D-Algorithm (1/4) h d' d i j e' n e k a g b c l f' f m Example 1 D’ G1

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Presentation transcript:

D-Algorithm (1/4) h d' d i j e' n e k a g b c l f' f m Example 1 D’ G1 Five logic values {0, 1, x, D, D’} Try to propagate fault effect thru G1  Set d to 1 h d' d Try to propagate fault effect thru G2  Set j,k,l,m to 1 1 i D’ G1 j 1 D e' 1 D’ ≠ n e G2 k a g D 1 b c Conflict at k  Backtrack ! l f' f m

D-Algorithm (2/4) h d' d i j e' n e k a g b c l f' f m Example 1 D’ G1 Five logic values {0, 1, x, D, D’} h Try to propagate fault effect thru G2  Set j,l,m to 1 d' d 1 i D’ G1 j 1 e' 1 n e G2 k D a g D 1 b D’ (next D-frontier chosen) c l Conflict at m  Backtrack ! f' f 1 m D’ ≠

D-Algorithm (3/4) h d' d i j e' n e k a g b c l f' f m Example Five logic values {0, 1, x, D, D’} Try to propagate fault effect thru G2  Set j,l to 1 h 1 d' d Fault propagation and line justification are both complete  A test is found ! 1 i D’ G1 j 1 e' 1 n e G2 k D a g D 1 b D’ c l This is a case of multiple path sensitization ! f' 1 f m D’ (next D-frontier chosen)

D-Algorithm (4/4) Decision Implication Comments a=0 Active the fault b=1 Unique D-drive c=1 g=D d=1 Propagate via i i=D’ d’=0 j=1 Propagate via n k=1 l=1 m=1 n=D e’=0 e=1 k=D’ Contradiction e=1 Propagate via k k=D’ e’=0 j=1 l=1 Propagate via n m=1 n=D f’=0 f=1 Contradiction m=D’ f=1 Propagate via m m=D’ f’=0 l=1 n=D

PODEM Algorithm (1/4) Example h d' d i j e' n e k a g b c l f' f m Select D-frontier G2 and set objective to (k,1)  e = 0 by backtrace break the sensitization across G2 (j=0) Backtrack ! h 1 d' d 1 i D’ G1 1 j e' n e G2 k a D 1 g b 1 c l f' f m

PODEM Algorithm (2/4) Example h d' d i j e' n e k a g b c l f' f m Select D-frontier G3 and set objective to (e,1) No backtrace is needed Success at G3 h 1 d' d 1 i D’ G1 1 j e' n e G2 k a D 1 g G3 b c l f' f m G4

PODEM Algorithm (3/4) Example h d' d i j e' n e k a g b c l f' f m Select D-frontier G4 and set objective to (f,1)  No backtrace is needed Succeed at G4 and G2 D appears at one PO A test is found !! h 1 d' d 1 i D’ G1 j 1 e' 1 n e G2 k D a g D 1 G3 b D’ c 1 l f' f m G4 D’

PODEM Algorithm (4/4) Objective PI assignment Implications D-frontier Comments a=0 a=0 h=1 g b=1 b=1 g c=1 c=1 g=D i,k,m d=1 d=1 d’=0 i=D’ k,m,n k=1 e=0 e’=1 Assignments need to be reversed during backtracking j=0 k=1 n=1 m no solutions!  backtrack e=1 e’=0 flip PI assignment j=1 n d e f f' e' d' h i j k l m g a b c 1 D’ D k=D’ m,n l=1 f=1 f’=0 l=1 m=D’ n=D