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Cluster-NMS #13
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Finally, we will get exactly the same results to Original NMS as long as vector |
How exactly does left multiplying diagonal matrix |
For example, let In our paper, the matrix
then do E×X. In practice, we use
then do element-wise multiplication with the upper triangular IoU matrix X. |
Why the extra and how do all those iterations converge to the original NMS result ? |
A diagonal matrix left multiplies another is equivalent to do row transformation (by Higher Algebra). So in practice, I replace it with element-wise multiplication for simplicity. Because it's faster than matrix multiplication. As for why the result of Cluster-NMS is equal to that of Original NMS, a simple case is provided here https://github.com/Zzh-tju/CIoU#description-of-cluster-nms-and-its-usage For mathematics detail, kindly refer to our paper. |
I may had missed something, but how is this (matrix in practice) being element-wise multiplication compared to the matrix given in the paper ? |
I am trying to understand Cluster-NMS operations.
The mathematical proof seems a bit complicated to follow and comprehend.
Why C1 does not change values ? In other words, why C1 == X ?
How to obtain b1 ?
Why is it
Cn = E x X
instead ofCn = E x Cn-1
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