maze

How to create maze walls in NetLogo?

我的未来我决定 提交于 2019-12-08 02:43:47
问题 I am trying to create a 5x5 grid with 2 exits and put some walls in it. In other words, I want to create a maze or a labyrinth. I was wondering if there is a way to make a border line thicker or change the colour of only one side of a patch. I want to put only one agent inside and let him find the exit by rewarding him with some points. (Q-learning algorithm) Does anyone have an idea? If this is not possible can you suggest comparable code please? Here is an example of what I want to create:

List of coordinates between irregular points in python

前提是你 提交于 2019-12-07 12:23:49
问题 Imagine we have two randomly selected points between 0 and 100 for both x and y. For example: (95,7), (35,6) Now using the simple pygame.draw.line() function we could easily draw a line between these points without any gaps. My question is, how could we find a list of all the coordinates in a single pixel thick line between the two points without any gaps in the line? Secondly, is this even possible? I am using this list of pixel for the crack maze algorithm that needs to "shoot" another

how to solve theta mazes using opencv python?

被刻印的时光 ゝ 提交于 2019-12-07 04:57:26
I have to find shortest path from the center of the maze to the outermost circle. I have to solve this problem using opencv and python I'm out! You can consider every white pixel in the image as a node of an undirected weighted graph. Every pixel (node) is connected to it's white neighbours. The weight of the edge connecting the two nodes is 1 in horizontal and vertical direction, and sqrt(2) (or simply 1.414 ) in diagonal direction. Than, since you know starting and ending point, you can run Dijkstra algorithm to find the shortest path between start and end. I used Rosetta Code implementation

Using a stack to traverse and solve a maze - Java

人盡茶涼 提交于 2019-12-07 04:35:49
问题 So I am trying to create a maze solver program that would solve a maze of X's and O's. What I would like to do is create a class of Points, so that I can create a 2-Dimensional array of Points which would allow printing to an output page as well as implementing the stack to be relatively simple. The simplest algorithm of the general idea I'd like to implement in the actual program itself I believe should be: 1) Move forward 2) Are you at a wall? 2a) If yes, turn left 3) Are you at the finish?

How to create maze walls in NetLogo?

时光怂恿深爱的人放手 提交于 2019-12-06 12:54:44
I am trying to create a 5x5 grid with 2 exits and put some walls in it. In other words, I want to create a maze or a labyrinth. I was wondering if there is a way to make a border line thicker or change the colour of only one side of a patch. I want to put only one agent inside and let him find the exit by rewarding him with some points. (Q-learning algorithm) Does anyone have an idea? If this is not possible can you suggest comparable code please? Here is an example of what I want to create: As asked, I've posted some of my work (although it seems inefficient to have done this manually). Here

maze problem and Recursive backtracker algorithm

孤者浪人 提交于 2019-12-06 11:45:39
i want to implement the Recursive backtracker algorithm to solve maze problem, but i cant understand 2.3 Command ("remove the wall between the current cell and the chosen cell") would any help me ? Mark the current cell as 'Visited' If the current cell has any neighbours which have not been visited Choose randomly one of the unvisited neighbours add the current cell to the stack remove the wall between the current cell and the chosen cell Make the chosen cell the current cell Recursively call this function else remove the last current cell from the stack Backtrack to the previous execution of

Prim's algorithm for generating a maze: Getting the neighbour cell

流过昼夜 提交于 2019-12-06 09:26:21
问题 I'm basing a maze generator program on the Prim's algorithm: This algorithm is a randomized version of Prim's algorithm. Start with a grid full of walls. Pick a cell, mark it as part of the maze. Add the walls of the cell to the wall list. While there are walls in the list: Pick a random wall from the list. If the cell on the opposite side isn't in the maze yet: Make the wall a passage and mark the cell on the opposite side as part of the maze. Add the neighboring walls of the cell to the

List of coordinates between irregular points in python

99封情书 提交于 2019-12-05 21:18:34
Imagine we have two randomly selected points between 0 and 100 for both x and y. For example: (95,7), (35,6) Now using the simple pygame.draw.line() function we could easily draw a line between these points without any gaps. My question is, how could we find a list of all the coordinates in a single pixel thick line between the two points without any gaps in the line? Secondly, is this even possible? I am using this list of pixel for the crack maze algorithm that needs to "shoot" another pixel while regarding any blocking walls that may interfere with its path. http://www.astrolog.org/labyrnth

Using a stack to traverse and solve a maze - Java

爱⌒轻易说出口 提交于 2019-12-05 08:47:18
So I am trying to create a maze solver program that would solve a maze of X's and O's. What I would like to do is create a class of Points, so that I can create a 2-Dimensional array of Points which would allow printing to an output page as well as implementing the stack to be relatively simple. The simplest algorithm of the general idea I'd like to implement in the actual program itself I believe should be: 1) Move forward 2) Are you at a wall? 2a) If yes, turn left 3) Are you at the finish? 3a) If no, go to 1 3b) If yes, solved But I'm having trouble coming up with a more in-depth algorithm,

Recursive Algorithm for 2D maze?

三世轮回 提交于 2019-12-05 04:48:08
问题 (This is not a duplicate) We have a 2D maze surrounded by X on all 4 sides and there are inner blocks too. All these characters of the maze is stored in 2D array. The program must find the path from start 'S' to goal 'G'. For this, a boolean method called 'solve(int row, int col) is uses and is initialized with row and column index of 'S'. The algorithm must be recursive. It should return true if its able to find the path to 'G' and false other wise. Here's how I tried to approach this