Problem 266: Pseudo Square Root

(see projecteuler.net/problem=266)

The divisors of 12 are: 1,2,3,4,6 and 12.
The largest divisor of 12 that does not exceed the square root of 12 is 3.
We shall call the largest divisor of an integer n that does not exceed the square root of n the pseudo square root (PSR) of n.
It can be seen that PSR(3102)=47.

Let p be the product of the primes below 190.
Find PSR(p) mod 10^16.

My Algorithm

The product of all primes below 190 is 2 * 3 * 5 * 7 * ... * 181 = 5397346292805549782720214077673687806275517530364350655459511599582614290.
That number is obviously too large for my poor C++ compiler ... but its logarithm isn't:
log(2) + log(3) + log(5) + ... + log(181) = log(53973462...2614290) approx 167.472

My program looks for an optimal subsets of those primes where
log(p_1) + log(p_2) + ... + log(p_n) approx 167.472 / 2 approx 83.736 = (because log(sqrt{x}) = log(x) / 2)

There are 2^42 approx 4.4 * 10^12 subsets. And again, I have a number that is too large for my computer ...
But there is hope - I split those 42 primes into two equally-sized sets:
the container left will be "responsible" for the first 21 primes, and right will control the next 21 primes.

Now there are 2^21 approx 2.1 million subsets of each left and right.
For each of those subsets I compute the logarithm of the product of its primes and a bitmask that indicates which primes were used.
If the n-th bit is set the n-th prime was used (that's the case for left) or the (n+21)-th prime was used (right).

For each value of left I need to find the largest value of right such that their sum doesn't exceed 83.736 (logRoot).
std::upper_bound can perform this task very fast if the input data is sorted (but it returns an iterator that is "one step too far").

Until now, all computations were based on logarithms. I could try e^{best sum} but rounding issues produce only garbage.
That's why I have to look at the bitmasks of the best combination of values from left and right.

I multiply all used primes according to leftMask and rightMask and repeatedly apply modulo 10^16.

Note

Programming languages with proper support for large number probably don't need to deal with logarithms.

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 upper limit for the prime numbers

This is equivalent to
echo 100 | ./266

Output:

(please click 'Go !')

Note: the original problem's input 190 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>
#include <cmath>
 
// all primes below 190
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, 101, 103, 107, 109, 113,
127, 131, 137, 139, 149, 151, 157, 163, 167, 173,
179, 181
};
 
// store a subsets logarithm and a bitmask of used primes
struct LogAndBitmask
{
// logarithm of product of used primes
double logarithm;
// bitmask of used primes
unsigned int bitmask;
// for std::sort and std::upper_bound
bool operator<(const LogAndBitmask& other) const
{
return logarithm < other.logarithm;
}
};
 
int main()
{
unsigned int maxPrime = 190;
std::cin >> maxPrime;
// keep only relevant primes
while (primes.back() > maxPrime)
primes.pop_back();
 
// compute logarithm of all primes
double logProduct = 0;
std::vector<double> logPrimes;
for (auto p : primes)
{
auto current = log(p);
logPrimes.push_back(current);
logProduct += current;
}
 
// log(sqrt(x)) = log(x) / 2
double logRoot = logProduct / 2;
 
// subdivides primes into two equally-sized sets (21 elements each)
const auto half = primes.size() / 2;
 
// generate all subsets of the lower half of the primes (left)
// and all subsets of the upper half of the primes (right)
// key is the logarithm of product of the used primes, value is the bitmask
// (if n-th bit is set, then the n-th prime is used)
std::vector<LogAndBitmask> right;
for (unsigned int bitmask = 0; bitmask < (1ULL << half); bitmask++)
{
// find logarithm for the upper half of primes
double logRight = 0;
for (unsigned int pos = 0; pos < half; pos++)
if ((bitmask & (1 << pos)) != 0)
logRight += logPrimes[pos + half];
 
// store logarithms and their generating bitmasks
if (logRight <= logRoot) // skip a few of the very large products
right.push_back({ logRight, bitmask});
}
 
// sort container / required for fast binary search (the algorithm behind std::upper_bound)
std::sort(right.begin(), right.end());
 
// find "best" combination of left and right:
// for each value of "left" look for best match in "right"
// keep the highest sum which doesn't exceed logRoot
double best = 0;
unsigned int leftMask = 0; // optimal bitmask for lower primes
unsigned int rightMask = 0; // optimal bitmask for larger primes
for (unsigned int bitmask = 0; bitmask < (1ULL << half); bitmask++)
{
// find logarithm for the same bitmask of the lower half of primes
double logLeft = 0;
for (unsigned int pos = 0; pos < half; pos++)
if ((bitmask & (1 << pos)) != 0)
logLeft += logPrimes[pos];
 
// how much should come from the bigger primes ?
LogAndBitmask missing = { logRoot - logLeft, 0 }; // bitmask doesn't matter, set to zero
 
// find best match
auto iteLogRight = std::upper_bound(right.begin(), right.end(), missing);
// ... which will slightly overshoot, therefore go back one step
iteLogRight--;
auto logRight = iteLogRight->logarithm;
 
// higher than before ?
if (best < logLeft + logRight)
{
best = logLeft + logRight;
// store bitmasks
leftMask = bitmask;
rightMask = iteLogRight->bitmask;
}
}
 
// last 16 digits of the product
unsigned long long result = 1;
// multiply all relevant primes (according to leftMask and rightMask)
const auto Modulo = 10000000000000000ULL;
// index of current prime
unsigned int currentPrime = 0;
 
// the first 21 primes ...
for (; currentPrime < half; currentPrime++)
{
// use that prime ?
if ((leftMask & 1) != 0)
{
result *= primes[currentPrime];
result %= Modulo;
}
// next bit
leftMask >>= 1;
}
// ... and the same for the next 21 primes
for (; currentPrime < primes.size(); currentPrime++)
{
// use that prime ?
if ((rightMask & 1) != 0)
{
result *= primes[currentPrime];
result %= Modulo;
}
// next bit
rightMask >>= 1;
}
 
// print result
std::cout << result << std::endl;
return 0;
}

This solution contains 16 empty lines, 35 comments and 4 preprocessor commands.

Benchmark

The correct solution to the original Project Euler problem was found in 0.7 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 36 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

July 3, 2017 submitted solution
July 3, 2017 added comments

Difficulty

65% Project Euler ranks this problem at 65% (out of 100%).

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