7

$\max(-y_i(w x_i), 0)$ is not partial derivable respect $w$ if $w x_i=0$. Loss functions are problematic when not derivable in some point, but even more when they are flat (constant) in some interval of the weights. Assume $y_i = 1$ and $w x_i < 0$ (that is, an error of type "false negative"). In this case, function $[y_i - \text{sign}(w x_i)]^2 ...


7

There does not appear to be a historical consensus on this. The Wikipedia page on the Perceptrons book (which does not come down on either side) gives an argument that the ability of MLPs to compute any Boolean function was widely known at the time (at the very least to McCulloch and Pitts). However, this page gives an account by someone present at the MIT ...


4

It can be done. The activation function of a neuron does not have to be monotonic. The activation that Rahul suggested can be implemented via a continuously differentiable function, for example $ f(s) = exp(-k(1-s)^2) $ which has a nice derivative $f'(s) = 2k~(1-s)f(s)$. Here, $s=w_0~x_0+w_1~x_1$. Therefore, standard gradient-based learning algorithms are ...


4

I assume you're talking about a perceptron threshold function. One definition of it with an explicit threshold is $$f(\textbf{x})= \begin{cases} 1& \text{if } \textbf{w}\cdot\textbf{x} > t\\ 0& \text{otherwise} \end{cases}.$$ Another form with a bias is $$f(\textbf{x})= \begin{cases} 1& \text{if } \textbf{w}\cdot\textbf{x} + b > 0\\ 0&...


3

Since we're dealing with real-values variables, it is almost certainly the case that the argument of the function will not be $0$. If you care strongly about that point, you can just use sub-gradients instead (and we do have sub-gradients for this function, so there is no problem).


3

The perceptron uses the Heaviside step (or sign) function as the activation function (so you are not free to use any activation function), while a GLM is a generalization of linear regression, where the link function can be, for example, the logit (which leads to the logistic regression), identity function (which leads to linear regression), and so on. The ...


2

Y is the desired output of the perceptron (often referred to as target) , for the given set of input vectors. Rationale behind Y.a<=0 : Prerequisite knowledge : A=A-B : Moves vector A away from direction of vector B A=A+B : Moves A in the direction of B A (.) B >0 ; A vector is directed acutely (<90 deg.) towards B vector A (.) B <0 ; A vector is ...


2

In Brief: re-train your dataset. I believe where you get lower accuracy scores, your model has not converged to the final state. duplicate your dataset multiple times and create a bigger one, then train your model with it. In Detail: number_of_samples_classified_correctly/total_number_of_samples(I'm not sure this should be the correct definition for ...


2

The main problems are that your activation function is not monotonic (as pointed out by csrev), and that it is not continuously differentiable. These make it very difficult / impossible to use standard gradient-based learning algorithms. So yes, there may exist a good solution of weight values, but it is very difficult to find or approximate those weight ...


2

Indeed I think the problem is with the way you've defined the activation function. By selecting it arbitrarily, you could solve many specific problems. In practice, activation functions used are monotonic. It keeps the error function convex at a per-layer level. In theory though I'm not sure exactly what Rosenblatt has claimed so it might be worth calling ...


2

I think your confusion comes from the fact that you are calling those two hidden nodes "perceptrons". You shouldn't call the hidden nodes in your network perceptrons. You should call them "nodes" or "neurons", although the term "multilayer perceptron" comes exactly from that. You should think of a perceptron as something like A perceptron simply computes a ...


2

The section of the book Perceptrons: An Introduction to Computational Geometry (expanded edition, third printing, 1988) that shows the limitations of the perceptron should be 11.8 The Nonseparable Case (p. 181), where the authors write There are many reasons for studying the operation of the perceptron learning program when there is no $\mathbf{A}^*$ with ...


2

Loss function is a function used to measure the loss. It is not used in any component of a neuron. It is used in updating the weights of the neuron i.e., in order to train the neuron. The contribution of a loss function is in the updation of $\bar{W}$. For a given $\bar{X}$ and $\bar{W}$, the neuron gives a post-action value $h$. But the desired output may ...


2

I am specifically asking about the probability that the value is 1 (that is, how sigmoid functions specifically check for this). They don't in general. In the quoted text, there is an explicit constraint that means this can be the case: If it is desirable to predict a probability of a binary class (emphasis mine). This means that the target value $y \in \{...


2

Circles: RETINA / $A_I$ (POJECTION AREA) / $A_{II}$ (ASSOCIATION AREA) Labels: (LOCALISED CONNECTIONS) / (RANDOM CONNECTIONS) / (RANDOM CONNECTIONS) again / RESPONSES


2

sign is not continuous and not differentiable. Let's say it is defined as follows: $$ \text{sing}(a) = \begin{cases} +1 & \text{if $a>0$}\\ -1 & \text{otherwise} \end{cases} $$ where $a = \textbf{w}^T\textbf{x}$ is a linear pre-activation function. The graph will look as follows: It is not continuous and, therefore, is not ...


2

Before proving that XOR cannot be linearly separable, we first need to prove a lemma: Lemma 1 Lemma: If 3 points are collinear and the middle point has a different label than the other two, then these 3 points cannot be linearly separable. Proof: Let us label the points as point $A$, $B$, and $C$. $A$ and $C$ have the same label, and $B$ has a different ...


2

It is important to note that the exact statement is the eqation given below can never be 0 for misclassified points in $ S^+$ $$ E(X) = (y - \text{sign}\{\overline{W} \cdot \overline{X}\}) $$ And $S+$ is defined as the set of all misclassified training points $X \in S$ that satisfy the condition $y(\overline{W} \cdot \overline{X})<0 $ which means that $y$ ...


1

In addition to those mentioned differences, a perceptron can be thought of as a standalone model (which is trained with a specific algorithm, the perceptron algorithm), while the artificial neuron (sometimes only referred to as neuron, in a similar way that an artificial neuron network is commonly abbreviated to neural network) is the smallest computational ...


1

Left-to right: red: SENSORY RECEPTOR OR blue: PROJECTION AREA blue: A-UNITS blue: R$_1$ red: BROKEN LINES SHOW blue: INHIBITORY CONNECTIONS blue: R$_2$


1

Assume we have a binary classification problem, which we want to solve with a simple single-layer perceptron. For a 2d space, a perceptron will have 2 inputs $x_1$ and $x_2$, and a bias denoted $x_0$ which is always $x_0=1$. It also has corresponding learnable weights $w_0$, $w_1$ and $w_2$. This can be vectorized: $$ \overline{x} = \begin{bmatrix} 1 \\ x_1 \...


1

The loss function is simply a way to measure how wrong a neural network is, it doesn't affect the output of the neuron. Say we have a neural network with 3 output neurons that attempts to classify images of cats, dogs, and humans. The output it gives is the confidence of the neural network's classification. For example if the output is [0, 0.2, 0.8] (0 being ...


1

I will first address your main question "Why did the development of neural networks stop between 50s and 80s?" In 40-50s there was a lot of progress (McCulloch and Pitts); the perceptron was invented (Rosenblatt). That gave rise to an AI hype giving many promises (exactly like today)! However, Minsky and Papert have proved in 1969 that a single-...


1

I will tell you my knowledge, correct me if I am wrong. Perceptron Learning Algorithm (PLA) is a simple method to solve the binary classification problem. Define a function: $$ f_w(x) = w^Tx + b $$ where $x \in \mathbb{R}^n$ is an input vector that contains data points and $w$ is a vector with the same dimension as $x$ which present for the parameters of our ...


1

In Single Perceptron / Multi-layer Perceptron(MLP), we only have linear separability because they are composed of input and output layers(some hidden layers in MLP) This is wrong. A multi-layer perceptron (i.e. a feed-forward neural network) with non-linear activation functions can perform non-linear classification and regression. In fact, an MLP with one ...


1

Anything that is not linearly separable cant be solved perceptrons, unless you use feature maps on data to map them to a higher dimension in which it is linearly separable. As a simple, concrete example, perceptron cant learn the XOR function. This page might help you further.


1

This is one of the main skills that separates someone with a deep understanding of, and experience in, machine learning learning, from a neophyte. There are several approaches: Try several methods, perhaps with automated hyperparameter optimization, and see if there's a big difference in typical model quality. This is pretty common if you don't have a lot ...


1

It seems I've solved the issue. There was several mistakes: 1. I've generated random weights from 0 to 1. As a result, too big numbers passed through softmax function (>10000), and the function wasn't calculated correctly. I divided each initial weight on the number of neurons in previous layer and solved the issue. 2. I've calculated separate delta for ...


1

The paper (or report) that formally introduced the perceptron is The Perceptron — A Perceiving and Recognizing Automaton (1957) by Frank Rosenblatt. If you read the first page of this paper, you can immediately understand that's the case. In particular, at some point (page 2, which corresponds to page 5 of the pdf), he writes Recent theoretical studies by ...


Only top voted, non community-wiki answers of a minimum length are eligible