Here is the problem and current code I am using.
Use the svd() function in MATLAB to compute , the rank-2 approximation of . Clearly state what is, rounded to 4 decimal places. Also, compute the root-mean square error (RMSE) between and . Which approximation is better, or ? Explain.
Solution:
%code
[U , S, V] = svd(A)
k = 2
V = V(:,1:k)
V = transpose(V)
Ak = U(:,1:k) .* S(1:k,1:k) .* V
diffA = A - A2
fro_norm = norm(diffA,'fro')
RMSE2 = (fro_norm)/sqrt(m*n)
However, when running, the line AK = . . . keeps giving an error because the matrix sizes are not compatible. So I understand that the matrix sizes need to match in order to do the multiplication, but I also know that the problem requires the following calculation requirements, that when k = 2, U has to use the first 2 columns, S has to use the first 2 rows and first 2 columns, and V has to be the transpose of V only using the first two columns.
I must be missing something in my understanding of the calculation, or creation of the sub k matrices. The matrix I have to use is a 3 x 3.
I defined an inline function f that takes as argument a (1,3) vector
a = [3;0.5;1];
b = 3 ;
f = #(x) x*a+b ;
Suppose I have a matrix X of size (N,3). If I want to apply f to each row of X, I can simply write :
f(X)
I verified that f(X) is a (N,1) vector such that f(X)(i) = f(X(i,:)).
Now, if I a add a quadratic term :
f = #(x) x*A*x' + x*a + b ;
the command f(X) raises an error :
Error using +
Matrix dimensions must agree.
Error in #(x) x*A*x' + x*a + b
I guess Matlab is considering the whole matrix X as the input to f. So it does not create a vector with each row, i, being equal to f(X(i,:)). How can I do it ?
I found out that there exist a built-in function rowfun that could help me, but it seems to be available only in versions r2016 (I have version r2015a)
That is correct, and expected.
MATLAB tries to stay close to mathematical notation, and what you are doing (X*A*X' for A 3×3 and X N×3) is valid math, but not quite what you intend to do -- you'll end up with a N×N matrix, which you cannot add to the N×1 matrix x*a.
The workaround is simple, but ugly:
f_vect = #(x) sum( (x*A).*x, 2 ) + x*a + b;
Now, unless your N is enormous, and you have to do this billions of times every minute of every day, the performance of this is more than acceptable.
Iff however this really and truly is your program's bottleneck, than I'd suggest taking a look at MMX on the File Exchange. Together with permute(), this will allow you to use those fast BLAS/MKL operations to do this calculation, speeding it up a notch.
Note that bsxfun isn't going to work here, because that does not support mtimes() (matrix multiplication).
You can also upgrade to MATLAB R2016b, which will have built-in implicit dimension expansion, presumably also for mtimes() -- but better check, not sure about that one.
I have a 151-by-151 matrix A. It's a correlation matrix, so there are 1s on the main diagonal and repeated values above and below the main diagonal. Each row/column represents a person.
For a given integer n I will seek to reduce the size of the matrix by kicking people out, such that I am left with a n-by-n correlation matrix that minimises the total sum of the elements. In addition to obtaining the abbreviated matrix, I also need to know the row number of the people who should be booted out of the original matrix (or their column number - they'll be the same number).
As a starting point I take A = tril(A), which will remove redundant off-diagonal elements from the correlation matrix.
So, if n = 4 and we have the hypothetical 5-by-5 matrix above, it's very clear that person 5 should be kicked out of the matrix, since that person is contributing a lot of very high correlations.
It's also clear that person 1 should not be kicked out, since that person contributes a lot of negative correlations, and thus brings down the sum of the matrix elements.
I understand that sum(A(:)) will sum everything in the matrix. However, I'm very unclear about how to search for the minimum possible answer.
I noticed a similar question Finding sub-matrix with minimum elementwise sum, which has a brute force solution as the accepted answer. While that answer works fine there it's impractical for a 151-by-151 matrix.
EDIT: I had thought of iterating, but I don't think that truly minimizes the sum of elements in the reduced matrix. Below I have a 4-by-4 correlation matrix in bold, with sums of rows and columns on the edges. It's apparent that with n = 2 the optimal matrix is the 2-by-2 identity matrix involving Persons 1 and 4, but according to the iterative scheme I would have kicked out Person 1 in the first phase of iteration, and so the algorithm makes a solution that is not optimal. I wrote a program that always generated optimal solutions, and it works well when n or k are small, but when trying to make an optimal 75-by-75 matrix from a 151-by-151 matrix I realised my program would take billions of years to terminate.
I vaguely recalled that sometimes these n choose k problems can be resolved with dynamic programming approaches that avoid recomputing things, but I can't work out how to solve this, and nor did googling enlighten me.
I'm willing to sacrifice precision for speed if there's no other option, or the best program will take more than a week to generate a precise solution. However, I'm happy to let a program run for up to a week if it will generate a precise solution.
If it's not possible for a program to optimise the matrix within an reasonable timeframe, then I would accept an answer that explains why n choose k tasks of this particular sort can't be resolved within reasonable timeframes.
This is an approximate solution using a genetic algorithm.
I started with your test case:
data_points = 10; % How many data points will be generated for each person, in order to create the correlation matrix.
num_people = 25; % Number of people initially.
to_keep = 13; % Number of people to be kept in the correlation matrix.
to_drop = num_people - to_keep; % Number of people to drop from the correlation matrix.
num_comparisons = 100; % Number of times to compare the iterative and optimization techniques.
for j = 1:data_points
rand_dat(j,:) = 1 + 2.*randn(num_people,1); % Generate random data.
end
A = corr(rand_dat);
then I defined the functions you need to evolve the genetic algorithm:
function individuals = user1205901individuals(nvars, FitnessFcn, gaoptions, num_people)
individuals = zeros(num_people,gaoptions.PopulationSize);
for cnt=1:gaoptions.PopulationSize
individuals(:,cnt)=randperm(num_people);
end
individuals = individuals(1:nvars,:)';
is the individual generation function.
function fitness = user1205901fitness(ind, A)
fitness = sum(sum(A(ind,ind)));
is the fitness evaluation function
function offspring = user1205901mutations(parents, options, nvars, FitnessFcn, state, thisScore, thisPopulation, num_people)
offspring=zeros(length(parents),nvars);
for cnt=1:length(parents)
original = thisPopulation(parents(cnt),:);
extraneus = setdiff(1:num_people, original);
original(fix(rand()*nvars)+1) = extraneus(fix(rand()*(num_people-nvars))+1);
offspring(cnt,:)=original;
end
is the function to mutate an individual
function children = user1205901crossover(parents, options, nvars, FitnessFcn, unused, thisPopulation)
children=zeros(length(parents)/2,nvars);
cnt = 1;
for cnt1=1:2:length(parents)
cnt2=cnt1+1;
male = thisPopulation(parents(cnt1),:);
female = thisPopulation(parents(cnt2),:);
child = union(male, female);
child = child(randperm(length(child)));
child = child(1:nvars);
children(cnt,:)=child;
cnt = cnt + 1;
end
is the function to generate a new individual coupling two parents.
At this point you can define your problem:
gaproblem2.fitnessfcn=#(idx)user1205901fitness(idx,A)
gaproblem2.nvars = to_keep
gaproblem2.options = gaoptions()
gaproblem2.options.PopulationSize=40
gaproblem2.options.EliteCount=10
gaproblem2.options.CrossoverFraction=0.1
gaproblem2.options.StallGenLimit=inf
gaproblem2.options.CreationFcn= #(nvars,FitnessFcn,gaoptions)user1205901individuals(nvars,FitnessFcn,gaoptions,num_people)
gaproblem2.options.CrossoverFcn= #(parents,options,nvars,FitnessFcn,unused,thisPopulation)user1205901crossover(parents,options,nvars,FitnessFcn,unused,thisPopulation)
gaproblem2.options.MutationFcn=#(parents, options, nvars, FitnessFcn, state, thisScore, thisPopulation) user1205901mutations(parents, options, nvars, FitnessFcn, state, thisScore, thisPopulation, num_people)
gaproblem2.options.Vectorized='off'
open the genetic algorithm tool
gatool
from the File menu select Import Problem... and choose gaproblem2 in the window that opens.
Now, run the tool and wait for the iterations to stop.
The gatool enables you to change hundreds of parameters, so you can trade speed for precision in the selected output.
The resulting vector is the list of indices that you have to keep in the original matrix so A(garesults.x,garesults.x) is the matrix with only the desired persons.
If I have understood you problem statement, you have a N x N matrix M (which happens to be a correlation matrix), and you wish to find for integer n where 2 <= n < N, a n x n matrix m which minimises the sum over all elements of m which I denote f(m)?
In Matlab it is fairly easy and fast to obtain a sub-matrix of a matrix (see for example Removing rows and columns from matrix in Matlab), and the function f is relatively inexpensive to evaluate for n = 151. So why can't you implement an algorithm that solves this backwards dynamically in a program as below where I have sketched out the pseudocode:
function reduceM(M, n){
m = M
for (ii = N to n+1) {
for (jj = 1 to ii) {
val(jj) = f(m) where mhas column and row jj removed, f(X) being summation over all elements of X
}
JJ(ii) = jj s.t. val(jj) is smallest
m = m updated by removing column and row JJ(ii)
}
}
In the end you end up with an m of dimension n which is the solution to your problem and a vector JJ which contains the indices removed at each iteration (you should easily be able to convert these back to indices applicable to the full matrix M)
There are several approaches to finding an approximate solution (eg. quadratic programming on relaxed problem or greedy search), but finding the exact solution is an NP-hard problem.
Disclaimer: I'm not an expert on binary quadratic programming, and you may want to consult the academic literature for more sophisticated algorithms.
Mathematically equivalent formulation:
Your problem is equivalent to:
For some symmetric, positive semi-definite matrix S
minimize (over vector x) x'*S*x
subject to 0 <= x(i) <= 1 for all i
sum(x)==n
x(i) is either 1 or 0 for all i
This is a quadratic programming problem where the vector x is restricted to taking only binary values. Quadratic programming where the domain is restricted to a set of discrete values is called mixed integer quadratic programming (MIQP). The binary version is sometimes called Binary Quadratic Programming (BQP). The last restriction, that x is binary, makes the problem substantially more difficult; it destroys the problem's convexity!
Quick and dirty approach to finding an approximate answer:
If you don't need a precise solution, something to play around with might be a relaxed version of the problem: drop the binary constraint. If you drop the constraint that x(i) is either 1 or 0 for all i, then the problem becomes a trivial convex optimization problem and can be solved nearly instantaneously (eg. by Matlab's quadprog). You could try removing entries that, on the relaxed problem, quadprog assigns the lowest values in the x vector, but this does not truly solve the original problem!
Note also that the relaxed problem gives you a lower bound on the optimal value of the original problem. If your discretized version of the solution to the relaxed problem leads to a value for the objective function close to the lower bound, there may be a sense in which this ad-hoc solution can't be that far off from the true solution.
To solve the relaxed problem, you might try something like:
% k is number of observations to drop
n = size(S, 1);
Aeq = ones(1,n)
beq = n-k;
[x_relax, f_relax] = quadprog(S, zeros(n, 1), [], [], Aeq, beq, zeros(n, 1), ones(n, 1));
f_relax = f_relax * 2; % Quadprog solves .5 * x' * S * x... so mult by 2
temp = sort(x_relax);
cutoff = temp(k);
x_approx = ones(n, 1);
x_approx(x_relax <= cutoff) = 0;
f_approx = x_approx' * S * x_approx;
I'm curious how good x_approx is? This doesn't solve your problem, but it might not be horrible! Note that f_relax is a lower bound on the solution to the original problem.
Software to solve your exact problem
You should check out this link and go down to the section on Mixed Integer Quadratic Programming (MIQP). It looks to me that Gurobi can solve problems of your type. Another list of solvers is here.
Working on a suggestion from Matthew Gunn and also some advice at the Gurobi forums, I came up with the following function. It seems to work pretty well.
I will award it the answer, but if someone can come up with code that works better I'll remove the tick from this answer and place it on their answer instead.
function [ values ] = the_optimal_method( CM , num_to_keep)
%the_iterative_method Takes correlation matrix CM and number to keep, returns list of people who should be kicked out
N = size(CM,1);
clear model;
names = strseq('x',[1:N]);
model.varnames = names;
model.Q = sparse(CM); % Gurobi needs a sparse matrix as input
model.A = sparse(ones(1,N));
model.obj = zeros(1,N);
model.rhs = num_to_keep;
model.sense = '=';
model.vtype = 'B';
gurobi_write(model, 'qp.mps');
results = gurobi(model);
values = results.x;
end
I have a vector A with size of 54000 x 1 and vector B with size of 54000 x 1 which is standard deviation of elements of A. From the other side, I have vector C with size of 300000 x 1. Now I want to find that each element of vector C correspondences to which row of vector A with accepted range 3*standard deviation? I have written the below code and it works fine for small vector but for large vector such that I have it is too too slow!!
for i=1:length(A)
L=A(i,1)- 3*B(i,1);
U=A(i,1)+ 3*B(i,1);
inds{i,1} = not(abs(sign(sign(L - C) + sign(U - C))));
end
Does anybody know how can I make this code faster or does anybody know another solution? THX.
MATLAB is an acronym for Matrix Laboratory and was developed to simply and speed up matrix(vector) calculations. Compared to C or any other programming language you can often skip the for loops when working with matrices. For your code you should be able to skip the for loop and do it as this:
L = A - 3*STD;
U = A + 3*STD;
inds = not(abs(sign(sign(L - C) + sign(U - C))));
And remember that i means the complex number in Matlab. Don't know if it affects speed though.
Edit:
Getting the result as a cell:
inds = num2cell(inds)
I need to create a polynomial of the form:
P(x) = q(1,1) + q(2,2)(x-z(1)) + q(3,3)(x-z(1))(x-z(2)) + --- + q(2n, 2n)(x-z(1))(x-z(2))...(x-z(2n)) NOTE: The indices of the equation have been shifted to accomodate MATLAB.
in MATLAB. Consult this link here specifically slides 15 and 16.
I have the matrix Q filled, so I have the diagonal, and I also have z(1:2n) filled.
I'm having a hard time figuring out a way to create a polynomial that I can graph this polynomial. I've tried to use a for loop to append each term to P(x), but it doesn't operate the way I thought it would.
So far, my code will calculate the coefficients (presented as Q(0,0) -> Q(2n+1, 2n+1) in the problem above) without a problem.
I'm having an issue with the construction of a degree n polynomial of the form described above. Plotting makes more sense now, create a vector x with evaluative values, and then run them through the polynomial "function" and plot the x vector against the resulting vector.
So I just need to create this polynomial.
I would use diag and cumprod to help you accomplish this. First use diag to extract the diagonals of your matrix Q. After, use cumprod to generate a vector of cumulative products.
How cumprod works on a vector is that for each element in the vector, the i'th element collects products from 1 up to the i'th element. As an example, if we had a vector V = [1 2 3 4 5], cumprod(V) would produce [1 2 6 24 120]. The 4th element (as an example) would be 1*2*3*4, representing the products from the 1st to the 4th element.
As such, this is the code that I would do:
qdiag = diag(Q);
xMinusZ = x - z; % Takes z and does x - z for every element in z
cumProdRes = cumprod(xMinusZ);
P = sum(qdiag .* [1;cumProdRes(1:end-1)]);
P should give you P(x) that you desired. Make sure that z is a column vector to make it compatible with the diagonals extracted from Q.
NB: I believe there is a typo in your equation. The last term of your equation (going with your convention) should have (x-z(2n-1)) and not (x-z(2n)). This is because the first term in your equation does not have z.
Here's an example. Let's suppose Q is defined
Q = [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16];
The vector z is:
z = [4;3;2;1];
Let's evaluate the function at x = 2
Extracting the diagonals of Q should give us Q = [1;6;11;16]. Subtract x from every element of z should give us:
xMinusZ = [-2;-1;0;1];
Using the equation that you have above, we have:
P = 1 + 6*(-2) + 11*(-2)*(-1) + 16*(-2)*(-1)*(0) = 11
This is what the code should give.
What if we want to do this for more than one value of x?
As you have stated in your post, you want to evaluate this for a series of x values. As such, you need to modify the code so that it looks like this (make sure that x is a column vector):
qdiag = diag(Q);
xMinusZ = repmat(x,1,length(z)) - repmat(z',length(z),1);
cumProdRes = cumprod(xMinusZ,2);
P = sum(repmat(qdiag',length(z),1).*[ones(length(z),1) cumProdRes(:,1:end-1)],2);
P should now give you a vector of outputs, and so if you want to plot this, simply do plot(x,P);