cannot run an if statement with "or" and "and" operators in it. I get an error saying "matrix dimensions must agree". I added a comment in a line where the error occurs.

number=4391
currentNum=number
numvec=zeros(4,1)
for b=1:4
currentRem=mod(currentNum,10);
numvec(b)=currentRem
currentNum=(currentNum-currentRem)/10
end
empty=zeros(4,1);
newnumvec=numvec
maximum=zeros(3,1);
for n=1:4
    for s=1:4
    if (newnumvec(1)>newnumvec(2:4)) | (newnumvec(n)>newnumvec((n+1):end) & newnumvec(n)>newnumvec(1:(n-1))) | (newnumvec(4)>newnumvec(1:3)) %here the error occurs
    maximum(s)=newnumvec(n)
    empty(n)=newnumvec(n)
    newnumvec=newnumvec-empty
    end %if
    end %for
end %for

Answer 1

[KALYAN ACHARJYA]

You are trying to compare two differnt data types-

Here

if (newnumvec(1)>newnumvec(2:4)) |....
        %what result ^ expected from this statement 
end

#See, you trying to compare 1>[9 3 4]..??

>> newnumvec(1)
ans =
     1
>> newnumvec(2:4)
ans =
     9
     3
     4

Same for others also.

[Arsen Korpetayev]

Thank you for your answer, Sir. However, when using inequality signs in an if statement between a scalar and a vector, Matlab percieves it as a command to compare each element of the vector with that scalar. This is exactly what I am trying to achieve. I have checked this theory with a simple code like this:

a=[1.5; 3]
b=1
if b<a(1:2)
    fprintf works
end

Therefore I am puzzled as where the problem with "matrix dimensions" lies and how it can approached. Many thanks for your answer nevertheless. Merry Christmas and a happy licking off leftover yogurt from a plastic package

[KALYAN ACHARJYA]

Here newnumvec have 4

>> whos newnumvec
  Name           Size            Bytes  Class     Attributes

  newnumvec      4x1                32  double     

But you trying to acesss n+1 in if statement

  newnumvec((n+1):end)) % For loop when n=1, then newnumvec((4+1):end))

whicj implies that newnumvec(5:end), how to get 5 index elemnet, newnumvec having 4 only.

Simmilar, when n=1 (first iterartion)

  newnumvec(1:(n-1))

which becomes newnumvec(1:(1-1))>> newnumvec(1:0), zero indexing concept is not allowed in Matlab.

newnumvec(1:0) statment is invalid

I have removed those n+1 and n-1 from the if statement, now it runs without coding error

number=4391
currentNum=number
numvec=zeros(4,1)
for b=1:4
currentRem=mod(currentNum,10);
numvec(b)=currentRem
currentNum=(currentNum-currentRem)/10
end
empty=zeros(4,1);
newnumvec=numvec
maximum=zeros(3,1);
for n=1:4
    for s=1:4
    if newnumvec(1)>newnumvec(2:4) | newnumvec(n)>newnumvec(n:end) & newnumvec(n)>newnumvec(1:n) | newnumvec(4)>newnumvec(1:3) %here the error occurs
    maximum(s)=newnumvec(n)
    empty(n)=newnumvec(n)
    newnumvec=newnumvec-empty
    end %if
    end %for
end %for

Hope it helps!

Answer 2

[Arsen Korpetayev]

Dear Kalyan, my sincere graditude for your response once again, although I don't know if you are getting paid for this. I see where the error was coming from, thank you for clarifying this. However, it's worth to note that a statement a(n)>a(n:end) will be false for any values, since a number cannot be greater than itself. I am closing this question, since I had relised that the above approach is not only inefficient, but seemingly incapable in its nature, because of the very discussed restrictions. Have a great winter season and don't forget to clean your coffee table. All the best, your sincere Arsen Korpetayev.

[KALYAN ACHARJYA]

I don't know if you are getting paid for this.

We are providing volunteering service here, if my anwer help to resolved the issue, you can give due credit by accepting the answer.

Cheers !