<< problem 34 - Digit factorials | Double-base palindromes - problem 36 >> |

# Problem 35: Circular primes

(see projecteuler.net/problem=35)

The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.

There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.

How many circular primes are there below one million?

# Algorithm

First, a a standard prime sieve finds all prime numbers up to our limit (1000000 by default) and keeps them in a `std::set`

.

Then each prime `x`

in `std::set`

is rotated by one digit to the right:

1. get the right-most digit:

`auto digit = rotated % 10;`

2. move all digits by one digit to the right ("erasing" the right-most digit):

`rotated /= 10;`

3. prepend the right-most digit:

` rotated += digit * shift;`

4. check whether rotated is part of our `std::set`

, too

5. if `rotated`

is equal to our initial value `x`

then we checked all rotations

The only point of interest is `shift`

which is a power of 10 such that 10^a = shift <= x <= 10^{a+1}.

E.g., if x = 3456 then shift = 1000.

## Modifications by HackerRank

We have to find the sum of all such prime numbers, not their count.

## Note

There are a few options to speed up the code:

1. All prime numbers are odd (except for 2): if `x != 2`

and any digit is even then this prime can't be circular.

2. We can simplify point 1 by noting that all single-digit primes are circular.

# 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 <iostream>
#include <set>
int main()
{
// highest number (1000000 in original problem)
unsigned int n;
std::cin >> n;
// precomputation: find all prime numbers up to n
std::set<unsigned int> primes;
primes.insert(2);
for (unsigned int i = 3; i <= n; i += 2)
{
bool isPrime = true;
// test against all prime numbers we have so far (in ascending order)
for (auto x : primes)
{
// divisible => not prime
if (i % x == 0)
{
isPrime = false;
break;
}
// prime is too large to be a divisor
if (x*x > i)
break;
}
// yes, we have a prime
if (isPrime)
primes.insert(i);
}
// now look at all primes
unsigned int sum = 0;
for (auto x : primes)
{
// move the right-most digit to the front of the number
// we need to know the "position" of the front-most digit:
// shift will be 1 for x = 1..9
// shift will be 10 for x = 10..99
// shift will be 100 for x = 100..999 and so on
unsigned int shift = 1;
while (x > shift * 10)
shift *= 10;
auto rotated = x;
do
{
// take right-most digit
auto digit = rotated % 10;
// remove it
rotated /= 10;
// and prepend it
rotated += digit * shift;
// rotated number not prime ?
if (primes.count(rotated) == 0)
break;
} while (rotated != x); // finished the circle ? (we have the initial number again)
// all rotations succeeded ?
#define ORIGINAL
#ifdef ORIGINAL
if (rotated == x)
sum++;
#else
if (rotated == x)
sum += x;
#endif
}
std::cout << sum << std::endl;
return 0;
}

This solution contains 10 empty lines, 17 comments and 6 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 | ./35`

Output:

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

__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 **0.16** seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.

Peak memory usage was about 6 MByte.

(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/euler035

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=35 - **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-35-circular-primes/ (written by Kristian Edlund)

Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p035.java (written by Nayuki)

Mathematica: github.com/nayuki/Project-Euler-solutions/blob/master/mathematica/p035.mathematica (written by Nayuki)

C: github.com/eagletmt/project-euler-c/blob/master/30-39/problem35.c (written by eagletmt)

Javascript: github.com/dsernst/ProjectEuler/blob/master/35 Circular primes.js (written by David Ernst)

Scala: github.com/samskivert/euler-scala/blob/master/Euler035.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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<< problem 34 - Digit factorials | Double-base palindromes - problem 36 >> |