How to resolve the algorithm Perfect numbers step by step in the CoffeeScript programming language

Published on 12 May 2024 09:40 PM

How to resolve the algorithm Perfect numbers step by step in the CoffeeScript programming language

Table of Contents

Problem Statement

Write a function which says whether a number is perfect.

A perfect number is a positive integer that is the sum of its proper positive divisors excluding the number itself. Equivalently, a perfect number is a number that is half the sum of all of its positive divisors (including itself).

Note:   The faster   Lucas-Lehmer test   is used to find primes of the form   2n-1,   all known perfect numbers can be derived from these primes using the formula   (2n - 1) × 2n - 1. It is not known if there are any odd perfect numbers (any that exist are larger than 102000). The number of   known   perfect numbers is   51   (as of December, 2018),   and the largest known perfect number contains  49,724,095  decimal digits.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Perfect numbers step by step in the CoffeeScript programming language

Source code in the coffeescript programming language

is_perfect_number = (n) ->
  do_factors_add_up_to n, 2*n
  
do_factors_add_up_to = (n, desired_sum) ->
  # We mildly optimize here, by taking advantage of
  # the fact that the sum_of_factors( (p^m) * x)
  # is (1 + ... + p^m-1 + p^m) * sum_factors(x) when
  # x is not itself a multiple of p.

  p = smallest_prime_factor(n)
  if p == n
    return desired_sum == p + 1

  # ok, now sum up all powers of p that
  # divide n
  sum_powers = 1
  curr_power = 1
  while n % p == 0
    curr_power *= p
    sum_powers += curr_power
    n /= p
  
  # if desired_sum does not divide sum_powers, we
  # can short circuit quickly
  return false unless desired_sum % sum_powers == 0
  
  # otherwise, recurse
  do_factors_add_up_to n, desired_sum / sum_powers

smallest_prime_factor = (n) ->
  for i in [2..n]
    return n if i*i > n
    return i if n % i == 0

# tests
do -> 
  # This is pretty fast...
  for n in [2..100000]
    console.log n if is_perfect_number n

  # For big numbers, let's just sanity check the known ones.
  known_perfects = [
    33550336
    8589869056
    137438691328
  ]
  for n in known_perfects
    throw Error("fail") unless is_perfect_number(n)
    throw Error("fail") if is_perfect_number(n+1)


  

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