# Calculation precision changed in 2020b?

26 views (last 30 days)
Hiroyuki Kato on 13 Oct 2020
Commented: Jan on 23 Oct 2020
I am encountering a precision error with Matlab2020b, which I did not have in version 2016b.
I have 78-dimension vector x (attached). if I do the following, even though the result should be 0, I get a complex number as a result from acos calculation:
> y = x;
> acos(dot(x,y)/sqrt(sum(x.^2)*sum(y.^2)))
ans = 0.0000e+00 + 2.1073e-08i
In Matlab2016b, I know that using "norm" function caused a precision error and acos(dot(x,y)/(norm(x)*norm(y)) gave a complex number.
Back then, the use of sqrt(sum(x.^2)*sum(y.^2)) was a recommended method to avoid this issue. (as summarized in this page: https://stackoverflow.com/questions/36093673/why-do-i-get-a-complex-number-using-acos)
This method has been working fine in 2016b, but now with exactly the same code I have the complex number issue coming back in 2020b.
Was there a change in the precision of calculation in the newer version of matlab? If so, is there any good work around to avoid this issue?
Thanks,
Hiroyuki

#### 1 Comment

Bruno Luong on 15 Oct 2020
"In Matlab2016b, I know that using "norm" function caused a precision error and acos(dot(x,y)/(norm(x)*norm(y)) gave a complex number.
Back then, the use of sqrt(sum(x.^2)*sum(y.^2)) was a recommended method to avoid this issue. (as summarized in this page: https://stackoverflow.com/questions/36093673/why-do-i-get-a-complex-number-using-acos)
This method has been working fine in 2016b, but now with exactly the same code I have the complex number issue coming back in 2020b."
Pure luck. None of the observation has rigorous justification.

Bruno Luong on 15 Oct 2020
Edited: Bruno Luong on 15 Oct 2020
This is a robust code.
theta = acos(max(min(dot(x,y)/sqrt(sum(x.^2)*sum(y.^2)),1),-1))
Note it returns 0 for x or y is 0. One might prefer NaN because correlation is undefined.

Jan on 20 Oct 2020
The ACOS function is numerically instable at 0 and pi.
SUM is instable at all. A trivial example: sum([1, 1e17, -1]) .There are different approaches to increase the accuracy of the summation, see https://www.mathworks.com/matlabcentral/fileexchange/26800-xsum
There is a similar approach for a stabilized DOT product, but the problem of ACOS will still exist. To determine the angle between two vectors, use a stable ATAN2 method, see https://www.mathworks.com/matlabcentral/answers/471918-angle-between-2-3d-straight-lines#answer_383392

Show 1 older comment
Jan on 21 Oct 2020
@Bruno: Sorry, my fault. I meant the "2*atan" approach, not "atan2":
x = rand(1, 78);
y = rand(1, 78);
angle1 = acos(dot(x, y) / (norm(x) * norm(y)))
angle2 = 2 * atan(norm(x * norm(y) - norm(x) * y) / ... % Typo fixed
norm(x * norm(y) + norm(x) * y))
Paul on 22 Oct 2020
I think you have an error in angle2. Should it not be:
angle2 = 2*atan(norm(norm(x)*y - norm(y)*x)/norm(norm(x)*y + norm(y)*x))
Jan on 23 Oct 2020
@Paul: Yes, there was a typo.

Uday Pradhan on 15 Oct 2020
Edited: Uday Pradhan on 16 Oct 2020
Hi Hiroyuki,
If you check (in R2020b):
>> X = dot(x,y) - sqrt(sum(x.^2)*sum(y.^2))
ans =
1.776356839400250e-15
where as in R2016b, we get:
>> dot(x,y) - sqrt(sum(x.^2)*sum(y.^2))
ans =
0
Hence, in R2020b, we get:
>> acos(X)
ans =
0.000000000000000e+00 + 2.107342425544702e-08i
This is because the numerator dot(x,y) is "greater" than the denominator sqrt(sum(x.^2)*sum(y.^2)) albeit by a very small margin and hence the fraction X becomes greater than 1 and thus acos(X) gives complex value.
To avoid this my suggestion would be to establish a threshold precision to measure equality of two variables, for example you could have a check function so that if abs(x-y) < 1e-12 then x = y
function [a,b] = check(x,y)
if abs(x-y) < 1e-12
a = x;
b = a;
end
end
Now, you can do [a,b] = check(x,y) and then call acos(a/b). This will also help in any other function where numerical precision can cause problems.
Another workaround can be found in this link : Determine the angle between two vectors.
Hope this helps!

Bruno Luong on 18 Oct 2020
Something weird that I don't quite understand. In v2020b, the dot code is
...
if isreal(a) && isreal(b)
c = a'*b;
return
end
elseif ~isequal(size(a),size(b))
error(message('MATLAB:dot:InputSizeMismatch'));
end
c = sum(conj(a).*b);
...
Why calculate using a'*b for a, b reals and sum(conj(a).*b) for complex?
If I was TMW I would code a'*b for both cases. It might have something to do with interleave complex internal storage, but I fail to see the advantage of using sum(conj(a).*b)
Paul on 18 Oct 2020
I will assume that the decision to change the implementation and yield a different answer was not undertaken lightly. I checked the release notes and did not see this change to dot, though the sum function did change in 2020B.
Paul on 19 Oct 2020
If you look further down in dot.m (2019a) to the section used when dim is specified, you will see that there is a path to
c = sum(conj(a).*b,dim)
even if a and b are both vectors and are isreal. For example
dot(1:3,1:3,1)

R2020b

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!