--- INSTITUTO SUPERIOR TÉCNICO
Sistemas de Apoio à Decisão Exame: 2 15 Julho 2010
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1. (6 PTE) Algumas perguntas fáceis para começar.
1) DW (2 pts) Assinale as seguintes frases como verdadeiras (V) ou falsas (F). Cada resposta correcta corresponde a uma cotação de +0.5 valor. Resposta errada -0,25. Se não responder, a cotação é 0.
(Indicate the question with (V) for true and (F) for false. Every correct answer has a value +0.5. False answer has a value -0.25. No answer the value 0.)
a. (0.5pts)
(max()+min())*2 é uma função de agregação algébrica distributive (max()+min())*2 is an algebraic distributive aggregation function
(F) every one got 0.25 for this answer!
b. (0,5pts)
Durante a análise de dados multidimensional todos os cuboids são materializados During multidimensional data analysis all cuboids are materialized
(F)
c. (0.5pts)
Um cuboid com dados mais finos da granulosidade pode ser gerado de coarser- granularity data cuboid
A cuboid with finer granularity data can be generated from coarser-granularity data cuboid
(F)
d. (0.5pts)
A matriz de co-variância é sempre simétrica
A covariance matrix is always symmetric (V)
2) DM (4pts) Assinale as seguintes frases como verdadeiras (V) ou falsas (F). Cada resposta correcta corresponde a uma cotação de +1 valor. Resposta errada -0,5. Se não responder a cotação é 0.
(Indicate the question with (V) for true and (F) for false. Every correct answer has a value +1. False answer has a value -0.5. No answer the value 0.)
(a) (1pts)
Complete linkage e Single linkage podem dar o resultado diferente quando usados na in tree clustering
Complete linkage and Single linkage may give different result when used in tree clustering (V)
(b) (1pts)
Sequence clustering sofre dos mesmos problemas que k-means. Qual é o número de k e de iniciação aleatória que conduz a resultados diferentes.
Sequence clustering suffers from the same problems as k-means. What is the number of k and random initialization that leads to different results.
(V)
(c) (1pts)
A multilayer Perceptron pode “overfit”.
A multilayer Perceptron may overfit.
(V)
(a) (1pts)
O Multilayer Perceptron com dez níveis escondidos e com função de activação linear é mais poderoso do que um Perceptron com um nível escondido e com função de activação linear Multilayer Perceptron with teen hidden layers and a linear activation function is more powerful then a Perceptron with one hidden layer and a linear activation function (F)
2. (3 pts) DW
Supor que um DW consiste em três dimensões, tempo, empregado e filial e one measure count. Supor que dimensão tempo tem a hierarquia dia, mês, quarter e ano. Supor dimensão empregado definido pelo attributes name e programming skills.
Suppose that a data warehouse consists of three dimensions, time, employee and branch and one measure count.
Suppose that the dimension time has the hierarchy day, month, quarter, year. Suppose the dimension employee is defined by the attributes name and programming skills.
(a) (1pts)
Apresente o modelo em estrela para esta DW.
Present a Star schema for a DW (b) (1pts)
Apresente o modelo “Snowflake” para esta DW.
Presen a Snowflake schema for a DW (c) (1pts)
Quantos cuboids existem dentro deste cubo?
How many cuboids are present in this cube?
(a)
(c)
L=5*2*2=20
3. (2 pts) fp-growth
Considere o seguinte conjunto de transacções:
(a) (1 pts) Construa uma árvore fp da base de dados de transacções com um suporte minimo de Min_sup=50%
Construct a fp-tree with Min_sup=50%
(a)
Min_Sup=50%=2.5 >= 3/5
Item Support
Beer "4/5"
Diaper "4/5"
Baby Powder "2/5"
Bread "1/5"
Umbrella "1/5"
Milk "2/5"
Detergent "1/5"
Sumol "1/5"
TID List of Items
1 Beer, Diaper, Baby Powder, Bread, Umbrella 2 Diaper, Baby Powder
3 Beer, Diaper, Milk 4 Diaper, Beer, Detergent 5 Beer, Milk, Sumol
fp-tree
(b) (1 pts) Determine conditonal pattern base.
Determine the conditional pattern base.
item cond.pattern base Diaper Beer:3
TID List of Items 1 Beer, Diaper 2 Diaper, 3 Beer, Diaper 4 Beer, Diaper, 5 Beer
4. (3 pts) ID3
Calcule a árvore de decisão para este conjunto de exemplos (com atributos: “Hair Length”,
“Weight” and “Age”) com target “Class” usando o algoritmo ID3. Indique os seus cálculos.
Calculate the decision tree using the examples given by the table (attributes: “Hair Length”,
“Weight” and “Age”) with target “Class”. Use the id3 algorithm. indicate the calculation steps.
p(M)=5/9, p(F)=4/9
I(table)= -5./9*log2(5./9)-4./9*log2(4./9)= 0.99108
E(Hair Length)=2/9 *(-1/2*log2(1/2) -1/2*log2(1/2))+2/9*(-1/2*log2(1/2) - 1/2*log2(1/2))=0.44444
Gain(Hair_Length)= 0.54664
E(Weight)=0 /*All wights are different*/
Gain(Wight)= 0.99108
E(Age)=0 /*All ages are different*/
Gain(Age)= 0.99108
done.
Alternative way, binary decision tree (not required here):
Continuous Valued Attributes (see slides) Create a discrete attribute to test continuous Temperature = 24.50C
(Temperature > 20.00C) = {true, false}
Where to set the threshold?
Temperatur 150C 180C 190C 220C 240C 270C
PlayTennis No No Yes Yes Yes No
Hair Length
0 1 2 4 6 6 8 10 10
M M M F F M F F M
Hair_Length >=2
E(Hair_Length>=2)=6/9*(-4/.6*log2(4./6)-2/.6*log2(2./6))= 6.1220 Weight
20 78 90 150 160 170 180 200 250
F F M F F M M M M
Weight >= 170
E(Weight>=170)=5./9*(-4./5*log2(4./5)-1./5*log2(1./5))= 0.40107 Age
1 8 10 34 36 38 41 45 70
F F M F M M F M M
Age<= 45
E(Age<=45)=7./9*(-4./7*log2(4./7)-3./7*log2(3./7))= 0.76629
5. (3 pts) Clustering
Dado o conjunto de dados
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x j = 0 0
, 1 0
, 2 2
com
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µ1= 1 1
,µ2= 2 2
e
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Σ1= 1 0 0 1
,Σ2 = 1 0 0 1
e P(C=i)=1 α=1. Execute um passo do algoritmo de clustering EM. Quais são os novos valores de
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µi,Σi,wi?
Given the data
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x j = 0 0
, 1 0
, 2 2
with
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µ1= 1 1
,µ2 = 2 2
e
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Σ1= 1 0 0 1
,Σ2 = 1 0 0 1
and P(C=i)=1 α=1. Preform on step of EM clustering algorithm. What are the values of
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µi,Σi,wi?
By Bayes’ rule pij=αP(xj|C=i)P(C=i)
pij=αP(xj|C=i)P(C=i)
Because we initialize our parameters arbitrary, we set P(C=i)=1 α=1, with the given data we can simplify d=2, and because of
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Σ1= 1 0 0 1
,Σ2 = 1 0 0 1
we get
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pi= pij
j=1 n
∑
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µi← pij x j pi
j=1 n
∑
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Σi← pij x j
x jT pi
j=1 n
∑
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P(x|C=i)= 1 (2π)d/ 2ΣC=i1/ 2
exp−1
2(x−µC=i)tΣC=i
−1(x−µC=i)
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P(x|C=i)= 1
(2π)exp−1
2(x−µC=i)t(x−µC=i)
p11= exp(-2./2)./(2*pi)= exp(-1)./(2*pi )=0.058550 p12= exp(-1./2)./(2*pi)= 0.096532
p13= exp(-2./2)./(2*pi)= exp(-1)./(2*pi )= 0.058550 p21= exp(-8./2)./(2*pi)= exp(-4)./(2*pi)= 0.0029150 p22= exp(-5./2)./(2*pi)= 0.013064
p23= exp(0)./(2*pi)= 0.15915
p1= 0.058550+0.096532+0.058550=0.21363 p2= 0.0029150+0.013064+0.15915=0.17513
µ1=([0 0]*0.058550+[1 0]*0.096532+[2 2]*0.058550)./0.21363 =[ 1 0.5814]
µ2=([0 0]* 0.0029150+[1 0]* 0.013064 +[2 2]* 0.15915)./ 0.17513 =[ 1.8921 1.817]
Σ1=([0 0]'*[0 0]*0.058550+[1 0]'*[1 0]*0.096532+[2 2]'*[2 2]*0.058550)./0.21363=
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Σ1= 1.5482 1.0963 1.0963 1.0963
Σ2= ([0 0]'*[0 0]* 0.0029150 +[1 0]'*[1 0]* 0.013064 +[2 2]'*[2 2]* 0.15915)./ 0.17513 =
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Σ2 = 3.7096 3.6350 3.6350 3.6350
Dados os pesos w1={w11=0,w12=0,w13=0,w14=0}, W1={W11=1,W21=1}, e a função de activação
Faça um gradiente descendente estocástico (uma etapa) com η=1 para o vector de entrada
x={2,0,1,0}={x1=21,x2=0,x3=1,x4=0} e alvo (target) t={1,1}, determine
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ΔWij e
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Δwjk para o primeiro passo de adaptação.
Given the wights w1={w11=0,w12=0,w13=0,w14=0}, W1={W11=1,W21=1}, and the activation function σ(x).
Perform one step of the stochastic gradient descent with η=1 with the input vector x={2,0,1,0}={x1=2,x2=0,x3=1,x4=0} eland the target t={1,1}, determine
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ΔWij e
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Δwjk after one adaptation step.
V1= 1./(1+exp(-(0)))= 0.5 O1= 1./(1+exp(-(0.5)))= 0.62246 O2= 1./(1+exp(-(0.5)))= 0.62246
W11=W21=1+0.088723*0.5=1.0444
δ1=(1-0.62246)* 0. 62246*(1-0.62246)= 0.088723 δ2=(1-0.62246)* 0. 62246*(1-0.62246)= 0.088723
δ1= 0.5*(1- 0.5)*(1*0.088723+1*0.088723)= 0.044361 w11=0+2*0.044361= 0.088722
w12=0+0*0.044361=0
w13=0+1*0.044361= 0.044361 w14=0+0*0.044361=0
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f(x)=σ(x)= 1 1+e(−x)
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ΔWij =(ti−oi)σ(neti)(1−σ(neti))Vj
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δ1=(t1−o1)σ(net1)(1−σ(net1)) ΔW1j =δ1Vj
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δ1=σ(net1)(1−σ(net1)) Wi1
i=1 2
∑ δi
