Pick nodes evenly distance from each other - igraph

In igraph is there a common way to pick such nodes in a graph so that they will be evenly (or as good as possible) spaced from each other?
I imagine it like some force-layout, but instead of doing it in 2D space it needs to be done in "graph-space".
In other words, need to pick a set B of nodes of graph A so that paths between closest neighbors of B will be as long as possible. B will be way smaller than A: something like 20 nodes in a 4 million nodes graph.

Related

Statement about mininum spanning tree

How do I disprove the following statement:
If an edge-weighted graph has a unique MST then its edge weights are distinct.
Is it too simple to make a graph that connects 3 vertices in a straight line and have the two edges to be distinct or is that not allowed because to have a unique tree you have to have more than one possible tree?
Also what does "its edge weights are distinct" actually mean, does it mean that no two edges are alike or that not all edges are alike?

Dividing a vector of points into two spaces

I have a memory-mapped file of many millions of 3D points as a STL vector using CGAL. Given an arbitrary plane that divides the dataset into approximately equal parts, I would like to sort the dataset such that all inside points are contiguous in the vector, and likewise the outside points. This process then needs to be repeated to an arbitrary depth, creating a non axis-aligned BSP tree.
Due to the size of the dataset I would like to do this in place if possible. I have a predicate functor that I use to create a filtered_iterator, but of course that doesn't sort the points, just skips non-matching ones. So I could create a second vector and copy the sorted points into that, and the re-use the original vector round-robin style, but I would like to avoid that if possible, if only to keep the iterators that mark the start and end of each space.
Of course, by invoking the question gods, I received direct communication from them almost as soon as I posted!
I had simply been blind to the STL algorithm partition which does exactly what I need.

Dealing with clusters when searching for points on map using mysql

I've found various questions with solutions similar to this problem but nothing quite on the money so far. Very grateful for any help.
I have a mysql (v.5.6.10) database with a single table called POSTS that stores millions upon millions of rows of lat/long points of interest on a map. Each point is classified as one of several different types. Each row is structured as id, type, coords:
id an unsigned bigint + primary key. This is auto incremented for each new row that is inserted.
type an unsigned tinyint used to encode the type of the point of interest.
coords a mysql geospatial POINT datatype representing the lat/long of the point of interest.
There is a SPATIAL index on 'coords'.
I need to find an efficient way to query the table and return up to X of the most recently-inserted points within a radius ("R") of a specific lat/long position ("Position"). The database is very dynamic so please assume that the data is radically different each time the table is queried.
If X is infinite, the problem is trivial. I just need to execute a query something like:
SELECT id, type, AsText(coords) FROM POSTS WHERE MBRContains(GeomFromText(BoundingBox, Position))
Where 'BoundingBox' is a mysql POLYGON datatype that perfectly encloses a circle of radius R from Position. Using a bounding box is, of course, not a perfect solution but this is not important for the particular problem that I'm trying to solve. I can order the results using "ORDER BY ID DESC" to retrieve and process the most-recently-inserted points first.
If X is less than infinite then I just need to modify the above to:
SELECT id, type, AsText(coords) FROM POSTS WHERE MBRContains(GeomFromText(BoundingBox, Position)) ORDER BY id DESC LIMIT X
The problem that I am trying to solve is how do I obtain a good representative set of results from a given region on the map when the points in that region are heavily clustered (for example, within cities on the map search region). For example:
In the example above, I am standing at X and searching for the 5 most-recently-inserted points of type black within the black-framed bounding box. If these points were all inserted in the cluster in the bottom right hand corner (let's assume that cluster is London) then my set of results will not include the black point that is near the top right of the search region. This is a problem for my application as I do not want users to be given the impression that there are no points of interest outside any areas where points are clustered.
I have considered a few potential solutions but I can't find one that works efficiently when the number of rows is huge (10s of millions). Approaches that I have tried so far include:
Dividing the search region into S number of squares (i.e., turning it into a grid) and searching for up to x/S points within each square - i.e., executing a separate mysql query for each square in the grid. This works OK for a small number of rows but becomes inefficient when the number of rows is massive as you need to divide the region into a large number of squares for the approach to work effectively. With only a small number of squares, you cannot guarantee that each square won't contain a densely populated cluster. A large number of squares means a large number of mysql searches which causes things to chug.
Adding a column to each row in the table that stores the distance to the nearest neighbour for each point. The nearest neighbour distance for a given point is calculated when the point is inserted into the table. With this structure, I can then order the search results by the nearest neighbour distance column so that any points that are in clusters are returned last. This solution only works when I'm searching for ALL points within the search region. For example, consider the situation in the diagram shown above. If I want to find the 5 most-recently-inserted points of type green, the nearest neighbour distance that is recorded for each point will not be correct. Recalculating these distances for each and every query is going to be far too expensive, even using efficient algorithms like KD trees.
In fact, I can't see any approach that requires pre-processing of data in table rows (or, put another way, 'touching' every point in the relevant search region dataset) to be viable when the number of rows gets large. I have considered algorithms like k-means / DBSCAN, etc. and I can't find anything that will work with sufficient efficiency given the use case explained above.
Any pearls? My intuition tells me this CAN be solved but I'm stumped so far.
Post-processing in that case seems more effective. Fetch last X points of a given type. Find if there is some clustering, for example: too many points too close together, relative to the distance of your point of view. Drop oldest of them (or these which are very close - may be your data is referencing a same POI). How much - up to you. Fetch next X points and see if there are some of them which are not in the cluster, or you can calculate a value for each of them based on remoteness and recentness and discard points according to that value.

Efficient represention for growing circles in 2D space?

Imagines there's a 2D space and in this space there are circles that grow at different constant rates. What's an efficient data structure for storing theses circles, such that I can query "Which circles intersect point p at time t?".
EDIT: I do realize that I could store the initial state of the circles in a spatial data structure and do a query where I intersect a circle at point p with a radius of fastest_growth * t, but this isn't efficient when there are a few circles that grow extremely quickly whereas most grow slowly.
Additional Edit: I could further augment the above approach by splitting up the circles and grouping them by there growth rate, then applying the above approach to each group, but this requires a bounded time to be efficient.
Represent the circles as cones in 3d, where the third dimension is time. Then use a BSP tree to partition them the best you can.
In general, I think the worst-case for testing for intersection is always O(n), where n is the number of circles. Most spacial data structures work by partitioning the space cleverly so that a fraction of the objects (hopefully close to half) are in each half. However, if the objects overlap then the partitioning cannot be perfect; there will always be cases where more than one object is in a partition. If you just think about the case of two circles overlapping, there is no way to draw a line such that one circle is entirely on one side and the other circle is entirely on the other side. Taken to the logical extreme, assuming arbitrary positioning of the circles and arbitrary radiuses, there is no way to partition them such that testing for intersection takes O(log(n)).
This doesn't mean that, in practice, you won't get a big advantage from using a tree, but the advantage you get will depend on the configuration of the circles and the distribution of the queries.
This is a simplified version of another problem I have posted about a week ago:
How to find first intersection of a ray with moving circles
I still haven't had the time to describe the solution that was expected there, but I will try to outline it here(for this simplar case).
The approach to solve this problem is to use a kinetic KD-tree. If you are not familiar with KD trees it is better to first read about them. You also need to add the time as additional coordinate(you make the space 3d instead of 2d). I have not implemented this idea yet, but I believe this is the correct approach.
I'm sorry this is not completely thought through, but it seems like you might look into multiplicatively-weighted Voronoi Diagrams (MWVDs). It seems like an adversary could force you into computing one with a series of well-placed queries, so I have a feeling they provide a lower-bound to your problem.
Suppose you compute the MWVD on your input data. Then for a query, you would be returned the circle that is "closest" to your query point. You can then determine whether this circle actually contains the query point at the query time. If it doesn't, then you are done: no circle contains your point. If it does, then you should compute the MWVD without that generator and run the same query. You might be able to compute the new MWVD from the old one: the cell containing the generator that was removed must be filled in, and it seems (though I have not proved it) that the only generators that can fill it in are its neighbors.
Some sort of spatial index, such as an quadtree or BSP, will give you O(log(n)) access time.
For example, each node in the quadtree could contain a linked list of pointers to all those circles which intersect it.
How many circles, by the way? For small n, you may as well just iterate over them. If you constantly have to update your spatial index and jump all over cache lines, it may end up being faster to brute-force it.
How are the centres of your circles distributed? If they cover the plane fairly evenly you can discretise space and time, then do the following as a preprocessing step:
for (t=0; t < max_t; t++)
foreach circle c, with centre and radius (x,y,r) at time t
for (int X = x-r; X < x+r; x++)
for (int Y = x-r; Y < y+r; y++)
circles_at[X][Y][T].push_back (&c)
(assuming you discretise space and time along integer boundaries, scale and offset however you like of course, and you can add circles later on or amortise the cost by deferring calculation for distant values of t)
Then your query for point (x,y) at time (t) could do a brute-force linear check over circles_at[x][y][ceil(t)]
The trade-off is obvious, increasing the resolution of any of the three dimensions will increase preprocessing time but give you a smaller bucket in circles_at[x][y][t] to test.
People are going to make a lot of recommendations about types of spatial indices to use, but I would like to offer a bit of orthogonal advice.
I think you are best off building a few indices based on time, i.e. t_0 < t_1 < t_2 ...
If a point intersects a circle at t_i, it will also intersect it at t_{i+1}. If you know the point in advance, you can eliminate all circles that intersect the point at t_i for all computation at t_{i+1} and later.
If you don't know the point in advance, then you can keep these time-point trees (built based on the how big each circle would be at a given time). At query time (e.g. t_query), find i such that t_{i-1} < t_query <= t_i. If you check all the possible circles at t_i, you will not have any false negatives.
This is sort of a hack for a data structure that is "time dynamics aware", but I don't know of any. If you have a threaded environment, then you only need to maintain one spacial index and be working on the next one in the background. It will cost you a lot of computation for the benefit of being able to respond to queries with low latency. This solution should be compared at the very least to the O(n) solution (go through each point and check if dist(point, circle.center) < circle.radius).
Instead of considering the circles, you can test on their bounding boxes to filter out the ones which do not contain the point. If your bounding box sides are all sorted, this is essentially four binary searches.
The tricky part is reconstructing the sorted sides for any given time, t. To do that, you can start off with the original points: two lists for the left and right sides with the x coordinate, and two lists for top and bottom with the y coordinates. For any time greater than 0, all the left side points will move to left, etc. You only need to check each location to the one next to it to obtain a points where the element and the one next to it are are swapped. This should give you a list of time points to modify your ordered lists. If you now sort these modification records by time, for any given starting time and an ending time you can extract all the modification records between the two, and apply them to your four lists in order. I haven't completely figured out the algorithm, but I think there will be edge cases where three or more successive elements can cross over exactly at the same time point, so you may need to modify the algorithm to handle those edge cases as well. Perhaps a list modification record that contains the position in list, and the number of records to reorder would suffice.
I think it's possible to create a binary tree that solves this problem.
Each branch should contain a growing circle, a static circle for partitioning and the latest time at which the partitioning circle cleanly partitions. Further more the growing circle that is contained within a node should always have a faster growing rate than either of it's child nodes' growing circles.
To do a query, take the root node. First check it's growing circle, if it contains the query point at the query time, add it to the answer set. Then, if the time that you're querying is greater than the time at which the partition line is broken, query both children, otherwise if the point falls within the partitioning circle, query the left node, else query the right node.
I haven't quite completed the details of performing insertions, (the difficult part is updating the partition circle so that the number of nodes on the inside and outside is approximately equal and the time when the partition is broken is maximized).
To combat the few circles that grow quickly case, you could sort the circles in descending order by rate of growth and check each of the k fastest growers. To find the proper k given t, I think you can perform a binary search to find the index k such that k*m = (t * growth rate of k)^2 where m is a constant factor you'll need to find by experimentation. The will balance the part the grows linearly with k with the part that falls quadratically with the growth rate.
If you, as already suggested, represent growing circles by vertical cones in 3d, then you can partition the space as regular (may be hexagonal) grid of packed vertical cylinders. For each cylinder calculate minimal and maximal heights (times) of intersections with all cones. If circle center (vertex of cone) is placed inside the cylinder, then minimal time is zero. Then sort cones by minimal intersection time. As result of such indexing, for each cylinder you’ll have the ordered sequence of records with 3 values: minimal time, maximal time and circle number.
When you checking some point in 3d space, you take the cylinder it belongs to and iterate its sequence until stored minimal time exceeds the time of the given point. All obtained cones, which maximal time is less than given time as well, are guaranteed to contain given point. Only cones, where given time lies between minimal and maximal intersection times, are needed to recalculate.
There is a classical tradeoff between indexing and runtime costs – the less is the cylinder diameter, the less is the range of intersection times, therefore fewer cones need recalculation at each point, but more cylinders have to be indexed. If circle centers are distributed non-evenly, then it may be worth to search better cylinder placement configuration then regular grid.
P.S. My first answer here - just registered to post it. Hope it isn’t late.

What is the time complexity of lookups in directed acyclic word graphs?

A directed acyclic word graph is a great data structure for certain tasks. I can't find any information on the time complexity of performing a lookup though.
I would guess it depends linearly on the average word length, and logarithmically on the number of words in the graph.
So is it O(L * log W), where W is the number of words and L is the average word length?
I think that complexity is just O(L). Number of operations is proportional to length of word and it does not matter how many entries structure have. (there might be differences based on implementation of node searching but that is in worst case and worst implementation just constant whit upper limit equal to size of alphabet)
I’d say it’s just O(L). For each lookup of a word of n characters, you always follow at most n edges, irrespective of how many other edges there are.
(That’s assuming a standard DAWG in which each node has outgoing edges for every letter of the alphabet, i.e. 26 for English. Even if you have fewer outgoing edges per node and therefore more levels, the number of edges to follow is still at most a constant multiple of n, so we still get O(L).)
How many words you already have in your structure seems to be irrelevant.
Even if, at each step, you perform a linear search for the correct edge to follow from the current node, this is still constant-time because the alphabet is bounded, and therefore so is the number of outgoing edges from each node.