<< problem 69 - Totient maximum Ordered fractions - problem 71 >>

# Problem 70: Totient permutation

Euler's Totient function, phi(n) (sometimes called the phi function), is used to determine the number of positive numbers
less than or equal to n which are relatively prime to n.
For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, phi(9) = 6.

The number 1 is considered to be relatively prime to every positive number, so phi(1) = 1.

Interestingly, phi(87109) = 79180, and it can be seen that 87109 is a permutation of 79180.

Find the value of n, 1 < n < 10^7, for which phi(n) is a permutation of n and the ratio dfrac{n}{phi(n)} produces a minimum.

# My Algorithm

The function phi(x) computes phi(x):

• divide x by all prime numbers
• count how many distinct prime numbers were divisors of x
The original formula for the totient is (see en.wikipedia.org/wiki/Euler's_totient_function):
result = x * (1 - 1/prime1) * (1 - 1/prime2) * (1 - 1/prime3) * ...
e.g. if x == 10
result = 10 * (1 - 1/2) * (1 - 1/5) = 4

When initializing result = x then the multiplication can be reduced to a subtraction:
whenever we find a prime factor p, then result -= result/p.
For phi(10) we have the prime factors 2 and 5
10 * (1 - 1/2) * (1 - 1/5)
= (10 * (1 - 1/2)) * (1 - 1/5)
= (10 - 10/2) * (1 - 1/5)result -= result/2
= 5 * (1 - 1/5)
= 5 - 5/5result -= result/5
= 4

In this iterative algorithm, result is becoming smaller and smaller after each step.
If it becomes obvious that the result is too small (minQuotient < x/result = result * minQuotient < x), then phi aborts.
This small optimization gives a 2.5x performance boost.

My fingerprint function was used in several Project Euler problems before, e.g. problem 49, problem 52, problem 62, ...

## Alternative Approaches

I bet you can severely speed up the program by incorporating some observations:

• minimizing n/phi(n) means maximizing phi(n)
• phi(n) is maximized for prime numbers
• if i is not a prime, then phi(n) is still big if it contains just a few prime factors and those are close to sqrt{n}

# 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 | ./70

Output:

(please click 'Go !')

Note: the original problem's input 10000000 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>

// precompute all primes less than sqrt(10^7)
std::vector<unsigned int> primes;

// return phi(x) if x/phi(x) <= minQuotient (else the result is undefined but > minQuotient)
unsigned int phi(unsigned int x, double minQuotient)
{
// 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;

// abort, this number can't be relevant because the quotient is already too high
if (result * minQuotient < x)
return result; // number is garbage but always >= its correct value
}

// prime number (result is still equal to x because we couldn't find any prime factors)
if (result == x)
return x - 1;
// (actually this case would be properly handled by the next if-clause, too)

// 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)
return result - result / reduced;
else
return result;
}

// count digits, two numbers have the same fingerprint if they are permutations of each other
unsigned long long fingerprint(unsigned int x)
{
unsigned long long result = 0;
while (x > 0)
{
auto digit = x % 10;
x /= 10;

unsigned long long shift = 1;
for (unsigned int i = 0; i < digit; i++)
shift *= 10;

result += shift;
}
return result;
}

int main()
{
unsigned int last;
std::cin >> last;

// step 1: the usual prime sieve
primes.push_back(2);
for (unsigned int i = 3; i*i < last; i += 2)
{
bool isPrime = true;

// test against all prime numbers we have so far (in ascending order)
for (auto p : primes)
{
// next prime is too large to be a divisor ?
if (p*p > i)
break;

// divisible ? => not prime
if (i % p == 0)
{
isPrime = false;
break;
}
}

// yes, we have a prime number
if (isPrime)
primes.push_back(i);
}

// step 2: analyze all phi(n)
unsigned int bestNumber  = 2;
double       minQuotient = 999999;
for (unsigned int n = 3; n < last; n++)
{
auto phi_n = phi(n, minQuotient);
double quotient = n / double(phi_n);
// already have a better quotient ?
if (minQuotient <= quotient)
continue;

// check if phi(n) is a permutation of n
if (fingerprint(phi_n) == fingerprint(n))
{
minQuotient = quotient;
bestNumber  = n;
}
}

// print result
std::cout << bestNumber << std::endl;
return 0;
}


This solution contains 19 empty lines, 24 comments and 2 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 1.5 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

March 11, 2017 submitted solution

# Hackerrank

My code solves 11 out of 11 test cases (score: 100%)

# Difficulty

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

Hackerrank describes this problem as easy.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

# 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 69 - Totient maximum Ordered fractions - problem 71 >>
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