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[BugFix] fast_any_tap boundary handling boundary handling in binary search
#1632
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. The name of the folder suggests that this case would reach max-iterations, right? Can we change it so we don't get confused? Same for below.
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. If we do intend to revisit the |
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Discussion needed wrt behaviour.
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. We originally talked about 2 edge cases in which the boundary handling needs an explicit choice:
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Indeed, we should have this discussion. As for min and max strategies, they are both 'unaffected' thanks to the explore_neighbourhood we put in place for the final wiggle.
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. I'm unresolving because I have a question that's likely related: The same test case for
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. I am in principle agreeing with your point. However, I would argue that we need to align the behavior of In short, we should revisit the |
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Pure binary search runs will not trigger max iteration error.
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so what does it return if there are 2 conflicting tap changers? Does it choose either of the settings? How does it choose which one is better?
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The built-in reset mechanism will handle that and the top layer optimization logic remains the same.