Implement this pseudo code with Java.
n is the size of the matrix for an n x n matrix.
k is the pivot.
i is the equation being eliminated.
j is the term you're working with.
Gaussian Elimination with Scaled Partial Pivoting
// Forward Elimination
function
SPPFwdElimination(coeff :
array(n,n), const :
vector(n), ind :
vector(n))
scaling := new
vector(n) // vector of scaling
factors
// Initialize index and scaling vectors
for i = 1 to
n
smax := 0
for j = 1 to
n
smax :=
max(smax,|coeff[i][j]|)
// find coefficient with greatest absolute value
end for
scaling[i] := smax
end for
for k = 1 to
n - 1
rmax := 0
maxInd := k
for i = k
to n
r :=
|coeff[ind[i]][k] /
scaling[ind[i]]| // ratio of
coefficient to scaling factor
if (r > rmax)
then
rmax := r
maxInd := i
end if
end for
swap(ind[maxInd],
ind[k])
for i = k + 1
to n
mult :=
coeff[ind[i]][k] /
coeff[ind[k]][k]
for j = k + 1
to n
coeff[ind[i]][j] :=
coeff[ind[i]][j] -
mult *
coeff[ind[k]][j]
end for
const[ind[i]] :=
const[ind[i]] - mult *
const[ind[k]]
end for
end for
end function
// Back Substitution
function SPPBackSubst(coeff :
array(n,n), const :
vector(n), sol :
vector(n), ind :
vector(n))
sol[n] :=
const[ind[n]] /
coeff[ind[n]][n]
for i = n - 1
to 1
sum :=
const[ind[i]]
for j = i + 1
to n
sum := sum -
coeff[ind[i]][j] *
sol[j]
end for
sol[i] := sum /
coeff[ind[i]][i]
end for
end function
// SPP Gaussian Algorithm
function SPPGaussian(coeff :
array(n,n), const :
vector(n))
sol := new
array(n,n)
ind := new
vector(n)
for i = 1 to
n
ind[i] := i
end for
call
SPPFwdElimination(coeff,const,ind)
call
SPPBackSubst(coeff,const,sol,ind)
end function
package chirag;
import java.util.*;
import java.lang.Math;
public class dog {
public static void SPPFwdElimination(int [][]coeff,int
b[],int [] ind,int n) {
int scaling[] =new int[n+1];
//Initialize index and scaling vectors
for(int i=1;i<=n;i++) {
int smax = 0;
for (int j =1;j<=n;j++) {
smax = Math.max(smax,Math.abs(coeff[i][j])); //
find coefficient with greatest absolute value
}
scaling[i]=smax;
}
for( int k=1;k<=n - 1;k++) {
int rmax=0;
int maxInd=k;
for (int i =k;i<=n;i++) {
int r = Math.abs(coeff[ind[i]][k] /
scaling[ind[i]]); // ratio of coefficient to scaling factor
if (r > rmax) {
rmax = r;
maxInd= i;
}
}
int temp=ind[maxInd];
ind[maxInd]=ind[k];
ind[k]=temp;
for(int i = k + 1;i<=n;i++) {
int mult = coeff[ind[i]][k] /
coeff[ind[k]][k];
for(int j = k + 1;j<=n;j++) {
coeff[ind[i]][j] = coeff[ind[i]][j] - mult *
coeff[ind[k]][j];
}
b[ind[i]] = b[ind[i]] - (mult*b[ind[k]]);
}
}
}
public static void SPPBackSubst(int [][]coeff,int
b[],int []sol,int ind[],int n) {
sol[n] = b[ind[n]] /
coeff[ind[n]][n];
for(int i = n - 1;i>=1;i--) {
int sum = b[ind[i]];
for(int j = i + 1;j<=n;j++) {
sum= sum - coeff[ind[i]][j] * sol[j];
}
sol[i] = sum / coeff[ind[i]][i];
}
}
public static void SPPGaussian(int [][]coeff,int []b,int n )
{
int []sol= new int [n+1];
int ind[] = new int[n+1];
for (int i =
1;i<n;i++) {
ind[i]=i;
}
SPPFwdElimination(coeff,b,ind,n);
SPPBackSubst(coeff,b,sol,ind,n);
}
public static void main(String[] args) {
int n;
Scanner sc = new
Scanner(System.in);
n=sc.nextInt();
int coeff[][]=new
int[n+1][n+1];
int []b=new int[n+1];
for(int i=1;i<=n;i++) {
for(int
j=1;j<=n;j++) {
coeff[i][j]=sc.nextInt();
}
b[i]=sc.nextInt();
}
SPPGaussian(coeff,b,n);
}
}
Implement this pseudo code with Java. n is the size of the matrix for an n...
Write a function that solves the matrix equation Ax = b using Gaussian Elimination. Your function should accept as input a n-by-n matrix A and an n-by-1 vector b, and it should produce a n-by-1 vector x that satisfies Ax = b. Gaussian Elimination has two parts: forwards elimination and backwards substitution. You'll need to use both to solve the problem. It's okay to rigidly follow the pseudocode in the book. Using C++ Don't just use a library call, even...
Using java fix the code I implemented so that it passes the JUnit Tests. MATRIX3 public class Matrix3 { private double[][] matrix; /** * Creates a 3x3 matrix from an 2D array * @param v array containing 3 components of the desired vector */ public Matrix3(double[][] array) { this.matrix = array; } /** * Clones an existing matrix * @param old an existing Matrix3 object */ public Matrix3(Matrix3 old) { matrix = new double[old.matrix.length][]; for(int i = 0; i <...
the code needs to be modifed. require a output for the
code
Java Program to Implement Insertion Sort import java.util.Scanner; /Class InsertionSort * public class Insertion Sort { /Insertion Sort function */ public static void sort( int arr) int N- arr.length; int i, j, temp; for (i-1; i< N; i++) j-i temp arrli; while (j> 0 && temp < arrli-1) arrli]-arrli-1]; j-j-1; } arrlj] temp; /Main method * public static void main(String [] args) { Scanner scan new Scanner( System.in...
The pscudocode shown below solves a system of n linear algebraic equations using Gauss-Jordan 125] elimination. DOFOR -1,n DOPOR 1 = k + 1,n + 1 END DO ae 1 DOFOR 1 = 1, n k THEN IF i DOFOR j- k+1,n+ 1 ENDDO END IF END DO END DO DOFOR m-1,n END DO Write a Matlab function program GaussJordan(A,n) which implements this algorithm and a) returns the solution. Here A is the augmented matrix consisting of the coefficient matrix...
When asked to describe an algorithm you are expected to give a
clear pseudo-code description of the algorithm
1. (10 pts) Here is a new sorting algorithm NewSort Suppose the original call made is NewSort(A,0,n-1) where A is an array integers. == void NewSort(int A[], int i, int j){ \\ sorts the subarray Aſi..j] if (j i+1) \\when there are only 2 elements if (A[i] > A[j]) swap(A,i,j) \\swaps A[i] and A[j] else { int k = (j-i+1)/3; NewSort(A,i,j-k); \\...
% Create a matrix which has 150 rows and 200 columns % Use the rand function to generate that matrix with radom values of entries % are between 0 and 1 a = rand(150,200); b = zeros(1,300); for i = 1:150 for j = 1:200 if rand() < 0.9 % 90% of rand() value less than 0.9 a(i,j) = 0; % Assign 0 for dead cells else a(i,j) = 1; % Assign 1 for living cells...
definition of Markov matrix
and related theorems are showed below
8.4.2Show that the matrix (8.4.21) is a Markov matrix which is not regular. Is A stable? Definition 8.7 Let A = (aij) A satisfies R(n, n) so that aij-0 for i, j = I, . . . , n. If j-1 243 8.4 Markov matrices that is, the components of each row vector in A sum up to 1, then A is called a Markov or stochastic matrix. If there...
Implement a rabin_hash method to make the following code work: int rabin_karp_batchmatch(int bsz, /* size of bitmap (in bits) to be used */ int k, /* chunk length to be matched */ const char *qs, /* query docoument (X)*/ int m, /* query document length */ const char *ts, /* to-be-matched document (Y) */ int n /* to-be-matched document length*/) { /*if the to-be-matched document is less than k length, return false*/ if (n < k) return 0; /*start our...
MATLAB
Write a code where the elements of a squared matrix size (order) n are equal to the sum of its row and column position (I + j)
I need a program in c++ the same as below code but I want it to prompt the user to enter the number of elements after that I want it to ask the user to enter the array elements #include<algorithm> #include<stdio.h> #include<string.h> #include<iostream> using namespace std; int a[50]={2,5,4,3}; bool x[100]; int N=4;//number of elements int k=10;//target sum int sum;//current target sum int cmp(const void *a,const void *b) { return *(int *)b-*(int *)a; } void backtrace(int n) { if(sum>k) return...