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Problem 214: Totient Chains

Let phi be Euler's totient function, i.e. for a natural number n, phi(n) is the number of k, 1 <= k <= n, for which gcd(k,n) = 1.

By iterating phi, each positive integer generates a decreasing chain of numbers ending in 1.
Here is a listing of all chains with length 4:
5,4,2,1
7,6,2,1
8,4,2,1
9,6,2,1
10,4,2,1
12,4,2,1
14,6,2,1
18,6,2,1

Only two of these chains start with a prime, their sum is 12.

What is the sum of all primes less than 40000000 which generate a chain of length 25?

My Algorithm

I copied phi() from problem 70 (and removed the minQuotient parameter). Then I copied a prime sieve from my toolbox.

The steps() function determines the totient chain length: it apply x = phi(x) until x is 1.

• abort if chain length exceeds 25 (maxSteps)
• if x is a power of two, that means only a single bit is set, then phi(x) = log_2(x) (shift right until that bit is in the right-most position)

Alternative Approaches

Instead of chasing down the chains you can go the other way around:

• set steps[0] = steps[1] = 1, then steps[x] = steps[phi(x)]
However, this requires 40 MByte RAM and isn't much faster than my approach (which needs about 5 MByte).

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 chain length and the maximum prime to be considered

This is equivalent to
echo "10 1000" | ./214

Output:

Note: the original problem's input 25 40000000 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>

std::vector<unsigned int> primes;

// return Euler totient (taken from problem 70 and slightly modified)
unsigned int phi(unsigned int x)
{
// totient function can be computed by finding all prime factors p
// and subtracting them from x
auto result  = x;
auto reduced = x;
for (auto p : primes)
{
// prime factors have to be p <= sqrt
if (p*p > reduced)
break;

// not a prime factor ...
if (reduced % p != 0)
continue;

// prime factors may occur multiple times, remove them all
do
{
reduced /= p;
} while (reduced % p == 0);

// but subtract from result only once
result -= result / p;
}

// we only checked prime factors <= sqrt(x)
// there might exist one (!) prime factor > sqrt(x)
// e.g. 3 is a prime factor of 6, and 3 > sqrt(6)
if (reduced > 1)
result -= result / reduced;

return result;
}

// count length of totient chain, abort if more than maxSteps
unsigned int steps(unsigned int x, unsigned int maxSteps)
{
unsigned int result = 1; // initial value is counted as a step, too

// if x is prime, then phi(x) = x - 1
x--;
result++;

// follow chain until we hit 1 or the chain becomes too long
while (x > 1 && result < maxSteps)
{
// one more step ...
x = phi(x);
result++;

// power of two ? (only a single bit set)
if ((x & (x - 1)) == 0)
{
// simple chain for powers of two:
// ... => 1024 => 512 => 256 => 128 => ... => 4 => 2 => 1
while (x > 1)
{
x >>= 1;
result++;
}
}
}

return result;
}

// sieve is copied from my toolbox

// odd prime numbers are marked as "true" in a bitvector
std::vector<bool> sieve;

// return true, if x is a prime number
bool isPrime(unsigned int x)
{
// handle even numbers
if ((x & 1) == 0)
return x == 2;

// lookup for odd numbers
return sieve[x >> 1];
}

// find all prime numbers from 2 to size
void fillSieve(unsigned int size)
{
// store only odd numbers
const unsigned int half = size >> 1;

// allocate memory
sieve.resize(half, true);
// 1 is not a prime number
sieve[0] = false;

// process all relevant prime factors
for (unsigned int i = 1; 2*i*i < half; i++)
// do we have a prime factor ?
if (sieve[i])
{
// mark all its multiples as false
unsigned int current = 3*i+1;
while (current < half)
{
sieve[current] = false;
current += 2*i+1;
}
}
}

int main()
{
unsigned long long result = 0;

unsigned int limit = 40000000;
unsigned int maxSteps = 25;
std::cin >> maxSteps >> limit;

// start the prime sieve
fillSieve(limit);

// convert sieve to a "denser" data structure for phi(x)
primes.push_back(2);

for (unsigned int i = 3; i < limit; i += 2)
{
// check bits in sieve
if (!isPrime(i))
continue;

// phi(i) needs only primes smaller than sqrt(i)
if (i*i <= limit)
primes.push_back(i);

// primes need to be quite large to generate a long chain ...
if (maxSteps == 25 && i < 9548417) // note: "magic constant" found in experiments
continue;

// compute chain
unsigned int current = steps(i, maxSteps + 1);
if (current == maxSteps)
result += i;
}

std::cout << result << std::endl;
return 0;
}


This solution contains 28 empty lines, 35 comments and 2 preprocessor commands.

Benchmark

The correct solution to the original Project Euler problem was found in 2.7 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 5 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

June 20, 2017 submitted solution

Difficulty

Project Euler ranks this problem at 40% (out of 100%).

Similar problems at Project Euler

Problem 351: Hexagonal orchards

Note: I'm not even close to solving all problems at Project Euler. Chances are that similar problems do exist and I just haven't looked at them.

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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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