<< problem 510 - Tangent Circles | First Sort I - problem 523 >> |

# Problem 518: Prime triples and geometric sequences

(see projecteuler.net/problem=518)

Let S(n) = sum{a+b+c} over all triples (a,b,c) such that:

- a, b, and c are prime numbers.

- a < b < c < n.

- a+1, b+1, and c+1 form a geometric sequence.

For example, S(100) = 1035 with the following triples:

(2, 5, 11), (2, 11, 47), (5, 11, 23), (5, 17, 53), (7, 11, 17),

(7, 23, 71), (11, 23, 47), (17, 23, 31), (17, 41, 97), (31, 47, 71), (71, 83, 97)

Find S(10^8).

# My Algorithm

This problem is somehow similar to problem 141:

I learnt from the 141 forum that a geometric progression can be represented as kyy, kxy, kxx where x > y and gcd(x, y) = 1

I wrote three nested loops for k, x and y which check whether kxx - 1, kxy - 1 and kyy - 1 are prime numbers.

My code contains a few optimizations:

- the loop nesting order x, k, y is roughly 30% faster than the more obvious x, y, k

- I deferred the gcd() test because of its slow modulo computations

- ... but early on do a quick bit test whether x and y are both even → because that implies gcd(x, y) > 1

## Note

You will find a little "debugging helper", too: I implemented a simple brute-force algorithm (see `bruteForce`

) which helped me

to locate two bugs I had in my first version of `count`

(which is the final/fast version).

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

This is equivalent to`echo 100 | ./518`

Output:

*Note:* the original problem's input `100000000`

__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.

#include <iostream>
#include <vector>
#include <cmath>
// ---------- standard prime sieve from my toolbox ----------

// 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;
// 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) + 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;
}
}
}
// greatest common divisor

template <typename T>
T gcd(T a, T b)
{
while (a != 0)
{
T c = a;
a = b % a;
b = c;
}
return b;
}
// ---------- problem specific code ----------

// slow approach

unsigned long long bruteForce(unsigned int limit)
{
unsigned long long sum = 0;
// look for geometric progression a+1, b+1, c+1 where all three are prime numbers
// iterate over all prime numbers a and c where a < c
for (unsigned int a = 2; a < limit; a++)
{
if (!isPrime(a))
continue;
for (auto c = a + 4; c < limit; c++)
{
if (!isPrime(c))
continue;
// (b+1)^2 = (a+1) * (c+1)
auto b2 = (a + 1) * (c + 1);
unsigned int b = sqrt(b2);
if (b * b != b2)
continue;
b--;
// of course: b must be prime, too
if (!isPrime(b))
continue;
// found one more solution
std::cout << a << " " << b << " " << c << std::endl;
sum += a + b + c;
}
}
return sum;
}
// find triples k*x*x, k*x*y, k*y*y whose predecessors are prime

unsigned long long count(unsigned int limit)
{
unsigned long long sum = 0;
// I tried a few variations and the following order of loops turned out to be the fastest:
// x => k => y
// the more obvious order x => y => k is about 30% slower on my computer
for (unsigned int x = 2; x*x < limit; x++)
for (unsigned int k = 1; k*x*x < limit; k++)
{
auto a = k*x*x - 1;
if (!isPrime(a))
continue;
for (unsigned int y = 1; y < x; y++)
{
// performance tweak: if both x and y are even then the gcd() test will fail
if ((x & 1) == 0 && (y & 1) == 0)
y++;
auto b = k*x*y - 1;
auto c = k*y*y - 1;
if (!isPrime(b))
continue;
if (!isPrime(c))
continue;
// avoid duplicate solutions
if (gcd(x, y) > 1)
continue;
// note: I delayed this gcd() for performance reasons
// it's more "expensive" to check right after the beginning of the y-loop
// found one more solution
sum += a + b + c;
}
}
return sum;
}
int main()
{
unsigned int limit = 100000000;
std::cin >> limit;
fillSieve(limit);
//std::cout << bruteForce(limit) << std::endl;
std::cout << count(limit) << std::endl;
return 0;
}

This solution contains 26 empty lines, 30 comments and 3 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.9 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.

Peak memory usage was about 8 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

September 2, 2017 submitted solution

September 2, 2017 added comments

# Difficulty

Project Euler ranks this problem at **20%** (out of 100%).

# Links

projecteuler.net/thread=518 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

# 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 |

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I scored 13,386 points (out of 15600 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.

# Copyright

I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.

All of my solutions can be used for any purpose and I am in no way liable for any damages caused.

You can even remove my name and claim it's yours. But then you shall burn in hell.

The problems and most of the problems' images were created by Project Euler.

Thanks for all their endless effort !!!

<< problem 510 - Tangent Circles | First Sort I - problem 523 >> |