<< problem 49 - Prime permutations Prime digit replacements - problem 51 >>

# Problem 50: Consecutive prime sum

The prime 41 can be written as the sum of six consecutive primes:
41 = 2 + 3 + 5 + 7 + 11 + 13

This is the longest sum of consecutive primes that adds to a prime below one-hundred.
The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.

Which prime, below one-million, can be written as the sum of the most consecutive primes?

# My Algorithm

The basic idea is pretty simple:

• generate a ton of prime numbers p
• for each sum sum_{x=i..j}{p_x} perform a primality test
• print the maximum sum that is prime
Initially I struggled a little bit to find a fast solution (especially fast enough for Hackerrank, where the sum may be up to 10^12).
Then my first brute force code revealed a few observations:
• if the sum of the first n primes is prime, then maybe the sum of the first n+1 primes isn't
• if the sum of the first n primes isn't prime, then maybe the sum of the first n+1 primes is
→ just keep going, no matter how many non-prime sums we have seen, eventually the sum will be prime again

Sometimes starting with the first prime 2 doesn't produce the highest sum. The problems mentions 953 which is 7 + 9 + 11 + ... + 89.
The surprising fact is that all "best" chains below 10^12 start with at most 131 (!). I can't explain why - that's just what I saw in my output !

My code generates prime numbers on-demand. Whenever the main loop runs out of primes, it calls morePrimes(x) which ensures that primes will contain at x prime numbers.
On top of that, primeSum[i] is the sum of the first i prime numbers (zero-based index), e.g. primeSum[2] = 2+3+5 = 10.

The interesting fact about primeSum is that the sum of the first x prime numbers excluding the initial y primes is primeSum[x] - primeSum[y].
For example, primeSum[23] - primeSum[2] = 963 - 10 = 953, that means there is chain containing 23-2=21 elements with a sum of 953.

A simple loop finds the largest sum which is below the target: primeSum[545] = 997661
If that number isn't prime (997661 = 7 * 359 * 397), then we look at its predecessor primeSum[544] and so on - until the sum is prime.
As I explained earlier, the best chain maybe doesn't start with the first prime.
Therefore we have to check primeSum[545] - primeSum[0] = 997659 as well, then try primeSum[544] - primeSum[0], ...
until we arrive at primeSum[545] - primeSum[31] because primes[31] = 131.

There are simple primality tests for such small number but they all fall apart when the sum is large (such as 10^12 in the Hackerrank version).
Take a look at my toolbox for inspiration.

## Modifications by HackerRank

It took my quite a while to come up with a fast and stable prime test.
Searching on the internet immediately brings up the Miller-Rabin test: en.wikipedia.org/wiki/Miller–Rabin_primality_test

Unfortunately, most C/C++ implementations either can't handle 64 bit numbers properly or are way to complex to fit in a few lines of code.
That's why had to write my own routine (of course inspired by looking at other sources).

Modular arithmetic was already used in problem 48, please see there for an explanation of mulmod and powmod.
My toolbox contains code for a 32 bit Miller-Rabin test where those two functions can be written in a much simpler way.

## Note

I have to admit that the mathematics of the Miller-Rabin test is not easy to understand for a non-mathematican like me:
I couldn't have written my code without these sources of inspiration:

# 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):

This is equivalent to
echo "1 1000" | ./50

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.

       #include <iostream>
#include <vector>

// return (a*b) % modulo
unsigned long long mulmod(unsigned long long a, unsigned long long b, unsigned long long modulo)
{
// fast path
if (a <= 0xFFFFFFF && b <= 0xFFFFFFF)
return (a * b) % modulo;

// we might encounter overflows (slow path)
unsigned long long result = 0;
unsigned long long factor = a % modulo;

// bitwise multiplication
while (b > 0)
{
// b is odd ? a*b = a + a*(b-1)
if (b & 1)
{
result += factor;
if (result >= modulo)
result %= modulo;
}

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

std::vector<unsigned int>       primes;
std::vector<unsigned long long> primeSum;

// make sure that at least "num" primes are available in "primes"
void morePrimes(unsigned int num)
{
if (primes.empty())
{
primes  .reserve(400000);
primeSum.reserve(400000);

primes.push_back(2);
primes.push_back(3);

primeSum.push_back(2);
}

for (auto i = primes.back() + 2; primes.size() <= num; i += 2)
{
bool isPrime = true;
// test against all prime numbers we have so far (in ascending order)
for (auto x : primes)
{
// prime is too large to be a divisor
if (x*x > i)
break;

// divisible => not prime
if (i % x == 0)
{
isPrime = false;
break;
}
}

// yes, we have a prime
if (isPrime)
primes.push_back(i);
}

for (auto i = primeSum.size(); i < primes.size(); i++)
primeSum.push_back(primeSum.back() + primes[i]);
}

int main()
{
// generate some primes
const unsigned int PrimesPerBatch = 10000;
morePrimes(PrimesPerBatch);

unsigned int tests;
std::cin >> tests;
while (tests--)
{
unsigned long long last = 1000000;
std::cin >> last;

unsigned long long best = 2; // highest prime sum
unsigned int maxLength  = 0; // longest chain (must add plus one)

// a brute-force search showed that all "good" chains start at 2..131
unsigned int start = 0; // primes[0] = 2
while (primes[start] <= 131 && primes[start] <= last)
{
unsigned long long subtract = 0;
if (start > 0)
subtract = primeSum[start - 1];

unsigned int pos = start + maxLength;
// find shortest chain whose sum exceeds the limit
while (primeSum[pos] - subtract <= last)
{
pos++;
// running out of prime numbers ? add more !
if (pos + 100 >= primes.size()) // plus 100 is probably too cautious
morePrimes(primes.size() + PrimesPerBatch);
}
pos--;

// chop off one prime number until the sum is prime, too
while (pos - start > maxLength)
{
unsigned long long sum = primeSum[pos] - subtract;
// yes, we have a good candidate (maybe better ones for other values of "start", though)
if (isPrime(sum))
{
maxLength = pos - start;
best = sum;
break;
}

pos--;
}

start++;
}

// if sum is > 0 then "length" didn't count the first element
if (best >= 2)
maxLength++;
std::cout << best << " " << maxLength << std::endl;
}

return 0;
}


This solution contains 38 empty lines, 46 comments and 2 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 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

February 27, 2017 submitted solution

# Hackerrank

My code solves 10 out of 10 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 5% (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 58: Spiral primes
Problem 60: Prime pair sets

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 49 - Prime permutations Prime digit replacements - problem 51 >>
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