# I wrote this code and it didn't work and its gave me " Unrecognized function or variable 'seq' " error , What I can do to sole this error??

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hothifa hasanat on 16 Sep 2021
Answered: Schubie on 29 Sep 2021
function [N,V,seq] = addition1_opt( n,m,P )
%UNTITLED Summary of this function goes here
% Detailed explanation goes here
%n = str2num(n);
%m = str2num(m);
%x = str2num(x);
'wait for initializing all sequence combinations ...';
L = ceil((6*(n-1)+1)/2);
N = (factorial(L)/factorial(L-m))^n;
C = nchoosek(1:L,m);
H = (factorial(L)/(factorial(m)*factorial(L-m)));
k=1;
done = false;
required_number_of_solutions = 10;
for i = 1: H
x_temp=perms(C(i,:));
for j=1: factorial(m)
x(k,:) = x_temp(j,:);
k=k+1;
end
end
x;
% for ro = 1:n
% mtx(ro,:) = x(
%
%
'Total number of sequences is'; k;
if n>4
end
if n==2
v=1;
for i=1:k-1
mtx(1,:) = x(i,:);
for j = 1:k-1
mtx(2,:) = x(j,:);
Test='F';
s=1;
rst(s)=0;
for co1 = 1:m
for co2 = 1:m
if co1 ~= co2
if sum(rst == mtx(1,co1) + mtx(2,co2))==0
Test='T';
rst(s) = mtx(1,co1) + mtx(2,co2);
s=s+1;
else
Test='F';
break;
end
end
end
if Test=='F'
break;
end
end
if Test=='T'
seq(:,:,v) = mtx;
v=v+1;
if v > required_number_of_solutions
done = true;
break;
end
if v>P
break;
end
end
clear rst
end
if done; break; end
if v>P
break;
end
end
end
if n==3
v=1;
for i=1:k-1
mtx(1,:) = x(i,:);
for j = 1:k-1
mtx(2,:) = x(j,:);
for l = 1:k-1
Test='F';
mtx(3,:) = x(l,:);
s=1;
rst=0;
mtx;
for co1 = 1:m
for co2 = 1:m
if co1 ~= co2
if sum(rst == mtx(1,co1) + mtx(2,co2))==0
Test='T';
rst(s) = mtx(1,co1) + mtx(2,co2);
s=s+1;
else
Test='F';
break;
end
end
end
if done; break; end
if Test=='F'
break;
end
end
if done; break; end
if Test=='F'
break;
end
for co2 = 1:m
for co3 = 1:m
if co2 ~= co3
if sum(rst == mtx(2,co2) + mtx(3,co3))==0
Test='T';
rst(s) = mtx(2,co2) + mtx(3,co3);
s=s+1;
else
Test='F';
break;
end
end
end
if done; break; end
if Test=='F'
break;
end
end
if done; break; end
% if Test=='F'
% break;
% end
if Test=='T'
seq(:,:,v) = mtx;
v=v+1;
if v > required_number_of_solutions
done = true;
break;
end
if v>P
break;
end
end
clear rst
end
if done; break; end
if v>P
break;
end
end
if done; break; end
if v>P
break;
end
end
end
if n==4
v=1;
for i=1:k-1
mtx(1,:) = x(i,:);
for j = 1:k-1
mtx(2,:) = x(j,:);
for l = 1:k-1
mtx(3,:) = x(l,:);
for h = 1:k-1
Test='F';
mtx(4,:) = x(h,:);
s=1;
rst=0;
mtx;
for co1 = 1:m
for co2 = 1:m
if co1 ~= co2
if sum(rst == mtx(1,co1) + mtx(2,co2))==0
Test='T';
rst(s) = mtx(1,co1) + mtx(2,co2);
s=s+1;
else
Test='F';
break;
end
end
end
if done; break; end
if Test=='F'
break;
end
end
if done; break; end
if Test=='F'
break;
end
for co2 = 1:m
for co3 = 1:m
if co2 ~= co3
if sum(rst == mtx(2,co2) + mtx(3,co3))==0
Test='T';
rst(s) = mtx(2,co2) + mtx(3,co3);
s=s+1;
else
Test='F';
break;
end
end
end
if done; break; end
if Test=='F'
break;
end
end
if done; break; end
if Test=='F'
break;
end
for co3 = 1:m
for co4 = 1:m
if co3 ~= co4
if sum(rst == mtx(3,co3) + mtx(4,co4))==0
Test='T';
rst(s) = mtx(3,co3) + mtx(4,co4);
s=s+1;
else
Test='F';
break;
end
end
end
if done; break; end
if Test=='F'
break;
end
end
if done; break; end
if Test=='T'
seq(:,:,v) = mtx;
v=v+1;
if v > required_number_of_solutions
done = true;
break;
end
end
if v>P
break;
end
clear rst
end
if done; break; end
if v>P
break;
end
end
if done; break; end
if v>P
break;
end
if done; break; end
end
if v>P
break;
end
end
end
'The unique sequences are:';
'! DONE !';

Rik on 16 Sep 2021
Edited: Rik on 16 Sep 2021
Apparently your code doesn't guarantee the creation of the seq variable. Since it only occurs in block starting with Test=='T', that must mean Test is not equal to 'T' at the correct times.
Because you decided to not use any comments, it is difficult to follow your code and give meaningful suggestions to fix the issue. Why don't you make sure you explain what your code is doing? Why is everything in one large function, instead of splitting up tasks in smaller (documented!) units?
Also, why don't you have a block at the start of your function that checks if the inputs are correct? If n is only allowed to be a positive integer below 5, check that and return a meaningful error.
hothifa hasanat on 17 Sep 2021
You have a bad language when you talked with me and you told me to ask my friend who wrote this for me and you dont know what is my situation that put me here to ask for help and I think this site is for begginer like me that ask proffesional like you to give them the help they need, but you gave a sermons. Any way I ask you to help me if you can, thanks you.🌹🌹

Image Analyst on 16 Sep 2021
I agree with Rik that it's an uncommented alphabet soup mess of code. Anyway, have the first few lines of your program be
function [N, V, seq] = addition1_opt(n, m, P)
% Create the output variables so we'll at least have something even if something goes wrong.
N = [];
V = [];
seq = [];
% Rest of code follows this.
So now seq will be defined. It's null, but it's still defined and you won't get that particular error about it not being defined, though you may get other errors of a different kind.
If you need more help, comment your code and tell us what you sent in for n, m, and P.
Image Analyst on 17 Sep 2021
Sorry but the uncommented code would take far longer for me to understand that I allow for Answers forums questions. But I'll show you how you can figure it out on your own:
Actually another way is to just put in comments. Often I find that when people put in comments they discover errors in the code/logic. So try that. Otherwise you just have to slog though it line by line with the debugger. But that's way too much work for Rik or I to donate to you - sorry.

Schubie on 29 Sep 2021
If you are still there...
I see a couple of issues with your program:
1.) There is a conflict between the variables "P" and "required_number_of_solutions" -> P comes from the input argument list and "required_number_of_solutions" is defined as "10", in line 14. Later in the program code both terms are used:
if v > required_number_of_solutions
and
if v>P
2.) Line 28 to 30 are:
if n>4
end
which does exactly nothing, but in the code there are three separate loops for "n==2", "n==3" and "n==4". Accordingly nothing happens if you enter a number for "n" which is not 2, 3 or 4. In particular, no "seq" will be calculated.
3.) In the output argument list there is an upper case "V", but in the code only lower case "v" is used.
4.) Lines like:
'Total number of sequences is'; k;
do absolutely nothing -> write
disp(['Total number of sequences is: ',num2str(k)]);
instead, if you want to have any feedback on the screen.
5.) There is no guarantee that the program will find a solution at all. In this case "seq" is not defined and that is the cause for your error message. You got a version from "Image Analyst" where he added "seq=[]" - this obviously suppresses the error message, which, however, indicated deeper problems with the program.
n+1.) I am not sure that the list is complete.
However, as a personal brain teaser, I deducted the original task from the submitted program and created a working version (at least it is working on my MATLAB Version 7.1.0.246 (R14))...
There isn't much left of the original program code, as I found it awkward to process different values for "n" separately. I took a recursive approach. So I don't know if it helps you anyway?
Best regards, Sebastian
%This program solves a task that user Hothifa Hasanat presented in the
%
%The task itself was deduced from the faulty original program:
%
%Our field of activity are matrices with n rows and m columns
%We are looking for arrangements in which all sums of each two numbers
%in consecutive rows and different columns are different.
%
%These solutions are stored in "seq".
%As soon as "p" solutions have been found, the program stops.
%
%Since the variable number of lines influences both the structure of the
%matrix and the evaluation, the task was solved by a recursive function.
%
%kind regards, Sebastian®
global nn x k v pp seq isum
disp(['number of rows n= ',num2str(n)]);
disp(['number of columns m= ',num2str(m)]);
disp(['number of Solutions: p= ',num2str(p)]);
disp(' ');
disp(['wait for initializing all combinations ...']);
L = ceil((6*(n-1)+1)/2);% % % % % % ??? where does this formula come from ???
N = (factorial(L)/factorial(L-m))^n;
H = nchoosek( L,m);%number of combinations to choose m out of L Elements
C = nchoosek(1:L,m);%all possible combinations of m out of L Elements
x=[];%build list of all possible rows:
for i=1:H;
x=[x;perms(C(i,:))];
end
k=size(x,1);%number of possible rows
disp(['Total number of possible rows is: ',num2str(k)]);
disp([num2str(k^n),' combinations of rows are possible: ',]);
disp(' ');
%invent an index matrix for summation
[na,nb]=meshgrid(1:m);fab=find(na~=nb);
isum=[na(fab),nb(fab)+m];
x=x.';%I prefer working with columns
seq=[];%empty matrix for solutions
v=0;%counts number of solutions
pp=p;%maximum number of solutions
nn=n;%required number of rows
%start recursion with 0 rows and empty matrix
%==============================
%==============================
V=v;%number of solutions found
disp([num2str(v),' solutions found:']);
%disp(seq);
%========================================================================
%Here comes the recursive function
global nn x k v pp seq isum
if v>=pp;return;end;
nloc=nloc+1;%now working in row "nloc"
%disp(['now working in row: ',num2str(nloc)]);
if nloc>=2;
ksum=[ksum;isum+(nloc-2)*size(x,1)];%append sum indices for next row
end
%------------------------------------------------------------------------
for i=1:k;%add and try all possible rows
mtxloc=[mtx,x(:,i)];
if v>=pp;return;end;
if (nloc<nn);%if not enough rows
%==========================================
%==========================================
else;%rows complete, therefore we check if mtx is a solution
if nloc>=2;%sums are only available with min 2 rows
sumlist=sum(mtxloc(ksum),2);%list of all required sums to check
if sum(diff(sort(sumlist))==0)==0;%all sums are different !!!!
disp(['solution nr: ',num2str(v)]);%display nr of solution
disp(mtxloc.');%display solution
v=v+1;seq(:,:,v)=mtxloc.';%store solution in seq()
end
end
end
end
%========================================================================