<< problem 193 - Squarefree Numbers | Investigating the behaviour of a recursively ... - problem 197 >> |

# Problem 196: Prime Triplets

(see projecteuler.net/problem=196)

Build a triangle from all positive integers in the following way:

1

*2**3*

4*5*6

*7*8910

*11*12*13*1415

16*17*18*19*2021

22*23*2425262728

*29*30*31*3233343536

*37*383940*41*42*43*4445

46*47*4849505152*53*5455

565758*59*60*61*6263646566

. . .

Each positive integer has up to eight neighbours in the triangle.

A set of three primes is called a prime triplet if one of the three primes has the other two as neighbours in the triangle.

For example, in the second row, the prime numbers 2 and 3 are elements of some prime triplet.

If row 8 is considered, it contains two primes which are elements of some prime triplet, i.e. 29 and 31.

If row 9 is considered, it contains only one prime which is an element of some prime triplet: 37.

Define S(n) as the sum of the primes in row n which are elements of any prime triplet.

Then S(8)=60 and S(9)=37.

You are given that S(10000)=950007619.

Find S(5678027) + S(7208785).

# My Algorithm

The last number of each row is a triangular number (en.wikipedia.org/wiki/Triangular number):

`getNumber`

returns the number located in column `x`

and row `y`

based on the formula for triangular numbers.

= T(y-1) + x = dfrac{(y-1)(y-1+1)}{2} + x = dfrac{y(y-1)}{2} + x

A triplet always fits in a 3x3 group. However, the center of those 3x3 doesn't need to be located in row n:

row n can be the top row, the center row or the bottom row of a 3x3 group.

Thus I scan through all 3x3 groups which are centered around row n-1, n and n+1.

`processLine`

creates an array `threePlus[]`

which is true for `x`

if the 3x3 group centered around `x`

contains at least 3 primes.

The next step is to walk through row n and add the current number `x`

to the result if:

- `x`

is a prime number

- any 3x3 group centered in the 3x3 group of `x`

has `threePlus[] = true`

My first idea was to include the Miller-Rabin primality test (from my toolbox) but it was too slow.

Then I wrote a segmented prime sieve similar to what you can find on my website create.stephan-brumme.com/eratosthenes/ (I called it "block-wise" algorithm).

## Note

My segmented sieve is a bitfield of all even numbers and its design follows the standard prime sieve from my toolbox.

# 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 "8 9" | ./196`

Output:

*Note:* the original problem's input `5678027 7208785`

__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 ----------
// note: a small tweak: fillSieve() aborts if sieve[] already has enough values

// 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;
// already existing ?
if (sieve.size() >= half)
return;
// 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;
}
}
}
// ---------- now problem-specific code ----------

// return number at position (x, y) where x <= y

unsigned long long getNumber(unsigned int x, unsigned int y)
{
// the last number in a line is a triangle number
// return x + T(y-1)
return x + y * (y - 1ULL) / 2;
}
std::vector<bool> segment;
unsigned long long segmentStart = 0;
// set segment[x] to true if x+from is prime

void fillSegmentedSieve(unsigned long long from, unsigned long long to)
{
// plain old sieve for all primes up to sqrt(to)
fillSieve(sqrt(to));
// first number covered by the segment
segmentStart = from | 1; // start with an odd number
// size of the segment
auto numValues = to - from + 1;
// assume all numbers are prime
segment.clear();
segment.resize(numValues + 1, true);
// cross off composites
for (unsigned long long p = 3; p*p <= to; p += 2)
if (isPrime(p))
{
// find smallest multiple in the segment
auto smallest = from - (from % p) + p;
// only odd multiples
if (smallest % 2 == 0)
smallest += p;
// walk through all odd multiples
for (size_t i = smallest; i <= to; i += 2*p)
segment[(i - segmentStart) / 2] = false;
}
}
// return true if number at position (x,y) is prime
// check boundaries, too (there parameter x can be negative)

bool isPrimeInSegment(int x, int y)
{
// out of bounds ?
if (x < 1 || x > y)
return false;
// check segmented sieve at that position
auto current = getNumber(x, y);
// reject all even number (except 2)
if (current % 2 == 0)
return current == 2;
// luokup
return segment[(current - segmentStart) / 2];
}
// return sum of all prime triplets in a certain line

unsigned long long processLine(unsigned int line)
{
// need to look two lines up and down
auto sieveFrom = getNumber(1, line - 2);
auto sieveTo = getNumber(1, line + 3) - 1;
// prevent line - 2 from becoming negative and producing strange results
if (line <= 2)
sieveFrom = 1;
// find all primes numbers for those 5 lines
fillSegmentedSieve(sieveFrom, sieveTo);
// find all primes with at least two direct neighbors that are prime, too
std::vector<bool> threePlus(segment.size(), false);
for (unsigned int y = line - 1; y <= line + 1; y++)
for (unsigned int x = 1; x <= y; x++)
{
// current number must be a prime
if (!isPrimeInSegment(x, y))
continue;
// count all primes in the 3x3 neighborhood (one step up,up-right,right,down-right,down, ...)
auto countPrimes = 0; // actually countPrimes is always at least 1 because there must be a prime at deltaX = deltaY = 0
for (int deltaX = -1; deltaX <= +1; deltaX++)
for (int deltaY = -1; deltaY <= +1; deltaY++)
if (countPrimes < 3 && isPrimeInSegment(x + deltaX, y + deltaY))
countPrimes++;
// at least three primes ?
threePlus[getNumber(x, y) - segmentStart] = (countPrimes >= 3);
}
// now look at the current line and compute sum of all triplets
unsigned long long sum = 0;
for (unsigned int x = 1; x <= line; x++)
{
// current number must be a prime
auto current = getNumber(x, line);
if (!isPrimeInSegment(x, line))
continue;
// look at 3x3 neighborhood whether at least one cell has threePlus[] = true
bool atLeastThree = false;
for (int deltaX = -1; deltaX <= +1; deltaX++)
for (int deltaY = -1; deltaY <= +1; deltaY++)
atLeastThree |= threePlus[getNumber(x + deltaX, line + deltaY) - segmentStart];
// found a triplet ?
if (atLeastThree)
sum += current;
}
return sum;
}
int main()
{
unsigned int one = 5678027;
unsigned int two = 7208785;
std::cin >> one >> two;
// fillSieve can re-use existing data if the second number not bigger than the first
if (one < two)
std::swap(one, two);
std::cout << processLine(one) + processLine(two) << std::endl;
return 0;
}

This solution contains 30 empty lines, 46 comments and 3 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.4 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.

Peak memory usage was about 11 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

August 19, 2017 submitted solution

August 19, 2017 added comments

# Difficulty

Project Euler ranks this problem at **65%** (out of 100%).

# Links

projecteuler.net/thread=196 - **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:*

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gray | problems are already solved but I haven't published my solution yet | |

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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 193 - Squarefree Numbers | Investigating the behaviour of a recursively ... - problem 197 >> |