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

朝陽科技大學 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.

朝陽科技大學 李麗華 教授 3 Perceptron Network Structure W 21 W 12 W 11 X1X1 X2X2 Y1Y1 f1f1 f2f2 f3f3 W 22 W 13 W 23

朝陽科技大學 李麗華 教授 4 Perceptron Training Process 1.Set up the network input nodes, layer(s), & connections 2.Randomly assign weights to W ij & bias: 3.Input training sets X i (preparing T j for verification ) 4.Training computation: 1 net j > 0 if Yj=Yj= 0 net j 0

朝陽科技大學 李麗華 教授 5 The training process 4.Training computation: If than: 5. repeat 3 ~5 until every input pattern is satisfied Update weights and bias :

朝陽科技大學 李麗華 教授 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, W ij, and the bias, θ j, is used to derive net j and, therefore, the output Y j can be obtained for pattern recognition.

Exercise1: Solving the OR problem Let the training patterns are as follows. X1X1 X2X2 f X1X1 X2X2 T Ｗ 11 Ｗ 21 X２X２ X１X１ Ｙ f

Let W 11 =1, W 21 =0.5, Θ =0.5 Let learning rate η=0.1 The initial net function is: net = W 11 X 1 + W 21 X 2 - Θ i.e. net = 1 X X 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) ΔW 11 =(0.1)(1)( 0)= 0, W 11 = W 11+ ΔW 11= 1 ΔW 12 =(0,1)(1)( 1)= 0.1, W 21 = =0.6 ΔΘ=-(0.1)(1)=-0.1 Θ= =0.4

Applying new weights to the net function: net=1X X 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= ØO.K. Now all the patterns are satisfied. Hence, the network is successfully trained for OR patterns. We get the final network is : net=1X X (This is not the only solution, other solutions are possible.)

The trained network is formed as follow: Θ=0.4 X２X２ X１X１ １ 0.6 Ｙ 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 as 1. f

朝陽科技大學 李麗華 教授 11 Exercise2: Solving the AND problem Let the training patterns are used as follow X1X1 X2X2 T f1f1 X2X2 X1X1

Sol: Let W 11 =0.5, W 21 =0.5, Θ=1, Let η=0.1 The initial net function is: net =0.5X X 21 – 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: ΔW 11 =(0,1)(1)( 1)= 0.1, W 11 =0.6 ΔW 21 =(0,1)(1)( 1)= 0.1, W 21 = =0.6, ΔΘ=-(0.1)(1)=-0.1, Θ=1-0.1=0.9

Applying new weights to the net function: net=0.6X X Verify the pattern (1,1) to see if it satisfies the expected output. (1,1),net= 0.3, Y= 1,δ= ØO.K. Since the previous patterns are not testified with the new weights, feed them again. (0,0), net=-0.9 Y=Ø,δ= ØO.K. (0,1),net=-0.3 Y= Ø,δ=ØO.K. (1,0),net=- 0.3, Y= Ø,δ= ØO.K. We can generate the pattern recognition function for OR pattern is: net= 0.6X X Θ =0.9 X２X２ X１X１ 0.6 Ｙ

朝陽科技大學 李麗華 教授 14 Example: Solving the XOR problem Let the training patterns are used as follow. X1X1 X2X2 T P2P2 P4P4 P3P3 f2f2 P1P1 X2X2 X1X1 f1f1 f3f3 f1f1 P2, P3 P4 f2f2 P1

Let W 11 =1.0, W 21 = -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. Θ3Θ3 f3f3 f1f1 f2f2 Ｗ 21 Ｗ 12 X2X2 X1X1 Ｗ 11 Ｗ 22 Θ1Θ1 Θ2Θ2

The following pattern is formed. So we solve f 1 as AND problem and f 2 as OR problem Assume we found the weights are: W 11 =0.3, W 21 =0.3, W 12 = 1, W 22 = 1 θ 1 =0.5 θ 2 =0.2 Therefore the network function for f1 & f2 is: f 1 = 0.3X X f 2 = 1 X X X1X1 X2X2 f1f1 f2f2 T3T

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 f 3., We can derive the result as following. f 3 =1 X X Θ =0.2 Θ =0.5 f1f Y f2f X２X２ X１X１ 1 Θ = -0.1

Perceptron Learning of the Logical AND Function

Perceptron Learning of the Logical OR Function

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