Exploiting Functional Dependence in Bayesian Network Inference Ji - PowerPoint PPT Presentation

Exploiting Functional Dependence in Bayesian Network Inference Jiˇ r´ ı Vomlel Laboratory for Inteligent systems University of Economics, Prague This presentation is available from: http://www.utia.cas.cz/vomlel/

Original model HV2 HV1 ACMI ACIM ACL ACD AD SB CMI CIM CL CD MT CP MAD MSB MC MMT1 MMT2 MMT3 MMT4

Evidence model of a task � 3 � 4 · 5 − 1 8 = 15 24 − 1 8 = 5 8 − 1 8 = 4 8 = 1 6 2 ⇔ MT & CL & ACL & SB & ¬ MMT 3 & ¬ MMT 4 & ¬ MSB T 1 CL ACL MT SB MMT3 MSB T1 MMT4 P(X1|T1) X1

Original model + one evidence model HV2 HV1 ACMI ACIM ACL ACD AD SB CMI CIM CL CD MT CP MAD MSB MC MMT1 MMT2 MMT3 MMT4 T1 X1

Total clique size 2500 2000 total clique size 1500 1000 500 0 0 1 2 3 4 number of solved tasks

Hierarchical evidence model (=parent divorcing) ACL CL MT SB MMT3 AUX2 MSB AUX1 AUX3 MMT4 CL ACL MT AUX5 SB MMT3 AUX4 MSB T1 MMT4 T1 P(X1|T1) X1 X1 − →

Total clique size 2500 no transformation parent divorcing 2000 total clique size 1500 1000 500 0 0 1 2 3 4 number of solved tasks

Factorized evidence model CL ACL MT SB MMT3 CL ACL MT SB MMT3 MSB B1 MMT4 MSB T1 MMT4 T1 P(X1|T1) X1 X1 − → ψ ( T 1 , MSB, SB, CL, ACL, MT, MMT 3 , MMT 4) =    ϕ ( T 1 , B 1) · ϕ ( B 1 , MSB ) · ϕ ( B 1 , SB ) · ϕ ( B 1 , CL ) · �  ϕ ( B 1 , ACL ) · ϕ ( B 1 , MT ) · ϕ ( B 1 , MMT 3) · ϕ ( B 1 , MMT 4) B 1

Functional dependence Y is functionally dependent on X 1 , . . . , X n if  if y = f ( x 1 , . . . , x n ) 1  ψ ( y, x 1 , . . . , x n ) = otherwise. 0  Example - a model with independence of causal influence C 1 C 2 . . . C n X 1 X 2 . . . X n f ( x 1 , . . . , x n ) Y

Factorization of MAX y = f ( x 1 , x 2 ) = max { x 1 , x 2 } P ( Y | X 1 , X 2 ) = � R h ( Y, R ) · g 1 ( X 1 , R ) · g 2 ( X 1 , R ) 0 +1 +2 0 +1 +2 0 +1 +2 0 +1 +2 0 1 = +1 1 1 1 +2 1 1 1 1 1 r 1 r 2 r 3 r 1 r 2 r 3 r 1 r 2 r 3 0 1 0 1 1 1 0 1 1 1 +1 -1 1 +1 1 1 +1 1 1 � r 1 ,r 2 ,r 3 ( × × ) +2 -1 1 +2 1 +2 1

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 ✄✁✄✁✄✁✄ ☎✁☎✁☎✁☎ ✄✁✄✁✄✁✄ ☎✁☎✁☎✁☎ ✄✁✄✁✄✁✄ ☎✁☎✁☎✁☎ ✄✁✄✁✄✁✄ ☎✁☎✁☎✁☎ ✄✁✄✁✄✁✄ ☎✁☎✁☎✁☎ ✄✁✄✁✄✁✄ ☎✁☎✁☎✁☎

Minimal base of rectangles (MBR) For a given partition Y 1 , . . . , Y q of X = × n i =1 X i find a set {R 1 , R 2 , . . . , R k } of minimal cardinality such that: • for j = 1 , . . . , k set R j is a rectangle, i.e. R j = × n i =1 D i , ∅ � = D i ⊆ X i , • each element Y ℓ , ℓ = 1 , 2 , . . . q of the partition can be generated from base {R 1 , . . . , R k } using operations � and � .

Factorization of MAX: problem reformulation Partition by values of Y : Y 1 = { (+1 , +1) } Y 2 = { (+1 , +2) , (+2 , +1) , (+2 , +2) } +1 +2 +3 Y 3 = { (+1 , +3) , (+2 , +3) , (+3 , +1) , 3 2 1 (+3 , +2)(+3 , +3) } +1 +1 +2 +3 Rectangular subspaces: +2 +2 +3 +2 R 1 = { (+1 , +1) } +3 +3 +3 +3 R 2 = { (+1 , +1) , (+1 , +2) , (+2 , +1) , (+2 , +2) } R 3 = X 1 × X 2 Y 1 = R 1 , Y 2 = R 2 � R 1 , and Y 3 = R 3 � R 1 .

Minimal base of rectangles for ADD 0 +1 +2 Y 1 = R 4 1 2 4 Y 2 = R 2 � R 4 � R 5 0 1 2 0 Y 3 = ( R 1 � ( R 2 � R 5 )) � ( R 3 � R 5 ) 5 1 2 3 +1 Y 4 = R 3 � R 5 � R 6 2 3 4 +2 Y 5 = R 6 6 3

Correspondence of MBR and factorization Y 1 = R 4 Y 2 = R 2 � R 4 � R 5 Y 3 = ( R 1 � ( R 2 � R 5 )) � ( R 3 � R 5 ) Y 4 = R 3 � R 5 � R 6 Y 5 = R 6 , Hidden variable B has one state for each rectangle h ( y, b ) R 1 R 2 R 3 R 4 R 5 R 6 g 1 ( x 1 , b ) , g 2 ( x 2 , b ) R 1 R 2 R 3 R 4 R 5 R 6 0 0 0 +1 0 0 y 1 0 +1 0 − 1 − 1 0 1 1 0 1 0 0 y 2 x 1 +1 − 1 − 1 0 +2 0 1 1 1 0 1 0 y 3 x 2 0 0 +1 0 − 1 − 1 1 0 1 0 0 1 y 4 x 3 0 0 0 0 0 +1 y 5

A Boolean function ( X 1 ∨ X 2 ) ⇒ ( X 2 ∧ X 3 ) Y = = ( ¬ X 1 ∧ ¬ X 2 ) ∨ ( X 2 ∧ X 3 ) 3 1 1 2 1 0 X 3 = 1 Y 0 = R 3 � ( R 2 � R 1 ) X 2 = 1 0 0 Y 1 = R 2 � R 1 X 3 = 0 1 0 X 2 = 0 1 X 1 = 0 X 1 = 1

Total clique size 2500 no transformation parent divorcing factorization 2000 total clique size 1500 1000 500 0 0 1 2 3 4 number of solved tasks