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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

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How to find the inverse of a Rectangular Matrix in C using GSL
Tag : c , By : jaredsmiller
Date : November 29 2020, 01:01 AM


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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]:
            add to potentialAreas: 
                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:
                        add to foundAreas:
                             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[0].Item1;
    int y1 = data[0].Item2;

    int x2 = data[1].Item1;
    int y2 = data[1].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)[2]
lin_indep_columns = [np.flatnonzero(U[i, :])[0] for i in range(U.shape[0])]

Find inverse of matrix using b^nth power


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