You can use a The Depth-First Search algorithm.
DFS, BFS
To Find the number in that matrix and check that number adjacency elements... import java.util.Scanner; public class FindAdjacencyMatrix { public static int[][] array1 = new int[30][30]; public static int i,j,num,m,n; public static void main(String args[]) { Scanner input = new Scanner(System.in); //------------------------------------------------------------------------------------------------- System.out.println("Enter the m ,n matrix"); m = input.nextInt(); n = input.nextInt(); //------------------------------------------------------------------------------------------------- System.out.println("Enter the matrix Element one by one:"); for(i = 0; i < m; i++) { for(j = 0; j < n; j++) { array1[i][j] = input.nextInt(); } } System.out.println("The Given Matrix is :"); for(i = 0; i < m; i++) { for(j = 0; j < n; j++) { System.out.print(" "+array1[i][j]); } System.out.print("\n"); } //------------------------------------------------------------------------------------------------- System.out.println("Find The Adjacency Elements for Given Number : "); System.out.println("Enter The Number : "); num = input.nextInt(); for(i = 0; i < m; i++) { for(j = 0; j < n; j++) { if(num == array1[i][j]) { System.out.println("Element is Found :"+num); findAdjacency(num,i,j); break; } } } //-------------------------------------------------------------------------------------- } private static void findAdjacency(int elem,int row,int col) { try { if( array1[row][col-1]!=-1) { System.out.println("Left Adjacency : "+array1[row][col-1]); } } catch(Exception e){ System.out.println(" Exception Throwing "); } try{ if(array1[row][col+1]!= -1) { System.out.println("Right Adjacency : "+array1[row][col+1]); } }catch(Exception e){ System.out.println(" Exception Throwing "); } try { if(array1[row-1][col]!= -1) { System.out.println("Top Adjacency : "+array1[row-1][col]); } } catch(Exception e){ System.out.println(" Exception Throwing "); } try { if(array1[row+1][col]!= -1) { System.out.println("Botto Adjacency : "+array1[row+1][col]); } } catch(Exception e){ System.out.println(" Exception Throwing "); } } //---------------------------------------------------------------------------------------------- }
To find the original matrix of an inverted matrix, simply invert it again. Consider A^-1^-1 = A^1 = A
You find the equation of a graph by finding an equation with a graph.
The Matrix Pill is a program to find the person still "jacked in" to the Matrix world
where can i find taco bell competitive profile matrix
Cayleys formula states that for a complete graph on nvertices, the number of spanning trees is n^(n-2). For a complete bipartite graph we can use the formula p^q-1 q^p-1. for the number of spanning trees. A generalization of this for any graph is Kirchhoff's theorem or Kirchhoff's matrix tree theorem. This theorem looks at the Laplacian matrix of a graph. ( you may need to look up what that is with some examples). For graphs with a small number of edges and vertices, you can find all the spanning trees and this is often quicker. There are also algorithms such as depth-first and breadth-first for finding spanning trees.
Invert rows and columns to get the transpose of a matrix
A non-square matrix cannot be inverted.
Each number in a matrix is called an element.
You can factorize the matrix using LU or LDLT factorization algorithm. inverse of a diagonal matrix (D) is really simple. To find the inverse of L, which is a lower triangular matrix, you can find the answer in this link.www.mcs.csueastbay.edu/~malek/TeX/Triangle.pdfSince (A T )-1 = (A-1 )T for all matrix, you'll just have to find inverse of L and D.