Non-minimal Diagnoses

1
Non-minimal Diagnoses

• Philippe Dague and Yuhong Yan
• NRC-IIT
• Philippe.dague_at_lipn.univ-paris13.fr
• Yuhong.yan_at_nrc.gc.ca

2
Diagnosis

• Consider only assignment AB(c) and AB(c) for
  diagnoses, the size of diagnostic space is 2n, n
  number of components
• Diagnostic space is structure by set inclusion as
  a lattice

3

• A principle of parsimony has been adopted by
  Reiter considering only minimal (for set
  inclusion) diagnoses
• Question Do these minimal diagnoses characterize
  all diagnoses?
• Expected answer yes, any superset of a diagnosis
  is a diagnosis as well (Minimal Diagnosis
  Hypothesis)
• This is verified for the polybox with correct
  mode, and the 3-inverter with correct and faulty
  modes (but with the unknown mode).

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Counter Example(1)exhaustive fault modes
I1
I2
1
0
Example 1.a)

• Assume the only fault modes are stuck at 0 and
  shorted (no unknown mode)
• Inverter(x)?AB(x) ? S0(x) ? Short(x)
• S0(x) ? out(x)0
• Short(x) ? out(x) in(x)
• Diagnoses minimal diagnoses
• I1 (stuck at 0 or shorted)
• I2 (shorted)
• But the superset I1,I2 is not a diagnosis
• Reason I2 cant be stuck at 0, so it should be
  shorted, but in this case out(I1)1 and I1 cant
  be stuck at 0 nor shorted

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Counter Example(1)exhaustive fault modes
Example 1,b) Polybox

• Suppose that in addition to correct modes, we
  have
• AB(adder) ? adder acts as multiplier
• Same observation as before F10, G12
• M1 is still a minimal diagnosis but the
  superset M1, A2 is not any more

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Counter example (2) Exoneration
0
0
Example 2.a)

• Exoneration correct mode expressed as necessary
  and sufficient condition of correctness
• 2-inverter
• Inverter(x) ? (AB(x) ? In(x) 0 ?
  Out(x)1?In(x) 1 ? Out(x)0)
• Minimal diagnosis
• But the supersets I1 and I2 are not
  diagnoses. Each inverter exonerates the other (is
  an alibi for the other)

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Counter example (2) Exoneration

• 3 light bulbs
• Bulb(x) ? voltage(x, on) ? AB(x) ? lit(x)
• Observation only B3 is lit
• B1, B2 is a minimal diagnosis. The superset
  B1, B2, B3 is not
• Reason B3 cant be faulty, as it is lit.

Example 2.b)
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Conclusion

• The minimal diagnosis hypothesis is not satisfied
  in general, as soon as exhaustive fault modes or
  sufficient condition of correctness exists
• So in the diagnostic space lattice, diagnoses are
  not characterized by minimal diagnoses
• Questions does a logical characterization of the
  diagnoses in the general case exist?
• Answer yes.
• For this, the notion of conflict has to be
  generalized

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Recall

• Notation for ??Components,
• D(?) ?AB(c)c? ? ???AB(c)c ?
  Components\?
• Definition a diagnosis is a D(?) such that SD ?
  OBS ? D(?) is satisfiable
• Definition minimal diagnosis is a diagnosis D(?)
  such that for no proper subset ? of ? is D(?) a
  diagnosis
• Definition a conflict as defined by Reiter
  (named from now a R-conflict) is a subset C of
  Components such that
• SD ? OBS ? ??AB(c)c ? C ?
• Logically it is equivalent to SD ? OBS
  ?AB(c)c ? C
• ( a disjunct of AB(c) is entailed by SD ? OBS)

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What appears in the counter example?

• 1.a (2-inverter) SD ? OBS AB(I1)?AB(I2)
• But also SD ? OBS AB(I1)??AB(I2)
• 1.b (polybox) SD ? OBS AB(M1)?AB(M2)
• and SD ? OBS AB(M1)?AB(M3)
• But also SD ? OBS AB(M2)?AB(M3)?AB(A2)
• 2.a SD ? OBS doesn't entail disjunct of AB but
• SD ? OBS AB(I1)??AB(I2)
• SD ? OBS ?AB(I1)?AB(I2)
• 2.b SD ? OBS AB(B1) and SD ? OBS AB(B2)
• but also SD ? OBS ?AB(B3)

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Extension conflict

• So the idea is to extend a conflict to any
  conjunct of AB(c) and AB(c) entailed by SD ? OBS
  .
• Definition An AB-literal is AB(c) or AB(c) for
  some c? Components.
• An AB-clause is a disjunction of AB-literals
  containing no complementary pair of AB-literals.
• A positive AB-clause is an AB-clause all of its
  literals are positive
• Definition A conflict of (SD, Components, OBS)
  is an AB-clause entailed by SD ? OBS.
• A positive conflict is a conflict which is a
  positive AB-clause
• Remark one can identify a positive conflict with
  an R-conflict

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Extension conflict (2)

• Definition a minimal conflict is a conflict no
  proper sub-clause of which is a conflict
• Example see 1.a) 1.b) 2.a) 2.b) (the right side
  formulas in slide 10 are the minimal conflicts)
• Remark one can identify a minimal positive
  conflict with a minimal R-conflict

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Extension conflict (3)

• Suppose ? is a set of first order sentences, a
  ground clause is an implicate of ? iff ? entails
  c. c is a prime implicate of ? iff no proper
  sub-clause of c in entailed by ?
• Minimal conflicts are AB-clauses which are prime
  implicates of SD ? OBS.
• Minimal conflicts can be computed by theorem
  prover or ATMS

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Extension conflict (4)

• Reiters property relating minimal diagnosis to
  minimal R-conflict can be reformulated.
• Property let ? be the set of positive minimal
  conflicts of (SD, Components, OBS) and
  ??Components, then D(?) is a minimal diagnosis
  iff ? is a minimal subset such that ??D(?) is
  satisfiable
• This property generalizes as
• Property let ? be the set of minimal conflict of
  (SD, Components, OBS) and ??Components, then D(?)
  is a diagnosis iff ??D(?) is satisfiable

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Characterizing minimal diagnoses from positive
minimal conflicts

• Def Suppose ? is a set of propositional
  formulas, a conjunction of literals ? (containing
  no pair of complementary literals) is an
  implicant of ? iff ? entails each formula of ?. ?
  is a prime implicant of ? iff no proper sub
  conjunction of ? is an implicant of ?.

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Characterizing minimal diagnoses from positive
minimal conflicts (2)

• The Reiters characteristics of minimal diagnoses
  as minimal hitting sets of the collection of
  minimal R-conflicts can be reformulated as
• Theorem D(?) is a minimal diagnosis of (SD,
  Components, OBS) iff ?AB(c)c? ? is a prime
  implicant of the set of the positive minimal
  conflicts of (SD, Components, OBS).

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When minimal diagnoses are enough to
characterizing all diagnoses?

• Theorem Minimal diagnosis hypothesis holds (i.e.
  D(?) is a diagnosis iff ??? with D(?) a minimal
  diagnosis) iff all minimal conflicts are positive
• Unfortunately there is no equivalent condition on
  the syntactic form of SD and OBS. But it exists
  sufficient conditions. We consider 2 of them

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the Ignorance of Abnormal Behaviour (IAB)

• Def the Ignorance of Abnormal Behaviour (IAB)
  condition holds iff in the clause form of SD?OBS
  every occurrence of an AB-predicate is positive
• Theorem If (SD, Components, OBS) satisfies the
  IAB condition, then MDH holds

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IAB(2)

• IAB is ensured, for example, if all sentence of
  SD where AB appears follow the schema
• ?AB(x)?P1(x)?P2(x)? ?Pn(x)?G1(x)? ?Gm(x)
• Where literals Pi(x) and Gj(x) do not mention AB
• i.e. when only necessary condition of correct
  behaviour are expressed
• Example
• ?AB(x)?transistor(x)?On(x)?off(x)?saturated(x)
• ?AB(x)?resistor(x)?ports(x,a b)?resistance(x)r
  ?v(x, a, b) r i(x,a)

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Limited Knowledge of Abnormal Behaviour (LKAB)

• Def the Limited Knowledge of Abnormal Behaviour
  (LKAB) condition holds iff ?(Cp, Cn, c),
  Cp?Components, Cn ? Components, Cp?Cn ?,
  c?Components, c?Cp,c?Cn, SD?OBS??AB(x)x?Cp ?
  ??AB(x)x?Cn satisfiable,
• SD?OBS?AB(c) satisfiable ? SD?OBS??AB(x)x?Cp?
  c? ??AB(x)x?Cn
• Remark IAB ? LKAB

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LKAB(2)

• LKAB is ensured, for example, if all sentences of
  SD where AB appears have one of the following two
  forms
• ?AB(x)?P1(x)?P2(x)? ?Pn(x)?G1(x)? ?Gm(x)
• AB(x)?P1(x)?P2(x)? ?Pn(x)?F1(x)? ?Fm(x)?U(x)
• Where Gi(x) describes a possible correct
  behaviour for x, Fi(x) describes a possible
  faulty behaviour for x, U(x) an unknown behaviour
• (Gi(x), Fi(x), U(x) only occur negatively in
  other clauses and U(x) only occurs in clauses
  expressing it is distinct of any Gi(x) and any
  Fi(x).)
• i.e. when only necessary conditions of correct
  behaviours and necessary condition of
  non-exhaustive faulty behaviours (with unknown
  mode) are expressed.

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LKAB(3)

• (see example in lecture diagnoses with fault
  modes).
• Theorem if (SD, Components, OBS) satisfies the
  LKAB condition and D(?) is a diagnosis, then
  D(?) is a diagnosis for every ? ? ?, such that
  for each c??, SD?OBS ?AB(c) is satisfiable

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Charactering Diagnoses from Minimal Conflicts

• Compact representation of diagnoses
• Example 1.b)
• AB(M1) ? ?AB(A2) ? K1(M2) ? K2(M3) ? K3(A1),
• where KiAB or ?AB
• they can be coded as AB(M1) ? ?AB(A2)

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Compact representation of diagnoses

• Definition A partial diagnosis for (SD,
  Components, OBS) is a satisfiable conjunction P
  of AB-literals such that for every satisfiable
  conjunction P of AB-literals containing P as
  sub-conjunction, SD?OBS ?P is satisfiable
• Remark if C, of size k, is the set of all
  components mentioned in P, the P?
  ?K(c)c?Components\C is a diagnosis, where each
  K(c) is AB(c) or ?AB(c). So P codes 2n-k
  diagnoses

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Kernel diagnosis

• It is natural to consider the minimal such
  partial diagnoses
• Definition A kernel diagnosis is a partial
  diagnosis whose no proper sub-conjunction is a
  diagnosis
• Property (Characterization of Diagnoses)
• D(?) is a diagnosis iff there is a kernel
  diagnosis which is a sub-conjunction of it

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Kernel Diagnoses (2) Examples

• 1.a) 2 kernel diagnoses
• AB(I1)??AB(I2) and ?AB(I1)?AB(I2)
• 1.b) 4 kernel diagnoses
• AB(M1)??AB(A2) AB(M1)?AB(M2)
• AB(M1)?AB(M3) AB(M2)?AB(M3)
• 2.a) 2 kernel diagnoses
• ?AB(I1)??AB(I2) AB(I1)?AB(I2)
• 2.b) 1 kernel diagnosis
• AB(B1)?AB(B2)??AB(B3)

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Theorem

• Theorem (Characterization of partial and kernel
  diagnoses from minimal conflicts)
• The partial diagnoses of (SD, Components, OBS)
  are the implicants of the minimal conflicts of
  (SD, Components, OBS)
• The kernel diagnoses of (SD, Components, OBS)
  are the prime implicants of the minimal conflicts
  of (SD, Components, OBS)
• The minimal diagnoses are the prime impliants of
  positive minimal conflicts
• Remark if all minimal conflicts are positive,
  there is a 1 to 1 correspondence between kernel
  diagnoses and minimal diagnoses
• ?AB(c)c?K ? ?AB(c)c?K ? ??AB(c)
  c?Components\K

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Exercise

• Full adder in Reiters paper (figure 1).
• Use kernel diagnosis to find diagnosis
• Use two-direction imply (?) in the model to find
  kernel diagnosis
• Add the axiom that all variables are Boolean
  (x0?x1), find kernel diagnosis