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Project Euler #6 - Sum Square Difference

Peter Kim Frank on May 30, 2019

Continuing the wonderful community solutions to Project Euler. This is Problem 6, sum square difference. The sum of the squares of the first ten...

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Luis Miguel • May 30 '19 • Edited

no loops!

python3

def sum_square_difference(n):
return (((n**2) * (n + 1)**2) / 4) - (n * (n + 1) * (2*n + 1) / 6)

print(sum_square_difference(100))

I dont know why when I tried to upload images they dont appear

Massimo Artizzu • May 30 '19

Aw yeah 👌

If some are wondering, that comes from well-known formulas for the sum of the first n integers and n squares. There's a generalization, too, but also a pretty common formula for the first n cubes.

A solid mathematical approach can really simplify programming challenges. It almost feels like cheating.

Nathan Tamez • May 30 '19

Definitely highlights the need for maths in software development.

Henry 👨‍💻 • May 30 '19

Woah, alright, you win 😲.
Also, there's a guide here to embed code and images in markdown (what dev.to uses).

Luis Miguel • May 30 '19

thanks, I'm going to check it.

Ali Spittel • May 30 '19 • Edited

Python!

def sum_square_difference(n):
    numbers = range(1, n+1)
    sum_squares = sum(i**2 for i in numbers)
    square_sum = sum(numbers)**2
    return square_sum - sum_squares

print(sum_square_difference(100))

Florian Rand • May 30 '19

I surrender!

Dídac • May 30 '19 • Edited

Ruby:

def sum_square_difference(n)
  numbers = (1..n)
  square_of_sum = numbers.sum ** 2
  sum_of_square = numbers.map { |n| n ** 2 }.sum
  square_of_sum - sum_of_square
end

puts sum_square_difference(10)

Nans Dumortier • May 30 '19

Ruby is so elegant 😍

Nans Dumortier • May 30 '19

Javascript !

const sumSquareDifference = (n) => {
  const numbers = [...Array(n + 1).keys()];
  const sumOfSquares = numbers.reduce((accumulator, number) => accumulator + (number ** 2));
  const squareOfSum = numbers.reduce((accumulator, number) => accumulator + number) ** 2;
  return squareOfSum - sumOfSquares;
}
console.log(sumSquareDifference(10));

Nathan Tamez • May 30 '19

Yours is good. But I have a feeling it uses more then one for/for each loop. May explain why it is a bit slower then mine. Not sure.

Nans Dumortier • May 30 '19

Yeah, also the fact that I'm iterating 2 times the same array !

Nathan Tamez • May 30 '19 • Edited

Here is my nodejs solution

function sumSquareDifference(min, max) {
    let sumOfSquares = 0;
    let squareOfSums = 0;
    for (i = min; i < max + 1; i++) {
        sumOfSquares += i ** 2;
        squareOfSums += i;
    }

    return (squareOfSums ** 2) - sumOfSquares;
}
const startTime = Date.now();
console.log(`The difference between the sum of the squares of the first one
 hundred natural numbers and the square of the sum,
is ${sumSquareDifference(1, 100)}`);
console.log(`Time Taken: ${Math.round(Date.now() - startTime)}ms`);

output

The difference between the sum of the squares of the first one hundred natural numbers and the square of the sum,is 25164150
Time Taken: 2ms

Nans Dumortier • May 30 '19 • Edited

Yours performs faster than the one I submitted, and still looks nice, well done !

EDIT: some of the lets could be replaced with const though 🙈

Nathan Tamez • May 30 '19

true

Olivier Picault • May 30 '19

From ProjectEuler:

Please do not deprive others of going through the same process by publishing your solution outside of Project Euler; for example, on other websites, forums, blogs, or any public repositories (e.g. GitHub), et cetera. Members found to be spoiling problems will have their accounts locked.

Jay • May 30 '19

Rust Solution: Playground

fn main() {
    println!("{}", get_difference(100));
}

fn get_difference(end: usize) -> u32 {
    ((1..=end).sum::<usize>().pow(2) - (1..=end).map(|n| n.pow(2)).sum::<usize>()) as u32
}

Henry 👨‍💻 • May 30 '19

My take on this problem in python:

def challenge():
    sumOfSquares = 0
    squareOfSums = 0
    difference = 0
    for i in range(101):
        sumOfSquares += i ** 2
        squareOfSums += i
    squareOfSums = squareOfSums ** 2
    difference =  squareOfSums - sumOfSquares
    print("The difference between the sum of squares and the square of sums up to 100 is {}".format(difference))

Output:

The difference between the sum of squares and the square of sums up to 100 is 25164150

Ademola John • May 18 '20 • Edited

Here is my Javascript Solution


function sumSquareDifference(n) {
  // Good luck!
  let a = 0,b=0,c=0;
  for(let i = 1;i<=n;i++){
    a+= (Math.pow(i,2));
    b+=i
  }
c=Math.pow(b,2)-a

console.log(a);
console.log(b);
  return c;
}

kip • May 30 '19 • Edited

Rust

fn get_sum_square_difference(end: usize) -> usize {
    let mut sum: usize = 0;
    let square_sum = (1..=end).fold(0usize, |fin, num| { sum += num; fin + num.pow(2) });

    sum.pow(2) - square_sum
}

David Skinner • May 30 '19

Go

goplay.space/#rQ0XekiCknZ

// func SumSquareDifference receives the max range
// and returns the difference of the sumSquare and the squareSum.
func SumSquareDifference(x int) int {
    sum := 0
    sumSquare := 0

    for i := 1; i <= x; i++ {
        sum += i
        sumSquare += i * i
    }

    squareSum := sum * sum

    return squareSum - sumSquare
}

You can see the complete program on the playground link.

I have included a block that confirms that my function provides the expected response.
I then start a timer so I can print the elapsed time to run the function.
After printing the results, I include a comment defining the Output, If the output changes it will generate an error on my computer.
The language is Go but it is more a C++ style, I am not the best Go programmer.
Code needs to be readable, compilers can do the optimization.

Florian Rand • May 30 '19 • Edited

Some fun!

// with for loop
func sumSqDiffLoop(n int) int {
    var sOsq int
    var sOts int
    for i := 1; i <= 100; i++ {
        sOsq += i * i
        sOts += i
    }
    return (sOts * sOts) - sOsq
}
// no loop - thanks to @lmbarr for his solution
func sumSqDiffNoLoop(n int) int {
    return (((n * n) * ((n + 1) * (n + 1))) / 4) - (n * (n + 1) * (2*n + 1) / 6)
}

// Let's make some benchmarks!

import "testing"

func BenchmarkSumSquareDiff(b *testing.B) {
    benchs := []struct {
        name string
        fun  func(int) int
    }{
        {"Sum Square Diff Loop", sumSqDiffLoop},
        {"Sum Square Diff no Loop", sumSqDiffNoLoop},
    }
        // Let's test it with 1000 instead of 100 :D
    input := 1000

    for _, bench := range benchs {
        b.Run(bench.name, func(b *testing.B) {
            for i := 0; i < b.N; i++ {
                bench.fun(input)
            }
        })
    }
}

Benchmark result:

❯ go test -bench=.

goos: linux
goarch: amd64
/Sum_Square_Diff_Loop-4             20000000          62.7 ns/op
/Sum_Square_Diff_no_Loop-4          500000000         3.58 ns/op
PASS
ok      _/h/f/s/g/src/project-euler/06  3.474s

Carlos E. Barboza • May 30 '19 • Edited

Swift:

func sumSquareDifference(range: Int) -> Int {
    var sumOfSquares = 0
    var squareOfSum = 0
    for number in 1..<range+1 { //Is there a better way? Other than range+1?
        sumOfSquares += (number*number)
        squareOfSum += number
    }
    squareOfSum = squareOfSum*squareOfSum
    return (squareOfSum - sumOfSquares)
}

print(sumSquareDifference(range:100))

Seiei Miyagi • May 30 '19

Ruby 2.7

p (1..100).then { @1.sum ** 2 - @1.sum { @1 ** 2 } }

hishamkhr • Mar 28 '20

hi
i try this:
sum of numbers= n(n+1)/2
then i get the square.
after that i found an equation about sum of squares:
(2*n^3 +3*n^2+n )/6
but i didn't get right result !!!!
so where i lost the control ?
thanks

Sagnik Ray • Jun 9 '20

public class ProjectEuler{
static int n = 100;
public static void main(String []args){
System.out.println(differencehelper());

}

static int sumsquare(int n){

    int sum = 0 ;

    sum = (n*(n+1)(2*n+1))/6;

    return sum;

}

static int squaresum(int n){

   int sum  = 0;

   sum = (n(n+1))/2;

   return (sum *sum);

}

static int differencehelper(){

    return squaresum(n) - sumsquare(n);

}

}

Easy Java Solution using Gauss' Formula Hope you liked it ;)

rahy-hub • Apr 9 '20

my code with C++

/*Sum square difference

Problem 6
The sum of the squares of the first ten natural numbers is,

12+22+...+102=385
The square of the sum of the first ten natural numbers is,

(1+2+...+10)2=552=3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025−385=2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.*/

include

using namespace std;
int square_sum(int s);

int main()
{
int limit=100 ,sum=0 ,sumQ=0;
for(int i=1;i<=limit;i++)
{
sumQ=sumQ+(i*i);
sum =sum+i;
}

cout<<"the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum is \n ";
cout<< square_sum(sum)-sumQ ;  // without function : cout<<sum*sum-sumQ
return 0;

}

int square_sum(int s)
{
return s*s;
}
Output >> 25164150

nurchikgit • Oct 13 '20

not bad)))

Mahmoud Bayazid • Mar 29 '20 • Edited

new_list = []

for i in range (1,101):
new_list.append(i^2)
a = sum(new_list)
b = sum(range(1,101))^2
print(b-a)

Mahmoud Bayazid • Mar 29 '20

new_list = []

for i in range (1,101):
new_list.append(i*2)
a = sum(new_list)
b = sum(range(1,101))*2
print(b-a)

Oleksii Filonenko • May 30 '19

Elixir:

(1..100 |> Enum.map(&:math.pow(&1, 2)) |> Enum.sum()) -
  (1..100 |> Enum.sum() |> :math.pow(2))