def SortCount(A): l = len(A) if l > 1: n = l//2 C = A[:n] D = A[n:] C, c = SortCount(A[:n]) D, d = SortCount(A[n:]) B, b = MergeCount(C,D) return B, b+c+d else: return A, 0 def MergeCount(A,B): count = 0 M = [] while A and B: if A[0] <= B[0]: M.append(A.pop(0)) else: count += len(A) M.append(B.pop(0)) M += A + B return M, count

## Monday, August 19, 2013

### Counting Inversions in Python

Below is an O(n log(n)) algorithm for counting the number of inversions in an array of distinct positive integers. We call the index pair (i,j) of an array A an inversion provided that i < j and A[i] > A[j]. The idea is to break the array A into a left half L:=A[:n] and a right half R:=A[:n] and then count the number of inversions in L, in R, and the number of split inversions (i,j) where i is less than n and j is greater than or equal to n. If L and R are already sorted then we can easily count the number of split inversions via a slight augmentation to the merge subroutine of merge-sort.

## Tuesday, August 6, 2013

### A Coin Tossing Game

As an example of the law of total probabilities, I'm going to explore a slightly generalized variant of an interview question I recently encountered. The basic scenario is to imagine two players, let's say Amanda and Bill, both of who have a weighted coin. They take turns tossing their coin and whoever obtains the first heads wins. Amanda's coin comes up heads with probability $\theta$ and Bill's with probability $\psi$. If Amanda tosses first, what's the probability she will win?

Subscribe to:
Posts (Atom)