Solution to problem 14 in Julia

src/Julia/Problem014.jl

@@ -0,0 +1,68 @@
+#=
+Created on 24 Jul 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 14 of Project Euler
+https://projecteuler.net/problem=14
+=#
+
+function chain_length(n, terms)
+    length = 0
+    while n != 1
+        if haskey(terms, n)
+            length += terms[n]
+            break
+        end
+        if n % 2 == 0
+            n = n / 2
+        else
+            n = 3n + 1
+        end
+        length += 1
+    end
+    return length
+end
+
+function Problem14()
+    #=
+    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.
+    =#
+
+    ans = 0
+    limit = 1_000_000
+    score = 0
+    terms = Dict()
+    for i in 1:limit
+        terms[i] = chain_length(i, terms)
+        if terms[i] > score
+            score = terms[i]
+            ans = i
+        end
+    end
+    return ans
+end
+
+
+println("Time to evaluate Problem 14:")
+@time Problem14()
+println("")
+println("Result for Problem 14: ", Problem14())