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1. In an error correction learning scenario, a sin...

Erstellt am: 31. Juli 2025

Beantwortet mit O4 Mini von Chat01

Frage

  1. In an error correction learning scenario, a single-layer perceptron is trained to classify inputs into two categories using a supervised learning approach. The initial weight vector is w=[0.5,−0.2], the learning rate is η=0.1, and the input vector is x=[1,2]. The desired output is d=1, but the actual output is calculated as y=w⋅x=0.5⋅1+(−0.2)⋅2=0.1. Compute the updated weight vector after one iteration of the error correction learning rule, given by wnew​=wold​+η(d−y)x. Show all steps of the calculation
    2. In a memory-based learning system, a neural network uses a Hebbian learning rule to update its synaptic weights. The weight matrix between two neurons is initially w=0.3. The pre-synaptic neuron has an activation of x=0.8, and the post-synaptic neuron has an activation of y=0.5. The learning rate is η=0.2. Using the Hebbian learning rule, Δw=η⋅x⋅y, calculate the updated weight after one iteration. Show all steps of the calculation.  
    3. In a competitive learning scenario, a neural network with three output neurons is trained using a winner-takes-all rule. The input vector is x=[0.6,0.8], and the initial weight vectors for the three neurons are w1​=[0.4,0.2], w2​=[0.7,0.5], and w3​=[0.3,0.9]. The learning rate is η=0.15. The neuron with the weight vector closest to the input (based on Euclidean distance) is updated using the rule wnew​=wold​+η(x−wold​). Calculate the Euclidean distances between the input and each weight vector, identify the winning neuron, and compute its updated weight vector after one iteration. Show all steps.
    4. In a Boltzmann learning system, a neural network has a unit with an initial weight w=0.4 connecting to another unit. The network uses a learning rule where the weight update is based on the difference in expected correlations, given by Δw=η(ρ+−ρ−), where ρ+=0.7 (clamped phase correlation) and ρ−=0.3 (free-running phase correlation), and the learning rate is η=0.1. Calculate the weight update Δw and the new weight after one iteration. Show all steps of the calculation

Gedanke

Working through weight updates

The user has asked for calculations related to weight updates. I'll provide the solutions step-by-step.

  1. First, for the simple update, the new weights are [0.59, -0.02].
  2. For Hebbian

Thought for a few seconds

Antwort

Here are each of the four updates worked out step by step:

1. Error‐correction learning (single‐layer perceptron)

  • Given: wold=[0.5, −0.2],x=[1, 2],d=1,η=0.1. \mathbf{w}_{\text{old}} = [0.5,\,-0.2],\quad \mathbf{x} = [1,\,2],\quad d = 1,\quad \eta = 0.1.
  • Compute actual output: y=wold⋅x=0.5⋅1+(−0.2)⋅2=0.5−0.4=0.1. y = \mathbf{w}_{\text{old}}\cdot \mathbf{x} = 0.5\cdot1 + (-0.2)\cdot2 = 0.5 - 0.4 = 0.1.
  • Error term: e=d−y=1−0.1=0.9. e = d - y = 1 - 0.1 = 0.9.
  • Weight change: Δw=η (d−y) x=0.1×0.9×[1,2]=[0.09,  0.18]. \Delta \mathbf{w} = \eta\,(d-y)\,\mathbf{x} = 0.1 \times 0.9 \times [1,2] = [0.09,\;0.18].
  • Updated weight vector: wnew=wold+Δw=[0.5, −0.2]+[0.09,  0.18]=[0.59, −0.02]. \mathbf{w}_{\text{new}} = \mathbf{w}_{\text{old}} + \Delta \mathbf{w} = [0.5,\,-0.2] + [0.09,\;0.18] = [0.59,\,-0.02].

2. Hebbian learning (memory‐based system)

  • Given: wold=0.3,x=0.8,y=0.5,η=0.2. w_{\text{old}} = 0.3,\quad x = 0.8,\quad y = 0.5,\quad \eta = 0.2.
  • Hebbian update rule: Δw=η x y=0.2×0.8×0.5=0.08. \Delta w = \eta\,x\,y = 0.2 \times 0.8 \times 0.5 = 0.08.
  • Updated weight: wnew=wold+Δw=0.3+0.08=0.38. w_{\text{new}} = w_{\text{old}} + \Delta w = 0.3 + 0.08 = 0.38.

3. Competitive (winner‐takes‐all) learning

  • Given:

    x=[0.6, 0.8],η=0.15, \mathbf{x} = [0.6,\,0.8],\quad \eta = 0.15,

    with initial weights

    w1=[0.4,0.2],w2=[0.7,0.5],w3=[0.3,0.9]. \mathbf{w}_1 = [0.4,0.2],\quad \mathbf{w}_2 = [0.7,0.5],\quad \mathbf{w}_3 = [0.3,0.9].

  • Compute Euclidean distances ∥x−wi∥\|\mathbf{x}-\mathbf{w}_i\|:

    1. (\displaystyle d_1 = \sqrt{(0.6-0.4)^2 + (0.8-0.2)^2} = \sqrt{0.2^2 + 0.6^2} = \sqrt{0.04 + 0.36} = \sqrt{0.40} \approx 0.632. )
    2. (\displaystyle d_2 = \sqrt{(0.6-0.7)^2 + (0.8-0.5)^2} = \sqrt{(-0.1)^2 + 0.3^2} = \sqrt{0.01 + 0.09} = \sqrt{0.10} \approx 0.316. )
    3. (\displaystyle d_3 = \sqrt{(0.6-0.3)^2 + (0.8-0.9)^2} = \sqrt{0.3^2 + (-0.1)^2} = \sqrt{0.09 + 0.01} = \sqrt{0.10} \approx 0.316. )

    Both neuron 2 and neuron 3 tie for the smallest distance (≈0.316). By convention (or breaking ties arbitrarily), we’ll choose neuron 2 as the “winner.”

  • Update rule for the winner (w2\mathbf{w}_2):

    w2new=w2old+η(x−w2old). \mathbf{w}_2^{\text{new}} = \mathbf{w}_2^{\text{old}} + \eta\bigl(\mathbf{x}-\mathbf{w}_2^{\text{old}}\bigr).

    First compute the difference:

    x−w2old=[0.6, 0.8]−[0.7, 0.5]=[−0.1, 0.3]. \mathbf{x} - \mathbf{w}_2^{\text{old}} = [0.6,\,0.8] - [0.7,\,0.5] = [-0.1,\,0.3].

    Then scale by η=0.15\eta=0.15:

    0.15×[−0.1, 0.3]=[−0.015, 0.045]. 0.15 \times [-0.1,\,0.3] = [-0.015,\,0.045].

    Finally add to the old w2\mathbf{w}_2:

    w2new=[0.7, 0.5]+[−0.015, 0.045]=[0.685, 0.545]. \mathbf{w}_2^{\text{new}} = [0.7,\,0.5] + [-0.015,\,0.045] = [0.685,\,0.545].

4. Boltzmann learning

  • Given: wold=0.4,ρ+=0.7,ρ−=0.3,η=0.1. w_{\text{old}} = 0.4,\quad \rho^+ = 0.7,\quad \rho^- = 0.3,\quad \eta = 0.1.
  • Update rule: Δw=η (ρ+−ρ−)=0.1×(0.7−0.3)=0.1×0.4=0.04. \Delta w = \eta\,\bigl(\rho^+ - \rho^-\bigr) = 0.1 \times (0.7 - 0.3) = 0.1 \times 0.4 = 0.04.
  • New weight: wnew=wold+Δw=0.4+0.04=0.44. w_{\text{new}} = w_{\text{old}} + \Delta w = 0.4 + 0.04 = 0.44.

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