Good morning everybody
I work to develop mathematical model to solve one of the industrial engineering problem but I have a problem in the write of the MATLAB code. So I simplified this problem in the following code. I need to merge all the result of X in one matrix after the for function used to use it in the next step (in this simple case this matrix will be 40*3)
LIST=randi([0,1],[4,3]);
for i = 1:10
j=i
V=randi([0,1],[4,3]);
for m = 1:4
for n = 1:2
if V(m,n)== 1;
X(m,n) = LIST(m,n);
elseif V(m,n)== 0;
X(m,n) = 2;
end
end
end
for m = 1:4
for n = 3
if V(m,n)== 1;
X(m,n) = LIST(m,n);
elseif V(m,n)== 0;
X(m,n) = 3;
end
end
end
X
end
Thank you for your time and your consideration
At each iteration of the outer loop (for i = 1:10) the values in the X matrix are overwritten.
In order to store all the values you need to increment the row value of the X matrix of the value of max limit of the second loop (for m = 1:4) that is 4.
You can do it by modifying the indexing of the X matrix as follows:
X(m+(i-1)*4,n)
You can make your script more "general" by identifying the limit evaluating it as the number of rows in the LIST matrix using the function size function:
[n_row,n_col]=size(LIST)
In this way
at the first iteration of the outer loop, the row index will range from 1 to 4.
at the second iteration itr will range from 1+(2-1)*4 to 4+(2-1)*4 that is from 5 to 8
and so on
This is the updated code
LIST=randi([0,1],[4,3]);
[n_row,n_col]=size(LIST)
for i = 1:10
j=i
V=randi([0,1],[4,3]);
% for m = 1:4
for m = 1:n_row
for n = 1:2
if V(m,n)== 1;
% X(m,n) = LIST(m,n);
X(m+(i-1)*n_row,n) = LIST(m,n);
elseif V(m,n)== 0;
% X(m,n) = 2
X(m+(i-1)*n_row,n) = 2;
end
end
end
% for m = 1:4
for m = 1:n_row
for n = 3
if V(m,n)== 1;
% X(m,n) = LIST(m,n)
X(m+(i-1)*n_row,n) = LIST(m,n);
elseif V(m,n)== 0;
% X(m,n) = 3;
X(m+(i-1)*n_row,n) = 3;
end
end
end
X
end
Hope this helps.
Qapla'
Related
I am trying to implement this alghoritm for finding a new constraint:
In my case we take only 3 natural numbers i.e 1,2, 3.
The sets associated with those natural numbers are M1, M2 and M3. Instead of the Newton Method in II(2), I chose a solver provided by Matlab fmincon.
Here is my code that is not working!
function[s_new]= checking2(M1,M2,M3,x)
M1=linspace(0,1,10)';
M2=linspace(0,1,100)';
M3=linspace(0,1,1000)';
bool1=0;
eta = 10^-8;
pocz=[];
max=-100;
x = [0.1,0.1]'; % warunek poczÄ…tkowy
A = [];
b = [];
Aeq = [];
beq = [];
Set=[0,1];
g = #(x,s) 5*x(1).^2.*sin(pi.*sqrt(s))./(1+s.^2) - x(2);
g_new = #(s) -g(x,s);
for i=1:length(M1)
if g(x,M1(i,:))>eta
s_new=M1(i,:);
bool1=1;
end
end
if ~bool1
for i=1:length(M1)
if g(x,M1(i,:))>max
pocz=M1(i,:);
max=g(x,M1(i,:));
end
end
if max<-eta
bool1=1;
end
end
if ~bool1
s_maybe = fmincon(g_new,pocz,A,b,Aeq,beq,min(Set),max(Set));
if g(x,s_maybe)>eta
s_new=s_maybe;
bool1=1;
end
end
if ~bool1
for i=1:length(M2)
if g(x,M2(i,:))>eta
s_new=M2(i,:);
bool1=1;
end
end
end
if ~bool1
for i=1:length(M2)
if g(x,M2(i,:))>max
pocz=M2(i,:);
max=g(x,M2(i,:));
end
end
if max<-eta
bool1=1;
end
end
if ~bool1
s_maybe = fmincon(g_new,pocz,A,b,Aeq,beq,min(Set),max(Set));
if g(x,s_maybe)>eta
s_new=s_maybe;
bool1=1;
end
end
if ~bool1
for i=1:length(M3)
if g(x,M3(i,:))>eta
s_new=M3(i,:);
bool1=1;
end
end
end
if ~bool1
s_new = 1;
end
disp(s_new);
The problem is:
Undefined function or variable 's_new'.
Error in checking2 (line 70)
disp(s_new);
So basically everything might be wrong, but I suppose it is something with fmincon.
EDIT:
The purpose of the alghoritm is to find a minimum of an objective function f(x), satisfying all the constraints g(x,s)<=0 for all s in S, where S is an infinite set (some interval in our case).
What my alghoritm does, at first it takes some finite subset of S and calculates the minimum of f on this set, then I am trying to update S with some s_new. This alghoritm that I am trying to implement is exactly the procedure for creating s_new. Then if it works properly, I will add s_new to my subset and calculate the minimum on the new set, and so on until g(x,s)<=eta, where eta is a small number.
I rewrite the algorithm, read through the comments
clc
clear
lb = 0;
ub = 1;
% Given
l = 3;
M1=linspace(lb,ub,10)';
M2=linspace(lb,ub,100)';
M3=linspace(lb,ub,1000)';
% one boolean value for each Matrix
bool = zeros(1,3);
eta = 10^-8;
% Used as fmincon initial starting guess
pocz = nan;
% Used to store the new finding s that fits all the conditions
s_new = nan;
% Fixed x
x = [0.1,0]';
% fmincon linear constraints
A = [];
b = [];
Aeq = [];
beq = [];
% Main function
g = #(x,s) 5*x(1).^2*sin(pi*sqrt(s))/(1+s.^2) - x(2);
% Optimization concerns s only, don't include x as x is fixed
g_new = #(s) -g(x,s);
% Assuming the maximum is reached at the upper bound, used in(II)(2)
max_s = ub;
maxfun = g(x, max_s);
% Use a cell, for each iteration use a specific matrix M
M = {M1, M2, M3};
for j = 1: length(M)
% used in (II)(1)
check = 0;
step = 1;
% (I) step 1
for i = 1:length(M{j})
% Stopping criteria
if g(x, M{j}(i)) > eta
s_new = M{j}(i);
bool(j) = 1;
break;
else
% Function maximum value for next step (II)
if maxfun < g(x, M{j}(i))
maxfun = g(x, M{j}(i));
% To be used in fmincon as pocz
max_s = M{j}(i);
end
end
% To be used in (II)(1)
if maxfun < -eta
check = 1;
end
end
% End of (I)
% Put (II)(1) here step 2
if ~bool(j) && check
step = step + 1;
% Stopping criteria
if step >= l
disp('S_new not defined');
break;
end
% otherwise go to the next M
end
% (II)(2) step 3
if ~bool(j)
step = step + 1;
if maxfun >= -eta && maxfun <= eta
pocz = max_s;
bool(j) = 1;
end
end
%% EDIT: if bool(j) changed to if ~bool(j)
% (II)(2) Continue
if ~bool(j)
s_maybe = fmincon(g_new,pocz,A,b,Aeq,beq,lb,ub);
% End of (II)(2)
% (II)(2)-1 step 4
step = step + 1;
if g(x, s_maybe) > eta
s_new = s_maybe;
bool(j) = 1;
end
% End of (II)(2)-1
end
% Put (II)(2) here step 5
if ~bool(j)
step = step + 1;
% Stopping criteria
if step >= l
disp('S_new not defined');
break;
end
% otherwise go to the next M
end
end
Working on an assignment in MATLAB and I can't seem to figure this problem out due to the arithmetic, I've been trying it for about 6 hours now
I need to create a loop that accepts user input > 1 (done) and loops through the following (m is input)
t1 = sqrt(m);
t2 = sqrt(m-sqrt(m));
t3 = sqrt(m-sqrt(m+sqrt(m)))
t4 = sqrt(m-sqrt(m+sqrt(m-sqrt(m))))
t5 = sqrt(m-sqrt(m+sqrt(m-sqrt(m+sqrt(m)))))
and so on until the new t value minus the old t value is < 1e-12
My current code is as follows
%Nested Radicals
clear all;
clc;
%User input for m
m = input('Please enter a value for m: ');
%Error message if m is less than 1
if m <= 1
fprintf('ERROR: m must be greater than 1\n')
m = input('Please enter a value for m: ');
end
%Error message if m is not an integer
if mod(m,1) ~= 0
fprintf('m must be an integer\n')
m = input('Please enter a value for m: \n');
end
%Nested things
t_old = m;
t_new = sqrt(m);
varsign = -1;
index = 1;
loop = true;
endResult = 1e-12;
sqrts = [sqrt(m), sqrt(m-sqrt(m)), sqrt(m-sqrt(m+sqrt(m))), sqrt(m-sqrt(m+sqrt(m-sqrt(m)))), sqrt(m-sqrt(m+sqrt(m-sqrt(m+sqrt(m)))))];
fprintf('m = %d\n',m)
fprintf('t1 = %14.13f\n', t_new')
while loop
if index ~= 1
curResult = abs(sqrts(1,index) - sqrts(1, index-1));
else
curResult = abs(sqrts(1, index));
end
if curResult > endResult
if index < 5
t_new = sqrts(1, index+1);
else
t_new = sqrts(1, index);
loop=false;
end
if index
fprintf('t%d = %14.13f\n', index, t_new)
end
else
fprintf('t%d = %14.13f\n', index, t_new);
break;
end
index = index + 1;
if index > 50
fprintf('t%d = %14.13f\n', index, t_new);
break;
end
end
Unless I'm very much mistaken, you can write the expression for t(n) as follows:
t(n) = sqrt(m-sqrt(m+t(n-2));
which makes it a lot easier to loop:
%Nested Radicals
clear all;
clc;
%User input for m
m = input('Please enter a value for m: ');
%Error message if m is less than 1
if m <= 1
fprintf('ERROR: m must be greater than 1\n')
m = input('Please enter a value for m:');
end
%Error message if m is not an integer
if mod(m,1) ~= 0
fprintf('m must be an integer\n')
m = input('Please enter a value for m:');
end
%Nested things
t_old = sqrt(m);
t_new = sqrt(m-sqrt(m));
threshold = 1e-12;
k = 3;
while abs(t_new - t_old) >= threshold
temp = sqrt(m-sqrt(m+t_old));
t_old = t_new;
t_new = temp;
k = k+1;
end
fprintf('t%d = %14.13f\n', k-2, t_old);
fprintf('t%d = %14.13f\n', k-1, t_new);
fprintf('t%d - t%d = %14.13f\n', k-2, k-1, t_old - t_new);
which gives for m=9 for example:
Please enter a value for m: 9
t17 = 2.3722813232696
t18 = 2.3722813232691
t17 - t18 = 0.0000000000005
I'm not sure what you're trying to do with the sqrts variable, you should be calculating each step on the fly in your loop, since you can't possibly know how deep you need to go
m = 5; % Get m however you want to
n = 0; % Iteration counter
tol = 1e-12 % Tolerance at which to stop
dt = 1; % initialise to some value greater than 'tol' so we can start the loop
% Loop until tn is less than tolerance. Would be sensible to add a condition on n,
% like "while tn > tol && n < 1000", so the loop doesn't go on for years if the
% condition takes a trillion loops to be satisfied
while dt > tol
% Set the value of the deepest nested expression
tn = sqrt(m);
% We know how many times take sqrt, so for loop our way out of the nested function
% Initially we want the sign to be -1, then +1, -1, ...
% This is achieved using ((-1)^ii)
for ii = 1:n
tn = sqrt(m + ((-1)^ii)*tn); % Calculate next nested function out
end
% Increment iteration number
n = n + 1;
dt = abs( t_old - tn );
t_old = tn;
end
I've not done any analysis on your function, so have no idea if it's guaranteed to converge to some value <1e-12. If it isn't then you definitely need to add some maximum iteration condition as I suggest in the comments above.
How can I speed up the following MATLAB code, using vectorization? Right now the single line in the loop is taking hours to run for the case upper = 1e7.
Here is the commented code with sample output:
p = 8;
lower = 1;
upper = 1e1;
n = setdiff(lower:upper,primes(upper)); % contains composite numbers between lower + upper
x = ones(length(n),p); % Preallocated 2-D array of ones
% This loop stores the unique prime factors of each composite
% number from 1 to n, in each row of x. Since the rows will have
% varying lengths, the rows are padded with ones at the end.
for i = 1:length(n)
x(i,:) = [unique(factor(n(i))) ones(1,p-length(unique(factor(n(i)))))];
end
output:
x =
1 1 1 1 1 1 1 1
2 1 1 1 1 1 1 1
2 3 1 1 1 1 1 1
2 1 1 1 1 1 1 1
3 1 1 1 1 1 1 1
2 5 1 1 1 1 1 1
For example, the last row contains the prime factors of 10, if we ignore the ones. I have made the matrix 8 columns wide to account for the many prime factors of numbers up to 10 million.
Thanks for any help!
This is not vectorization, but this version of the loop will save about half of the time:
for k = 1:numel(n)
tmp = unique(factor(n(k)));
x(k,1:numel(tmp)) = tmp;
end
Here is a quick benchmark for this:
function t = getPrimeTime
lower = 1;
upper = 2.^(1:8);
t = zeros(numel(upper),2);
for k = 1:numel(upper)
n = setdiff(lower:upper(k),primes(upper(k))); % contains composite numbers between lower to upper
t(k,1) = timeit(#() getPrime1(n));
t(k,2) = timeit(#() getPrime2(n));
disp(k)
end
p = plot(log2(upper),log10(t));
p(1).Marker = 'o';
p(2).Marker = '*';
xlabel('log_2(range of numbers)')
ylabel('log(time (sec))')
legend({'getPrime1','getPrime2'})
end
function x = getPrime1(n) % the originel function
p = 8;
x = ones(length(n),p); % Preallocated 2-D array of ones
for k = 1:length(n)
x(k,:) = [unique(factor(n(k))) ones(1,p-length(unique(factor(n(k)))))];
end
end
function x = getPrime2(n)
p = 8;
x = ones(numel(n),p); % Preallocated 2-D array of ones
for k = 1:numel(n)
tmp = unique(factor(n(k)));
x(k,1:numel(tmp)) = tmp;
end
end
Here's another approach:
p = 8;
lower = 1;
upper = 1e1;
p = 8;
q = primes(upper);
n = setdiff(lower:upper, q);
x = bsxfun(#times, q, ~bsxfun(#mod, n(:), q));
x(~x) = inf;
x = sort(x,2);
x(isinf(x)) = 1;
x = [x ones(size(x,1), p-size(x,2))];
This seems to be faster than the other two options (but is uses more memory). Borrowing EBH's benchmarking code:
function t = getPrimeTime
lower = 1;
upper = 2.^(1:12);
t = zeros(numel(upper),3);
for k = 1:numel(upper)
n = setdiff(lower:upper(k),primes(upper(k)));
t(k,1) = timeit(#() getPrime1(n));
t(k,2) = timeit(#() getPrime2(n));
t(k,3) = timeit(#() getPrime3(n));
disp(k)
end
p = plot(log2(upper),log10(t));
p(1).Marker = 'o';
p(2).Marker = '*';
p(3).Marker = '^';
xlabel('log_2(range of numbers)')
ylabel('log(time (sec))')
legend({'getPrime1','getPrime2','getPrime3'})
grid on
end
function x = getPrime1(n) % the originel function
p = 8;
x = ones(length(n),p); % Preallocated 2-D array of ones
for k = 1:length(n)
x(k,:) = [unique(factor(n(k))) ones(1,p-length(unique(factor(n(k)))))];
end
end
function x = getPrime2(n)
p = 8;
x = ones(numel(n),p); % Preallocated 2-D array of ones
for k = 1:numel(n)
tmp = unique(factor(n(k)));
x(k,1:numel(tmp)) = tmp;
end
end
function x = getPrime3(n) % Approach in this answer
p = 8;
q = primes(max(n));
x = bsxfun(#times, q, ~bsxfun(#mod, n(:), q));
x(~x) = inf;
x = sort(x,2);
x(isinf(x)) = 1;
x = [x ones(size(x,1), p-size(x,2))];
end
How would you code this in MATLAB?
This is what I've tried, but it doesn't work quite right.
function x = my_jacobi(A,b, tot_it)
%Inputs:
%A: Matrix
%b: Vector
%tot_it: Number of iterations
%Output:
%:x The solution after tot_it iterations
n = length(A);
x = zeros(n,1);
for k = 1:tot_it
for j = 1:n
for i = 1:n
if (j ~= i)
x(i) = -((A(i,j)/A(i,i)) * x(j) + (b(i)/A(i,i)));
else
continue;
end
end
end
end
end
j is an iterator of a sum over each i, so you need to change their order. Also the formula has a sum and in your code you're not adding anything so that's another thing to consider. The last thing I see that you're omitting is that you should save the previous state of xbecause the right side of the formula needs it. You should try something like this:
function x = my_jacobi(A,b, tot_it)
%Inputs:
%A: Matrix
%b: Vector
%tot_it: Number of iterations
%Output:
%:x The solution after tot_it iterations
n = length(A);
x = zeros(n,1);
s = 0; %Auxiliar var to store the sum.
xold = x
for k = 1:tot_it
for i = 1:n
for j = 1:n
if (j ~= i)
s = s + (A(i,j)/A(i,i)) * xold(j);
else
continue;
end
end
x(i) = -s + b(i)/A(i,i);
s = 0;
end
xold = x;
end
end
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