<< problem 510 - Tangent Circles Prime triples and geometric sequences - problem 518 >>

# Problem 516: 5-smooth totients

5-smooth numbers are numbers whose largest prime factor doesn't exceed 5.
5-smooth numbers are also called Hamming numbers.
Let S(L) be the sum of the numbers n not exceeding L such that Euler's totient function phi(n) is a Hamming number.
S(100)=3728.

# My Algorithm

The three fundamental properties of phi are (see en.wikipedia.org/wiki/Euler's_totient_function):
(1) phi(p) = p - 1 if p is prime
(2) phi(p^k) = p^{k-1} * (p - 1) if p is prime
(3) phi(xy) = phi(x) * phi(y) if x and y are coprime

A 5-smooth Hamming number can be written as 2^a 3^b 5^c.
From (1) follows that if the successor of a Hamming number is prime, then that prime has a totient which is a Hamming number:
(4) phi(2^a 3^b 5^c + 1) = 2^a 3^b 5^c if 2^a 3^b 5^c + 1 is prime

From (2) and (3) follows that the totient of each Hamming number is a Hamming number, too:
(5) phi(2^a 3^b 5^c) = phi(2^a) * phi(3^b) * phi(5^c)
(6) = 2^{a-1} * (2-1) * 3^{b-1} * (3-1) * 5^{c-1} * (5-1)
(7) = 2^{a-1} * 3^{b-1} * 2 * 5^{c-1} * 4
(8) = 2^{a+2} * 3^{b-1} * 5^{c-1}

And from (3) and (4) follows that if the totient of two primes p_1 and p_2 is a Hamming numbers each, then their product is a Hamming number as well:
(9) phi(p_1 p_2) = phi(p_1) * phi(p_2)
(10) = 2^{a_1} 3^{b_1} 5^{c_1} * 2^{a_2} 3^{b_2} 5^{c_2}
(11) = 2^{a_1 + a_2} 3^{b_1 + b_2} 5^{c_1 + c_2}

This applies not only to the product of two primes but also to the product of any number of (distinct) primes with a Hamming totient.
To avoid further confusion: let's assume that all those primes are bigger than 5.

Multiplying each such product by 2^k produces another Hamming number because:
(12) phi(2^k p_1 p_2) = phi(2^k) * phi(p) = 2^{k-1} * phi(p)
Multiplying each such product by 3^k produces another Hamming number because:
(13) phi(3^k p) = phi(3^k) * phi(p) = 3^{k-1} * 2 * phi(p)
Multiplying each such product by 5^k produces another Hamming number because:
(14) phi(5^k p) = phi(5^k) * phi(p) = 5^{k-1} * 2^2 * phi(p)

In general: since 2, 3 and 5 are coprime, multiplying each such product by 2^a 3^b 5^c produces another Hamming number (see (3) ).

The algorithm now looks as follows:

• find all Hamming numbers up to 10^12 and store them in hamming (3428 numbers)
• find all prime numbers that are the successor of a Hamming number and store them in primes (545 numbers)
• multiply all values of primes in every possible way, but each prime must appear at most once per product (and obviously the product must not exceed 10^12).
• multiply each such product by all Hamming numbers (again: not exceeding 10^12)

## Note

I wasn't sure whether I could have missed a few numbers whose totient is a Hamming number, too.
My program found the correct value for S(100) so I hoped for the best - and indeed, my result for S(10^12) was correct.

The main loop processes a queue of all prime number products.
Initially I was keen on keeping the queue sorted but soon discovered that it doesn't matter.
Even more, it doesn't matter which numbers you process first: processing the last (= largest) numbers first is faster and saves memory, too,
because the queue doesn't grow too much.

Almost all of my solutions which need a prime test either use a simple prime sieve or my implementation of the Miller-Rabin test.
Since the Miller-Rabin test is more than 100 lines long and not easily understandable, I often hesitate to include it.
This time I decided to give my wheel-based prime test a chance (see en.wikipedia.org/wiki/Wheel_factorization , it's part of my toolbox, too).

# 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 100 | ./516

Output:

Note: the original problem's input 1000000000000 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 <algorithm>

// ---------- finally I found a problem for my wheel-based prime test (waiting in my toolbox for ages ...) ----------

// wheel-based prime test
bool isPrime(unsigned long long x)
{
// prime test for 2, 3 and 5 and their multiples
if (x % 2 == 0 || x % 3 == 0 || x % 5 == 0)
return x == 2 || x == 3 || x == 5;

// wheel with size 30 (=2*3*5):
// test against 30m+1, 30m+7, 30m+11, 30m+13, 30m+17, 30m+19, 30m+23, 30m+29
// their deltas/increments are:
const unsigned int Delta[] = { 6, 4, 2, 4, 2, 4, 6, 2 };
unsigned long long i = 7;
// 7 belongs to the second test group
auto pos = 1;

// check numbers up to sqrt(x)
while (i*i <= x)
{
// not prime ?
if (x % i == 0)
return false;

// skip forward to next test divisor
i += Delta[pos];
// next delta/increment
pos = (pos + 1) & 7;
}

// passed all tests, must be a prime number
return x > 1;
}

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

// store all 3428 Hamming number smaller than 10^12
// and all 545 primes that are a successor of a Hamming number
std::vector<unsigned long long> hamming;
std::vector<unsigned long long> primes;

int main()
{
unsigned long long limit = 1000000000000; // 10^12
std::cin >> limit;

// iterate over all numbers 2^a * 3^b * 5^c
for (unsigned long long two = 1; two <= limit; two *= 2)
for (unsigned long long three = 1; two*three <= limit; three *= 3)
for (unsigned long long five = 1; two*three*five <= limit; five *= 5)
{
auto current = two * three * five;

// including 1 which is not divisible by 2, 3, or 5 but still 5-smooth
hamming.push_back(current);

if (isPrime(current + 1) && current > 5)
primes.push_back(current + 1);
}

// in ascending order ...
std::sort(hamming.begin(), hamming.end());
std::sort(primes.begin(),  primes.end());

// store all numbers where phi(x) is 5-smooth
// first=number, second=largest prime factor of that number
typedef std::pair<unsigned long long, unsigned long long> Candidate;
std::vector<Candidate> todo;
// "seed value"
todo.push_back({ 1, 1 });

unsigned long long sum = 0;
while (!todo.empty())
{
// pick next candidate
// note: it's much faster to pick larger candidates first to keep the queue short
auto number       = todo.back().first;
auto largestPrime = todo.back().second;
todo.pop_back();

// multiply with all Hamming numbers, each result's totient is a Hamming number, too
for (auto x : hamming)
{
auto next = x * number;
if (next > limit)
break;

sum += next;
}

// multiply with all primes that are larger than the currently largest prime
auto nextPrime = std::upper_bound(primes.begin(), primes.end(), largestPrime);
for (; nextPrime != primes.end(); nextPrime++)
{
auto next = *nextPrime * number;
if (next > limit)
break;

todo.push_back({ next, *nextPrime });
}
}

// same as modulo 2^32
sum &= 0xFFFFFFFF; // alternatively, I could have cast it to a 32-bit integer
std::cout << sum << std::endl;
return 0;
}


This solution contains 20 empty lines, 27 comments and 3 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.3 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

November 15, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 20% (out of 100%).

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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
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.

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