<< problem 117 - Red, green, and blue tiles Digit power sum - problem 119 >>

# Problem 118: Pandigital prime sets

Using all of the digits 1 through 9 and concatenating them freely to form decimal integers, different sets can be formed.
Interestingly with the set {2,5,47,89,631}, all of the elements belonging to it are prime.

How many distinct sets containing each of the digits one through nine exactly once contain only prime elements?

# Algorithm

First, I create a prime sieve for all prime numbers up to 100000000.
My isPrime can handle larger number, too, but has to revert to trial division (which is much, much slower).

The core routine is search which takes a vector digits and looks at all digits starting at position firstPos.
It appends all digits step-by-step to a local variable current and checks whether it is prime.
If yes, then the routine appends that prime number to merged and calls itself recursively.
If there are no more digits left in digits (that means firstPos == digits.size()) then the numbers in merged are a valid solution.

It took me some time to figure out that all numbers in merged must be in ascending order to avoid finding the same solution multiple times.

## Modifications by HackerRank

You are given a certain set of digits and have to find the sum of all solutions.
At the moment I have no idea why some test cases fail.

## Note

100,000,000 seems to be the "sweet spot" for my prime sieve. Higher values increase the time needed to fill the sieve significantly,
while lower values lead to a slower isPrime (because it has to use trial division more often).

# 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 <string>
#include <vector>
#include <algorithm>

#define ORIGINAL

// odd prime numbers are marked as "true" in a bitvector
std::vector<bool> sieve;
// return true, if x is a prime number
bool isPrime(unsigned int x)
{
// handle even numbers
if ((x & 1) == 0)
return x == 2;

// trial division for large numbers
if (x >= sieve.size() * 2)
{
for (unsigned int i = 3; i*i <= x; i += 2)
if (x % i == 0)
return false;
return true;
}

// lookup for odd numbers
return sieve[x >> 1];
}

// find all prime numbers from 2 to size
void fillSieve(unsigned int size)
{
// store only odd numbers
const unsigned int half = size >> 1;

// allocate memory
sieve.resize(half, true);
// 1 is not a prime number
sieve[0] = false;

// process all relevant prime factors
for (unsigned int i = 1; 2 * i*i < half; i++)
// do we have a prime factor ?
if (sieve[i])
{
// mark all its multiples as false
unsigned int current = 3 * i + 1;
while (current < half)
{
sieve[current] = false;
current += 2 * i + 1;
}
}
}

typedef std::vector<unsigned int> Digits;
std::vector<std::vector<unsigned int>> solutions;
void search(const Digits& digits, std::vector<unsigned int>& merged, size_t firstPos = 0)
{
// no more digits left => found a solution
if (firstPos == digits.size())
{
solutions.push_back(merged);
return;
}

// process one more digit at a time
unsigned int current = 0;
while (firstPos < digits.size())
{
// next digit
current *= 10;
current += digits[firstPos++];

// must be larger than its predecessor
if (!merged.empty() && current < merged.back())
continue;

// ... and prime, of course !
if (isPrime(current))
{
merged.push_back(current);
search(digits, merged, firstPos);
merged.pop_back();
}
}
}

int main()
{
// precompute primes (bigger primes are tested using trial division)
fillSieve(100000000);

unsigned int tests = 1;
std::cin >> tests;
while (tests--)
{
std::string strDigits = "123456789";
std::cin >> strDigits;

// convert to a sorted array/vector
Digits digits;
for (auto x : strDigits)
digits.push_back(x - '0');
std::sort(digits.begin(), digits.end());

// discard solutions from previous tests
solutions.clear();
do
{
// simple speed optimization: last digit must be odd
if (digits.back() % 2 == 0)
continue;

// let's go !
std::vector<unsigned int> merged;
search(digits, merged);
} while (std::next_permutation(digits.begin(), digits.end()));

#ifdef ORIGINAL
std::cout << solutions.size() << std::endl;
#else
// compute sum of each solution
std::vector<unsigned long long> sorted;
for (auto merged : solutions)
{
unsigned long long sum = 0;
for (auto x : merged)
sum += x;
sorted.push_back(sum);
}
// sort ascendingly
std::sort(sorted.begin(), sorted.end());

// remove duplicates (needed ?)
//auto garbage = std::unique(sorted.begin(), sorted.end());
//sorted.erase(garbage, sorted.end());

// and print all of them
for (auto x : sorted)
std::cout << x << std::endl;
std::cout << std::endl;
#endif
}

return 0;
}


This solution contains 21 empty lines, 31 comments and 8 preprocessor commands.

# 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 all digits to be used

This is equivalent to
echo "1 12345" | ./118

Output:

Note: the original problem's input 987654321 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 0.21 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 11 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

May 18, 2017 submitted solution

# Hackerrank

My code solves 9 out of 12 test cases (score: 60%)

I failed 3 test cases due to wrong answers and 0 because of timeouts

# Difficulty

Project Euler ranks this problem at 45% (out of 100%).

Hackerrank describes this problem as hard.

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.

projecteuler.net/thread=118 - 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-118-sets-prime-elements/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p118.java (written by Nayuki)

# 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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The 163 solved problems had an average difficulty of 22.2% at Project Euler and I scored 11,907 points (out of 13200) at Hackerrank's Project Euler+.
My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.
 << problem 117 - Red, green, and blue tiles Digit power sum - problem 119 >>
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