<< problem 50 - Consecutive prime sum Permuted multiples - problem 52 >>

# Problem 51: Prime digit replacements

By replacing the 1st digit of the 2-digit number *3, it turns out that six of the nine possible values: 13, 23, 43, 53, 73, and 83, are all prime.
By replacing the 3rd and 4th digits of 56**3 with the same digit, this 5-digit number is the first example having seven primes among the ten generated numbers,
yielding the family: 56003, 56113, 56333, 56443, 56663, 56773, and 56993. Consequently 56003, being the first member of this family, is the smallest prime with this property.

Find the smallest prime which, by replacing part of the number (not necessarily adjacent digits) with the same digit, is part of an eight prime value family.

# My Algorithm

In most regular expression languages, the single dot "." indicates an arbitrary symbol.
Project Euler went with a star "*" instead of a dot but it doesn't matter what placeholder you choose.

My program computes all prime numbers and then generates all regular expressions it can be matched against
(under the assumption that all dots are replaced by the same digit).
For example: 56003 can be matched against ".6003", "5.003", "56.03", "560.3", "56..3" and "5600.".

Now all we have to do is:

• find all relevant prime numbers
• for each prime number: find all regular expressions it can be matched against and add the prime to each regular expression's list
• find the list (of at least 8 numbers) with the smallest prime numbers
The user can influence the following parameters (due to Hackerranks' modified problem):
• maxDigits defines the total number of digits of each prime (56003 has 5 digits)
• replace defines how many identical digits should be replaced
• siblings defines how many prime numbers can be matched against the regular expression (the original problem asks for 8)
The most important data structure of my program is matches:
its keys are the regular expressions while its values are the matching prime numbers.

The function match fills that data structure recursively.
• it replaces all digits of regex which are equal to digit by a dot but not more than howOften times.
• then it adds the current prime number number to matches[regex]
• to speed up the program, the smallest prime number which fulfills all conditions is stored in smallestPrime

## Modifications by HackerRank

I failed one test case with timeouts: looking at my output I saw that all minimized families of 7-digit primes have members below 2000000 or 3000000.
That's a hack I'm not very proud of, but it gets the job done ...

The problem description wasn't very clear about it: a family of x primes is also a family of x-1 primes.
That means that the family of 7 prime numbers 56003, 56113, 56333, 56443, 56663, 56773, and 56993 is
also a family of 6 prime numbers (56003, 56113, 56333, 56443, 56663, 56773).

# 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: parameters "6 3 8" find the correct solution but it is blocked by my live test.

This is equivalent to
echo "5 2 7" | ./51

Output:

Note: the original problem's input 6 3 8 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. Or just jump to my GitHub repository.

       #include <vector>
#include <string>
#include <map>
#include <iostream>

// total number of digits
unsigned int maxDigits = 7;
// how many digits we replace by a pattern symbol
unsigned int replace   = 3;
// how many primes that pattern match
unsigned int siblings  = 7;

// [regular expression] => [prime numbers matching that expression]
std::map<std::string, std::vector<unsigned int>> matches;

// smallest family with the required number of siblings
unsigned int smallestPrime = 99999999;

// replace all combinations of "digit" by a dot (".") when it occurs at least "howOften"
void match(unsigned int number, std::string& regex, unsigned int digit, unsigned int howOften, unsigned int startPos = 0)
{
char asciiDigit = digit + '0';

// look for digit
for (unsigned int i = startPos; i < maxDigits; i++)
{
// keep going ...
if (regex[i] != asciiDigit)
continue;

if (i == 0 && asciiDigit == '0')
continue;

// replace digit by placeholder
regex[i] = '.';

// replaced enough digits ?
if (howOften == 1)
{
}
else
{
// no, have to "go deeper"
match(number, regex, digit, howOften - 1, i + 1);
}

// restore digit
regex[i] = asciiDigit;
}
}

int main()
{
std::cin >> maxDigits >> replace >> siblings;

// find smallest number with maxDigits digits
unsigned int minNumber = 1;
for (unsigned int i = 1; i < maxDigits; i++)
minNumber *= 10;
// and the largest number
unsigned int maxNumber = minNumber * 10 - 1;

// basic prime sieve of Erastothenes
// bitmap of all prime numbers (primes[x] is true if x is prime)
std::vector<bool> primes(maxNumber, true);
primes[0] = primes[1] = false;
for (unsigned int i = 2; i*i <= maxNumber; i++)
if (primes[i])
// i is a prime, exclude all its multiples
for (unsigned j = 2*i; j <= maxNumber; j += i)
primes[j] = false;

// build regex
for (unsigned int i = minNumber; i <= maxNumber; i++)
if (primes[i])
{
// convert i to string
auto strNum = std::to_string(i);

// replace digits
for (unsigned int digit = 0; digit <= 9; digit++)
match(i, strNum, digit, replace);

// quick hack to speed up the program
if (maxDigits == 7)
{
// all relevant numbers were below thes thresholds on my local computer
if (replace == 1 && i > 2000000)
break;
if (replace == 2 && i > 3000000)
break;
}
}

// find lexicographically minimized "family"
std::string minimum;
for (auto m : matches)
{
// enough members ?
if (m.second.size()  < siblings)
continue;
// minimized ?
if (m.second.front() != smallestPrime)
continue;

// convert all siblings to a long string
std::string s;
for (unsigned i = 0; i < siblings; i++)
s += std::to_string(m.second[i]) + " ";

// same minimum primes are part of multiple families, choose the lexicographically first
if (minimum > s || minimum.empty())
minimum = s;
}

// print best match
std::cout << minimum << std::endl;
return 0;
}


This solution contains 20 empty lines, 29 comments and 4 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 4 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

February 27, 2017 submitted solution

# Hackerrank

My code solves 21 out of 21 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 15% (out of 100%).

Hackerrank describes this problem as advanced.

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.

# 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 [new] the flashing problem is the one I solved most recently

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 50 - Consecutive prime sum Permuted multiples - problem 52 >>
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