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CS623: Introduction to Computing with Neural Nets (lecture-2)

Pushpak Bhattacharyya Computer Science and Engineering Department IIT Bombay. CS623: Introduction to Computing with Neural Nets (lecture-2). Recapituation. The human brain. Insula: Intuition & empathy. Pre-frontal cortex: Self control. Amygdala: Aggresion. Hippocampus:

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CS623: Introduction to Computing with Neural Nets (lecture-2)

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  1. Pushpak Bhattacharyya Computer Science and Engineering Department IIT Bombay CS623: Introduction to Computing with Neural Nets(lecture-2)

  2. Recapituation

  3. The human brain Insula: Intuition & empathy Pre-frontal cortex: Self control Amygdala: Aggresion Hippocampus: Emotional memory Anterior Cingulate Cortex: Anxiety Hypothalamus: Control of Hormones Pituitary gland: Mother instincts

  4. Functional Map of the Brain

  5. The Perceptron Model A perceptron is a computing element with input lines having associated weights and the cell having a threshold value. The perceptron model is motivated by the biological neuron. Output = y Threshold = θ w1 wn Wn-1 x1 Xn-1

  6. y 1 θ Σwixi Step function / Threshold function y = 1 for Σwixi >=θ =0 otherwise

  7. θ, ≤ θ, < w1 w2 wn w1 w2 wn . . . w3 . . . x1 x2 x3 xn x1 x2 x3 xn Perceptron Training Algorithm (PTA) Preprocessing: • The computation law is modified to y = 1 if ∑wixi > θ y = o if ∑wixi < θ  w3

  8. θ θ w1 w2 w3 wn wn w0=θ w1 w2 w3 . . . . . . x2 x3 xn x0= -1 x1 x2 x3 xn PTA – preprocessing cont… 2. Absorb θ as a weight  3. Negate all the zero-class examples x1

  9. Example to demonstrate preprocessing • OR perceptron 1-class <1,1> , <1,0> , <0,1> 0-class <0,0> Augmented x vectors:- 1-class <-1,1,1> , <-1,1,0> , <-1,0,1> 0-class <-1,0,0> Negate 0-class:- <1,0,0>

  10. Example to demonstrate preprocessing cont.. Now the vectors are x0 x1 x2 X1 -1 0 1 X2 -1 1 0 X3 -1 1 1 X4 1 0 0

  11. Perceptron Training Algorithm • Start with a random value of w ex: <0,0,0…> • Test for wxi > 0 If the test succeeds for i=1,2,…n then return w 3. Modify w, wnext = wprev + xfail

  12. Tracing PTA on OR-example w=<0,0,0> wx1 fails w=<-1,0,1> wx4 fails w=<0,0 ,1> wx2 fails w=<-1,1,1> wx1 fails w=<0,1,2> wx4 fails w=<1,1,2> wx2 fails w=<0,2,2> wx4 fails w=<1,2,2> success

  13. Theorems on PTA • The process will terminate • The order of selection of xi for testing and wnext does not matter.

  14. End of recap

  15. Proof of Convergence of PTA • Perceptron Training Algorithm (PTA) • Statement: Whatever be the initial choice of weights and whatever be the vector chosen for testing, PTA converges if the vectors are from a linearly separable function.

  16. Proof of Convergence of PTA • Suppose wn is the weight vector at the nth step of the algorithm. • At the beginning, the weight vector is w0 • Go from wi to wi+1 when a vector Xj fails the test wiXj > 0 and update wi as wi+1 = wi + Xj • Since Xjs form a linearly separable function,  w* s.t. w*Xj > 0 j

  17. Proof of Convergence of PTA • Consider the expression G(wn) = wn . w* | wn| where wn = weight at nth iteration • G(wn) = |wn| . |w*| . cos  |wn| where  = angle between wn and w* • G(wn) = |w*| . cos  • G(wn) ≤ |w*| ( as -1 ≤ cos ≤ 1)

  18. Behavior of Numerator of G wn . w* = (wn-1 + Xn-1fail ) . w* • wn-1 . w* + Xn-1fail . w* • (wn-2 + Xn-2fail ) . w* + Xn-1fail . w* ….. • w0 . w*+ ( X0fail + X1fail +.... + Xn-1fail ). w* w*.Xifail is always positive: note carefully • Suppose |Xj| ≥  , where  is the minimum magnitude. • Num of G ≥ |w0 . w*| + n  . |w*| • So, numerator of G grows with n.

  19. Behavior of Denominator of G • |wn| =  wn . wn •  (wn-1 + Xn-1fail )2 •  (wn-1)2+ 2. wn-1. Xn-1fail + (Xn-1fail )2 •  (wn-1)2+ (Xn-1fail )2 (as wn-1. Xn-1fail ≤ 0 ) •  (w0)2+ (X0fail )2 + (X1fail )2 +…. + (Xn-1fail )2 • |Xj| ≤  (max magnitude) • So, Denom ≤ (w0)2+ n2

  20. Some Observations • Numerator of G grows as n • Denominator of G grows as  n => Numerator grows faster than denominator • If PTA does not terminate, G(wn) values will become unbounded.

  21. Some Observations contd. • But, as |G(wn)| ≤ |w*| which is finite, this is impossible! • Hence, PTA has to converge. • Proof is due to Marvin Minsky.

  22. Convergence of PTA • Whatever be the initial choice of weights and whatever be the vector chosen for testing, PTA converges if the vectors are from a linearly separable function.

  23. θ w1 w2 w3 wn . . . x2 x3 xn Study of Linear Separability • W. Xj = 0 defines a hyperplane in the (n+1) dimension. => W vector and Xj vectors are perpendicular to each other. y x1

  24. Linear Separability X1+ + X2 + Xk+1 - Xk+ Positive set : w. Xj > 0 j≤k Xk+2 - - Xm - Negative set : w. Xj < 0 j>k Separating hyperplane

  25. Test for Linear Separability (LS) • Theorem: A function is linearly separable iff the vectors corresponding to the function do not have a Positive Linear Combination (PLC) • PLC – Both a necessary and sufficient condition. • X1, X2, … , Xm - Vectors of the function • Y1, Y2, … , Ym - Augmented negated set • Prepending -1 to the 0-class vector Xi and negating it, gives Yi

  26. X1 <0,0> + Y1 <-1,0,0> X2 <0,1> - Y2 <1,0,-1> X3 <1,0> - Y3 <1,-1,0> X4 <1,1> + Y4 <-1,1,1> Example (1) - XNOR • The set {Yi} has a PLC if ∑ Pi Yi = 0 , 1 ≤ i ≤ m • where each Pi is a non-negative scalar and • atleast one Pi > 0 • Example : 2 bit even-parity (X-NOR function)

  27. Example (1) - XNOR • P1 [ -1 0 0 ]T + P2 [ 1 0 -1 ]T + P3 [ 1 -1 0 ]T + P4 [ -1 1 1 ]T = [ 0 0 0 ]T • All Pi = 1 gives the result. • For Parity function, PLC exists => Not linearly separable.

  28. Xi Yi <0 0 0> - <1 0 0 0> <0 0 1> - <1 0 0 -1> <0 1 0> - <1 0 -1 0> <0 1 1> + <-1 0 1 1> <1 0 0> - <1 -1 0 0> <1 0 1> + <-1 1 0 1> <1 1 0> + <-1 1 1 0> <1 1 1> + <-1 1 1 1> Example (2) – Majority function • 3-bit majority function • Suppose PLC exists. Equations obtained are: • P1 + P2+ P3- P4 + P5- P6- P7 – P8 = 0 • -P5+ P6 + P7+ P8 = 0 • -P3+ P4 + P7+ P8 = 0 • -P2+ P4 + P6+ P8 = 0 • On solving, all Pi will be forced to 0 • 3 bit majority function • => No PLC => LS

  29. Limitations of perceptron • Non-linear separability is all pervading • Single perceptron does not have enough computing power • Eg: XOR cannot be computed by perceptron

  30. Solutions • Tolerate error (Ex: pocket algorithm used by connectionist expert systems). • Try to get the best possible hyperplane using only perceptrons • Use higher dimension surfaces Ex: Degree - 2 surfaces like parabola • Use layered network

  31. x1 x2 x1x2 0 0 0 0 1 1 1 0 0 1 1 0 Example - XOR • = 0.5 w1=1 w2=1  Calculation of XOR x1x2 x1x2 Calculation of x1x2 • = 1 w1=-1 w2=1.5 x2 x1

  32. Example - XOR • = 0.5 w1=1 w2=1 x1x2 1 1 x1x2 1.5 -1 -1 1.5 x2 x1

  33. Multi Layer Perceptron (MLP) • Question:- How to find weights for the hidden layers when no target output is available? • Credit assignment problem – to be solved by “Gradient Descent”

  34. Assignment • Find by solving linear inequalities the perceptron to solve the majority Boolean Function. • Then feed it to your implementation of the perceptron training algorithm and study its behaviour. • Download SNNS and WEKA packages and try your hand at the perceptron algorithm built in the packages.

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