Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Cartesian Product of any number of sets

  • Last Updated : 08 Jul, 2021

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']]


Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

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
            # product 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']]


My Personal Notes arrow_drop_up
Recommended Articles
Page :