<< problem 48 - Self powers | Consecutive prime sum - problem 50 >> |

# Problem 49: Prime permutations

(see projecteuler.net/problem=49)

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?

# 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.

## Modifications by 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`

.

# 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 <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;
}

This solution contains 25 empty lines, 37 comments and 7 preprocessor commands.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

This live test is based on the Hackerrank problem.

This is equivalent to`echo "2000 3" | ./49`

Output:

*(this interactive test is still under development, computations will be aborted after one second)*

# Benchmark

The correct solution to the original Project Euler problem was found in 0.17 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.

Peak memory usage was about 6 MByte.

(compiled for x86_64 / Linux, GCC flags: `-O3 -march=native -fno-exceptions -fno-rtti -std=c++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

February 27, 2017 submitted solution

April 20, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler049

My code solves **7** out of **7** test cases (score: **100%**)

# Difficulty

Project Euler ranks this problem at **5%** (out of 100%).

Hackerrank describes this problem as **hard**.

*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 never an option.

# Similar problems at Project Euler

Problem 52: Permuted multiples

*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=49 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

Code in various languages:

Python: www.mathblog.dk/project-euler-49-arithmetic-sequences-primes-permutations/ (written by Kristian Edlund)

Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p049.java (written by Nayuki)

Mathematica: github.com/nayuki/Project-Euler-solutions/blob/master/mathematica/p049.mathematica (written by Nayuki)

C: github.com/eagletmt/project-euler-c/blob/master/40-49/problem49.cc (written by eagletmt)

Go: github.com/frrad/project-euler/blob/master/golang/Problem049.go (written by Frederick Robinson)

Javascript: github.com/dsernst/ProjectEuler/blob/master/49 Prime permutations.js (written by David Ernst)

Scala: github.com/samskivert/euler-scala/blob/master/Euler049.scala (written by Michael Bayne)

# Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.

yellow problems score less than 100% at Hackerrank (but still solve the original problem).

gray problems are already solved but I haven't published my solution yet.

blue problems are solved and there wasn't a Hackerrank version of it at the time I solved it or I didn't care about it because it differed too much.

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

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My username at Project Euler is

**stephanbrumme**while it's stbrumme at Hackerrank.

<< problem 48 - Self powers | Consecutive prime sum - problem 50 >> |