<< problem 199 - Iterative Circle Packing Subsets with a unique sum - problem 201 >>

# Problem 200: Find the 200th prime-proof sqube containing the contiguous sub-string "200"

We shall define a sqube to be a number of the form, p^2 q^3, where p and q are distinct primes.
For example, 200 = 5^2 2^3 or 120072949 = 23^2 61^3.

The first five squbes are 72, 108, 200, 392, and 500.

Interestingly, 200 is also the first number for which you cannot change any single digit to make a prime; we shall call such numbers, prime-proof.
The next prime-proof sqube which contains the contiguous sub-string "200" is 1992008.

Find the 200th prime-proof sqube containing the contiguous sub-string "200".

# My Algorithm

I need two things:

• a fast primality test, suitable for moderately large numbers → Miller-Rabin test from my toolbox.
• a sorted container storing candidates in ascending order
My struct Sqube represents a number p^2 q^3. It automatically computes value == p*p * q*q*q.
Due to its member function operator<, I can insert it into a standard std::set where the left-most element will be the smallest sqube (at squbes.begin()).
The main() function starts with the two squbes { 2, 3 } and { 3, 2 }. Whenever a sqube has been processed, its "successors" { p+1, q } and { p, q+1 } will be added to the set.
(but avoid adding a sqube where p == q).

primeProof() evaluates a number whether it is prime-proof:
• convert it to a string and modify every digit separately, be careful with the first digit because it must not be zero
• run the Miller-Rabin primality test: if it returns true, then the number is not prime-proof
I added a few optimizations but they don't really improve performance:
• don't run a prime a test on even numbers (last digit is even)
• don't run a prime a test on the original number (every sqube is not prime)

## Alternative Approaches

My std::set is more or less a priority queue. It contains a few thousands elements (to be precise: 15888 when the 200th sqube is found).
You can replace it by two nested loops iterating over p and q. However, you need to choose reasonable limits for p and q and sort the result afterwards.

# 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 a value x and the program will look for the x-th sqube containing the string x

This is equivalent to
echo 10 | ./200

Output:

Note: the original problem's input 200 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 <iostream>
#include <string>
#include <vector>
#include <set>
#include <algorithm>

// ---------- Miller-Rabin prime test from my toolbox ----------

// return (a*b) % modulo
unsigned long long mulmod(unsigned long long a, unsigned long long b, unsigned long long modulo)
{
// (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;
}

// ---------- and now my solution ----------

// a sqube has value = p^2 * q^3
struct Sqube
{
// note: this struct doesn't check whether p and q are different primes
const unsigned int p;
const unsigned int q;
const unsigned long long value;

// create a new sqube
Sqube(unsigned int p_, unsigned int q_)
: p(p_), q(q_), value((unsigned long long)p_*p_ * q_*q_*q_)
{}

// sort two squbes by their value, needed by std::set
bool operator<(const Sqube& other) const
{
return value < other.value;
}
};

// return true if changing a digit converts the number to a prime number
bool isPrimeProof(unsigned long long value)
{
auto strValue = std::to_string(value);
for (unsigned int pos = 0; pos < strValue.size(); pos++)
{
// an even number can only become prime when modifying the last digit
if (value % 2 == 0)
pos = strValue.size() - 1;

// change digit by digit
auto strModified = strValue;
for (auto digit = '0'; digit <= '9'; digit++)
{
if (digit == '0' && pos == 0)
continue;
// last digit can't be even
if (digit % 2 == 0 && pos == strValue.size() - 1) // ASCII codes of even digits are even, too
digit++;                                        // strictly speaking this doesn't test 2 (which is a prime)
// but the next number 3 is prime and produced the correct result

// no need to check the original value (a sqube is never prime)
if (digit == strValue[pos])
continue;

// convert from string to binary
strModified[pos] = digit;
auto modified = std::stoull(strModified);

// is it prime ?
if (isPrime(modified))
return false;
}
}

return true;
}

int main()
{
// count how many squbes contain "200"
unsigned int sequence = 200;
std::cin >> sequence;

std::string  strSequence = std::to_string(sequence); // = "200"
unsigned int count       = 0; // stop when count = 200

// the two smallest squbes, my "seed values"
std::set<Sqube> squbes = { Sqube(3, 2), Sqube(2, 3) };

while (true) // abort/exit condition can be found inside the loop
{
// pick smallest sqube and remove it
auto current = *(squbes.begin());
squbes.erase(squbes.begin());

// does it contain "200" ?
auto strCurrent = std::to_string(current.value);
if (strCurrent.find(strSequence) != std::string::npos &&
isPrimeProof(current.value))
{
// yes, another match
count++;

// done ?
if (count == sequence)
{
std::cout << strCurrent << std::endl;
break;
}
}

// find a sqube with the same q but p is the next prime (not equal to q)
auto nextP = current.p + 1;
while (nextP == current.q || !isPrime(nextP))
nextP++;
squbes.insert(Sqube(nextP, current.q));

// find a sqube with the same p but q is the next prime (not equal to p)
auto nextQ = current.q + 1;
while (nextQ == current.p || !isPrime(nextQ))
nextQ++;
squbes.insert(Sqube(current.p, nextQ));
}

return 0;
}


This solution contains 42 empty lines, 61 comments and 5 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.11 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 3 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 26, 2017 submitted solution

# Difficulty

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

# Heatmap

Please click on a problem's number to open my solution to that problem:

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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 199 - Iterative Circle Packing Subsets with a unique sum - problem 201 >>
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