Calculation of T = N + 1 state of the 4 neuron Hopfield net.2

Exemplar 1

W12 = W23 = W34 = W24 =W13 = 1 * 0 = 0    W14 = 1 * 1 = 1

Exemplar 2

W12 = W14 = W34 = W24 =W13 = 1 * 0 = 0 W23 = 1 * 1 = 1

TRAINED NETWORK'S WEIGHTS : W14 = W23 = 1, ALL OTHERS ZERO

TEST PATTERN:

T + 1 STATES:

U1 = X2 * W12 + X3 * W13 + X4 * W14 = 1 * 0 + 1 * 0 + 0 * 1 = 0

U2 = X1 * W12 + X3 * W23 + X4 * W24 = 1 * 0 + 1 * 1 + 0 * 0 = 1

U3 = X2 * W23 + X1 * W13 + X4 * W34 = 1 * 1 + 1 * 0 + 0 * 0 = 1

U4 = X2 * W24 + X3 * W34 + X1 * W14 = 1 * 0 + 1 * 0 + 1 * 0 = 1

X1 = U1, X2 = U2, X3= U3, X4 = U4 using the step function for F

T + 2 STATES:

U1 = X2 * W12 + X3 * W13 + X4 * W14 = 1 * 0 + 1 * 0 + 1 * 1 = 1

U2 = X1 * W12 + X3 * W23 + X4 * W24 = 0 * 0 + 1 * 1 + 1 * 0 = 1

U3 = X2 * W23 + X1 * W13 + X4 * W34 = 1 * 1 + 0 * 0 + 1 * 0 = 1

U4 = X2 * W24 + X3 * W34 + X1 * W14 = 1 * 0 + 1 * 0 + 0 * 0 = 0

X1 = U1, X2 = U2, X3= U3, X4 = U4 using the step function for F

Hence the system oscilates between two solutions.  It is too small to be effective.