How do I declare varibles whose outputs are expected from Matlab simply without so many rules, like for example in c++?

[Below is my code its meant to do as follows, as long as k<NoS, it shall form Equations that at the end of the day are equivalent to k =NoS. Afterwards, it should sum them up, and do fminsearch for Xcal Ycal Zcal and t0. My problem is however I declare the variables Xcal,Ycal,Zcal and t0 whether using syms Xcal, Ycal,Zcal, t0 or Xcal=(1,1,'double') ...etc. Matlab always displays the following "error Conversion to double from function_handle is not possible.". I am new to matlab, but I think there are just too many rules about multiplication and division. Whichever way I do it, there are always errors chiefly '.*' is not allowed for cell variables or sometimes '*' operand is not allowed for this kind of variable and so on and so forth. I've written a separate function on a different.m.file which can run, but I am just hoping to keep things simple and operate within one script, using an anonymous function as indicated. Also I am not very farmiliar with how to call functions or variables across scripts. I am not sure if what I am doing is correct or could someone let me know what I am not doing right thanks]

.....
while k<NoS;
    k=1:NoS;
 Xcal={};
 Ycal={};
 Zcal={};
 t0={};
    eqn=zeros(1,NoS);
eqn(k)=........
    @(Xcal,Ycal,Zcal,t0)(sqrt(xj*xj-2*xj*Xcal+Xcal*Xcal+yj*yj-2*yj*Ycal+Ycal*Ycal+zj*zj-2*zj*Zcal+Zcal*Zcal)-c*tj+c*t0)
x0=[500,1000,50,5];
s2=sum(eqn(k));
soln =fminsearch(s2,x0)  
 break;
end

.....

[Stephen23]

https://en.wikipedia.org/wiki/XY_problem

https://xyproblem.info

Please explain what you are trying to achieve.

[Alphonce Owayo]

I need to minimize the sum of eqn(k) which are k=NoS in number. So Matlab should run fminsearch with the x0 values provided as a starting point. I need Xcal, Ycal, Zcal and t0 as the outputs when the sum is minimal. The problem is however I try declaring Xcal Ycal Zcal or t0,I run into problems of multiplication or division etc,of which the rules are not clear to me either, Or the errors such as the ones stated. I say so because were it in c++ I'd simply say float Xcal and from thereon I can do whatever I want with it without being told, 'this operation is not allowed' etc. At some point it gets frustrating.

[Walter Roberson]

    @(Xcal,Ycal,Zcal,t0)(sqrt(xj*xj-2*xj*Xcal+Xcal*Xcal+yj*yj-2*yj*Ycal+Ycal*Ycal+zj*zj-2*zj*Zcal+Zcal*Zcal)-c*tj+c*t0)

does not appear to vary with k?

[Alphonce Owayo]

It does because I just highlighted the part of the code with the issue, but A1 B1, C1 and D1 are arrays of size(k) : so xj=B1(k), yj=C1(k), zj=D1(k) and tj=A1(k)

Answer 1

[Walter Roberson]

You are trying to construct a vector of function handles. You cannot do that in MATLAB. This is because in MATLAB, () indicates both indexing and function invocation. If you had a vector eqn of function handles, would eqn(1) indicate indexing to retrieve the first function handle, or would it indicate that each element of the vector was to be invoked with parameter value 1?

Because of this, function handles must be stored in cell arrays, and then it becomes clear that {} indexes which one.

You also try to sum() the function handles. What is your expectation when you add function handles? Especially ones that might not have the same number of parameters or the same parameter names or which might have bound in different values of variables

A=1
f1 = @(x)x+A
A=2
f2 = @(x)x+A

f1+f2

is what exactly? It could not be value, but if it is a function handle then it has to bind in two different A simultaneously.

[Alphonce Owayo]

Ok, I get you a little but simply put how then can I do an fmin search for sum(k) =sum( sqrt(xj*xj-2*xj*Xcal+Xcal*Xcal+yj*yj-2*yj*Ycal+Ycal*Ycal+zj*zj-2*zj*Zcal+Zcal*Zcal)-c*tj+c*t0)). These are 8 vector equations in this case. I need sum of the k vectors to be minimum, while I get Xcal Ycal Zcal and t0 as outputs. The rest of those variables are known from the experiment, some are constants. The size of all the k vectors are the same. In this case k is 8, though it should be any number. Thanks.

[Walter Roberson]

NoS = 8;

A1 = randi(9, NoS, 1);
B1 = randi(9, NoS, 1);
C1 = randi(9, NoS, 1);
D1 = randi(9, NoS, 1);
c =  randi(20)/20;

syms Xcal Ycal Zcal t0
xj = B1(:);
yj = C1(:);
zj = D1(:)

zj = 8×1

3 3 2 3 2 9 1 2

tj = A1(:);

eqns = (sqrt(xj.*xj-2*xj*Xcal+Xcal*Xcal+yj.*yj-2*yj*Ycal+Ycal*Ycal+zj.*zj-2*zj*Zcal+Zcal*Zcal)-c*tj+c*t0)

eqns = 

s2 = sum(eqns);
F = matlabFunction(s2, 'vars', {[Xcal, Ycal, Zcal, t0]}, 'file', 's2.m', 'optimize', false)

F = function_handle with value:
    @s2

x0=[500,1000,50,5];
opts = optimset('fminsearch');
opts.MaxFunEvals = 1e5;
opts.MaxIter = 1e5;
[best, fval] = fminsearch(F, x0, opts);

 
Exiting: Maximum number of function evaluations has been exceeded
         - increase MaxFunEvals option.
         Current function value: -127680188996661018802776340601204504679739116743394470870553834000417957237043402105822935744474258656269534029514841447587798499464869380073561353531424768.000000 

You might notice that the current function value is rather negative. This is correct: if you lok at your equations, you can see that you can get arbitrarily negative values when t0 is negative.

Furthermore, we can see that the t0 contribution is the same for every equation, and appears only outside the sqrt(). Therefore you can split the sum into two pieces, one contributed by (NoS * c) * t0 and the other piece with the square roots and the -c*sum(tj). You can mininimize the two pieces independently.

[Alphonce Owayo]

Works like a charm thanks, although the values are a little crazy. I guess its the way I framed my problem. Is there a way to confine the sums to the neighbourhood of zero.? I can see this way the minimum value is not restricted so the solutions are out of this world.

[Alphonce Owayo]

I used nonlinlsq instead along those similar ideas and I am pretty happy with my answers now, thanks alot for your input Mr.Walter Robertson, I appreciate.

[Walter Roberson]

If you use fmincon() instead of fminsearch() then you can put in bounds.

If you are trying to minimize the sum, then that can be done too:

format long g
NoS = 8;

A1 = randi(9, NoS, 1);
B1 = randi(9, NoS, 1);
C1 = randi(9, NoS, 1);
D1 = randi(9, NoS, 1);
c =  randi(20)/20;

syms Xcal Ycal Zcal t0
xj = B1(:);
yj = C1(:);
zj = D1(:)

zj = 8×1

4 8 9 5 5 1 2 7

tj = A1(:);

eqns = (sqrt(xj.*xj-2*xj*Xcal+Xcal*Xcal+yj.*yj-2*yj*Ycal+Ycal*Ycal+zj.*zj-2*zj*Zcal+Zcal*Zcal)-c*tj+c*t0)

eqns = 

s2 = sum(eqns);
F = matlabFunction(s2.^2, 'vars', {[Xcal, Ycal, Zcal, t0]}, 'file', 's2.m', 'optimize', false)

F = function_handle with value:
    @s2

x0=[500,1000,50,5];
opts = optimset('fminsearch');
opts.MaxFunEvals = 1e5;
opts.MaxIter = 1e5;
[best, fval] = fminsearch(F, x0, opts);
best

best = 1×4

5.28673078857011 6.11901306252565 5.51087556249405 -4.35285398925722

fval

fval = 
      1.56106356119392e-13

[Alphonce Owayo]

{x1 =

-0.281080673516821

y1 =

19.0054637291946

z1 =

-9.49765205419143

ts =

-7.05830717920742e-09}

These are the typical range of my values except I realised fval =

354229770.329462 is very high.

Is it because of my inputs ?

typical xj; -25 -30 25 20 25 28 -25 -20

typical yj; 45 20 55 15 15 45 25 50

typicalzj; 0 0 0 0 40 40 40 40

typical tj; 0.000244 -0.00049 0 0.000732 0.070068 0.001221 -0.00049 0.054443

c= 4.3*10^6

[Walter Roberson]

format long g
NoS = 8;


syms Xcal Ycal Zcal t0
xj = [-25   -30    25    20    25    28   -25   -20];
yj = [45    20    55    15    15    45    25    50];
zj = [0  0    0    0    40   40    40   40];
tj = [0.000244  -0.00049  0  0.000732  0.070068  0.001221  -0.00049  0.054443];
c= 4.3*10^6

c = 
     4300000

eqns = (sqrt(xj.*xj-2*xj*Xcal+Xcal*Xcal+yj.*yj-2*yj*Ycal+Ycal*Ycal+zj.*zj-2*zj*Zcal+Zcal*Zcal)-c*tj+c*t0)

eqns = 

s2 = sum(eqns);
F = matlabFunction(s2.^2, 'vars', {[Xcal, Ycal, Zcal, t0]}, 'file', 's2.m', 'optimize', false)

F = function_handle with value:
    @s2

x0=[1,2,3,4];
opts = optimset('fminsearch');
opts.MaxFunEvals = 1e5;
opts.MaxIter = 1e5;
[best, fval] = fminsearch(F, x0, opts);
best

best = 1×4

1.05770074278451 2.89375508854424 3.52918424078698 0.0157047849691125

fval

fval = 
      2.49306076911199e-05

[Alphonce Owayo]

Great thanks, this has been helpful a great deal.

[Alphonce Owayo]

Hi Walter Robertson (plus anyone else who can lend a helping hand), your inputs have been very helpful and I am quite satisfied with my solutions, except one more issue.Just so you have a better understanding of my issue. This is an acoustic emission experiment (incase you are unfarmiliar with it, just ignore; the point is I got lots of data points), we use sensors to record the time the signals are received during an experiment. In the previous discussions we have just been dealing with one point, so you can imagine I got lots of points 1.048million to be exact. From your input ideas which were quite insightful,nevertheless I realised for different signals I kept getting the same solution then I realized the problem is beyond local optimization solvers. I worked my way to the global otimization solvers. In particular I used simulated annealing whose results are perfect except extremely slow with the current speed it would take 3 years to solve the about 1.05million signals I have, this is just a single experiment I got more. I tried exploring other relatively faster tools like pattern search but from our knowledge of the problems the solutions from pattern search are way off. Apart from improving computer specs (which I think might only increase the efficiency but is limited) is there any other way or ideas to radically improve the calculation speed or efficiency. Thanks

[Walter Roberson]

I am not clear on how having multiple points affects the equations ?

[Alphonce Owayo]

The equation and the answers are fine, the problem is it takes a long time to calculate. For example to calculate about 800 data points it has taken well over 48 hours. You can imagine I have 1.05 million data points from just one experiment. Like is there a way to make Matlab faster?

[Alphonce Owayo]

The other way would be to filter out my data and just anayze a frelatively fewer points (not my preferred option), since I am hoping to get as much information from my analysis as much as possible.

[Walter Roberson]

You said that the above, where we found XCal, YCal, ZCal, t0, represented just one point, and that you now have 1.5 million data points. I am not at all clear how the equations generalize? Are you looking for N objects, so finding XCal(N,1), YCal(N,1), ZCalk(N,1), t0(N,1), based upon best fit for the 1.5 million inputs?

If you are, then that would probably be a clustering task, possibly with a non-euclidean distance matrix.

Do you need to determine the optimum number of objects based upon the 1.5 million points? (Because theory says that the optimum number of clusters is equal to the number of unique points, unless there is a per-object cost.)

[Alphonce Owayo]

You said that the above, where we found XCal, YCal, ZCal, t0, represented just one point, and that you now have 1.5 million data points. Yes this is correct, see the code below dp=ne=1.05 million. Its automatic it reads the input from an excel file specially organized, then outputs it in a text file. Everything is fine I am quite happy with the solutions except the speed of calculations.

prompt0 = 'enter the No. of Data points... dp=';
dp = input(prompt0)
prompt1 = 'enter the No. of Sensors..NoS=';
NoS = input(prompt1)
ne=dp;
if NoS < 1;
disp ('Error,please check your input');
else
disp ('Input looks good,please proceed');
end
%A = 1:NoS;
i=1
 i=1;ne;
while i<ne;

    for i=1:ne;
        
        NoS==NoS;

 % Input sensor signal receipt time
disp("Import the times the sensor received the signals from 1 to NoS--");
ATM=[];
x=importdata('C:\Users\Alphonce\Desktop\J C& G Submission\Book2.xlsx');

ATM = x(i,1:NoS);

disp('Press enter two times to exit from Senszpos--');
while true
   % str=input('','s');
    % if isempty(str)
    if NoS-1==NoS-1;
        break;
   end
    [row,~,errmsg]=sscanf(str,'%f');  % getting row in column vector
   if ~isempty(errmsg)
    error(errmsg);
    end
    ATM=[ATM row];   % column catenation 
   
end
ATM=ATM';    % transpose for desired result
disp(ATM);
A1=ATM
A2 = [A1];

% input sensor x position

disp("Import sensor x pos from 1 to NoS--");
Sensxpos=[];
x1=importdata('C:\Users\Alphonce\Desktop\J C& G Submission\Book2.xlsx');
Sensxpos = x1(1:NoS,NoS+1);
while true
    %str=input('','s');
   %if isempty(str)
        if NoS-1==NoS-1;
        break;
   end
    [row,~,errmsg]=sscanf(str,'%f');  % getting row in column vector
    if ~isempty(errmsg)
    error(errmsg);
    end
    Sensxpos=[Sensxpos row];   % column catenation   
   
end
Sensxpos=Sensxpos';    % transpose for desired result
disp(Sensxpos);
B=Sensxpos
B1 = [B];

% input sensor y position

disp("Import sensor y pos from 1 to NoS--");
Sensypos=[];
x2=importdata('C:\Users\Alphonce\Desktop\J C& G Submission\Book2.xlsx');
Sensypos = x2(1:NoS,NoS+2);

while true
    %str=input('','s');
    %if isempty(str)
        if NoS-1==NoS-1;
        break;
    end
    [row,~,errmsg]=sscanf(str,'%f');  % getting row in column vector
    if ~isempty(errmsg)
    error(errmsg);
    end
   Sensypos=[Sensypos row];   % column catenation 
   
end
Sensypos=Sensypos';    % transpose for desired result
disp(Sensypos);
C=Sensypos
C1 = [C];

% input sensor z position
disp("Import sensor z pos from 1 to NoS--");
Senszpos=[];
x3=importdata('C:\Users\Alphonce\Desktop\J C& G Submission\Book2.xlsx');
Senszpos = x3(1:NoS,NoS+3);

while true
    % str=input('','s');
    % if isempty(str)
         if NoS-1==NoS-1;
        break;
   end
    [row,~,errmsg]=sscanf(str,'%f');  % getting row in column vector
    if ~isempty(errmsg)
    error(errmsg);
    end
    Senszpos=[Senszpos row];   % column catenation 
   
end
Senszpos=Senszpos';    % transpose for desired result
disp(Senszpos);
D=Senszpos
D1 = [D];

% speed of sound in concrete. (longitudinal)

Ela = 40.4e15; den= 2190; % nu = 0.17; % E in Gpa den in Kg/m^3
c = sqrt(Ela/den);
c
syms Xcal Ycal Zcal t0
xj = B1(:);
yj = C1(:);
zj = D1(:)
tj = A1(:);
eqns = (sqrt(xj.*xj-2*xj*Xcal+Xcal*Xcal+yj.*yj-2*yj*Ycal+Ycal*Ycal+zj.*zj-2*zj*Zcal+Zcal*Zcal)-c*tj+c*t0)
s2 = sum(eqns);
F = matlabFunction(s2.^2, 'vars', {[Xcal, Ycal, Zcal, t0]}, 'file', 's2.m', 'optimize', false)
x0=[0,0,0,0.01];
lb=[-35,-35,-20,-10];
ub=[35,35,20,10];
options = optimoptions('simulannealbnd','Display','Iter');
x = simulannealbnd(F,x0,lb,ub)
[x,fval,exitflag,output] = simulannealbnd(F,x0,lb,ub,options)
x1=x(1,1);
y1=x(1,2);
z1=x(1,3);
ts=x(1,4);
x1
y1
z1
ts
 fid=fopen('myresult4.txt','a');
 fprintf(fid,'%s\t',x1);
 fprintf(fid,'%s\t',y1);
 fprintf(fid,'%s\t',z1);
 fprintf(fid,'%s\n',ts);
fclose(fid);
if i>=ne;
    break;
end
    end
end

[Alphonce Owayo]

For more clarity in this case NoS=8, so in the excel file the times the sensor receives the signals are arranged from colum 1 to column 8, while the sensor x, y ,z positions are arranged in columns 9, 10 and 11 respectively (sensor positions do not change so, they just span from row 1 to row 8 in this case); Columns 1 to 8, will span to 1.05 million. So the code, is able to load and calculate on its own once you click run. My biggest issue now is the effciency of the calculation.

[Walter Roberson]

Your code is confused

i=1
 i=1;ne;
while i<ne;
    for i=1:ne;
        
        NoS==NoS;

At the end of the for i loop, i will be left as the last value it is assigned. Unless you modify i within the loop, that means that after for i=1:ne that i will be left as ne and then while i<ne would be false because i would equal ne. What is the point of having that while loop?

What is the point of having

i=1;ne;

which assigns 1 to i and prints nothing because of the semi-colon, then recalls ne and does no calculation with it, and then prints nothing because of the semi-colon ? The line could easily be mistaken for

i=1:ne;

but that would not be correct code because then i<ne would fail in the while

        NoS==NoS;

What is the point of that?

if NoS-1==NoS-1;

What is the point of that test? The only time the test can fail is if NoS is empty, or NoS is NaN.

[Alphonce Owayo]

Well I am still learning some of these things, thanks for pointing this out.

The idea is to set up a counter i from 1 to 1.5million

such that as long as i is less than 1.5 million it executes the while loop .

NoS==NoS; This is to ensure that while executing the loop you do not have to keep entering the value of NoS indicated in the input prompt1, once you start running its automatic.

if NoS-1==NoS-1; This is meant to let it know when to stop creating the array. Since the size of the array is determined by NoS. (its meant to say that if Columns(NoS-1) = NoS-1); stop picking the values and continue executing the rest of the code. For example if NoS is 8, it will pick from col 1 to col 8, it requires a condition such that it doesnt proceed to pick to col.9....

[Alphonce Owayo]

At the end of the for i loop, i will be left as the last value it is assigned. Unless you modify i within the loop, that means that after for i=1:ne that i will be left as ne and then while i<ne would be false because i would equal ne. This idea is kind of my point though, such that it assigns i=ne=1,2,3,4,5............1.5 million in turn each time returning Xcal Ycal Zcal and t0. So at the end of the day I got 1.5Million( Xcal, Ycal,Zcal and t0)s.

[Alphonce Owayo]

i<ne means, its been executed we already have its Xcal Ycal zcal and t0 recorded so its fine. This is why the condition to stop is if i>=ne(1.5million), break from the loop and stop. I have done that with a few data points, it seems to pick everything correctly and returns quite reasonable, but results but the speed/efficiency. i=1;ne, should be i=1:ne;