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Are there algorithms to produce 3 dimensional mazes? Essentially the same as a 2D maze but the Z depth axis can be traversed? The idea is still the same though, to get from Start to End. Could backtracking still be used?
Which algorithm should I use to generate a 3D maze?
See here. I mean that you can go into the cube too, not just iterate the faces of it.
I made 2d mazes a few years ago using Kruskal's Algorithm here. There should be no reason this couldn't work with the 3d case you described. Basically you'd consider a cell a cube, and have a large array that has (for every cells), 6 walls in the +/- x, y, and z directions. The algorithm initially starts with all walls everywhere and randomly makes walls disappear until every cell in the maze is connected.
I have the code for generating a 2D maze in, of all things, RPGLE (something I did as a self-exercise while learning the language). Because of the way I wrote it, about the only changes necessary for the general alogrithm would be to add the Z dimension as an additional dimension...
The entire thing is 20 pages long (although this includes input/output), so here's some code. You should be able to translate this into whatever language you need: I translated it from spaghetti-code BASIC (gotos were way overused here, yeah. But it was a fun exercise).
//set out maximum maze size
maximumMazeSquareCounter = mazeHorizontalSize * mazeVerticalSize + 1;
// generate a starting horizontal positiongetRandomNumber(seed : randomNumber);
currentHorizontalPosition = %inth(randomNumber * (mazeHorizontalSize - 1)) + 1;
currentVerticalPosition = 1;
mazeSquareCounter = 1;
// generate the top row of the maze (with entrance)
mazeTopRow = generateEntrance(currentHorizontalPosition);
//write to the printer file
writeMazeDataLine(mazeTopRow);
mazeSquareCounter += 1;
//set the first position in the maze(the entry square
setPathPoint(currentHorizontalPosition : currentVerticalPosition);
//do until we've reached every square in the maze
dou mazeSquareCounter >= maximumMazeSquareCounter;
//get the next available random direction
mazeDirection = getNextRandomDirection(getNextAvailableDirection(currentHorizontalPosition : currentVerticalPosition));
//select what to do by the returned results
select;
//when FALSE is returned - when the maze is trapped
when mazeDirection = FALSE;
//if not at the horizontal end of the maze
if currentHorizontalPosition <> mazeHorizontalSize;
//add one to the position
currentHorizontalPosition += 1;
//else if not at the vertical end of the maze
elseif currentVerticalPosition <> mazeVerticalSize;
//reset the horizontal position
currentHorizontalPosition = 1;
//increment the vertical position
currentVerticalPosition += 1;
//otherwise
else;
//reset both positions
currentHorizontalPosition = 1;
currentVerticalPosition = 1;
endif;
//when 'N' is returned - going up (other directions removed)
when mazeDirection = GOING_NORTH;
//set the point above current as visited
setPathPoint(currentHorizontalPosition : currentVerticalPosition - 1);
//set the wall point to allow passage
setWallDirection(currentHorizontalPosition : currentVerticalPosition : GOING_NORTH);
//change the position variable to reflect change
currentVerticalPosition -= 1;
//increment square counter
mazeSquareCounter += 1;
endsl;
enddo;
//generate a random exit
// get a random number
getRandomNumber(seed : randomNumber);
// set to the horzontal position
currentHorizontalPosition = %inth(randomNumber * (mazeHorizontalSize - 1)) + 1;
//set the vertical position
currentVerticalPosition = mazeVerticalSize;
//set wall to allow for exit
setWallDirection(currentHorizontalPosition : currentVerticalPosition : GOING_SOUTH);
The entire thing is backed by two two-dimensional arrays (well, the RPG equivalent): One for the walls that occupy the 'square', and the other for whether or not that square has been visited. The maze is created after every square has been visited. Garuanteed one-path only, worm-turns maze.
To make this three-dimensional, make it use three-dimensional arrays, and add the necessary dimension index.
I designed an algorithm some time ago for 2D mazes on a square grid, there is no reason why this shouldn't also work for a 3D maze on a cubic grid.
Start with a 3D grid initially fully populated with wall cells.
...
Start an agent at an edge of the grid, the agent travels in a straight line in the X, Y, Z, -X, -Y or -Z direction clearing wall as she travels.
Action 'N' has a small chance of occurring each step.
Action 'M' occurs when the cell directly in front of the agent is wall and the cell in front of that is empty.
'N' is a random choice of:
removing that agent
turning left or right 90 degrees
and creating an agent on the same square turned 90 degrees left, right or both (two agents).
'M' is a random choice of:
removing that agent
removing the wall in front of that agent and then removing that agent
and doing nothing, carrying on
turning left or right 90 degrees.
and creating an agent on the same square turned 90 degrees left, right or both (two agents).
The mazes are distinctive, and their character is highly flexible by adjusting the trigger for 'M' (to do with valid junctions) and by also adjusting the chances of 1 to 8 occurring. You may want to remove an action or two, or introduce your own actions, for example one to make a small clearing or sidestep one step.
The trigger for 'N' can also be another sort of randomness, for example the example below can be used to create fairly branchy mazes that still have some long straight parts.
float n = 1;
while (random_0_to_1 > 0.15)
{
n *= 1.2;
}
return (int)n;
Some small adjustments will be needed from my simple description, for example trigger for action 'M' will need to check the cells adjacent to the cells it checks as well depending on what sort of junctions are desirable.
Either 5 or 6 are needed for the maze to contain cycles and at least one alternative 'M' action to 5 and 6 is required for the maze to contain dead ends.
Some choices of chances/actions and 'M' triggers will tend to make mazes that don't work, for example are unsolvable or full of empty or wall cells, but many will produce consistently nice results.
Related
Here we have a box that is 4 * 7 and it can be filled with rectangles that are either 1 * 2 or 2 * 1. This depiction is from the book Competitive Programmer's Handbook.
To solve this problem most efficiently, the book mentions using the parts that can be in a particular row:
Since there are 4 things in this set, the maximum unique rows we can have is 4^m, where m is the number of columns. From each constructed row, we construct the next row such that it is valid. Valid means we cannot have vertical fragments out of order. Only if all vertical "caps" in the top row correspond to vertical "cups" in the bottom row and vice versa is the solution valid. (Obviously for the horizontal fragments, their construction is restricted in row creation itself, so it is not possible for there to be inter-row discrepancy here.)
The book then mysteriously says this:
It is possible to make the solution more efficient by using a more
compact representation for the rows. It turns out that it is
sufficient to know which columns of the previous row contain the upper
square of a vertical tile. Thus, we can represent a row using only
characters upper square of a vertical tile and □, where □ is a combination of characters lower vertical square, left horizontal square and right horizontal square.
Using this representation, there are only 2^m distinct rows and the
time complexity is O(n2^(2m)).
Why does this simple square work? How would you know if there is a horizontal box underneath the top vertical fragment? How would you know left and right horizontal fragments are aligned? It breaks my mind why this is possible. Does anyone know?
Here is my ad hoc C++ implementation for 2 by n matrix, which would not work in this case, but I was trying to abstract it:
int ways[251];
int f(int n){
if (ways[n] != 1) return ways[n];
return (ways[n] = f(n-1) + f(n-2));
}
int main(){
ways[0] = 1;
ways[1] = 1;
for (int i = 2; i <= 250; i++){
ways[i] = -1;
cout << f(250) << '\n';
}
}
Situation:
Given some points with coordinate (x, y)
Range 0 < x < 100,000,000 and 0 < y <100,000,000
I have to find smallest square which contains at least N no of points on its edge and inside it.
I used vector to store coordinates and searched all squares with side length minLength upto side length maxLength (Appling Brute Force in relevant space)
struct Point
{
int x;
int y;
};
vector<Point> P;
int minLength = sqrt(N) - 1;
int maxLength = 0;
// bigx= largest x coordinate of any point
// bigy= largest y coordinate of any point
// smallx= smallest x coordinate of any point
// smally= smallest y coordinate of any point
(bigx - smallx) < (bigy - smally) ? maxLength = (bigx - smallx) : maxLength = (bigy - smally);
For each square I looked up, traversed through complete vector to see if at least N points are on its edge and inside it.
This was quite time inefficient.
Q1. What data structure should I use to improve time efficiency without changing Algorithm I used?
Q2. Efficient Algorithm for this problem?
There are points on 2 opposite edges - if not, you could shrink the square by 1 and still contain the same number of points. That means the possible coordinates of the edges are limited to those of the input points. The input points are probably not on the corners, though. (For a minimum rectangle, there would be points on all 4 edges as you can shrink one dimension without altering the other)
The next thing to realize is that each point divides the plane in 4 quadrants, and each quadrant contains a number of points. (These can add up to more than the total number of points as the quadrants have one pixel overlap). Lets say that NW(p) is the number of points to the northwest of point p, i.e. those that have x>=px and y>=py. Then the number of points in a square is NW(bottomleft) + NW(topright) - NW(bottomright) - NW(topleft).
It's fairly easy to calculate NW(p) for all input points. Sort them by x and for equal x by y. The most northwestern point has NW(p)==0. The next point can have NW(p)==1 if it's to the southeast of the first point, else it has NW(p)==0. It's also useful to keep track of SW(p) in this stage, as you're working through the points from west to east and they're therefore not sorted north to south. Having calculated NW(p), you can determine the number of points in a square S in O(1)
Recall that the square size is restricted by by the need to have points on opposite edges. Assume the points are on the left (western) and right edge - you still have the points sorted by x order. Start by assuming the left edge is at your leftmost x coordinate, and see what the right edge must be to contain N points. Now shift the left edge to the next x coordinate and find a new right edge (and thus a new square). Do this until the right edge of the square is the rightmost point.
Its also possible that the square is constrained in y direction. Just sort the points in y direction and repeat, then choose the smallest square between the two outcomes.
Since you're running linearly through the points in x and y direction, that part is just O(N) and the dominant factor is the O(N log N) sort.
Look at http://en.wikipedia.org/wiki/Space_partitioning for algorithms that use the Divide-and-Conquer technique to solve this. This is definitely solvable in Polynomial time.
Another variant algorithms can be on the following lines.
Generate a vornoi-diagram on the points to get neighbour information. [ O(nlog(n)) ]
Now use Dynamic Programming, the DP will be similar to the problem of finding the maximum subarray in a 2D array. Here instead of the sum of numbers, you will keep count of points before it.
2.a Essentially a recursion similar to this will hold. [ O(n) ]
Number of elements in square from (0,0) to (x,y ) = (Number of elems
from square (0,0 to ((x-1),y))+ (Number of elems in square 0,0 - ( x, y-1))
- (Number of elems in (0,0)-((x-1),(y-1)))
Your recurrence will have to change for all the points on its neighbourhood and to the left and above, instead of just the points above and left as above.
Once the DP is ready, you can query the points in a sqare in O(1).
Another O(n^2) loop to find from all possible combinations and find the least square.
You can even greedily start from the smallest squares first, that way you can end your search as soon as you find a suitable square..
The rtree allows spatial searching, but doesn't have stl implementation, although sqlite would allow binding. This can answer "get all points within range", "k nearest neighbours"
Finding a region which has the most dense data, is a problem similar to clustering.
Iterating over the points and finding the N nearest entries to each point. Then generate the smallest circle - centre would be the Max(x) - min(x), Max(y) - min(y). A square can be formed which contains all the neighbours, and would be somewhere between 2r length and 2sqrt(r) length sides compared to circle.
Time taken O(x) to build structure
O(X N log(X)) to search for smallest cluster
Note: There are a bunch of answers for your second question (which will probably reap bigger benefits), but I'm only referring to your first one, i.e. what data to use without changing the algorithm.
There, I think that your choice using a vector is already pretty good, because in general vectors offer the best payload/overhead ratio and also the fastest iteration. In order to find out specific bottlenecks, use a profiler, otherwise you are only guessing. With large vectors, there are a few things to avoid though:
Overallocation, this wastes space.
Underallocation, this causes copying when the vector is grown to the necessary size.
Copying.
I am trying to make a simulation of a rabbit population. these rabbits are printed on a grid say 50x50 on a console app.
I have several questions which would really help.
If I want to keep a record of all the bunnehs with their individual characteristics, is a List the best way to record this data?
For random movement, I generate a random move (-1,0,+1) on the x-axis and on the y-axis.
If i want to avoid collision against other existing rabbits, is calculating the new x,y and comparing them against the list the only way.
like:
foreach(Bunny bunny2 in lst)
if (bunny2.x == newbunny.x && bunny2.y == newbunny.y)
then move newbunny again
Thank you for your time
Based on what you're talking about, I'd use a 2 dimensional to hold all of your bunnies. e.g.
Bunny[49][49] grid; //make a 2D array of bunny objects
This way you can have your data structure perfectly represent your system. Then you could actually use the indices (x, y) of the grid to map to the 2D array. For example this would get you the bunny at x = 3, y = 17:
Bunny myBunny = grid[2][16]
Also this would ease in collision detection, as you can just check if there is a bunny in the relevant grid space you want to move to, so if the bunny in the previous example wanted to go x-1 spaces, then you could check if that grid coordinate had a null value:
if(grid[1][16] == null) { //then bunny can go there
//code to move bunny
grid[2][16] = null //remove bunny from old grid space
} else { //buny can't go there
//code
}
NB! In order for this to work, you need to make sure you take the bunny OUT of the grid space it is moving from, as done in the previous example. That way you maintain where there are no bunnies versus where they are.
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I want to know the accurate method to find the center of an arbitrary shape in 3D.
In the figure, I have explained 3 cases.
In first case how can we compute the center of arbitrary points ? One idea is summation of all points divided by total number of points. Is it the only method and Is it accurate ?
2nd is how to compute the approximate center of irregular arbitrary shape in 3D? Is this the same case of 1 ?
How can we compute the center line of the bended/curved tube that composed of arbitrary vertices ? For this problem we have to solve first two cases I think so ?
The solution really depends on what you are actually looking for.
If you seek the average position of all points in a shape, then averaging them does indeed give you that. But it can be far off from the point which you'd intuitively say is "in the middle". For example, consider a box where one side has twice as many vertices as the opposite side. The average position would be on the half of that side, not in the middle of the box.
More likely, I'd say you are looking for the point defined by calculating the max and min bounds in each dimension and then averaging the two. Because you tagged this with C++, here's some example code:
// Define max and min
double max[DIMENSIONALITY];
double min[DIMENSIONALITY];
// Init max and min to max and min acceptable values here. (see numeric_limits)
// Find max and min bounds
for(size_t v_i = 0; v_i < num_vertices; ++v_i)
{
for(int dim = 0; dim < DIMENSIONALITY; ++dim)
{
if(shape[v_i][dim] < min[dim]) min[dim] = shape[v_i][dim];
if(shape[v_i][dim] > max[dim]) max[dim] = shape[v_i][dim];
}
}
// Calculate middle
double middle[DIMENSIONALITY];
for(int dim = 0; dim < DIMENSIONALITY; ++dim)
middle[dim] = 0.5 * (max[dim] + min[dim]);
For either solution, the dimensionality of the problem doesn't matter.
Edit:
As pointed out in the comment below, this may result in a middle point which lies outside of the shape itself. If you need a point which lies inside the shape, an alternative approach must be used. A simple solution could be to use ray-marching across each axis.
I have searched Google and Stackoverflow for this question, but I still don't understand how a minimax function works.
I found the wikipedia entry has a pseudocode version of the function:
function integer minimax(node, depth)
if node is a terminal node or depth <= 0:
return the heuristic value of node
α = -∞
for child in node: # evaluation is identical for both players
α = max(α, -minimax(child, depth-1))
return α
Several other minimax functions I found with Google are basically the same thing; I'm trying to implement this in C++, and this is what I have come up with so far:
double miniMax(Board eval, int iterations)
{
//I evaluate the board from both players' point of view and subtract the difference
if(iterations == 0)
return boardEval(eval, playerNumber) - boardEval(eval, opponentSide());
/*Here, playerTurn tells the findPossibleMoves function whose turn it is;
I mean, how do you generate a list of possible moves if you don't even know
whose turn it's supposed to be? But the problem is, I don't see where I can
get playerTurn from, as there are only 2 parameters in all the examples of
minimax I've seen*/
vector<int> moves = eval.findPossibleMoves(playerTurn);
//I'm assuming -∞ in the wikipedia article means a very low number?
int result = -999999999;
//Now I run this loop to evaluate each possible move
/*Also, the Lua example in the wiki article has
alpha = node.player==1 and math.max(alpha,score) or math.min(alpha,score)
Is alpha a boolean there?!*/
for(int i = 0; i * 2 < moves.size(); i++)
{
//I make a copy of the board...
Board temp = eval;
/*and make the next possible move... once again playerTurn crops up, and I
don't know where I can get that variable from*/
temp.putPiece(moves[i * 2], moves[i * 2 + 1], playerTurn);
/*So do I create a function max that returns the bigger of two doubles?*/
result = max(result, -miniMax(temp, iterations - 1));
}
return result;
/*So now I've returned the maximum score from all possible moves within a certain
# of moves; so how do I know which move to make? I have the score; how do I know
which sequence of moves that score belongs to?*/
}
As you can see, I'm pretty confused about this minimax function. Please at the very least give me some hints to help me with this.
Thanks! :)
That sample from Wikipedia is doing NegaMax with Alpha/Beta pruning.
You may be helped by getting the naming straight:
The basis is MiniMax, a literal implementation would involve 2 methods that take turns (mutually recursive), 1 for each side.
Lazy programmers turn this into NegaMax, one method with a strategically placed - operator.
Alpha/Beta pruning is keeping track of a Window of best moves (over multiple depths) to detect dead branches.
Your playerTurn is used to determine whose turn it is . In NegaMax you can derive this from the depth (iterations) being odd or even. But it would be easier to use 2 parameters (myColor, otherColor) and switch them at each level.
Your miniMax() function should remember the best move it found so far. So instead of this code:
/*So do I create a function max that returns the bigger of two doubles?*/
result = max(result, -miniMax(temp, iterations - 1));
You should do something like this:
/*So do I create a function max that returns the bigger of two doubles?*/
double score = -miniMax(temp, iterations - 1);
if (score > result)
{
result = score;
bestMove = i;
}
Of course, you need a variable "bestMove" and a way to return the best move found to the caller.
Add the playerTurn variable as an argument to miniMax, and call miniMax which the current player's move initially and recursively.
Also, opponentSide needs to be a function of playerTurn.
A good place to start with game tree searching is the chess programming wiki. For your question about the move: I think it is most common to have two max-functions. The difference between the two max functions is that one returns only the score and the other returns the score and the best move. A recursive call order would be like following:
maxWithBestMoveReturn(...) --> min(...) --> max(...) --> min(...)
There are some good papers with pseudocode for the Alpha Beta algorithm:
TA Marsland - Computer Chess and Search
J Schaeffer - The games Computers (and People) Play
To your question in the comment: and math.max(alpha,score) or math.min(alpha,score) Is alpha a boolean there?!
No alpha is a window bound in a alpha beta algorithm. The alpha value gets updated with a new value. Because alpha and beta are swapped with the recursive call of the negamax-Function the alpha variable refers to the beta variable in the next recursive call.
One note to the playerTurn variable: The minimax or alpha-beta algorithm doesn't need this information. So i would give the information -- who's next --, into the Board-Structure. The functions findPossibleMoves and boardEval get all information they need from the Board-Structure.
One note to the recursive break condition: If i understand your code right, then you only have the one with iterations == o. I think this means the algorithm has reached the desired depth. But what if there are no possible moves left befor the algorithm reaches this depth. Maybe you should write following:
vector<int> moves = findPossibleMoves(...);
if (!moves.size())
return boardEval(...);
In your pseudocode, the node variable has to contain all the information about the current board position (or whatever). This information would include whose turn it is to move.