Proposal by for how to solve all our RANSAC problems #84
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Co-authored-by: minnerbe <innerbergerm@hhmi.org>
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Linking the previous discussion: #82 |
axtimwalde
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axtimwalde
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Does not address circles of more than two as discussed in #82 (or I am missing something).
| { | ||
| ++i; | ||
| continue A; | ||
| break; // the while loop |
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this now goes back to the for loop, should be equivalent to continue A: since there is no code after the while-loop anymore.
| // make sure we do not only keep the last model, since the | ||
| // second last can be the better (defined as num inliers) one | ||
| // now we need to make sure it is enough inliers | ||
| if ( isGood && m.betterThan( copy ) && tempInliers.size() >= minNumInliers ) |
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This only works for circles between two models, do we have a guarantee that there aren't circles with three or four models?
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it should work for all cases, it just makes sure we always remember the best model (as long as |inliers| >= minNumInliers, which can grow in the while-loop. The break condition is still the old one <, which worked well.
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lets find time to discuss this in person, we thought everything is addressed ... but we may miss something |
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The returned set is actually good, since the best RANSAC model can be retrieved - even though it cannot be re-computed from the points itself - we made this more clear in the docs (this was the original issue from 4 years ago) |
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@axtimwalde code is self-explanatory ;)