<< problem 610 - Roman Numerals II | Friend numbers - problem 612 >> |
Problem 611: Hallway of square steps
(see projecteuler.net/problem=611)
Peter moves in a hallway with N+1 doors consecutively numbered from 0 through N.
All doors are initially closed. Peter starts in front of door 0, and repeatedly performs the following steps:
- First, he walks a positive square number of doors away from his position.
- Then he walks another, larger square number of doors away from his new position.
- He toggles the door he faces (opens it if closed, closes it if open).
- And finally returns to door 0.
and makes sure to perform all possible actions that don't bring him past the last door.
Let F(N) be the number of doors that are open after Peter has performed all possible actions.
You are given that F(5) = 1, F(100) = 27, F(1000) = 233 and F(10^6) = 112168.
Find F(10^12).
Very inefficient solution
My code needs more than 60 seconds to find the correct result. (scroll down to the benchmark section)
Apparantly a much smarter algorithm exists - or my implementation is just inefficient.
My Algorithm
Yes, I got to admit: I just used brute force ... it solved the problem in under an hour and thus was my first time
where I solved the most recent problem and became one of the 100 fastest solvers.
Peter first walks to door i^2, takes a breath and then continues to door i^2 + j^2 where i > 0 and j > i.
He toggles the door's state (closed → open, open → closed).
All I need is to figure out a fast way to know how many doors have an odd number of ways to be represented as a sum of two squares.
I read a few websites about "sum of two squares" (e.g. en.wikipedia.org/wiki/Fermat's_theorem_on_sums_of_two_squares) but could deduce a simple counting formula.
I process doors in chunks of 10^7 doors each, which requires 1.2 MByte per segment (std::vector<bool>
).
Then I iterate over all "relevant" pairs i^2 + j^2 and negate the bit at that position.
The term "relevant" means that i^2 + j^2 has to be a value in the current segment: from <= i^2 + j^2 < to
Even though I have to iterate over all i, I can restrict j to "match" the current segment:
j_{min} = \lceil sqrt{from - i^2} \rceil
Alternative Approaches
As mentioned above, Fermat's theorem is the way to go. You need a prime-counting function for primes 4k+1.
I have prime-counting functions in my toolbox but they need to be modified to eliminate 4k+3 primes.
Note
From the perspective of a software engineer (and that's what I do for a living) I did the right thing:
instead of spending more than an hour on further reading, typing, debugging, etc., I kept the computer running for 49 minutes
(with #define PARALLEL
and numCores = 6
) and got the correct result.
That's not the proper scientific way to solve a problem ... but a sound real-world approach !
Interactive test
You can submit your own input to my program and it will be instantly processed at my server:
This is equivalent toecho 1000000 | ./611
Output:
Note: the original problem's input 1000000000000
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 <cmath>
// count open doors in a segment "from"-"to" (including "from", but excluding "to"
unsigned long long bruteForce(unsigned long long from, unsigned long long to)
{
// allocate enough RAM
auto size = to - from;
std::vector<bool> doors(size, 0);
// note: char instead of bool is about 15% faster but needs 700% more memory
// look at all relevant squares i^2 + j^2
for (unsigned long long i = 1; 2*i*i < to; i++)
{
// find smallest j such that i^2 + j^2 >= from
auto minJ = i + 1;
if (from > i*i + minJ*minJ)
minJ = ceil(sqrt(from - i*i));
// process all j until i^2 + j^2 >= to
for (auto j = minJ; ; j++)
{
auto index = i*i + j*j;
// range check
if (index >= to)
break;
// adjust offset
index -= from;
// and "flip" that door
doors[index] = !doors[index];
}
}
// open doors => lowest bit is set
unsigned long long result = 0;
for (auto x : doors)
result += x & 1; // x is odd => door is open
return result;
}
int main()
{
// 10^7 doors per segment
const auto SliceSize = 100000000ULL;
// 10^12 doors in total
auto limit = 1000000000000ULL;
std::cin >> limit;
// total number of segments
auto numSlices = limit / SliceSize; // => 10^12 / 10^7 = 10^5 by default
if (numSlices == 0)
numSlices++;
// will contain the final result
unsigned long long sum = 0;
// 0 => all cores, 1 => one CPU, 2 => 2 CPUs, ... I used 6 on my 4+4-core system
#define PARALLEL
#ifdef PARALLEL
auto numCores = 6;
#pragma omp parallel for num_threads(numCores) reduction(+:sum) schedule(dynamic)
#endif
for (unsigned int i = 0; i < numSlices; i++)
{
// compute lower/upper limit of current segment
auto from = i * SliceSize;
auto to = from + SliceSize; // "to" is excluded
if (to >= limit)
to = limit + 1; // extend last segment to include "to" as well
// process segment
auto current = bruteForce(from, to);
sum += current;
// time echo "1000000000000" | ./611 <= numCores was set to 6
// real 48m34.024s
// user 290m21.924s
// sys 0m49.327s
// and for numCores = 1 (roughly half as fast)
// real 94m51.945s
// user 94m42.514s
// sys 0m8.803s
}
std::cout << sum << std::endl;
return 0;
}
This solution contains 14 empty lines, 25 comments and 7 preprocessor commands.
Benchmark
The correct solution to the original Project Euler problem was found in 5692 seconds (exceeding the limit of 60 seconds).
The code can be accelerated with OpenMP but the timings refer to the single-threaded version on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 15 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
October 10, 2017 submitted solution
October 10, 2017 added comments
Links
projecteuler.net/thread=611 - the best forum on the subject (note: you have to submit the correct solution first)
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 610 - Roman Numerals II | Friend numbers - problem 612 >> |