Hi everyone this is What I did to carry out an iteration method(gauss seidel) and I want when iteration number greater than 30 it will stop and generate the corresponding result up to 30 iteration. But I wonder why the output result were so weird and I try to check the value on the command window by typing x_ans(:,1) it gives me the correct value. It really made me frustrated why the generate result were not the same. Or any other circumstance or function can be used to set for not converging condition. Sincerely thanks in advance for every single help.
clear;clc
A = [2 8 3 1;0 2 -1 4;7 -2 1 2;-1 0 5 2]
B = [-2;4;3;5]
Es = 1e-5
n = length(B);
x = zeros(n,1);
Ea = ones(n,1);
iter = 0;
while max(Ea) >= Es
if iter <= 30
iter = iter + 1;
x_old = x;
for i = 1:n
j = 1:n;
j(i) = [];
x_cal = x;
x_cal(i) = [];
x(i) = (B(i) - sum(A(i,j) * x_cal)) / A(i,i);
end
else
break
end
x_ans(:,iter) = x;
Ea(:,iter) =abs(( x - x_old) ./ x);
end
result = [1:iter; x_ans; Ea]'
I've gone through the formulas and they are all OK. On a side note, the sum is not necessary. The problem lies with your input data - try reordering! check for example the following, which works
A = [7 -2 1 2;
2 8 3 1;
-1 0 5 2;
0 2 -1 4;]
B = [3;-2;5;4]
see the wiki under convergence.
Related
I have a MATLAB matrix like below:
column no: 1 2 3 4 5 6
matrix elements 1 1 2 3 6 2
Column numbers represent node ID and elements of the matrix represent the node towards which that node points. Please help me find hop count from a particular node to node 1. I have written the following code but it doesn't solve the problem.
x = ones(1, n);
checkbit = zeros(1, n);
nodedest = [1 1 2 3 6 2];
hopcount = zeros(1, n);
for i = 1:n
for j = 1:n
if nodedest(j) == 1 && checkbit(j) == 0
hopcount(j) = hopcount(j) + 1;
checkbit(j) = 1;
else
x(j) = nodedest(j);
end
if x(j) ~= 1
hopcount(j) = hopcount(j) + 1;
x(j) = nodedest(x(j));
end
end
end
You are looking for a breadth-first search to find the shortest path in your graph. Without touching the data in any way, you can do this in O(n) time per node, given the tree-like structure of your graph:
nodedest = [1 1 2 3 6 2];
hopcount = zeros(1, 6);
for n = 2:6
k = n
while k ~= 1
hopcount(n) = hopcount(n) + 1
k = nodedest(k)
end
end
If you are willing to reverse the sense of your edges (introducing a one-to-many relationship), you could accomplish the same thing in one pass, reducing the entire algorithm from O(n2) to O(n) time complexity. The trade-off would be that memory complexity would increase from O(1) to O(n):
nodedest = [1 1 2 3 6 2];
% Reverse the input
nodesource = cell(1, 6);
nodesource(:) = {[]}
for n = 2:6
k = nodedest(n);
nodesource{k} = [nodesource{k} n];
end
% implement bfs, using the assumption that the graph is a simple tree
hopcount = zeros(1, 6);
cache = [1];
hops = 0;
while ~isempty(cache)
next = []
for c = cache
hopcount(c) = hops;
next = [next nodesource(c)]
end
hops = hops + 1;
cache = next
end
I am trying to implement the paper detection of copy move forgery using histogram of oriented gradients.
The algorithm is:
Divide the image into overlapping blocks.
Calculate feature vectors for each block and store them in a matrix.
Sorting the matrix lexicographically
Using block matching to identify forged regions.
https://www.researchgate.net/publication/276518650_Detection_of_copy-move_image_forgery_using_histogram_of_orientated_gradients
I am stuck with the 3rd step and can't proceed.
The code I have implemented is:
clc;
clear all;
close all;
%read image
img = imread('006_F.png');
img=rgb2gray(img);
img=imresize(img, 1/4);
figure(1);
imshow(img);
b=16; %block size
nrc=5; %no. of rows to check
td=416; %threshold
[r, c]=size(img);%Rows and columns;
column=(r-b+1)*(c-b+1);
M= zeros(column,4);
Mi = zeros(1,2);
i=1;
disp('starting extraction of features');
for r1 = 1:r-b+1
for c1 = 1:c-b+1
% Extract each block
B = img(r1:r1+b-1,c1:c1+b-1);
features = extractHOGFeatures(B);%extracting features
M(i, :) = features;
Mi(i,:) = [r1 c1];
i=i+1;
end
end
[S, index] = sortrows(M , [ 1 2 3 4]);
P= zeros(1,6);
b2=r-b+1;
disp('Finding Duplicates');
for i = 1:column
iv = index(i);
xi=mod(iv,b2) + 1;
yi=ceil(iv/b2);
j = i+1;
while j < column && abs(i - j) < 5
jv=index(j);
xj=mod(jv,b2) + 1;
yj=ceil(jv/b2);
z=sqrt(power(xi-xj,2) + power(yi-yj,2));
% only process those whose size is above Nd
if z > 16
offset = [xi-xj yi-yj];
P = [P;[xi yi xj yj xi-xj yi-yj]];
end
j = j + 1;
end
end
rows = size(P,1);
P(:,6) = P(:,6) - min(P(:,6));
P(:,5) = P(:,5) - min(P(:,5));
maxValP = max(P(:,6)) + 1;
P(:,5) = maxValP .* P(:,5) + P(:,6);
mostfrequentval = mode(P(:,5));
disp('Creating Image');
idx = 2;
% Create a copy of the image and mask it
RI = img;
while idx < rows
x1 = P(idx,1);
y1 = P(idx,2);
x2 = P(idx,3);
y2 = P(idx,4);
if (P(idx,5) == mostfrequentval)
RI(y1:y1,x1:x1) = 0;
RI(y2:y2,x2:x2) = 0;
end
idx = idx + 1;
end;
After going through some references indicated in the paper you are working on (ref. [8] and [20]):
The lexicographic sorting is the equivalent of the alphabetical one, for numbers i.e., [1 1 1 1] < [1 1 2 1] < [2 3 4 5] < [2 4 4 5]
So, in your case, you case use the function sortrows() in the following way:
A = [1 1 1 1;1 1 1 2;1 1 1 4;1 2 2 2; 1 2 2 1; 1 4 6 3; 2 3 4 5; 2 3 6 6]; % sample matrix
[B,idx] = sortrows(A,[1 2 3 4]); % Explicit notation but it is the Matlab default setting so equivalent to sortrows(A)
It means: Sort the rows of A by first looking at the first column and, in case of equality, looking at the second one, and so on.
If your are looking for a reverse order, you specify '-' before the number of the column.
So in the end, your code is good and if the results are not as expected it has to come from another step of the implementation...
Edit: the parameter idx records the original index of the sorted rows.
I'm trying to write my own program to sort vectors in matlab. I know you can use the sort(A) on a vector, but I'm trying to code this on my own. My goal is to also sort it in the fewest amount of swaps which is kept track of by the ctr variable. I find and sort the min and max elements first, and then have a loop that looks at the ii minimum vector value and swaps it accordingly.
This is where I start to run into problems, I'm trying to remove all the ii minimum values from my starting vector but I'm not sure how to use the ~= on a vector. Is there a way do this this with a loop? Thanks!
clc;
a = [8 9 13 3 2 8 74 3 1] %random vector, will be function a once I get this to work
[one, len] = size(a);
[mx, posmx] = max(a);
ctr = 0; %counter set to zero to start
%setting min and max at first and last elements
if a(1,len) ~= mx
b = mx;
c = a(1,len);
a(1,len) = b;
a(1,posmx) = c;
ctr = ctr + 1;
end
[mn, posmn] = min(a);
if a(1,1) ~= mn
b = mn;
c = a(1,1);
a(1,1) = b;
a(1,posmn) = c;
ctr = ctr + 1;
end
ii = 2; %starting at 2 since first element already sorted
mini = [mn];
posmini = [];
while a(1,ii) < mx
[mini(ii), posmini(ii - 1)] = min(a(a~=mini))
if a(1,ii) ~= mini(ii)
b = mini(ii)
c = a(1,ii)
a(1,ii) = b
a(1,ii) = c
ctr = ctr + 1;
end
ii = ii + 1;
end
I am unable to get converging values using a Gauss-Seidel algorithm
Here is the code:
A = [12 3 -5 2
1 6 3 1
3 7 13 -1
-1 2 -1 7];
b = [2
-3
10
-11];
ep = 1e-8;
[m, n] = size(A);
[n, p] = size(b);
x = zeros(n, 1001);
x(:, 1) = []
for k=0:1000
ka = k + 1;
if ka == 1001
break;
end
xnew = zeros(n,1);
for i=1:n
sum = 0;
j = 1;
while j < i
s1 = s1 + A(i,j) * x(j, ka + 1);
j = j + 1;
end
j = i + 1;
while j <= n
sum = sum + A(i,j) * x(j, ka);
j = j + 1;
end
xnew(i) = (b(i) - sum) / A(i, i);
% if result is within error bounds exit loop
if norm(b - A * xnew, 2) < ep * norm(b, 2)
'ending'
break
end
end
x(:,ka + 1) = xnew;
end
I cannot get the A * xnew to converge on b what am I doing wrong?
I have tried running this changing the syntax several times, but I keep getting values that are way off.
Thanks!
Gabe
You have basically two problems with your code:
(1) You are using two different variables "sum" and "s1". I replaced it by mySum. By the way, dont use "sum", since there is a matlab function with this name.
(2) I think there is also a problem in the update of x;
I solved this problem and I also tried to improve your code:
(1) You dont need to save all "x"s;
(2) It is better to use a "while" than a for when you dont know how many iterations you need.
(3) It is good to use "clear all" and "close all" in general in order to keep your workspace. Sometimes old computations may generate errors. For instance, when you use matrices with different sizes and the same name.
(4) It is better to use dot/comma to separate the lines of the matrices
You still can improve this code:
(1) You can test if A is square and if it satisfies the conditions necessary to use this numerical method: to be positive definite or to be diagonally dominant.
clear all
close all
A = [12 3 -5 2;
1 6 3 1;
3 7 13 -1;
-1 2 -1 7];
b = [2;
-3;
10;
-11];
ep = 1e-8;
n = length(b); % Note this method only works for A(n,n)
xNew=zeros(n,1);
xOld=zeros(n,1);
leave=false;
while(~leave)
xOld=xNew;
for i=1:n
mySum = 0;
j = i + 1;
while j <= n
mySum = mySum + A(i,j) * xOld(j,1);
j = j + 1;
end
j = 1;
while j < i
mySum = mySum + A(i,j) * xNew(j,1);
j = j + 1;
end
mySum=b(i,1)-mySum;
xNew(i,1) = mySum / A(i, i);
end
if (norm(b - A * xNew, 2) < ep * norm(b, 2))
disp('ending');
leave=true;
end
xOld = xNew;
end
xNew
I saw other topics about this error but I couldn't figure it out. The error "In an assignment A(I) = B, the number of elements in B and I must be the same" occurs at the second for loop. How can I change my code to avoid this error?
h1 = [70 31.859 15 5.774 3.199 2.15 1.626];
h2 = [31.859 15 5.774 3.199 2.15 1.626 1.415];
b = [1253 1253 1253 1253 1253 1253 1253];
R = [455.4 425.6 377.6 374.9 371.3 273.7 268.3];
r = [0.5448714286 0.5291754292 0.6150666667 0.4459646692 0.3279149734 0.2437209302 0.1297662977];
k = [200 200 200 200 200 200 200];
s = sqrt(r/(1-r));
v2 = [20 0 0 0 0 0 0];
v1 = [0 0 0 0 0 0 0];
Ch1 = [0 0 0 0 0 0 0];
Ch2 = [0 0 0 0 0 0 0];
C = [100 100 100 100 100 100 100];
F = b .* k .* sqrt(R-(h1-h2))- R.*sin((acos((R-((h1-h2)./2))./R))) .* (pi/2) .* (1./sqrt(r./(1-r))) .* (atan(sqrt(r./(1-r))))-(pi/4) - (1./(sqrt(r./(1-r)) .* sqrt(h2./R))).* log((h2+R.*((sqrt(h1./R).*tan(1/2 .* atan(sqrt(r./(1-r)).*sqrt(h1./r).*log(1./(1-k))))).^2).*sqrt(1-r))./h2)
M = -R.*R.*(k./2).*(.2*(sqrt(h2./R)*tan(0.5*(atan(s)))-(pi/8).*sqrt(h2./R).*log(1./1-r)))-(acos((R-((h1-h2)./2))./R))
for i=1:6
v1(i) = ((v2(i)*h2)/h1);
v2(i+1) = v1(i);
end
vr = ((v1.*h1)./h2)./(((tan(0.5.*((atan(s)))-(pi/8).*sqrt(h2./R).*log(1./(1-r)))).^(2))+1)
%--------------------------------------------------------------------------
% Calculating E
w = (((2.*R.*h2).^(3/2))./(300.*(b.^2)))
if (w <= (3*10^-4));
E = ((0.0821.*((log(w))^2))+(1.25.*log(w))+4.89)
end
if ((3*10^-4)<= w <= (2.27*10^-3));
E = ((0.0172.*((log(w)).^2))+(0.175.*log(w))+0.438)
end
if (w > (2.27*10^-3))
E = 0.01
end
%--------------------------------------------------------------------------
% Calculating Ch:
y = ((((2.*R).^(0.5)).*((h2).^(1.5)))./(b.^2))
N1 = (0.5-(1/pi).*atan((log(y)+8.1938)./(1.1044)))
N = ((h2./h1).*N1)
for i=1:1;7
Ch2(i) = (h2.*((N.*((Ch1(i)./h2)-(C./h2)))+(C/h2)))
Ch1(i+1) = Ch2(i)
end
DeltaStrain = (E.*((Ch2./h2)-(Ch1./h1)))
if DeltaStrain > 0;
Stepp = ((2/pi).*(sqrt(DeltaStrain))))
Control = 2;
else
Stepp = ((2/pi).*(sqrt(-DeltaStrain))
Control = 0;
end
In the line
Ch2(i) = (h2.*((N.*((Ch1(i)./h2)-(C./h2)))+(C/h2)))
h2 is a vector, and Ch2(i) is a scalar. You cannot assign the value of a vector to a scalar. I suspect you want to replace the entire for loop. Right now you have
for i=1:1;7
Ch2(i) = (h2.*((N.*((Ch1(i)./h2)-(C./h2)))+(C/h2)))
Ch1(i+1) = Ch2(i)
end
(?? what is the meaning of 1:1;7? Is that a typo? I am thinking you want 1:7...
Since you seem to be using the result of one loop to change the value of Ch1 which you are using again in the next loop, it may be tricky to vectorize; but I wonder what you are expecting the output to be, since you really do have a vector as the result of the RHS of the equation. I can't be sure if you want to compute the result for one element at a time, or whether you want to compute vectors (and end up appending results to Ch1 and Ch2). The following line would run without throwing an error - but it may not be the calculation you want. Please clarify what you are hoping to achieve if this is an incorrect guess.
for i = 1:7
Ch2(i) = h2.*(N.*((Ch1(i) - C(i))./h2(i))) + C(i)./h2(i);
Ch1(i+1) = Ch2(i);
end