SLIDE 1
Original model
ACD HV2 HV1 AD SB CMI CIM CL CD MT MMT1 MMT2 MMT3 MMT4 MC MAD MSB ACMI ACIM ACL
CP
Exploiting Functional Dependence in Bayesian Network Inference Ji - - PowerPoint PPT Presentation
Exploiting Functional Dependence in Bayesian Network Inference Ji - - PowerPoint PPT Presentation
Exploiting Functional Dependence in Bayesian Network Inference Ji r Vomlel Laboratory for Inteligent systems University of Economics, Prague This presentation is available from: http://www.utia.cas.cz/vomlel/ Original model HV2 HV1
SLIDE 2
SLIDE 3
Evidence model of a task
3 4 · 5 6
- − 1
8 = 15 24 − 1 8 = 5 8 − 1 8 = 4 8 = 1 2
T1 ⇔ MT & CL & ACL & SB & ¬MMT3 & ¬MMT4 & ¬MSB
X1 P(X1|T1) MSB SB CL ACL MMT3 MT MMT4 T1
SLIDE 4
Original model + one evidence model
MMT1 T1 MMT2 MMT3 MMT4 MC MAD MSB ACMI ACIM ACL ACD CD MT CL CIM CMI SB AD HV1 HV2 X1
CP
SLIDE 5
Total clique size
500 1000 1500 2000 2500 1 2 3 4 total clique size number of solved tasks
SLIDE 6
Hierarchical evidence model (=parent divorcing)
X1 P(X1|T1) MSB SB CL ACL MMT3 MT MMT4 T1
− →
MSB X1 SB CL ACL MT MMT3 MMT4 AUX3 AUX2 AUX1 AUX4 AUX5 T1
SLIDE 7
Total clique size
500 1000 1500 2000 2500 1 2 3 4 total clique size number of solved tasks no transformation parent divorcing
SLIDE 8
Factorized evidence model
X1 P(X1|T1) MSB SB CL ACL MMT3 MT MMT4 T1
− →
X1 MMT4 T1 B1 MSB SB CL ACL MT MMT3
ψ(T1, MSB, SB, CL, ACL, MT, MMT3, MMT4) =
- B1
ϕ(T1, B1) · ϕ(B1, MSB) · ϕ(B1, SB) · ϕ(B1, CL)· ϕ(B1, ACL) · ϕ(B1, MT) · ϕ(B1, MMT3) · ϕ(B1, MMT4)
SLIDE 9
Functional dependence
Y is functionally dependent on X1, . . . , Xn if ψ(y, x1, . . . , xn) = 1 if y = f(x1, . . . , xn)
- therwise.
Example - a model with independence of causal influence f(x1, . . . , xn) . . . . . . C1 C2 Cn Xn X2 X1 Y
SLIDE 10
Factorization of MAX
y = f(x1, x2) = max {x1, x2} P(Y |X1, X2) =
R h(Y, R) · g1(X1, R) · g2(X1, R)
+1 +2 +1 +2 +1 +2 +1 +2 1 +1 1 1 1 +2 1 1 1 1 1 = r1 r2 r3 1
- r1,r2,r3(
+1
- 1
1 +2
- 1
1 r1 r2 r3 1 1 1 × +1 1 1 +2 1 r1 r2 r3 1 1 1 × +1 1 1 ) +2 1
SLIDE 11
Proper difference and disjunctive union
- If A ⊇ B then proper difference of A and B is defined as
A B = {x ∈ A ∧ x ∈ B}.
✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂A B
- If A ∩ B = ∅ then disjunctive union of A and B is defined as
A B = {x ∈ A ∨ x ∈ B}.
✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✁✁✁ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✂✁✂✁✂✁✂ ✄✁✄✁✄✁✄ ✄✁✄✁✄✁✄ ✄✁✄✁✄✁✄ ✄✁✄✁✄✁✄ ✄✁✄✁✄✁✄ ✄✁✄✁✄✁✄ ✄✁✄✁✄✁✄ ✄✁✄✁✄✁✄ ✄✁✄✁✄✁✄ ☎✁☎✁☎✁☎ ☎✁☎✁☎✁☎ ☎✁☎✁☎✁☎ ☎✁☎✁☎✁☎ ☎✁☎✁☎✁☎ ☎✁☎✁☎✁☎ ☎✁☎✁☎✁☎ ☎✁☎✁☎✁☎ ☎✁☎✁☎✁☎B A
SLIDE 12
Minimal base of rectangles (MBR)
For a given partition Y1, . . . , Yq of X = ×n
i=1Xi
find a set {R1, R2, . . . , Rk} of minimal cardinality such that:
- for j = 1, . . . , k set Rj is a rectangle,
i.e. Rj = ×n
i=1Di, ∅ = Di ⊆ Xi,
- each element Yℓ, ℓ = 1, 2, . . . q of the partition can be generated
from base {R1, . . . , Rk} using operations and .
SLIDE 13
Factorization of MAX: problem reformulation
+3 +2 +2 +3 +3 +3 +3
1
+2 +1 +1 +2 +3 +1 +2 +3
3 2
Partition by values of Y : Y1 = { (+1, +1) } Y2 = { (+1, +2), (+2, +1), (+2, +2) } Y3 = { (+1, +3), (+2, +3), (+3, +1), (+3, +2)(+3, +3) } Rectangular subspaces: R1 = { (+1, +1) } R2 = { (+1, +1), (+1, +2), (+2, +1), (+2, +2) } R3 = X1 × X2
Y1 = R1, Y2 = R2 R1, and Y3 = R3 R1.
SLIDE 14
Minimal base of rectangles for ADD
2 2 3 3 4 2 1
+1
1 2 4 5 6 3
1
+2 +1 +2
Y1 = R4 Y2 = R2 R4 R5 Y3 = (R1 (R2 R5)) (R3 R5) Y4 = R3 R5 R6 Y5 = R6
SLIDE 15
Correspondence of MBR and factorization
Y1 = R4 Y2 = R2 R4 R5 Y3 = (R1 (R2 R5)) (R3 R5) Y4 = R3 R5 R6 Y5 = R6 , Hidden variable B has one state for each rectangle h(y, b) R1 R2 R3 R4 R5 R6 y1 +1 y2 +1 −1 −1 y3 +1 −1 −1 +2 y4 +1 −1 −1 y5 +1 g1(x1, b), g2(x2, b) R1 R2 R3 R4 R5 R6 x1 1 1 1 x2 1 1 1 1 x3 1 1 1
SLIDE 16
A Boolean function
Y = (X1 ∨ X2) ⇒ (X2 ∧ X3) = (¬X1 ∧ ¬X2) ∨ (X2 ∧ X3)
3 2 1
X1 = 1 1 1 1 1 X2 = 0 X2 = 1 X3 = 0 X3 = 1 X1 = 0
Y0 = R3 (R2 R1) Y1 = R2 R1
SLIDE 17