Problem 308: An amazing Prime-generating Automaton

(see projecteuler.net/problem=308)

A program written in the programming language Fractran consists of a list of fractions.

The internal state of the Fractran Virtual Machine is a positive integer, which is initially set to a seed value.
Each iteration of a Fractran program multiplies the state integer by the first fraction in the list which will leave it an integer.

For example, one of the Fractran programs that John Horton Conway wrote for prime-generation consists of the following 14 fractions:

dfrac{17}{91}, dfrac{78}{85}, dfrac{19}{51}, dfrac{23}{38}, dfrac{29}{33}, dfrac{77}{29}, dfrac{95}{23}, dfrac{77}{19}, dfrac{1}{17}, dfrac{11}{13}, dfrac{13}{11}, dfrac{15}{2}, dfrac{1}{7}, dfrac{55}{1}

Starting with the seed integer 2, successive iterations of the program produce the sequence:
15, 825, 725, 1925, 2275, 425, ..., 68, 4, 30, ..., 136, 8, 60, ..., 544, 32, 240, ...

The powers of 2 that appear in this sequence are 2^2, 2^3, 2^5, ...
It can be shown that all the powers of 2 in this sequence have prime exponents and that all the primes appear as exponents of powers of 2, in proper order!

If someone uses the above Fractran program to solve Project Euler Problem 7 (find the 10001st prime), how many iterations would be needed until the program produces 2^{10001st \space prime} ?

My Algorithm

The crucial discovery is that all fractions can be factorized such that the largest prime factor is 29 and no prime appears as a square (or cube).
The sequence becomes:

dfrac{17}{7 * 13}, dfrac{2 * 3 * 13}{5 * 17}, dfrac{19}{3 * 17}, dfrac{23}{2 * 19}, dfrac{29}{3 * 11}, dfrac{7 * 11}{29}, dfrac{5 * 19}{23}, dfrac{7 * 11}{19}, dfrac{1}{17}, dfrac{11}{13}, dfrac{13}{11}, dfrac{3 * 5}{2}, dfrac{1}{7}, dfrac{5 * 11}{1}

The 10001st prime needs 10001 bits to be represented - way beyond the 64 bits of unsigned long long.
However, I can factorize the current number as well and it's initial value is 2 = 2^1.
I could include all the other (non-existing) prime factors, too, so that
2 = 2^1 * 3^0 * 5^0 * 7^0 * 11^0 * 13^0 * 17^0 * 19^0 * 23^0 * 29^0

Each fraction increments or decrements some exponents. No exponent must become negative because then it wouldn't be an integer anymore.
Whenever all exponents except the exponent of 2 are zero, then I found another prime and it's the exponent of 2.

I wrote a brute-force function enumerate() that finds the 100th prime in a few seconds (matching the values of OEIS A007547).
But more important, I included a "visualization" of each step which revealed:

These are the 19 steps for the first prime (I replaced zeros by an underline to emphasize the patterns):

 2357111317192329
step 0:1_________
step 1:_11_______
step 2:_12_1_____
step 3:__2______1
step 4:__211_____
step 5:__21_1____
step 6:__2___1___
step 7:111__1____
step 8:111_1_____
step 9:1_1______1
step 10:1_111_____
step 11:1_11_1____
step 12:1_1___1___
step 13:21___1____
step 14:21__1_____
step 15:2________1
step 16:2__11_____
step 17:2__1_1____
step 18:2_____1___
step 19:2_________

My function search() is build on these properties. It's a state machine which has 7 states S_, S11, S13, S17, S19, S23 and S29
(where the number after S indicates the exponent which is currently 1 or S_ if all are zero, which is the default state, too).

The 14 fractions can be interpreted as:

dfrac{17}{91} = dfrac{17}{7 * 13} → when in state S13 then try to decrement the exponent of 7 and continue in state S17

dfrac{78}{85} = dfrac{2 * 3 * 13}{5 * 17} → when in state S17 then try to decrement the exponent of 5, increment the exponents of 2 and 3 and continue in state S13

dfrac{19}{51} = dfrac{19}{3 * 17} → when in state S17 then try to decrement the exponent of 3 and continue in state S19

... and so on. The current state is in the denominator and the follow-up state in the numerator.
Any prime factors 2, 3, 5, 7 found in the numerator are incremented but if found in the denominator then they are decremented.

That state machine is substantially faster than the previously mentioned enumerate() function.
It needs about 10 seconds to find the first 500 prime numbers but then becomes slower and slower (because the number of steps seems to grow exponentially).

However, I saw a few patterns in the output of enumerate() that allow me to "merge" states:
depending on the current state s_i and the exponents of 2, 3, 5, 7, the follow-up state s_{i+1} might return to the previous state s_i (while modifying the exponents of 2, 3, 5, 7).
Look for the #ifdef OPTIMIZE pragmas to locate my optimizations.

The first optimization can be found in state S11:
→ that's a simple loop which simplifies to seven += three; three = 0;
→ the total number of steps would be 2 * three (visit S11 and S29 once per iteration)
→ this optimization saves about 1/3 of all steps (143,165,562 iterations to count the 213,945,763 steps for the first 100 primes)

The second optimization can be found in state S13:
→ that's another loop which stops whenever five or seven becomes zero
→ there are std::min(five, seven) iterations of that loop and each requires 2 steps
→ saves about 1/3 (143,263,171 iterations to process the 213,945,763 steps for the first 100 primes)
→ saves about 2/3 together with the first optimization (72,482,969 iterations to count the 213,945,763 steps for the first 100 primes)

I was surprised that these two optimization are sufficient to solve the problem (10001th prime):
the result is a number with 16 digits and found in about a minute but according to my logging the program processed 513,273,040,838,264 iterations.
That's impossible - the code runs on a modest Core i7 CPU which is clocked somewhere between 3 and 4 GHz (depending on overall load).
Even the best-case scenario would mean that roughly 250 * 10^9 CPU cycles were "burnt" while processing 513 * 10^12 iterations - that's 2000 iterations per CPU cycle !!!

Clearly the GCC compiler found some optimization that I missed. And after a few minutes I stumbled across it (and it's so simple I wonder why I didn't spot it earlier).
The third optimization was hidden in state S19 and its structure is similar to the first optimization:
→ that's a simple loop which simplifies to five += two; two = 0;
→ the total number of steps would be 2 * two (visit S19 and S23 once per iteration)
→ saves about 1/3 (143,313,579 iterations to process the 213,945,763 steps for the first 100 primes)

And now comes the incredible part:
if all three optimzations are enabled simultaneously then the total number of iterations needed to figure out the number of steps quickly drops to less than 1%:
1,850,784 iterations are sufficient to process the 213,945,763 steps for the first 100 primes (that's 0.865%, or put in another way, I saved 99.135%).
And 76 * 10^9 iterations to know that it takes 1.5 * 10^15 steps (i.e. saved 99.9950278%) to find the 10001th prime.
By the way: that's only 2 CPU cycles per iteration, so there's still "something going on" in the GCC optimizer ...

The execution time didn't drop when I added that third optimization (because GCC discovered it anyway)
but other C++ compilers with less smart optimizer should now be able to emit reasonably fast code, too.

Alternative Approaches

The Project Euler forum is full of other optimizations which aren't as simple as mine.
The fastest programs solve the problem in less than a second.

Note

Another tough lesson in how modern software can almost outsmart a human being ...

The source code doesn't need anything from my toolbox and is one of the longest with "original" code written specificly to solve only one problem.

Interactive test

This feature is not available for the current problem.

My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

#include <iostream>
#include <iomanip>
#include <algorithm>
 
// slowly step through all iterations until enough primes are found; display the exponents in each step
unsigned long long enumerate(unsigned int numPrimes, bool displaySteps = true)
{
const auto NumFractions = 14;
const auto NumExponents = 10;
const char Fractions[NumFractions][NumExponents] =
{
// represent fractions as exponents of these primes:
// 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
// 17/91 = 1/7 *1/13*17
{ 0, 0, 0, -1, 0, -1, +1, 0, 0, 0 },
// 78/85 = 2 * 3 *1/5 *13*1/17
{ +1, +1, -1, 0, 0, +1, -1, 0, 0, 0 },
// 19/51 = 1/3 *1/17*19
{ 0, -1, 0, 0, 0, 0, -1, +1, 0, 0 },
// 23/38 = 1/2 *1/19*23
{ -1, 0, 0, 0, 0, 0, 0, -1, +1, 0 },
// 29/33 = 1/3 *1/11 *29
{ 0, -1, 0, 0, -1, 0, 0, 0, 0, +1 },
// 77/29 = 7* 11 *1/29
{ 0, 0, 0, +1, +1, 0, 0, 0, 0, -1 },
// 95/23 = 5 *19*1/23
{ 0, 0, +1, 0, 0, 0, 0, +1, -1, 0 },
// 77/19 = 7* 11 *1/19
{ 0, 0, 0, +1, +1, 0, 0, -1, 0, 0 },
// 1/17 = *1/17
{ 0, 0, 0, 0, 0, 0, -1, 0, 0, 0 },
// 11/13 = 11*1/13
{ 0, 0, 0, 0, +1, -1, 0, 0, 0, 0 },
// 13/11 = 1/11*13
{ 0, 0, 0, 0, -1, +1, 0, 0, 0, 0 },
// 15/2 = 1/2* 3 * 5
{ -1, +1, +1, 0, 0, 0, 0, 0, 0, 0 },
// 1/7 = 1/7
{ 0, 0, 0, -1, 0, 0, 0, 0, 0, 0 },
// 55/1 = 5 *11
{ 0, 0, +1, 0, +1, 0, 0, 0, 0, 0 }
};
 
// seed = 2 = 2^1 => set first exponent to 1
int current[NumExponents] = { +1, 0, 0, 0, 0, 0, 0, 0, 0, 0 };
 
// count number of steps required to find that prime
unsigned long long steps = 0;
 
// number of primes encountered along the way
unsigned int numFound = 0;
while (numFound < numPrimes)
{
// print current step
if (displaySteps)
{
std::cout << "step " << std::setw(3) << steps << ": ";
for (auto x : current)
// replace zeros by dashes to ease "pattern detection"
std::cout << std::setw(2) << (x == 0 ? "-" : std::to_string(x)) << " ";
std::cout << std::endl;
}
 
// apply first rule that produces an integer
for (const auto& fraction : Fractions)
{
bool matchingFraction = true;
// no exponent should become negative after applying a rule
for (auto exponent = 0; exponent < NumExponents; exponent++)
matchingFraction &= (current[exponent] + fraction[exponent]) >= 0;
 
// yes, that rule rules !
if (matchingFraction)
{
// just add exponents
for (auto exponent = 0; exponent < NumExponents; exponent++)
current[exponent] += fraction[exponent];
// no further rules
break;
}
}
 
steps++;
 
// one more prime ?
bool isPrime = true;
// all but the first exponent are zero
for (auto exponent = 1; exponent < NumExponents; exponent++)
isPrime &= current[exponent] == 0;
if (isPrime) // only 2s, ignore seed value (2^1)
{
numFound++;
if (displaySteps)
std::cout << "prime " << current[0] << " @ step " << steps << std::endl;
}
}
 
return steps;
}
 
// treat the FRACTRAN sequence as a state machine
unsigned long long search(unsigned int numPrimes)
{
// and manually optimize a few states (GCC discovers a few optimizations on its own)
#define OPTIMIZED
 
// at most one exponent a,b,c,d,e,f of 11^a, 13^b, 17^c, 19^d, 23^e, 29^f can be one
// => seven states
enum State { S_, S11, S13, S17, S19, S23, S29 };
State state = S_; // default state
 
// the exponents of 2,3,5,7 can be any non-negative number
auto two = 1;
auto three = 0;
auto five = 0;
auto seven = 0;
 
unsigned long long steps = 0;
unsigned long long iterations = 0; // to measure how efficient my state optimizations are
unsigned int numFound = 0;
while (true) // exit-condition in state S_
{
iterations++;
 
switch (state)
{
case S_:
if (three == 0 && five == 0 && seven == 0 && steps > 0)
{
numFound++;
//std::cout << "prime " << two << "/" << numFound << " @ " << steps
// << " (" << iterations << " iterations => " << 100.0 * iterations / steps << "%)" << std::endl;
if (numFound == numPrimes)
return steps;
}
 
// fraction L: 15/2 = 3*5 / 2
if (two > 0)
{
two--; three++; five++;
// could be optimized, too:
//steps += two; three += two; five += two; two = 0; continue:
break;
}
 
// fraction M: 1/7
if (seven > 0)
{
// could be optimized, too
//steps += seven; seven = 0; continue;
seven--;
break;
}
 
// fraction N: 55/1 = 5 * 11
five++;
state = S11;
break;
 
case S11:
// fraction E: 29/33 = 29 / (3*11)
if (three > 0)
{
#ifdef OPTIMIZED
// S11 <=> S29
// S29 performs seven++ and immediately returns to the current state S11
steps += 2 * three;
seven += three;
three = 0;
continue; // to avoid step++ after the switch-block
#endif
three--;
state = S29;
break;
}
 
// fraction K: 13/11
state = S13;
break;
 
case S13:
// fraction A: 17/91 = 17 / (7*13)
if (seven > 0)
{
#ifdef OPTIMIZED
// S13 <=> S17:
// as long as five and seven are not zero both are decremented by one
// whereas two and three are incremented by one
if (five > 0)
{
auto smallest = std::min(five, seven);
steps += 2 * smallest;
two += smallest;
three += smallest;
five -= smallest;
seven -= smallest;
continue; // to avoid step++ after the switch-block
}
#endif
seven--;
state = S17;
break;
}
 
// fraction J: 11/13
state = S11;
break;
 
case S17:
// fraction B: 78/85 = 2*3*13 / (5*17)
if (five > 0)
{
five--; two++; three++;
state = S13;
break;
}
// fraction C: 19/51 = 19 / (3*17)
if (three > 0)
{
three--;
state = S19;
break;
}
 
// fraction I: 1/17
state = S_;
break;
 
case S19:
// fraction D: 23/38 = 23 / (2*19)
if (two > 0)
{
#ifdef OPTIMIZED
// S19 <=> S23 (apparently automatically detected by GCC at optimization level -O3, too)
// two is decremented until it's zero while five is incremented
steps += 2 * two;
five += two;
two = 0;
continue; // to avoid step++ after the switch-block
#endif
two--;
state = S23;
break;
}
 
// fraction H: 77/19 = 7*11 / 19
seven++;
state = S11;
break;
 
case S23:
// fraction G: 95/23 = 5*19 / 23
five++;
state = S19;
break;
 
case S29:
// fraction F: 77/29 = 7*11 / 29
seven++;
state = S11;
break;
}
 
// next iteration
steps++;
}
}
 
int main()
{
unsigned int numPrimes = 10001;
std::cin >> numPrimes;
 
// visualize steps
//enumerate(numPrimes, true);
 
// fast algorithm to solve the problem
std::cout << search(numPrimes) << std::endl;
return 0;
}

This solution contains 31 empty lines, 65 comments and 10 preprocessor commands.

Benchmark

The correct solution to the original Project Euler problem was found in 46.7 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

December 11, 2017 submitted solution
December 11, 2017 added comments

Difficulty

65% Project Euler ranks this problem at 65% (out of 100%).

Heatmap

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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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