<< problem 412 - Gnomon numbering Prime connection - problem 425 >>

# Problem 418: Factorisation triples

Let n be a positive integer. An integer triple (a, b, c) is called a factorisation triple of n if:

1 <= a <= b <= c
a * b * c = n.

Define f(n) to be a + b + c for the factorisation triple (a, b, c) of n which minimises c / a.
One can show that this triple is unique.

For example, f(165) = 19, f(100100) = 142 and f(20!) = 4034872.

Find f(43!).

# My Algorithm

Even though 43! is a pretty large number, its factorization is extremely simple: just factorize all numbers 1, 2, 3, ..., 43.
Then add all exponents of their prime factors:
43! = 2^39 * 3^19 * 5^9 * 7^6 * 11^3 * 13^3 * 17^2 * 19^2 * 23 * 29 * 31 * 37 * 41 * 43

The number of divisors is (see en.wikipedia.org/wiki/Divisor_function):
(39 + 1) * (19 + 1) * (9 + 1) * (6 + 1) * (3 + 1) * (3 + 1) * (2 + 1) * (2 + 1) * (1 + 1) * (1 + 1) * (1 + 1) * (1 + 1) * (1 + 1) * (1 + 1)
= 40 * 20 * 10 * 7 * 4 * 4 * 3 * 3 * 2 * 2 * 2 * 2 * 2 * 2
= 516096000

If c / a is minimal then a, b and c should be pretty close to each other.
That means all three variables are approx sqrt{43!} with a < sqrt{43!} and c > sqrt{43!}.

I started with the assumption 0.99 * sqrt{43!} < a < sqrt{43!} and found tons of valid triples.
Even 0.9999 * sqrt{43!} < a < sqrt{43!} and sqrt{43!} < c < 1.0001 * sqrt{43!} is "good" enough.

Thus only about 1000 divisors are left and I can compare them in a brute-force way (see search()):

• for each a and c compute b = 43! / (a*c) by manipulating the exponents of the prime factors
• if such b exists then check if c / a is smaller than anything seen before

## Note

Most of the time is spent in findCandidates() which extracts these 1000 relevant divisors.

Smaller factorials than 43! need a wider corridor than 0.0001 to find a valid solution and
that's what the second command-line parameter is for: e.g. for 20! you have to enter "20 0.004" (but even "20 1" is quite fast).

# 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 "20 0.004" | ./418

Output:

Note: the original problem's input 43 0.0002 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 <map>
#include <cmath>

// all prime factors of 43!
std::vector<unsigned char> primes = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 };
std::vector<unsigned char> factors; // = { 39, 19, 9, 6, 3, 3, 2, 2, 1, 1, 1, 1, 1, 1 };

// store all relevant divisors of 43!
std::map<unsigned long long, std::vector<unsigned char>> candidates;

// fill container "factors" with prime exponents of 43!
void factorizeFactorial(unsigned int factorial)
{
factors.resize(primes.size(), 0);
// each number 2..43
for (unsigned int i = 2; i <= factorial; i++)
{
// find their prime factors ...
auto reduce = i;
for (unsigned int j = 0; j < primes.size(); j++)
{
auto p = primes[j];
while (reduce % p == 0)
{
factors[j]++;
reduce /= p;
}
}
}
}

// find all divisors close to sqrt3(43!) in a recursive way
void findCandidates(std::vector<unsigned char>& exponents, unsigned int pos,
unsigned long long current, unsigned long long atLeast, unsigned long long atMost)
{
// multiplied all primes ?
if (pos == exponents.size())
{
// must be within range of relevant divisors
if (current < atLeast || current > atMost)
return;

// store it
candidates[current] = exponents;
return;
}

// try all exponents of the current prime
auto maxExponent = exponents[pos];
for (exponents[pos] = 0; exponents[pos] <= maxExponent; exponents[pos]++)
{
if (exponents[pos] > 0)
{
// fast exit if too big
if (current * primes[pos] > atMost)
break;

current *= primes[pos];
}

// go deeper ...
findCandidates(exponents, pos + 1, current, atLeast, atMost);
}
// restore previous value
exponents[pos] = maxExponent;
}

// find a+b+c, where root3 is an approximation of sqrt3(43!)
unsigned long long search(double root3)
{
// a+b+c
unsigned long long result = 0;

auto bestRatio = candidates.rbegin()->first / double(candidates.begin()->first);
// the initial numbers probably don't "match"

// a < sqrt3(43!) and c > sqrt3(43!)
auto mid = candidates.lower_bound(root3);
// compare all pairs a,c
for (auto a = candidates.begin(); a != mid; a++)
for (auto c = mid; c != candidates.end(); c++)
{
// too far apart ?
if (a->first * bestRatio < c->first)
break;

// calculate b based on the unused prime exponents
bool isValid = true;
unsigned long long b = 1;
for (unsigned int i = 0; i < primes.size() && isValid; i++)
{
// find "remaining" exponent for each prime
auto used = a->second[i] + c->second[i];
// make sure we don't consume "too many" exponents
isValid = used <= factors[i];

// calculate b
while (used < factors[i])
{
b *= primes[i];
used++;
}
}

// no valid b ?
if (!isValid)
continue;

// verify that a <= b <= c
if (b < a->first || b > c->first)
continue;

// compare ratio against the previous record
auto ratio = c->first / double(a->first);
if (bestRatio > ratio)
{
bestRatio = ratio;
result = a->first + b + c->first;
}

// a,b,c was a valid triple => if c increments then c/a worsens
break;
}

// a+b+c
return result;
}

int main()
{
unsigned int limit = 43;
std::cin >> limit;

// factorize 43!
factorizeFactorial(limit);

// cheating ?! I optimized the search range manually for 43!
double maxRatio = 0.0002;
std::cin >> maxRatio;
auto atLeast = 1 - maxRatio / 2;
auto atMost  = 1 + maxRatio / 2;

// approximation of 43!
double factorial = 1;
for (unsigned int i = 2; i <= limit; i++)
factorial *= i;

// a,b,c will be close to sqrt3(43!)
auto root3 = pow(factorial, 1.0/3);
findCandidates(factors, 0, 1, root3 * atLeast, root3 * atMost);

// let's find the best triple !
std::cout << search(root3) << std::endl;
return 0;
}


This solution contains 24 empty lines, 35 comments and 4 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 1.0 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

October 4, 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 412 - Gnomon numbering Prime connection - problem 425 >>
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