CSC411Artificial Intelligence1 Chapter 9 Reasoning in Uncertain Situations Contents Uncertain situations Non-monotonic logic and reasoning Certainty Factor algebra Fuzzy logic and reasoning Dempster-Shafer theory of evidence Bayesian belief network Markov models

CSC411Artificial Intelligence2 Traditional Logic Based on predicate logic Three important assumptions: –Predicate descriptions are sufficient w.r.t. to the domain –Information is consistent –Knowledge base grows monotonically

CSC411Artificial Intelligence3 Non-monotonic Logic Addresses the three assumptions of traditional logic –Knowledge is incomplete No knowledge about p: true or false? Prolog – closed world assumption –Knowledge is inconsistent Based on how the world usually works Most birds fly, but Ostrich doesn’t –Knowledge base grows non-monotonically New observation may contradict the existing knowledge, thus the existing knowledge may need removal. Inference based on assumptions, how come if the assumptions are later shown to be incorrect Three modal operators are introduced

CSC411Artificial Intelligence4 Unless Operator New information may invalidate previous results Implemented in TMS – Truth Maintenance Systems to keep track of the reasoning steps and preserve the KB consistency Introduce Unless operator –Support inferences based on the belief that its argument is not true –Consider p(X) unless q(X)  r(X) If p(X) is true and not believe q(X) true then r(X) p(Z) r(W)  s(W) From above, conclude s(X). Later, change believe or find q(X) true, what happens? Retract r(X) and s(X) –Unless deals with believe, not truth Either unknown or believed false Believed or known true –Monotonocity

CSC411Artificial Intelligence5 Is-consistent-with Operator M When reason, make sure the premises are consistent Format: M p – p is consistent with KB Consider –X good_student(X)  M study_hard(X)  graduates(X) –For all X who is a good student, if the fact that X studies hard is consistent with KB, then X will graduate –Not necessary to prove that X study hard. How to decide p is consistent with KB –Negation as failure –Heuristic-based and limited search

CSC411Artificial Intelligence6 Default Logic Introduce a new format of inference rules: –A(Z)  :B(Z)  C(Z) –If A(Z) is provable, and it is consistent with what we know to assume B(Z), then conclude C(Z) Compare with is-consistent-with operator –Similar –Difference is the reasoning method In default logic, new rules are used to infer sets of plausible extensions –Example: X good_student(X)  :study_hard(X)  graduates(X) Y party(Y)  :not(study_hard(Y))  not(graduates(X))

CSC411Artificial Intelligence7 Stanford Certainty Factor Algebra Measure of confidence or believe Summation may not be 1 Simple case: –Confidence for: MB(H|E) –Confidence against: MD(H|E) –Properties: 1>MB(H|E)>0 while MD(H|E)=0, or 1>MD(H|E)>0 while MB(H|E)=0 –Put together CF(H|E) = MB(H|E) – MD(H|E) 1 > CF(H|E) > -1

CSC411Artificial Intelligence8 CF Combination Premises combination –CF( P and Q) = min(CF(P), CF(Q)) –CF( P or Q) = max(CF(P), CF(Q)) Rule CF: each rule has a confidence measure CF propagation –Rule R: P  Q with CF=CF(R) –CF(Q) = CF(P)CF(R) Rule combination –Rules R1: P1  Q: CF1(Q) = CF(P1)xCF(R1) – R2: P2  Q: CF2(Q) = CF(P2)xCF(R2) –CF(Q) = CF1+CF2 – (CF1xCF2) if both positive CF1+CF2 + (CF1xCF2) if both negative (CF1+CF2)/(1-min(|CF1|,|CF2|)) otherwise

CSC411Artificial Intelligence9 Fuzzy Sets Classic sets –Completeness: x in either A or ¬A –Exclusive: can not be in both A and ¬A Fuzzy sets –Violate the two assumptions –Possibility theory -- measure of confidence or believe –Probability theory – randomness –Process imprecision –Introduce membership function –Believe xA in some degree between 0 and 1, inclusive

CSC411Artificial Intelligence10 The fuzzy set representation for “small integers.”

CSC411Artificial Intelligence11 A fuzzy set representation for the sets short, medium, and tall males.

CSC411Artificial Intelligence12 Fuzzy Set Operations Fuzzy set operations are defined as the operations of membership functions Complement: ¬A = C –mC = 1 – mA Union: A  B =C –mC = max(mA, mB) Intersection: A  B = C –mC = min(mA, mB) Difference: A – B = C –mC = max(0, mA-mB)

CSC411Artificial Intelligence13 Fuzzy Inference Rules Rule format and computation –If x is A and y is B then z is C mC(z) = min(mA(x), mB(y)) –If x is A or y is B then z is C mC(z) = max(mA(x), mB(y)) –If x is not A then z is C mC(z) = 1 – mA(x)

CSC411Artificial Intelligence14 The inverted pendulum and the angle θ and dθ/dt input values.

CSC411Artificial Intelligence15 The fuzzy regions for the input values θ (a) and dθ/dt (b). N – Negative, Z – Zero, P – Positive

CSC411Artificial Intelligence16 The fuzzy regions of the output value u, indicating the movement of the pendulum base: Negative Big, Negative, Zero, Positive, Positive Big.

CSC411Artificial Intelligence17 The fuzzificzation of the input measures X 1 = 1: mZ(X 1 ) = mP(X 1 ) = 0.5, mN(X 1 ) = 0 X 2 = -4: mZ(X 2 ) = 0.2, mN(X 2 ) = 0.8, mP(X 2 ) = 0

CSC411Artificial Intelligence18 The Fuzzy Associative Matrix (FAM) for the pendulum problem. The input values are on the left and top. Fuzzy Rules:

CSC411Artificial Intelligence19 The fuzzy consequents (a) and their union (b). The centroid of the union (-2) is the crisp output.

CSC411Artificial Intelligence20 Dempster-Shafer Theory Probability theory limitation –Assign a single number to measure any situation, no matter how it is complex –Cannot deal with missing evidence, heuristics, and limited knowledge Dempster-Shafer theory –Extend probability theory –Consider a set of propositions as a whole –Assign a set of propositions an interval [believe, plausibility] to constraint the degree of belief for each individual propositions in the set –The belief measure bel is in [0,1] 0 – no support evidence for a set of propositions 1 – full support evidence for a set of propositions –The plausibility of p, pl(p) = 1 – bel(not(p)) Reflect how evidence of not(p) relates to the possibility for belief in p Bel(not(p))=1: full support for not(p), no possibility for p Bel(not(p))=0: no support for not(p), full possibility for p Range is also in [0,1]

CSC411Artificial Intelligence21 Properties of Dempster-Shafer Initially, no support evidence for either competing hypotheses, say h1 and h2 –Dempster-Shafer: [bel, pl] = [0, 1] –Probability theory: p(h1)=p(h2)=0.5 Dempster-Shafer belief functions satisfy weaker axioms than probability function Two fundamental ideas: –Obtaining belief degrees for one question from subjective probabilities for related questions –Using Dempster rule to combine these belief degrees when they are based on independent evidence

CSC411Artificial Intelligence22 An Example Two persons M and B with reliabilities detect a computer and claim the result independently. How you believe their claims? Question (Q): detection claim Related question (RQ): detectors’ reliability Dempster-Shafer approach –Obtain belief degrees for Q from subjective (prior) probabilities for RQ for each person –Combine belief degrees from two persons Person M: –reliability 0.9, unreliability 0.1 –Claim h1 –Belief degree of h1 is bel(h1)=0.9 –Belief degree of not(h1) is bel(not(h1))=0.0, different from probability theory, since no evidence supporting not(h1) –pl(h1) = 1 – bel(not(h1)) = 1-0 =1 –Thus belief measure for M claim h1 is [0.9, 1] Person B: –Reliability 0.8, unreliability 0.2 –Claim h2 –bel(h2) =0.8, bel(not(h2))=0, pl(h2)=1-bel(not(h2))=1-0 –Belief measure for B claim h2 is [0.8,1]

CSC411Artificial Intelligence23 Combining Belief Measure Set of propositions: M claim h1 and B claim h2 –Case 1: h1 = h2 Reliability M and B: 09x0.8=0.72 Unreliability M and B: 0.1x0.2=0.02 The probability that at least one of two is reliable: =0.98 Belief measure for h1=h2 is [0.98,1] –Case 2: h1 = not(h2) Cannot be both correct and reliable At least one is unreliable –Reliable M and unreliable B: 0.9x(1-0.8)=0.18 –Reliable B and unreliable M: 0.8x(1-0.1)=0.08 –Unreliable M and B: (1-0.9)x(1-0.8)=0.02 –At least one is unreliable: =0.28 Given at least one is unreliable, posterior probabilities Given at least one is unreliable, posterior probabilities –Reliable M and unreliable B: 0.18/0.28=0.643 –Reliable B and unreliable M: 0.08/0.28=0.286 Belief measure for h1 –Bel(h1)=0.643, bel(not(h1))=bel(h2)=0.286 –Pl(h1)=1-bel(not(h1))= =0.714 –Belief measure: [0.643, 0.714] Belief measure for h2 –Bel(h2)=0.286, bel(not(h2))=bel(h1)=0.683 –Pl(h2)=1-bel(not(h2))= =0.317 –Belief measure: [0.286, 0.317]

CSC411Artificial Intelligence24 Dempster’s Rule Assumption: –probable questions are independent a priori –As new evidence collected and conflicts, independency may disappear Two steps 1.Sort the uncertainties into a priori independent pieces of evidence 2.Carry out Dempster rule Consider the previous example –After M and B claimed, a repair person is called to check the computer, and both M and B witnessed this. –Three independent items of evidence must be combined Not all evidence is directly supportive of individual elements of a set of hypotheses, but often supports different subsets of hypotheses, in favor of some and against others

CSC411Artificial Intelligence25 General Dempster’s Rule Q – an exhaustive set of mutually exclusive hypotheses Z – a subset of Q M – probability density function to assign a belief measure to Z M n (Z) – belief degree to Z, where n is the number of sources of evidences

CSC411Artificial Intelligence26 Bayesian Belief Network A computational model for reasoning to the best explanation of a data set in the uncertainty context Motivation –Reduce the number of parameters of the full Bayesian model –Show how the data can partition and focus reasoning –Avoid use of a large joint probability table to compute probabilities for all possible events combination Assumption –Events are either conditionally independent or their correlations are so small that they can be ignored Directed Graphical Model –The events and (cause-effect) relationships form a directed graph, where events are vertices and relationships are links

CSC411Artificial Intelligence27 The Bayesian representation of the traffic problem with potential explanations. The joint probability distribution for the traffic and construction variables The Traffic Problem Given bad traffic, what is the probability of road construction? p(C|T)=p(C=t, T=t)/(p(C=t, T=t)+p(C=f, T=t))=.3/(.3+.1)=.75

CSC411Artificial Intelligence28 An Example Traffic problem –Events: Road construction C Accident A Orange barrels B Bad traffic T Flashing lights L –Joint probability P(C,A,B,T,L)=p(C)*p(A|C)*p(B|C,A)*p(T|C,A,B)*p(L|C,A,B,T) Number of parameters: 2^5=32 –Reduction Assumption: Parameters are only dependent on parents Calculation of joint probability –P(C,A,B,T,L)=p(C)*p(A)*p(B|C)*p(T|C,A)*p(L|A) –Number of parameters: =20

CSC411Artificial Intelligence29 BBN Definition Links represent conditional probabilities for causal influence These influences are directed: presence of some event causes other events These influences are not circular Thus a BBN is a DAG: Directed Acyclic Graph

CSC411Artificial Intelligence30 Discrete Markov Process Finite state machine –A graphical representation –State transition depends on input stream –States and transitions reflect properties of a formal language Probabilistic finite state machine –A finite state machine –Transition function represented by a probability distribution on the current state Discrete Markov process (chain, machine) –A specialization of probabilistic finite state machine –Ignores its input values

CSC411Artificial Intelligence31 A Markov state machine or Markov chain with four states, s 1,..., s 4 At any time the system is in one of distinct states The system undergoes state change or remain Divide time into discrete intervals: t1, t2, …, tn Change state according to the probability distribution of each state S(t) – the actual state at time t p(S(t)) = p(S(t)|S(t-1), s(t-2), s(t-3), …) First-order markov chain –Only depends on the direct predecessor state –P(S(t)) = p(S(t)|S(t-1))

CSC411Artificial Intelligence32 Observable Markov Model Assume p(S(t)|S(t-1)) is time invariant, that is, transition between specific states retains the same probabilistic relationship State transition probability a ij between s i and s j : –a ij =p(S(t)=s i |S(t-1)=s j ), 1<=i,j<=N –If i=j, no transition (remain the same state) –Properties: a ij >=0,  i a ij =1

CSC411Artificial Intelligence33 S1 – sun S2 – cloudy S3 – fog S4 – precipitation Time intervals: noon to noon Question: suppose that today is sunny, what is the probability of the next five days being sunny, sunny, cloudy, cloudy, precipitation?