Is max operation differentiable in Pytorch? - optimization

I am using Pytorch to training some neural networks. The part I am confused about is:
prediction = myNetwork(img_batch)
max_act = prediction.max(1)[0].sum()
loss = softcrossentropy_loss - alpha * max_act
In the above codes, "prediction" is the output tensor of "myNetwork".
I hope to maximize the larget output of "prediction" over a batch.
For example:
[[-1.2, 2.0, 5.0, 0.1, -1.5] [9.6, -1.1, 0.7, 4,3, 3.3]]
For the first prediction vector, the 3rd element is the larget, while for the second vector, the 1st element is the largets. And I want to maximize "5.0+9.6", although we cannot know what index is the larget output for a new input data.
In fact, my training seems to be successful, because the "max_act" part was really increased, which is the desired behavior to me. However, I heard some discussion about whether max() operation is differentiable or not:
Some says, mathmatically, max() is not differentiable.
Some says, max() is just an identity function to select the largest element, and this largest element is differentiable.
So I got confused now, and I am worried if my idea of maximizing "max_act" is wrong from the beginning.
Could someone provide some guidance if max() operation is differentiable in Pytorch?

max is differentiable with respect to the values, not the indices. It is perfectly valid in your application.
From the gradient point of view, d(max_value)/d(v) is 1 if max_value==v and 0 otherwise. You can consider it as a selector.
d(max_index)/d(v) is not really meaningful as it is a discontinuous function, with only 0 and undefined as possible gradients.

Related

Tensorflow Tensorboard - should I follow the "smooth" value or the "Value"?

I am using TF tensorboard to monitor the training progress for a model. I am getting a bit confused because I am seeing the two points that represent the validation loss value showing a different direction:
Time=13:30 Smoothed=18.33 Value=15.41..........
Time=13:45 Smoothed=17.76 Value=16.92
In this case, is the validation loss increasing or decreasing? thanks!
As I cannot put figures in the comments, have a look at this graph.
If you watch the falling slope between x = 50 and x = 100, you will see that locally, the real values increase at some points (usually after downward spikes). So you could conclude that your function values are increasing. But at a larger scope you will see that the function values are decreasing. The smoothing helps you to get make the interpretation easier, but does not return exact values.
Coming back to the local example, it would give you the insight that the overall trend is a decreasing function, but it does not provide accurate loss values.

Machine learning: why the cost function does not need to be derivable?

I was playing around with Tensorflow creating a customized loss function and this question about general machine learning arose to my head.
My understanding is that the optimization algorithm needs a derivable cost function to find/approach a minimum, however we can use functions that are non-derivable such as the absolute function (there is no derivative when x=0). A more extreme example, I defined my cost function like this:
def customLossFun(x,y):
return tf.sign(x)
and I expected an error when running the code, but it actually worked (it didn't learn anything but it didn't crash).
Am I missing something?
You're missing the fact that the gradient of the sign function is somewhere manually defined in the Tensorflow source code.
As you can see here:
def _SignGrad(op, _):
"""Returns 0."""
x = op.inputs[0]
return array_ops.zeros(array_ops.shape(x), dtype=x.dtype)
the gradient of tf.sign is defined to be always zero. This, of course, is the gradient where the derivate exists, hence everywhere but not in zero.
The tensorflow authors decided to do not check if the input is zero and throw an exception in that specific case
In order to prevent TensorFlow from throwing an error, the only real requirement is that you cost function evaluates to a number for any value of your input variables. From a purely "will it run" perspective, it doesn't know/care about the form of the function its trying to minimize.
In order for your cost function to provide you a meaningful result when TensorFlow uses it to train a model, it additionally needs to 1) get smaller as your model does better and 2) be bounded from below (i.e. it can't go to negative infinity). It's not generally necessary for it to be smooth (e.g. abs(x) has a kink where the sign flips). Tensorflow is always able to compute gradients at any location using automatic differentiation (https://en.wikipedia.org/wiki/Automatic_differentiation, https://www.tensorflow.org/versions/r0.12/api_docs/python/train/gradient_computation).
Of course, those gradients are of more use if you've chose a meaningful cost function isn't isn't too flat.
Ideally, the cost function needs to be smooth everywhere to apply gradient based optimization methods (SGD, Momentum, Adam, etc). But nothing's going to crash if it's not, you can just have issues with convergence to a local minimum.
When the function is non-differentiable at a certain point x, it's possible to get large oscillations if the neural network converges to this x. E.g., if the loss function is tf.abs(x), it's possible that the network weights are mostly positive, so the inference x > 0 at all times, so the network won't notice tf.abs. However, it's more likely that x will bounce around 0, so that the gradient is arbitrarily positive and negative. If the learning rate is not decaying, the optimization won't converge to the local minimum, but will bound around it.
In your particular case, the gradient is zero all the time, so nothing's going to change at all.
If it didn't learn anything, what have you gained ? Your loss function is differentiable almost everywhere but it is flat almost anywhere so the minimizer can't figure out the direction towards the minimum.
If you start out with a positive value, it will most likely be stuck at a random value on the positive side even though the minima on the left side are better (have a lower value).
Tensorflow can be used to do calculations in general and it provides a mechanism to automatically find the derivative of a given expression and can do so across different compute platforms (CPU, GPU) and distributed over multiple GPUs and servers if needed.
But what you implement in Tensorflow does not necessarily have to be a goal function to be minimized. You could use it e.g. to throw random numbers and perform Monte Carlo integration of a given function.

taking the gradient in Tensorflow, tf.gradient

I am using this function of tensorflow to get my function jacobian. Came across two problems:
The tensorflow documentation is contradicted to itself in the following two paragraph if I am not mistaken:
gradients() adds ops to the graph to output the partial derivatives of ys with respect to xs. It returns a list of Tensor of length len(xs) where each tensor is the sum(dy/dx) for y in ys.
Blockquote
Blockquote
Returns:
A list of sum(dy/dx) for each x in xs.
Blockquote
According to my test, it is, in fact, return a vector of len(ys) which is the sum(dy/dx) for each x in xs.
I do not understand why they designed it in a way that the return is the sum of the columns(or row, depending on how you define your Jacobian).
How can I really get the Jacobian?
4.In the loss, I need the partial derivative of my function with respect to input (x), but when I am optimizing with respect to the network weights, I define x as a placeholder whose value is fed later, and weights are variable, in this case, can I still define the symbolic derivative of function with respect to input (x)? and put it in the loss? ( which later when we optimize with respect to weights will bring second order derivative of the function.)
I think you are right and there is a typo there, it was probably meant to be "of length len(ys)".
For efficiency. I can't explain exactly the reasoning, but this seems to be a pretty fundamental characteristic of how TensorFlow handles automatic differentiation. See issue #675.
There is no straightforward way to get the Jacobian matrix in TensorFlow. Take a look at this answer and again issue #675. Basically, you need one call to tf.gradients per column/row.
Yes, of course. You can compute whatever gradients you want, there is no real difference between a placeholder and any other operation really. There are a few operations that do not have a gradient because it is not well defined or not implemented (in which case it will generally return 0), but that's all.

Tensorflow: opt.compute_gradients() returns values different from the weight difference of opt.apply_gradients()

Question: What is the most efficient way to get the delta of my weights in the most efficient way in a TensorFlow network?
Background: I've got the operators hooked up as follows (thanks to this SO question):
self.cost = `the rest of the network`
self.rmsprop = tf.train.RMSPropOptimizer(lr,rms_decay,0.0,rms_eps)
self.comp_grads = self.rmsprop.compute_gradients(self.cost)
self.grad_placeholder = [(tf.placeholder("float", shape=grad[1].get_shape(), name="grad_placeholder"), grad[1]) for grad in self.comp_grads]
self.apply_grads = self.rmsprop.apply_gradients(self.grad_placeholder)
Now, to feed in information, I run the following:
feed_dict = `training variables`
grad_vals = self.sess.run([grad[0] for grad in self.comp_grads], feed_dict=feed_dict)
feed_dict2 = `feed_dict plus gradient values added to self.grad_placeholder`
self.sess.run(self.apply_grads, feed_dict=feed_dict2)
The command of run(self.apply_grads) will update the network weights, but when I compute the differences in the starting and ending weights (run(self.w1)), those numbers are different than what is stored in grad_vals[0]. I figure this is because the RMSPropOptimizer does more to the raw gradients, but I'm not sure what, or where to find out what it does.
So back to the question: How do I get the delta on my weights in the most efficient way? Am I stuck running self.w1.eval(sess) multiple times to get the weights and calc the difference? Is there something that I'm missing with the tf.RMSPropOptimizer function.
Thanks!
RMSprop does not subtract the gradient from the parameters but use more complicated formula involving a combination of:
a momentum, if the corresponding parameter is not 0
a gradient step, rescaled non uniformly (on each coordinate) by the square root of the squared average of the gradient.
For more information you can refer to these slides or this recent paper.
The delta is first computed in memory by tensorflow in the slot variable 'momentum' and then the variable is updated (see the C++ operator).
Thus, you should be able to access it and construct a delta node with delta_w1 = self.rmsprop.get_slot(self.w1, 'momentum'). (I have not tried it yet.)
You can add the weights to the list of things to fetch each run call. Then you can compute the deltas outside of TensorFlow since you will have the iterates. This should be reasonably efficient, although it might incur an extra elementwise difference, but to avoid that you might have to hack around in the guts of the optimizer and find where it puts the update before it applies it and fetch that each step. Fetching the weights each call shouldn't do wasteful extra evaluations of part of the graph at least.
RMSProp does complicated scaling of the learning rate for each weight. Basically it divides the learning rate for a weight by a running average of the magnitudes of recent gradients of that weight.

What is the best way to implement weight constraints in TensorFlow?

Suppose we have weights
x = tf.Variable(np.random.random((5,10)))
cost = ...
And we use the GD optimizer:
upds = tf.train.GradientDescentOptimizer(lr).minimize(cost)
session.run(upds)
How can we implement for example non-negativity on weights?
I tried clipping them:
upds = tf.train.GradientDescentOptimizer(lr).minimize(cost)
session.run(upds)
session.run(tf.assign(x, tf.clip_by_value(x, 0, np.infty)))
But this slows down my training by a factor of 50.
Does anybody know a good way to implement such constraints on the weights in TensorFlow?
P.S.: in the equivalent Theano algorithm, I had
T.clip(x, 0, np.infty)
and it ran smoothly.
You can take the Lagrangian approach and simply add a penalty for features of the variable you don't want.
e.g. To encourage theta to be non-negative, you could add the following to the optimizer's objective function.
added_loss = -tf.minimum( tf.reduce_min(theta),0)
If any theta are negative, then add2loss will be positive, otherwise zero. Scaling that to a meaningful value is left as an exercise to the reader. Scaling too little will not exert enough pressure. Too much may make things unstable.
As of TensorFlow 1.4, there is a new argument to tf.get_variable that allows to pass a constraint function that is applied after the update of the optimizer. Here is an example that enforces a non-negativity constraint:
with tf.variable_scope("MyScope"):
v1 = tf.get_variable("v1", …, constraint=lambda x: tf.clip_by_value(x, 0, np.infty))
constraint: An optional projection function to be applied to the
variable
after being updated by an Optimizer (e.g. used to implement norm
constraints or value constraints for layer weights). The function must
take as input the unprojected Tensor representing the value of the
variable and return the Tensor for the projected value
(which must have the same shape). Constraints are not safe to
use when doing asynchronous distributed training.
By running
sess.run(tf.assign(x, tf.clip_by_value(x, 0, np.infty)))
you are consistently adding nodes to the graph and making it slower and slower.
Actually you may just define a clip_op when building the graph and run it each time after updating the weights:
# build the graph
x = tf.Variable(np.random.random((5,10)))
loss = ...
train_op = tf.train.GradientDescentOptimizer(lr).minimize(loss)
clip_op = tf.assign(x, tf.clip(x, 0, np.infty))
# train
sess.run(train_op)
sess.run(clip_op)
I recently had this problem as well. I discovered that you can import keras which has nice weight constraint functions as use them directly in the kernen constraint in tensorflow. Here is an example of my code. You can do similar things with kernel regularizer
from keras.constraints import non_neg
conv1 = tf.layers.conv2d(
inputs=features['x'],
filters=32,
kernel_size=[5,5],
strides = 2,
padding='valid',
activation=tf.nn.relu,
kernel_regularizer=None,
kernel_constraint=non_neg(),
use_bias=False)
There is a practical solution: Your cost function can be written by you, to put high cost onto negative weights. I did this in a matrix factorization model in TensorFlow with python, and it worked well enough. Right? I mean it's obvious. But nobody else mentioned it so here you go. EDIT: I just saw that Mark Borderding also gave another loss and cost-based solution implementation before I did.
And if "the best way" is wanted, as the OP asked, what then? Well "best" might actually be application-specific, in which case you'd need to try a few different ways with your dataset and consider your application requirements.
Here is working code for increasing the cost for unwanted negative solution variables:
cost = tf.reduce_sum(keep_loss) + Lambda * reg # Cost = sum of losses for training set, except missing data.
if prefer_nonneg: # Optionally increase cost for negative values in rhat, if you want that.
negs_indices = tf.where(rhat < tf.constant(0.0))
neg_vals = tf.gather_nd(rhat, negs_indices)
cost += 2. * tf.reduce_sum(tf.abs(neg_vals)) # 2 is a magic number (empirical parameter)
You are free to use my code but please give me some credit if you choose to use it. Give a link to this answer on stackoverflow.com please.
This design would be considered a soft constraint, because you can still get negative weights, if you let it, depending on your cost definition.
It seems that constraint= is also available in TF v1.4+ as a parameter to tf.get_variable(), where you can pass a function like tf.clip_by_value. This seems like another soft constraint, not hard constraint, in my opinion, because it depends on your function to work well or not. It also might be slow, as the other answerer tried the same function and reported it was slow to converge, although they didn't use the constraint= parameter to do this. I don't see any reason why one would be any faster than the other since they both use the same clipping approach. So if you use the constraint= parameter then you should expect slow convergence in the context of the original poster's application.
It would be nicer if also TF provided true hard constraints to the API, and let TF figure out how to both implement that as well as make it efficient on the back end. I mean, I have seen this done in linear programming solvers already for a long time. The application declares a constraint, and the back end makes it happen.