1 - 1 = -0 on Matlab why? How to fix that?

I am working with a matrix that has two columns.

I writing a code to do columns 1 minus columns 2. And the matlab display the result which is all fine.

However, some rows have the same values; for example;

x = [1 2 3; 2 2 5; 3 2 6]

y = [ 4 2 6; 1 2 7; 5 2 3]

w=x(:,2)-y(:,2)

% now on this case I get

w =

0

0

0

% however, the code that I am working with has one row1 column 1 that has value of 1.0700 and row1 column 2 has value 1.0700

now this give an output of -0.0000

I was wondering if there is a solution for this issue.

Now, I am doing some operation to column 2 and my guess is that there is some rounding issue.

I do look at both columns before I do the subtraction between columns 1 & 2 and I see that both values are 1.0700 so where does this rounding proplem is coming from.

[per isakson]

Could it have something to do with floating point precision? See What Every Programmer Should Know About Floating-Point Arithmetic. Is 1.0700 the result of a computation?

[Steven Shaaya]

So column 1 has fixed values. Column 2 values are the result of a computation.

So how to make matlab deal with column 2 as fixed values as well?

[Stephen23]

"However, some rows have the same values"

In fact they don't have the same values.

"however, the code that I am working with has one row1 column 1 that has value of 1.0700 and row1 column 2 has value 1.0700 now this give an output of -0.0000"

Because the values are not the same. This is easy to demonstrate:

format short
X = [pi+1,1.07]

X = 1×2

4.1416 1.0700

Y = [pi,1.07+eps(2)]

Y = 1×2

3.1416 1.0700

X-Y

ans = 1×2

1.0000 -0.0000

X(2)-Y(2)

ans = -4.4409e-16

"I was wondering if there is a solution for this issue."

What is the issue?

You subtract two different values and so MATLAB correctly gives you the non-zero difference (as the closest representable binary floating point number). That is the expected behavior.

"I do look at both columns before I do the subtraction between columns 1 & 2 and I see that both values are 1.0700 so where does this rounding proplem is coming from"

It comes from the values of the numbers. Simply "looking" at two numbers displayed to five signficant figures will not tell you if they differ after the twelfth significant figure. You will need to either display them to a much higher precision or (better) look at their exact hexadecimal representation to see if they are exactly the same.

Most likely your algorithm is not robust for binary floating point numbers: assuming exact equivalence of binary floating point numbers in your algorithm is most likely a bug.

[Hiro Yoshino]

As long as you are using floating point numbers, you cannot avoid this phenomenon unless you perform computation using integers and convert the results to the floating point numbers.

This is because of the rounding issue, in other words, the approximations your computer makes.

Please imagine that 1/3 should be rounded up to be treated in the digital world.

You should go through Floating Point numbers and apply some techniques that make your situation better (not perfect though).

I guess eps command would help you understand further.

[Steven Shaaya]

@Stephen

Well, I do not know if -0.000 is an acceptable result for my assignment or not.

I was just wondering if there is an easy fix for this because I know that and from my lectures, that -0.0000 does not affect the system that I am working with.

The problem comes to my code because I am using the if statement to determine if w = x-y is less than zero or greater and since I know -0.0000 difference does not affect the system then I want to ignore this result and take it as it is a positive number or 0.

Thank you

[Walter Roberson]

No, if you need to test whether x-y is less than zero, then you should accept the fact that it is less than zero in your case.

If you are in a situation where what you care about is whether the difference affects the system then you should use a different test, as close to the boundary as feasible for whether the difference affects the system. If the boundary is -1022/4096 then use that.

[per isakson]

"I do not know if -0.000 is an acceptable result for my assignment or not." There is no perfect solution. Maybe you interpret the assignment too literally.

"if w = x-y is less than zero or greater" IMO the best solution is to handle the uncertainty around zero in this if-statement. And explain in a comment.

[Steven Shaaya]

Can I make the matlab round the values of culmun 2 to four decimal places, so that the value of column 2 will be exactly 1.0700 and I don't have to worry about 10^-16 rounding issue anymore? So I can round the values before using "If statement".

I know how to round values in matlab when using fprintf function

but I am not sure how to do that for a value; for example, say

X = 0.90000000001 % how to round this value to four decimal places without using fprintf

Thank you all

[Steven Shaaya]

Thank you all again for your help.

I solved the issue by using "round" function. I did work and -0.0000 is not there anymore!

Answer 1

[Walter Roberson]

num2hex(x(1,1:2))

You will find that they have slightly different internal representations.

[Steven Shaaya]

I still have the same issue

[Walter Roberson]

Do the numbers show up with the same hex representation, or with different hex representations?

If they are different hex representations then that proves that they are not the same number.

I suggest that you use https://www.mathworks.com/matlabcentral/fileexchange/22239-num2strexact-exact-version-of-num2str to display the exact value of each of the values.

If you have Mac then you can get away with something like

fprintf('%.999g\n', x(1,1:2))

Historically fprintf() produced incorrect outputs on Windows; I do not know if that has been changed on Windows.

[Steven Shaaya]

No, they do not have the same values.

there is a difference of 10^-16.