My script sees the value 2 different then 2.0000 which is a problem

hello, i am facing a problem in my script, where MATLAB is seeing the value A=2.000 > B=2 . i have checked the value A to see if it belongs to a value slightly larger then 2 by applying

floor(A) and ceil(A) which both give a result of 2.

what is the problem in your opinion and how could i fix it? Thank you!

Answer 1

[Guillaume]

You cannot compare floating point numbers obtained through different methods with ==. This is true for all computer languages. This is due to the fact that many numbers cannot be represented exatly as floating point numbers (0.1 is one of them) and therefore small errors accumulate differently depending on the operations you perform.

For more details, see why is 0.3-0.2-0.1 not equal to zero. From your floor test you can see than optimal_distance is slightly smaller than 2 while temp_distance is either exactly 2 or slightly larger than 2. By slightly, we're talking of a difference most likely around or smaller than 1e-16.

The proper way to compare floating point number is to compare their difference to a small enough tolerance for your calculatios. In your case:

abs(optimal_distance - optimal_distance) < 1e-10

is probably appropriate.

[Ibrahim DAWI]

You are right, thank you for your answer it was very helpful!

Answer 2

[Sayyed Ahmad]

Look at Operator Precedence in matlab.

    A=2000;
    B=2;
    C=A>B;

[Ibrahim DAWI]

i dont't think i explained my self well, the following code should explain my question:

temp_distance(9)

ans =

     2

K>> optimal_distance(9)

ans =

    2.0000

K>> temp_distance(9)==optimal_distance(9)

ans =

     0

K>> floor(optimal_distance(9))

ans =

     1

K>> floor(temp_distance(9))

ans =

     2

[Guillaume]

Sayyed Ahmad's comment mistakenly posted as an answer moved here. Please use comment on this Answer instead of starting a new answer.

you have to look inside your function to find the solution.

This codes will simulate your function output values

a=2
b=2-eps
a==b

answer:

a =

       2

b =

      2.0000

ans =

    logical

     0

[Dennis]

try:

if abs(temp_distance(9)-optimal_distance(9))<0.001

[Guillaume]

The tolerance (the 0.001, see discussion in my answer) needs to be defined appropriately for the magnitude of the numbers compared and the possible accumulated error of the calculations that produced these numbers.

As we don't know anything about these calculations, we can't be sure of anything but if 0.001 out of 2 is around the magnitude of the accumulated error I'd say that you've got big problems. It is most likely that a smaller tolerance would be more appropriate.