If else if statement problem

function I = compvec(x,y)
%Calculates the angle
c = ((dot(x,y))/((sqrt(x(1)^2 + x(2)^2)) * (sqrt(y(1)^2 + y(2)^2))))
if x==0 | y==0
    disp('Linearly dependent vectors (one of them is zero).r')
    disp('{x, y} is a linearly dependent set.')

%Problem occurs here, if c = 1, else statement is displayed instead
elseif c == 1 || c == -1
    disp('Linearly dependent vectors (both non-zero).')
    disp('{x, y} is a linearly dependent set.')

elseif c == 0
    disp('Linearly independent vectors (orthogonal).')
    disp('{x, y} is a linearly independent set.')
else
    disp('{x, y} is a linearly independent set.')
end

Answer 1

[David Sanchez]

You can solve the issue rounding the value of c:

%Problem occurs here, if c = 1, else statement is displayed instead
elseif round(c) == 1 || round(c) == -1
    disp('Linearly dependent vectors (both non-zero).')
    disp('{x, y} is a linearly dependent set.')

elseif round(c) == 0
    disp('Linearly independent vectors (orthogonal).')
    disp('{x, y} is a linearly independent set.')
else
    disp('{x, y} is a linearly independent set.')
end

[Lord Chinex]

what if I had a c = 0.5? Wouldn't it be rounded incorrectly to 1.

Answer 2

[Julia]

This could be due to rounding issues. It is very unlikely that you get the exact value 1 in Matlab.

Answer 3

[Roger Stafford]

There is more than one thing amiss in this piece of code. Probably the problem you have in mind is caused by your 'if' statement:

if x==0 | y==0

Both x and y are (presumably) two-element vectors, so the logical expression

x==0 | y==0

also has two logical elements. One of these is "x(1)==0|y(1)==0" and the other is "x(2)==0|y(2)==0". The 'if' will execute if both of these are true, and that is clearly not what you intended. For example the 'if' will execute in case x(1)==0 and y(2)==0, even though neither the x vector nor the y vector is a zero vector. What you should have written is:

 if (x(1)==0 and x(2)==0) | (y(1)==0 and y(2)==0)

or more compactly

 if all(x==0) | all(y==0)

You could even do:

 if norm(x)==0 | norm(y) == 0

My next objection is requiring exact equality in

 elseif c == 1 | c == -1

You could well have essentially parallel vectors but due to round-off errors in the computation of c, there would not be exact equality to 1 or -1. You need to allow a tolerance for a small deviation from exact equality here. The same applies to the test "elseif c == 0" for orthogonality.

One final complaint. You take the test for zero vector too late to avoid getting a NaN for an answer in computing c. You should have made this test beforehand to avoid a possible NaN.

(Note: As you undoubtedly realize, you are computing the cosine of the angle rather than the angle between those two vectors.)

[Roger Stafford]

Also I should point out that

   abs(x(1)*y(2)-x(2)*y(1)) < tol

and

   abs(dot(x,y)) < tol

where 'tol' is a very small tolerance for roundoff errors, are good tests for vectors x and y being parallel and orthogonal, respectively.