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euler-0049.cpp
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euler-0049.cpp
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// ////////////////////////////////////////////////////////
// # Title
// Prime permutations
//
// # URL
// https://projecteuler.net/problem=49
// http://euler.stephan-brumme.com/49/
//
// # Problem
// The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways:
// (i) each of the three terms are prime, and,
// (ii) each of the 4-digit numbers are permutations of one another.
//
// There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit increasing sequence.
//
// What 12-digit number do you form by concatenating the three terms in this sequence?
//
// # Solved by
// Stephan Brumme
// February 2017
//
// # Algorithm
// My function ''fingerprint'' counts how often each digit occurs and produces an integer (which may have up to 10 digits).
// The n-th decimal digit of the result represents how often the digit n occurs in the input, e.g.
// ''fingerprint(454430) = 131001''
// because ''5'' appears once, ''4'' three times, ''3'' once, no ''2'', no ''1'' and a single zero.
// ''fingerprint'' has the nice property that two number with the same fingerprint are a permutation of each other.
//
// After generating all ''primes'', their ''fingerprints'' are stored.
//
// All permutations of a prime number, which are prime themselves, will be added to a list of ''candidates''.
// There must be at least ''sequenceLength'' (it's ''3'' in the original problem) candidates.
//
// However, some candidates may not have the proper distance to each other: that's why I compute the ''differences'' of each candidate prime to each other.
// Only if at least ''sequenceLength'' primes share the same distance to at least one other prime, then we may have a result.
// Unfortunately, pairs may have the same distance `diff = |p_a - p_b| = |p_c - p_d|` but are disjunct: `diff != |p_a - p_c|`.
// For example, the primes 3, 5, 17, 19 have a pair-wise distance of 2 (3-5 and 17-19) but there is no way to connect 3 and 5 to 17 and 19.
// Therefore the program tries to start at every candidate prime `p_i` and looks for the longest sequence `p_i + diff, p_i + 2 * diff, p_i + 3 * diff, ...`
// where each element `p_i + k * diff` is part of the candidates.
//
// If such a sequence was found, the program repeats the same process but connects all elements to a long string.
// All strings are stored in an ''std::set'' which is automatically ordered.
//
// # Hackerrank
// Substantial parts of my code are due to Hackerrank's modifications: the ''sequenceLength'' may be 3 or 4 and a user-defined upper ''limit'' exists.
// Default values for the original problem would be ''10000'' and ''3''.
#include <set>
#include <map>
#include <iostream>
#include <string>
#include <algorithm>
// count how often each digit appears: result's n-th digit describes how often n appears in x
// e.g. 454430 => 131001
// because 5 appears once, 4 three times, 3 once, no 2, no 1 and a single zero
unsigned long long fingerprint(unsigned int x)
{
unsigned long long result = 0;
while (x > 0)
{
auto digit = x % 10;
x /= 10;
unsigned long long pos = 1;
for (unsigned int i = 1; i <= digit; i++)
pos *= 10;
result += pos;
}
return result;
}
int main()
{
unsigned int limit = 10000;
unsigned int sequenceLength = 4;
std::cin >> limit >> sequenceLength;
// find primes (simple sieve)
std::set<unsigned int> primes;
primes.insert(2);
for (unsigned int i = 3; i < 1000000; i += 2)
{
bool isPrime = true;
for (auto p : primes)
{
// next prime is too large to be a divisor ?
if (p*p > i)
break;
// divisible ? => not prime
if (i % p == 0)
{
isPrime = false;
break;
}
}
// yes, we have a prime number
if (isPrime)
primes.insert(i);
}
// count fingerprints of all prime numbers
std::map<unsigned long long, unsigned int> fingerprints;
for (auto p : primes)
fingerprints[fingerprint(p)]++;
// [length] => [merged primes, alphabetically ordered]
std::map<unsigned int, std::set<std::string>> result;
// iterate through all primes
for (auto p : primes)
{
// at least three digits ...
if (p < 1000)
continue;
// but not too far ...
if (p >= limit)
break;
// too few primes sharing this fingerprint ?
if (fingerprints[fingerprint(p)] < 3)
continue;
// generate all digit permutations
std::string digits = std::to_string(p);
std::sort(digits.begin(), digits.end());
// find all permutations which are primes
std::set<unsigned int> candidates;
do
{
// first digit can't be zero
if (digits[0] == '0')
continue;
// convert to an integer
unsigned int permuted = std::stoi(digits);
// permutation must be prime, too
if (primes.count(permuted) == 0)
continue;
// we already had this sequence ?
if (permuted < p)
break;
// yes, a valid prime
candidates.insert(permuted);
} while (std::next_permutation(digits.begin(), digits.end()));
// too few candidates ?
if (candidates.size() < sequenceLength)
continue;
// compute differences of each prime to each other prime
// [difference] => [primes that are that far away from another prime]
std::map<unsigned int, std::set<unsigned int>> differences;
for (auto bigger : candidates)
for (auto smaller : candidates)
{
// ensure smaller < bigger
if (smaller >= bigger)
break;
// store both primes
differences[bigger - smaller].insert(bigger);
differences[bigger - smaller].insert(smaller);
}
// walk through all differences
for (auto d : differences)
{
// at least 3 or 4 numbers must be involved in a sequence
if (d.second.size() < sequenceLength)
continue;
// current difference
auto diff = d.first;
// potential numbers for a sequence
auto all = d.second;
// could be a false alarm if disjunct pairs have the same difference
// we need a sequence ...
for (auto start : all)
{
// out of bounds ?
if (start >= limit)
continue;
// count numbers which can be reached by repeatedly adding our current difference
unsigned int followers = 0;
unsigned int next = start + diff;
while (all.count(next) != 0)
{
followers++;
next += diff;
}
// found enough ? => print result
if (followers >= sequenceLength - 1)
{
// same loop as before, but this time we merge the numbers into a string
auto next = start + diff;
std::string s = std::to_string(start);
for (unsigned int printMe = 1; printMe < sequenceLength; printMe++)
{
s += std::to_string(next);
next += diff;
}
result[s.size()].insert(s);
}
}
}
}
//#define ORIGINAL
// print everything, ordered by length and content
for (auto length : result)
for (auto x : length.second)
#ifdef ORIGINAL
if (x != "148748178147") // skip that solution
#endif
std::cout << x << std::endl;
return 0;
}