i'm searching for a algorithm that take a matrix (in fact, a double entry array) and return an array of matrix that:
is square (WIDTH = HEIGHT)
all of the element in the matrix has the same value.
I don't know if that is clear, so imagine that you have a image made of pixels that is red, blue or green and i want to get an array that contained the least possible squares. Like the pictures shows
EDIT:
Ok, maybe it's not clear: I've a grid of element that can have some values like that:
0011121
0111122
2211122
0010221
0012221
That was my input, and i want in output somethings like that:
|0|0|111|2|1|
|0|1|111|22|
|2|2|111|22|
|00|1|0|22|1|
|00|1|2|22|1|
When each |X| is an array that is a piece of the input array.
My goal is to minimize the number of output array
This problem does not seem to have an efficient solution.
Consider a subset of instances of your problem defined as follows:
There are only 2 values of matrix elements, say 0 and 1.
Consider only matrix elements with value 0.
Identify each matrix element m_ij with a unit square in a rectangular 2D grid whose lower left corner has the coordinates (i, n-j).
The set of unit squares SU chosen this way must be 'connected' and must not have 'holes'; formally, for each pair of units squares (m_ij, m_kl) \in SU^2: (i, j) != (k, l) there is a sequence <m_ij = m_i(0)j(0), m_i(1)j(1), ..., m_i(q)j(q) = m_kl> of q+1 unit squares such that (|i(r)-i(r+1)| = 1 _and_ j(r)=j(r+1)) _or_ (i(r)=i(r+1) _and_ |j(r)-j(r+1)| = 1 ); r=0...q (unit squares adjacent in the sequence share one side), and the set SUALL of all unit squares with lower left corner coordinates from the integers minus SU is also 'connected'.
Slicing matrices that admit for this construction into a minimal number of square submatrices is equivalent to tiling the smallest orthogonal polygon enclosing SU ( which is the union of all elements of SU ) into the minimum number of squares.
This SE.CS post gives the references (and one proof) that show that this problem is NP-complete for integer side lengths of the squares of the tiling set.
Note that according to the same post, a tiling into rectangles runs in polynomial time.
Some hints may be useful.
For representation of reduced matrix, maybe a vector is better because it's needed to be stored (start_x,start_y,value ... not sure if another matrix very useful).
Step 1: loop on x for n occurrences (start with y=0)
Step 2: loop on y for/untill n occurrences. Most of cases here will be m lees then n.
(case m greater then n excluded since cannot do a square) Fine, just keep the min value[m]
Step 3: mark on vector (start_x,start_y, value)
Repeat Step 1-3 from x=m until end x
Step 4: End x, adjust y starting from most left_x found(m-in vector, reiterate vector).
...
keep going till end matrix.
Need to be very careful of how boundary are made(squares) in order to include in result full cover of initial matrix.
Reformulate full-initial matrix can be recomposed exactly from result vector.
(need to find gaps and place it on vector derived from step_4)
Note ! This is not a full solution, maybe it's how to start and figure out on each steps what is to be adjusted.
Related
I have to write a Java method that calculates this math function:
The function itself is not a problem but I need some advice.
C(u, v) is a two-dimensional array where I will save the result.
f(x, y) is a two-dimensional array from where I get the values I need to calculate C(u, v)
Obviously, I have to assume that the lengths of u and v are different (same for x and y). So, although the two summations are x = 0 to N-1 and y = 0 to N-1 (which would suggest that the matrix is the same size), I think it would be better to treat it as if it had different sizes.
So :
x will be the lines (and I will find them with matrix.length)
y will be the columns (and I will find them with matrix [i].lenth)
(the two points above will also be reflected on the dimensions of the matrix C)
2N will be the total number of elements in the matrix
Did I think it right?
you seem to be right, except for the 2N part,
2N is not the total number of elements, if the matrix has is N by M size then the total number of elements is N*M.
since it says on the equation (2N)then it probably means (N+M) or it could be (2N) or (2M), you need to be sure.
I was presented with the following Backtracking/Seating problem:
Assume you have a room of size k * k, meaning you have exactly k * k seats available and an n amount of people you have to seat. Furthermore, every person has their own specific radius in which they really don't want anyone else to sit in.
The assignment is to write an algorithm in pseudocode or pure text form to find any seating plan (not necessarily perfectly optimal) through backtracking in which every person is not within someone else's radius or indicate that such a seating plan is not possible. Note that the time complexity of the algorithm is not of importance.
To further illustrate this, assume we are given k = 8, meaing we have a total of 8x8 seats. For people n=3 we have Red with a radius of 3, Blue with a radius of 2 and Yellow with a radius of 1.
In the following picture we have an example of an invalid seating plan, because Blue is clearly within Red's radius. One of the many possible valid seating plans for this would be Red at seat (0,0), Blue at seat (4,0) and Yellow at seat (7,0).
Where I'm stuck:
My attempt/pseudocode assumes that an array of n people is an array of radii, sorted by their size in descending order.
We then iterate over the x and y coordinates in respective for loops. We place the largest Element at (0,0) and then pick the next element/radius and check if we can place it in the next (x,y). If person 1 is within person 2's radius or vice-versa, the position is invalid and we move on to the next (x,y). If the position is valid, we move on to the next element/radius and repeat that process until we exhausted all radii/people in n.
Even though this might even work in actual code, I fail to understand how I should implement the required backtracking aspect of my pseudocode. I do believe I understand the basics of how backtracking works but I just can't seem to figure out how I can implement that process into this problem.
I have a 2D array of size n representing n number of points in the 3D space, position[][] for XYZ (e.g. position[0][0] is X, position[0][1] is Y, and position[0][2] is Z coordinate of point 0.
What I need to do is to do clustering on the points, so to have n/k number of clusters of size k so that each cluster consists of the k closest points in the 3D space. For instance, if n=100 and k=5, I want to have 20 clusters of 5 points which are the closest neighbors in space.
How can I achieve that? (I need pseudo-code. For snippets preferably in Java)
What I was doing so far was a simple sorting based on each component. But this does NOT give me necessarily the closest neighbors.
Sort based on X (position[0][0])
Then sort based on Y (position[0][1])
Then sort based on Z (position[0][2])
for (int i=0; i<position.length; i++){
for (int j=i+1; j<position.length; j++){
if(position[i][0] > position[i+1][0]){
swap (position[i+1][0], position[i][0]);
}
}
}
// and do this for position[i][1] (i.e. Y) and then position[i+2][2] (i.e. Z)
I believe my question slightly differs from the Nearest neighbor search with kd-trees because neighbors in each iteration should not overlap with others. I guess we might need to use it as a component, but how, that's the question.
At start you do not have a octree but list of points instead like:
float position[n][3];
So to ease up the clustering and octree creation you can use 3D point density map. It is similar to creating histogram:
compute bounding box of your points O(n)
so process all points and determine min and max coordinates.
create density map O(max(m^3,n))
So divide used space (bbox) into some 3D voxel grid (use resolution you want/need) do a density map like:
int map[m][m][m]`
And clear it with zero.
for (int x=0;x<m;x++)
for (int y=0;y<m;y++)
for (int z=0;z<m;z++)
map[x][y][z]=0;
Then process all points determine its cell position from x,y,z and increment it.
for (int i=0;i<n;i++)
{
int x=(m-1)*(position[i][0]-xmin)/(xmax-xmin);
int y=(m-1)*(position[i][1]-ymin)/(ymax-ymin);
int z=(m-1)*(position[i][2]-zmin)/(zmax-zmin);
map[x][y][z]++;
// here you can add point i into octree belonging to leaf representing this cell
}
That will give you low res density map. The higher number in cell map[x][y][z] the more points are in it which means a cluster is there and you can also move point to that cluster in your octree.
This can be recursively repeated for cells that have enough points. To make your octree create density map 2x2x2 and recursively split each cell until its count is lesser then threshold or cell size is too small.
For more info see similar QAs
Finding holes in 2d point sets? for the density map
Effective gif/image color quantization? for the clustering
What you what is not Clustering. From what you said, I think you want to divided your N points into N/k groups, with each group have k points, while keeping the points in each cluster are closest in the 3D space.
Think an easy example, if you want to do the same thing one one dimension, that is, just sort the numbers, and the first k points into cluster 1, the second k points into cluster 2, and so on.
Then return the 3D space problem, the answer is the same. Just first find the point with minimum x-axis, y-axis and z-axis, altogether with its closest k-1 points into Cluster 1. Then for the lest points, find the minimum x-axis, y-axis and z-axis points, and k-1 closest points not clustered into Cluster 2, and so on.
Above process will get your results, but that maybe not meaningful in practice, maybe cluster algorithms such as k-means could help you.
I work on a project, for this project I need to generate a square random invertible matrix.
I found out how to generate a square random matrix, still I want to be sure that this is an invertible one, without having to compute the determinant or to generate this matrix multiple times, can you please give me a tip?
One way is to generate the SVD of the matrix. That is you generate 'random' (square) orthogonal matrices U and V, and a 'random' diagonal matrix S, and then compute
M = U*S*V'
Note that every matrix has an SVD
As long as none of the diagonal elements of S are 0, M will be invertible. Many routines that deal with invertible matrices are sensitive to the condition number of the matrix; errors tend to increase as the condition number gets larger. The condition number of M is the same as the condition numner of S which is the largest (by absolute value) diagonal element of S divided by the smallest (by absolute value). You may want to control this. One way is to generate the elements of S to be uniform in say [lo,hi] and then randomly set the sign.
One way to generate 'random' orthogonal matrices is to generate then as a product of 'random' Householder reflections, that is matrices of the form
R_v = 1 - 2*v*v'/(v'*v)
where v is a 'random' vector.
Every n by n orthogonal matrix can be written as a product of n Householder reflections.
All this is not as computationally severe as it at first might look. Due to the special form of the reflectors it is straightforward to write routines that compute
R_u*M and M*R_v'
in M using only n extra storage and being O( n * n)
So one scheme would be
Generate S
Repeat n times
Generate random non zero vector u
Update S to be R_u*S
Generate random non zero vector v
Update S to be S*R_v'
An LU decomposition might work.
Generate two matrices, L, which is lower triangular with all entries above the main diagonal zero and, U, an upper triangular matrix with entries below the main diagonal zero. Then form a matrix A=LU.
The determinant of either L or U is just the product of entries down the main diagonal so you just need to ensure none of these are zero. The determinant of A is the product of the two determinants.
I need to find "holes" in a 2D grid in java - can you point me toward the best sort of algorithm to do this?
With the input of the following points:
5,3
5,4
8,4
5,5
6,3
7,3
7,4
6,5
I need to figure out the positions of the "hole" or surrounded space in this grid. I'm a bit lost as to how to do this.
Plot of the points:
Assuming each point is 1x1
This is basically a blob extraction algorithm + a bit extra. Do this:
1) Find the westmost, eastmost, northmost and southmost any solid is placed. Remember them as xmin xmax ymin ymax.
2) Allocate an 2d array of integers (initialized to 0) with those dimensions, and place all solid points in it as the value -1.
3) Make a counter initialized to 1. Scan the 2d array. Every time you find a point that is 0, set it to counter and floodfill counters onto every adjacent point that is not a -1 until you've run out of points to floodfill onto. (To do a floodfill, one way is to keep a set of all points you haven't floodfilled all the neighbours of yet, and iterate over these, adding new points to the set until the set is exhausted -> nothing left to floodfill onto.) Now increment the counter and continue.
4) When you've scanned the whole grid, scan the perimeter. Every time you see a non -1 on the perimeter, mark that blob as not being surrounded (by having an array of bools as long as the number of blobs you found).
5) Every numbered blob you have not marked is surrounded.
Read about blob extraction here: http://en.wikipedia.org/wiki/Blob_extraction