Detect sign change in Matlab
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K
le 11 Déc 2012
Réponse apportée : Sindar
le 29 Oct 2019
Hi everyone, I have one question. I have two values : prior_value and current_value. Is there any easy way or any function that can be used to detect sign change between this two values? positive to negative or the other way. Thank you.
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Jan
le 11 Déc 2012
Modifié(e) : Jan
le 11 Déc 2012
Either:
if ~isequal(sign(previous), sign(current))
Or the trivial:
if previos * current < 0
In both cases the 0 must be considered also, e.g. by "<= 0".
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Derek O'Connor
le 18 Déc 2012
Modifié(e) : Derek O'Connor
le 18 Déc 2012
@Jan, the statement ``if previos * current < 0 is prone to overflow and underflow which will give the wrong result'', contains an implicit question: do you recommend a statement that is prone to overflow and underflow which will give the wrong result? Sorry for the idiomatic English.
Your argument ``But how often does this happen in a real problem?'' reminds me of Intel's argument that the FDIV BUG didn't matter because it happened very rarely. IBM's analysis found that it happened many thousands of times per day, contrary to Intel's rarely. Remember, this error-prone sign test could be buried deep in a large program, called millions of times, and you have no idea what values are thrown at it.
For a nice history of the Pentium FDIV bug see here:
and for the sign test see here:
In a previous comment you say ``Nothing is wrong with the comparison of the SIGNs [ if sign(previous) == sign(current) ], see my first suggestion. It makes it only harder to handle the cases of zero values.''
How about if sign(1/previous) == sign(1/current) ?
Mathematically sign(x) = sign(1/x), x ~= 0, so this test handles all non-zero values along with signed zeros.
The difficulties in defining machine precision are discussed by Nick Higham and Cleve Moler here:
Jan
le 18 Déc 2012
@Derek: I do not think that this dicussion matchs the complexity level of the OP's question. If a user asks for different signs of two values, the topic "a*b<0" should be mentioned. I hope, that the OP feels encouraged to find such trivial solutions by his own in the future.
There is absolutely no doubt, that a multiplication can lead to numerical difficulties. Neither the famous FDIV story nor a discussion about common errors is required to prove this.
sign(1/x) will suffer from 1/realmin ~= realmax.
Sindar
le 29 Oct 2019
A method safe against overflow.
counting zeros as not a sign change
Sign changes:
if sign(prior_value)*sign(current_value) == -1
Sign stays the same:
if sign(prior_value)*sign(current_value) ~= -1
counting zeros as a sign change (even 0,0)
Sign changes:
if sign(prior_value)*sign(current_value) ~= 1
Sign stays the same:
if sign(prior_value)*sign(current_value) == 1
In all expressions, replace '*' with '.*' to allow prior_value and current_value to be vectors (or matrices), and return a vector of locations where the sign changes. If you want to check whether all the signs change, wrap with all(). Wrap with any() to check if any of the signs change.
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