How to generate a matrix with number of lines inserted from keyboard

I have to generate a matrix with i lines and j=i+1 collumns whose elements:

=2 if i=j;

=-1 if | i -j |=1;

=0 otherwise.

Here's my attempt at solving this problem:

prompt='Insert number of lines';
i=input(prompt)
j=i+1
% getting the number of lines and collumns
if i==j 
A=2*ones(i,j);
end
% checking the first condition for i and j
else if abs(i-j)==1
A=-1*ones(i,j);
end
%  checking the second condition for i and j
else
A=zeros(i,j)
% checking the third condition for i and j
end
end
A
% displaying matrix A

And I get this error when I try run it:

else if abs(i-j)==1
 |
Error: Illegal use of reserved keyword "else".

I don't understand what I'm doing wrong.

Answer 1

[Star Strider]

You had too many end statements. They were defining very short if blocks, and so some of those blocks did not begin correctly as a result.

Your code still has problems (it is not doing what you want it to). Since this appears to be a homework assignment, I will defer to you to get it running correctly.

This runs:

if i==j
    A=2*ones(i,j);
    % checking the first condition for i and j
elseif abs(i-j)==1
    A=-1*ones(i,j);
    %  checking the second condition for i and j
else
    A=zeros(i,j)
    % checking the third condition for i and j
end

[Fiddl Sticks]

It's not homework.

I've skipped most of the Matlab classes at uni and now I'm trying to prepare for the exam using the course given by the teacher which also has problems in it.

The code you provided does seem to work but since you pointed out that it still has problems I'd like to know what problems does it still have more exactly.

In the course up to the point where I had to solve this problem there was nowhere specified how to insert numbers from keyboard and the examples for if statements as well as for, while and switch loops didn't contain it either. It's not unlikely for something I'm missing to not be in there.

[Star Strider]

O.K. If it’s not homework, here’s my solution:

i = 5;
j = i + 1;
A = zeros(i,j);
for k1 = 1:i
    for k2 = 1:j
        if k1==k2
            A(k1,k2) = 2;
            % checking the first condition for i and j
        elseif abs(k1-k2)==1
            A(k1,k2) = -1;
        end
    end
end

I will let your explore it to see how it works.

I used a specific value for ‘i’ so I didn’t have to type it in each time.

The input call you used should work to collect ‘i’ from your Command Window. (Another option — and the one I prefer — is the inputdlg function. You then need to recover the returned value from the cell array of strings that it produces, however it’s more flexible.)

[Fiddl Sticks]

I think that the first code does what it is supposed to do and this one doesn't.

First of all i's value is nowhere specified in the problem therefore inserting i from keyboard seems like the logical thing to do instead of giving it a base value.

Next the matrix with i lines and j collums is supposed to have it's elemts =2 if i=j, =-1 if | i-j |=1 and 0 otherwise. Your method doesn't respect this condition as the result is

A =

     2    -1     0     0     0     0
    -1     2    -1     0     0     0
     0    -1     2    -1     0     0
     0     0    -1     2    -1     0
     0     0     0    -1     2    -1

I'll try to figure out how to use inputlg properly even though it seems like a headache.

Thanks for the answers.

[Star Strider]

I gave ‘i’ a specific value simply because I didn’t want to type ‘i’ in from the keyboard every time while I tested my code. Use input or inputdlg if you want keyboard input for it.

That said, an alternate way of computing ‘A’ gives the same result as my double-loop code:

i = 5;
j = i + 1;
A = zeros(i,j);

[K1,K2] = ndgrid(1:i, 1:j);
A = 2*(K1==K2);
A = -1*(abs(K1-K2) == 1) + A

producing:

A =
     2    -1     0     0     0     0
    -1     2    -1     0     0     0
     0    -1     2    -1     0     0
     0     0    -1     2    -1     0
     0     0     0    -1     2    -1

so this is either correct, or I made the same error twice.

[Fiddl Sticks]

Well the use of loops seems uneccessary since the problem doesn't mention any specific value for i or any specific set of values, but it also doesn't mention if all of the matrix's elements should equal only to -1, 2 or 0 for theyr specific conditions so I guess the problem's not clear enough.

Still if I wanted to get a matrix with all of it's elements equal to -1, 2 or 0 for theyr specific condition what would I have to change in your code?

[Star Strider]

It depends on the conditions.

The ‘logical indexing’ approach I used here would likely work.

For example to set all elements of ‘A’ equal to 5 for K1 > K2:

i = 5;
j = i + 1;
A = zeros(i,j);

A = 5*(K1 > K2)

produces:

A =
     0     0     0     0     0     0
     5     0     0     0     0     0
     5     5     0     0     0     0
     5     5     5     0     0     0
     5     5     5     5     0     0

[Fiddl Sticks]

But that's not all elements of A equal to 5.

This would be all elements of A equal to 5:

A =

     5     5     5     5     5     5
     5     5     5     5     5     5
     5     5     5     5     5     5
     5     5     5     5     5     5
     5     5     5     5     5     5

[Star Strider]

In my example, I did not set all the elements equal to 5. I set only the elements where K1>K2 equal to 5. That was the point of the example.

Perhaps I should have phrased it: ‘... to set all elements of ‘A’ for K1 > K2 equal to 5’