Reasoning Under Uncertainty: Variable Elimination Example
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1 Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29 March 26, 2007 Textbook 9.5 Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 1
2 Lecture Overview 1 Recap 2 More Variable Elimination 3 Variable Elimination Example Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 2
3 Summing out variables Our second operation: we can sum out a variable, say X 1 with domain {v 1,..., v k }, from factor f(x 1,..., X j ), resulting in a factor on X 2,..., X j defined by: ( X 1 f)(x 2,..., X j ) = f(x 1 = v 1,..., X j ) + + f(x 1 = v k,..., X j ) Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 3
4 Multiplying factors Our third operation: factors can be multiplied together. The product of factor f 1 (X, Y ) and f 2 (Y, Z), where Y are the variables in common, is the factor (f 1 f 2 )(X, Y, Z) defined by: (f 1 f 2 )(X, Y, Z) = f 1 (X, Y )f 2 (Y, Z). Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 4
5 Probability of a conjunction Suppose the variables of the belief network are X 1,..., X n. {X 1,..., X n } = {Q} {Y 1,..., Y j } {Z 1,..., k } Q is the query variable {Y 1,..., Y j } is the set of observed variables {Z 1,..., k } is the rest of the variables, that are neither queried nor observed What we want to compute: P (Q, Y 1 = v 1,..., Y j = v j ) We can compute P (Q, Y 1 = v 1,..., Y j = v j ) by summing out each variable Z 1,..., Z k We sum out these variables one at a time the order in which we do this is called our elimination ordering. P (Q, Y 1 = v 1,..., Y j = v j ) = Z k Z 1 P (X 1,..., X n ) Y1 = v 1,...,Y j = v j. Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 5
6 Probability of a conjunction What we know: the factors P (X i px i ). Using the chain rule and the definition of a belief network, we can write P (X 1,..., X n ) as n i=1 P (X i px i ). Thus: P (Q, Y 1 = v 1,..., Y j = v j ) = Z k P (X 1,..., X n ) Y1 = v 1,...,Y j = v j. Z 1 = n P (X i px i ) Y1 = v 1,...,Y j = v j. Z k Z 1 i=1 Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 6
7 Computing sums of products Computation in belief networks thus reduces to computing the sums of products. It takes 14 multiplications or additions to evaluate the expression ab + ac + ad + aeh + afh + agh. How can this expression be evaluated more efficiently? factor out the a and then the h giving a(b + c + d + h(e + f + g)) this takes only 7 multiplications or additions ( How can we compute n Z 1 i=1 P (X i px i ) efficiently? Factor out those terms that don t involve Z 1 : )( ) P (X i px i ) P (X i px i ) i Z 1 {X i} px i (terms that do not involve Z i ) Z 1 i Z 1 {X i} px i (terms that involve Z i ) Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 7
8 Lecture Overview 1 Recap 2 More Variable Elimination 3 Variable Elimination Example Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 8
9 Summing out a variable efficiently To sum out a variable Z j from a product f 1,..., f k of factors: Partition the factors into those that don t contain Z j, say f 1,..., f i, those that contain Z j, say f i+1,..., f k We know: f 1 f k = (f 1 f i ) f i+1 f k. Z j Zj ( ) f Zj i+1 f k is a new factor; let s call it f. Now we have: Z j f 1 f k = f 1 f i f. Store f explicitly, and discard f i+1,..., f k. Now we ve summed out Z j. Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 9
10 Variable elimination algorithm To compute P (Q Y 1 = v 1... Y j = v j ): Construct a factor for each conditional probability. Set the observed variables to their observed values. For each of the other variables Z i {Z 1,..., Z k }, sum out Z i Multiply the remaining factors. Normalize by dividing the resulting factor f(q) by Q f(q). Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 10
11 Lecture Overview 1 Recap 2 More Variable Elimination 3 Variable Elimination Example Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 11
12 Variable elimination example A Compute P (G H = h 1 ). Elimination order: A, C, E, H, I, B, D, F P P (G, H) = A,B,C,D,E,F,I P (A, B, C, D, E, F, G, H, I) P P (G, H) = A,B,C,D,E,F,I P (A) P (B A) P (C) P (D B, C) P (E C) P (F D) P (G F, E) P (H G) P (I G) B D F C E G H I Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 12
13 Variable elimination example Compute P (G H = h 1 ). Elimination order: A, C, E, H, I, B, D, F P P (G, H) = A,B,C,D,E,F,I P (A) P (B A) P (C) P (D B, C) P (E C) P (F D) P (G F, E) P (H G) P (I G) P Eliminate A: P (G, H) = B,C,D,E,F,I f1(b) P (C) P (D B, C) P (E C) P (F D) P (G F, E) P (H G) P (I G) A P f 1(B) := a dom(a) P (A = a) P (B A = a) B D F C E G H I Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 12
14 Variable elimination example Compute P (G H P = h 1 ). Elimination order: A, C, E, H, I, B, D, F P (G, H) = B,C,D,E,F,I f1(b) P (C) P (D B, C) P (E C) P (F D) P (G F, E) P (H G) P (I G) Eliminate C: P (G, H) = P B,D,E,F,I f1(b) f2(b, D, E) P (F D) P (G F, E) P (H G) P (I G) A B D F C E P f 1(B) := a dom(a) P P (A = a) P (B A = a) f 2(B, D, E) := c dom(c) P (C = c) P (D B, C = c) P (E C = c) G H I Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 12
15 Variable elimination example Compute P (G H = h 1 ). Elimination order: A, C, E, H, I, B, D, F P(G, H) = f1(b) f2(b, D, E) P (F D) P (G F, E) P (H G) P (I G) B,D,E,F,I Eliminate E: P P (G, H) = B,D,F,I f1(b) f3(b, D, F, G) P (F D) P (H G) P (I G) A B D F C E P f 1(B) := a dom(a) P P (A = a) P (B A = a) f 2(B, D, E) := c dom(c) P P (C = c) P (D B, C = c) P (E C = c) f 3(B, D, F, G) := e dom(e) f2(b, D, E = e) P (G F, E = e) G H I Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 12
16 Variable elimination example Compute P (G H = h 1 ). Elimination order: A, C, E, H, I, B, D, F P P (G, H) = B,D,F,I f1(b) f3(b, D, F, G) P (F D) P (H G) P (I G) Observe H = h 1: P P (G, H = h 1) = B,D,F,I f1(b) f3(b, D, F, G) P (F D) f4(g) P (I G) A B D F C E P f 1(B) := a dom(a) P P (A = a) P (B A = a) f 2(B, D, E) := c dom(c) P P (C = c) P (D B, C = c) P (E C = c) f 3(B, D, F, G) := e dom(e) f2(b, D, E = e) P (G F, E = e) f 4(G) := P (H = h 1 G) G H I Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 12
17 Variable elimination example Compute P (G H = h 1 ). Elimination order: A, C, E, H, I, B, D, F P P (G, H = h 1) = B,D,F,I f1(b) f3(b, D, F, G) P (F D) f4(g) P (I G) Eliminate I: P P (G, H = h 1) = B,D,F f1(b) f3(b, D, F, G) P (F D) f4(g) f5(g) A B D F C E P f 1(B) := a dom(a) P P (A = a) P (B A = a) f 2(B, D, E) := c dom(c) P P (C = c) P (D B, C = c) P (E C = c) f 3(B, D, F, G) := e dom(e) f2(b, D, E = e) P (G F, E = e) f 4(G) := P (H = h 1 G) P f 5(G) := i dom(i) P (I = i G) G H I Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 12
18 Variable elimination example Compute P (G H = P h 1 ). Elimination order: A, C, E, H, I, B, D, F P (G, H = h 1) = B,D,F f1(b) f3(b, D, F, G) P (F D) f4(g) f5(g) Eliminate B: P P (G, H = h 1) = D,F f6(d, F, G) P (F D) f4(g) f5(g) A B D F C E P f 1(B) := a dom(a) P P (A = a) P (B A = a) f 2(B, D, E) := c dom(c) P P (C = c) P (D B, C = c) P (E C = c) f 3(B, D, F, G) := e dom(e) f2(b, D, E = e) P (G F, E = e) f 4(G) := P (H = h 1 G) P f 5(G) := i dom(i) P P (I = i G) f 6(D, F, G) := b dom(b) f1(b = b) f3(b = b, D, F, G) G H I Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 12
19 Variable elimination example Compute P (G H = P h 1 ). Elimination order: A, C, E, H, I, B, D, F P (G, H = h 1) = D,F f6(d, F, P G) P (F D) f4(g) f5(g) Eliminate D: P (G, H = h 1) = F f7(f, G) f4(g) f5(g) A B D F H G C E I P f 1(B) := a dom(a) P P (A = a) P (B A = a) f 2(B, D, E) := c dom(c) P P (C = c) P (D B, C = c) P (E C = c) f 3(B, D, F, G) := e dom(e) f2(b, D, E = e) P (G F, E = e) f 4(G) := P (H = h 1 G) P f 5(G) := i dom(i) P P (I = i G) f 6(D, F, G) := P b dom(b) f1(b = b) f3(b = b, D, F, G) f 7(F, G) := d dom(d) f6(d = d, F, G) P (F D = d) Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 12
20 Variable elimination example Compute P (G H = P h 1 ). Elimination order: A, C, E, H, I, B, D, F P (G, H = h 1) = F f7(f, G) f4(g) f5(g) Eliminate F : P (G, H = h 1) = f 8(G) f 4(G) f 5(G) A B D F H G C E I P f 1(B) := a dom(a) P P (A = a) P (B A = a) f 2(B, D, E) := c dom(c) P P (C = c) P (D B, C = c) P (E C = c) f 3(B, D, F, G) := e dom(e) f2(b, D, E = e) P (G F, E = e) f 4(G) := P (H = h 1 G) P f 5(G) := i dom(i) P P (I = i G) f 6(D, F, G) := P b dom(b) f1(b = b) f3(b = b, D, F, G) f 7(F, G) := d dom(d) f6(d = d, F, G) P (F D = d) P f 8(G) := f dom(f ) f7(f = f, G) Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 12
21 Variable elimination example Compute P (G H = h 1 ). Elimination order: A, C, E, H, I, B, D, F P (G, H = h 1) = f 8(G) f 4(G) f 5(G) Normalize: P (G H = h 1) = P (G,H = h 1 ) Pg dom(g) P (G,H = h 1) A B D F H G C E I P f 1(B) := a dom(a) P P (A = a) P (B A = a) f 2(B, D, E) := c dom(c) P P (C = c) P (D B, C = c) P (E C = c) f 3(B, D, F, G) := e dom(e) f2(b, D, E = e) P (G F, E = e) f 4(G) := P (H = h 1 G) P f 5(G) := i dom(i) P P (I = i G) f 6(D, F, G) := P b dom(b) f1(b = b) f3(b = b, D, F, G) f 7(F, G) := d dom(d) f6(d = d, F, G) P (F D = d) P f 8(G) := f dom(f ) f7(f = f, G) Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 12
22 What good was Conditional Independence? That s great... but it looks incredibly painful for large graphs. And... why did we bother learning conditional independence? Does it help us at all? Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 13
23 What good was Conditional Independence? That s great... but it looks incredibly painful for large graphs. And... why did we bother learning conditional independence? Does it help us at all? yes we use the chain rule decomposition right at the beginning Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 13
24 What good was Conditional Independence? That s great... but it looks incredibly painful for large graphs. And... why did we bother learning conditional independence? Does it help us at all? yes we use the chain rule decomposition right at the beginning Can we use our knowledge of conditional independence to make this calculation even simpler? Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 13
25 What good was Conditional Independence? That s great... but it looks incredibly painful for large graphs. And... why did we bother learning conditional independence? Does it help us at all? yes we use the chain rule decomposition right at the beginning Can we use our knowledge of conditional independence to make this calculation even simpler? yes there are some variables that we don t have to sum out intuitively, they re the ones that are pre-summed-out in our tables example: summing out I on the previous slide Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 13
26 One Last Trick One last trick to simplify calculations: we can repeatedly eliminate all leaf nodes that are neither observed nor queried, until we reach a fixed point. Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 14
27 One Last Trick One last trick to simplify calculations: we can repeatedly eliminate all leaf nodes that are neither observed nor queried, until we reach a fixed point. alarm fire smoke Can we justify that on a threenode graph Fire, Alarm, and Smoke when we ask for: P (F ire)? Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 14
28 One Last Trick One last trick to simplify calculations: we can repeatedly eliminate all leaf nodes that are neither observed nor queried, until we reach a fixed point. alarm fire smoke Can we justify that on a threenode graph Fire, Alarm, and Smoke when we ask for: P (F ire)? P (F ire Alarm)? Reasoning Under Uncertainty: Variable Elimination Example CPSC 322 Lecture 29, Slide 14
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