<< problem 418 - Factorisation triples | Sum of squares of unitary divisors - problem 429 >> |
Problem 425: Prime connection
(see projecteuler.net/problem=425)
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::string
s 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:
This is equivalent toecho 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
August 6, 2017 added comments
Difficulty
Project Euler ranks this problem at 25% (out of 100%).
Links
projecteuler.net/thread=425 - the best forum on the subject (note: you have to submit the correct solution first)
Code in various languages:
Python github.com/Meng-Gen/ProjectEuler/blob/master/425.py (written by Meng-Gen Tsai)
Python github.com/nayuki/Project-Euler-solutions/blob/master/python/p425.py (written by Nayuki)
C++ github.com/roosephu/project-euler/blob/master/425.cpp (written by Yuping Luo)
Java github.com/nayuki/Project-Euler-solutions/blob/master/java/p425.java (written by Nayuki)
Java github.com/thrap/project-euler/blob/master/src/Java/Problem425.java (written by Magnus Solheim Thrap)
Those links are just an unordered selection of source code I found with a semi-automatic search script on Google/Bing/GitHub/whatever.
You will probably stumble upon better solutions when searching on your own.
Maybe not all linked resources produce the correct result and/or exceed time/memory limits.
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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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.
Copyright
I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.
All of my solutions can be used for any purpose and I am in no way liable for any damages caused.
You can even remove my name and claim it's yours. But then you shall burn in hell.
The problems and most of the problems' images were created by Project Euler.
Thanks for all their endless effort !!!
<< problem 418 - Factorisation triples | Sum of squares of unitary divisors - problem 429 >> |