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?

My 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.

Interactive test

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

Input data (separated by spaces or newlines):

This is equivalent to
echo 100 | ./35

Output:

(please click 'Go !')

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)

My code

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

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 <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.

Benchmark

The correct solution to the original Project Euler problem was found in 0.16 seconds on an 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=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 24, 2017 submitted solution
April 6, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler035

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

Difficulty

5% 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
  the flashing problem is the one I solved most recently

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The 289 solved problems (level 11) had an average difficulty of 32.1% at Project Euler and
I scored 13,486 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.

more about me can be found on my homepage, especially in my coding blog.
some names mentioned on this site may be trademarks of their respective owners.
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