<< problem 54 - Poker hands Powerful digit sum - problem 56 >>

# Problem 55: Lychrel numbers

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?

# My Algorithm

Intermediate numbers may become quite huge (as hinted in the problem), therefore I decided to store all numbers in a BigNumber,
which is a std::vector where each digit is stored as a single element (lowest digits first), e.g. 3401 = { 1, 0, 4, 3 }.

Then, reversing the digits becomes super-easy:
auto reverse = number;
std::reverse(reverse.begin(), reverse.end());

And checking whether we have a palindrome
if (number == reverse)

Adding the digits is based on the basic algorithm taught in school.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

This live test is based on the Hackerrank problem.

Input data (separated by spaces or newlines):

This is equivalent to
echo 130 | ./55

Output:

(this interactive test is still under development, computations will be aborted after one second)

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

The code contains #ifdefs to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

       #include <vector>
#include <map>
#include <algorithm>
#include <iostream>

//#define ORIGINAL

// a sequence of digits, lowest digits first, e.g. 3401 = { 1, 0, 4, 3 }
typedef std::vector<unsigned int> BigNumber;

// count all numbers converging to BigNumber
std::map<BigNumber, unsigned int> finalNumber;

// return true if x is a Lychrel number, stop after maxIterations (and return true if "indecisive")
bool isLychrel(unsigned int x, unsigned int maxIterations)
{
// split integer into its digit, store each digit in a separate cell
BigNumber number;
while (x > 0)
{
number.push_back(x % 10);
x /= 10;
}

// try to find a palindrome in the first 60 iterations
for (unsigned int iteration = 0; iteration < maxIterations; iteration++)
{
auto reverse = number;
std::reverse(reverse.begin(), reverse.end());

// check if palindrome
#ifdef ORIGINAL
if (iteration > 0) // originally, the initial number is allowed to be a palindrome
#endif
// no, can't be a Lychrel number
if (number == reverse)
{
finalNumber[number]++;
return false;
}

auto sum = number;
unsigned int carry = 0;
for (size_t digit = 0; digit < number.size(); digit++)
{
// get digit "from the other end"
sum[digit] += reverse[digit] + carry;

// overflow ?
if (sum[digit] >= 10)
{
sum[digit] -= 10;
carry = 1;
}
else
{
carry = 0;
}
}
if (carry > 0)
sum.push_back(carry);

number = std::move(sum);
}

// yes, we have a Lychrel number
return true;
}

int main()
{
// consider a number to be a Lychrel number if no palindrome after that many iterations
#ifdef ORIGINAL
unsigned int iterations = 50;
#else
unsigned int iterations = 60;
#endif

unsigned int maxNumber = 10000;
std::cin >> maxNumber;

// count all Lychrel number
unsigned int count = 0;
for (unsigned int i = 0; i <= maxNumber; i++)
if (isLychrel(i, iterations))
count++;

#ifdef ORIGINAL
std::cout << count << std::endl;

#else

unsigned int bestCount = 0;
BigNumber    bestNumber;
// find number most converged to
for (auto f : finalNumber)
if (bestCount < f.second)
{
bestCount  = f.second;
bestNumber = f.first;
}

// print single digits, highest digits first (they were stored last in BigNumber)
std::reverse(bestNumber.begin(), bestNumber.end());
for (auto digit : bestNumber)
std::cout << digit;
std::cout << " " << bestCount << std::endl;
#endif

return 0;
}


This solution contains 18 empty lines, 17 comments and 12 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.01 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

# Changelog

February 27, 2017 submitted solution

# Hackerrank

My code solves 6 out of 6 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 5% (out of 100%).

Hackerrank describes this problem as easy.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

# Heatmap

Please click on a problem's number to open my solution to that problem:

 green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.
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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

 << problem 54 - Poker hands Powerful digit sum - problem 56 >>
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