Given a grid of points I'm trying to find the path between two of them.
Like in this picture: I'd need to find the points for the yellow line:
What are the best methods / algorithms I can use?
Thanks
Check out the A* algorithm
It's what's used in many video games for pathfinding problems, and can be built out to be very robust.
Dijkstra's algorithm can be a good start.
You haven't exactly defined how you want to use diagonal lines so you will have to write the final function as you need it, i suppose taking the path with shortest length of those that use diagonals, noting that path from a>c is shorter than path a>b>c for a,b,c in path
grid = [[False]*16 for i in range(16)]
#mark grid of walls
def rect(p1,p2):
x1, y1 = p1
x2, y2 = p2
for x in range(x1, x2+1):
for y in range(y1, y2+1):
yield (x, y)
rects = [((1,2),(5,5)),
((5,5),(14,15)),
((11,5),(11,11)),
((5,11),(11,11)),
((4,7),(5,13)),
((5,13),(13,13))]
for p1,p2 in rects:
for point in rect(p1,p2):
x,y = point
grid[x][y] = True
start = (1,2)
end = (12,13)
assert(grid[start[0]][start[1]])
assert(grid[end[0]][end[1]])
def children(parent):
x,y = parent
surrounding_points = ((x1,y1) for x1 in range(x-1,x+2) for y1 in range(y-1,y+2) if x1>0 and y<15)
for x,y in surrounding_points:
if grid[x][y]:
#not a wall
grid[x][y] = False
#set as wall since we have been there already
yield x,y
path = {}
def bfs(fringe):
if end in fringe:
return
new_fringe = []
for parent in fringe:
for child in children(parent):
path[child] = parent
new_fringe.append(child)
del fringe
if new_fringe:
bfs(new_fringe)
bfs([start])
def unroll_path(node):
if node != start:
return unroll_path(path[node]) + [node]
else:
return [start]
path = unroll_path(end)
def get_final_path_length(path):
#return length of path if using straight lines
for i in range(len(path)):
for j in range(i+1,len(path)):
#if straight line between pathi and pathj
return get_final_path(path[j+1:]) + distance_between_i_and_j
Related
I'm looking for an enumeration algorithm to search through a 3D array "sphering" around a given starting point.
Given an array a of size NxNxN where each N is 2^k for some k, and a point p in that array. The algorithm I'm looking for should do the following: If a[p] satisfies a certain predicate, the algorithm stops and p is returned. Otherwise the next point q is checked, where q is another point in the array that is the closest to p and hasn't been visited yet. If that doesn't match either, the next q'is checked an so on until in the worst case the whole array has been searched.
By "closest" here the perfect solution would be the point q that has the smallest Euclidean distance to p. As only discrete points have to be considered, perhaps some clever enumeration algorithm woukd make that possible. However, if this gets too complicated, the smallest Manhattan distance would be fine too. If there are several nearest points, it doesn't matter which one should be considered next.
Is there already an algorithm that can be used for this task?
You can search for increasing squared distances, so you won't miss a point. This python code should make it clear:
import math
import itertools
# Calculates all points at a certain distance.
# Coordinate constraint: z <= y <= x
def get_points_at_squared_euclidean_distance(d):
result = []
x = int(math.floor(math.sqrt(d)))
while 0 <= x:
y = x
while 0 <= y:
target = d - x*x - y*y
lower = 0
upper = y + 1
while lower < upper:
middle = (lower + upper) / 2
current = middle * middle
if current == target:
result.append((x, y, middle))
break
if current < target:
lower = middle + 1
else:
upper = middle
y -= 1
x -= 1
return result
# Creates all possible reflections of a point
def get_point_reflections(point):
result = set()
for p in itertools.permutations(point):
for n in range(8):
result.add((
p[0] * (1 if n % 8 < 4 else -1),
p[1] * (1 if n % 4 < 2 else -1),
p[2] * (1 if n % 2 < 1 else -1),
))
return sorted(result)
# Enumerates all points around a center, in increasing distance
def get_next_point_near(center):
d = 0
points_at_d = []
while True:
while not points_at_d:
d += 1
points_at_d = get_points_at_squared_euclidean_distance(d)
point = points_at_d.pop()
for reflection in get_point_reflections(point):
yield (
center[0] + reflection[0],
center[1] + reflection[1],
center[2] + reflection[2],
)
# The function you asked for
def get_nearest_point(center, predicate):
for point in get_next_point_near(center):
if predicate(point):
return point
# Example usage
print get_nearest_point((1,2,3), lambda p: sum(p) == 10)
Basically you consume points from the generator until one of them fulfills your predicate.
This is pseudocode for a simple algorithm that will search in increasing-radius spherical husks until it either finds a point or it runs out of array. Let us assume that condition returns either true or false and has access to the x, y, z coordinates being tested and the array itself, returning false (instead of exploding) for out-of-bounds coordinates:
def find_from_center(center, max_radius, condition) returns a point
let radius = 0
while radius < max_radius,
let point = find_in_spherical_husk(center, radius, condition)
if (point != null) return point
radius ++
return null
the hard part is inside find_in_spherical_husk. We are interested in checking out points such that
dist(center, p) >= radius AND dist(center, p) < radius+1
which will be our operating definition of husk. We could iterate over the whole 3D array in O(n^3) looking for those, but that would be really expensive in terms of time. A better pseudocode is the following:
def find_in_spherical_husk(center, radius, condition)
let z = center.z - radius // current slice height
let r = 0 // current circle radius; maxes at equator, then decreases
while z <= center + radius,
let z_center = (z, center.x, point.y)
let point = find_in_z_circle(z_center, r)
if (point != null) return point
// prepare for next z-sliced cirle
z ++
r = sqrt(radius*radius - (z-center.z)*(z-center.z))
the idea here is to slice each husk into circles along the z-axis (any axis will do), and then look at each slice separately. If you were looking at the earth, and the poles were the z axis, you would be slicing from north to south. Finally, you would implement find_in_z_circle(z_center, r, condition) to look at the circumference of each of those circles. You can avoid some math there by using the Bresenham circle-drawing algorithm; but I assume that the savings are negligible compared with the cost of checking condition.
given a grid of paths with different width, how can i find a path which leads to the end point?
The path is going to be represented by a two dimentional array where 0 means cannot be walk on, 1 means it is walkable, 2 represents starting point and 3 represents end point. Consider the following example:
21111111100000
00000011000000
00001111111111
00001111100111
00001110000101
00001111100113
in the above example the width of a path varies from 1 to 3, and there exists many solutions which would lead to the end point. I want to find one path which leads to it and the path does not have to be the shortest one (should not be the longest one either). The width of each path is unknown which means the grid could be all "1"s except the starting and end point.
Edited: The path should not contain uneccessary "wasted" walk meaning that if a vertical path has width 2 the result should not just walk down the path and then take one step right then walk all the way up
I agree with Calumn: DFS is the simplest approach here. Here is a simple solution in python-like pseudocode. It will print the solution as a sequence of 'L','R',U','D' to indicate left,right,up, or down.
def flood(x,y,story):
if (visited[x][y] or map[x][y]=='0'): return;
visited[x][y]=True;
if (map[x][y]=='3'):
print 'done. The path is: '+story
return
if (x<len(a[0])): flood(x+1,y,story+'R')
if (y<len(a)): flood(x,y+1,story+'D')
if (x>0): flood(x-1,y,story+'L')
if (y>0): flood(x,y-1,story+'U')
def solve(map):
visited = array_of_false_of_same_size_as(map)
x,y = find_the_two(map)
flood(x,y,'')
The optimization of making it stop as soon as it finds a solution is left as an exercise to the reader (you could make flood return a boolean to indicate if it found something, or use a global flag).
(p.s. I made this answer community wiki since I'm just clarifying Calumn's answer. I can't claim much credit)
Breadth-First Search version, also in Python
For what it's worth, and just to show that breadth-first search is not that complicated, an actual runnable program in Python:
def find(grid, xstart=0, ystart=0):
# Maps (xi,yi) to (x(i-1), y(i-1))
prev = {(xstart, ystart):None}
# Prepare for the breadth-first search
queue = [(xstart, ystart)]
qpos = 0
# Possibly enqueue a trial coordinate
def enqueue(prevxy, dx, dy):
x = prevxy[0] + dx
y = prevxy[1] + dy
xy = (x, y)
# Check that it hasn't been visited and the coordinates
# are valid and the grid position is not a 0
if (xy not in prev
and x >= 0 and x < len(grid)
and y >= 0 and y < len(grid[x])
and grid[x][y] != 0):
# Record the history (and the fact that we've been here)
prev[xy] = prevxy
# If we found the target, signal success
if grid[x][y] == 3:
return xy
# Otherwise, queue the new coordinates
else:
queue.append(xy)
return None
# The actual breadth-first search
while qpos < len(queue):
xy = queue[qpos]
qpos += 1
found = ( enqueue(xy, 1, 0)
or enqueue(xy, 0, 1)
or enqueue(xy, -1, 0)
or enqueue(xy, 0, -1))
if found: break
# Recover the path
path = []
while found:
path.append(found)
found = prev[found]
path.reverse()
return path
# Test run
grid = [ [2,1,1,1,1,1,1,1,1,0,0,0,0,0]
, [0,0,0,0,0,0,1,1,0,0,0,0,0,0]
, [0,0,0,0,1,1,1,1,1,1,1,1,1,1]
, [0,0,0,0,1,1,1,1,1,0,0,1,1,1]
, [0,0,0,0,1,1,1,0,0,0,0,1,0,1]
, [0,0,0,0,1,1,1,1,1,0,0,1,1,3]
]
for x, y in find(grid): grid[x][y]='*'
print '\n'.join(''.join(str(p) for p in line) for line in grid)
Output:
*******1100000
000000*1000000
000011******11
00001111100*11
00001110000*01
00001111100***
I have a rectangular area R in the coordinate system and a set of points P lying inside R. All sides are parallel to the axes and all points are integers. I want to divide R into smaller rectangles in such a way that
(a) the sides of the rectangles either stick to the sides of R or include at least one point from P and
(b) within each rectangle there is exactly one point from P
I have to find the smallest amount of rectangles which would cover all points from P. An example is drawn here: http://i5.minus.com/jC5LnVhjk6soT.png The purple line would indicate an incorrect division because the upper rectangle does not include a point from P. The blue line, however, is quite alright, because both rectangles have a point from P within, so the correct output would be: 2, because this is the minimum amount of rectangles.
Does anyone have an idea of an algorithm/method to find the smallest number?
According to your specifications, I end up with this recursive algorithm : (pseudocode~ruby implementation)
def resolve R, P
if P.empty?
return nil
elsif P.size == 1
return 1
end
results = []
P.each do |p|
rect1, rect2 = split_vertical(R, P, p)
s1 = split_resolve(rect1, rect2)
rect1, rect2 = split_horizontal(R, P, p)
s2 = split_resolve(rect1, rect2)
results.push [s1, s2].min
end
return results.min
end
def split_resolve rect1, rect2
sum1 = resolve(rect1.R, rect1.P)
sum2 = resolve(rect2.R, rect2.P)
if !sum1.nil? and !sum2.nil?
return sum1 + sum2
else
return nil
end
end
Functions split_vertical and split_horizontal split the area R with a vertical and an horizontal line passing through the point p.
You can also optimize this algorithm using dynamic programming. You can store the results of sub-rectangles without computing it an other time. This happend when severals points stands on the same line.
ps : Don't copy the raw source code, you may have some problems with the nil idiom. It's just a pseudocode example for the whole process.
I'm implementing the Bentley-Ottmann Algorithm in Lua for finding intersecting points in a polygon using the pseudo code located here.
I'm relatively new to implementing algorithms so I couldn't understand all parts of it. Here's my code so far:
local function getPolygonIntersectingVertices( poly )
-- initializing and sorting X
local X = {}
for i = 1, table.getn( poly ) do
if i == 1 then
table.insert( X, { x = poly[i].x, y = poly[i].y, endpoint = 'left' } )
elseif i == table.getn( poly ) then
table.insert( X, { x = poly[i].x, y = poly[i].y, endpoint = 'right' } )
else
table.insert( X, { x = poly[i].x, y = poly[i].y, endpoint = 'right' })
table.insert( X, { x = poly[i].x, y = poly[i].y, endpoint = 'left' })
end
end
local sortxy = function( a, b )
if a.x < b.x then return true
elseif a.x > b.x then return false
elseif a.y <= b.y then return true
else return false end
end
table.sort( X, sortxy )
-- Main loop
local SL = {}
local L = {}
local E
local i = 1
while next(X) ~= nil do
E = { x = X[i].x, y = X[i].y, endpoint = X[i].endpoint }
if E.endpoint == 'left' then
-- left endpoint code here
elseif E.endpoint == 'right' then
-- right endpoint code here
else
end
table.remove( X, i )
end
return L
end
My polygon is a table using this structure: { { x = 1, y = 3 }, { x = 5, y = 6 }, ... }
How do I determine "the segment above segE in SL;" and "the segment below segE in SL;" and what to do if the sweep line (SL) is empty? Also when inserting I into X, should I mark it with endpoint = 'intersect' and append it to the end so when the loop comes to this part goes into the "else" statement of the main loop or I've got the whole algorithm wrong?
It would be perfect in someone can show me a link with a simple implementation in Python, Ruby, etc. as I find it hard to follow the pseudo code and match it with the C++ example.
Your reference link fails from my location. I will reference the Wikipedia article, which is reasonably good.
How do I determine "the segment above segE in SL;" and "the segment below segE in SL;"
The algorithm requires a BST for current scan line intersections sorted on a key of y, i.e. in order vertically. So the segment above is the BST successor and the one below is the BST predecessor. Finding the predecessor and successor of a given node in a BST is standard stuff. The predecessor of key K is the rightmost node left of K. The successor is the leftmost node right of K. There are several ways of computing these. The simplest is to use parent pointers to walk back up and then down the tree from K. A stack-based iterator is another.
what to do if the sweep line (SL) is empty?
Keep processing the event queue. An empty sweep line just means no segments are crossing at its current x location.
Also when inserting I into X, should I mark it with endpoint = 'intersect' and append it to the end ...?
The event queue must remain sorted on the x-coordinate of points. When you insert an intersection it must be in x-coordinate order, too. It must be marked as an intersection because intersections are processed differently from endpoints. It will be processed in due course when it's the first remaining item in x order.
Note that Bentley Ottman - just as nearly all geometric algorithms - is notoriously subject to horrendous failures due to floating point inaccuracy. Also, the algorithm is normally given with a "general position" assumption, which lets out all the nasty cases of vertical edges, point-edge coincidence, edge-edge overlaps, etc. My strongest recommendation is to use rational arithmetic. Even then, getting a fully robust, correct implementation is a significant achievement. You can tell this by the very small number of free implementations!
Say I'm building a board game with a hextile grid, like Settlers of Catan:
Note that each vertex and edge may have an attribute (a road and settlement above).
How would I make a data structure which represents this board? What are the patterns for accessing each tile's neighbors, edges and vertices?
Amit Patel has posted an amazing page on this topic. It's so comprehensive and wonderful that it needs to be the definitive answer to this question: Hexagonal Grids
Such a grid can be represented in a two-dimensional array:
If
2
7 3
1
6 4
5
is the number one with its neighbors in the hex grid, then you can put this into a 2D array like so:
2 3
7 1 4
6 5
Obviously neighbor-ness is determined in this grid not only by being horizontally or vertically adjacent but also using one diagonal.
You can use a graph too, if you like to, though.
This article goes through how to set up a Isomeric/Hexagonal grid game. I recommend you have a look at the Forcing Isometric and Hexagonal Maps onto a Rectangular Grid section and the the movement section. Although it is different from what you are looking for it may help you formulate how to do what you want.
I've dealt a lot with hexes. In cases like this, you track each of the 6 points for the borders of the hex. This lets you draw it quite easily.
You would have a single array of objects that represent hexes. Each of these hex objects also has 6 "pointers" (or an index to another array) pointing to another array of "sides". Same thing for "vertices". Of course the vertices would have 3 pointers to the adjoining hexes, and the sides would have 2.
So, a hex may be something like:
X, Y, Point(6), Vertices(6), Sides(6)
Then you have a Hex array, vertice array, and side array.
Then it is pretty simple to find the vertices/sides for a hex, or whatever.
When I say pointer it could just as easily be an integer pointing to the element in the vertice or side array or whatever. And of course arrays could be lists or whatever.
You could create a 2D array and then consider the valid positions as:
On even-numbered rows (0,2,4,...): the odd numbered cells.
On odd-numbered rows (1,3,5,...): the even numbered cells.
For each cell, its neighbors would be:
Same column, 2 rows up
Same column, 2 rows down
1 left + 1 up
1 left + 1 down
1 right + 1 up
1 right + 1 down
Illustration:
The x marks are hexes. x that are diagonal to each other are neighbors. | connects vertical neighbors.
2
7 3
1
6 4
5
You can also try to 'flat' rows of your map. For this example it would be:
2
7 1 3
6 5 4
Its sometimes more useful to have rows in one row:P
I would suggest something like the following (I'll use Delphi-style declarations):
type
THexEdge = record
Hexes: array[1..2] of Integer; // Index of adjoining hexes.
// Other edge stuff goes here.
end;
THexVertex = record
Hexes: array[1..3] of Integer; // Index of adjoining hexes.
// Other vertex stuff goes here.
end;
THex = record
Edges: array[1..6] of Integer; // Index of edge.
Vertices: array[1..6] of Integer; // Index of vertex.
// Other hex stuff goes here.
end;
var
Edges: array of THexEdge;
Vertices: array of THexVertex;
HexMap: array of THex;
Each hex has six edges and six vertices. Each edge keeps track of its two adjoining hexes, and each vertex keeps track of its three adjoining hexes (hexes on the edges of the map will be a special case).
There are many things that you could do a different way of course. You could use pointers rather than arrays, you could use objects rather than records, and you could store your hexes in a two-dimensional array as other answerers have suggested.
Hopefully, that might give you some ideas about one way to approach it though.
We implemented a Settlers of Catan AI for a class project, and modified code from this answer (which was buggy) to create a Board with constant time random access to vertices and edges. It was a fun problem, but the board took a lot of time, so in case anyone is still looking for a simple implementation here is our Python code:
class Board:
# Layout is just a double list of Tiles, some will be None
def __init__(self, layout=None):
self.numRows = len(layout)
self.numCols = len(layout[0])
self.hexagons = [[None for x in xrange(self.numCols)] for x in xrange(self.numRows)]
self.edges = [[None for x in xrange(self.numCols*2+2)] for x in xrange(self.numRows*2+2)]
self.vertices = [[None for x in xrange(self.numCols*2+2)] for x in xrange(self.numRows*2+2)]
for row in self.hexagons:
for hexagon in row:
if hexagon == None: continue
edgeLocations = self.getEdgeLocations(hexagon)
vertexLocations = self.getVertexLocations(hexagon)
for xLoc,yLoc in edgeLocations:
if self.edges[xLoc][yLoc] == None:
self.edges[xLoc][yLoc] = Edge(xLoc,yLoc)
for xLoc,yLoc in vertexLocations:
if self.vertices[xLoc][yLoc] == None:
self.vertices[xLoc][yLoc] = Vertex(xLoc,yLoc)
def getNeighborHexes(self, hex):
neighbors = []
x = hex.X
y = hex.Y
offset = 1
if x % 2 != 0:
offset = -1
if (y+1) < len(self.hexagons[x]):
hexOne = self.hexagons[x][y+1]
if hexOne != None: neighbors.append(hexOne)
if y > 0:
hexTwo = self.hexagons[x][y-1]
if hexTwo != None: neighbors.append(hexTwo)
if (x+1) < len(self.hexagons):
hexThree = self.hexagons[x+1][y]
if hexThree != None: neighbors.append(hexThree)
if x > 0:
hexFour = self.hexagons[x-1][y]
if hexFour != None: neighbors.append(hexFour)
if (y+offset) >= 0 and (y+offset) < len(self.hexagons[x]):
if (x+1) < len(self.hexagons):
hexFive = self.hexagons[x+1][y+offset]
if hexFive != None: neighbors.append(hexFive)
if x > 0:
hexSix = self.hexagons[x-1][y+offset]
if hexSix != None: neighbors.append(hexSix)
return neighbors
def getNeighborVertices(self, vertex):
neighbors = []
x = vertex.X
y = vertex.Y
offset = -1
if x % 2 == y % 2: offset = 1
# Logic from thinking that this is saying getEdgesOfVertex
# and then for each edge getVertexEnds, taking out the three that are ==vertex
if (y+1) < len(self.vertices[0]):
vertexOne = self.vertices[x][y+1]
if vertexOne != None: neighbors.append(vertexOne)
if y > 0:
vertexTwo = self.vertices[x][y-1]
if vertexTwo != None: neighbors.append(vertexTwo)
if (x+offset) >= 0 and (x+offset) < len(self.vertices):
vertexThree = self.vertices[x+offset][y]
if vertexThree != None: neighbors.append(vertexThree)
return neighbors
# used to initially create vertices
def getVertexLocations(self, hex):
vertexLocations = []
x = hex.X
y = hex.Y
offset = x % 2
offset = 0-offset
vertexLocations.append((x, 2*y+offset))
vertexLocations.append((x, 2*y+1+offset))
vertexLocations.append((x, 2*y+2+offset))
vertexLocations.append((x+1, 2*y+offset))
vertexLocations.append((x+1, 2*y+1+offset))
vertexLocations.append((x+1, 2*y+2+offset))
return vertexLocations
# used to initially create edges
def getEdgeLocations(self, hex):
edgeLocations = []
x = hex.X
y = hex.Y
offset = x % 2
offset = 0-offset
edgeLocations.append((2*x,2*y+offset))
edgeLocations.append((2*x,2*y+1+offset))
edgeLocations.append((2*x+1,2*y+offset))
edgeLocations.append((2*x+1,2*y+2+offset))
edgeLocations.append((2*x+2,2*y+offset))
edgeLocations.append((2*x+2,2*y+1+offset))
return edgeLocations
def getVertices(self, hex):
hexVertices = []
x = hex.X
y = hex.Y
offset = x % 2
offset = 0-offset
hexVertices.append(self.vertices[x][2*y+offset]) # top vertex
hexVertices.append(self.vertices[x][2*y+1+offset]) # left top vertex
hexVertices.append(self.vertices[x][2*y+2+offset]) # left bottom vertex
hexVertices.append(self.vertices[x+1][2*y+offset]) # right top vertex
hexVertices.append(self.vertices[x+1][2*y+1+offset]) # right bottom vertex
hexVertices.append(self.vertices[x+1][2*y+2+offset]) # bottom vertex
return hexVertices
def getEdges(self, hex):
hexEdges = []
x = hex.X
y = hex.Y
offset = x % 2
offset = 0-offset
hexEdges.append(self.edges[2*x][2*y+offset])
hexEdges.append(self.edges[2*x][2*y+1+offset])
hexEdges.append(self.edges[2*x+1][2*y+offset])
hexEdges.append(self.edges[2*x+1][2*y+2+offset])
hexEdges.append(self.edges[2*x+2][2*y+offset])
hexEdges.append(self.edges[2*x+2][2*y+1+offset])
return hexEdges
# returns (start, end) tuple
def getVertexEnds(self, edge):
x = edge.X
y = edge.Y
vertexOne = self.vertices[(x-1)/2][y]
vertexTwo = self.vertices[(x+1)/2][y]
if x%2 == 0:
vertexOne = self.vertices[x/2][y]
vertexTwo = self.vertices[x/2][y+1]
return (vertexOne, vertexTwo)
def getEdgesOfVertex(self, vertex):
vertexEdges = []
x = vertex.X
y = vertex.Y
offset = -1
if x % 2 == y % 2: offset = 1
edgeOne = self.edges[x*2][y-1]
edgeTwo = self.edges[x*2][y]
edgeThree = self.edges[x*2+offset][y]
if edgeOne != None: vertexEdges.append(edgeOne)
if edgeTwo != None: vertexEdges.append(edgeTwo)
if edgeThree != None: vertexEdges.append(edgeThree)
return vertexEdges
def getHexes(self, vertex):
vertexHexes = []
x = vertex.X
y = vertex.Y
xOffset = x % 2
yOffset = y % 2
if x < len(self.hexagons) and y/2 < len(self.hexagons[x]):
hexOne = self.hexagons[x][y/2]
if hexOne != None: vertexHexes.append(hexOne)
weirdX = x
if (xOffset+yOffset) == 1: weirdX = x-1
weirdY = y/2
if yOffset == 1: weirdY += 1
else: weirdY -= 1
if weirdX >= 0 and weirdX < len(self.hexagons) and weirdY >= 0 and weirdY < len(self.hexagons):
hexTwo = self.hexagons[weirdX][weirdY]
if hexTwo != None: vertexHexes.append(hexTwo)
if x > 0 and x < len(self.hexagons) and y/2 < len(self.hexagons[x]):
hexThree = self.hexagons[x-1][y/2]
if hexThree != None: vertexHexes.append(hexThree)
return vertexHexes
I am sitting here "in my free time coding for fun" with hexes. And it goes like this... I will tell you what it looks like in words.
Hexagon: it has six neighbour hexagons. It can deliver the reference for each neighbouring hex tile. It can tell you what it consists of(water ,rock, dust). It can connect itself to others and vice versa. It can even automatically connect the others surrounding him to create a greater field and or making sure all fields can be adressed by its neighbours.
A building references up to three roads and three Hex Tiles. They can tell you which they are.
A road references two hexes and other roads when they are adressed by neighbouring tiles. They can tell which tiles that are and which roads or buildings they connect to.
This is just an idea how I would work on it.