I'm having some trouble finding the right approach to coding this.
Take a random-generated 2d array, about 50x50 with each cell having a value 1~99.
Starting at a random position "Green", and the goal is to surround the target "Red" with the lowest amount of actions.
Moving to a neighboring cell takes 1~99 actions depending on it's value.
example small array with low values:
[
Currently the best idea i have is, generate 4 sets of checkpoints based on the diagonals of the target and then using a lot of Dijkstra's to find a path that goes through all of them, as well as the starting point.
One problem i have is this very quickly becomes an extreme numbers of paths.
FROM any starting point "NorthWest-1 to NW-20" TO any ending point in "NE-1 to NE-20", is 400 possibilities. Adding the 3rd and 4th diagonal to that becomes 400 * 20 * 20.
Another problem using diagonal checkpoints is that the problem is not [shortest path from green to a diagonal (orange path)]
[
but rather from "green to any point on the path around red".
Current pseudocode;
take 2 sets of diagonals nearest to Green/start
find the shortest path that connects those diagonals while going through Green
(backtracking through the path is free)
draw a line starting from the target point, in-between the 2 connected diagonals,
set those cells to value infinite to force going around them (and thus around the target)
find the shortest path connecting the now-seperated diagonals
Unfortunately this pseudocode already includes some edge cases where the 'wall' blocks the most efficient path.
If relevant, this will be written in javascript.
Edit, as an edge case it could spiral the target before surrounding, though extremely rare
Edit2; "Surround" means disconnect the target from the rest of the field, regardless of how large the surrounded area is, or even if it includes the starting point (eg, all edges are 0)
Here is another larger field with (probably) optimal path, and 2 fields in text-form:
https://i.imgur.com/yMA14sS.png
https://pastebin.com/raw/YD0AG6YD
For short, let us call paths that surround the target fences. A valid fence is a set of (connected) nodes that makes the target disconnected from the start, but does not include the target. A minimal fence is one that does so while having a minimal cost. A lasso could be a fence that includes a path to the start node. The goal is to build a minimal-cost lasso.
A simple algorithm would be to use the immediate neighborhood of the target as a fence, and run Dijkstra to any of those fence-nodes to build a (probably non-optimal) lasso. Note that, if optimal answers are required, the choice of fence actually influences the choice of path from the start to the fence -- and vice-versa, the choice of path from start to fence can influence how the fence itself is chosen. The problem cannot be split neatly into two parts.
I believe that the following algorithm will yield optimal fences:
Build a path using Dijkstra from start to target (not including the end-points). Let us call this the yellow path.
Build 2 sets of nodes, one on each side of this yellow path, and neighboring it. Call those sets red and blue. Note that, for any given node that neighbors the path, it can either be part of the path, blue set, red set, or is actually an end-point.
For each node in the red set, run Dijkstra to find the shortest path to a node in the blue set that does not cross the yellow path.
For each of those previous paths, see which is shortest after adding the (missing) yellow-path bit to connect the blue and red ends together.
The cost is length(yellowPath) * cost_of_Dijkstra(redStart, anyBlue)
To make a good lasso, it would be enough to run Dijkstra from the start to any fence node. However, I am unsure of whether the final lasso will be optimal or not.
You might want to consider the A* search algorithm instead, you can probably adjust the algorithm to search for all 4 spots at once.
https://en.wikipedia.org/wiki/A*_search_algorithm
Basically A* expands Dijkstra's algorithm by focusing it's search on spots that are "closer" to the destination.
There are a number of other variations for search algorithms that may be more useful for your situation as well in the "Also See" section, though some of them are more suited for video game path planning rather than 2D grid paths.
Edit after reviewing question again:
Seems each spot has a weight. This makes the distance calculation a bit less straightforward. In this case, I would treat it as an optimization. For the heuristic cost function, it may be best to just use the most direct path (diagonal) to the goal as the heuristic cost, and then just use A* search to try to find an even better path.
As for the surround logic. I would treat that as it's own logic and a separate step (likely the second step). Find least cost path to the target first. Then find the cheapest way to surround the path. Honestly, the cheapest way to surround a point is probably worth it's own question.
Once you have both parts, it should be easy enough to merge the two. There will be some point where the two first overlap and that is where they are merged together.
Related
I'm making a game engine for a board game called Blockade and right now I'm trying to generate all legal moves in a position. The rules aren't exactly the same as the actual game and they don't really matter. The gist is: the board is a matrix and you move a pawn and place a wall every move.
In short, I have to find whether or not a valid path exists from every pawn to every goal after every potential legal move (imagine a pawn doesn't move and a wall is just placed), to rule out illegal moves. Or rather, if I simplify it to a subproblem, whether or not the removal of a few edges (placing a wall) removes all paths to a node.
Brute-forcing it would take O(k*n*m), where n and m are the board dimensions and k is the number of potential legal moves. Searching for a path (worst case; traversing most of the board) is very expensive, but I'm thinking with dynamic programming or some other idea/algorithm it can be done faster since the position is the same the wall placement just changes, or rather, in graph terms, the graph is the same which edges are removed is just changed. Any sort of optimization is welcome.
Edit:
To elaborate on the wall (blockade). A wall is two squares wide/tall (depending on whether it's horizontal or vertical) therefore it will usually remove at least four edges, eg:
p | r
q | t
In this 2x2 matrix, placing a wall in the middle (as shown) will remove jumping from and to:
p and t, q and r, p and r, and q and t
I apologize ahead of time if I don't fully understand your question as it is asked; there seems to be some tacit contextual knowledge you are hinting at in your question with respect to knowledge about how the blockade game works (which I am completely unfamiliar with.)
However, based on a quick scan on wikipedia about the rules of the game, and from what I gather from your question, my understanding is that you are effectively asking how to ensure that a move is legal. Based on what I understand, an illegal move is a wall/blockade placement that would make it impossible for any pawn to reach its goal state.
In this case, I believe a workable solution that would be fairly efficient would be as follows.
Define a path tree of a pawn to be a (possibly but not necessarily shortest) path tree from the pawn to each reachable position. The idea is, you want to maintain a path tree for every pawn so that it can be updated efficiently with every blockade placement. What is described in the previous sentence can be accomplished by observing and implementing the following:
when a blockade is placed it removes 2 edges from the graph, which can sever up to (at most) two edges in all your existing path trees
each pawn's path tree can be efficiently recomputed after edges are severed using the "adoption" algorithm of the Boykov-Komolgrov maxflow algorithm.
once every pawns path tree is recomputed efficiently, simply check that each pawn can still access its goal state, if not mark the move as illegal
repeat for each possible move (reseting graphs as needed during the search)
Here are resources on the adoption algorithm that is critical to doing what is described efficiently:
open-source implementation as part of the BK-maxflow: https://www.boost.org/doc/libs/1_66_0/libs/graph/doc/boykov_kolmogorov_max_flow.html
implementation by authors as part of BK-maxflow: https://pub.ist.ac.at/~vnk/software.html
detailed description of adoption (stage) algorithm of BK maxflow algorithm: section 3.2.3 of https://www.csd.uwo.ca/~yboykov/Papers/pami04.pdf
Note reading the description of the adopton algorithm included in the last
bullet point above would be most critical to understanding how to adopt
orphaned portions of your path-tree efficiently.
In terms of efficiency of this approach, I believe on average you should expect on average O(1) operations for each adopted edge, meaning this approach should take about O(k) time to compute where k is the number of board states which you wish to compute for.
Note, the pawn path tree should actually be a reverse directed tree rooted at the goal nodes, which will allow the computation to be done for all legal pawn placements given a blockade configuration.
A few suggestions:
To check if there's a path from A to B after ever
Every move removes a node from the graph/grid. So what we want to know is if there are critical nodes on the path from A to B (single points that could be blocked to break the path. This is a classic flow problem. For this application you want to set the vertex capacity to 1 and push 2 units of flow (basically just to verify that there are at least 2 paths). If there are 2 paths, no one block can disconnect you from the destination. You can optimize it a bit by using an implicit graph, but if you're new to this maybe create the graph to visualize it better. This should be O(N*M), the size of your grid.
Optimizations
Since this is a game, you know that the setup doesn't change dramatically from one step to another. So, you can keep track of the two paths. If the blocade is not placed on any of the paths, you can ignore it. You already have 2 paths to destination.
If the block does land on one of the paths, cancel only that path and then look for another (reusing the one you already have).
You can also speed up the pawn movement. This can be a bit trick, but what you want is to move the source. I'm assuming the pawn moves only a few cells at a time, maybe instead of finding completely new paths, you can simply adjust them to connect to the new position, speeding up the update.
I'm developing a game similar to Pacman: consider this maze:
Each white square is a node from the maze where an object located at P, say X, is moving towards node A in the right-to-left direction. X cannot switch to its opposite direction unless it encounters a dead-end such as A. Thus the shortest path joining P and B goes through A because X cannot reverse its direction towards the rightmost-bottom node (call it C). A common A* algorithm would output:
to get to B from P first go rightward, then go upward;
which is wrong. So I thought: well, I can set the C's visited attribute to true before running A* and let the algorithm find the path. Obviously this method doesn't work for the linked maze, unless I allow it to rediscover some nodes (the question is: which nodes? How to discriminate from useless nodes?). The first thinking that crossed my mind was: use the previous method always keeping track of the last-visited cell; if the resulting path isn't empty, you are done. Otherwise, when you get to the last-visited dead-end, say Y, (this step is followed by the failing of A*) go to Y, then use standard A* to get to the goal (I'm assuming the maze is connected). My questions are: is this guaranteed to work always? Is there a more efficient algorithm, such as an A*-derived algorithm modified to this purpose? How would you tackle this problem? I would greatly appreciate an answer explaining both optimal and non-optimal search techniques (actually I don't need the shortest path, a slightly long path is good, but I'm curious if such an optimal algorithm running as efficiently as Dijkstra's algorithm exists; if it does, what is its running time compared to a non-optimal algorithm?)
EDIT For Valdo: I added 3 cells in order to generalize a bit: please tell me if I got the idea:
Good question. I can suggest the following approach.
Use Dijkstra (or A*) algorithm on a directed graph. Each cell in your maze should be represented by multiple (up to 4) graph nodes, each node denoting the visited cell in a specific state.
That is, in your example you may be in the cell denoted by P in one of 2 states: while going left, and while going right. Each of them is represented by a separate graph node (though spatially it's the same cell). There's also no direct link between those 2 nodes, since you can't switch your direction in this specific cell.
According to your rules you may only switch direction when you encounter an obstacle, this is where you put links between the nodes denoting the same cell in different states.
You may also think of your graph as your maze copied into 4 layers, each layer representing the state of your pacman. In the layer that represents movement to the right you put only links to the right, also w.r.t. to the geometry of your maze. In the cells with obstacles where moving right is not possible you put links to the same cells at different layers.
Update:
Regarding the scenario that you described in your sketch. It's actually correct, you've got the idea right, but it looks complicated because you decided to put links between different cells AND states.
I suggest the following diagram:
The idea is to split your inter-cell AND inter-state links. There are now 2 kinds of edges: inter-cell, marked by blue, and inter-state, marked by red.
Blue edges always connect nodes of the same state (arrow direction) between adjacent cells, whereas red edges connect different states within the same cell.
According to your rules the state change is possible where the obstacle is encountered, hence every state node is the source of either blue edges if no obstacle, or red if it encounters an obstacle (i.e. can't emit a blue edge). Hence I also painted the state nodes in blue and red.
If according to your rules state transition happens instantly, without delay/penalty, then red edges have weight 0. Otherwise you may assign a non-zero weight for them, the weight ratio between red/blue edges should correspond to the time period ratio of turn/travel.
I'm implementing a robot to be able to solve any maze (where the robot only has front sensors, but I make it scan the surroundings), and I was able to get it to turn the maze into a map where 0 represents walls, and 1 represents roads, with possibly slanted roads. Now, the robot is not fast at turning, but fairly fast at moving down a straight line. Therefore, a normal shortest path algorithm through the somewhat slanted hallway would be slow, although the paths are wide enough for it.
For example, we find
0001111111000
0011111110000
0111111100000
1111111000000
1111110000000
As a possible map. I'd like the robot to recognize that it can walk diagonally, or even just go straight up then right then right again, instead of turning every time in a normal shortest path algorithm.
Any ideas? Also, a complete algorithm change is welcome too - I'm fairly new to this.
I've faced similar problem some time ago.
You can assign weights to surrounding cells and less weight to the front cell, thus making a weight graph that is made during the movement.
I used Dijkstra algorithm with weights of 2 for surrounding cells and weight 1 for the front cell, you must pass direction of robot to Dijkstra and when adding them to the priority queue, and when extracting cells from the queue add the neighbors with respect to the direction saved in the extracted cell.
Then make the move and then recompute the modified Dijkstra for finding the nearest unseen cell.
I am looking for an algorithm that could find a random cycle in a graph from a node while that cycle is traversing around another nodes (area). For example, from the green star on the left of the image, finds a random cycle that goes around the red-star nodes.
Given that you are looking for a path that is "random", but still fairly close to minimal perimeter around the "red star", you could try this:
First, we need to choose the direction we are going. Let us decide on clockwise, and we start at point S.
Second, calculate the shortest path around the red star, including the point S. I am not going into details here (e.g. what if it is concave) since this is another question. Also, notice that deciding on S is already taking away from the randomness of the algorithm.
While choosing the path, keep 3 parameters (forward, left, right) that present the weight in the random choice of the next move. The difference in the outcome will be largely determined by the handling of theses parameters. You could always keep the weight equal, and then you might never get back to the start point, and even if you do, you don't know that the red star is inside.
To fix this, check the position with the minimal path calculated before.
If you are on it, then right = 0. Also, the directions going away from the minimal path could get less and less chances the further you are from it.
Hope this was helpful.
Imagine I am implementing Dijkstra's algorithm at a park. There are points and connections between those points; these specify valid paths the user can walk on (e.g. sidewalks).
Now imagine that the user is on the grass (i.e. not on a path) and wants to navigate to another location. The problem is not in Dijkstra's algorithm (which works fine), the problem is determining at which vertex to begin.
Here is a picture of the problem: (ignore the dotted lines for now)
Black lines show the edges in Dijkstra's algorithm; likewise, purple circles show the vertices. Sidewalks are in gray. The grass is, you guessed it, green. The user is located at the red star, and wants to get to the orange X.
If I naively look for the nearest vertex and use that as my starting point, the user is often directed to a suboptimal path, that involves walking further away from their destination at the start (i.e. the red solid path).
The blue solid path is the optimal path that my algorithm would ideally come up with.
Notes:
Assume no paths cross over other paths.
When navigating to a starting point, the user should never cross over a path (e.g. sidewalk).
In the image above, the first line segment coming out of the star is created dynamically, simply to assist the user. The star is not a vertex in the graph (since the user can be anywhere inside the grass region). The line segment from the star to a vertex is simply being displayed so that the user knows how to get to the first valid vertex in the graph.
How can I implement this efficiently and correctly?
Idea #1: Find the enclosing polygon
If I find the smallest polygon which surrounds my starting point, I can now create new paths for Dijkstra's algorithm from the starting point (which will be added as a new vertex temporarily) to each of the vertices that make up the polygon. In the example above, the polygon has 6 sides, so this would mean creating 6 new paths to each of its vertices (i.e. the blue dotted lines). I would then be able to run Dijkstra's algorithm and it would easily determine that the blue solid line is the optimal path.
The problem with this method is in determining which vertices comprise the smallest polygon that surrounds my point. I cannot create new paths to each vertex in the graph, otherwise I will end up with the red dotted lines as well, which completely defeats the purpose of using Dijkstra's algorithm (I should not be allowed to cross over a sidewalk). Therefore, I must take care to only create paths to the vertices of the enclosing polygon. Is there an algorithm for this?
There is another complication with this solution: imagine the user now starts at the purple lightning bolt. It has no enclosing polygon, yet the algorithm should still work by connecting it to the 3 points at the top right. Again, once it is connected to those, running Dijkstra's is easy.
Update: the reason we want to connect to one of these 3 points and not walk around everything to reach the orange X directly is because we want to minimize the walking done on unpaved paths. (Note: This is only a constraint if you start outside a polygon. We don't care how long you walk on the grass if it is within a polygon).
If this is the correct solution, then please post its algorithm as an answer.
Otherwise, please post a better solution.
You can start off by running Dijkstra from the target to find its distance to all vertices.
Now let's consider the case where you start "inside" the graph on the grass. We want to find all vertices that we can reach via a straight line without crossing any edge. For that we can throw together all the line segments representing the edges and the line segments connecting the start point to every vertex and use a sweep-line algorithm to find whether the start-vertex lines intersect any edge.
Alternatively you can use any offline algorithm for planar point location, those also work with a sweep line. I believe this is in the spirit of the more abstract algorithm proposed in the question in that it reports the polygon that surrounds the point.
Then we just need to find the vertex whose connection line to the start does not intersect any edge and the sum d(vertex, target) + d(vertex, start) is minimum.
The procedure when the vertex is outside the graph is somewhat underspecified, but I guess the exact same idea would work. Just keep in mind that there is the possibility to walk all around the graph to the target if it is on the border, like in your example.
This could probably be implemented in O((n+m) log m) per query. If you run an all-pairs shortest path algorithm as a preprocessing step and use an online point location algorithm, you can get logarithmic query time at the cost of the space necessary to store the information to speed up shortest path queries (quadratic if you just store all distance pairs).
I believe simple planar point location works just like the sweep line approaches, only with persistent BSTs to store all the sweepline states.
I'm not sure why you are a bothering with trying to find a starting vertex when you already have one. The point you (the user) are standing at is another vertex in of itself. So the real question now is to find the distance from your starting point to any other point in the enclosing polygon graph. And once you have that, you can simply run Dijkstra's or another shortest path algorithm method like A*, BFS, etc, to find the shortest path to your goal point.
On that note, I think you are better off implementing A* for this problem because a park involves things like trees, playgrounds, ponds (sometimes), etc. So you will need to use a shortest path algorithm that takes these into consideration, and A* is one algorithm that uses these factors to determine a path of shortest length.
Finding distance from start to graph:
The problem of finding the distance from your new vertex to other vertices can be done by only looking for points with the closest x or y coordinate to your start point. So this algorithm has to find points that form a sort of closure around the start point, i.e. a polygon of minimum area which contains the point. So as #Niklas B suggested, a planar point algorithm (with some modifications) might be able to accomplish this. I was looking at the sweep-line algorithm, but that only works for line segments so that will not work (still worth a shot, with modifications might be able to give the correct answer).
You can also decide to implement this algorithm in stages, so first, find the points with the closest y coordinate to the current point (Both negative and positive y, so have to use absolute value), then among those points, you find the ones with the closest x coordinate to the current point and that should give you the set of points that form the polygon. Then these are the points you use to find the distance from your start to the graph.