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.

initial

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.

jumps

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:

Input data (separated by spaces or newlines):

This is equivalent to
echo 10 | ./321

Output:

(please click 'Go !')

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

30% Project Euler ranks this problem at 30% (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
  the flashing problem is the one I solved most recently
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The 299 solved problems (that's level 11) had an average difficulty of 32.4% 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.

more about me can be found on my homepage, especially in my coding blog.
some names mentioned on this site may be trademarks of their respective owners.
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