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

# Problem 196: Prime Triplets

Build a triangle from all positive integers in the following way:
1
23
456
78910
1112131415
161718192021
22232425262728
2930313233343536
373839404142434445
46474849505152535455
5657585960616263646566
. . .

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:

Input data (separated by spaces or newlines):
Note: Enter two lines a and b and the program will print S(a)+S(b).

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. Or just jump to my GitHub repository.

       #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;

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
// 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 an 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=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

August 19, 2017 submitted solution

# Difficulty

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

# 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.

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