23

This small instability at the end of convergence is a feature of Adam (and RMSProp) due to how it estimates mean gradient magnitudes over recent steps and divides by them. One thing Adam does is maintain a rolling geometric mean of recent gradients and squares of the gradients. The squares of the gradients are used to divide (another rolling mean of) the ...


9

I think the answer to this question is weight sharing in convolutional layers, which you don't have in fully-connected ones. In convolutional layers you only train the kernel, which is then convolved with the input of that layer. If you make the input larger, you still would use the same kernel, only the size of the output would also increase accordingly. ...


6

I agree with Dawny33, choosing learning rate only scales w. While training of Perceptron we are trying to determine minima and choosing of learning rate helps us determine how fast we can reach that minima. If we choose larger value of learning rate then we might overshoot that minima and smaller values of learning rate might take long time for convergence. ...


6

It's actually not true. CNN's don't have to have a fixed-size input. It is possible to build CNN architectures that can handle variable-length inputs. Most standard CNNs are designed for a fixed-size input, because they contain elements of their architecture that don't generalize well to other sizes, but this is not inherent. For example, standard CNN ...


5

You are correct that stacking two layers with a linear activation function on top of each other does not do anything that a single layer could not do (i.e. it is still a linear combination of terms). This changes, once you use other activation functions. Then, once you combine neurons from the previous layer in the next layer e.g. as $w_0 + w_1 f(e_1) +w_2 ...


5

Is it true that linear classifiers differ only in the Learning algorithm, but do they do the same during Prediction y = w1*x1 + w2*x2 + ... + c? Yes, all parametric linear classifiers try to predict the weights for the same equation $y = w1*x1 + w2*x2 + ... + c$ and differ in how they try to optimize the cost function. If I used one method for ...


4

C1 and C2 are coefficients for L1 and L2 regularization, respectively. You can use the same definition of params_space regardless of the optimization algorithm.


4

I managed to find the Bishop's version by unearthing! the Rosenblatt's 1962 Principles of neurodynamics, Page 110 book, so the Wikipedia's version must be the alternative one. It is worth noting that the book also has a chapter on error back-propagation (page 292), which resembles the Wikipedia version, but I think it is not exactly the same.


4

When the data is linearly inseparable, we use MLP. Here what is meant by "data"--is it the response or the input feature that is linearly inseparable? This means that a linear function of the input features is unable to separate the response. To answer your question a bit more directly: Given only a linear function of the inputs, the response is the thing ...


4

First of all, let me add this schema which I think is quite nice to understand the transition and improvement from the initial Rosenblatt's perceptron and the Adaline algorithm: In Adaline, provided that the cost function (your y(t)-s(t)) is differentiable, the weights can be updated and there is no restriction of y and s having the same sign: the objective ...


3

The Adaline (Adaptive Linear Element) and the Perceptron are both linear classifiers when considered as individual units. They both take an input, and based on a threshold, output e.g. either a 0 or a 1. The main difference between the two, is that a Perceptron takes that binary response (like a classification result) and computes an error used to update ...


3

Suppose bias as a threshold. Using threshold, your activation function moves across the $x$ axis which may get complicated. Consequently, people usually use the bias term and always centre the activation function which is the step function at zero. There is nothing wrong in both cases.


3

After some study, I figured out the answer and want to share with people if someone also finds it helpful. The loss function of Perceptron is hinge loss or $J(w) = max(0, -yw^Tx)$. Adding a constant to the loss function does not change the function value, as it does not change the sign of the decision. In other words $J_2(w) = max(0, -\alpha yw^Tx) = J(...


3

Yes, the perceptron learning algorithm is a linear classifier. If your data is separable by a hyperplane, then the perceptron will always converge. It will never converge if the data is not linearly separable. In practice, the perceptron learning algorithm can be used on data that is not linearly separable, but some extra parameter must be defined in order ...


3

The algorithm works by adding or subtracting the feature vector to/from the weight vector. If you only add/subtract parts of the feature vector your a not guaranteed to always nudge the weights in the right direction, which could mess with the convergence of the procedure. The idea is that in the weight space every input vector is a hyperplane. You need to ...


3

With regard to the single-layered perceptron (e.g. as described in wikipedia), for every initial weights vector $w_0$ and training rate $\eta>0$, you could instead choose $w_0'=\frac{w_0}{\eta}$ and $\eta'=1$. For the same training set, training a perceptron with $w_0,\eta$ would be identical to training with $w_0',\eta'$, in the sense that: Both ...


3

A Multilayer Perceptron changes your weights by an algorithm called "backpropagation". This algorithm uses gradient descent combined with a learning rate to change every weight in your MLP. Basically the backpropagation functions by chaining together all functions that are called when calculating the output of one particular node (so chaining together all ...


3

The two pictures you show illustrate how to interprete one perceptron and a MLP consisting of 3 layers. Let us discuss the geometry behind one perceptron first, before explaining the image. We consider a perceptron with $n$ inputs. Thus let $\mathbf{x} \in \mathbb{R}^{n}$ be the input vector, $\mathbf{w} \in \mathbb{R}^{n}$ be the weights, and let $b \in \...


2

The choice of learning rate m does not matter because it just changes the scaling of w. I agree that it is just the scaling of w which is done by the learning rate. Having said that, as I have explained in this answer, the magnitude of learning rate does play a part in the accuracy of the perceptron.


2

The Perceptron's output $f$ is $$ f(\overline\theta \cdot \overline{x}) = \begin{cases} 1 &\text{ if } \overline{\theta }\cdot \overline{x} > 0 \\ 0 &\text{ if } \overline{\theta }\cdot \overline{x} \le 0\end{cases} $$ Here, $\overline{x} = (1, x_1, \dots, x_n)$ where $(x_1, \dots, x_n)$ is the input vector. You can see that the output only ...


2

Your conclusion is correct. Note that the classifier is of the form of $$f(x; \theta) = \operatorname{sign}(\theta^Tx)$$ while the update rule is $$\theta^{(k+1)}=\theta^{(k)}+\eta y_tx_t$$ which only occurs when there is a misclassification and we only care about the in the classification but $$\operatorname{sign}(\theta^Tx)=\operatorname{sign}(\...


2

The brief answers are: 1, 2. No. They depend on different subsets of examples. 3, 4. Yes, but only if it separates classes linearly and you are extremely lucky. Otherwise no. 5, 6. No, because SVM and LDA find only one solution, but perceptron can find many. Now let me explain. Decision boundaries of classicl SVM and LDA are calculated offline based ...


2

Two points about the whole thing You did not test yet. The point behind the training process is to make machine able to learn from the data conditioned on the ability of generalizing this to predicting samples which it has not seen before. Otherwise, the good training is actually overfitting. So here you trained on X and you need to create new samples and ...


2

You are right, the least square solution need not be unique as you have illustrated. In general, we are trying to solve for $w$ in the system $$Xw=t$$ which of course need not even be consistent (meaning has a solution). In the event that it has a solution, the uniqueness is also not guaranteed if the matrix $X$ has non-zero vector in the nullspace of $X$....


2

Would my code be faster if I rewrite it with matrices? Without seeing the code it's impossible to know, but very likely. Also, I would never model single neurons. Too much overhead without any use. Model layers instead. how can I get my performance to be comaparable to that of sklearn Sklearn is open source. Read the code: https://github.com/scikit-...


2

Rather than viewing your data as X = [[0.8, 0.1], [0.7, 0.2], [0.9, 0.3], [0.3, 0.8], [0.1, 0.7], [0.1, 0.9]] Y = [-1, -1, -1, 1, 1, 1] You could have treat the problem as X = [[1, 0.8, 0.1], [1, 0.7, 0.2], [1, 0.9, 0.3], [1, 0.3, 0.8], [1, 0.1, 0.7], [1, 0.1, 0.9]] Y = [-1, -1, -1, 1, 1, 1] That is to append a $1$ in every single entry of $X$. The ...


2

SVMs were developed by Vapnik (1995,1998), based on the structural risk minimization principle (Vapnik, 1982) from statistical learning theory. The complexity of the class of functions performing classification or regression and the algorithm’s generalizability are related. The Vapnik-Chervonenkis (VC) theory provides a general measure of complexity and ...


1

The behavior appears to actually depend on the learning rate $\eta$; a smaller $\eta$ affects which points are misclassified in the next iteration, which affects the weight update more than just by the simple scaling you alluded to. With appropriately small learning rates though, it seems you are guaranteed convergence to some local minimum, if you avoid ...


1

The differences between the Perceptron and Adaline: The Perceptron uses the class labels to learn model coefficients. Adaline uses continuous predicted values (from the net input) to learn the model coefficients, which is more “powerful” since it tells us by “how much” the model is right or wrong.


1

Firstly, in all the linear separator algorithms such as linear regression, logistic regression and the perceptron, adding the bias is as simple as adding a feature column consisting of all 1's. Then the third weight that will be trained will act as the bias $b$. I have some working code for a multi-class perceptron First let's generate some artificial data ...


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