If I have 3 different matrices, one for rotation (R), one for translation(T), and one for scaling (S), how to achieve the reverse effect of these matrices by manipulating the one caused it?
What I gathered so far is if I transposed the rotation matrix, I will achieve what I desire (is this correct?). And what about the other two?
And if there is a common way, are there any special cases where these ways will not suffice?
The inverse of the rotation matrix R is indeed its transpose RT.
The inverse of the scaling matrix S is easy as it contains only diagonal elements (the first three rows, as the last row is always equals to (0 0 0 1))
So you just replace each diagonal si with 1/si.
Finally the translation matrix T is an identity matrix with the translation vector on the last column. The inverse is achieved by replacing those elements with their negative.
Also the inverse of the product of 3 matrices, is the reverse product
(S T R)-1 = R-1 T-1 S-1
This is for a geophysical analysis program I am creating. I already have code to do all this, but I am looking for inspirations and ideas (good datastructures and algorithms).
What I want to model:
Velocity as a function of depth (z)
The model is built up from multiple layers (<10)
Every layer is accessible by an index going from 0 for the top most layer to n for the bottom most layer
Every layer has velocity as a linear function of depth (gradient a_k and axis intercept b_k of the kth layer)
Every layer has a top and bottom depth (z_k-1 and z_k)
The model is complete, there is no space between layers. The point directly between two layers belongs to the lower layer
Requirements:
Get velocity at an arbitrary depth within the model. This will be done on the order of 1k to 10k times, so it should be well optimized.
Access to the top and bottom depths, gradients and intercepts of a layer by the layer index
What I have so far:
I have working Python code where every layer is saved as a numpy array with the values of z_k (bottom depth), z_k-1 (top depth), a_k (velocity gradient) and b_k (axis intercept). To evaluate the model at a certain depth, I get the layer index (, use that to get the parameters of the layer and pass them to a function that evaluates the linear velocity gradient.
So you have piecewise linear dependence, where z-coordinates of pieces ends go irrregular, and want to get function value at given z.
Note that there is no sense to use binary search for 10 pieces (3-4 rounds of BS might be slower than 9 simple comparisons).
But what precision have your depth queries? Note that you can store a table both for 1-meter resolution and for 1 millimeter too - only 10^7 entries provide O(1) access to any precalculated velocity value
For limited number of pieces it is possible to make long formula (involving integer division) but results perhaps should be slower.
Example for arbitrary three-pieces polyline with border points 2 and 4.5:
f = f0 + 0.2*int(z/2.0)*(z-2.0) + 0.03*int(z/4.5)*(z-4.5)
I'm trying to come up with a way to arrive at a "score" based on an integer number of "points" that is adjustable using a small number (3-5?) of parameters. Preferably it would be simple enough to reasonably enter as a function/calculation in a spreadsheet for tuning the parameters by the "designer" (not a programmer or mathematician). The first point has the most value and eventually additional points have a fixed or nearly fixed value. The transition from the initial slope of point value to final slope would be smooth. See example shapes below.
Points values are always positive integers (0 pts = 0 score)
At some point, curve is linear (or nearly), all additional points have fixed value
Preferably, parameters are understandable to a lay person, e.g.: "smoothness of the curve", "value of first point", "place where the additional value of points is fixed", etc
For parameters, an example of something ideal would be:
Value of first point: 10
Value of point #: 3 is: 5
Minimum value of additional points: 0.75
Exact shape of curve not too important as long as the corner can be more smooth or more sharp.
This is not for a game but more of a rating system with multiple components (several of which might use this kind of scale) will be combined.
This seems like a non-traditional kind of question for SO/SE. I've done mostly financial software in my career, I'm hoping there some domain wisdom for this kind of thing I can tap into.
Implementation of Prune's Solution:
Google Sheet
Parameters:
Initial value (a)
Second value (b)
Minimum value (z)
Your decay ratio is b/a. It's simple from here: iterate through your values, applying the decay at each step, until you "peg" at the minimum:
x[n] = max( z, a * (b/a)^n )
// Take the larger of the computed "decayed" value,
// and the specified minimum.
The sequence x is your values list.
You can also truncate intermediate results if you want integers up to a certain point. Just apply the floor function to each computed value, but still allow z to override that if it gets too small.
Is that good enough? I know there's a discontinuity in the derivative function, which will be noticeable if the minimum and decay aren't pleasantly aligned. You can adjust this with a relative decay, translating the exponential decay curve from y = 0 to z.
base = z
diff = a-z
ratio = (b-z) / diff
x[n] = z + diff * ratio^n
In this case, you don't need the max function, since the decay has a natural asymptote of 0.
Grid Illumination: Given an NxN grid with an array of lamp coordinates. Each lamp provides illumination to every square on their x axis, every square on their y axis, and every square that lies in their diagonal (think of a Queen in chess). Given an array of query coordinates, determine whether that point is illuminated or not. The catch is when checking a query all lamps adjacent to, or on, that query get turned off. The ranges for the variables/arrays were about: 10^3 < N < 10^9, 10^3 < lamps < 10^9, 10^3 < queries < 10^9
It seems like I can get one but not both. I tried to get this down to logarithmic time but I can't seem to find a solution. I can reduce the space complexity but it's not that fast, exponential in fact. Where should I focus on instead, speed or space? Also, if you have any input as to how you would solve this problem please do comment.
Is it better for a car to go fast or go a long way on a little fuel? It depends on circumstances.
Here's a proposal.
First, note you can number all the diagonals that the inputs like on by using the first point as the "origin" for both nw-se and ne-sw. The diagonals through this point are both numbered zero. The nw-se diagonals increase per-pixel in e.g the northeast direction, and decreasing (negative) to the southwest. Similarly ne-sw are numbered increasing in the e.g. the northwest direction and decreasing (negative) to the southeast.
Given the origin, it's easy to write constant time functions that go from (x,y) coordinates to the respective diagonal numbers.
Now each set of lamp coordinates is naturally associated with 4 numbers: (x, y, nw-se diag #, sw-ne dag #). You don't need to store these explicitly. Rather you want 4 maps xMap, yMap, nwSeMap, and swNeMap such that, for example, xMap[x] produces the list of all lamp coordinates with x-coordinate x, nwSeMap[nwSeDiagonalNumber(x, y)] produces the list of all lamps on that diagonal and similarly for the other maps.
Given a query point, look up it's corresponding 4 lists. From these it's easy to deal with adjacent squares. If any list is longer than 3, removing adjacent squares can't make it empty, so the query point is lit. If it's only 3 or fewer, it's a constant time operation to see if they're adjacent.
This solution requires the input points to be represented in 4 lists. Since they need to be represented in one list, you can argue that this algorithm requires only a constant factor of space with respect to the input. (I.e. the same sort of cost as mergesort.)
Run time is expected constant per query point for 4 hash table lookups.
Without much trouble, this algorithm can be split so it can be map-reduced if the number of lampposts is huge.
But it may be sufficient and easiest to run it on one big machine. With a billion lamposts and careful data structure choices, it wouldn't be hard to implement with 24 bytes per lampost in an unboxed structures language like C. So a ~32Gb RAM machine ought to work just fine. Building the maps with multiple threads requires some synchronization, but that's done only once. The queries can be read-only: no synchronization required. A nice 10 core machine ought to do a billion queries in well less than a minute.
There is very easy Answer which works
Create Grid of NxN
Now for each Lamp increment the count of all the cells which suppose to be illuminated by the Lamp.
For each query check if cell on that query has value > 0;
For each adjacent cell find out all illuminated cells and reduce the count by 1
This worked fine but failed for size limit when trying for 10000 X 10000 grid
I have got a 3D grid (voxels), where some of the voxels are filled, and some are not. The 3D grid is sparsely filled, so I have got a set filledVoxels with coordinates (x, y, z) of the filled voxels. What I am trying to do is find out is for each filled voxel, how many neighboring voxels are filled too.
Here is an example:
filledVoxels contains the voxels (1, 1, 1), (1, 2, 1), and (1, 3, 1).
Therefore, the neighbor counts are:
(1,1,1) has 1 neighbor
(1,2,1) has 2 neighbors
(1,3,1) has 1 neighbor.
Right now I have this algorithm:
voxelCount = new Map<Voxel, Integer>();
for (voxel v in filledVoxels)
count = checkAllNeighbors(v, filledVoxels);
voxelCount[v] = count;
end
checkAllNeighbors() looks up all 26 surrounding voxels. So in total I am doing 26*filledVoxels.size() lookups, which is quite slow.
Is there any way to cut down the number of required lookups? When you look at the above example you can see that I am checking the same voxels several times, so it might be possible to get rid of lookups with some clever caching.
If this helps in any way, the voxels represent a voxelized 3D surface (but there might be holes in it). I usually want to get a list of all voxels that have 5 or 6 neighbors.
You can transform your voxel space into a octree in which every node contains a flag that specifies whether it contains filled voxels at all.
When a node does not contain filled voxels, you don't need to check any of its descendants.
I'd say if each of your lookups is slow (O(size)), you should optimize it by binary search in an ordered list (O(log(size))).
The constant 26, I wouldn't worry much. If you iterate smarter, you could cache something and have 26 -> 10 or something, I think, but unless you have profiled the whole application and found out decisively that it is the bottleneck I would concentrate on something else.
As ilya states, there's not much you can do to get around the 26 neighbor look-ups. You have to make your biggest gains in efficiently identifying whether a given neighbor is filled or not. Given that the brute force solution is essentially O(N^2), you have a lot of possible ground to gain in that area. Since you have to iterate over all filled voxels at least once, I would take an approach similar to the following:
voxelCount = new Map<Voxel, Integer>();
visitedVoxels = new EfficientSpatialDataType();
for (voxel v in filledVoxels)
for (voxel n in neighbors(v))
if (visitedVoxels.contains(n))
voxelCount[v]++;
voxelCount[n]++;
end
next
visitedVoxels.add(v);
next
For your efficient spatial data type, a kd-tree, as Zifre suggested, might be a good idea. In any case, you're going to want to reduce your search space by binning visited voxels.
If you're marching along the voxels one at a time, you can keep a lookup table corresponding to the grid, so that after you've checked it once using IsFullVoxel() you put the value in this grid. For each voxel you're marching in you can check if its lookup table value is valid, and only call IsFullVoxel() it it isn't.
OTOH it seems like you can't avoid iterating over all neighboring voxels, either using IsFullVoxel() or the LUT. If you had some more a priori information it could help. For instance, if you knew that there were at most x neighboring filled voxels, or you knew that there were at most y neighboring filled voxels in each direction. For instance, if you know you're looking for voxels with 5 to 6 neighbors, you can stop after you've found 7 full neighbors or 22 empty neighbors.
I'm assuming that a function IsFullVoxel() exists that returns true if a voxel is full.
If most of the moves in your iteration were to neighbors, you could reduce your checking by around 25% by not looking back at the ones you just checked before you made the step.
You may find a Z-order curve a useful concept here. It lets you (with certain provisos) keep a sliding window of data around the point you're currently querying, so that when you move to the next point, you don't have to throw away many of the queries you've already performed.
Um, your question is not very clear. I'm assuming you just have a list of the filled points. In that case, this is going to be very slow, because you have to iterate through it (or use some kind of tree structure such as a kd-tree, but this will still be O(log n)).
If you can (i.e. the grid is not too big), just make a 3d array of bools. 26 lookups in a 3d array shouldn't really take that long (and there really is no way to cut down on the number of lookups).
Actually, now that I think of it, you could make it a 3d array of longs (64 bits). Each 64 bit block would hold 64 (4 x 4 x 4) voxels. When you are checking the neighbors of a voxel in the middle of the block, you could do a single 64 bit read (which would be much faster).
Is there any way to cut down the number of required lookups?
You will, at minimum, have to perform at least 1 lookup per voxel. Since that's the minimum, then any algorithm which only performs one lookup per voxel will meet your requirement.
One simplistic idea is to initialize an array to hold the count for each voxel, then look at each voxel and increment the neighbors of that voxel in the array.
Pseudo C might look something like this:
#define MAXX 100
#define MAXY 100
#define MAXZ 100
int x, y, z
char countArray[MAXX][MAXY][MAXZ];
initializeCountArray(MAXX, MAXY, MAXZ); // Set all array elements to 0
for(x=0; x<MAXX; x++)
for(y=0;y<MAXY;y++)
for(z=0;z<MAXZ;z++)
if(VoxelExists(x,y,z))
incrementNeighbors(x,y,z);
You'll need to write initializeCountArray so it sets all array elements to 0.
More importantly you'll also need to write incrementNeighbors so that it won't increment outside the array. A slight speed increase here is to only perform the above algorithm on all voxels on the interior, then do a separate run on all the outside edge voxels with a modified incrementNeighbrs routine that understands there won't be neighbors on one side.
This algorithm results in 1 lookup per voxel, and at most 26 byte additions per voxel. If your voxel space is sparse then this will result in very few (relative) additions. If your voxel space is very dense, you might consider reversing the algorithm - initialize the array to the value of 26 for each entry, then decrement the neighbors when a voxel doesn't exist.
The results for a given voxel (ie, how many neighbors do I have?) reside in the array. If you need to know how many neighbors voxel 2,3,5 has, just look at the byte in countArray[2][3][5].
The array will consume 1 byte per voxel. You could use less space, and possibly increase the speed a little bit by packing the bytes.
There are better algorithms if you know details about your data. For instance, a very sparse voxel space will benefit greatly from an octree, where you can skip large blocks of lookups when you already know there are no filled voxels inside. Most of these algorithms, however, would still require at least one lookup per voxel to fill their matrix, but if you are performing several operations then they may benefit more than this one operation.