CODEWITHSUNDEEP
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Kuma
Kuma

Reputation: 123

Function to calculate the difference between sum of squares and square of sums

I am trying to Write a function called sum_square_difference which takes a number n and returns the difference between the sum of the squares of the first n natural numbers and the square of their sum.

I think i know how to write a function that defines the sum of squares

def sum_of_squares(numbers):
    total = 0
    for num in numbers:
        total += (num ** 2)
    return(total)

I have tried to implement a square of sums function:

def square_sum(numbers):  
    total = 0
    for each in range: 
        total = total + each
    return total**2

I don't know how to combine functions to tell the difference and i don't know if my functions are correct.

Any suggestions please? I am using Python 3.3

Thank you.

Upvotes: 5

Views: 8234

Answers (5)

Raza Hussain
Raza Hussain

Reputation: 762

In Ruby language you can achieve this in this way

def diff_btw_sum_of_squars_and_squar_of_sum(from=1,to=100) # use default values from 1..100. 
((1..100).inject(:+)**2) -(1..100).map {|num| num ** 2}.inject(:+)
end

diff_btw_sum_of_squars_and_squar_of_sum #call for above method

Upvotes: 2

Volatility
Volatility

Reputation: 32300

The function can be written with pure math like this:

The formula

Translated into Python:

def square_sum_difference(n):
    return int((3*n**2 + 2*n) * (1 - n**2) / 12)

The formula is a simplification of two other formulas:

def square_sum_difference(n):
    return int(n*(n+1)*(2*n+1)/6 - (n*(n+1)/2)**2)

n*(n+1)*(2*n+1)/6 is the formula described here, which returns the sum of the squares of the first n natural numbers.

(n*(n+1)/2))**2 uses the triangle number formula, which is the sum of the first n natural numbers, and which is then squared.

This can also be done with the built in sum function. Here it is:

def sum_square_difference(n):
    r = range(1, n+1)  # first n natural numbers
    return sum(i**2 for i in r) - sum(r)**2

The range(1, n+1) produces an iterator of the first n natural numbers.

>>> list(range(1, 4+1))
[1, 2, 3, 4]

sum(i**2 for i in r) returns the sum of the squares of the numbers in r, and sum(r)**2 returns the square of the sum of the numbers in r.

Upvotes: 11

mlwn
mlwn

Reputation: 1187

# As beta says, # (sum(i))^2 - (sum(i^2)) is very easy to calculate :) # A = sum(i) = i*(i+1)/2 # B = sum(i^2) = i*(i+1)*(2*i + 1)/6 # A^2 - B = i(i+1)(3(i^2) - i - 2) / 12 # :) # no loops... just a formula !**

Upvotes: 5

Beta
Beta

Reputation: 99084

This is a case where it pays to do the math beforehand. You can derive closed-form solutions for both the sum of the squares and the square of the sum. Then the code is trivial (and O(1)).

Need help with the two solutions?

Upvotes: 3

Roland Smith
Roland Smith

Reputation: 43495

def sum_square_difference(n):
    r = range(1,n+1)
    sum_of_squares =  sum(map(lambda x: x*x, r))
    square_sum = sum(r)**2
    return sum_of_squares - square_sum

Upvotes: 2

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