<< problem 181 - Investigating in how many ways objects of two ... Maximum product of parts - problem 183 >>

# Problem 182: RSA encryption

The RSA encryption is based on the following procedure:
Generate two distinct primes p and q.
Compute n = pq and phi = (p-1)(q-1).
Find an integer e, 1 < e < phi, such that gcd(e, phi)=1.

A message in this system is a number in the interval [0, n-1].
A text to be encrypted is then somehow converted to messages (numbers in the interval [0, n-1]).
To encrypt the text, for each message, m, c = m^e mod n is calculated.

To decrypt the text, the following procedure is needed: calculate d such that ed = 1 mod phi, then for each encrypted message, c, calculate m = c^d mod n.

There exist values of e and m such that m^e mod n = m.
We call messages m for which m^e mod n = m unconcealed messages.

An issue when choosing e is that there should not be too many unconcealed messages.
For instance, let p = 19 and q = 37.
Then n = 19 * 37 = 703 and phi = 18 * 36 = 648.
If we choose e = 181, then, although gcd(181, 648) = 1 it turns out that all possible messages
m (0 <= m <= n-1) are unconcealed when calculating m^e mod n.
For any valid choice of e there exist some unconcealed messages.
It's important that the number of unconcealed messages is at a minimum.

Choose p = 1009 and q = 3643.
Find the sum of all values of e, 1 < e < phi(1009, 3643) and gcd(e, phi)=1, so that the number of unconcealed messages for this value of e is at a minimum.

# My Algorithm

This problem bugged me for a while. However, searching the internet for "rsa fixed points" brought up a link to a discussion on StackOverflow
(see math.stackexchange.com/questions/1298664/rsa-fixed-point) which contained a simple formula:
(1 + gcd(e-1, p-1)) * (1 + gcd(e-1, q-1))

All I have to do is to enumerate all encrypted message (called encrypted in my code) in a for-loop:

• if the formula returns a value I already had then add encrypted to sum
• else "reset" sum to the current value of encrypted

## Note

The programming part isn't very exciting. Due the huge number of calls to isCoprime (and gcd) I researched ways to speed up these functions.

• isCoprime becomes about 30% if I check the lowest bits of its parameters: if both are zero then the parameters are both even and can't be coprime
• using special CPU instructions (ctz on Intel CPUs) gives gcd a performance boost of about 30% as well
→ this code path is protected by #ifdef GNUC because Visual C++ has differed names and calling conventions for their intrinsics

I updated my toolbox accordingly.

# 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 two distinct p and q less than 2000.

This is equivalent to
echo "19 37" | ./182

Output:

Note: the original problem's input 1009 3643 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 <numeric>

// I optimized the following two routines which originated in my toolbox

// greatest common divisor
template <typename T>
T gcd(T a, T b)
{
// C++17 will have std::gcd(a, b) which is most likely as fast as my implementation

#ifdef __GNUC__
// and          https://github.com/lemire/Code-used-on-Daniel-Lemire-s-blog/blob/master/2013/12/26/gcd.cpp
if (a == 0)
return b;
if (b == 0)
return a;

auto shift = __builtin_ctz(a | b);
a >>= __builtin_ctz(a);
do
{
b >>= __builtin_ctz(b);
if (a > b)
std::swap(a, b);

b -= a;
} while (b != 0);
return a << shift;

#else

// standard GCD
while (a != 0)
{
T c = a;
a = b % a;
b = c;
}
return b;
#endif
}

// return true if a and b are coprime
template <typename T>
bool isCoprime(T a, T b)
{
// fast reject if both are even (=> gcd(a,b) >= 2)
if (((a|b) & 1) == 0)
return false;

return gcd(a, b) == 1;
}

int main()
{
// problem's constants
unsigned int p = 1009;
unsigned int q = 3643;
std::cin >> p >> q;

// compute phi according to problem statement
auto phi = (p - 1) * (q - 1);

unsigned int best = 0xFFFFFFFF;
// sum of all e when best is minimized
unsigned long long sum = 0;

// iterate over all messages
for (unsigned int encryption = 0; encryption < phi; encryption++)
{
// must be coprime
if (!isCoprime(encryption, phi))
continue;

auto badP = gcd(p - 1, encryption - 1) + 1;
auto badQ = gcd(q - 1, encryption - 1) + 1;

// same number of unconcealed messages ? add all of them
if (best == numPlaintext)
sum += encryption;
// improved (=lower) number of unconcealed messages ? reset sum
else if (best >  numPlaintext)
{
best = numPlaintext;
sum  = encryption;
}
}

// display result
std::cout << sum << std::endl;
return 0;
}


This solution contains 16 empty lines, 18 comments and 5 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.20 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

September 3, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 60% (out of 100%).

# 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 181 - Investigating in how many ways objects of two ... Maximum product of parts - problem 183 >>
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