<< problem 35 - Circular primes | Truncatable primes - problem 37 >> |

# Problem 36: Double-base palindromes

(see projecteuler.net/problem=36)

The decimal number, 585 = 10010010012 (binary), is palindromic in both bases.

Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2.

(Please note that the palindromic number, in either base, may not include leading zeros.)

# Algorithm

My short function `num2str`

converts a number `x`

into a `std::string`

with `base`

.

It's more or less the same code I used in previous problems when I wanted to extract right-most digit of a number.

Note that the result "grows" from right-to-left, that's why I always insert a single char at position 0.

The function `isPalindrome`

uses the STL function `std::reverse`

to efficiently reverse a string in-place.

If it's identical to the original string, then we have a palindrome.

The main loop checks number with base 10 first, because they are shorter than numbers with base 2 to 9.

In most cases we can abort earlier and save a little time.

## Modifications by HackerRank

Base can be anything from 2 to 9.

## Note

Of course you can speed up the whole process considerably:

1. generate all numbers from `1`

to `y`

where `y`

has half as many digits as `x`

.

2. then convert to an `std::string`

and append its reverse.

3. convert that string to an integer `x`

and test with `isPalindrome(num2str(x, 2))`

whether `x`

's binary representation is a palindrome, too.

This way you have to check 1000 instead of 1000000 numbers.

I haven't benchmark it but my suspicion is that if `num2str`

appends digits (instead of prepending) and reverses the string as a final step,

then we might see a speed-up as well.

# My code

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

#include <iostream>
#include <string>
#include <algorithm>
// base: decimal=10, binary=2

std::string num2str(unsigned int x, unsigned int base)
{
std::string result;
while (x > 0)
{
auto digit = x % base;
x /= base;
result.insert(0, 1, char(digit + '0'));
}
return result;
}
// true if string is a palindrome

bool isPalindrome(const std::string& s)
{
auto other = s;
std::reverse(other.begin(), other.end());
return other == s;
}
// I was quite surprised that the basic brute-force approach is sufficiently fast ...

int main()
{
unsigned int limit, base; // limit = 1000000 and base = 2
std::cin >> limit >> base;
unsigned int sum = 0;
for (unsigned int x = 1; x < limit; x++)
if (isPalindrome(num2str(x, 10)) && // palindrome in base 10 ?
isPalindrome(num2str(x, base))) // palindrome in base k ?
// yes, double palindrome
sum += x;
std::cout << sum << std::endl;
return 0;
}

This solution contains 5 empty lines, 4 comments and 3 preprocessor commands.

# Interactive test

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

This is equivalent to`echo "100 2" | ./36`

Output:

*Note:* the original problem's input `1000000 2`

__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)*

# Benchmark

The correct solution to the original Project Euler problem was found in 0.31 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 24, 2017 submitted solution

April 6, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler036

My code solves **15** out of **15** 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=36 - **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-36-palindromic-base-10-2/ (written by Kristian Edlund)

Haskell: github.com/nayuki/Project-Euler-solutions/blob/master/haskell/p036.hs (written by Nayuki)

Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p036.java (written by Nayuki)

Mathematica: github.com/nayuki/Project-Euler-solutions/blob/master/mathematica/p036.mathematica (written by Nayuki)

C: github.com/eagletmt/project-euler-c/blob/master/30-39/problem36.c (written by eagletmt)

Javascript: github.com/dsernst/ProjectEuler/blob/master/36 Double-base palindromes.js (written by David Ernst)

Scala: github.com/samskivert/euler-scala/blob/master/Euler036.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 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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My username at Project Euler is

**stephanbrumme**while it's stbrumme at Hackerrank.

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