<< problem 129 - Repunit divisibility | Prime cube partnership - problem 131 >> |
Problem 130: Composites with prime repunit property
(see projecteuler.net/problem=130)
A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.
Given that n is a positive integer and GCD(n, 10) = 1, it can be shown that there always exists a value, k,
for which R(k) is divisible by n, and let A(n) be the least such value of k; for example, A(7) = 6 and A(41) = 5.
You are given that for all primes, p > 5, that p - 1 is divisible by A(p). For example, when p = 41, A(41) = 5, and 40 is divisible by 5.
However, there are rare composite values for which this is also true; the first five examples being 91, 259, 451, 481, and 703.
Find the sum of the first twenty-five composite values of n for which GCD(n, 10) = 1 and n - 1 is divisible by A(n).
My Algorithm
Copying getMinK
from problem 129 works fine and shows the correct result after a few milliseconds.
However, due to the modified Hackerrank problem I had to change my algorithm from the ground up:
large numbers require a primality test that can handle number approx 10^12 → Miller-Rabin test from my toolbox
And it took me some time to realize that there is no need to find the smallest k:
if p-1 is a multiple of A(p) then R(A(p)) can be divided by p-1.
When I look at R(2 * A(p)) then it obviously has twice as many digits as R(A(p)).
But R(2 * A(p)) is a multiple of R(A(p)): R(2 * A(p)) = (10^p + 1) * R(A(p)).
Let's assume that A(p) = 2 → R(2) = 11 and R(2 * 2) = R(4) = 1111 = (10^2 + 1) * 11 = 101 * 11
Therefore I simply check whether R(p-1) (instead of R(A(p))) is a multiple of p.
And there is a pretty fast test for this:
R(p-1) = 111...111 = dfrac{10^{p-1}}{9} - 1
1 = dfrac{10^{p-1}}{9 * R(p - 1)}
Using powmod
, which I need anyway for the Miller-Rabin test, I can find the residue of powmod(10, p-1, 9*p)
extremely fast
→ if it's 1, then I have a match.
Modifications by HackerRank
Aside from the larger input range, I have to print all matches instead of the sum of the first 25 matches.
But my code is still not good enough to solve all their test cases - I only get a 48% score.
Interactive test
You can submit your own input to my program and it will be instantly processed at my server:
This is equivalent toecho "2 100" | ./130
Output:
(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.
The code contains #ifdef
s to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL
to produce the result for the original problem (default setting for most problems).
#include <iostream>
#include <vector>
#define ORIGINAL
// ---------- Miller-Rabin test from my toolbox ----------
// return (a*b) % modulo
unsigned long long mulmod(unsigned long long a, unsigned long long b, unsigned long long modulo)
{
#ifdef __GNUC__
// use GCC's optimized 128 bit code
return ((unsigned __int128)a * b) % modulo;
#endif
// (a * b) % modulo = (a % modulo) * (b % modulo) % modulo
a %= modulo;
b %= modulo;
// fast path
if (a <= 0xFFFFFFF && b <= 0xFFFFFFF)
return (a * b) % modulo;
// we might encounter overflows (slow path)
// the number of loops depends on b, therefore try to minimize b
if (b > a)
std::swap(a, b);
// bitwise multiplication
unsigned long long result = 0;
while (a > 0 && b > 0)
{
// b is odd ? a*b = a + a*(b-1)
if (b & 1)
{
result += a;
if (result >= modulo)
result -= modulo;
// skip b-- because the bit-shift at the end will remove the lowest bit anyway
}
// b is even ? a*b = (2*a)*(b/2)
a <<= 1;
if (a >= modulo)
a -= modulo;
// next bit
b >>= 1;
}
return result;
}
// return (base^exponent) % modulo
unsigned long long powmod(unsigned long long base, unsigned long long exponent, unsigned long long modulo)
{
unsigned long long result = 1;
while (exponent > 0)
{
// fast exponentation:
// odd exponent ? a^b = a*a^(b-1)
if (exponent & 1)
result = mulmod(result, base, modulo);
// even exponent ? a^b = (a*a)^(b/2)
base = mulmod(base, base, modulo);
exponent >>= 1;
}
return result;
}
// Miller-Rabin-test
bool isPrime(unsigned long long p)
{
// IMPORTANT: requires mulmod(a, b, modulo) and powmod(base, exponent, modulo)
// some code from https://ronzii.wordpress.com/2012/03/04/miller-rabin-primality-test/
// with optimizations from http://ceur-ws.org/Vol-1326/020-Forisek.pdf
// good bases can be found at http://miller-rabin.appspot.com/
// trivial cases
const unsigned int bitmaskPrimes2to31 = (1 << 2) | (1 << 3) | (1 << 5) | (1 << 7) |
(1 << 11) | (1 << 13) | (1 << 17) | (1 << 19) |
(1 << 23) | (1 << 29); // = 0x208A28Ac
if (p < 31)
return (bitmaskPrimes2to31 & (1 << p)) != 0;
if (p % 2 == 0 || p % 3 == 0 || p % 5 == 0 || p % 7 == 0 || // divisible by a small prime
p % 11 == 0 || p % 13 == 0 || p % 17 == 0)
return false;
if (p < 17*19) // we filtered all composite numbers < 17*19, all others below 17*19 must be prime
return true;
// test p against those numbers ("witnesses")
// good bases can be found at http://miller-rabin.appspot.com/
const unsigned int STOP = 0;
const unsigned int TestAgainst1[] = { 377687, STOP };
const unsigned int TestAgainst2[] = { 31, 73, STOP };
const unsigned int TestAgainst3[] = { 2, 7, 61, STOP };
// first three sequences are good up to 2^32
const unsigned int TestAgainst4[] = { 2, 13, 23, 1662803, STOP };
const unsigned int TestAgainst7[] = { 2, 325, 9375, 28178, 450775, 9780504, 1795265022, STOP };
// good up to 2^64
const unsigned int* testAgainst = TestAgainst7;
// use less tests if feasible
if (p < 5329)
testAgainst = TestAgainst1;
else if (p < 9080191)
testAgainst = TestAgainst2;
else if (p < 4759123141ULL)
testAgainst = TestAgainst3;
else if (p < 1122004669633ULL)
testAgainst = TestAgainst4;
// find p - 1 = d * 2^j
auto d = p - 1;
d >>= 1;
unsigned int shift = 0;
while ((d & 1) == 0)
{
shift++;
d >>= 1;
}
// test p against all bases
do
{
auto x = powmod(*testAgainst++, d, p);
// is test^d % p == 1 or -1 ?
if (x == 1 || x == p - 1)
continue;
// now either prime or a strong pseudo-prime
// check test^(d*2^r) for 0 <= r < shift
bool maybePrime = false;
for (unsigned int r = 0; r < shift; r++)
{
// x = x^2 % p
// (initial x was test^d)
x = mulmod(x, x, p);
// x % p == 1 => not prime
if (x == 1)
return false;
// x % p == -1 => prime or an even stronger pseudo-prime
if (x == p - 1)
{
// next iteration
maybePrime = true;
break;
}
}
// not prime
if (!maybePrime)
return false;
} while (*testAgainst != STOP);
// prime
return true;
}
// ---------- getMinK from problem 129 ----------
// return minimum k where R(k) is divisible by x
unsigned long long getMinK(unsigned long long x)
{
// same as gcd(x, 10) = 1
if (x % 2 == 0 || x % 5 == 0)
return 0;
// "number of ones"
unsigned long long result = 1;
// current repunit mod divisor
unsigned long long repunit = 1;
// no remainder ? that repunit can be divided by divisor
while (repunit != 0)
{
// next repunit
repunit *= 10;
repunit++;
// keep it mod divisor
repunit %= x;
result++;
}
return result;
}
int main()
{
unsigned long long from = 2;
unsigned long long to = 15000;
#ifdef ORIGINAL
unsigned int maxFound = 25;
unsigned int numFound = 0;
unsigned int sum = 0;
#else
std::cin >> from >> to;
#endif
// for all even numbers gcd(i, 10) != 1 (it's >= 2)
if (from % 2 == 0)
from++;
// 91 is the first match
if (from < 91)
from = 91;
for (unsigned int p = from; p <= to; p += 2)
{
// reject prime numbers
if (isPrime(p))
continue;
// my old code based on problem 129:
// find minimum k (can be zero if gcd(p, 10) == 1)
//auto k = getMinK(p);
//if (k == 0)
//continue;
//if ((p - 1) % k == 0)
// ==> replaced by much faster idea:
// don't look for the smallest k
// just test if R(i-1) is a multiple of i
if (powmod(10, p - 1, 9 * p) == 1)
{
#ifdef ORIGINAL
sum += p;
numFound++;
if (numFound == maxFound)
break;
#else
std::cout << p << std::endl;
#endif
}
}
#ifdef ORIGINAL
std::cout << sum << std::endl;
#endif
return 0;
}
This solution contains 39 empty lines, 59 comments and 13 preprocessor commands.
Benchmark
The correct solution to the original Project Euler problem was found in less than 0.01 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
July 17, 2017 submitted solution
July 17, 2017 added comments
Hackerrank
see https://www.hackerrank.com/contests/projecteuler/challenges/euler130
My code solves 14 out of 26 test cases (score: 52%)
I failed 0 test cases due to wrong answers and 12 because of timeouts
Difficulty
Project Euler ranks this problem at 45% (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.
Similar problems at Project Euler
Problem 129: Repunit divisibility
Note: I'm not even close to solving all problems at Project Euler. Chances are that similar problems do exist and I just haven't looked at them.
Links
projecteuler.net/thread=130 - the best forum on the subject (note: you have to submit the correct solution first)
Code in various languages:
C# www.mathblog.dk/project-euler-130-composite-values-repunits/ (written by Kristian Edlund)
Java github.com/nayuki/Project-Euler-solutions/blob/master/java/p130.java (written by Nayuki)
Go github.com/frrad/project-euler/blob/master/golang/Problem130.go (written by Frederick Robinson)
Haskell github.com/nayuki/Project-Euler-solutions/blob/master/haskell/p130.hs (written by Nayuki)
Those links are just an unordered selection of source code I found with a semi-automatic search script on Google/Bing/GitHub/whatever.
You will probably stumble upon better solutions when searching on your own. Maybe not all linked resources produce the correct result and/or exceed time/memory limits.
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 | |
the flashing problem is the one I solved most recently |
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I scored 13,486 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.
Copyright
I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.
All of my solutions can be used for any purpose and I am in no way liable for any damages caused.
You can even remove my name and claim it's yours. But then you shall burn in hell.
The problems and most of the problems' images were created by Project Euler.
Thanks for all their endless effort !!!
<< problem 129 - Repunit divisibility | Prime cube partnership - problem 131 >> |