News for hackers. Stuff that matters.
Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by jl2718 on Saturday March 07, 2020 @12:52PM (#22512895)
This is really not a new idea for anybody that has studied numerical optimization or done a lot of much simpler linear modeling. It has to do with the gradients, which are basically random steps in high-dimensional spaces unless you align with a ‘large’ eigenvector of the Hessian. This explanation didn’t go over well at all with the ML cargo cultists until someone gave it a name. Interesting how things work.
Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by stared on Saturday March 07, 2020 @12:58PM (#22512938)
Could you point to concrete papers, showing that empirically?

(Many things were rediscovered over and over, sure.)

> random steps in high-dimensional spaces unless you align with a ‘large’ eigenvector of the Hessian

In this case, the main observation is different than this one, and has much more to do with sparsity (and an exponentially growing number of connections with each layer).

Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by lsorber on Saturday March 07, 2020 @02:07PM (#22513350)
Nocedal has some papers in this direction.
Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by moultano on Saturday March 07, 2020 @04:35PM (#22514287)
I don't see any connection between what you've written and the lottery ticket hypothesis. Could you elaborate? (In particular, the core of the lottery ticket hypothesis is that layers make the nets combinatorial, and so I don't see how anything from simple linear modelling should transfer.)
Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by jl2718 on Saturday March 07, 2020 @09:29PM (#22515924)
RELU training is basically the same as active sets. You call it nonlinearity. I call it a constrained sub-problem. The point is that it is always linear for the active variables. If you’re talking only about gradients, you’re not doing anything non-linear.
Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by moultano on Saturday March 07, 2020 @09:40PM (#22515966)
What does that have to do with the lottery ticket hypothesis?
Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by jl2718 on Sunday March 08, 2020 @12:18AM (#22516473)
Because the opposite hypothesis, that 'winning tickets' can emerge from 'bad tickets' using stochastic gradient descent, implies that randomness doesn't push you back into the null space of the generalization curvature, which you can empirically observe that it almost always will, except for the same eigenvectors which tend to reinforce where there is strong alignment already, iteration after iteration. If you train a lot of small problems you can watch this happen just by looking at the angles between each update. Maybe that's the wrong explanation; I don't know. The practitioner take-away is that some initializations are just lucky/unlucky and you learn this quickly if you start a lot of small problems from scratch. Try stochastic BFGS and the frustration forces some inquiry. The paper is very experimental, which is great, but the conjecture itself is pointing roughly at some feature of SGD (i.e. "SGD seeks out and trains a subset of well-initialized weights.") without offering much direct explanation, for which I'm offering a hand-wavy viewpoint that also comes more from experience/frustration trying to optimize in ill-conditioned local curvatures. There may be a combinatorial way to look at 'lottery tickets', and that may be more productive than looking at it as eigenbasis pursuit, but it seems to me that it certainly might have something to do with that. Either way, the terms are irrelevant, and if it is true now, it was true before it got a name.
Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by 0-_-0 on Sunday March 08, 2020 @06:40AM (#22517389)
Let me get this straight: You're saying that for most weights, the gradient won't point towards the minimum (which makes sense since they are highly correlated), but for certain groups of weights the gradient is well aligned with the eigenvector, and that group makes a large step toward the optimum and becomes the "winning ticket".

Well, maybe, but won't the gradient become misaligned just after the first step?

Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by sdenton4 on Saturday March 07, 2020 @03:53PM (#22513958)
It's also fun to consider that most neural network architectures are mostly piecewise linear: relu(Mx + b) for standard feedforward or convolutional layers... So one would expect a lot of knowledge and folklore from (piecewise) linear optimization to carry over nicely.

IIRC, ResNet was a good example of that: It's more efficient to learn relu(x + Mx + b) than relu(Mx + b)

One difficulty, though, is that proofs generally /don't/ carry over, and IME a lot of 'classical' methods constrain themselves to provable operations... So at times it can be hard to tell the difference between 'useful folklore' and 'tweak that makes the proofs work.'

I've been delving into sparse coding literature from about ten years ago, and there's a lot of this kind of difficulty... Interestingly, the best sparse coding architectures ended up being very very similar to shallow neural networks. There's a nice 'deep ksvd denoising' paper from a few months back which improves handily on the older sparse coding architectures by bringing in a smattering of ideas from the DNN age; the rant at the end is makes the case for building these 'blended' architectures to get the best of both worlds. https://arxiv.org/abs/1909.13164

I tend to think that the DNN architectures beat the provable-domain architectures for Good Reasons, but the USEFUL response is to build awesome blended architectures that maybe go a long way towards bringing down model sizes and complexity, while increasing explainability.

Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by thesz on Saturday March 07, 2020 @06:22PM (#22514967)
If I am remember correctly, residual layer has form of x+nonlin(Mx+b), not one you provided.

I also found (in my "learn DL" experiments) that for ReLU the x - relu(Mx + b) version works better (trains faster and achieves better accuracy).

Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by bonoboTP on Saturday March 07, 2020 @07:00PM (#22515153)
Plus or minus shouldn't make any difference as long as your input distribution is symmetric around zero.

A simple transformation shows you can get the same effect by just flipping the input the output and the weights. The weights are initialized from a symmetric distribution, so the only difference may come from having the input nonevenly distributed around zero.

y = x - relu(Mx + b)

-y = -x + relu((-M)(-x) + b)

y' = x' + relu(M'x' + b)

Have you tried normalizing your input to have zero mean?

Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by thesz on Sunday March 08, 2020 @08:21AM (#22517674)
ReLU(-x) is not equal to ReLU(x).

The output after ReLU is all non-negatives. So subtraction, actually, does a correction on input.

Yes, I tried normalization on input values and normalization and cross-correlation reduction of outputs of affine transformations. They all have separate positive effects on speed of training and final accuracy.

Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by bonoboTP on Sunday March 08, 2020 @11:02AM (#22518333)
I know relu(-x) is not relu(x). I think the equations I wrote are correct. My point is, your version is not fundamentally different because we can get your effect by using the standard version and flipping the input to the network, the M matrix and then the output will come out with the sign flipped as well. If you put a sign-flipped fully conn layer and a softmax on this (for example) the result will be the same.

The only reason that yours might be better is an asymmetric distribution of x o around 0. If you flip the sign on x, you should get the same benefits.

To summarize: your net is not exactly the same as the usual one, but if you train your version on x and I train the usual version with -x, our results will be indistinguishable.

Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by thesz on Monday March 09, 2020 @10:19AM (#22525513)
No, you can't "flip" input because it is, most probably, output of ReLUs which are all non-negative. In that case you have to learn flipping and all that correction at the Mx+b part, which is, let's say, in the range of tens of thousands of parameters.
Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by bonoboTP on Monday March 09, 2020 @11:31PM (#22532519)
I meant the input input. The first input. Just flip the fist input to the net. Then the flip will automatically cascade through all layers. The M matrices are initialized from a distribution that is symmetric around zero. They don't need to be explicitly flipped, it's not a "correction" in that sense. As long as you only have this type of modules and normal non-residual ReLU layers, it will be equivalent. The net is not equivalent but if you include the random init in your analysis, then you can see there will be a bijection.

I don't think I can explain this well over comments, think about what happens if you initialize your net from scratch. "Self-contained though experiment: Does it make a difference if you multiply all weights by -1 directly after init? " Once that is clear in itself, think of what happens if you init and flip all your very first inputs, the inputs that you feed to the start of the network.

Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by thesz on Tuesday March 10, 2020 @04:08AM (#22533619)
So I made simple experiment in ghci (the "$" operators is a function application: f $ g $ x === f(g(x))):

  Prelude> let relu x = (x + abs x)/2
  Prelude> let res x = x - relu x
  Prelude> let f x = res $ res x
  Prelude> map f [-2..2]
  [-2.0,-1.0,0.0,0.0,0.0]
  Prelude> let res x = x + relu (negate x)
  Prelude> let f x = res $ res x
  Prelude> map f [-2..2]
  [0.0,0.0,0.0,1.0,2.0]
  Prelude> let f x = res $ res $ res x
  Prelude> map f [-2..2]
  [0.0,0.0,0.0,1.0,2.0]
  Prelude> let res x = x - relu x
  Prelude> let f x = res $ res $ res x
  Prelude> map f [-2..2]
  [-2.0,-1.0,0.0,0.0,0.0]
The networks pass different inputs, actually.

I have to add that you cannot have zero mean outputs of residual layer with the definition res x = x + relu (Ax + b). In that case outputs will have non-zero mean and subsequent layers will have to correct for that.

Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by bonoboTP on Tuesday March 10, 2020 @07:57AM (#22534647)
I never said relu(-anything). Please read the equations I wrote above with extreme care to where the minus signs are. They are at precise locations. I also emphasize that my explanation only works if you take into account the fact that the net is randomly initialized. I wrote a quoted sentence in my prev comment which was carefully formulated to build up the right intuition. I don't think I can explain it better.

But I'll try: the sequence of actions (random init, teain your residual net, test your net) is indistinguishable in effect from (random init, train usual net on negated starting input, and test on negated input). The second version will be indistinguishable from running a repeated experiment of the first type with a new random init.

Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by thesz on Tuesday March 10, 2020 @03:12PM (#22538546)
Yes, I read that. The problem is that you need to train the bias to get comparable results.

Also, please note that you cannot have zero mean outputs of residual in the res x = x + relu (Ax + b) and you obviously can have zero mean outputs in the subtraction case (res x = x - relu (Ax + b)).

The very fact that in one case you have zero mean outputs and in other case you don't brings me to necessity to point you to SELU: https://towardsdatascience.com/selu-make-fnns-great-again-sn...

This SELU paper demonstrates, in my opinion, the benefits of having zero mean outputs.

(I have to say that in my experiments SELU was not all that beneficial, but other means that bring zero means into existence were)

I think that residual neural network is capable to route around the case of having to learn non-zero means in inputs. So you are right in stating that these two cases will be indistinguishable. I just have to say that having subtraction instead of addition helps neural net to train faster and get better accuracy just because training process have less things to learn.

Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by bonoboTP on Wednesday March 11, 2020 @12:25AM (#22542975)
I'm fairly sure I was right in the above analysis which would imply the minus version cannot have any substantial advantages.

> please note that you cannot have zero mean outputs of residual in the res x = x + relu (Ax + b) and you obviously can have zero mean outputs in the subtraction case (res x = x - relu (Ax + b)

This is incorrect. You seem to assume x is positive an therefore adding something relu'd onto it will take it further from zero, while your subtractive one can pull it towards and beyond zero. The problem in this reasoning is that in your version you sequentially subtract the residuals so you have the exact symmetric effect, getting further away from zero.

It's like a left hand and a right hand. Not the same, but have the same effect.

I said all I could at this point. If you still have your experiments set up, just try negating your input to the network and initialize randomly. The accuracies observed will be indistinguishable from using your variant. The network is not the same but the training procedure yields a sample from the same distribution.

Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by thesz on Monday March 16, 2020 @04:12PM (#22597069)
Sorry. I consulted the code and it turned out I used relu(Ax + b) - x as a modified residual layer.

I hope it clears things up.

Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by sdenton4 on Saturday March 07, 2020 @06:42PM (#22515073)
Correct; the idea is to take Identity + Tweak, so the Relu should hit the 'tweak.'
Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by srean on Sunday March 08, 2020 @11:57AM (#22518671)
Hey sdenton did you work on or publish that harmonic analysis of correlations that you had mentioned long time ago. I am still quite curious about that. If polishing it for ICML etc is too much of a downer, you can upload it to arxiv. I would love to read that.
Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by sdenton4 on Sunday March 08, 2020 @03:02PM (#22519832)
Hey, send me an email (should be in my profile, or easy to guess: it's a gmail account) and I'll see what I can dig up. :)

I've admittedly been pretty terrible about actually publishing... But definitely have notes and slide decks that could probably be put together into Something.

Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by nograpes on Saturday March 07, 2020 @12:59PM (#22512948)
Do you have any good papers or textbooks where I can learn about this from a numerical optimization perspective?
Re:Proving the Lottery Ticket Hypothesis: Pruning is All You... by adampk on Saturday March 07, 2020 @01:10PM (#22512999)
https://www.amazon.com/Numerical-Optimization-Operations-Fin...

seems to be "the" numerical optimization textbook