I would like to write a Matlab code to calculate the following:
\sum_{k=0}^{N-1} \frac{1}{k!} \sum_{i=0}^{k} {k \choose i}(a-1)^{k-i} a^k
and my code is:
N = 3;
a = [3 4];
for k = 0:N-1
f = 0;
for i = 0:k
f = f + nchoosek(k,i).* a.^k .* (a-1).^(k-i);
end
sumoff = sum(f);
all = (( 1./ (factorial(k))).*sumoff);
end
overall= sum(all);
'all' variable gives different value when it is inside the for loop rather than outside. But I want it to calculate when k = 0:N-1. What am I doing wrong?
Thank you.
The issue is your current code overwrites all on every iteration. Moving it outside the loop also doesn't work because you'll only save the result of the last iteration.
To save the all of every iteration, define all as a vector and then assign each intermediate result into that vector:
N = 3;
a = [3 4];
% preallocate a vector for `all`
all = nan(N-1, 1);
for k = 0:N-1
f = 0;
for i = 0:k
f = f + nchoosek(k,i) .* a.^k .* (a-1).^(k-i);
end
sumoff = sum(f);
% assign your intermediate result into the `all` vector
all(k+1) = ((1./(factorial(k))) .* sumoff);
end
overall = sum(all);
I have the following DP which I am applying on a binarized image (either 0 or 1) in Matlab
[x, y] = size(img);
dp = zeros(x, y);
dp(1,:) = img(1,:);
dp(:,1) = img(:,1);
for i = 2:x
for j = 2:y
if img(i, j) == 0
dp(i, j) = min([dp(i, j - 1), dp(i - 1, j), dp(i - 1, j - 1)]) + 1;
end
end
end
The code for large x and y takes a lot of time maybe because of the if condition and using for loops instead of writing vectorized code.
Can anyone optimize it.?
Or is there any approach which optimizes the above code by exploiting the fact that the matrix img contains either 0 or 1 (fewer 1s than 0s).
Also is it possible to somehow use parallel for loops to speed up.?
As far as I am aware, you cannot really speed up this computation in general. But if you know that there are only very few entries where img(i,j)==0 following approach might save you a little bit of time:
[x, y] = size(img);
dp = zeros(x, y);
dp(1,:) = img(1,:);
dp(:,1) = img(:,1);
[i, j] = find(img(2:end, 2:end) == 0); % Extract only these pixels where we actually need to do something
i = i + 1; %correct for removing the first row and column
j = j + 1;
for k = 1:numel(i);
dp(i(k), j(k)) = min([dp(i(k), j(k) - 1), dp(i(k) - 1, j(k)), dp(i(k) - 1, j(k) - 1)]) + 1;
end
Can you help me please translate this equation into a code?
I tried this code but its only the first part
theSum = sum(M(:, y) .* S(:, y) ./ (1 + K(:, y)))
EDIT: sorry, I was having a brainfart. The answer below makes no assumptions on the nature of M, K, etc, which is why I recommended functions like that. But they're clearly matrices. I'll make another answer, I'll leave this here for reference though in case it's useful
I would start by making the M, K, X, L, and O expressions into simple functions, so that you can easily call them as M(z,y), X(z,y) (or X(z,j) depending on the input you need ) etc.
Then you will convert each summation into a for loop and collect the result (you can think about vectorisation later, right now focus on translating the problem). The double summation is essentially a nested for loop, where the result of the inner loop is used in the outer one at each outer iteration.
So your end result should look something like:
Summation1 = 0;
for z = 1 : Z
tmp = M(z,y) / K(z,y) * (X(z,y) / (1 + L(z,y));
Summation1 = Summation1 + tmp;
end
Summation2 = 0;
for j = 1 : Y
if j ~= y
for z = 1 : Z
tmp = (M(z,j) * X(z,j) * O(j)) / (K(z,j)^2 * (1 + L(z,j)) * X(z,y);
Summation2 = Summation2 + tmp;
end
end
end
Result = Summation1 - Summation2;
(Btw, this assumes that all operations are on scalars. If M(z,y) outputs a vector, adjust for elementwise operations appropriately)
IF M, K, etc are all matrices, and all operations are expected to be element-wise, then this is a vectorised approach for this equation.
Left summation is
S1 = M(1:Z,y) ./ K(1:Z,y) .* X(1:Z,y) ./ (1 + L(1:Z,y));
S1 = sum(S1);
Right summation is (assuming (O is a horizontal vector)
S2 = M(1:Z, 1:Y) .* X(1:X, 1:Y) .* repmat(O(1:Y), [Z,1]) ./ ...
(K(1:Z, 1:Y) .^ 2 .* (1 + L(1:Z, 1:Y))) .* X(1:Z, 1:Y);
S2(:,y) = []; % remove the 'y' column from the matrix
S2 = sum(S2(:)); % add all elements
End result: S1 - S2
this is lambda version vectorized:
equation = #(y,M,K,X,L,O) ...
sum(M(:,y)./K(:,y).*X(:,y)./(1+L(:,y))) ...
-sum(sum( ...
bsxfun( ...
#times ...
,M(:,[1:y-1,y+1:end]) ...
.* X(:,[1:y-1,y+1:end]) ...
.* O(:,[1:y-1,y+1:end]) ...
./ (K(:,[1:y-1,y+1:end]) .^ 2 ...
.*(1+ L(:,[1:y-1,y+1:end]))) ...
,X(:,y) ...
) ...
));
%%% example:
y = 3;
Y = 5;
Z = 10;
M = rand(Y, Z);K = rand(Y, Z);X = rand(Y, Z);L = rand(Y, Z);O = rand(Y, Z);
equation(y,M,K,X,L,O)
this is my first time here so I hope that someone can help me.
I'm trying to implementing the Gauss-Seidel method and the power method using a matrix with the storage CSR or called Morse storage. Unfortunately I can't manage to do better then the following codes:
GS-MORSE:
function [y] = gs_morse(aa, diag, col, row, nmax, tol)
[n, n] = size(A);
y = [1, 1, 1, 1];
m = 1;
while m < nmax,
for i = 1: n,
k1 = row(i);
k2 = row(i + 1) - 1;
for k = k1: k2,
y(i) = y(i) + aa(k) * x(col(k));
y(col(k)) = y(col(k)) + aa(k) * diag(i);
end
k2 = k2 + 1;
y(i) = y(i) + aa(k) * diag(i);
end
if (norm(y - x)) < tol
disp(y);
end
m = m + 1;
for i = 1: n,
x(i) = y(i);
end
end
POWER-MORSE:
I was able only to implement the power method but I don't understand how to use the former matrix... so my code for power method is:
function [y, l] = potencia_iterada(A, v)
numiter=100;
eps=1e-10;
x = v(:);
y = x/norm(x);
l = 0;
for k = 1: numiter,
x = A * y;
y = x / norm(x);
l0 = x.' * y;
if abs(l0) < eps
return
end
l = l0;
end
Please anyone can help me for completing these codes or can explain me how can I do that? I really don't understand how to do. Thank you very much
here is my code where i have one error regarding array index out out of bound. plzz help me to rectify it
I = imread('E:\degraded images\village.jpg');
imshow(I)
I = im2double(I);
I = log(1 + I);
M = 2*size(I,1) + 1;
N = 2*size(I,2) + 1;
sigma = 10;
[X, Y] = meshgrid(1:N,1:M);
centerX = ceil(N/2);
centerY = ceil(M/2);
gaussianNumerator = (X - centerX).^2 + (Y - centerY).^2;
H = exp(-gaussianNumerator./(2*sigma.^2));
H = 1 - H;
imshow(H,'InitialMagnification',25)
H = fftshift(H);
If = fft2(I, M, N);
Iout = real(ifft2(H.*If)); ** here the code has error . ??? Error using ==> times Number of array dimensions must match for binary array op.**
H is 2-D while If is 3-D. You can use repmat with H or subset If. I don't know which one is correct for your situation. For instance,
rempat( H, [1, 1, 3 ] ) .* If;
or
H .* If(:,:,ind); % ind is the index of the 2-D array you want to subset