Calculation precision changed in 2020b?

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Hiroyuki Kato
Hiroyuki Kato 2020-10-13
评论： Jan 2020-10-23
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
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Bruno Luong
Bruno Luong 2020-10-15
编辑：Bruno Luong 2020-10-15
"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.

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采纳的回答

Bruno Luong
Bruno Luong 2020-10-15
编辑：Bruno Luong 2020-10-15
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.

更多回答（2 个）

Jan
Jan 2020-10-20
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
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Paul
Paul 2020-10-22
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))

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Uday Pradhan
Uday Pradhan 2020-10-15
编辑：Uday Pradhan 2020-10-16
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!
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Paul
Paul 2020-10-18
编辑：Paul 2020-10-18
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
Paul 2020-10-19
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)

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