<< problem 124 - Ordered radicals Cuboid layers - problem 126 >>

# Problem 125: Palindromic sums

The palindromic number 595 is interesting because it can be written as the sum of consecutive squares: 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2.

There are exactly eleven palindromes below one-thousand that can be written as consecutive square sums, and the sum of these palindromes is 4164.
Note that 1 = 0^2 + 1^2 has not been included as this problem is concerned with the squares of positive integers.

Find the sum of all the numbers less than 108 that are both palindromic and can be written as the sum of consecutive squares.

# My Algorithm

My function isPalindrome determines whether it's parameter is a palindrome or not:

• iteratively remove the lowest digit and append it to reverse
• if at the end reverse == x then we have a palindrome
A small optimization is to check the right-most digit for zero because then it can't be a palindrome.

Two nested loops iterate over all possible sequences.
My first solution wasn't correct because I didn't notice that sums may appear multipe times (there are two "collisions" for the original problem).
All palindromes are temporarily stored in solutions and then I remove all duplicates (see oeis.org/A267600).

## Alternative Approaches

There is a closed formula for a sequence sum_{a=i..j}{a^2} but I'm not sure whether it would be faster because my incremental computation of current is extremely efficient.

## Modifications by HackerRank

The upper limit can be defined as well as the distance between consecutive "numbers" (what I call stride):
instead of a^2 + (a+1)^2 + (a+2)^2 + ... the terms are a^2 + (a+stride)^2 + (a+2 * stride)^2 + ...

# Interactive test

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

Number of test cases (1-5):

Input data (separated by spaces or newlines):
Note: Enter the limit and the difference between consecutive numbers (1 for the original problem)

This is equivalent to
echo "1 1000 1" | ./125

Output:

(please click 'Go !')

Note: the original problem's input 100000000 1 cannot be entered
because just copying results is a soft skill reserved for idiots.

(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.

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

// return true if x is a palindrome
bool isPalindrome(unsigned int x)
{
auto reduced = x / 10;
auto reverse = x % 10;
// fast exit: a trailing zero can't create a palindrome
if (reverse == 0)
return false;

while (reduced > 0)
{
// chop off the lowest digit and append it to "reverse"
reverse *= 10;
reverse += reduced % 10;
reduced /= 10;
}

// palindrome ? both must be equal
return reverse == x;
}

int main()
{
unsigned int tests = 1;
std::cin >> tests;
while (tests--)
{
unsigned int limit  = 100000000;
unsigned int stride = 1; // distance between consecutive square numbers

std::cin >> limit >> stride;

std::vector<unsigned int> solutions;
for (unsigned long long first = 1; 2*first*first < limit; first++)
{
auto next = first + stride;
// sum of a^2 + b^2 + ...
auto current = first * first + next * next;
// still within the limit ?
while (current < limit)
{
// check
if (isPalindrome(current))
solutions.push_back(current);

// add one element to the sequence
next    += stride;
current += next * next;
}
}

// sort ...
std::sort(solutions.begin(), solutions.end());
// .. and remove duplicates
auto garbage = std::unique(solutions.begin(), solutions.end());
solutions.erase(garbage, solutions.end());

// count all solutions
unsigned long long sum = 0;
for (auto x : solutions)
sum += x;

std::cout << sum << std::endl;
}

return 0;
}


This solution contains 12 empty lines, 11 comments and 3 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in less than 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

May 17, 2017 submitted solution
May 17, 2017 added comments

# Hackerrank

My code solves 35 out of 35 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 25% (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 124 - Ordered radicals Cuboid layers - problem 126 >>
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