EltonTom EltonTom - 6 months ago 22
Ruby Question

Optimization for finding perfect-square algorithm

The question I'm working on is:


Find which sum of squared factors are a perfect square given a specific range.
So if the range was (1..10) you would get each number's factors (all factors for 1, all factors for 2, all factors for 3 ect..) Square those factors, then add them together. Finally check if that sum is a perfect square.


I am stuck on refactoring/optimization because my solution is too slow.

Here is what I came up with:

def list_squared(m, n)
ans = []
range = (m..n)

range.each do |i|
factors = (1..i).select { |j| i % j == 0 }
squares = factors.map { |k| k ** 2 }
sum = squares.inject { |sum,x| sum + x }
if sum == Math.sqrt(sum).floor ** 2
all = []
all += [i, sum]
ans << all
end
end

ans
end


This is an example of what I would put in the method:

list_squared(1, 250)


And then the desired output would be an array of arrays with each array containing the number whose sum of squared factors was a perfect square and the sum of those squared factors:

[[1, 1], [42, 2500], [246, 84100]]

Answer

I would start by introducing some helper methods (factors and square?) to make your code more readable.

Furthermore I would reduce the number of ranges and arrays to improve memory usage.

require 'prime'

def factors(number)
  [1].tap do |factors|
    primes = number.prime_division.flat_map { |p, e| Array.new(e, p) }
    (1..primes.size).each do |i| 
      primes.combination(i).each do |combination| 
        factor = combination.inject(:*)
        factors << factor unless factors.include?(factor)
      end
    end
  end
end

def square?(number)
  square = Math.sqrt(number)
  square == square.floor
end

def list_squared(m, n)
  (m..n).map do |number|
    sum = factors(number).inject { |sum, x| sum + x ** 2 }
    [number, sum] if square?(sum)
  end.compact
end

list_squared(1, 250)

A benchmark with a narrow range (up to 250) shows only a minor improvement:

require 'benchmark'
n = 1_000

Benchmark.bmbm(15) do |x|
  x.report("original_list_squared :") { n.times do; original_list_squared(1, 250); end }
  x.report("improved_list_squared :") { n.times do; improved_list_squared(1, 250); end }
end

# Rehearsal -----------------------------------------------------------
# original_list_squared :   2.720000   0.010000   2.730000 (  2.741434)
# improved_list_squared :   2.590000   0.000000   2.590000 (  2.604415)
# -------------------------------------------------- total: 5.320000sec

#                               user     system      total        real
# original_list_squared :   2.710000   0.000000   2.710000 (  2.721530)
# improved_list_squared :   2.620000   0.010000   2.630000 (  2.638833)

But a benchmark with a wider range (up to 10000) shows a much better performance than the original implementation:

require 'benchmark'
n = 10

Benchmark.bmbm(15) do |x|
  x.report("original_list_squared :") { n.times do; original_list_squared(1, 10000); end }
  x.report("improved_list_squared :") { n.times do; improved_list_squared(1, 10000); end }
end

# Rehearsal -----------------------------------------------------------
# original_list_squared :  36.400000   0.160000  36.560000 ( 36.860889)
# improved_list_squared :   2.530000   0.000000   2.530000 (  2.540743)
# ------------------------------------------------- total: 39.090000sec

#                               user     system      total        real
# original_list_squared :  36.370000   0.120000  36.490000 ( 36.594130)
# improved_list_squared :   2.560000   0.010000   2.570000 (  2.581622)

tl;dr: The bigger the N the better performs my code compared to the original implementation...