1 Variable Elimination in Bayesian Networks

Transcription

1 CS 440: Introduction to Artificial Inttelligence Lecture date: 12 Feruary 2008 Instructor: Eyal Amir Guest Lecturer: Mark Richards 1 Variale Elimination in Bayesian Networks Review: Inference Given a joint proaility distriution over variales a set of variales X X 1, X 2,..., X n, we can make inferences of the form (Y Z), where Y X is the set of query variales, Z X are the evidence variales. The other variales H (those not mentioned in the query) are called hidden variales. To e clear, X Y Z H. The most naive and expensive way to do inference is to use the full joint proaility distriution and sum out the hidden variales. By the product rule, P (Y Z)P (Z) P (Y, Z). (Note that this is true for any distriution. This does not have anything to do with independence.) So to answer the query P (Y Z), we can compute P (Y,Z) P (Z). Note that P (Y, Z) H P (Y, Z, H). From the view of the full joint proaility tale, we are summing the proailities for all of the entries in the tale that match the values of the query variales and evidence variales (this includes entries for all of the cominations of values for the hidden variales) and dividing y the sum of all the entries that match the values of the evidence variales. (Note that the values in the first sum are a suset of the values in the latter sum.) This method is straightforward conceptually ut has many prolems. First, you have to compute and store the values in the full joint distriution. The numer of entries is i Options(X i), where Options(X i ) is the numer of different values that variale X i may have. So for n Boolean variales, the full joint has 2 n entries. For large networks, it is unreasonale to store/compute this many values. Furthermore, if a query requires summing out most of the variales, the time complexity is O(2 n ) as well. Normalizing Suppose we have variales A and B. We want to compute P (A ). We use to denote B true. The result of our query P (A, ) is a two-element tale that specifies P (a, ) and P ( a, ).Since P (A )P () P (A, ), we can compute P (A ) P (A,) P (). But we also know that P () P (a, ) + P ( a, ). Since we need oth P (a, ) and P (, ) to answer the query, this may actually e the est way to compute P (). The way this is expressed in the ook is that P (A ) αp (A, ), where α is a normalizing constant that makes the values in the tale P (A ) sum to 1. (In this case, α is P (), ut the idea is that it doesn t matter what it is; we just know that we need to normalize to make sure that the values in P (A ) sum to 1. Enumeration Using the urglar alarm network, suppose we wish to compute the proaility that there is an earthquake, given that oth John and Mary call: P (E j, m) αp (E, j, m)

2 We need to sum over all the values of the hidden vars: αp (E, j, m) a P (E, j, m,, a) This may e a slight ause of notation. The lower case j, m mean that JohnCalls true and MaryCalls true. The lowercase, a mean that we are summing over all values of the variales B, A. So now we have a sum over entries from the full joint. We can retrieve these using our Bayesian network structure: P (E, j, m,, a) a P ()P (E)P (a, E)P (j a)p (m a) a In general, sums of this form could take O(n2 n ) time to compute. There may e n values to multiply together to compute each product term, and up to O(2 n ) total terms to sum up. We can save some computations y pushing the s inward as much as possile: P ()P (E)P (a, E)P (j a)p (m a) P (E) a P () a P (a, E)P (j a)p (m a) We need to do this sum oth for E true and E false. For the first case: ( ) P () (P (a, e)p (j a)p (m a) + P ( a, e)p (j a)p (m a)) + P (e) P ( ) (P (a, e)p (j a)p (m a) + P ( a, e)p (j a)p (m a)) Even here, we can see some wasted computations: the product P (j a)p (m a) is computed twice, and so is P (j a)p (m a). In this example, we do two extra multiplications, which is no ig deal. But in a large network, this kind of waste can e deilitating. Variale Elimination (We follow Section 14.4 in Russell-Norvig with some additional details.) Here we introduce a variale elimination algorithm that will help us avoid the duplicate computations; it is a form of dynamic programming. We introduce tales called factors to help us do the ookkeeping. Initially, we will have one factor for each CPT term in our expression: E f E (E) T.002 F.998 α P (E) }{{} E T.001 F.999 P () P (a, E) P (j a) P (m a) }{{}}{{}}{{}}{{} a B A J M A B E f A (A, B, E) T T T.95 T T F.94 T F T.29 T F F.001 F T T.05 F T F.06 F F T.71 F F F.999 A f J (A) T.9 F.05 A f M (A) T.7 F.01

3 The initial tales store the values from the CPTs that we will need over the course of the query evaluation. The factor for E, denoted f E (E), is a two-element vector ecause we will e computing oth P (e, j, m) and P ( e, j, m) (we will normalize at the end to compute P (e j, m) and P ( e j, m)). The suscript E names the factor, and the (E) means that the value of E varies in the factor. The factor for B is also a two-element vector f B (B), ecause we will need to sum out the variale B. The factor f A (A, B, E) requires elements ecause we will need all cominations of values for A, B, and E in the expression. And even though J and M oth have parents in the network, the factors f J (A) and f M (A) have only two elements ecause throughout the expression, we are only interested in cases where JohnCalls true and MaryCalls true. There are two operations that we need to e ale to perform on factors: multiplication and summing out. The multiplication operation is similar to a natural join on dataase tales, as opposed to a matrix multiplication or an element-y-element multiplication. The set of variales in the product of two factors is the union of the sets of variales in the operands. In the product factor, there will e one entry for every comination of values for variales in the resulting set. The value of each entry is the product of the entries in the original tales that match the values for all relevant variales. For example, consider the product f JM (A) f J (A)f M (A): A f JM (A) T.9 *.7 F.05 *.01 A f J (A) T.9 F.05 A f M (A) T.7 F.01 The union of variales in the operands is A. The entry for T in the resulting tale is the product of the entries where A T rue in the operands. Now consider f AJM (A, B, E) f A (A, B, E)f JM (A). A B E f AJM (A, B, E) T T T.95 *.63 T T F.94 *.63 T F T.29 *.63 T F F.001 *.63 F T T.05 *.0005 F T F.06 *.0005 F F T.71 *.0005 F F F.999 *.0005 A B E f AJM (A, B, E) T T T.95 T T F.94 T F T.29 T F F.001 F T T.05 F T F.06 F F T.71 F F F.999 A f JM (A) T.63 F.0005 The entry for A T, B T, E T in f AJM (A, B, E) is the product of corresponding entries from f A (A, B, E) where A T, B T, E T and from f J M(A) where A T (since B, E do not appear in f J M(A). In general, we have f(x 1,..., X j, Y 1,..., Y k, Z 1..., Z l ) f 1 (X 1,..., X j, Y 1,..., Y k )f 2 (Y 1,..., Y k, Z 1,..., Z l ) The factors f 1 and f 2 share common variales Y. If f 1 has a total of j + k variales and f 2 has k + l variales, then the product will have j + k + l variales. Each entry in the result will have all the variales X Y Z and the entry for a particular valuation of X, Y, Z will e the product of the entry in f 1 that has the same values for X, Y and the entry in f 2 that has the same values for Y, Z.

4 The other operation we need for factors is summing out. If, for example, we sum out A from f AJM (A, B, E), the resulting factor will e named f ĀJM (B, E) (with the ar on the A in the suscript denoting that A was summed out). There is one entry in f ĀJM (B, E) for each comination of values of B, E in f AJM (B, E). The value of the entry is the sum of all entries in f AJM (B, E) that have the same values of B, E (ut differing values of A): B E f ĀJM (B, E) T T.95 * * T F.94 * * F T.29 * * F F.001 * * We are now ready to compute the answer to our query: αp (E) P () a P (a, e)p (j a)p (m a) αf E (E) B f B(B) a f A(A, B, E)f J (A)f m (A) αf E (E) B f B(B) a f A(A, B, E)f JM (A) αf E (E) B f B(B) a f AJM(A, B, E) αf E (E) B f B(B)f ĀJM (B, E) αf E (E) B fbājm(b, E) αf E (E)f BĀJM (E) αf E BĀJM (E) We have already computed f ĀJM, so we can proceed from step 5, where we compute fbājm(b, E): B E f B ĀJM (B, E) T T.5985 *.001 T F.5922 *.001 F T.1830 *.999 F F *.999 T.001 F.999 B E f ĀJM (B, E) T T.5985 T F.5922 F T.1830 F F Next, we sum out B to produce f BĀJM (E). E f BĀJM (E) T F * Now we compute f E BĀJM (E): E f BĀJM (E) T.1834 * F * E f E (E) T.002 F.998 E f BĀJM (E) T.1834 F

5 So we have αf E BĀJM (E) α , ,.8228, or in other words, P (e j, m) Another Example Using the Bayesian Network from Homework 2 Prolem 3, we will compute P (E d): P (E d) αp (E, d) α a,,c P (a,, c, E, d) α a P (a) c P (c a) P ()P (d )P (E, c) α a f A(A) c f C(A, C) f B(B)f D (B)f E (B, C, E) α a f A(A) c f C(A, C) f BDE(B, C, E) α a f A(A) c f C(A, C)f BDE (C, E) α a f A(A) c f C BDE (A, C, E) α a f A(A)f C BDE (A, E) α a f A C BDE (A, E) αf Ā C BDE (E) The only variale in the expression that is fixed is D. So the initial factors are: A f A (A) T.3 F.7 A C f C (A, C) T T.2 T F.8 F T.4 F F.6 T.8 F.2 B f D (B) T.1 F.6 B C E f E (B, C, E) T T T.5 T T F.5 T F T.9 T F F.1 F T T.4 F T F.6 F F T.7 F F F.3 We can compute f BDE (B, C, E) from the rightmost three factors in one step: B C E f BDE (B, C, E) T T T.8 *.1 *.5.04 T T F.8 *.1 *.5.04 T F T.8 *.1 * T F F.8 *.1 * F T T.2 *.6 * F T F.2 *.6 * F F T.2 *.6 * F F F.2 *.6 * T.8 F.2 B f D (B) T.1 F.6 B C E f E (B, C, E) T T T.5 T T F.5 T F T.9 T F F.1 F T T.4 F T F.6 F F T.7 F F F.3 Now we sum out B:

6 C E f BDE (C, E) T T T F F T F F Next, we compute f C BDE (A, C, E): A C E f C BDE (A, C, E) T T T.2 * T T F.2 * T F T.8 * T F F.8 * F T T.4 * F T F.4 * F F T.6 * F F F.6 * A C f C (A, C) T T.2 T F.8 F T.4 F F.6 C E f BDE (C, E) T T.088 T F.112 F T.156 F F.044 Now we sum out C: A E f C BDE (A, E) T T T F F T F F Next, we compute f A C BDE (A, E): A E f A C BDE (A, E) T T.3 * T F.3 * F T.7 * F F.7 * A f A (A) T.3 F.7 A E f C BDE (A, E) T T.1424 T F.0576 F T.1288 F F.0712 Then we sum out A E f Ā C BDE (A, E) T F Finally, αf Ā C BDE (E) α.13288, , So P ( e d) In these small-network examples, it may seem like the variale elimination algorithm actually requires more work than enumeration. But in some large networks, the computational savings can e large. The complexity of variale elimination depends on the size of the largest factor that must

7 e created during the computation. This in turn depends on the order in which the variales are eliminated. In polytree networks with a consistent ordering of the variales (i.e., a parent comes efore any of its children), the time and space complexity is linear in the size of the network.