Euler 014 ruby Solution

Problem

The following iterative sequence is defined for the set of positive integers:

\(n → n/2\) (n is even)

\(n → 3n + 1\) (n is odd)

Using the rule above and starting with 13, we generate the following sequence:

\[13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1\]

It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.

Which starting number, under one million, produces the longest chain?

NOTE: Once the chain starts the terms are allowed to go above one million.

Answer: 837799

Solution

euler014.rb

#!/usr/bin/env ruby
$cache = []

def collatz_sequence_length(start, options = {})
  options = { cache: true }.merge(options)

  chain_length = 1
  current_value = start

  while 1 != current_value
    unless $cache[current_value].nil?
      chain_length += $cache[current_value]
      break
    end

    chain_length += 1

    if current_value.even?
      current_value /= 2
    else
      current_value = (current_value * 3) + 1
    end
  end

  $cache[start] = chain_length if options[:cache]
  chain_length
end

def longest_collatz_sequence_process(options = {})
  longest_starting_number = 1

  max_chain_length = -1

  2.upto(1_000_000) do |i|
    chain_length = collatz_sequence_length(i, options)
    if chain_length > max_chain_length
      longest_starting_number = i
      max_chain_length = chain_length
    end
  end
  longest_starting_number
end

def longest_collatz_sequence
  longest_collatz_sequence_process
  # require 'benchmark'
  # Benchmark.measure { longest_collatz_sequence_process(cache: false) }
  # Cache on: 3.460000   0.110000   3.570000 (  3.580396)
  # Cache off: 22.350000   0.200000  22.550000 ( 22.635814)
end

puts longest_collatz_sequence if __FILE__ == $PROGRAM_NAME

Euler 014 ruby Solution

Problem

Solution

See Also