Discussion on: Daily Challenge #7 - Factorial Decomposition

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Woo 3 days in a row!

I'm not too good when it comes to maths and I'm sure there are loads of more efficient algorithms for finding prime factors, but I think this works!

require 'prime'
def decomp n
  return "1" if n == 1
  counts = Hash.new 0
  (n / 2).ceil.downto(1).each do |x|
    next if !Prime.prime? x
    until n % x > 0 or n == 0
      n /= x
      counts[x] += 1
    end
  end
  counts.to_a.reverse.map(&proc {|pair| pair[1] > 1 ? "#{pair[0]}^#{pair[1]}" : pair[0]}).join " * "
end
def test n
  fact = Math.gamma(n + 1).to_i
  puts "n = #{n}; n! = #{fact}; decomp(n!) = #{decomp(fact)}"
end
(1..10).each {|x| test x}

Output:

n = 1; n! = 1; decomp(n!) = 1
n = 2; n! = 2; decomp(n!) = 2
n = 3; n! = 6; decomp(n!) = 2 * 3
n = 4; n! = 24; decomp(n!) = 2^3 * 3
n = 5; n! = 120; decomp(n!) = 2^3 * 3 * 5
n = 6; n! = 720; decomp(n!) = 2^4 * 3^2 * 5
n = 7; n! = 5040; decomp(n!) = 2^4 * 3^2 * 5 * 7
n = 8; n! = 40320; decomp(n!) = 2^7 * 3^2 * 5 * 7
n = 9; n! = 362880; decomp(n!) = 2^7 * 3^4 * 5 * 7
n = 10; n! = 3628800; decomp(n!) = 2^8 * 3^4 * 5^2 * 7

Spacie • Jul 4 '19 • Edited

Turns out there's actually a built in method for this ... So just the formatting I guess...

require 'prime'
def decomp n
  fact = Math.gamma(n + 1).to_i
  "n = #{n}; n! = #{fact}; decomp(n!) = #{n == 1 ? "1" : Prime.prime_division(fact).map(&proc {|pair| pair[1] > 1 ? "#{pair[0]}^#{pair[1]}" : pair[0]}).join(" * ")}"
end
(1..10).each {|x| puts decomp x}