<< problem 230 - Fibonacci Words The Race - problem 232 >>

# Problem 231: The prime factorisation of binomial coefficients

The binomial coefficient ^10 C_3 = 120.
120 = 23 * 3 * 5 = 2 * 2 * 2 * 3 * 5, and 2 + 2 + 2 + 3 + 5 = 14.
So the sum of the terms in the prime factorisation of ^10 C_3 is 14.

Find the sum of the terms in the prime factorisation of ^20000000 C_15000000.

# My Algorithm

Aside from a standard prime sieve, the main work is done in add(n):

The binomial coefficient is: ^n{C_k} = {{n}choose{k}} = dfrac{n!}{(n-k)! k!}
where the factorial is n! = 1 * 2 * 3 * 4 * ... * n, e.g. 10! = 1 * 2 * 3 * 4 * ... * 10.

However, the problem statement doesn't ask for the factorial but the sum of prime factors of all elements in the factorial sequence.
That means add(10!) = 1 + 2 + 3 + (2 + 2) + 5 + (2 + 3) + 7 + (2 + 2 + 2) + (3 + 3) + (2 + 5) = 45.

You can see that every second number contains prime factor 2, every third prime factor 3, every fifth prime factor 5 ... and so on.
There are \lfloor frac{10}{2} \rfloor = 5 elements which contain prime factor 2 at least once. That's a sum of 2 * 5 = 10.
There are \lfloor frac{10}{2^2} \rfloor = 2 elements which contain prime factor 2 at least twice. That's a sum of 2 * 2 = 4 on top.
There are \lfloor frac{10}{2^3} \rfloor = 1 elements which contain prime factor 2 at least three times. That's a sum of 2 * 1 = 2 on top.
The sum of all prime factors 2 is 10 + 4 + 2 = 16.

Following the same logic, but this time prime factor 3:
There are \lfloor frac{10}{3} \rfloor = 3 elements which contain prime factor 3 at least once. That's a sum of 3 * 3 = 9.
There are \lfloor frac{10}{3^2} \rfloor = 1 elements which contain prime factor 3 at least twice. That's a sum of 3 * 1 = 3.
The sum of all prime factors 3 is 9 + 3 = 12.

Now the algorithm becomes apparent:

• iterate over all potential prime factors p
• compute how many numbers contain p, add p * count to the sum
• compute how many numbers contain p^2, add p * count to the sum
• compute how many numbers contain p^3, add p * count to the sum
• ...

## Alternative Approaches

My first approach was to generate a huge look-up table:

• if a number i is a prime number, set sum[i] = i
• then for each number i, multiply it by all primes p and set sum[i * p] = sum[i] + p
• add all sums[] up to i, that's simply sums[i] += sums[i - 1] because all i are processed in increasing order
That way more than 160 MByte RAM were used - which would have been a new record for my Project Euler solutions.
I felt very uncomfortable of that memory consumption for such a "small" problem. And it was about 10x slower.
So I spent a whole commuter ride thinking about the problem and came up with the current solution ...
Note: read access to that precomputed table is extremely fast ! If you need to query heaps of factorial prime sums then
my first approach might be an actually viable solution.

# 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):

This is equivalent to
echo "10 3" | ./231

Output:

Note: the original problem's input 20000000 15000000 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 <vector>

// odd prime numbers are marked as "true" in a bitvector
std::vector<bool> sieve;
// collection of all primes from the sieve (I do this because it's faster)
std::vector<unsigned int> primes = { 2 };

// 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;

// 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;
}
}
}

// ----- here the main algorithm starts -----

// return sum of all prime factors of n!
unsigned long long add(unsigned int n)
{
unsigned long long sum = 0;
for (auto p : primes)
{
// prime too large ? => done
if (p > n)
return sum;

unsigned long long multipleP = p;
// initial count
unsigned long long count = n / multipleP;
do
{
sum += p * count;

// increase exponent by one, that means p^i => p^(i+1)
multipleP *= p;
// update count for next iteration
count = n / multipleP;
} while (count > 0); // multipleP > n
}
return sum;
}

// ----- below is my first approach -----

// my first attempt: compute a huge lookup table where
// (code not used anymore)
std::vector<unsigned long long> sums;
void generateTable(unsigned int limit)
{
sums.resize(limit + 1, 0);
for (unsigned int i = 2; i < sums.size(); i++)
{
// prime number ? its only prime factor is the number itself
if (sums[i] == 0)
sums[i] = i;

// multiply with all prime numbers
for (auto p : primes)
{
// too large ?
if (i * p >= sums.size())
break;

sums[i * p] = sums[i] + p;
}

// add all prime factors of (i-1)! to i
// => that's the sum of prime factors of i!
sums[i] += sums[i - 1];
}
}

int main()
{
// read input values n >= k
unsigned int n = 20000000;
unsigned int k = 15000000;
std::cin >> n >> k;

// generate all prime numbers
fillSieve(n);

// extract prime numbers
for (unsigned int i = 3; i <= n; i += 2)
if (isPrime(i))
primes.push_back(i);

// older, slower algorithm
//generateTable(n);
//std::cout << sums[n] - (sums[n - k] + sums[k]) << std::endl;

return 0;
}


This solution contains 22 empty lines, 35 comments and 3 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.14 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 13 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

June 28, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 40% (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 230 - Fibonacci Words The Race - problem 232 >>
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