Feat: Project Euler Problem 14

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domix80 2023-02-19 14:05:13 +01:00
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* [Sol2](https://github.com/TheAlgorithms/Ruby/blob/master/project_euler/problem_004/sol2.rb)
* Problem 005
* [Sol1](https://github.com/TheAlgorithms/Ruby/blob/master/project_euler/problem_005/sol1.rb)
* Problem 014
* [Sol1](https://github.com/TheAlgorithms/Ruby/blob/master/project_euler/problem_014/sol1.rb)
* Problem 020
* [Sol1](https://github.com/TheAlgorithms/Ruby/blob/master/project_euler/problem_020/sol1.rb)
* Problem 021

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#Problem 14: https://projecteuler.net/problem=14
#Problem Statement:
#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?
def solution?()
index_best_result = 0
for num in 2..1000000
index_candidate = 0
n = num
while n > 1
if n%2 == 0
n = n / 2
else
n = (3*n) + 1
end
index_candidate +=1
end
if index_best_result < index_candidate
index_best_result = index_candidate
value = num
end
end
result = value
end
answer = solution?()
p answer