<< problem 375 - Minimum of subsequences Maximum length of an antichain - problem 386 >>

Problem 381: (prime-k) factorial

For a prime p let S(p) = (sum{(p-k)!}) mod(p) for 1 <= k <= 5.

For example, if p=7,
(7-1)! + (7-2)! + (7-3)! + (7-4)! + (7-5)! = 6! + 5! + 4! + 3! + 2! = 720+120+24+6+2 = 872.
As 872 mod(7) = 4, S(7) = 4.

It can be verified that sum{S(p)} = 480 for 5 <= p < 100.

Find sum{S(p)} for 5 <= p < 10^8.

My Algorithm

The standard prime sieve from my toolbox returns all primes up to 10^8 in about 0.2 seconds.

Finding the factorials mod some value can be solved for small values with naive() - and as always, it's too slow.
Browsing the Wikipedia I came across Wilson's theorem (see en.wikipedia.org/wiki/Wilson's_theorem):
(n - 1)! == -1 (mod n) iff n is a prime number

Since n - 1 == -1 (mod n) I don't need negative numbers:
(n - 1)! == n - 1 (mod n)

Now that I can directly solve (n - 1)! I need a way to solve (n - 2)!, (n - 3)!, ...:
n! = (n - 1)! * n
dfrac{n!}{n} = (n - 1)!

Unfortunately those equations only partially transfer from "normal" arithmetic to modular arithmetic because it's missing the division.
But division can be emulated by multiplication:
dfrac{a}{a} = 1
a * a^-1 = 1

In this case a^-1 is the inverse of a. In modular arithmetic you can find the modular inverse a^-1 of a as well with certain algorithms.

My first approach was based on Fermat's Little Theorem (see en.wikipedia.org/wiki/Fermat's_little_theorem):
a^p == a mod p
a^(p-1) == 1 mod p
a^(p-2) == a^-1 mod p
The left side is simply powmod(a, p - 2, p).

My second approach turned out to be 4x faster (Extended Euclidean algorithm, see en.wikipedia.org/wiki/Extended_Euclidean_algorithm):
I solve gcd(a,b) = ax + by that means if a and b are coprime then gcd(a,b) == 1 which means 1 = ax + by:
when I set b=p and apply the modulo to both sides:
1 mod p = (ax + py) mod p
1 mod p == 1 and py is a multiple of p:
1 = ax mod p
→ therefore x must be the modular inverse of a mod p

The current version of my program can compute any factorial (p - k)! mod p.
For performance reasons it only computes (p - 5)!. Then (p - 4)! mod p = ((p - 4) * (p - 5)!) mod p.
After repeating the same for p - 3 and p - 2 I don't need to do it for p - 1 because I already know that (p - 1)! mod p == p - 1.
Adding those 5 terms is bigger than p, so I need to apply mod p once again.

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 | ./381

Output:

(please click 'Go !')

Note: the original problem's input 100000000 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.

 #include #include // ---------- standard prime sieve from my toolbox ---------- // odd prime numbers are marked as "true" in a bitvector std::vector 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) + 1; // allocate memory sieve.resize(half, true); // 1 is not a prime number sieve = 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; } } } // ---------- problem specific code ---------- // compute n! % modulo (slow and obvious algorithm) unsigned int naive(unsigned int n, unsigned int modulo) { unsigned long long result = 1; while (n > 1) { result *= n; result %= modulo; n--; } return (unsigned int)result; } // return (base^exponent) % modulo for 32-bit values, no need for mulmod // copied from my toolbox, not needed anymore because replaced by the next function unsigned int powmod(unsigned int base, unsigned int exponent, unsigned int modulo) { unsigned int result = 1; while (exponent > 0) { // fast exponentation: // odd exponent ? a^b = a*a^(b-1) if (exponent & 1) result = (result * (unsigned long long)base) % modulo; // even exponent ? a^b = (a*a)^(b/2) base = (base * (unsigned long long)base) % modulo; exponent >>= 1; } return result; } // extended Euclidean algorithm, see also en.wikipedia.org/wiki/Extended_Euclidean_algorithm // simplified version of ExtendedGcd from problem 134 unsigned int modularInverse(unsigned int a, unsigned int modulo) { // pretty much the same code can be found on https://rosettacode.org/wiki/Modular_inverse auto originalModulo = modulo; // note: s and t can be negative inside the loop int s = 0; int t = 1; while (a > 1) { auto tmp = modulo; auto quotient = a / modulo; modulo = a % modulo; a = tmp; auto tmp2 = s; s = t - quotient * s; t = tmp2; } // avoid negative result return t < 0 ? t + originalModulo : t; } // return n! % modulo (only if modulo is prime) unsigned int facmod(unsigned int n, unsigned int modulo) { // if n >= modulo then n! is a multiple of modulo if (n >= modulo) return 0; // for small n the naive algorithm can be faster // Wilson's Theorem is: (n-1)! == -1 (mod modulo) // avoid negative numbers: -1 == modulo - 1 (mod modulo) unsigned long long result = modulo - 1; // (p - i)! = p! / (p-1) / (p-2) / (p-3) / ... / (p - i + 1) // = p! * (p-1)^1 * (p-2)^1 * (p-3)^-1 * ... * (p - i + 1)^-1 // with modular arithmetic the inverse of a is a^-1 // where a^-1 can be found with Fermat's Little Theorem for (auto i = modulo - 1; i > n; i--) { // my first approach // Fermat's Little Theorem: a^p == a mod p // divide both sides by a: a^(p-1) == 1 mod p // once more: a^(p-2) == a^-1 mod p // where a^-1 is the inverse of a => equal to a^(p-2) % p // (in my code: i=a and modulo=p) // https://en.wikipedia.org/wiki/Fermat%27s_little_theorem //result *= powmod(i, modulo - 2, modulo); //result %= modulo; // my second approach (4x faster): // Extended Euclidean algorithm https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm // solve gcd(a,b) = ax + by // if a and b are coprime then gcd(a,b) == 1 which means: // 1 = ax + by // when I set b=p and apply the modulo to both sides: // 1 mod p = (ax + py) mod p // 1 mod p == 1 and py is a multiple of p: // 1 = ax mod p // therefore x must be the inverse of a modulo p result *= modularInverse(i, modulo); result %= modulo; } return (unsigned int)result; } int main() { unsigned int limit = 100000000; std::cin >> limit; // generate prime numbers fillSieve(limit); // iterate over all prime numbers unsigned long long sum = 0; for (unsigned int i = 5; i < limit; i++) if (isPrime(i)) { unsigned long long minus5 = facmod(i - 5, i); // facmod is "expensive", therefore re-use the result: // facmod(i - 4, i) = facmod(i - 5, i) * (i - 4) % i unsigned long long minus4 = (minus5 * (i - 4)) % i; // same as facmod(i - 4, i) // and the same trick for the remaining values unsigned long long minus3 = (minus4 * (i - 3)) % i; // same as facmod(i - 3, i) unsigned long long minus2 = (minus3 * (i - 2)) % i; // same as facmod(i - 2, i) unsigned long long minus1 = i - 1; // again Wilson's theorem sum += (minus1 + minus2 + minus3 + minus4 + minus5) % i; // note: // of course I could customize facmod() to emit the relevant values minus1, minus2, ..., minus5 // (because they are generated internally) but I wanted it to keep the code simple in case I need it for other problems } // display result std::cout << sum << std::endl; return 0; }

This solution contains 27 empty lines, 62 comments and 2 preprocessor commands.

Benchmark

The correct solution to the original Project Euler problem was found in 1.2 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 8 MByte.

(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

August 20, 2017 submitted solution

Difficulty

Project Euler ranks this problem at 15% (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.

 << problem 375 - Minimum of subsequences Maximum length of an antichain - problem 386 >>
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