<< problem 247 - Squares under a hyperbola Prime Subset Sums - problem 249 >>

# Problem 248: Numbers for which Euler's totient function equals 13!

The first number n for which phi(n)=13! is 6227180929.

Find the 150,000th such number.

# My Algorithm

There are two basic properties of Euler's totient function phi(x) (see en.wikipedia.org/wiki/Euler's_totient_function):
(1) phi(p) = p - 1 if p is a prime
(2) phi(a b) = phi(a) phi(b) if a and b are coprime

Every number can be split uniquely into its prime factors.
My program transforms phi(x) = 13! such that phi(x) = phi(d_1 * d_2 * ... * d_n) = phi(d_1) * phi(d_2) * ... * phi(d_n) = D_1 * D_2 * ... * D_n.

If I find all divisors of 13! and partition them in such a way that 13! = D_1 * D_2 * ... D_n I can apply rule (1) and/or (2).
For example, if for some divisor I can show that D_i + 1 is prime then according to (1):
(3) D_i + 1 = phi(p_1) + 1 = (p_i - 1) = p_i
So I have found a prime factor D_i + 1 of a potential number n where phi(n) = 13!

Factorizing 13! gives us:
(4) 13! = 2^10 * 3^5 * 5^2 * 7 * 11 * 13

Now I iterate over all divisors of 13! and check whether D_i + 1 is prime (see findCandidates).
All those numbers are stored in the container candidates.

Then I combine the candidates until the product of their totients yields 13! (see search).
By the way: I needed a few minutes to realize that some candidates can appear multiple times in the product.

The problem statement says that 6227180929 is the smallest number where phi(x) = 13!:
(5) 6227180929 = 66529 * 93601 = (66528 + 1) * (93600 + 1)
(6) 66528 = 2^5 * 3^3 * 7 * 11
(7) 93600 = 2^5 * 3^2 * 5^2 * 13
(8) 66528 * 93600 = 6227020800 = 13!

And indeed, my program finds 13!'s divisors 66528 and 93600 (among others) because their successor is a prime and combines them.

Sorting those roughly 180000 numbers finishes almost instantly but I went the crazy way and preferred
std::nth_element because it is a little bit more efficient - but you can't measure the performance advantage ...

## Note

A huge part of my code consists of my standard Miller-Rabin prime test which I just copied from my toolbox.

# 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 "13 1" | ./248

Output:

Note: the original problem's input 13 150000 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>
#include <map>
#include <algorithm>

// ---------- Miller-Rabin prime test from my toolbox ----------

// return (a*b) % modulo
unsigned long long mulmod(unsigned long long a, unsigned long long b, unsigned long long modulo)
{
// (a * b) % modulo = (a % modulo) * (b % modulo) % modulo
a %= modulo;
b %= modulo;

// fast path
if (a <= 0xFFFFFFF && b <= 0xFFFFFFF)
return (a * b) % modulo;

// we might encounter overflows (slow path)
// the number of loops depends on b, therefore try to minimize b
if (b > a)
std::swap(a, b);

// bitwise multiplication
unsigned long long result = 0;
while (a > 0 && b > 0)
{
// b is odd ? a*b = a + a*(b-1)
if (b & 1)
{
result += a;
if (result >= modulo)
result -= modulo;
// skip b-- because the bit-shift at the end will remove the lowest bit anyway
}

// b is even ? a*b = (2*a)*(b/2)
a <<= 1;
if (a >= modulo)
a -= modulo;

// next bit
b >>= 1;
}

return result;
}

// return (base^exponent) % modulo
unsigned long long powmod(unsigned long long base, unsigned long long exponent, unsigned long long modulo)
{
unsigned long long result = 1;
while (exponent > 0)
{
// fast exponentation:
// odd exponent ? a^b = a*a^(b-1)
if (exponent & 1)
result = mulmod(result, base, modulo);

// even exponent ? a^b = (a*a)^(b/2)
base = mulmod(base, base, modulo);
exponent >>= 1;
}
return result;
}

// Miller-Rabin-test
bool isPrime(unsigned long long p)
{
// IMPORTANT: requires mulmod(a, b, modulo) and powmod(base, exponent, modulo)

// some code from             https://ronzii.wordpress.com/2012/03/04/miller-rabin-primality-test/
// with optimizations from    http://ceur-ws.org/Vol-1326/020-Forisek.pdf
// good bases can be found at http://miller-rabin.appspot.com/

// trivial cases
const unsigned int bitmaskPrimes2to31 = (1 << 2) | (1 << 3) | (1 << 5) | (1 << 7) |
(1 << 11) | (1 << 13) | (1 << 17) | (1 << 19) |
(1 << 23) | (1 << 29); // = 0x208A28Ac
if (p < 31)
return (bitmaskPrimes2to31 & (1 << p)) != 0;

if (p % 2 == 0 || p % 3 == 0 || p % 5 == 0 || p % 7 == 0 || // divisible by a small prime
p % 11 == 0 || p % 13 == 0 || p % 17 == 0)
return false;

if (p < 17 * 19) // we filtered all composite numbers < 17*19, all others below 17*19 must be prime
return true;

// test p against those numbers ("witnesses")
// good bases can be found at http://miller-rabin.appspot.com/
const unsigned int STOP = 0;
const unsigned int TestAgainst1[] = { 377687, STOP };
const unsigned int TestAgainst2[] = { 31, 73, STOP };
const unsigned int TestAgainst3[] = { 2, 7, 61, STOP };
// first three sequences are good up to 2^32
const unsigned int TestAgainst4[] = { 2, 13, 23, 1662803, STOP };
const unsigned int TestAgainst7[] = { 2, 325, 9375, 28178, 450775, 9780504, 1795265022, STOP };

// good up to 2^64
const unsigned int* testAgainst = TestAgainst7;
// use less tests if feasible
if (p < 5329)
testAgainst = TestAgainst1;
else if (p < 9080191)
testAgainst = TestAgainst2;
else if (p < 4759123141ULL)
testAgainst = TestAgainst3;
else if (p < 1122004669633ULL)
testAgainst = TestAgainst4;

// find p - 1 = d * 2^j
auto d = p - 1;
d >>= 1;
unsigned int shift = 0;
while ((d & 1) == 0)
{
shift++;
d >>= 1;
}

// test p against all bases
do
{
auto x = powmod(*testAgainst++, d, p);
// is test^d % p == 1 or -1 ?
if (x == 1 || x == p - 1)
continue;

// now either prime or a strong pseudo-prime
// check test^(d*2^r) for 0 <= r < shift
bool maybePrime = false;
for (unsigned int r = 0; r < shift; r++)
{
// x = x^2 % p
// (initial x was test^d)
x = mulmod(x, x, p);
// x % p == 1 => not prime
if (x == 1)
return false;

// x % p == -1 => prime or an even stronger pseudo-prime
if (x == p - 1)
{
// next iteration
maybePrime = true;
break;
}
}

// not prime
if (!maybePrime)
return false;
} while (*testAgainst != STOP);

// prime
return true;
}

// ---------- problem-specific code ----------

// will be 13!
unsigned long long factorial = 1;
// factorized 13! such such factors = 10; etc.
std::map<unsigned int, unsigned int> factors;
// all primes where prime-1 is a divisor of 13!
std::set<unsigned int> candidates;
// all x where phi(x) = 13!
std::vector<unsigned long long> results;

// "build" all numbers using the prime factors in "factors"
// if number + 1 is prime then store as a candidate
void findCandidates(unsigned long long number = 1, unsigned int lastPrime = 1)
{
// all primes consumed ?
auto nextPrime = factors.upper_bound(lastPrime);
if (nextPrime == factors.end())
{
// check if successor is prime
if (isPrime(number + 1))
candidates.insert(number + 1);
return;
}

// process all combinations recursively
auto base     = nextPrime->first;
auto exponent = nextPrime->second;
for (unsigned int i = 0; i <= exponent; i++)
{
findCandidates(number, base);
number *= base;
}
}

// look for all number with phi(x) = 13!
void search(unsigned long long number = 1, unsigned long long phi = 1,
unsigned int largestPrime = 1)
{
// find next prime (or increase the exponent of the latest prime)
auto from = candidates.lower_bound(largestPrime);
for (auto i = from; i != candidates.end(); i++)
{
// current prime
auto current = *i;

// include one more prime factor
auto nextNumber = number * current;

// phi(prime)   = prime - 1
auto nextPhi = phi * (current - 1);
// phi(prime^x) = phi(prime^(x-1)) * prime
if (current == largestPrime)
nextPhi += phi; // phi * current = phi * (current - 1) + phi

// too large ?
if (nextPhi >  factorial)
break;
// match ?
if (nextPhi == factorial)
{
results.push_back(nextNumber);
break;
}

// keep going if 13! is still reachable
if (factorial % nextPhi == 0)
search(nextNumber, nextPhi, current);
}
}

int main()
{
unsigned int thirteen = 13;
unsigned int index = 150000;
std::cin >> thirteen >> index;

// compute 13!
factorial = 1;
for (unsigned int i = 2; i <= thirteen; i++)
factorial *= i;

// and factorize it
auto reduce = factorial;
for (unsigned int i = 2; i <= thirteen; i++)
while (reduce % i == 0)
{
factors[i]++;
reduce /= i;
}

// generate all primes whose predecessor is a divisor of 13!
findCandidates();

// find all numbers with eulerphi() = 13!
search();

// array is zero-indexed
index--;
// partial sort: only need the 150000th to be at its correct position
auto pos = results.begin() + index;
std::nth_element(results.begin(), pos, results.end());
// (I could have used std::sort as well, but std::nth_element is a tiny bit faster in this case)
std::cout << *pos << std::endl;
return 0;
}


This solution contains 38 empty lines, 64 comments and 5 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 4 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

November 12, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 70% (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 247 - Squares under a hyperbola Prime Subset Sums - problem 249 >>
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