Imagine you have the following formula:
a=4*b*c^2
Is there any way in Matlab to program this is in a way that the if 2 of 3 variables are provided, Matlab will solve and provide the missing one?
Because the only alternative I am seeing is using switch-case and solving the equation myself.
if isempty(a)
switchVar=1
elseif isempty(b)
switchVar=2;
else
switchVar=3;
end
switch switchVar
case 1
a=4*b*c^2;
case 2
b=a/4/c^2;
case 3
c=sqrt(a/4/b);
end
Thanky you very much in advance!
For a numeric (rather than symbolic) solution...
You can do this with some faffing around and anonymous functions. See the comments for details:
% For the target function: 0 = 4*b*c - a
% Let x = [a,b,c]
% Define the values we know about, i.e. not "c"
% You could put any values in for the known variables, and NaN for the unknown.
x0 = [5, 10, NaN];
% Define an index for the unknown, and then clear any NaNs
idx = isnan(x0);
x0(idx) = 0;
% Make sure we have 1 unknown
assert( nnz( idx ) == 1 );
% Define a function which handles which elements of "x"
% actually influence the equation
X = #(x,ii) ( x*idx(ii) + x0(ii)*(~idx(ii)) );
% Define the function to solve, 0 = 4*b*c - a = 4*x(2)*x(3)^2 - x(1) = ...
f = #(x) 4 * X(x,2) * X(x,3).^2 - X(x,1);
% Solve using fzero. Note this requires an initial guess
x0(idx) = fzero( f, 1 );
We can check these results are correct by plotting the function for a range of c values, and checking the intersection with the x-axis aligns to the output x0(3):
c = -1:0.01:1;
y = 4*x(2)*c.^2-x(1);
figure
hold on
plot(t,y);
plot(t,y.*0,'r');
plot(x0(3),0,'ok','markersize',10,'linewidth',2)
Note that there were 2 valid solutions, since this is a quadratic. The initial condition provided to fzero will largely dictate which solution is found.
Edit:
You can condense this down a bit with some tweaks to my earlier syntax:
% Define all initial conditions. This includes known variable values
% i.e. a = 5, b = 10
% As well as the initial guess for unknown variable values
% i.e. c = 1 (maybe? ish?)
x0 = [5, 10, 1];
% Specify the index of the unknown variable in x
idx = 3;
% Define the helper function which handles the influence of each variable
X = #(x,ii) x*(ii==idx) + x0(ii)*(ii~=idx);
% Define the function to solve, as before
f = #(x) 4 * X(x,2) * X(x,3).^2 - X(x,1);
% Solve
x0(idx) = fzero( f, x0(idx) )
This approach has the benefit that you can just change idx (and re-run the definition steps for X and f) to switch the variable of choice!
First, specify the given known variables
Then rewrite the equation as 0 = 4*b*c - a
Finally use solve to find the missing value
Code is as follows
syms a b c
% define known variable
a = 2; c = 5;
% equation rewritten
f = 4*b*c^2 - a == 0;
missing_value = solve(f);
Related
I am trying to implement a simplex algorithm following the rules I was given at my optimization course. The problem is
min c'*x s.t.
Ax = b
x >= 0
All vectors are assumes to be columns, ' denotes the transpose. The algorithm should also return the solution to dual LP. The rules to follow are:
Here, A_J denotes columns from A with indices in J and x_J, x_K denotes elements of vector x with indices in J or K respectively. Vector a_s is column s of matrix A.
Now I do not understand how this algorithm takes care of condition x >= 0, but I decided to give it a try and follow it step by step. I used Matlab for this and got the following code.
X = zeros(n, 1);
Y = zeros(m, 1);
% i. Choose starting basis J and K = {1,2,...,n} \ J
J = [4 5 6] % for our problem
K = setdiff(1:n, J)
% this while is for goto
while 1
% ii. Solve system A_J*\bar{x}_J = b.
xbar = A(:,J) \ b
% iii. Calculate value of criterion function with respect to current x_J.
fval = c(J)' * xbar
% iv. Calculate dual solution y from A_J^T*y = c_J.
y = A(:,J)' \ c(J)
% v. Calculate \bar{c}^T = c_K^T - u^T A_K. If \bar{c}^T >= 0, we have
% found the optimal solution. If not, select the smallest s \in K, such
% that c_s < 0. Variable x_s enters basis.
cbar = c(K)' - c(J)' * inv(A(:,J)) * A(:,K)
cbar = cbar'
tmp = findnegative(cbar)
if tmp == -1 % we have found the optimal solution since cbar >= 0
X(J) = xbar;
Y = y;
FVAL = fval;
return
end
s = findnegative(c, K) %x_s enters basis
% vi. Solve system A_J*\bar{a} = a_s. If \bar{a} <= 0, then the problem is
% unbounded.
abar = A(:,J) \ A(:,s)
if findpositive(abar) == -1 % we failed to find positive number
disp('The problem is unbounded.')
return;
end
% vii. Calculate v = \bar{x}_J / \bar{a} and find the smallest rho \in J,
% such that v_rho > 0. Variable x_rho exits basis.
v = xbar ./ abar
rho = J(findpositive(v))
% viii. Update J and K and goto ii.
J = setdiff(J, rho)
J = union(J, s)
K = setdiff(K, s)
K = union(K, rho)
end
Functions findpositive(x) and findnegative(x, S) return the first index of positive or negative value in x. S is the set of indices, over which we look at. If S is omitted, whole vector is checked. Semicolons are omitted for debugging purposes.
The problem I tested this code on is
c = [-3 -1 -3 zeros(1,3)];
A = [2 1 1; 1 2 3; 2 2 1];
A = [A eye(3)];
b = [2; 5; 6];
The reason for zeros(1,3) and eye(3) is that the problem is inequalities and we need slack variables. I have set starting basis to [4 5 6] because the notes say that starting basis should be set to slack variables.
Now, what happens during execution is that on first run of while, variable with index 1 enters basis (in Matlab, indices go from 1 on) and 4 exits it and that is reasonable. On the second run, 2 enters the basis (since it is the smallest index such that c(idx) < 0 and 1 leaves it. But now on the next iteration, 1 enters basis again and I understand why it enters, because it is the smallest index, such that c(idx) < 0. But here the looping starts. I assume that should not have happened, but following the rules I cannot see how to prevent this.
I guess that there has to be something wrong with my interpretation of the notes but I just cannot see where I am wrong. I also remember that when we solved LP on the paper, we were updating our subjective function on each go, since when a variable entered basis, we removed it from the subjective function and expressed that variable in subj. function with the expression from one of the equalities, but I assume that is different algorithm.
Any remarks or help will be highly appreciated.
The problem has been solved. Turned out that the point 7 in the notes was wrong. Instead, point 7 should be
I am trying to compute the value of this integral using Matlab
Here the other parameters have been defined or computed in the earlier part of the program as follows
N = 2;
sigma = [0.01 0.1];
l = [15];
meu = 4*pi*10^(-7);
f = logspace ( 1, 6, 500);
w=2*pi.*f;
for j = 1 : length(f)
q2(j)= sqrt(sqrt(-1)*2*pi*f(j)*meu*sigma(2));
q1(j)= sqrt(sqrt(-1)*2*pi*f(j)*meu*sigma(1));
C2(j)= 1/(q2(j));
C1(j)= (q1(j)*C2(j) + tanh(q1(j)*l))/(q1(j)*(1+q1(j)*C2(j)*tanh(q1(j)*l)));
Z(j) = sqrt(-1)*2*pi*f(j)*C1(j);
Apprho(j) = meu*(1/(2*pi*f(j))*(abs(Z(j))^2));
Phi(j) = atan(imag(Z(j))/real(Z(j)));
end
%integration part
c1=w./(2*pi);
rho0=1;
fun = #(x) log(Apprho(x)/rho0)/(x.^2-w^2);
c2= integral(fun,0,Inf);
phin=pi/4-c1.*c2;
I am getting an error like this
could anyone help and tell me where i am going wrong.thanks in advance
Define Apprho in a separate *.m function file, instead of storing it in an array:
function [ result ] = Apprho(x)
%
% Calculate f and Z based on input argument x
%
% ...
%
meu = 4*pi*10^(-7);
result = meu*(1/(2*pi*f)*(abs(Z)^2));
end
How you calculate f and Z is up to you.
MATLAB's integral works by calling the function (in this case, Apprho) repeatedly at many different x values. The x values called by integral don't necessarily correspond to the 1: length(f) values used in your original code, which is why you received errors.
I have a function replace_me which is defined like as: function w = replace_me(v,a,b,c). The first input argument v is a vector, while a, b, and c are all scalars. The function replaces every element of v that is equal to a with b and c. For example, the command
x = replace_me([1 2 3],2,4,5); returns x as [1 4 5 3].
The code that I have created is
function w = replace_me(v,a,b,c)
[row,column]=size(v);
new_col=column+1;
w=(row:new_col);
for n=(1:column)
if a==v(n)
v(n)=b;
o=n;
d=n-1;
u=n+1;
for z=1:d
w(z)=v(z);
end
for z=u:column
w(z+1)=v(z);
end
w(o)=b;
w(o+1)=c;
end
end
end
It works perfectly fine for x = replace_me([1 2 3],2,4,5); I get required output but when I try x = replace_me([1 2 3], 4, 4, 5) my function fails.
To resolve this problem I want to use an if else statements having conditions that if a is equal to any element of vector v we would follow the above equation else it returns back the vector.
I tried to use this as if condition but it didn't worked
if v(1:column)==a
Any ideas
I'm not entirely sure if I understand what you are trying to achieve, but form what I understand you're looking for something like this:
function [v] = replace_me(v,a,b,c)
v = reshape(v,numel(v),1); % Ensure that v is always a column vector
tol = 0.001;
aPos = find( abs(v-a) < tol ); % Used tol to avoid numerical issues as mentioned by excaza
for i=numel(aPos):-1:1 % Loop backwards since the indices change when inserting elements
index = aPos(i);
v = [v(1:index-1); b; c; v(index+1:end)];
end
end
function w = move_me(v,a)
if nargin == 2
w=v(v~=a);
w(end+1:end+(length(v)-length(w)))=a;
elseif isscalar(v)
w=v;
else
w=v(v~=0);
w(end+1)=0;
end
end
It has been very difficult to use google, MATLAB documentation, I've spent a few hours, and I cannot learn how to
x = 1
y = x
x = 10
y
ans = 10
what happens instead is:
x = 1
y = x
x = 10
y
ans = 1
The value of x is stored into y. But I want to dynamically update the value of y to equal x.
What operation do I use to do this?
Thanks.M
Matlab is 99% a pass-by-value environment, which is what you have just demonstrated. The 1% which is pass-by-reference is handles, either handle graphics (not relevant here) or handle classes, which are pretty close to what you want.
To use a handle class to do what you describe, put the following into a file call RefValue.
classdef RefValue < handle
properties
data = [];
end
end
This creates a "handle" class, with a single property called "data".
Now you can try:
x = RefValue;
x.data = 1;
y = x;
x.data = 10;
disp(y.data) %Displays 10.
you can try something of the following;
x=10;
y='x'
y
y =
x
eval(y)
x =
10
You can also define an implicit handle on x by defining a function on y and referring to it:
x = 1;
y = #(x) x;
y(x) % displays 1
x = 10;
y(x) % displays 10
In MATLAB, this is not possible. However, there are many ways to get similar behavior. For example, you could have an array a = [1, 5, 3, 1] and then index it by x and y. For x = 2, you could assign a(x) = 7, y = x, and once you change a(x) = 4, a(y) == 4.
So indexing may be the fastest way to emulate references, but if you want some elegant solution, you could go through symbolic variables as #natan points out. What's important to take from this is that there are no pointers in MATLAB.
I want to vectorize the following MATLAB code. I think it must be simple but I'm finding it confusing nevertheless.
r = some constant less than m or n
[m,n] = size(C);
S = zeros(m-r,n-r);
for i=1:m-r+1
for j=1:n-r+1
S(i,j) = sum(diag(C(i:i+r-1,j:j+r-1)));
end
end
The code calculates a table of scores, S, for a dynamic programming algorithm, from another score table, C.
The diagonal summing is to generate scores for individual pieces of the data used to generate C, for all possible pieces (of size r).
Thanks in advance for any answers! Sorry if this one should be obvious...
Note
The built-in conv2 turned out to be faster than convnfft, because my eye(r) is quite small ( 5 <= r <= 20 ). convnfft.m states that r should be > 20 for any benefit to manifest.
If I understand correctly, you're trying to calculate the diagonal sum of every subarray of C, where you have removed the last row and column of C (if you should not remove the row/col, you need to loop to m-r+1, and you need to pass the entire array C to the function in my solution below).
You can do this operation via a convolution, like so:
S = conv2(C(1:end-1,1:end-1),eye(r),'valid');
If C and r are large, you may want to have a look at CONVNFFT from the Matlab File Exchange to speed up calculations.
Based on the idea of JS, and as Jonas pointed out in the comments, this can be done in two lines using IM2COL with some array manipulation:
B = im2col(C, [r r], 'sliding');
S = reshape( sum(B(1:r+1:end,:)), size(C)-r+1 );
Basically B contains the elements of all sliding blocks of size r-by-r over the matrix C. Then we take the elements on the diagonal of each of these blocks B(1:r+1:end,:), compute their sum, and reshape the result to the expected size.
Comparing this to the convolution-based solution by Jonas, this does not perform any matrix multiplication, only indexing...
I would think you might need to rearrange C into a 3D matrix before summing it along one of the dimensions. I'll post with an answer shortly.
EDIT
I didn't manage to find a way to vectorise it cleanly, but I did find the function accumarray, which might be of some help. I'll look at it in more detail when I am home.
EDIT#2
Found a simpler solution by using linear indexing, but this could be memory-intensive.
At C(1,1), the indexes we want to sum are 1+[0, m+1, 2*m+2, 3*m+3, 4*m+4, ... ], or (0:r-1)+(0:m:(r-1)*m)
sum_ind = (0:r-1)+(0:m:(r-1)*m);
create S_offset, an (m-r) by (n-r) by r matrix, such that S_offset(:,:,1) = 0, S_offset(:,:,2) = m+1, S_offset(:,:,3) = 2*m+2, and so on.
S_offset = permute(repmat( sum_ind, [m-r, 1, n-r] ), [1, 3, 2]);
create S_base, a matrix of base array addresses from which the offset will be calculated.
S_base = reshape(1:m*n,[m n]);
S_base = repmat(S_base(1:m-r,1:n-r), [1, 1, r]);
Finally, use S_base+S_offset to address the values of C.
S = sum(C(S_base+S_offset), 3);
You can, of course, use bsxfun and other methods to make it more efficient; here I chose to lay it out for clarity. I have yet to benchmark this to see how it compares with the double-loop method though; I need to head home for dinner first!
Is this what you're looking for? This function adds the diagonals and puts them into a vector similar to how the function 'sum' adds up all of the columns in a matrix and puts them into a vector.
function [diagSum] = diagSumCalc(squareMatrix, LLUR0_ULLR1)
%
% Input: squareMatrix: A square matrix.
% LLUR0_ULLR1: LowerLeft to UpperRight addition = 0
% UpperLeft to LowerRight addition = 1
%
% Output: diagSum: A vector of the sum of the diagnols of the matrix.
%
% Example:
%
% >> squareMatrix = [1 2 3;
% 4 5 6;
% 7 8 9];
%
% >> diagSum = diagSumCalc(squareMatrix, 0);
%
% diagSum =
%
% 1 6 15 14 9
%
% >> diagSum = diagSumCalc(squareMatrix, 1);
%
% diagSum =
%
% 7 12 15 8 3
%
% Written by M. Phillips
% Oct. 16th, 2013
% MIT Open Source Copywrite
% Contact mphillips#hmc.edu fmi.
%
if (nargin < 2)
disp('Error on input. Needs two inputs.');
return;
end
if (LLUR0_ULLR1 ~= 0 && LLUR0_ULLR1~= 1)
disp('Error on input. Only accepts 0 or 1 as input for second condition.');
return;
end
[M, N] = size(squareMatrix);
if (M ~= N)
disp('Error on input. Only accepts a square matrix as input.');
return;
end
diagSum = zeros(1, M+N-1);
if LLUR0_ULLR1 == 1
squareMatrix = rot90(squareMatrix, -1);
end
for i = 1:length(diagSum)
if i <= M
countUp = 1;
countDown = i;
while countDown ~= 0
diagSum(i) = squareMatrix(countUp, countDown) + diagSum(i);
countUp = countUp+1;
countDown = countDown-1;
end
end
if i > M
countUp = i-M+1;
countDown = M;
while countUp ~= M+1
diagSum(i) = squareMatrix(countUp, countDown) + diagSum(i);
countUp = countUp+1;
countDown = countDown-1;
end
end
end
Cheers