Associative Memory Find an associative matrix that will store the stimulus-respo

Question

Associative Memory Find an associative matrix that will store the stimulus-response pair s=[.8 0 .6] T=[2 2] Find an associative matrix that will store the stimulus pair s=[.6 0 -8] T=[0 2] Sum the two matrices from parts (a) and (b) and compute the responses of the three stimulus input vectors: s=[0 1 0] s=.8 1 .6] s=[.1 0 .7] Linear Threshold Units For each of the following tasks, sketch the task in stimulus space, and determine whether the task is linearly separable (LS). lf it is LS, find weights and a threshold of a unit that computes the task. If it is not LS, find a feed-forward network that will compute the task, including all weight and threshold values. Hidden Unit Representations A hidden unit space is presented below (left) for a solution to the XOR task using backprop, where the input representations are A =(0, 0), B= (0, 1), C=(1, 0), and D=(1, 1). Sketch the corresponding stimulus space including the discrimination boundaries for the hidden units. Indicate the directions of the corresponding weight vectors. Another hidden unit space is presented below (right) for a network with two output units and the same input patterns as part (a). Discrimination boundaries (dashed lines) are shown for the two output units. One is solving the xOR task. What is the other task?

Explanation / Answer

% Logistics Map % Classic chaos example. Plots semi-stable values of % x(n+1) = r*x(n)*(1-x(n)) as r increases to 4. % % Michael Hanchak, Dayton OH, USA, 2011 clear scale = 10000; % determines the level of rounding maxpoints = 200; % determines maximum values to plot N = 3000; % number of "r" values to simulate a = 2.0; % starting value of "r" b = 4; % final value of "r"... anything higher diverges. rs = linspace(a,b,N); % vector of "r" values M = 500; % number of iterations of logistics equation % Loop through the "r" values for j = 1:length(rs) r=rs(j); % get current "r" x=zeros(M,1); % allocate memory x(1) = 0.5; % initial condition (can be anything from 0 to 1) for i = 2:M, % iterate x(i) = r*x(i-1)*(1-x(i-1)); end % only save those unique, semi-stable values out{j} = unique(round(scale*x(end-maxpoints:end))); end % Rearrange cell array into a large n-by-2 vector for plotting data = []; for k = 1:length(rs) n = length(out{k}); data = [data; rs(k)*ones(n,1),out{k}]; end % Plot the data figure(97);clf h=plot(data(:,1),data(:,2)/scale,'k.'); set(h,'markersize',1) axis tight set(gca,'units','normalized','position',[0 0 1 1]) set(gcf,'color','white') axis off

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