<< problem 267 - Billionaire Sum of Squares - problem 273 >>

# Problem 268: Counting numbers with at least four distinct prime factors less than 100

It can be verified that there are 23 positive integers less than 1000 that are divisible by at least four distinct primes less than 100.

Find how many positive integers less than 10^16 are divisible by at least four distinct primes less than 100.

# My Algorithm

There are 25 primes less than 100.
I start a loop running from 0 to 2^25. I multiply all the n-th primes if the n-th bit of the counter (named mask) is set.
If mask has at least four bits set then the product of those primes needs to be further analyzed:

• product can exceed 10^16 and cause strange results, reject all those large products
• there are 10^16 / product numbers divisible by product
I have to be careful to avoid counting the same numbers twice:
for example 2310 is divisible by 2,3,5,7 and 11 (in fact 2*3*5*7*11 = 2310). But 2310 is also divisible by (2,3,5,7), (2,3,5,11), (2,5,7,11) and (3,5,7,11).
That means (2,3,5,7,11) "overlaps" all combinations of its primes where I take one prime away.
The tetrahedral numbers can compute this "overlap" (see en.wikipedia.org/wiki/Tetrahedral_number) together with
the inclusion-exclusion principle (see en.wikipedia.org/wiki/Inclusion–exclusion_principle).

In plain English: I add the count of numbers divisible by 4 primes, subtract the count divisible by 5 primes, add 6s, subtract 7s, etc.
→ if numBits is even, then add else subtract, the fast way is just to look at the lowest bit of the numBits counter

# 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 "4 25 1000" | ./268

Output:

(please click 'Go !')

Note: the original problem's input 4 25 10000000000000000 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>

// all 25 primes less than 100
std::vector<unsigned int> primes = {  2,  3,  5,  7, 11,
13, 17, 19, 23, 29,
31, 37, 41, 43, 47,
53, 59, 61, 67, 71,
73, 79, 83, 89, 97 };

// count numbers with at least "minPrimes" prime factors
// slow version - helping me while debugging the fast code below
unsigned int bruteForce(unsigned int limit, unsigned int minPrimes)
{
unsigned int result = 0;
for (unsigned int i = 1; i <= limit; i++)
{
// count distinct prime factors
unsigned int numPrimeFactors = 0;
for (auto p : primes)
if (i % p == 0)
numPrimeFactors++;

// at least four prime factors ?
if (numPrimeFactors >= minPrimes)
result++;
}
return result;
}

// return the n-th tetrahedral number
unsigned int tetrahedral(unsigned int n)
{
// see https://en.wikipedia.org/wiki/Tetrahedral_number
return n * (n + 1) * (n + 2) / 6;
}

int main()
{
// at least 4 primes, search up to 10^16
unsigned int   minPrimes = 4;
unsigned int   numPrimes = primes.size();
unsigned long long limit = 10000000000000000ULL;

std::cin >> minPrimes >> numPrimes >> limit;

// find result for small limits
//std::cout << bruteForce(limit, minPrimes) << std::endl;

// precompute tetrahedral numbers
std::vector<unsigned long long> count(numPrimes + 1, 0);
for (unsigned int i = minPrimes; i < count.size(); i++)
count[i] = tetrahedral(i - minPrimes + 1); // same as choose(i - 1, minPrimes - 1)

unsigned long long sum = 0;
unsigned int maxMask = (1 << numPrimes) - 1;
for (unsigned int mask = 0; mask < maxMask; mask++)
{
// multiply all primes
unsigned long long product = 1;
// but don't exceed 10^16
bool tooLarge = false;
// at least four bits must be set (=> pick at least four different primes)
unsigned int numBits = 0;
// look at each prime
for (unsigned int current = 0; current < numPrimes; current++)
{
// is this prime used ?
unsigned int bitpos = 1 << current;
if ((mask & bitpos) == 0)
continue;

// include this prime factor
numBits++;
product *= primes[current];

// avoid too large products (beware of 64 bit overflows ...)
if (product > limit)
{
tooLarge = true;

// speed optimization:
// if the next mask has the same number of bits set (or more), then it will exceed the limit as well
// because the primes are getting larger
// => get closer to the next number with the less bits set
// add the lowest set bit to the current number
// => worst case: same number of bits if the last bit is a single bit
// => best  case: several bits next to the lowest bit are set to and all flip to zeros (and a zero becomes a one)
auto withLowestBit = mask & (mask - 1);
auto lowestBit     = mask - withLowestBit;
mask += lowestBit;
// hope for the best case ...
while (mask & (lowestBit << 1))
{
lowestBit <<= 1;
mask += lowestBit;
}

mask--; // because the for-loop will add 1 in each iteration
break;
}
}

// skip if too few primes or their product became too large
if (tooLarge || numBits < minPrimes)
continue;

// count numbers divisible by the current primes
auto divisible = limit / product;
// they were part of products with less primes, too, hence compute the "overlap"
divisible *= count[numBits];

// add if even number of primes, else subtract (inclusion-exclusion principle)
if (numBits & 1)
sum -= divisible;
else
sum += divisible;
}

// that's it !
std::cout << sum << std::endl;
return 0;
}


This solution contains 17 empty lines, 31 comments and 2 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.9 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

September 27, 2017 submitted solution
September 27, 2017 added comments

# Difficulty

Project Euler ranks this problem at 70% (out of 100%).

# Heatmap

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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
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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 267 - Billionaire Sum of Squares - problem 273 >>
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