Discussion
The weight w for the autoassociative BAM set of vectors \((x_{1}, x_{2},....., x_{L})\) is computed as ........................... .
Answer: A
The key gives (A) \(\sum x_{i}x_{i}^{t}\).
An autoassociative bidirectional associative memory (BAM) stores each pattern against itself. Its weight matrix is the sum of the outer products of the stored vectors, \(W = \sum_{i=1}^{L} x_{i}x_{i}^{t}\). With each \(x_{i}\) a column vector of \(n\) elements, \(x_{i}x_{i}^{t}\) is an \(n \times n\) matrix, so W is \(n \times n\) and can recall a stored pattern from a noisy copy.
Option B gives the same matrix if the vectors are written as rows, so it depends on a convention that the question does not state, but with the usual column-vector reading it is the scalar \(x_{i}^{t}x_{i}\) and cannot be a weight matrix. Options C and D divide one vector by another, which is not defined for vectors.
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