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Cartesian Product of any number of sets
  • Last Updated : 28 Aug, 2018

Given N number of sets. The task is to write a program to perform cartesian product of all the sets in a given order.


1st set: 1 2
2nd set: A 
3rd set: x 
4th set: 5 6
[['1', 'A', 'x', '5'],
 ['1', 'A', 'x', '6'],
 ['2', 'A', 'x', '5'],
 ['2', 'A', 'x', '6']]

1st set: 1 2
2nd set: A 
3rd set: x y z 
[['1', 'A', 'x'],
 ['1', 'A', 'y'],
 ['1', 'A', 'z'],
 ['2', 'A', 'x'],
 ['2', 'A', 'y'],
 ['2', 'A', 'z']]

Approach: The approach is to compute the product of set-1 and set-2 at the beginning and then the resultant of set-1 and set-2 will have a product with set-3 and then the resultant of set-1, set-2, set-3 will have a Cartesian product with set-4 and so on till set-n.

Below is the implementation of the above approach.

# Python program for cartesian
# product of N-sets 
# function to find cartesian product of two sets 
def cartesianProduct(set_a, set_b):
    result =[]
    for i in range(0, len(set_a)):
        for j in range(0, len(set_b)):
            # for handling case having cartesian
            # prodct first time of two sets
            if type(set_a[i]) != list:         
                set_a[i] = [set_a[i]]
            # coping all the members
            # of set_a to temp
            temp = [num for num in set_a[i]]
            # add member of set_b to 
            # temp to have cartesian product     
    return result
# Function to do a cartesian 
# product of N sets 
def Cartesian(list_a, n):
    # result of cartesian product
    # of all the sets taken two at a time
    temp = list_a[0]
    # do product of N sets 
    for i in range(1, n):
        temp = cartesianProduct(temp, list_a[i])
# Driver Code
list_a = [[1, 2],          # set-1
          ['A'],          # set-2
          ['x', 'y', 'z']]   # set-3
# number of sets
n = len(list_a) 
# Function is called to perform 
# the cartesian product on list_a of size n 
Cartesian(list_a, n)
[[1, 'A', 'x'],
 [1, 'A', 'y'],
 [1, 'A', 'z'], 
 [2, 'A', 'x'], 
 [2, 'A', 'y'], 
 [2, 'A', 'z']]

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