ecen689-613-spring09-handout7

ecen689-613-spring09-handout7 - x e P ( x ) ; for ( x V al...

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ECEN 689-613: SP TP PROB GRAPHICAL MODELS, Spring‘09 Handout #7 Pearl’s Message-Passing Algorithm (pseudocode) Date – Apr 21, 2009 (Tue) Consider a Bayesian Network over a set of nodes X with a DAG G that is a singly connected directed graph, or a polytree. Let E ∈ X be a set of evidence variables such that E = e . For every random variable X ∈ X and for all values x V al ( X ) , we can compute λ ( x ) , π ( x ) , and P ( x | e ) using the following algorithm. function initialize network { E = ; e = ; for ( X ∈ X ) { for ( x V al ( X )) { λ ( x ) = 1; } for ( Z Pa ( X )) { for ( z V al ( Z )) { λ X ( z ) = 1 ; } } for ( Y Ch ( X )) { for ( x V al ( X )) { π Y ( x ) = 1 ; } } } for ( every root R ∈ X ) { for ( r V al ( R )) { π ( r ) = P ( r ) ; P ( r | e ) = P ( r ) ; } for ( W Ch ( R )) { send π msg ( R,W ) ; } } }
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function update network ( V = ˆ v ) // New instantiated variable V = ˆ v { E = E ∪ { V } ; e = e ∪ { ˆ v } ; for ( v V al ( V )) { if ( v = ˆ v ) { λ ( v ) = 1; π ( v ) = 1; P ( v | e ) = 1; } else { λ ( v ) = 0; π ( v ) = 0; P ( v | e ) = 0; } } for ( Z Pa ( V Z / E ) { send λ msg ( V,Z ) ; } for ( Y Ch ( V )) { send π msg ( V,Y ) ; } } function send λ msg ( Y,X ) // Y (child) X (parent) { for ( x V al ( X )) { // W 1 ,...,W k are other parents of Y λ Y ( x ) = y h w 1 ,...,w k ± P ( y | x,w 1 ,...,w k ) Q k i =1 π Y ( w i ) ²i λ ( y ) ; λ ( x ) = Q U Ch ( X ) λ U ( x ) ; e P ( x ) = λ ( x ) π ( x ) ; } α =
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Unformatted text preview: x e P ( x ) ; for ( x V al ( X )) { P ( x | e ) = 1 e P ( x ) ; } for ( Z Pa ( X ) && Z / E ) { send msg ( X,Z ) ; } for ( U Ch ( X )- { Y } ) { send msg ( X,U ) ; } } function send msg ( Z,X ) // Z (parent) X (child) { for ( z V al ( Z )) { X ( z ) ( z ) Q U Ch ( Z )-{ X } U ( z ) ; } if ( X / E ) { for ( x V al ( X )) { // Z 1 ,...,Z k are parents of X ( x ) = z 1 ,...,z k P ( x | z 1 ,...,z k ) Q k i =1 X ( z i ) e P ( x ) = ( x ) ( x ) ; } = x e P ( x ) ; for ( x V al ( X )) { P ( x | e ) = 1 e P ( x ) ; } for ( Y Ch ( X )) { send msg ( X,Y ) ; } } if ( x s.t. ( x ) 6 = 1) { for ( W Pa ( X )- { Z } && W / E ) { send msg ( X,W ) ; } } }...
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ecen689-613-spring09-handout7 - x e P ( x ) ; for ( x V al...

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