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Resolving Rule Conflicts

Resolving Rule Conflicts. Assign explicit priorities to rules Specificity of a rule’s antecedents R1: IF car will not start THEN battery is dead R2: IF car will not start AND headlights will not work THEN battery is dead Order in which rule are entered in kbase

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Resolving Rule Conflicts

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  1. Resolving Rule Conflicts • Assign explicit priorities to rules • Specificity of a rule’s antecedents R1: IF car will not start THEN battery is dead R2: IF car will not start AND headlights will not work THEN battery is dead • Order in which rule are entered in kbase • timestamp, rule entered first given priority • Order of rules in kbase • Recency of facts entered • real-time ES

  2. Certainty factors • higher CF rule fires first • rule that gives higher confidence to goal being pursued

  3. Certainty Theory • Representing uncertain evidence • Representing uncertain rules • Combining evidence from multiple sources eg. IF A and B THEN X cf 0.8 IF C THEN X cf 0.7 What is the certainty of X? • -1<= CF <= 1 • -100 <= CF <= 100 • Used in • rule conclusions, values of variables in premise, answers to user queries

  4. Certainty theory • Commutative property • if more than one rule gathers information, then the combined CF value can not be dependent upon the order of the processing of the rules • Asymptotic property • the certainty model should incrementally add belief to a hypothesis as new positive evidence is obtained; however, unless we encounter some evidence that absolutely confirms the hypothesis, we cannot be totally certain. • Thus, confirming evidence increases out belief, but unless absolute certainty is found, cf approaches 1, but never equals 1.

  5. Rule: IF E THEN H CF(Rule) CF(H,E) = CF(E) * CF(Rule) Example: IF econ-two-years = strong THEN likelihood-of-inflation = strong CF 40 Given econ-two-years = strong with cf=70 CF(likelihood-of-inflation = strong ) = (40 * 70)/100 = 28

  6. Conjunctive: IF E1 and E2 and … and En THEN H CF(Rule) CF(H, E1 and E2 and … and En) = min{CF(Ei)}* CF(Rule) • Disjunctive IF E1 or E2 or … or En THEN H CF(Rule) CF(H, E1 or E2 or … or En) = max{CF(Ei)}* CF(Rule)

  7. Certainty propagation in similarly concluded rules R1: IF E1 THEN H CF1 R2: IF E2 THEN H CF2 (supporting evidence increases our belief) CFcombine(CF1,CF2) = CF1 + CF2(1 - CF1), when both > 0 = (CF1+ CF2) / (1 - min(|CF1|, |CF2|), when one < 0 = CF1+ CF2(1+ CF1), when both < 0.

  8. Premise with AND (conjunctive) Example: IF economy-two-years = strong AND availability-of-investment-capital = low THEN likelihood-of-inflation = strong CF of condition = min(cf1, cf2) • Premise with AND (conjunctive) Example: IF economy-two-years = poor OR unemployment-outlook = poor THEN economic-outlook = poor CF of condition = max(cf1, cf2)

  9. Premise with both AND and OR Example: IF has-credit-card = yes (cf= 80) OR cash = ok (cf = 90) AND payments = ok (cf = 85) THEN approval = ok (has-credit-card = yes AND payments = ok) [min(80,85)] OR (cash = ok AND payments = ok) [min(90,85)] CF = 80 + 85 - (80*85)/100 = 97 (some systems use this) CF = max(80,85) = 85 (some systems use this)

  10. Example: R1: IF weatherman says it will rain THEN it will rain CF 0.8 R2: IF farmer says it will rain THEN it will rain CF 0.8 Case (a): Weatherman and farmer are certain in rain CF(E1) = CF(E2) = 1.0 CF(H, E1) = CF(E1) * CF(Rule1) = 1.0*0.8 = 0.8 CF(H, E2) = CF(E2) * CF(Rule2) = 1.0*0.8 = 0.8 CFcombine(CF1, CF2) = CF1+ CF2(1- CF1) = 0.8+0.8(1-0.8) = 0.96 CF of a hypothesis which is supported by more than one rule, can be incrementally increased by supporting evidence from both rules.

  11. Case (b):Weatherman certain in rain, farmer certain in no rain CF(E1) = 1.0, CF(E2) = -1.0 CF1 = 0.8, CF2 = -0.8 CFcombine(CF1, CF2) = (0.8 + (-0.8)) / (1 - min(0.8, 0.8)} = 0 (unknown) Case (b): CF(E1) = -0.8, CF(E2) = -0.6 CFcombine(CF1, CF2) = -0.64 - 0.48(1 - 0.64) = -0.81 Incremental decrease in certainty from more than one source of disconfirming evidence Case (c): CFcombine(CF1, CF2, CF3,….) = 0.999 = CFold Single piece of disconfirming evidence CFnew = -0.8 Cfcombine = (0.999 - 0.8) / (1 - 0.8) = 0.995 Single piece of disconfirming evidence does not have a major impact on many pieces of confirming evidence.

  12. Certainty Threshold • CF-condition < Threshold => rule fails to fire Rule condition fails if CF(premise) < Threshold But if rule fires, and after that CF(conclusion) < Threshold the conclusion will still be asserted.

  13. Interpreting CF values Definitely not -1.0 Almost certainly not -0.8 Maybe 0.4 Probably not -0.6 Probably 0.6 Maybe not -0.4 Almost certainly 0.8 Unknown -0.2 to 0.2 Definitely 1.0 • Acquiring CF from experts • Prompting users for CF values

  14. Example R1: IFthe weather looks lousy (E1) OR I am in a lousy mood (E2) THEN I shouldn’t go to the ball game CF 0.9 (H1) R2: IF I believe it is going to rain (E3) THEN the weather looks lousy CF 0.8 (E1) R3: IF I believe it is going to rain (E3) AND the weatherman says it is going to rain (E4) THEN I am in a lousy mood CF 0.9 (E2) R4: IF the weatherman says it is going to rain (E4) THEN the weather looks lousy CF 0.7 (E1) R5: IF the weather looks lousy (E1) THEN I am in a lousy mood CF 0.95 (E2)

  15. Assume user enters following facts: • I believe it is going to rain, CF(E3) = 0.95 • Weatherman says it is going to rain, CF(E4) = 0.8 • Goal: I shouldn’t go to the ball game (H1) • Step 1: Pursue R1: Pursue premise of R1 “weather looks lousy” (E1): R2 and R4 • Step 2: Pursue R2 CF(E1, E3) = CF(E3) * CF(Rule R2) = 0.8*0.95 = 0.76 • Step 3: Pursue R4 CF(E1, E4) = CF(E4)*CF(Rule R4) = 0.7*0.85=0.60 • Step 4: Combine evidence for E1 CF(E1) = CF(E1,E3) + CF(E1,E4)*(1 - CF(E1,E3)) = 0.76 + 0.60* (1.0 - 0.76) = 0.90

  16. Step 5: Pursue premise 2 of Rule 1 (E2) : R3 and R5 • Step 6: Pursue R5 CF(E2, E1) = CF(E1)* CF(Rule R5) = 0.95*0.9 = 0.86 • Step 7: Pursue R3 CF(E2, E3 and E4) = min{CF(E3), CF(E4)} * CF(Rule R3) = min{0.95,0.85}*0.9 = 0.77 • Step 8: Combine evidence for E2: “I am in a lousy mood” CF(E2) = CF(E2, E1) + CF(E2, E3 and E4) ( 1 - CF(E2, E1)) = 0.86 + 0.77 ( 1- 0.86) = 0.97 • Step 9: return to Rule R1 From steps 4 and 9 CF(H1, E1 or E2) = max{CF(E1), CF(E2)} * CF(Rule R1) = max{0.9,0.97}*0.9 = 0.87 = CF(I shouldn’t go to the ball game) Conclusion: I almost definitely shouldn’t go to the ball game.

  17. Controlling search with CF • Meta rules • Example: IF CF(Problem is in electrical system) < 0.5 THEN PURSUE problem is fuel system

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