<< problem 348 - Sum of a square and a cube | Hexagonal orchards - problem 351 >> |
Problem 349: Langton's ant
(see projecteuler.net/problem=349)
An ant moves on a regular grid of squares that are coloured either black or white.
The ant is always oriented in one of the cardinal directions (left, right, up or down) and moves from square to adjacent square according to the following rules:
- if it is on a black square, it flips the color of the square to white, rotates 90 degrees counterclockwise and moves forward one square.
- if it is on a white square, it flips the color of the square to black, rotates 90 degrees clockwise and moves forward one square.
My Algorithm
I had no idea how to solve this problem until I read the Wikipedia page about Langton's ant: en.wikipedia.org/wiki/Langton's ant
One thing was especially interesting: the initial movement of the ant may be chaotic, but after a while a certain pattern develops which repeats after 104 steps.
My program simulates the movement of the ant on a 64x64 grid (starting in the middle).
It counts the number of black squares and every 104 steps (= 1 cycle) it computes the delta compared to the number of black squares 104 steps ago.
As soon as I observe the same difference over at least 10 cycles I can easily find out how many cycles are needed for 10^18 steps (minus the steps already taken).
To simplify my computation I don't check the difference at the start of a cycle but after the 40th steps of a cycle because 10^18 mod 104 = 40.
Note
The recurring pattern appears after about 10000 steps.
Interactive test
You can submit your own input to my program and it will be instantly processed at my server:
This is equivalent toecho 10 | ./349
Output:
Note: the original problem's input 1000000000000000000
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.
#include <iostream>
#include <vector>
int main()
{
auto limit = 1000000000000000000LL;
std::cin >> limit;
// colors encoded with a single bit
const bool White = false;
const bool Black = true;
// 128x128 squares (nice power of two with sufficient squares for the first 10000 steps)
const unsigned int Size = 128;
// 2D grid stored as a 1D array: pos = y * Size + x;
std::vector<bool> grid(Size * Size, White);
// initial position of the ant
unsigned int x = Size / 2;
unsigned int y = Size / 2;
// delta movement when ant moves Up, Left, Down, Right
const short DeltaX[] = { 0, +1, 0, -1 };
const short DeltaY[] = { +1, 0, -1, 0 };
// direction 0..3 (see DeltaX and DeltaY)
short direction = 0; // any direction is fine
// a pattern with cycle length 104 emerges after a while (see https://en.wikipedia.org/wiki/Langton%27s_ant )
const auto Cycle = 104;
const auto Remainder = limit % Cycle; // = 40
// number of black squares
auto count = 0LL;
// and the value of "count" 104 steps ago
auto last = count;
// keep track of the deltas (which are "count-last")
std::vector<int> lastDeltas = { 0 };
// assume a "steady state" if the most recent 10 cycles have the same deltas
const unsigned int StopIfSameDeltas = 10;
// number of steps taken
auto steps = 0;
while (steps < limit)
{
// check every 104 steps whether the "steady state" was achieved
if (steps % Cycle == Remainder)
{
// change of Black squares
auto diff = int(count - last);
lastDeltas.push_back(diff);
last = count;
// need a few samples ...
if (lastDeltas.size() >= StopIfSameDeltas)
{
bool samesame = true;
// always the same diff during the most recent cycles ?
for (auto scan = lastDeltas.rbegin(); scan != lastDeltas.rbegin() + StopIfSameDeltas; scan++)
if (*scan != diff)
{
// nope, found a different difference :-(
samesame = false;
break;
}
// yes, entered "steady state"
if (samesame)
{
// determine number of cycles left
auto remainingCycles = (limit - steps) / Cycle;
// and multiply by 12
count += remainingCycles * diff;
// stop searching
break;
}
}
}
// ant moves one step
auto pos = y * Size + x;
if (grid[pos] == White)
{
// flip square
grid[pos] = Black;
count++;
// counter-clockwise
direction = (direction + 1) % 4;
}
else // Black
{
// flip square
grid[pos] = White;
count--;
// clockwise
direction = (direction + 4 - 1) % 4; // plus 4 avoids negative numbers
}
// "That's one small step for an ant, one giant leap for antkind"
x += DeltaX[direction];
y += DeltaY[direction];
steps++;
}
// display result
std::cout << count << std::endl;
return 0;
}
This solution contains 17 empty lines, 28 comments and 2 preprocessor commands.
Benchmark
The correct solution to the original Project Euler problem was found in less than 0.01 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(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
July 19, 2017 submitted solution
July 19, 2017 added comments
Difficulty
Project Euler ranks this problem at 35% (out of 100%).
Links
projecteuler.net/thread=349 - 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/349.py (written by Meng-Gen Tsai)
Python github.com/steve98654/ProjectEuler/blob/master/349.py
C++ github.com/roosephu/project-euler/blob/master/349.cpp (written by Yuping Luo)
Java github.com/thrap/project-euler/blob/master/src/Java/Problem349.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 a new problem is published | |
the flashing problem is the one I solved most recently |
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I scored 13,486 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 348 - Sum of a square and a cube | Hexagonal orchards - problem 351 >> |