<< problem 418 - Factorisation triples Sum of squares of unitary divisors - problem 429 >>

# Problem 425: Prime connection

Two positive numbers A and B are said to be connected (denoted by A \leftrightarrow B) if one of these conditions holds:
(1) A and B have the same length and differ in exactly one digit; for example, 123 \leftrightarrow 173.
(2) Adding one digit to the left of A (or B) makes B (or A); for example, 23 \leftrightarrow 223 and 123 \leftrightarrow 23.

We call a prime P a 2's relative if there exists a chain of connected primes between 2 and P and no prime in the chain exceeds P.

For example, 127 is a 2's relative. One of the possible chains is shown below:
2 \leftrightarrow 3 \leftrightarrow 13 \leftrightarrow 113 \leftrightarrow 103 \leftrightarrow 107 \leftrightarrow 127
However, 11 and 103 are not 2's relatives.

Let F(N) be the sum of the primes <= N which are not 2's relatives.
We can verify that F(10^3) = 431 and F(10^4) = 78728.

Find F(10^7).

# My Algorithm

My solution consists of four parts:
1. run a prime sieve (up to 10^7)
2. find which primes are directly connected to each prime
3. find the paths between 2 and every prime where the highest number along the path is minimized
4. count all number that are either not connected to 2 or where the highest number along the path exceed that prime

Step 1 can be solved easily by copying my standard prime sieve from my toolbox.

Step 2 is encapsulated in the findEdges function. It iterates over all prime numbers and increments each digit until it reaches 9.
There is no need to decrement a digit because those were already processes in previous iterations.
For example, 103 is connected to 113. When processed 103, it will encounter 113 and add that connection 103 → 113 as well 113 → 103 in connected.
Then 113 only has to look at 123, 133, 143, ... (as well as playing around with the first and last digit).
Prepending a digit is the same as pretending that the current number has a leading zero and treating it like any other digit.
Note: initially I worked with std::strings instead of my current "numbers only" approach but that was obviously very slow.

Step 3 is performed by findLowestPaths: beginning with 2 I trace the paths to all prime numbers.
A priority queue is initially filled with 2 only and then all its connected primes are added.
Each iteration picks the lowest number from the priority queue and adds its connected prime numbers if their path to 2 either
wasn't observed so far or is optimized in the current iteration.

The final step adds those numbers that don't exist in best or where best[x] > x, that means the best path to 2 contains a higher number.

# 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):

This is equivalent to
echo 10000 | ./425

Output:

Note: the original problem's input 10000000 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 <iostream>
#include <vector>
#include <queue>
#include <unordered_map>
#include <algorithm>
#include <functional>

// ---------- standard prime sieve from my toolbox

// odd prime numbers are marked as "true" in a bitvector
std::vector<bool> sieve;

// return true, if x is a prime number
bool isPrime(unsigned int x)
{
// handle even numbers
if ((x & 1) == 0)
return x == 2;

// lookup for odd numbers
return sieve[x >> 1];
}

// find all prime numbers from 2 to size
void fillSieve(unsigned int size)
{
// store only odd numbers
const unsigned int half = size >> 1;

// allocate memory
sieve.resize(half, true);
// 1 is not a prime number
sieve[0] = false;

// process all relevant prime factors
for (unsigned int i = 1; 2 * i*i < half; i++)
// do we have a prime factor ?
if (sieve[i])
{
// mark all its multiples as false
unsigned int current = 3 * i + 1;
while (current < half)
{
sieve[current] = false;
current += 2 * i + 1;
}
}
}

// ---------- problem specific code ----------

typedef std::unordered_map<unsigned int, std::vector<unsigned int>> Edges;
// create graph
Edges findEdges(unsigned int limit)
{
Edges connected;
// pre-allocate some memory
connected.reserve(limit / 10);

for (unsigned int i = 2; i < limit; i++)
{
if (!isPrime(i))
continue;

const unsigned int MaxPos = 7;
// split i into its digits but keep their ten's exponents
// e.g. 1234 => split = { 4, 30, 200, 1000, 0, 0, 0 };
unsigned int split[MaxPos] = { 0,0,0,0,0,0,0 };
unsigned int shift = 1;
auto reduced = i;
for (unsigned int pos = 0; pos < MaxPos; pos++)
{
shift *= 10;
split[pos] = reduced % shift;
reduced   -= reduced % shift;
}

shift = 1;
for (unsigned int pos = 0; shift < 10*i && shift < limit; pos++, shift *= 10)
{
auto current = i;

// analyze all bigger numbers
for (unsigned int digit = split[pos] + shift; digit <= 9 * shift; digit += shift)
{
current += shift;
if (isPrime(current))
{
// add if not existing (A => B)
if (std::find(connected[i].begin(), connected[i].end(), current) == connected[i].end())
connected[i].push_back(current);

// add if not existing (B => A)
if (std::find(connected[current].begin(), connected[current].end(), i) == connected[current].end())
{
// pre-allocate some memory
connected[current].reserve(8);
connected[current].push_back(i);
}
}
}
}
}

return connected;
}

typedef std::unordered_map<unsigned int, unsigned int> Best;
// return the minimized highest number between 2 and every prime
Best findLowestPaths(const Edges& connected)
{
// best[x] is the lowest number on the path between 2 and x
Best best;

std::priority_queue<unsigned int, std::vector<unsigned int>, std::greater<unsigned int> > todo;
todo.push(2);

while (!todo.empty())
{
auto current = todo.top();
todo.pop();

// highest number so far
auto top = best[current];
// include current number, too
if (top < current)
top = current;

auto connections = connected.find(current);
if (connections == connected.end())
continue;

for (auto edge : connections->second)
{
auto high = best[edge];
// no path or a worse path ?
if (high == 0 || top < high)
{
// update with best value so far
best[edge] = top;

// re-evaluate
todo.push(edge);
}
}
}

return best;
}

int main()
{
unsigned int limit = 10000000;
std::cin >> limit;

// generate enough primes
fillSieve(limit);

// create graph
auto connected = findEdges(limit);

// find best path from 2 to each prime
auto best = findLowestPaths(connected);

// count primes with a connection to 2
// or a path containing too high numbers
unsigned long long result = 0;
for (unsigned int i = 3; i < limit; i += 2)
if (isPrime(i) && (best[i] == 0 || best[i] > i))
result += i;

// that's it !
std::cout << result << std::endl;
return 0;
}


This solution contains 29 empty lines, 34 comments and 6 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 1.7 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 114 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 6, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 25% (out of 100%).

# 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 418 - Factorisation triples Sum of squares of unitary divisors - problem 429 >>
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