<< 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
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
hasthreePlus[] = true
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 toecho "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;
// 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 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
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)
Code in various languages:
C++ github.com/Meng-Gen/ProjectEuler/blob/master/196.cc (written by Meng-Gen Tsai)
C++ github.com/roosephu/project-euler/blob/master/196.cpp (written by Yuping Luo)
C github.com/LaurentMazare/ProjectEuler/blob/master/e196.c (written by Laurent Mazare)
Java github.com/HaochenLiu/My-Project-Euler/blob/master/196.java (written by Haochen Liu)
Java github.com/thrap/project-euler/blob/master/src/Java/Problem196.java (written by Magnus Solheim Thrap)
Go github.com/frrad/project-euler/blob/master/golang/Problem196.go (written by Frederick Robinson)
Mathematica github.com/steve98654/ProjectEuler/blob/master/196.nb
Those links are just an unordered selection of source code I found with a semi-automatic search script on Google/Bing/GitHub/whatever.
You will probably stumble upon better solutions when searching on your own.
Maybe not all linked resources produce the correct result and/or exceed time/memory limits.
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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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.
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 !!!
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