<< problem 213 - Flea Circus Crack-free Walls - problem 215 >>

# 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.

       #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.

# 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.

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