Perceptron Homework Assume a 3 input perceptron plus bias (it - - PowerPoint PPT Presentation

CS 478 - Perceptrons 1

Perceptron Homework

l

Assume a 3 input perceptron plus bias (it outputs 1 if net > 0, else 0)

l

Assume a learning rate c of 1 and initial weights all 1: Δwi = c(t – z) xi

l

Show weights after each pattern for just one epoch

l

Training set 1 0 1 -> 0 1 1 0 -> 0 1 0 1 -> 1 0 1 1 -> 1 Pattern Target Weight Vector Net Output ΔW 1 0 1 1 1 1 1 1 3 1

• 1 0 -1 -1

1 1 0 1 0 1 0 0 1 1

• 1-1 0 -1

1 0 1 1 1

• 1 0 0 -1
• 2

1 0 1 1 0 1 1 1 1 0 0 1 0 1 1 0 0 0 0 0 0 1 0

Quadric Machine Homework

l

Assume a 2 input perceptron expanded to be a quadric perceptron (it outputs 1 if net > 0, else 0). Note that with binary inputs of -1, 1, that x2 and y2 would always be 1 and thus do not add info and are not needed (they would just act like to more bias weights)

l

Assume a learning rate c of .4 and initial weights all 0: Δwi = c(t – z) xi

l

Show weights after each pattern for one epoch with the following non-linearly separable training set.

l

Has it learned to solve the problem after just one epoch?

l

Which of the quadric features are actually needed to solve this training set?

CS 478 - Regression 2

x y Target

• 1
• 1
• 1

1 1 1

• 1

1 1 1

Quadric Machine Homework

l wx represents the weight for feature x l Assume a 2 input perceptron expanded to be a quadric perceptron (it

• utputs 1 if net > 0, else 0)

l Assume a learning rate c of .4 and initial weights all 0: Δwi = c(t – z) xi l Show weights after each pattern for one epoch with the following non-

linearly separable training set.

l Has it learned to solve the problem after just one epoch? l Which of the quadric features are actually needed to solve this training

set?

CS 478 - Regression 3

x y xy wx wy wxy bias Δwx Δwx Δwxy Δbias net Target Output

• 1
• 1

1

• 1

1

• 1
• .4

.4

• .4

.4 1 1

• 1
• 1
• .4

.4

• .4

.4 .4

• .4
• .4

.4 1 1 1 1

• .8

.8

• .8
• .8

.8

Quadric Machine Homework

l wx represents the weight for feature x l Assume a 2 input perceptron expanded to be a quadric perceptron (it

• utputs 1 if net > 0, else 0)

l Assume a learning rate c of .4 and initial weights all 0: Δwi = c(t – z) xi l Show weights after each pattern for one epoch with the following non-

linearly separable training set.

l Has it learned to solve the problem after just one epoch? - Yes l Which of the quadric features are actually needed to solve this training

set? – Really only needs feature xy.

CS 478 - Regression 4

x y xy wx wy wxy bias Δwx Δwx Δwxy Δbias net Target Output

• 1
• 1

1

• 1

1

• 1
• .4

.4

• .4

.4 1 1

• 1
• 1
• .4

.4

• .4

.4 .4

• .4
• .4

.4 1 1 1 1

• .8

.8

• .8
• .8

.8

Linear Regression Homework

l Assume we start with all weights as 0 (don’t forget the

bias)

l What are the new weights after one iteration through the

following training set using the delta rule with a learning rate of .2

CS 478 - Homework 5

x y Target .3 .8 .7

• .3

1.6

• .1

.9 1.3

Linear Regression Homework

l Assume we start with all weights as 0 (don’t forget the bias) l What are the new weights after one iteration through the

following training set using the delta rule with a learning rate c = .2

CS 478 - Homework 6

x y Target Net w1 w2 Bias .3 .8 .7 .042 .112 .140

• .3

1.6

• .1

.307 .066

• .018

.059 .9 1.3 .118 .279

• .018

.295

0 + .2(.7 – 0).3 = .042

Δwi = c(t − net)xi

Logistic Regression Homework

l You don’t actually have to come up with the weights for this one,

though you could quickly by using the closed form linear regression approach

l Sketch each step you would need to learn the weights for the following

data set using logistic regression

l Sketch how you would generalize the probability of a heart attack

given a new input heart rate of 60

CS 478 - Homework 7

Heart Rate Heart Attack 50 Y 50 N 50 N 50 N 70 N 70 Y 90 Y 90 Y 90 N 90 Y 90 Y

Logistic Regression Homework

1.

Calculate probabilities or odds for each input value

2.

Calculate the log odds

3.

Do linear regression on these 3 points to get the logit line

4.

To find the probability of a heart attack given a new input heart rate of 60, just calculate p = elogit(60)/(1+elogit(60)) where logit(60) is the value on the logit line for 60

CS 478 - Homework 8

Heart Rate Heart Attack 50 Y 50 N 50 N 50 N 70 N 70 Y 90 Y 90 Y 90 N 90 Y 90 Y

Heart Rate Heart Attacks Total Patients Probability: # attacks/ Total Patients Odds: p/(1-p) = # attacks/ # not Log Odds: ln(Odds)

50 1 4 .25 .33

• 1.11

70 1 2 .5 1 90 4 5 .8 4 1.39

CS 478 – Backpropagation 9

BP-1) A 2-2-1 backpropagation model has initial weights as shown. Work through one cycle of learning for the f ollowing pattern(s). Assume 0 momentum and a learning constant of 1. Round calculations to 3 significant digits to the right of the decimal. Give values for all nodes and links for activation, output, error signal, weight delta, and final weights. Nodes 4, 5, 6, and 7 are just input nodes and do not have a sigmoidal output. For each node calculate the following (show necessary equati on for each). Hint: Calculate bottom-top-bottom. a =

• =

= w = w =

1 2 3 7 +1 4 +1 6 5

a) All weights initially 1.0 Training Patterns 1) 0 0 -> 1 2) 0 1 -> 0

CS 478 – Backpropagation 10

BP-1) net2 = wi xi = (1*0 + 1*0 + 1*1) = 1 net3 = 1

• 2 = 1/(1+e-net) = 1/(1+e-1) = 1/(1+.368) = .731
• 3 = .731
• 4 = 1

net1 = (1*.731 + 1*.731 + 1) = 2.462

• 1 = 1/(1+e-2.462)= .921

1 = (t1 - o1) o1 (1 - o1) = (1 - .921) .921 (1 - .921) = .00575 w21 = j oi = 1 o2 = 1 * .00575 * .731 = .00420 w31 = 1 * .00575 * .731 = .00420 w41 = 1 * .00575 * 1 = .00575 2 = oj (1 - oj) k wjk = o2 (1 - o2) 1 w21 = .731 (1 - .731) (.00575 * 1) = .00113 3 = .00113 w52 = j oi = 2 o5 = 1 * .00113 * 0 = 0 w62 = 0 w72 = 1 * .00113 * 1 = .00113 w53 = 0 w63 = 0 w73 = 1 * .00113 * 1 = .00113 1 2 3 7 +1 4 +1 6 5

CS 478 – Backpropagation 11 Second pass for 0 1 -> 0 Modified Weights: w21 = 1.0042 w31 = 1.0042 w41 = 1.00575 w52 = 1 w62 = 1 w72 = 1.00113 w53 = 1 w63 = 1 w73 = 1.00113 net2 = wi xi = (1*0 + 1*1 + 1*1.00113) = 2.00113 net3 = 2.00113

• 2 = 1/(1+e-net) = 1/(1+e-2.00113) = .881
• 3 = .881
• 4 = 1

net1 = (1.0042*.881 + 1.0042*.881 + 1.00575*1) = 2.775

• 1 = 1/(1+e-2.775)= .941

1 = (t1 - o1) o1 (1 - o1) = (0 - .941) .941 (1 - .941) = -.0522 w21 = j oi = 1 o2 = 1 * -.0522 * .881 = -.0460 w31 = 1 * -.0522 * .881 = -.0460 w41 = 1 * -.0522 * 1 = -.0522 2 = oj (1 - oj) k wjk = o2 (1 - o2) 1 w21 = .881 (1 - .881) (-.0522 * 1.0042) = -.00547 3 = -.00547 w52 = j oi = 2 o5 = 1 * -.00547* 0 = 0 w62 = 1 * (-.00547) * 1 = -.00547 w72 = 1 * (-.00547) * 1 = -.00547 w53 = 0 w63 = -.00547 w73 = 1 * (-.00547) * 1 = -.00547 w21 = 1.0042 - .0460 = .958 w31 = 1.0042 - .0460 = .958 w41 = 1.00575 - .0522 = .954 w52 = 1 + 0 = 1 w62 = 1 - .00547 = .995 w72 = 1.00113 - .00547 = .996 w53 = 1 + 0 = 1 w63 = 1 - .00547 = .995 w73 = 1.00113 - .00547 = .996

PCA Homework

CS 478 - Homework 12 Original Data x y p1 .2

• .3

p2

• 1.1

2 p3 1

• 2.2

p4 .5

• 1

p5

• .6

1 mean

• .1

Terms m 5 Number of instances in data set n 2 Number of input features p 1 Final number of principal components chosen

• Use PCA on the given data set to get a transformed

data set with just one feature (the first principal component (PC)). Show your work along the way.

• Show what % of the total information is contained in

the 1st PC.

• Do not use a PCA package to do it. You need to go

through the steps yourself, or program it yourself.

• You may use a spreadsheet, Matlab, etc. to do the

arithmetic for you.

• You may use any web tool or Matlab to calculate the

eigenvectors from the covariance matrix.

PCA Homework

CS 478 - Homework 13 Original Data x y p1 .2

• .3

p2

• 1.1

2 p3 1

• 2.2

p4 .5

• 1

p5

• .6

1 mean

• .1

Zero Centered Data x y p1 .2

• .2

p2

• 1.1

2.1 p3 1

• 2.1

p4 .5

• .9

p5

• .6

1.1 mean Covariance Matrix x y .715

• 1.39
• 1.39

2.72 EigenVectors x y Eigenvalue

• .456
• .890

3.431

• .890
• .456

.0037

% total info in 1st principal component 3.431/(3.431 + . 0037) = 99.89%

Matrix A – p × n x y 1st PC

• .456
• .890

Terms m 5 Number of instances in data set n 2 Number of input features p 1 Final number of principal components chosen Matrix B = Transposed zero centered Training Set p1 p2 p3 p4 p5 x .2

• 1.1

1 .5

• .6

y

• .2

2.1

• 2.1
• .9

1.1 A × B New Data Set 1st PC p1 .0870 p2

• 1.368

p3 1.414 p4 0.573 p5

• 0.710

Decision Tree Homework

l Info(S) = - 2/9·log22/9 - 4/9·log24/9 -3/9·log23/9 = 1.53 – Not necessary unless you want to calculate information gain l Starting with all instances, calculate gain for each attribute l Let’s do Meat: l InfoMeat(S) = 4/9·(-2/4log22/4 - 2/4·log22/4 - 0·log20/4) +

5/9·(-0/5·log20/5 - 2/5·log22/5 - 3/5·log23/5) = .98

– Information Gain is 1.53 - .98 = .55 l Finish this level, find best attribute and split, and then find the

best attribute for at least the left most node at the next level

– Assume sub-nodes are sorted alphabetically left to right by attribute

CS 478 - Decision Trees 14 Meat N,Y Crust D,S,T Veg N,Y Quality B,G,Gr

Y Thin N Great N Deep N Bad N Stuffed Y Good Y Stuffed Y Great Y Deep N Good Y Deep Y Great N Thin Y Good Y Deep N Good N Thin N Bad

Info(S) = − pi

i=1 |C|

∑

log2(pi)

InfoA(S) = | Sj | | S | Info(Sj)

i= j |A|

∑

= − | Sj | | S | ⋅ pi

i=1 |C|

∑

log2(pi)

j=1 |A|

∑

Decision Tree Homework

l InfoMeat(S) = 4/9·(-2/4log22/4 - 2/4·log22/4 - 0·log20/4) +

5/9·(-0/5·log20/5 - 2/5·log22/5 - 3/5·log23/5) = .98

l InfoCrust(S) = 4/9·(-1/4log21/4 - 2/4·log22/4 - 1/4·log21/4) +

2/9·(-0/2·log20/2 - 1/2·log21/2 - 1/2·log21/2) + 3/9·(-1/3·log21/3 - 1/3·log21/3 - 1/3·log21/3) = 1.41

l InfoVeg(S) = 5/9·(-2/5log22/5 - 2/5·log22/5 - 1/5·log21/5) +

4/9·(-0/4·log20/4 - 2/4·log22/4 - 2/4·log22/4) = 1.29

l Attribute with least remaining info is Meat

CS 478 - Decision Trees 15 Meat N,Y Crust D,S,T Veg N,Y Quality B,G,Gr

Y Thin N Great N Deep N Bad N Stuffed Y Good Y Stuffed Y Great Y Deep N Good Y Deep Y Great N Thin Y Good Y Deep N Good N Thin N Bad

Info(S) = − pi

i=1 |C|

∑

log2(pi)

InfoA(S) = | Sj | | S | Info(Sj)

i= j |A|

∑

= − | Sj | | S | ⋅ pi

i=1 |C|

∑

log2(pi)

j=1 |A|

∑

Decision Tree Homework

l Left most node will be Meat = N containing 4 instances l InfoCrust(S) = 1/4·(-1/1log21/1 - 0/1·log20/1 - 0/1·log20/1) +

1/4·(-0/1·log20/1 - 1/1·log21/1 - 0/1·log20/1) + 2/4·(-1/2·log21/2 - 1/2·log21/2 - 0/2·log20/2) = .5

l InfoVeg(S) = 2/4·(-2/2log22/2 - 0/2·log20/2 - 0/2·log20/2) +

2/4·(-0/2·log20/2 - 2/2·log22/2 - 0/2·log20/2) = 0

l Attribute with least remaining info is Veg which will split

into two pure nodes

CS 478 - Decision Trees 16 Meat N,Y Crust D,S,T Veg N,Y Quality B,G,Gr

N Deep N Bad N Stuffed Y Good N Thin Y Good N Thin N Bad

Info(S) = − pi

i=1 |C|

∑

log2(pi)

InfoA(S) = | Sj | | S | Info(Sj)

i= j |A|

∑

= − | Sj | | S | ⋅ pi

i=1 |C|

∑

log2(pi)

j=1 |A|

∑

k-Nearest Neighbor Homework

l Assume the following training set. l Assume a new point (.5, .2)

– For nearest neighbor distance use Manhattan distance – What would the output be for 3-nn with no distance weighting?

Show work and vote.

– What would the output be for 3-nn with distance weighting? Show

work and vote.

CS 478 - Homework 17

x y Target .3 .8 A

• .3

1.6 B .9 B 1 1 A

k-Nearest Neighbor Homework

l Assume the following training set. l Assume a new point (.5, .2)

– For nearest neighbor distance use Manhattan distance – What would the output be for 3-nn with no distance weighting?– A

wins with vote 2/3

– What would the output be for 3-nn with distance weighting? A has

vote 1/.82 + 1/1.32 = 2.15, B wins with vote 1/.42 = 6.25

CS 478 - Homework 18

x y distance Target .3 .8 .2 + .6 = .8 A

• .3

1.6 .8 + 1.4 = 2.2 B .9 .0 .4 + .2 = .6 B 1 1 .5 + .8 = 1.3 A

RBF Homework

l Assume you have an RBF with

– Two inputs – Three output classes A, B, and C (linear units) – Three prototype nodes at (0,0), (.5,1) and (1,.5) – The radial basis function of the prototype nodes is

l max(0, 1 – Manhattan distance between the prototype node and the

instance) – Assume no bias and initial weights of .6 into output node A, -.4

into output node B, and 0 into output node C

– Assume top layer training is the delta rule with LR = .1

l Assume we input the single instance .6 .8

– Which class would be the winner? – What would the weights be updated to if it were a training instance

• f .6 .8 with target class B?

CS 478 - Homework 19

RBF Homework: Instance .6 .8 -> B

CS 478 - Homework 20

Node Distance Activation Prototype 1 .6 + .8 = 1.4 Max(0, 1 – 1.4) = 0 Prototype 2 .1 + .2 = .3 .7 Prototype 3 .4 + .3 = .7 .3 Node Net Value Output A 0*.6 + .7*.6 + .3*.6 = .6 Winner Output B 0*-.4 + .7*-.4 + .3*-.4 = -.4 Output C 0*0 + .7*0 + .3*0 = 0 Weight Delta Values Δwp1,A = .1(0 – .6)0 = 0 Δwp2,A = .1(0 – .6).7 = -.042 Δwp3,A = .1(0 – .6).3 = -.018 Δwp1,B = .1(1 – -.4)0 = 0 Δwp2,B = .1(1 – -.4).7 = .098 Δwp3,B = .1(1 – -.4).3 = .042 Δwp1,C = .1(0 – 0)0 = 0 Δwp2,C = .1(0 – 0).7 = 0 Δwp3,C = .1(0 – 0).3 = 0

Size (B, S) Color (R,G,B) Output (P,N) B R P S B P S B N B R N B B P B G N S B P

CS 478 - Bayesian Learning 21

vNB = argmax

v j ∈V

P(v j) P(ai |v j)

i

∏

For the given training set: 1. Create a table of the statistics needed to do Naïve Bayes 2. What would be the output for a new instance which is Small and Blue? 3. What is the Naïve Bayes value and the normalized probability for each

• utput class (P or N) for this case
• f Small and Blue?

Naïve Bayes Homework

Size (B, S) Color (R,G,B) Output (P,N) B R P S B P S B N B R N B B P B G N S B P

CS 478 - Bayesian Learning 22

vNB = argmax

v j ∈V

P(v j) P(ai |v j)

i

∏

What do we need?

P(P) 4/7 P(N) 3/7 P(Size=B|P) 2/4 P(Size=S|P) 2/4 P(Size=B|N) 2/3 P(Size=S|N) 1/3 P(Color=R|P) 1/4 P(Color=G|P) 0/4 P(Color=B|P) 3/4 P(Color=R|N) 1/3 P(Color=G|N) 1/3 P(Color=B|N) 1/3

What is our output for a new instance which is Small and Blue? vP = P(P)* P(S|P)*P(B|P) = 4/7*2/4*3/4= .214 vN = P(N)* P(S|N)*P(B|N) = 3/7*1/3*1/3= .048 Normalized Probabilites: P= .214/(.214+.048) = .817 N = .048/(.214+.048) = .183

HAC Homework

l For the data set below show all iterations (from 5 clusters

until 1 cluster remaining) for HAC single link. Show work. Use Manhattan distance. In case of ties go with the cluster containing the least alphabetical instance. Show the dendrogram for the HAC case, including properly labeled distances on the vertical-axis of the dendrogram.

CS 478 - Clustering 23

Pattern x y a .8 .7 b

• .1

.2 c .9 .8 d .2 e .2 .1

HAC Homework

CS 478 - Clustering 24

Pattern x y a .8 .7 b

• .1

.2 c .9 .8 d .2 e .2 .1 a b c d e a 1.4 .2 1.3 1.2 b 1.6 .1 .4 c 1.5 1.4 d .3 e Closest clusters distance 1 5 separate clusters 0 2 b and d .1 3 a and c .2 4 {b,d} and e

.3

5 {b,d,e} and {a,c}

1.2

Distance Matrix

k-means Homework

l For the data below, show the centroid values and which

instances are closest to each centroid after centroid calculation for two iterations of k-means using Manhattan distance

l By 2 iterations I mean 2 centroid changes after the initial

centroids

l Assume k = 2 and that the first two instances are the initial

centroids

CS 478 - Clustering 25

Pattern x y a .9 .8 b .2 .2 c .7 .6 d

• .1
• .6

e .5 .5

k-means Homework

l For the data below, show the centroid values and which

instances are closest to each centroid after centroid calculation for two iterations of k-means using Manhattan distance

l By 2 iterations I mean 2 centroid changes after the initial

centroids

l Assume k = 2 and that the first two instances are the initial

centroids

CS 478 - Clustering 26

Pattern x y a .9 .8 b .2 .2 c .7 .6 d

• .1
• .6

e .5 .5 Iteration Centroid 1 and instances Centroid 2 and instances .9, .8 {a, c} .2, .2 {b, d, e} 1 .8, .7 {a, c, e} .2, .033 {b, d} 2 .7, .633 {a, c, e} .05, -.2 {b, d} 3 .7, .633 {a, c, e} .05, -.2 {b, d}