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# Problem 214: Totient Chains

(see projecteuler.net/problem=214)

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.

E.g. if we start with 5 the sequence 5,4,2,1 is generated.

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.

I added two performance optimizations:

- 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).

# 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.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

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)*

# Benchmark

The correct solution to the original Project Euler problem was found in 2.7 seconds on a 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=c++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

June 20, 2017 added comments

# Difficulty

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

# Links

projecteuler.net/thread=214 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

# Heatmap

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Look at my progress and performance pages to get more details.

My username at Project Euler is

**stephanbrumme**while it's stbrumme at Hackerrank.

# Copyright

I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.

All of my solutions can be used for any purpose and I am in no way liable for any damages caused.

You can even remove my name and claim it's yours. But then you shall burn in hell.

The problems and most of the problems' images were created by Project Euler.

Thanks for all their endless effort.

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