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Problem 321: Swapping Counters
(see projecteuler.net/problem=321)
A horizontal row comprising of 2n + 1 squares has n red counters placed at one end and n blue counters at the other end,
being separated by a single empty square in the centre. For example, when n = 3.
A counter can move from one square to the next (slide) or can jump over another counter (hop) as long as the square next to that counter is unoccupied.
Let M(n) represent the minimum number of moves/actions to completely reverse the positions of the coloured counters; that is, move all the red counters to the right and all the blue counters to the left.
It can be verified M(3) = 15, which also happens to be a triangle number.
If we create a sequence based on the values of n for which M(n) is a triangle number then the first five terms would be:
1, 3, 10, 22, and 63, and their sum would be 99.
Find the sum of the first forty terms of this sequence.
My Algorithm
I spent most of the time figuring out how M(n) can be computed: my function countMoves()
returns the optimal value for small n.
Basically this function performs a breadth-first search on all allowed moves.
The first ten results are:
nM(n)
13
28
315
424
535
648
763
880
999
10120
Apparently M(n) is always the predecessor of a perfect square. To be more specific:
(1) M(n) = (n + 1)^2 - 1
Now that I know how to compute M(n) (even for large n) I have to find out which M(n) are triangle numbers T(t) = t(t + 1) / 2:
(2) (n + 1)^2 - 1 = t(t + 1) / 2
If you uncomment #define SLOWSEARCH
in main()
then a brute-force search enumerates the first 20 solutions in a few seconds.
However, finding all 40 solutions would take ages.
The equation (2) can be changed to:
(3) 2(n + 1)^2 - 2 = t(t + 1)
(4) 2n^2 + 4n = t^2 + t
(5) 0 = t^2 + t - 2n^2 - 4n
If I replace x = t and y = n:
(6) 0 = x^2 - 2y^2 + x - 4y
then it's a Diophantine equation that can be solved with a tool (I prefer www.alpertron.com.ar/QUAD.HTM):
(7) 0 = ax^2 + bxy + cy^2 + dx + ey + f
(8) a = 1 and b = 0 and c = -2 and d = 1 and e = -4 and f = 0
Solutions can be found iteratively:
(10) X_{n+1} = P * X_n + Q * Y_n + K
(11) Y_{n+1} = R * X_n + S * Y_n + L
And the factors are (printed by the solver):
(12) P = 3 and Q = 4 and K = 5 and R = 2 and S = 3 and L = 3
My "seed" values were obtained from the brute-force search:
(13) x_1 = 2 and y_1 = 1
Unfortunately I saw that (10), (11) and (12) only showed every second solution (in comparison to my brute-force search):
the iterations produced x_3 and y_3, x_5 and y_5, x_7 and y_7, ... and missed all values with an even index.
Since I knew the second solution (that means x_2 and y_2) from my brute-force search, I tried them and voila: I got x_4 and y_4, x_6 and y_6, x_8 and y_8, ...
In the end I have two pairs (x1
, y1
) and (x2
, y2
). I don't fully understand the underlying problem, but my solution works.
Note
Often I take a look at other solutions (especially in the Project Euler forum) and I'm quite surprised that everybody seems to immediately know
that M(n) = n * (n + 2) (which is the same as (n + 1)^2 - 1). That's not obvious to me at all.
Most 30% problems are much easier than this "Swapping Counters" problem which I only managed to solve in a trial-and-error way.
And I don't like these Diophantine equations because I have to make use of a dedicated solver (in my case www.alpertron.com.ar/QUAD.HTM) which is a non-trivial program on its own.
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 | ./321
Output:
Note: the original problem's input 40
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>
#include <set>
#include <string>
#include <algorithm>
// find the the length of shortest move sequence
// too slow for anything > 10
unsigned int countMoves(unsigned int stonesPerColor)
{
const auto Red = 'R';
const auto Blue = 'B';
const auto Empty = '.';
const auto Length = 2 * stonesPerColor + 1;
// create initial state
const auto initial = std::string(stonesPerColor, Red ) + Empty + std::string(stonesPerColor, Blue);
// and final state
const auto final = std::string(stonesPerColor, Blue) + Empty + std::string(stonesPerColor, Red);
// breadth search
std::vector<std::string> last = { initial };
std::set <std::string> alreadySeen = { initial };
unsigned int numMoves = 0;
while (true)
{
std::vector<std::string> next;
for (const auto& current : last)
{
if (current == final)
return numMoves;
// locate the empty cell
size_t pos = 0;
while (current[pos] != Empty)
pos++;
// generate all possible moves
if (pos >= 2)
{
// jump two positions to the right
auto jumpTwoRight = current;
std::swap(jumpTwoRight[pos - 2], jumpTwoRight[pos]);
// accept only positions that I haven't seen before
if (alreadySeen.count(jumpTwoRight) == 0)
{
next.push_back(jumpTwoRight);
alreadySeen.insert(jumpTwoRight);
}
}
if (pos >= 1)
{
// move one position to the right
auto moveOneRight = current;
std::swap(moveOneRight[pos - 1], moveOneRight[pos]);
// accept only positions that I haven't seen before
if (alreadySeen.count(moveOneRight) == 0)
{
next.push_back(moveOneRight);
alreadySeen.insert(moveOneRight);
}
}
if (pos < Length - 1)
{
// move one position to the left
auto moveOneLeft = current;
std::swap(moveOneLeft[pos + 1], moveOneLeft[pos]);
// accept only positions that I haven't seen before
if (alreadySeen.count(moveOneLeft) == 0)
{
next.push_back(moveOneLeft);
alreadySeen.insert(moveOneLeft);
}
}
if (pos < Length - 2)
{
// jump two positions to the left
auto jumpTwoLeft = current;
std::swap(jumpTwoLeft[pos + 2], jumpTwoLeft[pos]);
// accept only positions that I haven't seen before
if (alreadySeen.count(jumpTwoLeft) == 0)
{
next.push_back(jumpTwoLeft);
alreadySeen.insert(jumpTwoLeft);
}
}
}
// next move
numMoves++;
last = std::move(next);
}
// never reached
return 0;
}
int main()
{
unsigned int numValues = 40;
std::cin >> numValues;
// count moves via brute-force
//#define COUNTMOVES
#ifdef COUNTMOVES
for (unsigned int stonesPerColor = 1; stonesPerColor <= numValues; stonesPerColor++)
std::cout << "M(" << stonesPerColor << ")=" << countMoves(stonesPerColor) << std::endl;
#endif
// solve t * (t + 1) / 2 = (n + 1)^2 - 1 via brute-force
//#define SLOWSEARCH
#ifdef SLOWSEARCH
unsigned long long t = 1;
unsigned long long n = 1;
// only feasible for the first 20 solutions
unsigned int i = 1;
unsigned int m = 0;
while (i <= numValues)
{
auto triangle = t * (t + 1) / 2;
auto moves = (n + 1) * (n + 1) - 1;
// is triangle number too small ?
if (triangle < moves)
t++;
// is triangle number too large ?
else if (triangle > moves)
n++;
else // if (triangle == moves)
{
// found a match
m += n;
std::cout << "M(" << i << ")=" << m << std::endl;
i++; t++; n++;
}
}
#endif
// solve ax^2 + bxy + cy^2 + dx + ey + f = 0
// a = 1;
// b = 0;
// c = -2;
// d = 1;
// e = -4;
// f = 0;
// inital solution
unsigned long long x1 = 2;
unsigned long long y1 = 1;
unsigned long long x2 = 5;
unsigned long long y2 = 3;
unsigned long long sum = y1 + y2;
for (unsigned int n = 3; n <= numValues; n++)
{
// https://www.alpertron.com.ar/QUAD.HTM
// a = 1;
// b = 0;
// c = -2;
// d = 1;
// e = -4;
// f = 0;
const auto P = 3;
const auto Q = 4;
const auto K = 5;
const auto R = 2;
const auto S = 3;
const auto L = 3;
// update first pair x,y
auto nextX1 = P*x1 + Q*y1 + K;
auto nextY1 = R*x1 + S*y1 + L;
x1 = nextX1;
y1 = nextY1;
sum += y1;
if (++n > numValues)
break;
// update second pair x,y
auto nextX2 = P*x2 + Q*y2 + K;
auto nextY2 = R*x2 + S*y2 + L;
x2 = nextX2;
y2 = nextY2;
sum += y2;
}
// print result
std::cout << sum << std::endl;
return 0;
}
This solution contains 19 empty lines, 43 comments and 9 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
August 30, 2017 submitted solution
August 30, 2017 added comments
Difficulty
Project Euler ranks this problem at 30% (out of 100%).
Links
projecteuler.net/thread=321 - the best forum on the subject (note: you have to submit the correct solution first)
Code in various languages:
Go github.com/frrad/project-euler/blob/master/golang/Problem321.go (written by Frederick Robinson)
Mathematica github.com/frrad/project-euler/blob/master/mathematica/Problem321.m (written by Frederick Robinson)
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 |
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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 !!!
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