Given that η refers to learning rate and xi refers to the ith input to the…
2025
Given that η refers to learning rate and xi refers to the ith input to the neuron, Which of the followings most suitably describes the weight updation rule of a Kohonen SOM? (where 'j' refers to the jth neuron in the lattice)
- A.
wij(new)=(1−η)∗wij(old)+η∗xi
- B.
wij(new)=η∗wij(old)+(1−η)∗xi
- C.
wij(new)=(1−η)∗wij(old)+xi
- D.
wij(new)=wij(old)+η∗xi
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Correct answer: A
Key idea: the SOM weight update moves each neuron's weight vector toward the input vector by a fraction determined by the learning rate and the neighborhood function.
General vector form: w_j(new) = w_j(old) + η * h_j * (x − w_j(old)), where h_j is the neighborhood function (depends on distance from the winner).
Component-wise (i-th component): w_ij(new) = w_ij(old) + η * h_j * (x_i − w_ij(old)).
For the winning neuron the neighborhood function equals 1, so w_ij(new) = w_ij(old) + η * (x_i − w_ij(old)) = (1−η) * w_ij(old) + η * x_i.
Therefore the expression (1−η) * w_ij(old) + η * x_i correctly represents the updated weight for the winning neuron.
Common mistakes to avoid:
Using swapped coefficients like η * w_ij(old) + (1−η) * x_i is incorrect because it does not equal the standard η * (x_i − w_ij(old)) update.
Omitting η on the input term (giving (1−η) * w_ij(old) + x_i) removes control over step size and causes overly large updates.
Adding η * x_i directly to the old weight without subtracting η * w_ij(old) (i.e., w_ij(old) + η * x_i) is not equivalent to moving toward x_i by a controlled fraction of the difference.
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