<< problem 201 - Subsets with a unique sum Generalised Hamming Numbers - problem 204 >>

Problem 203: Squarefree Binomial Coefficients

The binomial coefficients ^nC_k can be arranged in triangular form, Pascal's triangle, like this:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
.........

It can be seen that the first eight rows of Pascal's triangle contain twelve distinct numbers: 1, 2, 3, 4, 5, 6, 7, 10, 15, 20, 21 and 35.

A positive integer n is called squarefree if no square of a prime divides n. Of the twelve distinct numbers in the first eight rows of Pascal's triangle,
all except 4 and 20 are squarefree. The sum of the distinct squarefree numbers in the first eight rows is 105.

Find the sum of the distinct squarefree numbers in the first 51 rows of Pascal's triangle.

My Algorithm

A number in Pascal's triangle in row n and column k has the value \binom{n}{k}.
There are two ways to compute this value:

1. \binom{n}{k} = frac{n!}{k!(n-k)!}

2. \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}

The first equation tells me that each potential prime factor doesn't exceed 51 because the largest number contained in any of the factorials is 51.
Therefore isSquarefree performs a trial division by p^2 where p are the prime numbers between 2 and 51.
(Actually I was too lazy to include a proper prime sieve and divide by all numbers between 2 and 51. There is no measureable performane loss.)

The second equation is useful to compute a row based on the previous row. This way I avoid potential overflows of large factorials, too.

The triangle is symmetric, therefore I check only the left half. And unique keeps track of all numbers I have seen so far.

Note

My code is far from optimal:

• isSquarefree checks numbers which are not prime, too, which is redundant
• if a number appears multiple times in the triangles, isSquarefree doesn't need to be called

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 the number of rows of the triangle

This is equivalent to
echo 8 | ./203

Output:

(please click 'Go !')

Note: the original problem's input 51 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 <set>

// return true if x cannot be divided by p^2 for all primes 2 <= p <= maxSquare
bool isSquarefree(unsigned long long x, unsigned int maxSquare)
{
// instead of a proper prime sieve, just perform trial division of all numbers
for (unsigned int p = 2; p <= maxSquare; p++)
if (x % (p*p) == 0)
return false;

// yes, squarefree
return true;
}

int main()
{
// size of the triangle
unsigned int numRows = 51;
std::cin >> numRows;

// all squarefree numbers
std::set<unsigned long long> squareFree = { 1 };

// initial row
std::vector<unsigned long long> current = { 1 };
// ... and compute all further rows
for (unsigned int row = 1; row < numRows; row++)
{
// last and first element is always 1
std::vector<unsigned long long> next(current.size() + 1, 1);

// fill remaining cells
for (unsigned int column = 1; column < next.size() - 1; column++)
next[column] = current[column - 1] + current[column];

// symmetric: check only half of the triangle, skip borders, too (always 1)
for (unsigned int i = 1; i <= next.size() / 2; i++)
{
auto x = next[i];
if (isSquarefree(x, numRows))
squareFree.insert(x); // std::set prevents duplicates
}

// next iteration
current = std::move(next);
}

// find sum of all squarefree numbers
unsigned long long sum = 0;
for (auto x : squareFree)
sum += x;

// display result
std::cout << sum << std::endl;
return 0;
}


This solution contains 10 empty lines, 13 comments and 3 preprocessor commands.

Benchmark

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

May 31, 2017 submitted solution

Difficulty

Project Euler ranks this problem at 25% (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.

 << problem 201 - Subsets with a unique sum Generalised Hamming Numbers - problem 204 >>
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