The Problem
I have a bit array which represents a 2-dimensional "map" of "tiles". This image provides a graphical example of the bits in the bit array:
I need to determine how many contiguous "areas" of bits exist in the array. In the example above, there are two such contiguous "areas", as illustrated here:
Tiles must be located directly N, S, E or W of a tile to be considered "contiguous". Diagonally-touching tiles do not count.
What I've Got So Far
Because these bit arrays can become relatively large (several MB in size), I have intentionally avoided using any sort of recursion in my algorithm.
The pseudo-code is as follows:
LET S BE SOURCE DATA ARRAY
LET C BE ARRAY OF IDENTICAL LENGTH TO SOURCE DATA USED TO TRACK "CHECKED" BITS
FOREACH INDEX I IN S
IF C[I] THEN
CONTINUE
ELSE
SET C[I]
IF S[I] THEN
EXTRACT_AREA(S, C, I)
EXTRACT_AREA(S, C, I):
LET T BE TARGET DATA ARRAY FOR STORING BITS OF THE AREA WE'RE EXTRACTING
LET F BE STACK OF TILES TO SEARCH NEXT
PUSH I UNTO F
SET T[I]
WHILE F IS NOT EMPTY
LET X = POP FROM F
IF C[X] THEN
CONTINUE
ELSE
SET C[X]
IF S[X] THEN
PUSH TILE NORTH OF X TO F
PUSH TILE SOUTH OF X TO F
PUSH TILE WEST OF X TO F
PUSH TILE EAST OF X TO F
SET T[X]
RETURN T
What I Don't Like About My Solution
Just to run, it requires two times the memory of the bitmap array it's processing.
While extracting an "area", it uses three times the memory of the bitmap array.
Duplicates exist in the "tiles to check" stack - which seems ugly, but not worth avoiding the way I have things now.
What I'd Like To See
Better memory profile
Faster handling of large areas
Solution / Follow-Up
I re-wrote the solution to explore the edges only (per #hatchet 's suggestion).
This was very simple to implement - and eliminated the need to keep track of "visited tiles" completely.
Based on three simple rules, I can traverse the edges, track min/max x & y values, and complete when I've arrived at the start again.
Here's the demo with the three rules I used:
One approach would be a perimeter walk.
Given a starting point anywhere along the edge of the shape, remember that point.
Start the bounding box as just that point.
Walk the perimeter using a clockwise rule set - if the point used to get to the current point was above, then first look right, then down, then left to find the next point on the shape perimeter. This is kind of like the simple strategy of solving a maze where you continuously follow a wall and always bear to the right.
Each time you visit a new perimeter point, expand the bounding box if the new point is outside it (i.e. keep track of the min and max x and y.
Continue until the starting point is reached.
Cons: if the shape has lots of single pixel 'filaments', you'll be revisiting them as the walk comes back.
Pros: if the shape has large expanses of internal occupied space, you never have to visit them or record them like you would if you were recording visited pixels in a flood fill.
So, conserves space, but in some cases at the expense of time.
Edit
As is so often the case, this problem is known, named, and has multiple algorithmic solutions. The problem you described is called Minimum Bounding Rectangle. One way to solve this is using Contour Tracing. The method I described above is in that class, and is called Moore-Neighbor Tracing or Radial Sweep. The links I've included for them discuss them in detail and point out a problem I hadn't caught. Sometimes you'll revisit the start point before you have traversed the entire perimeter. If your start point is for instance somewhere along a single pixel 'filament', you will revisit it before you're done, and unless you consider this possibility, you'll stop prematurely. The website I linked to talks about ways to address this stopping problem. Other pages at that website also talk about two other algorithms: Square Tracing, and Theo Pavlidis's Algorithm. One thing to note is that these consider diagonals as contiguous, whereas you don't, but that should just something that can be handled with minor modifications to the basic algorithms.
An alternative approach to your problem is Connected-component labeling. For your needs though, this may be a more time expensive solution than you require.
Additional resource:
Moore Neighbor Contour Tracing Algorithm in C++
I actually got a question like this in an interview once.
You can pretend the array is a graph and the connected nodes are the adjacent ones. My algo would involves going 1 to the right until you find a marked node. When you find one do a breadth first search which runs in O(n) and avoids recursion. When the BFS returns keep searching from where you left off and if the node has already been marked by one of the previous BFS's you obviously don't need to search. I wasn't sure if you wanted to actually return the number of objects found, but it's easy to keep track by just incrementing a counter when you hit the first marked square.
Generally when you do a flood fill type algorithm you are placed in a spot and asked to fill. Since this is finding all the filled regions one way you would want to optimize it is to avoid rechecking the already marked nodes from previous BFS's, unfortunately at the moment I cannot think of a way to do that.
One hacky way to reduce memory consumption would be too store a short[][] instead of a boolean. Then use this scheme to avoid making a whole second 2d-array
unmarked = 0, marked = 1, checked and unmarked = 3, checked and marked = 3
This way you can check the status of an entry by its value and avoid making a second array.
Related
I have any number of points on an imaginary 2D surface. I also have a grid on the same surface with points at regular intervals along the X and Y access. My task is to map each point to the nearest grid point.
The code is straight forward enough until there are a shortage of grid points. The code I've been developing finds the closest grid point, displaying an already mapped point if the distance will be shorter for the current point.
I then added a second step that compares each mapped point to another and, if swapping the mapping with another point produces a smaller sum of the total mapped distance of both points, I swap them.
This last step seems important as it reduces the number crossed map lines. (This would be used to map points on a plate to a grid on another plate, with pins connecting the two, and lines that don't cross seem to have a higher chance that the pins would not make contact.)
Questions:
Can anyone comment on my thinking that if the image above were truly optimized, (that is, the mapped points--overall--would have the smallest total distance), then none of the lines were cross?
And has anyone seen any existing algorithms to help with this. I've searched but came up with nothing.
The problem could be approached as a variation of the Assignment Problem, with the "agents" being the grid squares and the points being the "tasks", (or vice versa) with the distance between them being the "cost" for that agent-task combination. You could solve with the Hungarian algorithm.
To handle the fact that there are more grid squares than points, find a bounding box for the possible grid squares you want to consider and add dummy points that have a cost of 0 associated with all grid squares.
The Hungarian algorithm is O(n3), perhaps your approach is already good enough.
See also:
How to find the optimal mapping between two sets?
How to optimize assignment of tasks to agents with these constraints?
If I understand your main concern correctly, minimising total length of line segments, the algorithm you used does not find the best mapping and it is clear in your image. e.g. when two line segments cross each other, simple mathematic says that if you rearrange their endpoints such that they do not cross, it provides a better total sum. You can use this simple approach (rearranging crossed items) to get better approximation to the optimum, you should apply swapping for more somehow many iterations.
In the following picture you can see why crossing has longer length than non crossing (first question) and also why by swapping once there still exists crossing edges (second question and w.r.t. Comments), I just drew one sample, in fact one may need many iterations of swapping to get non crossed result.
This is a heuristic algorithm certainly not optimum but I expect to be very good and efficient and simple to implement.
I am making a simple game and stumbled upon this problem. Assume several points in 2D space. What I want is to make points close to each other interact in some way.
Let me throw a picture here for better understanding of the problem:
Now, the problem isn't about computing the distance. I know how to do that.
At first I had around 10 points and I could simply check every combination, but as you can already assume, this is extremely inefficient with increasing number of points. What if I had a million of points in total, but all of them would be very distant to each other?
I'm trying to find a suitable data structure or a way to look at this problem, so every point can only mind their surrounding and not whole space. Are there any known algorithms for this? I don't exactly know how to name this problem so I can google exactly what I want.
If you don't know of such known algorighm, all ideas are very welcome.
This is a range searching problem. More specifically - the 2-d circular range reporting problem.
Quoting from "Solving Query-Retrieval Problems by Compacting Voronoi Diagrams" [Aggarwal, Hansen, Leighton, 1990]:
Input: A set P of n points in the Euclidean plane E²
Query: Find all points of P contained in a disk in E² with radius r centered at q.
The best results were obtained in "Optimal Halfspace Range Reporting in Three Dimensions" [Afshani, Chan, 2009]. Their method requires O(n) space data structure that supports queries in O(log n + k) worst-case time. The structure can be preprocessed by a randomized algorithm that runs in O(n log n) expected time. (n is the number of input points, and k in the number of output points).
The CGAL library supports circular range search queries. See here.
You're still going to have to iterate through every point, but there are two optimizations you can perform:
1) You can eliminate obvious points by checking if x1 < radius and if y1 < radius (like Brent already mentioned in another answer).
2) Instead of calculating the distance, you can calculate the square of the distance and compare it to the square of the allowed radius. This saves you from performing expensive square root calculations.
This is probably the best performance you're gonna get.
This looks like a nearest neighbor problem. You should be using the kd tree for storing the points.
https://en.wikipedia.org/wiki/K-d_tree
Space partitioning is what you want.. https://en.wikipedia.org/wiki/Quadtree
If you could get those points to be sorted by x and y values, then you could quickly pick out those points (binary search?) which are within a box of the central point: x +- r, y +- r. Once you have that subset of points, then you can use the distance formula to see if they are within the radius.
I assume you have a minimum and maximum X and Y coordinate? If so how about this.
Call our radius R, Xmax-Xmin X, and Ymax-Ymin Y.
Have a 2D matrix of [X/R, Y/R] of double-linked lists. Put each dot structure on the correct linked list.
To find dots you need to interact with, you only need check your cell plus your 8 neighbors.
Example: if X and Y are 100 each, and R is 1, then put a dot at 43.2, 77.1 in cell [43,77]. You'll check cells [42,76] [43,76] [44,76] [42,77] [43,77] [44,77] [42,78] [43,78] [44,78] for matches. Note that not all cells in your own box will match (for instance 43.9,77.9 is in the same list but more than 1 unit distant), and you'll always need to check all 8 neighbors.
As dots move (it sounds like they'd move?) you'd simply unlink them (fast and easy with a double-link list) and relink in their new location. Moving any dot is O(1). Moving them all is O(n).
If that array size gives too many cells, you can make bigger cells with the same algo and probably same code; just be prepared for fewer candidate dots to actually be close enough. For instance if R=1 and the map is a million times R by a million times R, you wouldn't be able to make a 2D array that big. Better perhaps to have each cell be 1000 units wide? As long as density was low, the same code as before would probably work: check each dot only against other dots in this cell plus the neighboring 8 cells. Just be prepared for more candidates failing to be within R.
If some cells will have a lot of dots, each cell having a linked list, perhaps the cell should have an red-black tree indexed by X coordinate? Even in the same cell the vast majority of other cell members will be too far away so just traverse the tree from X-R to X+R. Rather than loop over all dots, and go diving into each one's tree, perhaps you could instead iterate through the tree looking for X coords within R and if/when you find them calculate the distance. As you traverse one cell's tree from low to high X, you need only check the neighboring cell to the left's tree while in the first R entries.
You could also go to cells smaller than R. You'd have fewer candidates that fail to be close enough. For instance with R/2, you'd check 25 link lists instead of 9, but have on average (if randomly distributed) 25/36ths as many dots to check. That might be a minor gain.
I am looking into a Planar Point Location, and right now thinking about make use of The Slabs approach.
I've read through a couple of articles:
Planar Point Location Using Persistent Search Trees
Persistent Data Structures
Point location on Wikipedia
I understand the concept behind a persistent balanced binary search tree, but let's omit them in this question, I don't really care about extra storage space overhead. The thing with the articles is they are discussing how speed can be improved, but don't explain basic stuff. For example:
Could you please correct me if I am wrong:
We draw a line across all intersections.
Now, we have slabs that are split by the segments at different angles.
Every intersection with any line is considered a vertex.
Slabs are sorted by order in a binary search tree (let's omit a partially persistent bst)
Somehow, sectors are sorted in those respective BST's, even though the segments dividing them are almost always at an angle. Does each node has to carry a definition of area?
Please refer to this example image:
Questions:
How would I actually figure out if the point lies in Node c and not in Node b? Is it somehow via area?
Do I instead need arrange my nodes to contain information about the segments? I could then check if the query point lies above a segment (if that's how I should determine my sectors)? If so, would I then search through a Polygon List after, to see which polygon this particular segment belongs to?
Maybe I need to store BST for each line and not a slab?
Would I then have to look at 2 BST's belonging to line on the left, and the 2nd one of the line to the right from the vertex? I could then sort the vertexes by y coordinate in each tree and return the y coordinate of a vertex (the end of a segment) right below my query point. Having done this for left and right line, I would then do a comparison to see if the names of the segment those vertexes come from actually match.
However, this will not give me the right answer, since even if the names do match, I might be below or above the segment (if I am close to it). Also, this implies I have to do 3 binary searches (1 for lines, 1 for the y-coordinate on left line, 1 for the right one), and the books says I only need to do 2 searches (1 for slab, 2nd for sector).
Could someone please point me in the direction to do it?
I probably just missed some essential thought or something.
Edit:
Here is another good article, that explains the solution to the problem, however, I don't quite understand how I would achieve the following:
"Consider any query point q ∈ R2. To find the face of G that contains q, we first use binary search with the x-coordinate of q to find the vertical slab s that contains q. Given s, we use binary search with the y-coordinate of q to find the edges of Es between which q lies. "
How exactly to find those 2 edges? Is it as simple as checking if the point lies below the segment? However, this seems like a complicated check to do (and expensive), as we descend down the tree, inspecting the other nodes.
Well, this was asked a year ago so I guess you're not going to be around to accept the answer. It's a fun question, though, so here goes anyway...
Yes, you can divide the plane into slabs using vertical lines through every intersection. The important result is that all line segments that touch a slab will go all the way through, and the order of those line segments will be the same across the entire slab.
In each slab, you make a binary search tree of those line segments, ordered by their order in the slab. The areas that you call "nodes" are the leaves of the tree, and the line segments are the internal nodes. I will call the leaves "areas" to avoid ambiguity. Since the order of the line segments is the same across the entire slab, the order in this tree is valid for every x coordinate in the slab.
to determine the area that contains any point, you find the slab that contains the point, and then do a simple search in the BST to find the area.
Lets say you're looking for point C, and you're at the node in slab 2 that contains (x2,y2) - (x3,y4). Determine whether C is above or below this line segment, and then take that branch in the tree.
For example, if C is (cx,cy) then the y coordinate of the segment at x=cx is:
testy = ((cx-x2)/(x3-x2)) * (y4-y2) + y2
If cy < testy, then take the upward branch, otherwise take the downward branch. (assuming positive Y goes downward).
When you get to a leaf, that's the area that contains C. You're done.
Now... Making a whole new tree for every slab takes a lot of space, since you can have N^2 intersection points and therefore N^2 slabs. Storing a separate tree in every slab would take O(N^3) space.
Using, say, persistent red-black trees, however, the slabs can actually share a lot of nodes. Using a simple purely functional implementation like Chris Okasaki's red-black tree takes O(log N) space per intersection for a total of O(N^2*log N) space.
I seem to remember that there is also a constant-space-per-change persistent red-black tree that would give you O(N^2), but I don't have a reference.
Note that for most real-life scenarios there are closer to O(N) intersections, because lines don't cross very much, but it's still nice to save the space by using a persistent tree.
I do not know how to describe the goal succinctly, which may be why I haven't been able to find an applicable algorithm despite ample searching, but a picture shows it clearly:
Given the state of items in the grid at the left, does anyone know of an algorithm for efficiently finding the ending positions shown in the grid at right? In this case all the items have "fallen" "down", but the direction of course is arbitrary. The point is just that:
There are a collection of items of arbitrary shapes, but all composed of contiguous squares
Items cannot overlap
All items should move the maximum distance in a given direction until they are touching a wall, or they are touching another item which [...is touching another item ad infinitum...] is touching a wall.
This is not homework, I'm not a student. This is for my own interest in geometry and programming. I haven't mentioned the language because it doesn't matter. I can implement whatever algorithm in the language I'm using for the specific project I'm working on. A useful answer could be described in words or code; it's the ideas that matter.
This problem could probably be abstracted into some kind of graph (in the mathematical sense) of dependencies and slack space, so perhaps an algorithm aimed at minimizing lag time could be adapted.
If you don't know the answer but are about to try to make up an algorithm on the spot, just remember that there can be circular dependencies, such as with the interlocking pink (backwards) "C" and blue "T" shapes. Parts of T are below C, and parts of C are below T. This would be even more tricky if interlocking items were locked through a "loop" of several pieces.
Some notes for an applicable algorithm: All the following are very easy and fast to do because of the way I've built the grid object management framework:
Enumerate the individual squares within a piece
Enumerate all pieces
Find the piece, if any, occupying a specific square in the overall grid
A note on the answer:
maniek hinted it first, but bloops has provided a brilliant explanation. I think the absolute key is the insight that all pieces moving the same amount maintain their relationship to each other, and therefore those relationships don't have to be considered.
An additional speed-up for a sparsely populated board would be to shift all pieces to eliminate rows that are completely empty. It is very easy to count empty rows and to identify pieces on one side ("above") an empty row.
Last note: I did in fact implement the algorithm described by bloops, with a few implementation-specific modifications. It works beautifully.
The Idea
Define the set of frozen objects inductively as follows:
An object touching the bottom is frozen.
An object lying on a frozen object is frozen.
Intuitively, exactly the frozen objects have reached their final place. Call the non-frozen objects active.
Claim: All active objects can fall one unit downwards simultaneously.
Proof: Of course, an active object will not hit another active object, since their relative position with respect to each other does not change. An active object will also not hit a frozen object. If that was so, then the active object was, in fact, frozen, because it was lying on a a frozen object. This contradicts our assumption.
Our algorithm's very high-level pseudo-code would be as follows:
while (there are active objects):
move active objects downwards simultaneously until one of them hits a frozen object
update the status (active/frozen) of each object
Notice that at least one object becomes frozen in each iteration of the while loop. Also, every object becomes frozen exactly once. These observations would be used while analyzing the run-time complexity of the actual algorithm.
The Algorithm
We use the concept of time to improve the efficiency of most operations. The time is measured starting from 0, and every unit movement of the active objects take 1 unit time. Observe that, when we are at time t, the displacement of all the objects currently active at time t, is exactly t units downward.
Note that in each column, the relative ordering of each cell is fixed. One of the implications of this is that each cell can directly stop at most one other cell from falling. This observation could be used to efficiently predict the time of the next collision. We can also get away with 'processing' every cell at most once.
We index the columns starting from 1 and increasing from left to right; and the rows with height starting from 1. For ease of implementation, introduce a new object called bottom - which is the only object which is initially frozen and consists of all the cells at height 0.
Data Structures
For an efficient implementation, we maintain the following data structures:
An associative array A containing the final displacement of each cell. If a cell is active, its entry should be, say, -1.
For each column k, we maintain the set S_k of the initial row numbers of active cells in column k. We need to be able to support successor queries and deletions on this set. We could use a Van Emde Boas tree, and answer every query in O(log log H); where H is the height of the grid. Or, we could use a balanced binary search tree which can perform these operations in O(log N); where N is the number of cells in column k.
A priority queue Q which will store the the active cells with its key as the expected time of its future collision. Again, we could go for either a vEB tree for a query time of O(log log H) or a priority queue with O(log N) time per operation.
Implementation
The detailed pseudo-code of the algorithm follows:
Populate the S_k's with active cells
Initialize Q to be an empty priority queue
For every cell b in bottom:
Push Q[b] = 0
while (Q is not empty):
(x,t) = Q.extract_min() // the active cell x collides at time t
Object O = parent_object(x)
For every cell y in O:
A[y] = t // freeze cell y at displacement t
k = column(y)
S_k.delete(y)
a = S_k.successor(y) // find the only active cell that can collide with y
if(a != nil):
// find the expected time of the collision between a and y
// note that both their positions are currently t + (their original height)
coll_t = t + height(a) - height(y) - 1
Push/update Q[a] = coll_t
The final position of any object can be obtained by querying A for the displacement of any cell belonging to that object.
Running Time
We process and freeze every cell exactly once. We perform a constant number of lookups while freezing every cell. We assume parent_object lookup can be done in constant time. The complexity of the whole algorithm is O(N log N) or O(N log log H) depending on the data structures we use. Here, N is the total number of cells across all objects.
And now something completely different :)
Each piece that rests on the ground is fixed. Each piece that rests on a fixed piece is fixed. The rest can move. Move the unfixed pieces 1 square down, repeat until nothing can move.
Okay so this appears to be as follows -
we proceed in step
in each step we build a directed graph whose vertices are the object set and whose edges are as follows =>
1) if x and y are two objects then add an edge x->y if x cannot move until y moves. Note that we can have both x->y and y->x edges.
2) further there are objects which can no longer move since they are at the bottom so we colour their vertices as blue. The remaining vertices are red.
2) in the directed graph we find all the strongly connected components using Kosaraju's algorithm/Tarjan's algorithm etc. (In case you are not familiar with SCC then they are extremely powerful technique and you can refer to Kosaraju's algorithm.) Once we get the SCCs we shrink them to a small vertex i.e. we replace the SCC by a single vertex while preserving all external(to SCC) edges. Now if any of the vertex in SCC is blue then we color the new vertex as blue else it is red. This implies that if one object cannot move in SCC then none can.
3) the graph you get will be a directed acyclic graph and you can do a topological sort. Traverse the vertex in increasing order of top numbering and as long as you see a red colour vertex and move the objects represented by the vertex.
continue this step until you can no longer move any vertex in step 3.
If two objects A and B overlap then we say that they are inconsistent relative to each other. For proof of correctness argue the following lemmas
1) "if I move an SCC then none of the object in it cause inconsistency among themselves."
2) "when I move an object in step 3 then I do not cause inconsistencies"
The challenge for you now will be to formally prove the correctness and find suitable data structures to solve it in efficient. Let me know if you need any help.
I haven't cleared all the details, but I think the following seems like a somewhat systematic approach:
Looking at the whole picture as a graph, what you need is a topologic sort of all the vertices - i.e. the items. Item A should be before item B in the sorting if any part of A is below any part of B. Then, when you have the items sorted topologically, you can just iterate through them in this order and determine the positions - as all the items below the current one already have fixed positions.
In order to be able to do the topologic sort, you need an acyclic graph. You can then use some of the algorithms for Strongly Connected Components to find all the cycles and compress them into single vertices. Then you can perform a topsort on the resulting graph.
To find the positions of the pieces within a SCC: first consider it as one big piece and determine where it will end up. This will determine some fixed pieces that cannot move anymore. Remove them and repeat the same procedure for the rest of the pieces in this SCC (if any) to find out their final positions.
The third part is the only one that seems computationally intensive - if you have a very complicated structure for the pieces, but it should still be more optimal than trying to move the pieces one square of the grid at a time.
EDITED several times. I think this is all you need to do:
Find all the pieces that can only fall mutually dependent on each other and combine them into an equivalent larger piece (e.g. the T and backwards C in your picture.)
Iterate through all the pieces, moving them the maximum direction down before they hit something. Repeat until nothing moves.
I am now looking for an elegant algorithm to recursively find neighbors of neighbors with the geohashing algorithm (http://www.geohash.org).
Basically take a central geohash, and then get the first 'ring' of same-size hashes around it (8 elements), then, in the next step, get the next ring around the first etc. etc.
Have you heard of an elegant way to do so?
Brute force could be to take each neighbor and get their neighbors simply ignoring the massive overlap. Neighbors around one central geohash has been solved many times (here e.g. in Ruby: http://github.com/masuidrive/pr_geohash/blob/master/lib/pr_geohash.rb)
Edit for clarification:
Current solution, with passing in a center key and a direction, like this (with corresponding lookup-tables):
def adjacent(geohash, dir)
base, lastChr = geohash[0..-2], geohash[-1,1]
type = (geohash.length % 2)==1 ? :odd : :even
if BORDERS[dir][type].include?(lastChr)
base = adjacent(base, dir)
end
base + BASE32[NEIGHBORS[dir][type].index(lastChr),1]
end
(extract from Yuichiro MASUI's lib)
I say this approach will get ugly soon, because directions gets ugly once we are in ring two or three. The algorithm would ideally simply take two parameters, the center area and the distance from 0 being the center geohash only (["u0m"] and 1 being the first ring made of 8 geohashes of the same size around it (=> [["u0t", "u0w"], ["u0q", "u0n"], ["u0j", "u0h"], ["u0k", "u0s"]]). two being the second ring with 16 areas around the first ring etc.
Do you see any way to deduce the 'rings' from the bits in an elegant way?
That depends on what you mean by Neighbor. I'm assuming this is being used in the context of a proximity search. In that case I would think that your best bet would be to search from the outermost ring inward.
Assume you can find the outermost set (longest proximity) in the searchable Universe. Store that as the new Universe and then find the next inner set in that Universe. This search should subtract that inner set from the Universe. Store the old Universe (the outermost ring) and repeat this process until you hit the center. Each search after the first one will reduce your search area and give you a ring.
Start by constructing the sides immediately around the central geohash i.e. top, right, bottom and left, initially each of these will only comprise a single geohash and a corner. Then recursively iterate the sides using the adjacent function with direction as corresponds that side (i.e. expand left for the left side) whilst maintaining an appropriate result set and the sides for the next iteration. You also need to handle the diagonal/corner geohash for each side (e.g. left-top for left, top-right for top, if using a clockwise association). For an example of the procedure see this implementation I did in Lua or Javascript (but with additional functionality), which start with a call to Grid().