Problem
https://projecteuler.net/problem=14
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
#!/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