Fast Differentiable Matrix Sqrt Root

Related tags

Deep Learning matrix-factorization neural-style-transfer vision-transformer decorrelated-bn global-covariance-pooling
Overview

Fast Differentiable Matrix Sqrt Root

Geometric Interpretation of Matrix Square Root and Inverse Square Root

This repository constains the official Pytorch implementation of ICLR 22 paper "Fast Differentiable Matrix Square Root".

Usages

Check torch_utils.py for the implementation. Minimal exemplery usage is given as follows:

# Import and define function
from torch_utils import *
FastMatSqrt=MPA_Lya.apply
FastInvSqrt=MPA_Lya_Inv.apply

# For any batched matrices, compute their square root or inverse square root:
rand_matrix = torch.randn(5,32,32)
rand_cov = rand_matrix.bmm(rand_matrix.transpose(1,2))
rand_cov_sqrt = FastMatSqrt(rand_cov)
rand_inv_sqrt = FastInvSqrt(rand_cov)

Computer Vision Experiments

All the codes for the following experiments are available:

Citation

Please consider citing our paper if you think the code is helpful to your research.

@inproceedings{song2022fast,
  title={Fast Differentiable Matrix Square Root},
  author={Song, Yue and Sebe, Nicu and Wang, Wei},
  booktitle={ICLR},
  year={2022}
}

Contact

If you have any questions or suggestions, please feel free to contact me

[email protected]

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Comments
  • A question of compute P and Q weithou P = I - P, Q = I - Q

    A question of compute P and Q weithou P = I - P, Q = I - Q

    (https://github.com/KingJamesSong/FastDifferentiableMatSqrt/blob/466c9bf5f7ca5c3a405e104197e63f20b1c70114/torch_utils.py#L28-L40)

    why not $p \underline{} sqrt = \mathbf{I} - p \underline{} sqrt,q \underline{} sqrt=\mathbf{I} - q \underline{} sqrt $ This seems inconsistent with the formula in the paper

    opened by HongLouyemeng 4
  • Inverse of singular matrix doesn't exist, so what does the solution represent ??

    Inverse of singular matrix doesn't exist, so what does the solution represent ??

    Hi,

    I was using the code for the inversion of a covariance matrix, but the covariance matrix I have created is from a single data point i.e [email protected] where A is a vector of shape (n,1).

    So [email protected] is a rank1 matrix of shape (n,n). So the inverse should not exist for this matrix. But, when calculating the inverse of this matrix, the method provides a solution.

    So I wanted to ask what is this solution. Since the method is iterative, so it approximates the result, but to what result is it approximating ??

    opened by lavish619 2
Owner
YueSong
Ph.D. student in Computer Vision and Numerical Analysis
YueSong
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