# Project Euler #6 - Sum Square Difference

### Peter Kim Frank twitter logo github logo May 30 '19・1 min read

Project Euler (7 Part Series)

Continuing the wonderful community solutions to Project Euler.

This is Problem 6, sum square difference.

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.

DISCUSS (27) 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.

Definitely highlights the need for maths in software development.

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));
``````

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.

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

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
``````

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

EDIT: some of the `let`s could be replaced with `const` though 🙈

true

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)
``````

Ruby is so elegant 😍

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.

:)

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
``````

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
}
``````

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
``````

# 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.

Ruby 2.7

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

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))
``````

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