# Problem 146: Numbers for which no three consecutive digits have a sum greater than a given value

The smallest positive integer n for which the numbers n^2+1, n^2+3, n^2+7, n^2+9, n^2+13, and n^2+27 are consecutive primes is 10.
The sum of all such integers n below one-million is 1242490.

What is the sum of all such integers n below 150 million?

# My Algorithm

A primitive approach is:
if ( isPrime(n*n + 1) &&
 isPrime(n*n + 3) &&
 !isPrime(n*n + 5) &&
 isPrime(n*n + 7) &&
 isPrime(n*n + 9) &&
 !isPrime(n*n + 11) &&
 isPrime(n*n + 13) &&
 !isPrime(n*n + 15) &&
 !isPrime(n*n + 17) &&
 !isPrime(n*n + 19) &&
 !isPrime(n*n + 21) &&
 !isPrime(n*n + 23) &&
 !isPrime(n*n + 25) &&
 isPrime(n*n + 27))

My Miller-Rabin primality tests easily finds the correct answer - but is much too slow (it takes several minutes).
That primality test was more or less copied from my toolbox. I only added an optional parameter fastCheckAgainstSmallPrimes (see more on that later).

Most of the code in main tries is about speeding up the whole algorithm.

n^2+1 can only be prime if n^2 is an even number (technically that's not true for n=1 but we know that n >= 10).

Another significant insight is that 5 is the only prime where the last digit is 5.
In order to ensure that n^2+1, n^2+3, n^2+7 and n^2+9 are prime (remember: n^2 is even), the last digit of n^2 must be a zero.
And if the last digit of n^2 is zero, then the last digit of n must be zero, too.
Therefore I only need to look at all n which are multiples of 10 (increment = 10).

My default implementation of isPrime performs a trial division against small primes because that's usually much faster than
the full-blown Miller-Rabin test and already eliminates many candidates.
However, since I have to check many numbers simultaneously, I made that trial division optional and perform it for all six potential primes at once.
Only if all pass that test, then I start the Miller-Rabin test.

I have to admit that initially my trial division only considered prime numbers up to 17. On the www.mathblog.dk blog I saw that using many more primes can be faster.
That "trick" made my program about 10x faster because isPrime is called only 11552 times instead of more than 650000 times.

## Modifications by HackerRank

The "good" offsets 1, 3, 7, 9, 13, 27 are replaced by arbitrary six numbers (but each is smaller than 40).
It took me some time to realize that these offsets can be either all even, too.
The search space is reduced from 150 million to 10 million, too.

I have no idea why one Hackerrank test case fails.

# 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 1000000 | ./146

Output:

Note: the original problem's input 150000000 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. Or just jump to my GitHub repository.

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 <vector>

#define ORIGINAL

// 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;
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;
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, bool fastCheckAgainstSmallPrimes = true)
{
// IMPORTANT: requires mulmod(a, b, modulo) and powmod(base, exponent, modulo)

// 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
if (fastCheckAgainstSmallPrimes)
{
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 = mulmod(x, x, 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;
}

int main()
{
// generate all primes up to 500
std::vector<unsigned int> primes = { 2 };
for (unsigned int i = 3; i < 500; 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);
}

unsigned int tests = 1;
#ifndef ORIGINAL
std::cin >> tests;
#endif
while (tests--)
{
// largest number to be considered
unsigned int limit = 150*1000000;
std::cin >> limit;

// always six consective increments where n^2 + good[i] must be prime
std::vector<unsigned int> good = { 1, 3, 7, 9, 13, 27 }; // will be overwritten by Hackerrank input
#ifndef ORIGINAL
for (auto& x : good)
std::cin >> x;
#endif

bool sameParity = true;
unsigned int parity = good.front() % 2;
for (auto x : good)
sameParity &= x % 2 == parity;
if (!sameParity)
{
std::cout << "0" << std::endl;
continue;
}

// collect all numbers between 1 and the largest element of good which are not part of good
// => n^2 + bad[i] must not be prime
for (unsigned int i = parity; i < good.back(); i += 2)
{
bool isGood = false;
for (auto x : good)
isGood |= x == i;

if (!isGood)
}

// performance tweak: if good[] contains 1, 3, 7 and 9 then n must be a multiple of 10
unsigned int increment = 2;
unsigned int start     = 1 - parity;
increment = 10;

// sum of all matching numbers
unsigned long long sum = 0;
for (unsigned int n = start; n < limit; n += increment)
{
auto square = n * (unsigned long long)n;

#ifdef ORIGINAL
// n^2 + 3, n^2 + 7 and n^2 + 13 must be primes
// => n^2 must not be divisible by 3, 7 or 13
if (square % 3 == 0 || square % 7 == 0 || square % 13 == 0)
continue;
#endif

// check all numbers against small primes
bool ok = true;
for (auto p : primes)
{
for (auto check : good)
{
auto current = square + check;
// not prime if remainder is zero
if (current != p && current % p == 0)
{
// note: for n=10, current can be part of primes[],
// that's why I check "current != p" to prevent false negatives
ok = false;
break;
}
}

// failed (= not prime) ?
if (!ok)
break;
}

// use slower Miller-Rabin if still ok
for (auto x : good)
if (ok)
ok &=  isPrime(square + x, false);
if (ok)
ok &= !isPrime(square + x, true);

// passed all tests ?
if (ok)
sum += n;
}

// display result
std::cout << sum << std::endl;
}

return 0;
}


This solution contains 42 empty lines, 60 comments and 11 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

June 16, 2017 submitted solution

# Hackerrank

My code solves 15 out of 26 test cases (score: 56%)

I failed 3 test cases due to wrong answers and 8 because of timeouts

# Difficulty

Project Euler ranks this problem at 50% (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 [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.

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.
thanks to the KaTeX team for their great typesetting library !