-Artificial Neural Network- Chapter 3 Perceptron

1. -Artificial Neural Network-Chapter 3 Perceptron 朝陽科技大學 資訊管理系 李麗華 教授

2. Introduction of Perceptron In 1957, Rosenblatt and several other researchers developed perceptron, which used the similar network as proposed by McCulloch, and the learning rule for training network to solve pattern recognition problem. (*) But, this model was later criticized by Minsky who proved that it cannot solve the XOR problem.

3. W11 f1 W13 X1 W12 f3 Y1 W21 W23 f2 X2 W22 Perceptron Network Structure

4. 1 net j > 0 Yj= if 0 net j 0 Perceptron Training Process • Set up the network input nodes, layer(s), & connections • Randomly assign weights to Wij & bias: • Input training sets Xi (preparing Tj for verification ) • Training computation:

5. If than: Update weights and bias : The training process • Training computation: 5. repeat 3 ~5 until every input pattern is satisfied

6. The recall process After the network has trained as mentioned above, any input vector X can be send into the perceptron network. The trained weights, Wij, and the bias, θj , is used to derive netj and, therefore, the output Yj can be obtained for pattern recognition.

7. X１ Ｗ11 f Ｙ X2 X２ Ｗ21 X1 f Exercise1: Solving the OR problem • Let the training patterns are as follows.

8. Let W11=1, W21=0.5, Θ=0.5 Let learning rate η=0.1 • The initial net function is: net = W11•X1+ W21•X2 -Θ i.e. net = 1•X11 + 0.5•X21 - 0.5 • Feed the input pattern into network one by one (0,0), net= -0.5, Y=0, (T-Y)= 0 O.K. (0,1), net= 0, Y= 0, (T-Y)=1-Ø=1 Need to update weight (1,0), net= 0.5, Y=1, (T-Y)= 0 O.K. (1,1), net= 1, Y=1, (T-Y)= 0 O.K. • update weights for pattern (0,1) ΔW11=(0.1)(1)( 0)= 0, W11= W11+ΔW11=1 ΔW12=(0,1)(1)( 1)= 0.1, W21=0.5+0.1=0.6 ΔΘ=-(0.1)(1)=-0.1 Θ=0.5-0.1=0.4

9. Applying new weights to the net function: net=1•X1+0.6•X2-0.4 • Verify the pattern (0,1) to see if it satisfies the expected output. (0,1), net= 0.2, Y= 1, (T-Y)=Ø O.K. • Feed the next input pattern, again, one by one (1,0), net= 0.6, Y=1, (T-Y)= Ø O.K. (1,1), net= 1.2, Y=1, (T-Y)= Ø O.K. • Since the first pattern(0,0) has not been testified with the new weights, feed again. (0,0), net=-0.4, Y=Ø, δ= Ø O.K. • Now all the patterns are satisfied. Hence, the network is successfully trained for OR patterns. We get the final network is : net=1•X1+0.6•X2-0.4 (This is not the only solution, other solutions are possible.)

10. X１ １ Ｙ X２ 0.6 Θ=0.4 • The trained network is formed as follow: f • Recall process: • Once the network is trained, we can apply any two element vectors as a pattern and feed the pattern into the network for recognition. For example, we can feed (1,0) into to the network • (1,0), net= 0.6, Y=1 • Therefore, this pattern is recognized as1.

11. X2 f1 X1 Exercise2: Solving the AND problem • Let the training patterns are used as follow

12. Sol: Let W11=0.5, W21=0.5, Θ=1, Let η=0.1 • The initial net function is: net=0.5X11+0.5X21 – 1 • Feed the input pattern into network one by one (0,0), net=-1, Y=Ø, (T-Y)= Ø, O.K. (0,1), net=-0.5, Y= Ø, (T-Y)= Ø, O.K. (1,0), net=- 0.5, Y= Ø, (T-Y)= Ø, O.K. (1,1), net= Ø, Y= Ø, (T-Y)= 1, Need to update weight • update weights for pattern (1,1) which does not satisfying the expected output: ΔW11=(0,1)(1)( 1)= 0.1, W11=0.6 ΔW21=(0,1)(1)( 1)= 0.1, W21=0.5+0.1=0.6, ΔΘ=-(0.1)(1)=-0.1, Θ=1-0.1=0.9

13. 0.6 X１ Ｙ Θ=0.9 X２ 0.6 • Applying new weights to the net function: net=0.6X1 + 0.6X2 - 0.9 • Verify the pattern (1,1) to see if it satisfies the expected output. (1,1), net= 0.3, Y= 1, δ= Ø O.K. • Since the previous patterns are not testified with the new weights, feed them again. (0,0), net=-0.9 Y=Ø, δ= Ø O.K. (0,1), net=-0.3 Y= Ø, δ=Ø O.K. (1,0), net=- 0.3, Y= Ø, δ= Ø O.K. • We can generate the pattern recognition function for OR pattern is: net= 0.6X1 + 0.6X2 - 0.9

14. X2 P4 P2 f1 P1 f2 X1 f2 P3 P2, P3 P4 f3 f1 P1 Example: Solving the XOR problem • Let the training patterns are used as follow.

15. Θ1 Ｗ11 f1 X1 Ｗ12 f3 Ｗ21 f2 X2 Ｗ22 Θ3 Θ2 • Let W11=1.0, W21= -1.0, Θ=0 If we choose one layer network, it will be proved that the network cannot be converged. This is because the XOR problem is a non-linear problem, i.e., one single linear function is not enough to recognize the pattern. Therefore, the solution is to add one hidden layer for extra functions.

16. The following pattern is formed. • So we solve f1 as AND problem and f2 as OR problem • Assume we found the weights are: W11=0.3, W21=0.3, W12= 1, W22= 1 θ1=0.5 θ2=0.2 • Therefore the network function for f1 & f2 is: f1 = 0.3X11+ 0.3X21 - 0.5 f2 = 1•X12+ 1•X22 - 0.2

17. Θ=0.5 0.3 X１ f1 1 1 Y 0.3 -0.5 X２ f2 Θ= -0.1 1 Θ=0.2 • Now we need to feed the input one by one for training the network for f1 and f2 seprearately. This is to satisfiying the expected output for f1 using T1 and for f2 using T2. • Finally, we use the output of f1 and f2 as input pattern to train the network for f3. ,We can derive the result as following. f3=1•X13 - 0.5X23 + 0.1

18. Perceptron Learning of the Logical AND Function

19. Perceptron Learning of the Logical OR Function

20. Perceptron Attempt at Learning the Logical XOR Function  The one-layer network cannot solve the XOR problem…..the network goes to the infinite loop