Confused on how this code to compute the powerset works
By : user2418962
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further I found this function on geeksforgeeks to find all subsets of a given set. I'm just not sure what the if statement in the nested for loop is checking. I understand that its using the bitwise AND operator, but I'm confused as to how it helps know which elements to include or not include during any iteration. , If there are three items in a powerset, there are 2^3 combinations. code :
a, b, c
===
[]
[a]
[b]
[a, b]
[c]
[a, c]
[b, c]
[a, b, c]
a, b, c
===
[0, 0, 0] > [0*a, 0*b, 0*c] = []
[1, 0, 0] > [1*a, 0*b, 0*c] = [a]
[0, 1, 0] > [0*a, 1*b, 0*c] = [b]
[1, 1, 0] > [1*a, 1*b, 0*c] = [a, b]
[0, 0, 1] > [0*a, 0*b, 1*c] = [c]
[1, 0, 1] > [1*a, 0*b, 1*c] = [a, c]
[0, 1, 1] > [0*a, 1*b, 1*c] = [b, c]
[1, 1, 1] > [1*a, 1*b, 1*c] = [a, b, c]

Powerset algorithm in Python: difference between + and append on lists
By : user3455653
Date : March 29 2020, 07:55 AM
may help you . The problem with your algorithm is that lists are mutable, and you are creating a list full of references to the same list. Any time you append to one of them, you are appending to all of them, because there is only one of them. They are all the same list, you just have more than one reference to it. Imagine having a list of 10 copies of somebody's phone number; if you call up the first phone number and ask them to put on a hat, then when you call the second phone number, the person on the other end will already be wearing a hat, because it is the same person. If you call each number and say "put on a hat" each time, you'll end up with a list of 10 phone numbers for one person wearing 10 hats, when you actually wanted phone numbers for 10 people wearing one hat each. code :
def powerset(array):
# a list containing an empty tuple
powerset = [()]
for num in array:
for i in range(len(powerset)):
curr_subset = powerset[i]
# append one more number to the tuple
curr_subset += (num,)
powerset.append(curr_subset)
return powerset
>>> powerset([1, 2, 3])
[(), (1,), (2,), (1, 2), (3,), (1, 3), (2, 3), (1, 2, 3)]

How to compute line segment intersections on a map
By : Alok
Date : March 29 2020, 07:55 AM
may help you . I'm assuming the algorithm you're using is the one for finding line intersections on a Cartesian coordinate system (i.e. a plane). Unfortunately, the earth is not a plane (or even a sphere) so using that algorithm will introduce error. Google Maps uses an ellipsoid (specifically WGS84) to approximate the surface of the earth, so you'll need an algorithm for finding the intersections of arcs on an ellipsoid. This page might contains some helpful information: http://mathhelpforum.com/calculus/90196pointintersectiontwolines.html

How to compute the intersections and unions of two arrays in Hive?
By : darcy
Date : March 29 2020, 07:55 AM
I hope this helps . Your problem solution is here. Go to the githubLink, there is lot of udfs are created by klout. Download, crate the JAR and add the JAR in the hive. Example

How to compute fast the containment order of all intersections of a collection of sets
By : Jessitowers
Date : March 29 2020, 07:55 AM
around this issue Here's a function that takes advantage of the assumption that the lists in the input are the facets of an abstract polytope. Instead of taking all intersections of collections of facets, this function assumes the input is a complete list of Mfaces (polytopes of rank M) within a polytope of rank M + 1. It then performs a loop in which each iteration takes a complete list of Mfaces and produces a complete list of (M1)faces, while accumulating all the containment pairs for those two lists of faces. The main loop of the function intersects each pair of Mfaces and builds a structure listing each intersection and the Mfaces that contained it. These intersections include all the (M1)faces, but also include some lowerrank faces. The lowerrank faces can be identified by observing that each of them is a subset of an (M1)face, so any intersection which is a subset of another intersection is eliminated.

