Optimization for finding perfect-square algorithm

Question

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 woul

Answer 1

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 my code performs compared to the original implementation...

Answer 2

The trick that frequently solves questions like this is to switch from trial division to a sieve. In Python (sorry):

def list_squared(m, n):
    factor_squared_sum = {i: 0 for i in range(m, n + 1)}
    for factor in range(1, n + 1):
        i = n - n % factor  # greatest multiple of factor less than or equal to n
        while i >= m:
            factor_squared_sum[i] += factor ** 2
            i -= factor
    return {i for (i, fss) in factor_squared_sum.items() if isqrt(fss) ** 2 == fss}


def isqrt(n):
    # from http://stackoverflow.com/a/15391420
    x = n
    y = (x + 1) // 2
    while y < x:
        x = y
        y = (x + n // x) // 2
    return x

The next optimization is to step factor only to isqrt(n), adding the factor squares in pairs (e.g., 2 and i // 2).

Answer 3

One way to make it more efficient is to use Ruby's built-in method Prime::prime_division.

For any number n, if prime_division returns an array containing a single element, that element will be [n,1] and n will have been shown to be prime. That prime number has factors n and 1, so must be treated differently than numbers that are not prime.

require 'prime'

def list_squared(range)
  range.each_with_object({}) do |i,h|
    facs = Prime.prime_division(i)
    ssq = 
    case facs.size
    when 1 then facs.first.first**2 + 1
    else facs.inject(0) { |tot,(a,b)| tot + b*(a**2) }
    end
    h[i] = facs if (Math.sqrt(ssq).to_i)**2 == ssq
  end
end

list_squared(1..10_000)
  #=> { 1=>[], 48=>[[2, 4], [3, 1]], 320=>[[2, 6], [5, 1]], 351=>[[3, 3], [13, 1]],
  #     486=>[[2, 1], [3, 5]], 1080=>[[2, 3], [3, 3], [5, 1]],
  #     1260=>[[2, 2], [3, 2], [5, 1], [7, 1]], 1350=>[[2, 1], [3, 3], [5, 2]],
  #     1375=>[[5, 3], [11, 1]], 1792=>[[2, 8], [7, 1]], 1836=>[[2, 2], [3, 3], [17, 1]],
  #     2070=>[[2, 1], [3, 2], [5, 1], [23, 1]], 2145=>[[3, 1], [5, 1], [11, 1], [13, 1]],
  #     2175=>[[3, 1], [5, 2], [29, 1]], 2730=>[[2, 1], [3, 1], [5, 1], [7, 1], [13, 1]],
  #     2772=>[[2, 2], [3, 2], [7, 1], [11, 1]], 3072=>[[2, 10], [3, 1]],
  #     3150=>[[2, 1], [3, 2], [5, 2], [7, 1]], 3510=>[[2, 1], [3, 3], [5, 1], [13, 1]],
  #     4104=>[[2, 3], [3, 3], [19, 1]], 4305=>[[3, 1], [5, 1], [7, 1], [41, 1]],
  #     4625=>[[5, 3], [37, 1]], 4650=>[[2, 1], [3, 1], [5, 2], [31, 1]],
  #     4655=>[[5, 1], [7, 2], [19, 1]], 4998=>[[2, 1], [3, 1], [7, 2], [17, 1]],
  #     5880=>[[2, 3], [3, 1], [5, 1], [7, 2]], 6000=>[[2, 4], [3, 1], [5, 3]],
  #     6174=>[[2, 1], [3, 2], [7, 3]], 6545=>[[5, 1], [7, 1], [11, 1], [17, 1]],
  #     7098=>[[2, 1], [3, 1], [7, 1], [13, 2]], 7128=>[[2, 3], [3, 4], [11, 1]],
  #     7182=>[[2, 1], [3, 3], [7, 1], [19, 1]], 7650=>[[2, 1], [3, 2], [5, 2], [17, 1]],
  #     7791=>[[3, 1], [7, 2], [53, 1]], 7889=>[[7, 3], [23, 1]],
  #     7956=>[[2, 2], [3, 2], [13, 1], [17, 1]],
  #     9030=>[[2, 1], [3, 1], [5, 1], [7, 1], [43, 1]],
  #     9108=>[[2, 2], [3, 2], [11, 1], [23, 1]], 9295=>[[5, 1], [11, 1], [13, 2]],
  #     9324=>[[2, 2], [3, 2], [7, 1], [37, 1]]} 

This calculation took approximately 0.15 seconds.

For i = 6174

 (2**1) * (3**2) * (7**3) #=> 6174

and

 1*(2**2) + 2*(3**2) + 3*(7**2) #=> 169 == 13*13