Problem 131: Prime cube partnership

(see projecteuler.net/problem=131)

There are some prime values, p, for which there exists a positive integer, n, such that the expression n^3 + n^2 p is a perfect cube.

For example, when p = 19, 8^3 + 8^2 * 19 = 123.

What is perhaps most surprising is that for each prime with this property the value of n is unique, and there are only four such primes below one-hundred.

How many primes below one million have this remarkable property?

My Algorithm

The equation n^3 + n^2 p = k^3 can be rewritten:
n^3 (1 + dfrac{p}{n}) = k^3

Which becomes:
n * sqrt[3]{1 + dfrac{p}{n}} = k

And:
n * sqrt[3]{dfrac{n+p}{n}} = k

The only way that n and k are integers is when the cube root is rational:
sqrt[3]{dfrac{n+p}{n}} = sqrt[3]{dfrac{a^3}{b^3}} = dfrac{a}{b}

Then the equation would be:
n * dfrac{a}{b} = k

The new variables a and b are:
a^3 = n+p
b^3 = n

Solving for p:
p = a^3 - n = a^3 - b^3

The binomial expansion a^3 - b^3 = (a - b)(a^2 + ab + b^2) tells us:
p = (a - b)(a^2 + ab + b^2)

All values a, b and p must be integers. Moreover, p must be a prime. Remember: a prime number can only be factorized into two numbers: 1 and itself.
Obviously a - b < a^2 + ab + b^2 (for positive values of a and b). Then it follows for the factors 1 and p:
1 = a - b
p = a^2 + ab + b^2

The first equation is interesting: a and b are consecutive numbers:
a = b + 1

Which can be inserted in the second equation:
p = a^2 + a(a+1) + (a+1)^2
p = a^2 + a^2 + a + a^2 + 2a + 1
p = 3a^2 + 3a + 1

Whenever p = 3a^2 + 3a + 1 is prime, then a valid value was found.
I didn't see it at first, but (a + 1)^3 - a^3 = 3a^2 + 3a + 1, too → p is the difference of two consecutive cubes.

This time I use my Wheel-based primality test (because it's the fastest and I haven't used it in a long time).
It's part of my toolbox.

Modifications by HackerRank

Hackerrank has a huge amount of input values: I have to split my program into two parts. First, I find all solutions up to a limit,
then I scan those solutions and print the result.

This time all solutions are generated in ascending order. std::lower_bound finds the largest matching prime below the input very fast (binary search) and returns its position.
For example: querying 100 returns position 3 because 61 is the largest prime below 100. std::distance returns 4 because index counting starts at zero.

My primality test is too slow to handle all primes below 25 * 10^12. About 10^12 is the most I can process in two seconds.

Note

I increased limit to 10^8 for the live test. You only need limit = 1000000 for the original problem.

Interactive test

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

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 100" | ./131

Output:

(please click 'Go !')

Note: the original problem's input 1000000 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 <algorithm>
 
// wheel-based prime test from my toolbox
bool isPrime(unsigned int x)
{
// prime test for 2, 3 and 5
if (x % 2 == 0 || x % 3 == 0 || x % 5 == 0)
return x == 2 || x == 3 || x == 5;
 
// wheel with size 30 (=2*3*5):
// test against 30m+1, 30m+7, 30m+11, 30m+13, 30m+17, 30m+19, 30m+23, 30m+29
// their deltas/increments are:
const unsigned char Delta[] = { 6, 4, 2, 4, 2, 4, 6, 2 };
// start with 7, which is 30*0+7
unsigned int i = 7;
// 7 belongs to the second test group
unsigned int pos = 1;
 
// check numbers up to sqrt(x)
while (i*i <= x)
{
// not prime ?
if (x % i == 0)
return false;
 
// skip forward to next test divisor
i += Delta[pos];
// next delta/increment
pos = (pos + 1) & 7;
}
 
// passed all tests, must be a prime number
return x > 1;
}
 
int main()
{
// find all matching primes up to this limit
unsigned int limit = 100000000;
 
// store all matching primes
std::vector<unsigned int> matches;
 
// start with a=1
unsigned int a = 1;
while (true)
{
// (a+1)^3 - a^3 = 3a^2 + 3a + 1
auto p = 3*a*a + 3*a + 1;
// too big ?
if (p >= limit)
break;
 
// found one more prime ?
if (isPrime(p))
matches.push_back(p);
 
// keep going ...
a++;
}
 
// process STDIN
unsigned int tests = 1;
std::cin >> tests;
while (tests--)
{
std::cin >> limit;
// find highest position located before the limit (matches are sorted, should use binary search)
auto lower = std::lower_bound(matches.begin(), matches.end(), limit);
// count number of primes
auto result = std::distance(matches.begin(), lower);
std::cout << result << std::endl;
}
 
return 0;
}

This solution contains 12 empty lines, 22 comments and 3 preprocessor commands.

Benchmark

The correct solution to the original Project Euler problem was found in 0.01 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

June 27, 2017 submitted solution
June 27, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler131

My code solves 4 out of 7 test cases (score: 50%)

I failed 0 test cases due to wrong answers and 3 because of timeouts

Difficulty

40% Project Euler ranks this problem at 40% (out of 100%).

Hackerrank describes this problem as hard.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

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
  the flashing problem is the one I solved most recently

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The 289 solved problems (level 11) had an average difficulty of 32.1% at Project Euler and
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.

more about me can be found on my homepage, especially in my coding blog.
some names mentioned on this site may be trademarks of their respective owners.
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