The picture shows minimised case of my assignment, I try to minimize sum of all "y" values with using Matlab. These y values are changing with X matrix. So, y values are the functions of X matrix.
X variables are binary numbers and the sum of consecutive two numbers in the second row must be greater than or equal to 1. In addition, sum of consecutive three numbers in the third row must be greater than or equal to 1.
How can I solve this problem? Thanks for your help.
This appears to be a mixed integer linear programming problem. (All linear constraints, and a binary one)
These can be solved with intlinprog.
[x,fval] = intlinprog(f,intcon,A,b,Aeq,beq,lb,ub)
Related
Please see the following issue:
P=rand(4,4);
for i=1:size(P,2)
for j=1:size(P,2)
[r,p]=corr(P(:,i),P(:,j))
end
end
Clearly, the loop will cause the number of correlations to be doubled (i.e., corr(P(:,1),P(:,4)) and corr(P(:,4),P(:,1)). Does anyone have a suggestion on how to avoid this? Perhaps not using a loop?
Thanks!
I have four suggestions for you, depending on what exactly you are doing to compute your matrices. I'm assuming the example you gave is a simplified version of what needs to be done.
First Method - Adjusting the inner loop index
One thing you can do is change your j loop index so that it only goes from 1 up to i. This way, you get a lower triangular matrix and just concentrate on the values within the lower triangular half of your matrix. The upper half would essentially be all set to zero. In other words:
for i = 1 : size(P,2)
for j = 1 : i
%// Your code here
end
end
Second Method - Leave it unchanged, but then use unique
You can go ahead and use the same matrix like you did before with the full two for loops, but you can then filter the duplicates by using unique. In other words, you can do this:
[Y,indices] = unique(P);
Y will give you a list of unique values within the matrix P and indices will give you the locations of where these occurred within P. Note that these are column major indices, and so if you wanted to find the row and column locations of where these locations occur, you can do:
[rows,cols] = ind2sub(size(P), indices);
Third Method - Use pdist and squareform
Since you're looking for a solution that requires no loops, take a look at the pdist function. Given a M x N matrix, pdist will find distances between each pair of rows in a matrix. squareform will then transform these distances into a matrix like what you have seen above. In other words, do this:
dists = pdist(P.', 'correlation');
distMatrix = squareform(dists);
Fourth Method - Use the corr method straight out of the box
You can just use corr in the following way:
[rho, pvals] = corr(P);
corr in this case will produce a m x m matrix that contains the correlation coefficient between each pair of columns an n x m matrix stored in P.
Hopefully one of these will work!
this works ?
for i=1:size(P,2)
for j=1:i
Since you are just correlating each column with the other, then why not just use (straight from the documentation)
[Rho,Pval] = corr(P);
I don't have the Statistics Toolbox, but according to http://www.mathworks.com/help/stats/corr.html,
corr(X) returns a p-by-p matrix containing the pairwise linear correlation coefficient between each pair of columns in the n-by-p matrix X.
I want to compare two image histograms. They are as follows:
h1 --> double valued 1 dimension vector .4096 in length.
h2 --> double valued 1 dimension vector .4096 in length.
I am using this matlab function here:
http://clickdamage.com/sourcecode/code/compareHists.m
It is as follows:
% s = compareHists(h1,h2)
% returns a histogram similarity in the range 0..1
%
% Compares 2 normalised histograms using the Bhattacharyya coefficient.
% Assumes that sum(h1) == sum(h2) == 1
%
function s = compareHists(h1,h2)
s = sum(sum(sum(sqrt(h1).*sqrt(h2))));
My question is :
Is there a need for multiple sums?
Even if there is only one sum in the above equation, it would suffice..right?
like this: sum(sqrt(h1).*sqrt(h2)) --> ?
Can some one please explain the code above? Also, tell me if I use a single sum will it be all right?
I tried both ways and got the same answer for two image histograms. I did this with only two histograms not more and hence want to be sure.
Thanks!
In general, sum does the sum along one dimension only. If you want to sum along multiple dimensions you either
use sum several times; or
use linear indexing to reduce to a single dimension and then use sum once: sum(sqrt(h1(:)).*sqrt(h2(:))).
In your case, if there's only one dimension, yes, a single sum would suffice.
You are right. Only one sum is needed. However, if either h1 or h2 is a multidimensional matrix, then you may want to sum as many as the dimensions. For example:
A=magic(4); % a 4 by 4 matrix of magic numbers.
sum(A) % returns [34,34,34,34], i.e. the sum of elements in each column.
sum(sum(A)) % returns 136, i.e. the sum of all elements in A.
I believe the code you downloaded originaly was written to handle multiple histograms stacked as columns of a matrix. This is (IMHO) the reason for the multiple sums.
In your case you can leave it with only one sum.
You can do even better - without any sum
Hover here to see the answer
s = sqrt(h1(:)')*sqrt(h2(:));
The trick is to use vector multiplication!
I don't see any points in 3 sums too, but if you have not a vector with histogram but a matrix you will need 2 sums like this sum(sum(sqrt(h1).*sqrt(h2))) to compare them. First one will calculate the sum of the rows, the second - the sum of the columns.
I have a set of samples, S, and I want to find its PDF. The problem is when I use ksdensity I get values greater than one!
[f,xi] = ksdensity(S)
In array f, most of the values are greater than one! Would you please tell me what the problem can be? Thanks for your help.
For example:
S=normrnd(0.3035, 0.0314,1,1000);
ksdensity(S)
ksdensity, as the name says, estimates a probability density function over a continuous variable. Probability densities can be larger than 1, they can actually have arbitrary values from zero upwards. The constraint on probabilities is that their sum over an exhaustive range of possibilities has to be 1. For probability densities, the constraint is that the integral over the whole range of values is 1.
A crude approximation of an integral of the pdf estimated by ksdensity can be obtained in Matlab like this:
sum(f) * min(diff(xi))
assuming that the values in xi are equally spaced. The value of this expression should be approximately 1.
If in your application you believe this approximation is not close enough to 1, you might want to specify the grid of estimation points (second parameter pts) such that the spacing is finer or the range is wider than the one automatically generated by ksdensity.
I have to two evenly sized very large vectors (columns) A and B. I would like to divide vector A by vector B. This will give me a large matrix AxB filled with zeros, except the last column. This column contains the values I'm interested in. When I simple divide the vectors in a Matlab script, I run out of memory. Probably because the matrix AxB becomes very large. Probably I can prevent this from happening by repeating the following:
calculating the first row of matrix AxB
filter the last value and put it into another vector C.
delete the used row of matrix AxB
redo step 1-4 for all rows in vector A
How can I make a loop which does this?
You're question doesn't make it clear what you are trying to do, although it sounds like you want to do an element wise division.
Try:
C = A./B
"Matrix product AxB" and "dividing vectors" are distinct operations.
If we understood this correctly, what you do want to calculate is "C = last column from AxB", such that:
lastcolsel=zeros(size(B,2),1)
C=(A*B)*lastcolsel
If that code breaks your memory limit, recall that matrix product is associative (MxN)xP = Mx(NxP). Simplifying your example, we get:
lastcolsel=zeros(size(B,2),1)
simplifier=B*lastcolsel
C=A*simplifier
As an input I have two number x and y. x>y.
I want to create exactly y non-zero random number which their sum will be equal to x. I know randi([min max]) function . Can you help me?
If I got it right, you want something like this:
data = rand(1,y);
data = data * x / sum(data);
data will contain exactly y positive uniformly distributed numbers which sum equals to x.
Check out the file random vectors generator with fixed sum in Matlab FEX. I believe this will answer your question.
Leonid's approach will certainly generate a set of random numbers that have the correct sum, but it won't select uniformly over the allowed space. If this is important, an approach that will work is the following:
(with x = 1):
Generate Y-1 random numbers uniformly over [0,1].
Sort the Y-1 numbers from smallest to largest. Call these {y1,...,y_{N-1}}
Take as the Y random numbers the set {y_1-0 ,y_2-y1,...,1-y_{N-1}} == {n_1,... n_Y}.
These n_i clearly sum to one. It is easy to prove uniformity by considering the probability for a given realization of the n_i.