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

# Problem 55: Lychrel numbers

(see projecteuler.net/problem=55)

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?

# 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.

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

The code contains `#ifdef`

s 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;
}
// add the reverse
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
{
// add one more digit
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.

# 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.

This is equivalent to`echo 130 | ./55`

Output:

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

# Benchmark

The correct solution to the original Project Euler problem was found in **0.01** seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.

(compiled for x86_64 / Linux, GCC flags: `-O3 -march=native -fno-exceptions -fno-rtti -std=c++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

April 21, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler055

My code solved **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 never an option.

# Links

projecteuler.net/thread=55 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

Code in various languages:

Python: www.mathblog.dk/project-euler-55-how-many-lychrel-numbers-are-there-below-ten-thousand/ (written by Kristian Edlund)

Haskell: github.com/nayuki/Project-Euler-solutions/blob/master/haskell/p055.hs (written by Nayuki)

Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p055.java (written by Nayuki)

Go: github.com/frrad/project-euler/blob/master/golang/Problem055.go (written by Frederick Robinson)

Javascript: github.com/dsernst/ProjectEuler/blob/master/55 Lychrel numbers.js (written by David Ernst)

Scala: github.com/samskivert/euler-scala/blob/master/Euler055.scala (written by Michael Bayne)

# Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.

yellow problems score less than 100% at Hackerrank (but still solve the original problem).

gray problems are already solved but I haven't published my solution yet.

blue problems are already solved and there wasn't a Hackerrank version of it (at the time I solved it) or I didn't care about it because it differed too much.

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

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