  C RUBY-ON-RAILS MYSQL ASP.NET DEVELOPMENT RUBY .NET LINUX SQL-SERVER REGEX WINDOWS ALGORITHM ECLIPSE VISUAL-STUDIO STRING SVN PERFORMANCE APACHE-FLEX UNIT-TESTING SECURITY LINQ UNIX MATH EMAIL OOP LANGUAGE-AGNOSTIC VB6 MSBUILD # How to find the inverse of a Rectangular Matrix in C using GSL ## How to find the inverse of a Rectangular Matrix in C using GSL Tag : c , By : jaredsmiller Date : November 29 2020, 01:01 AM Boards Message : You Must Login Or Sign Up to Add Your Comments .

## How to find same-value rectangular areas of a given size in a matrix most efficiently?

Tag : algorithm , By : bashmish
Date : March 29 2020, 07:55 AM
hop of those help? You can do it in O(n^2) relatively easy.
First, some-precalculations. For each cell in matrix, calculate how many consecutive cells below it have the same number.
``````0 0 0 0 0 2 2
1 1 2 2 2 1 1
0 0 1 1 1 0 0
1 1 0 0 0 1 1
0 0 0 0 0 0 0
``````
``````0 0 0 0 0 0 0
-------------
4 4 2 2 2 0 0
4 4 2 2 2 0 0
0 0 2 2 2 1 1
-------------
0 0 0 0 0 1 1
``````
``````int current_width = 0;
for (int j = 0; j < matrix.width; ++j) {
if (a[i][j] < height - 1) {
// this column has different numbers in it, no game
current_width = 0;
continue;
}

if (current_width > 0) {
// this column should consist of the same numbers as the one before
if (matrix[i][j] != matrix[i][j - 1]) {
current_width = 1; // start streak anew, from the current column
continue;
}
}

++current_width;
if (current_width >= width) {
// we've found a rectangle!
}
}
``````

## find all rectangular areas with certain properties in a matrix

Tag : algorithm , By : Josh
Date : March 29 2020, 07:55 AM
may help you . I finally found a solution that works almost in linear time (there is a small factor depending on the number of found areas). I think this is the fastest possible solution.
Inspired by this answer: https://stackoverflow.com/a/7353193/145999 (pictures also taken from there)
`````` foundAreas = new list()

For each row y backwards:
potentialAreas = new List()
for each column x:
if M2[x,y]>M2[x-1,y]:
new Area(left=x,height=M2[x,y],hasOne=M1[x,y],hasTop=false)
if M2[x,y]<M2[x-1,y]:
for each area in potentialAreas:
if area.hasTop and area.hasOne<area.height:
new Box(Area.left,y-area.height,x,y)
if M2[x,y]==0: delete all potentialAreas
else:
find the area in potentialAreas with height=M2[x,y]
or the one with the closest bigger height: set its height to M2[x,y]
delete all potentialAreas with a bigger height

for each area in potentialAreas:
if area.hasOne>M1[x,y]: area.hasOne=M1[x,y]
if M2[x,y+1]==0: area.hasTop=true
``````

## How to find whether a 2D array (rectangular matrix) have a straight line

Tag : java , By : JackIT
Date : March 29 2020, 07:55 AM
seems to work fine Well, let's start from two degenerated cases:
if you have 0 points, answer as you wish (either there's a line, or there's no such line) if you have 1 or 2 points, the answer is always yes
``````      (y1 - y2) * (xi - x2) == (yi - y2) * (x1 - x2)
(y2*x1 - y1*x2) * (xi - x2) == (y2*xi - yi*x2) * (x1 - x2)
``````
``````   (2, 4)
(4, 3)
(6, 2)
(8, 1)
``````
``````private static bool SameLine(IEnumerable<Tuple<int, int>> points) {
if (null == points)
return true;

Tuple<int, int>[] data = points.ToArray();

// i = 2 - first two points are always on the same line
for (int i = 2; i < data.Length; ++i) {
int x1 = data.Item1;
int y1 = data.Item2;

int x2 = data.Item1;
int y2 = data.Item2;

int xi = data[i].Item1;
int yi = data[i].Item2;

// y = k * x + b where k = infinity
if (x1 == x2) {
if (xi != x1)
return false;

continue;
}

// Same k in y = k * x + b
if (!((y1 - y2) * (xi - x2) == (yi - y2) * (x1 - x2)))
return false;

// Same b in y = k * x + b
if (!((y2 * x1 - y1 * x2) * (xi - x2) == (y2 * xi - yi * x2) * (x1 - x2)))
return false;
}

return true;
}
``````
``````  Tuple<int, int>[] test = new Tuple<int, int>[] {
new Tuple<int, int>(2, 4),
new Tuple<int, int>(4, 3),
new Tuple<int, int>(6, 2),
new Tuple<int, int>(8, 1),
};

// Same line
Console.Write(SameLine(test) ? "Same line" : "different lines");
``````

## How can I find a basis for the column space of a rectangular matrix?

Tag : development , By : Francesco
Date : March 29 2020, 07:55 AM
Does that help One way is to use the LU decomposition. The factor U will be of the same size as your matrix, but will be upper-triangular. In each row of U, pick the first nonzero element: these are pivot elements, which belong to linearly independent columns. A self-contained example:
``````import numpy as np
from scipy.linalg import lu
A = np.array([[1, 2, 3], [2, 4, 2]])     # example for testing
U = lu(A)
lin_indep_columns = [np.flatnonzero(U[i, :]) for i in range(U.shape)]
``````

## Find inverse of matrix using b^nth power

Tag : c , By : hammer_1968
Date : March 29 2020, 07:55 AM 