<< problem 57 - Square root convergents XOR decryption - problem 59 >>

# Problem 58: Spiral primes

Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.

37 36 35 34 33 32 31
38 17 16 15 14 13 30
39 18 5 4 3 12 29
40 19 6 1 2 11 28
41 20 7 8 9 10 27
42 21 22 23 24 25 26
43 44 45 46 47 48 49

It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that
8 out of the 13 numbers lying along both diagonals are prime; that is, a ratio of frac{8}{13} approx 62%.

If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed.
If this process is continued, what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%?

# Algorithm

My solution consists almost entirely of "old code":
- the Miller-Rabin primality test from problem 50 ...
- which in turn requires modular arithmetic from problem 48

That code can be found in my toolbox, too.

## Modifications by HackerRank

My portable mulmod was a little bit too slow for Hackerrank, therefore I added the "GCC"-Hack:
the GCC-extension __int128 simplifies mulmod to a one-line function (which is much faster, too).

# My code

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

       #include <iostream>

// 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)
{
// 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
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;
}

int main()
{
unsigned int percentage = 10;
std::cin >> percentage;

// the lower right diagonal contains only squares
unsigned long long numPrimes  = 0;
unsigned long long sideLength = 1;
unsigned long long diagonals  = 1;
do
{
sideLength += 2;
diagonals  += 4;

unsigned long long lowerRight = sideLength * sideLength;
unsigned long long lowerLeft  = lowerRight - (sideLength - 1);
unsigned long long upperLeft  = lowerLeft  - (sideLength - 1);
unsigned long long upperRight = upperLeft  - (sideLength - 1);

// no need to test lowerRight since it's a square and never prime
if (isPrime(lowerLeft)  != 0)
numPrimes++;
if (isPrime(upperLeft)  != 0)
numPrimes++;
if (isPrime(upperRight) != 0)
numPrimes++;
} while (numPrimes * 100 / diagonals >= percentage);

std::cout << sideLength << std::endl;
return 0;
}


This solution contains 28 empty lines, 40 comments and 3 preprocessor commands.

# 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 60 | ./58

Output:

(please click 'Go !')

(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 0.02 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
(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

February 28, 2017 submitted solution
April 24, 2017 added comments

# Hackerrank

My code solved 9 out of 9 test cases (score: 100%)

# Difficulty

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 never an option.

# Similar problems at Project Euler

Problem 50: Consecutive prime sum
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.

# Links

projecteuler.net/thread=58 - 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-58-primes-diagonals-spiral/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p058.java (written by Nayuki)
Go: github.com/frrad/project-euler/blob/master/golang/Problem058.go (written by Frederick Robinson)
Scala: github.com/samskivert/euler-scala/blob/master/Euler058.scala (written by Michael Bayne)

# 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 already 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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 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200
The 133 solved problems had an average difficulty of 16.9% at Project Euler and I scored 11,174 points (out of 12300) at Hackerrank's Project Euler+.
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