As seen during the lecture, the complexity of the forward pass is \(O\left(LK^{2}\right)\).
In the lecture, Nicolas showed how to compute the derivatives jointly in an efficient manner (back propagation) and this also has efficiency \(O\left(LK^{2}\right)\).
My question is, what would be the complexity if we did not compute the derivative jointly? \(O\left(LK^{3}\right)\)?
Someone correct me if I am wrong but I would say \(\mathcal{O}(L^2 K ^2)\) in that case, because it would mean that for each layer, we compute all the intermediate derivatives
Complexity
As seen during the lecture, the complexity of the forward pass is \(O\left(LK^{2}\right)\).
In the lecture, Nicolas showed how to compute the derivatives jointly in an efficient manner (back propagation) and this also has efficiency \(O\left(LK^{2}\right)\).
My question is, what would be the complexity if we did not compute the derivative jointly? \(O\left(LK^{3}\right)\)?
Thank you in advance :))
1
Hello,
Someone correct me if I am wrong but I would say \(\mathcal{O}(L^2 K ^2)\) in that case, because it would mean that for each layer, we compute all the intermediate derivatives
1
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