN cable that I know is correct
because my pc is dual boot and i can connect in my windows
how to fix it?