Exam1                                                                                      CO 40503                                                                       Spring 2006

Name_______________________

 

 

1.              Answer T or F to the following questions (10 points)

 

F               a.     AI is field of research concerned with the development of hardware

 

T              b.    A trained STM has a low entropy value

 

F               c.     In CLS, it is necessary to perform a complex algorithm to obtain the R/P (Reward/Punishment) compensation

 

T              d.    Backprogation is the R/P mechanism in Neural Networks

 

F               e.     Associative Memory is a slow but very accurate classifier

 

T          f.     To select the next option in a CLS, the use of the maximum assumes better results

 

T              g.  In a neural network, the formula  ¶d(i,<t>) = α*ea(j)*a(h)+β*¶(i,<t-1>) includes a correction for the momentum

 

F               h.    It is possible to write STM files using a Java Applet

 

F               i.    The p = p + § + (1-P)(P) compensates more at the beginning

 

T                k   A Neural Network is a slow learning algorithm, however it produces a classifier than can generalize and extrapolate well

 

2.             Choose the correct option  (10 points)

 

     j__  The difference between Information and Knowledge in  AI is similar to the difference between

 

                         a.  Find and Select     b.    Find and Search   c.   Search and Select           d.    Get and Put

   

     k__  Donald Michie, Peter Bock have developed algorithms for

 

                  a.  Neural Networks   b. Associative Memory  c.  Expert Systems    d.  Collective Learning

 

    l__ Rummelhart  is well known for his work on

 

                                     a.  Neural Networks   b. Associative Memory  c.  Expert Systems    d.  Collective Learning

 

    m__ The following formula is generally used to to obtain faster learning at the beginning

 

            a.   P = p + (1-p)(1-p)§        b.    p   =  p + (1-p)p §        c.     p   = p -        d.   p = p*(1-§)                     

 

    n__  A non learner random player can be obtained in an STM if you use

 

             a.   p = 0           b.     § = 0          c.   p = 1        d.    § = 1       


3. Given the following STM for Exclusive OR Learner (20 points)

 

 

STM

0

1

00

0.5

0.5

01

0.5

0.5

10

0.5

0.5

11

0.5

0.5

 

 

  a) Using a single steps process using the following random numbers

     (0.56,   0.45,  0.23, .076, 0.12, 0.67)   perform the first six iterations with the following four inputs  (00,  01, 10, 11, 00,11) with the following truth vector  ( 0,1,1,0,0,0 ). Show the result of the STM after the sixth iteration. Use extra paper to neatly show all your computations  (USING BETA = 0.8)

 

STM

0

1

00

0.90 

0.10 

01

0.10 

0.90 

10

0.10 

0.90 

11

0.90 

0.10 

 

 

b) Selecting the Maximum, again in single steps, perform the first six iterations with the following four inputs  (00,  01, 10, 11, 00,11)  with the same truth vector  ( 0,1,1,0,0,0 ). Show the result of the STM after the sixth iteration. Use extra paper to neatly show all your computations  (USING BETA = 0.8)

 

00

0.90 

0.10 

01

0.10 

0.90 

10

0.10 

0.90 

11

0.90 

0.10 

 

 

c)  Any comments about both methods?

 

 

In this both results are the same since we are using single non collective learning

 

4.  Consider again the exclusive Or function, using the following Neural network formulas, perform the first two iterations (propagation, backpropagation) with the following inputs  (00,  01)  with the same truth vector  ( 0,1). Assuming the initial seed values given below, present the final values Use extra paper to neatly show all your computations

 

 

Propagation

 

 

 

Backpropagation

 

 

Input or Intermediate Axon :  a(h) 

 

alpha

0.9

 

dendritic weight:  d(i) : weight for  a(h) in n(j)

general output error :  e(s) = (V(s)-a(s))

 

neuronal summation :  n(j) =  ·d(i)*a(i)

 

intermediate error: e(i) = ·(ea(pos)*d(pos))

neuronal threshold  u(j) :  Fire constant

 

axon error  :  ea(j) = a(j)*(1-a(j))*e(j)

 

Output axon :  a(j) =  Round<  1/(1 + exp(-n(j)-u(j) )

dendritic correction: ¶d(i,<t>) = α*ea(j)*a(h)+β*¶(i,<t-1>)

Output axon  :  a(j) : fire if n(j) > - u(j)

 

new dendritic value  d(i,<t+1>)  = d(i,<t>) +   ¶d(i,<t>)

 

 

 

 

 

Propagation

 

 

 

 

 

d (i)

a(h)

n(j) 

T(j)

a(j) T

a(j) Sigmoid

a(1)

1.00

0.00

0.00

0.00

0.00

0.50

a(2)

1.00

1.00

1.00

0.00

1.00

0.73

 

 

 

 

 

 

 

 

d(1)

-5.05

0.00

0.00

 

 

 

d(2)

-3.05

1.00

-3.05

 

 

 

T(3)

4.94

1.00

4.94

 

 

 

a(3)

 

 

-3.05

4.94

1.00

0.87

 

 

 

 

 

 

 

 

d(11)

-3.95

0.00

0.00

 

 

 

d(22)

-1.95

1.00

-1.95

 

 

 

d(33)

-8.00

1.00

-8.00

 

 

 

T(4)

6.06

1.00

6.06

 

 

 

a(4)

 

 

-9.95

6.06

0.00

0.02

 

 

 

Backpropagation

 

 

 

 

a(j) f

e(j) f

ea(j)

¶(i,<t>)

¶(i,<t-1>)

d(i,<t+1>)

 

a(4)

0.54

-0.54

-1.34E-01

 

 

 

 

u(4)

 

 

 

-0.07

-0.07

6.00

 

d(33)

 

 

 

0.00

0.00

-7.98

 

d(22)

 

 

 

-0.05

-0.05

-1.99

 

d(11)

 

 

 

-0.05

-0.05

-4.01

 

 

 

 

 

 

 

 

 

a(3)

0.14

1.07

1.32E-01

 

 

 

 

u(3)

 

 

 

0.07

0.07

5.04

 

d(2)

 

 

 

0.05

0.05

-2.99

 

d(1)

 

 

 

0.05

0.05

-4.96

 

 

 

 

 

 

 

 

 

a(2)

0.73

-0.14

-2.79E-02

-0.01

-0.01

0.99

 

a(1)

0.73

-0.13

-2.58E-02

-0.01

-0.01

0.99