Does AStar grid algorithm only handle square grids? - c++

Looking for some clarification as I cannot seem to get an answer. When writing an astar algorirthm for grids, I was wondering if it was meant to work with any size rectangle or just perfectly square grids?
If there is a specific method for handling the heuristic for rectangles, what is it?
If people need to know I’m writing it in C++ for use in UE4.
Thanks everyone!

No, A* doesn't need a grid at all. You can use any placement of nodes, and as long as your heuristic is admissible, A* should work.
In fact, if you can guarantee your heuristic is admissible (i.e., it is guaranteed to never overestimate the distance), your nodes don't actually need a position at all. Of course, many actual applications do have nodes with particular locations, and Euclidean distance is a convenient admissible heuristic.
The only thing you need to do to make a rectangular grid work is to make sure your metrics are correct. That is: your node-to-node distances and your Euclidean distance calculation must correctly reflect the horizontal and vertical spacing of your grid.

Related

Given 2 points with known speed direction and location, compute a path composed of (circle) arcs

So, I have two points, say A and B, each one has a known (x, y) coordinate and a speed vector in the same coordinate system. I want to write a function to generate a set of arcs (radius and angle) that lead A to status B.
The angle difference is known, since I can get it by subtracting speed unit vector. Say I move a certain distance with (radius=r, angle=theta) then I got into the exact same situation. Does it have a unique solution? I only need one solution, or even an approximation.
Of course I can solve it by giving a certain circle and a line(radius=infine), but that's not what I want to do. I think there's a library that has a function for this, since it's quite a common approach.
A biarc is a smooth curve consisting of two circular arcs. Given two points with tangents, it is almost always possible to construct a biarc passing through them (with correct tangents).
This is a very basic routine in geometric modelling, and it is indispensable for smoothly approximating an arbirtrary curve (bezier, NURBS, etc) with arcs. Approximation with arcs and lines is heavily used in CAM, because modellers use NURBS without a problem, but machine controllers usually understand only lines and arcs. So I strongly suggest reading on this topic.
In particular, here is a great article on biarcs on biarcs, I seriously advice reading it. It even contains some working code, and an interactive demo.

All k nearest neighbors in 2D, C++

I need to find for each point of the data set all its nearest neighbors. The data set contains approx. 10 million 2D points. The data are close to the grid, but do not form a precise grid...
This option excludes (in my opinion) the use of KD Trees, where the basic assumption is no points have same x coordinate and y coordinate.
I need a fast algorithm O(n) or better (but not too difficult for implementation :-)) ) to solve this problem ... Due to the fact that boost is not standardized, I do not want to use it ...
Thanks for your answers or code samples...
I would do the following:
Create a larger grid on top of the points.
Go through the points linearly, and for each one of them, figure out which large "cell" it belongs to (and add the points to a list associated with that cell).
(This can be done in constant time for each point, just do an integer division of the coordinates of the points.)
Now go through the points linearly again. To find the 10 nearest neighbors you only need to look at the points in the adjacent, larger, cells.
Since your points are fairly evenly scattered, you can do this in time proportional to the number of points in each (large) cell.
Here is an (ugly) pic describing the situation:
The cells must be large enough for (the center) and the adjacent cells to contain the closest 10 points, but small enough to speed up the computation. You could see it as a "hash-function" where you'll find the closest points in the same bucket.
(Note that strictly speaking it's not O(n) but by tweaking the size of the larger cells, you should get close enough. :-)
I have used a library called ANN (Approximate Nearest Neighbour) with great success. It does use a Kd-tree approach, although there was more than one algorithm to try. I used it for point location on a triangulated surface. You might have some luck with it. It is minimal and was easy to include in my library just by dropping in its source.
Good luck with this interesting task!

How can I solve Codeforces Beta round#12 Problem D?

Click here to view the problem.
I can't come to a solution better than O(n^2) but with n<=500000 this won't work!
My Idea is to sort them by (beauty+intellect+richness) and test any of them with those after it.
Please help!
If you restrict the problem to two attributes (e.g. only B_i and R_i, just for illustration purposes), you can think of these attributes as points in a 2D plane. For each point (corresponding to a Lady) you'll have to count the number of points in the (semi-infinite) rectangle 'right and above' the given point.
I think a faster than O(n^2) solution would involve a range tree although I have not thought about the details. See also an illustration here.
EDIT: and you would store (or update while building) the number of points 'below' each node with the node so you can e.g. easily get the number of points below or above the splitting point of a given node.
I think you can solve this in O(n log n) by considering each lady as a point in 3-space and computing the convex hull of the points (see, e.g., here). Then, any point inside the hull is a potential suicide case.

Is there a data structure with these characteristics?

I'm looking for a data structure that would allow me to store an M-by-N 2D matrix of values contiguously in memory, such that the distance in memory between any two points approximates the Euclidean distance between those points in the matrix. That is, in a typical row-major representation as a one-dimensional array of M * N elements, the memory distance differs between adjacent cells in the same row (1) and adjacent cells in neighbouring rows (N).
I'd like a data structure that reduces or removes this difference. Really, the name of such a structure is sufficient—I can implement it myself. If answers happen to refer to libraries for this sort of thing, that's also acceptable, but they should be usable with C++.
I have an application that needs to perform fast image convolutions without hardware acceleration, and though I'm aware of the usual optimisation techniques for this sort of thing, I feel a specialised data structure or data ordering could improve performance.
Given the requirement that you want to store the values contiguously in memory, I'd strongly suggest you research space-filling curves, especially Hilbert curves.
To give a bit of context, such curves are sometimes used in database indexes to improve the locality of multidimensional range queries (e.g., "find all items with x/y coordinates in this rectangle"), thereby aiming to reduce the number of distinct pages accessed. A bit similar to the R-trees that have been suggested here already.
Either way, it looks that you're bound to an M*N array of values in memory, so the whole question is about how to arrange the values in that array, I figure. (Unless I misunderstood the question.)
So in fact, such orderings would probably still only change the characteristics of distance distribution.. average distance for any two randomly chosen points from the matrix should not change, so I have to agree with Oli there. Potential benefit depends largely on your specific use case, I suppose.
I would guess "no"! And if the answer happens to be "yes", then it's almost certainly so irregular that it'll be way slower for a convolution-type operation.
EDIT
To qualify my guess, take an example. Let's say we store a[0][0] first. We want a[k][0] and a[0][k] to be similar distances, and proportional to k, so we might choose to interleave the storage of first row and first column (i.e. a[0][0], a[1][0], a[0][1], a[2][0], a[0][2], etc.) But how do we now do the same for e.g. a[1][0]? All the locations near it in memory are now taken up by stuff that's near a[0][0].
Whilst there are other possibilities than my example, I'd wager that you always end up with this kind of problem.
EDIT
If your data is sparse, then there may be scope to do something clever (re Cubbi's suggestion of R-trees). However, it'll still require irregular access and pointer chasing, so will be significantly slower than straightforward convolution for any given number of points.
You might look at space-filling curves, in particular the Z-order curve, which (mostly) preserves spatial locality. It might be computationally expensive to look up indices, however.
If you are using this to try and improve cache performance, you might try a technique called "bricking", which is a little bit like one or two levels of the space filling curve. Essentially, you subdivide your matrix into nxn tiles, (where nxn fits neatly in your L1 cache). You can also store another level of tiles to fit into a higher level cache. The advantage this has over a space-filling curve is that indices can be fairly quick to compute. One reference is included in the paper here: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.30.8959
This sounds like something that could be helped by an R-tree. or one of its variants. There is nothing like that in the C++ Standard Library, but looks like there is an R-tree in the boost candidate library Boost.Geometry (not a part of boost yet). I'd take a look at that before writing my own.
It is not possible to "linearize" a 2D structure into an 1D structure and keep the relation of proximity unchanged in both directions. This is one of the fundamental topological properties of the world.
Having that that, it is true that the standard row-wise or column-wise storage order normally used for 2D array representation is not the best one when you need to preserve the proximity (as much as possible). You can get better result by using various discrete approximations of fractal curves (space-filling curves).
Z-order curve is a popular one for this application: http://en.wikipedia.org/wiki/Z-order_(curve)
Keep in mind though that regardless of which approach you use, there will always be elements that violate your distance requirement.
You could think of your 2D matrix as a big spiral, starting at the center and progressing to the outside. Unwind the spiral, and store the data in that order, and distance between addresses at least vaguely approximates Euclidean distance between the points they represent. While it won't be very exact, I'm pretty sure you can't do a whole lot better either. At the same time, I think even at very best, it's going to be of minimal help to your convolution code.
The answer is no. Think about it - memory is 1D. Your matrix is 2D. You want to squash that extra dimension in - with no loss? It's not going to happen.
What's more important is that once you get a certain distance away, it takes the same time to load into cache. If you have a cache miss, it doesn't matter if it's 100 away or 100000. Fundamentally, you cannot get more contiguous/better performance than a simple array, unless you want to get an LRU for your array.
I think you're forgetting that distance in computer memory is not accessed by a computer cpu operating on foot :) so the distance is pretty much irrelevant.
It's random access memory, so really you have to figure out what operations you need to do, and optimize the accesses for that.
You need to reconvert the addresses from memory space to the original array space to accomplish this. Also, you've stressed distance only, which may still cause you some problems (no direction)
If I have an array of R x C, and two cells at locations [r,c] and [c,r], the distance from some arbitrary point, say [0,0] is identical. And there's no way you're going to make one memory address hold two things, unless you've got one of those fancy new qubit machines.
However, you can take into account that in a row major array of R x C that each row is C * sizeof(yourdata) bytes long. Conversely, you can say that the original coordinates of any memory address within the bounds of the array are
r = (address / C)
c = (address % C)
so
r1 = (address1 / C)
r2 = (address2 / C)
c1 = (address1 % C)
c2 = (address2 % C)
dx = r1 - r2
dy = c1 - c2
dist = sqrt(dx^2 + dy^2)
(this is assuming you're using zero based arrays)
(crush all this together to make it run more optimally)
For a lot more ideas here, go look for any 2D image manipulation code that uses a calculated value called 'stride', which is basically an indicator that they're jumping back and forth between memory addresses and array addresses
This is not exactly related to closeness but might help. It certainly helps for minimation of disk accesses.
one way to get better "closness" is to tile the image. If your convolution kernel is less than the size of a tile you typical touch at most 4 tiles at worst. You can recursively tile in bigger sections so that localization improves. A Stokes-like (At least I thinks its Stokes) argument (or some calculus of variations ) can show that for rectangles the best (meaning for examination of arbitrary sub rectangles) shape is a smaller rectangle of the same aspect ratio.
Quick intuition - think about a square - if you tile the larger square with smaller squares the fact that a square encloses maximal area for a given perimeter means that square tiles have minimal boarder length. when you transform the large square I think you can show you should the transform the tile the same way. (might also be able to do a simple multivariate differentiation)
The classic example is zooming in on spy satellite data images and convolving it for enhancement. The extra computation to tile is really worth it if you keep the data around and you go back to it.
Its also really worth it for the different compression schemes such as cosine transforms. (That's why when you download an image it frequently comes up as it does in smaller and smaller squares until the final resolution is reached.
There are a lot of books on this area and they are helpful.

How do we find a biggest white rectangle in a n x n bitmap ?

Any idea on how to solve such problems (in C++)-
like which is the best Algorithm to use.
Say you have a n x n rectangular area black and white (O and 1) pixels and you're looking for the biggest white rectangle in this area.
I would write something simple like below:
first pass: create a set of 1 line segments for each pixel row.
second pass aggregate rectangles:
for each segment iterate on rows to find the largest rectangle containing it.
if you use another segment in the process mark it as used, not need to try it again
at any point keep only the largest rectangle found
That's only a first draft of a possible solution. It should be rewritten using a more formal algorithmic syntax and many details should be provided. Each step hides pitfalls to avoid if you want to be efficient. But it should not be too hard to code.
If I did not missed something, what I described above should basically be O(n4) in the worst case, with the first pass O(n2) used to find horizontal segments (could be quite fast with a very small loop) and the second pass probably much less thant O(n4) in practice (depends on segment size, really is nb_total_segment x nb_segment_per_line x nb_overlapping_segment).
That looks not bad to me. Can't see any obvious way to do it with better O complexity, (but of course there may be some way, O(n4) is not that good).
If you provide some details on input structure and expected result it may even be some fun to code.
What you ask for is known as blob filtering on the computer vision world.
http://en.wikipedia.org/wiki/Blob_extraction
http://en.wikipedia.org/wiki/Connected_component_labeling
http://www.aforgenet.com/framework/features/blobs_processing.html