Skip to content

EHN: A divide-and-conquer, and brute-force algorithms for array inversions co… #1133

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Merged
merged 3 commits into from
Aug 15, 2019
Merged
Changes from all commits
Commits
File filter

Filter by extension

Filter by extension

Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
173 changes: 173 additions & 0 deletions divide_and_conquer/inversions.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,173 @@
from __future__ import print_function, absolute_import, division

"""
Given an array-like data structure A[1..n], how many pairs
(i, j) for all 1 <= i < j <= n such that A[i] > A[j]? These pairs are
called inversions. Counting the number of such inversions in an array-like
object is the important. Among other things, counting inversions can help
us determine how close a given array is to being sorted

In this implementation, I provide two algorithms, a divide-and-conquer
algorithm which runs in nlogn and the brute-force n^2 algorithm.

"""


def count_inversions_bf(arr):
"""
Counts the number of inversions using a a naive brute-force algorithm

Parameters
----------
arr: arr: array-like, the list containing the items for which the number
of inversions is desired. The elements of `arr` must be comparable.

Returns
-------
num_inversions: The total number of inversions in `arr`

Examples
---------

>>> count_inversions_bf([1, 4, 2, 4, 1])
4
>>> count_inversions_bf([1, 1, 2, 4, 4])
0
>>> count_inversions_bf([])
0
"""

num_inversions = 0
n = len(arr)

for i in range(n-1):
for j in range(i + 1, n):
if arr[i] > arr[j]:
num_inversions += 1

return num_inversions


def count_inversions_recursive(arr):
"""
Counts the number of inversions using a divide-and-conquer algorithm

Parameters
-----------
arr: array-like, the list containing the items for which the number
of inversions is desired. The elements of `arr` must be comparable.

Returns
-------
C: a sorted copy of `arr`.
num_inversions: int, the total number of inversions in 'arr'

Examples
--------

>>> count_inversions_recursive([1, 4, 2, 4, 1])
([1, 1, 2, 4, 4], 4)
>>> count_inversions_recursive([1, 1, 2, 4, 4])
([1, 1, 2, 4, 4], 0)
>>> count_inversions_recursive([])
([], 0)
"""
if len(arr) <= 1:
return arr, 0
else:
mid = len(arr)//2
P = arr[0:mid]
Q = arr[mid:]

A, inversion_p = count_inversions_recursive(P)
B, inversions_q = count_inversions_recursive(Q)
C, cross_inversions = _count_cross_inversions(A, B)

num_inversions = inversion_p + inversions_q + cross_inversions
return C, num_inversions


def _count_cross_inversions(P, Q):
"""
Counts the inversions across two sorted arrays.
And combine the two arrays into one sorted array

For all 1<= i<=len(P) and for all 1 <= j <= len(Q),
if P[i] > Q[j], then (i, j) is a cross inversion

Parameters
----------
P: array-like, sorted in non-decreasing order
Q: array-like, sorted in non-decreasing order

Returns
------
R: array-like, a sorted array of the elements of `P` and `Q`
num_inversion: int, the number of inversions across `P` and `Q`

Examples
--------

>>> _count_cross_inversions([1, 2, 3], [0, 2, 5])
([0, 1, 2, 2, 3, 5], 4)
>>> _count_cross_inversions([1, 2, 3], [3, 4, 5])
([1, 2, 3, 3, 4, 5], 0)
"""

R = []
i = j = num_inversion = 0
while i < len(P) and j < len(Q):
if P[i] > Q[j]:
# if P[1] > Q[j], then P[k] > Q[k] for all i < k <= len(P)
# These are all inversions. The claim emerges from the
# property that P is sorted.
num_inversion += (len(P) - i)
R.append(Q[j])
j += 1
else:
R.append(P[i])
i += 1

if i < len(P):
R.extend(P[i:])
else:
R.extend(Q[j:])

return R, num_inversion


def main():
arr_1 = [10, 2, 1, 5, 5, 2, 11]

# this arr has 8 inversions:
# (10, 2), (10, 1), (10, 5), (10, 5), (10, 2), (2, 1), (5, 2), (5, 2)

num_inversions_bf = count_inversions_bf(arr_1)
_, num_inversions_recursive = count_inversions_recursive(arr_1)

assert num_inversions_bf == num_inversions_recursive == 8

print("number of inversions = ", num_inversions_bf)

# testing an array with zero inversion (a sorted arr_1)

arr_1.sort()
num_inversions_bf = count_inversions_bf(arr_1)
_, num_inversions_recursive = count_inversions_recursive(arr_1)

assert num_inversions_bf == num_inversions_recursive == 0
print("number of inversions = ", num_inversions_bf)

# an empty list should also have zero inversions
arr_1 = []
num_inversions_bf = count_inversions_bf(arr_1)
_, num_inversions_recursive = count_inversions_recursive(arr_1)

assert num_inversions_bf == num_inversions_recursive == 0
print("number of inversions = ", num_inversions_bf)


if __name__ == "__main__":
main()