# If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.
#
# Not all numbers produce palindromes so quickly. For example,
#
# 349 + 943 = 1292,
# 1292 + 2921 = 4213
# 4213 + 3124 = 7337
#
# That is, 349 took three iterations to arrive at a palindrome.
#
# Although no one has proved it yet, it is thought that some numbers, like 196,
# never produce a palindrome. A number that never forms a palindrome through
# the reverse and add process is called a Lychrel number. Due to the
# theoretical nature of these numbers, and for the purpose of this problem, we
# shall assume that a number is Lychrel until proven otherwise. In addition you
# are given that for every number below ten-thousand, it will either (i) become
# a palindrome in less than fifty iterations, or, (ii) no one, with all the
# computing power that exists, has managed so far to map it to a palindrome. In
# fact, 10677 is the first number to be shown to require over fifty iterations
# before producing a palindrome: 4668731596684224866951378664 (53 iterations,
# 28-digits).
#
# Surprisingly, there are palindromic numbers that are themselves Lychrel
# numbers; the first example is 4994.
#
# How many Lychrel numbers are there below ten-thousand?


using ProjectEulerSolutions


# Cycle through list from 11 to n to calculate if a value and its reverse sum
# is a palindrome or not.  Trying a dynamic programming-type answer would not
# help much since the span of the possible numbers to store is quite large.
function p055solution(n_max::Integer=100)::Integer
    count = 0
    count_thresh = 50
    # We start with Int128, but we might do better with BigInt in some choices
    # of n_max.
    if n_max < 11_000
        n_start = Int128(10) # Single digit numbers are already palindromes
    else    
        n_start = BigInt(10)
    end
    for n in n_start:n_max
        index = 0
        n2 = reverse_sum(n)
        while(!ispalindrome_integer(n2))
            n2 = reverse_sum_integer(n2)
            index += 1
            if (index == count_thresh)
                count += 1
                break
            end
        end
    end
   
    return count
end


p055 = Problems.Problem(p055solution)

Problems.benchmark(p055, 10_000)
