<< problem 59 - XOR decryption Cyclical figurate numbers - problem 61 >>

# Problem 60: Prime pair sets

The primes 3, 7, 109, and 673, are quite remarkable. By taking any two primes and concatenating them in any order the result will always be prime.
For example, taking 7 and 109, both 7109 and 1097 are prime. The sum of these four primes, 792, represents the lowest sum for a set of four primes with this property.

Find the lowest sum for a set of five primes for which any two primes concatenate to produce another prime.

# My Algorithm

Let's skip forward to main: a simple prime sieve finds all primes below 20000 (or whatever the user input is).
Then each prime a is matched against each larger prime b > a: if a merged with b is prime and b merged with a is prime, then a and b are a prime pair.

When looking at my code, a is actually named smallPrime and b is named largePrime.
After calling the function match (see below) all primes pairs of smallPrime are stored in candidates.
Now each element of candidates definitely forms a prime pair with smallPrime.

The original problem asks for a set of five primes, but the Hackerrank problem extends this to 3, 4 and 5 primes.
The functions checkTriple, checkQuadruple and checkQuintuple follow the same idea:
match each candidate against each other using nested loops. If 3, 4 or 5 primes match simultanueously, then insert the sum into sums
(which is a simple container keeping track of all results).

At the end of the problem, sums is sorted and displayed.

## Helper functions

merge concatenate two numbers a and b: the smallest shift = 10^x is found such that shift > b.
Then the result ist a * shift + b. You can do the same pretty easy with strings but that's much slower.
match checks whether the concatenated numbers ab and ba are prime (using the Miller-Rabin test I copied from problem 50).

## Modifications by HackerRank

As explained before, the prime sets may contain either (at least) 3, 4 or 5 primes.
Note: do not eliminate duplicate sums from the output. This was actually pretty unclear to me and took me quite some time to figure out.

## Note

That's the longest code I had to write for a Project Euler problem so far !
(at least I could copy about 50% from previous problems ...)

# 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 "700 4" | ./60

Output:

(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>

// will be the result
std::vector<unsigned int> sums;

// return (a*b) % modulo
unsigned long long mulmod(unsigned long long a, unsigned long long b, unsigned long long modulo)
{
#ifdef __GNUC__
// use GCC's optimized 128 bit code
return ((unsigned __int128)a * b) % modulo;
#endif

// (a * b) % modulo = (a % modulo) * (b % modulo) % modulo
a %= modulo;
b %= modulo;

// fast path
if (a <= 0xFFFFFFF && b <= 0xFFFFFFF)
return (a * b) % modulo;

// we might encounter overflows (slow path)
// the number of loops depends on b, therefore try to minimize b
if (b > a)
std::swap(a, b);

// bitwise multiplication
unsigned long long result = 0;
while (a > 0 && b > 0)
{
// b is odd ? a*b = a + a*(b-1)
if (b & 1)
{
result += a;
if (result >= modulo)
result -= modulo;
// skip b-- because the bit-shift at the end will remove the lowest bit anyway
}

// b is even ? a*b = (2*a)*(b/2)
a <<= 1;
if (a >= modulo)
a -= modulo;

// next bit
b >>= 1;
}

return result;
}

// return (base^exponent) % modulo
unsigned long long powmod(unsigned long long base, unsigned long long exponent, unsigned long long modulo)
{
unsigned long long result = 1;
while (exponent > 0)
{
// fast exponentation:
// odd exponent ? a^b = a*a^(b-1)
if (exponent & 1)
result = mulmod(result, base, modulo);

// even exponent ? a^b = (a*a)^(b/2)
base = mulmod(base, base, modulo);
exponent >>= 1;
}
return result;
}

// Miller-Rabin-test
bool isPrime(unsigned long long p)
{
// some code from             https://ronzii.wordpress.com/2012/03/04/miller-rabin-primality-test/
// with optimizations from    http://ceur-ws.org/Vol-1326/020-Forisek.pdf
// good bases can be found at http://miller-rabin.appspot.com/

// trivial cases
const unsigned int bitmaskPrimes2to31 = (1 <<  2) | (1 <<  3) | (1 <<  5) | (1 <<  7) |
(1 << 11) | (1 << 13) | (1 << 17) | (1 << 19) |
(1 << 23) | (1 << 29); // = 0x208A28Ac
if (p < 31)
return (bitmaskPrimes2to31 & (1 << p)) != 0;

if (p %  2 == 0 || p %  3 == 0 || p %  5 == 0 || p % 7 == 0 || // divisible by a small prime
p % 11 == 0 || p % 13 == 0 || p % 17 == 0)
return false;

if (p < 17*19) // we filtered all composite numbers < 17*19, all others below 17*19 must be prime
return true;

// test p against those numbers ("witnesses")
// good bases can be found at http://miller-rabin.appspot.com/
const unsigned int STOP = 0;
const unsigned int TestAgainst1[] = { 377687, STOP };
const unsigned int TestAgainst2[] = { 31, 73, STOP };
const unsigned int TestAgainst3[] = { 2, 7, 61, STOP };
// first three sequences are good up to 2^32
const unsigned int TestAgainst4[] = { 2, 13, 23, 1662803, STOP };
const unsigned int TestAgainst7[] = { 2, 325, 9375, 28178, 450775, 9780504, 1795265022, STOP };

// good up to 2^64
const unsigned int* testAgainst = TestAgainst7;
// use less tests if feasible
if (p < 5329)
testAgainst = TestAgainst1;
else if (p < 9080191)
testAgainst = TestAgainst2;
else if (p < 4759123141ULL)
testAgainst = TestAgainst3;
else if (p < 1122004669633ULL)
testAgainst = TestAgainst4;

// find p - 1 = d * 2^j
auto d = p - 1;
d >>= 1;
unsigned int shift = 0;
while ((d & 1) == 0)
{
shift++;
d >>= 1;
}

// test p against all bases
do
{
auto x = powmod(*testAgainst++, d, p);
// is test^d % p == 1 or -1 ?
if (x == 1 || x == p - 1)
continue;

// now either prime or a strong pseudo-prime
// check test^(d*2^r) for 0 <= r < shift
bool maybePrime = false;
for (unsigned int r = 0; r < shift; r++)
{
// x = x^2 % p
// (initial x was test^d)
x = powmod(x, 2, p);
// x % p == 1 => not prime
if (x == 1)
return false;

// x % p == -1 => prime or an even stronger pseudo-prime
if (x == p - 1)
{
// next iteration
maybePrime = true;
break;
}
}

// not prime
if (!maybePrime)
return false;
} while (*testAgainst != STOP);

// prime
return true;
}

// merge two numbers, "append their digits"
unsigned long long merge(unsigned long long a, unsigned long long b)
{
// merge(12, 34) => 1234
unsigned long long shift = 10;
while (shift <= b)
shift *= 10;
return a * shift + b;
}

// true if a and b can be merged in any way and the result is still a prime
bool match(unsigned long long a, unsigned long long b)
{
return isPrime(merge(a, b)) && isPrime(merge(b, a));
}

// find all triplets:
// all numbers in "candidates" already match with first
// now we have to check every numbers in "candidates" against each other
void checkTriple(unsigned int first, const std::vector<unsigned int>& candidates)
{
for (size_t index2 = 0; index2 < candidates.size(); index2++)
for (size_t index3 = index2 + 1; index3 < candidates.size(); index3++)
// match ?
if (match(candidates[index2], candidates[index3]))
{
// append sum to result set
auto sum = first + candidates[index2] + candidates[index3];
sums.push_back(sum);
}
}

// find all quadruples, same idea as checkTriple, but this time 1+3 number must match
void checkQuadruple(unsigned int first, const std::vector<unsigned int>& candidates)
{
for (size_t index2 = 0; index2 < candidates.size(); index2++)
for (size_t index3 = index2 + 1; index3 < candidates.size(); index3++)
{
// not even a triple ?
if (!match(candidates[index2], candidates[index3]))
continue;

// match fourth number
for (size_t index4 = index3 + 1; index4 < candidates.size(); index4++)
if (match(candidates[index2], candidates[index4]) &&
match(candidates[index3], candidates[index4]))
{
// append sum to result set
auto sum = first + candidates[index2] + candidates[index3] + candidates[index4];
sums.push_back(sum);
}
}
}

// find all quintuples, same idea as above, just nested one level deeper
void checkQuintuple(unsigned int first, const std::vector<unsigned int>& candidates)
{
for (size_t index2 = 0; index2 < candidates.size(); index2++)
for (size_t index3 = index2 + 1; index3 < candidates.size(); index3++)
{
// not even a triple ?
if (!match(candidates[index2], candidates[index3]))
continue;

for (size_t index4 = index3 + 1; index4 < candidates.size(); index4++)
{
// not even a quadruple ?
if (!match(candidates[index2], candidates[index4]) ||
!match(candidates[index3], candidates[index4]))
continue;

// match fifth number
for (size_t index5 = index4 + 1; index5 < candidates.size(); index5++)
if (match(candidates[index2], candidates[index5]) &&
match(candidates[index3], candidates[index5]) &&
match(candidates[index4], candidates[index5]))
{
// append sum to result set
auto sum = first + candidates[index2] + candidates[index3] +
candidates[index4] + candidates[index5];
sums.push_back(sum);
}
}
}
}

int main()
{
// all primes are below this threshold
unsigned int maxPrime = 20000;
// number of primes in a group
unsigned int size     =     5;
std::cin >> maxPrime >> size;

// all primes that can be part of a result set
std::vector<unsigned int> primes;
// find all primes up to n (20000 at most)
// note: 2 is deliberately excluded because "any combination must be a prime"
// => but any number where we concat 2 to the end can't be prime
for (unsigned int i = 3; i < maxPrime; i += 2)
{
bool isPrime = true;
for (auto p : primes)
{
if (p*p > i)
break;
if (i % p == 0)
{
isPrime = false;
break;
}
}

if (isPrime)
primes.push_back(i);
}

for (size_t i = 0; i < primes.size(); i++)
{
auto smallPrime = primes[i];
// no prime number ends with 5 (except 5 itself) => just a simple performance tweak
if (smallPrime == 5)
continue;

// find all larger primes that can be paired with "smallPrime"
std::vector<unsigned int> candidates;
for (size_t j = i + 1; j < primes.size(); j++)
{
auto largePrime = primes[j];
if (match(smallPrime, largePrime))
candidates.push_back(largePrime);
}

// all other candidates must be "pairable" to each other, too
if (size == 3)
checkTriple(smallPrime, candidates);
else if (size == 4)
else // size == 5
checkQuintuple(smallPrime, candidates);
}

// print all sums in ascending order, some sums may occur multiple times
std::sort(sums.begin(), sums.end());
for (auto s : sums)
std::cout << s << std::endl;
}


This solution contains 38 empty lines, 65 comments and 5 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 1.7 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

March 1, 2017 submitted solution

# Hackerrank

My code solves 15 out of 15 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 20% (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 rarely an option.

# Similar problems at Project Euler

Problem 50: Consecutive prime sum
Problem 58: Spiral primes

Note: I'm not even close to solving all problems at Project Euler. Chances are that similar problems do exist and I just haven't looked at them.

# 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 59 - XOR decryption Cyclical figurate numbers - problem 61 >>
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