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

# Problem 70: Totient permutation

(see projecteuler.net/problem=70)

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.

# 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 - 5/5 → `result -= 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)

# 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>
// 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.

# 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 100 | ./70`

Output:

*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)*

# Benchmark

The correct solution to the original Project Euler problem was found in **1.51** seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.

(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

March 11, 2017 submitted solution

May 2, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler070

My code solved **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 never an option.

# Links

projecteuler.net/thread=70 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

Code in various languages:

Python: www.mathblog.dk/project-euler-70-investigate-values-of-n-for-which-φn-is-a-permutation-of-n/ (written by Kristian Edlund)

Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p070.java (written by Nayuki)

Go: github.com/frrad/project-euler/blob/master/golang/Problem070.go (written by Frederick Robinson)

Scala: github.com/samskivert/euler-scala/blob/master/Euler070.scala (written by Michael Bayne)

# Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.

yellow problems score less than 100% at Hackerrank (but still solve the original problem).

gray problems are already solved but I haven't published my solution yet.

blue problems are already solved and there wasn't a Hackerrank version of it (at the time I solved it) or I didn't care about it because it differed too much.

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