<< problem 291 - Panaitopol Primes | Zeckendorf Representation - problem 297 >> |
Problem 293: Pseudo-Fortunate Numbers
(see projecteuler.net/problem=293)
An even positive integer N will be called admissible, if it is a power of 2 or its distinct prime factors are consecutive primes.
The first twelve admissible numbers are 2,4,6,8,12,16,18,24,30,32,36,48.
If N is admissible, the smallest integer M > 1 such that N+M is prime, will be called the pseudo-Fortunate number for N.
For example, N=630 is admissible since it is even and its distinct prime factors are the consecutive primes 2,3,5 and 7.
The next prime number after 631 is 641; hence, the pseudo-Fortunate number for 630 is M=11.
It can also be seen that the pseudo-Fortunate number for 16 is 3.
Find the sum of all distinct pseudo-Fortunate numbers for admissible numbers N less than 10^9.
My Algorithm
The problem description starts with "an even positive integer [...]" and continues with "[...] its distinct prime factors are consecutive primes.".
Therefore any admissible number must be a multiple of 2 and its prime factors must start with 2, 3, 5, 7, ... without any gap.
Its general form is: 2^a * 3^b * 5^c * ... where a exponent is only zero if all following exponents are zero, too.
The highest relevant base (prime) is 23 because 2^1 * 3^1 * 5^1 * 7^1 * 9^1 * 11^1 * 13^1 * 17^1 * 19^1 * 23^1 = 223092870 < 10^9.
Extending the series with 29^1 would exceed 10^9.
In step 1 of my algorithm I create all such admissible numbers up to 10^9 in ascending order:
- put a 2 in the automatically sorted
todo
container, its highest prime factor is 2 (index 0 offactors
) - then repeat until
todo
is empty: - pick the lowest number from
todo
and store it inadmissible
- multiply the same number by all values of
factor[0 ... index+1]
(those are the prime factors) and store them intodo
(if < 10^9)
Step 3 needs these prime numbers to run a primality test on each admissible number's right neighbors.
Admissible numbers are always even, therefore only the numbers +3, +5, +7, +9, ... have to be checked.
Note
There are a few optimization tricks:
- some admissible numbers are smaller than the prime numbers generated in step 2 → simple linear search in
primes
- for this specific range, not all primes in step 2 are actually needed → it's sufficient to have all primes up to sqrt(2.8 * 10^8) → 30% faster
Interactive test
You can submit your own input to my program and it will be instantly processed at my server:
This is equivalent toecho 50 | ./293
Output:
Note: the original problem's input 1000000000
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 <map>
#include <set>
int main()
{
unsigned int limit = 1000000000;
std::cin >> limit;
// ---------- step 1: find all admissible numbers ----------
// generate all numbers 2^a * 3^b * 5^c * 7^d * 11^e ...
const unsigned int NumFactors = 9;
const unsigned int factors[NumFactors] = { 2,3,5,7,11,13,17,19,23 };
// product of all factors is just under 10^9, including the next prime (29) would be bigger than 10^9
// store all admissible numbers
std::vector<unsigned int> admissible;
// [number] => [index of its highest prime factor]
std::map<unsigned int, unsigned char> todo = { { 2, 0 } };
while (!todo.empty())
{
// pick smallest number and remove from to-do list
auto current = *(todo.begin());
todo.erase(todo.begin());
auto number = current.first;
auto maxPrime = current.second;
// add to admissible list
admissible.push_back(number);
// multiply by all already used prime numbers and the next one
for (unsigned char i = 0; i <= maxPrime + 1; i++) // can start with i = maxPrime, too
{
// only up to 9 primes are relevant
if (i == NumFactors)
break;
// put on to-do list if below 10^9
auto next = number * (unsigned long long)factors[i];
if (next < limit)
todo[next] = std::max(maxPrime, i);
}
}
// note: only 6656 admissible numbers
// ---------- step 2: generate prime numbers up to sqrt(10^9) ----------
// find at least one prime factor for each number below 10^9
// => all up to sqrt(10^9) which is about 31622
// and a few more to be on the safe side ... not needed in this case, but safety first :-)
auto primeLimit = limit + 1000;
// but I can cheat and reduce primeLimit as long as I get the correct result
// => makes the program about 30% faster
//primeLimit = 271623361;
std::vector<unsigned int> primes = { 2 };
// simply trial-division prime generator
for (unsigned int i = 3; i*i <= primeLimit; i += 2)
{
bool isPrime = true;
// test against all prime numbers we have so far (in ascending order)
for (auto x : primes)
{
// prime is too large to be a divisor
if (x*x > i)
break;
// divisible => not prime
if (i % x == 0)
{
isPrime = false;
break;
}
}
// yes, we have a prime
if (isPrime)
primes.push_back(i);
}
// only 1911 out of the 3401 primes are actually needed in step 3
// ---------- step 3: look for each admissible number's fortunate number ----------
// std::set automatically filters duplicates
std::set<unsigned int> fortunate;
size_t pos = 0;
for (auto x : admissible)
{
// every admissible number is even and m > 1
// => therefore the first candidate is at m = 3
const unsigned int MinDistance = 3;
// fast check for small primes
if (x <= primes.back() - MinDistance)
{
while (primes[pos] < x + MinDistance)
pos++;
auto distance = primes[pos] - x;
// might be a duplicate but std::set makes sure that all values are unique
fortunate.insert(distance);
continue;
}
// every prime is odd (except for 2)
// => I look at every second number
for (unsigned int m = MinDistance; ; m += 2)
{
auto candidate = x + m;
// is it a prime ?
bool isPrime = true;
// trial division
for (auto p : primes) // actually I could skip p = 2 because all candidates are odd
if (candidate % p == 0)
{
// it's a composite number
isPrime = false;
break;
}
// no, keep going ...
if (!isPrime)
continue;
// yes, a prime
auto distance = candidate - x;
// might be a duplicate but std::set ensures that all values are unique
fortunate.insert(distance);
break;
}
}
// add all fortunate numbers
unsigned int sum = 0;
for (auto x : fortunate)
sum += x;
// display result
std::cout << sum << std::endl;
return 0;
}
This solution contains 26 empty lines, 40 comments and 4 preprocessor commands.
Benchmark
The correct solution to the original Project Euler problem was found in 0.10 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(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
July 26, 2017 submitted solution
July 26, 2017 added comments
Difficulty
Project Euler ranks this problem at 25% (out of 100%).
Links
projecteuler.net/thread=293 - the best forum on the subject (note: you have to submit the correct solution first)
Heatmap
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I scored 13,486 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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