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abcdefguan's solution

to Difference Of Squares in the OCaml Track

Published at Aug 29 2019 · 0 comments
Instructions
Test suite
Solution

Note:

This exercise has changed since this solution was written.

Find the difference between the square of the sum and the sum of the squares of the first N natural numbers.

The square of the sum of the first ten natural numbers is (1 + 2 + ... + 10)² = 55² = 3025.

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

Hence the difference between the square of the sum of the first ten natural numbers and the sum of the squares of the first ten natural numbers is 3025 - 385 = 2640.

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Source

Problem 6 at Project Euler http://projecteuler.net/problem=6

test.ml

open OUnit2
open Difference_of_squares

let ae exp got _test_ctxt = assert_equal exp got

let square_of_sum_tests = [
  "square of sum 1" >::
  ae 1 (square_of_sum 1);
  "square of sum 5" >::
  ae 225 (square_of_sum 5);
  "square of sum 100" >::
  ae 25502500 (square_of_sum 100);
]


let sum_of_squares_tests = [
  "sum of squares 1" >::
  ae 1 (sum_of_squares 1);
  "sum of squares 5" >::
  ae 55 (sum_of_squares 5);
  "sum of squares 100" >::
  ae 338350 (sum_of_squares 100);
]


let difference_of_squares_tests = [
  "difference of squares 1" >::
  ae 0 (difference_of_squares 1);
  "difference of squares 5" >::
  ae 170 (difference_of_squares 5);
  "difference of squares 100" >::
  ae 25164150 (difference_of_squares 100);
]

let () =
  run_test_tt_main (
    "difference of squares tests" >:::
    List.concat [square_of_sum_tests; sum_of_squares_tests; difference_of_squares_tests]
  )
let square_of_sum n = 
    n * n * (n+1) * (n+1) / 4

let sum_of_squares n =
    n * (n+1) * (2 * n + 1) / 6

let difference_of_squares n =
    square_of_sum n - sum_of_squares n

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