Reasoning Under Uncertainty: Variable Elimination

Transcription

1 Reasoning Under Uncertainty: Variable Elimination CPSC 322 Uncertainty 7 Textbook 10.5 Reasoning Under Uncertainty: Variable Elimination CPSC 322 Uncertainty 7, Slide 1

2 Lecture Overview 1 Recap 2 Variable Elimination 3 Variable Elimination Example Reasoning Under Uncertainty: Variable Elimination CPSC 322 Uncertainty 7, Slide 2

3 Factors and Assigning Variables A factor is a representation of a function from a tuple of random variables into a number. A factor denotes a distribution over the given tuple of variables in some (unspecified) context We can make new factors out of an existing factor Our first operation: assign some or all of the variables of a factor. f(x 1 = v 1, X 2,..., X j ), where v 1 dom(x 1 ), is a factor on X 2,..., X j. f(x 1 = v 1, X 2 = v 2,..., X j = v j ) is a number that is the value of f when each X i has value v i. The former is also written as f(x 1, X 2,..., X j ) X1 = v 1,...,X j = v j Reasoning Under Uncertainty: Variable Elimination CPSC 322 Uncertainty 7, Slide 3

4 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 CPSC 322 Uncertainty 7, Slide 4

5 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 CPSC 322 Uncertainty 7, Slide 5

6 Lecture Overview 1 Recap 2 Variable Elimination 3 Variable Elimination Example Reasoning Under Uncertainty: Variable Elimination CPSC 322 Uncertainty 7, Slide 6

7 Probability of a conjunction Suppose the variables of the belief network are X 1,..., X n. What we want to compute: the factor P (X q, X o1 = v 1,..., X oj = v j ) We can compute P (X q, X o1 = v 1,..., X oj = v j ) by summing out the variables X s1,..., X sk = {X 1,..., X n } \ {X q, X o1,..., X oj }. We sum out these variables one at a time the order in which we do this is called our elimination ordering. P (X q, X o1 = v 1,..., X oj = v j ) = P (X 1,..., X n ) Xo1 = v 1,...,X oj = v j. X sk X s1 Reasoning Under Uncertainty: Variable Elimination CPSC 322 Uncertainty 7, Slide 7

8 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 (X q, X o1 = v 1,..., X oj = v j ) = P (X 1,..., X n ) Xo1 = v 1,...,X oj = v j. X sk X s1 = n P (X i px i ) Xo1 = v 1,...,X oj = v j. X sk X s1 i=1 Reasoning Under Uncertainty: Variable Elimination CPSC 322 Uncertainty 7, Slide 8

9 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? Reasoning Under Uncertainty: Variable Elimination CPSC 322 Uncertainty 7, Slide 9

10 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 Reasoning Under Uncertainty: Variable Elimination CPSC 322 Uncertainty 7, Slide 9

11 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 X s1 n i=1 P (X i px i ) efficiently? Reasoning Under Uncertainty: Variable Elimination CPSC 322 Uncertainty 7, Slide 9

12 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 X s1 i=1 P (X i px i ) efficiently? Factor out those terms that don t involve X s1 : )( ) P (X i px i ) P (X i px i ) i X s1 {X i} px i (terms that do not involve X si ) X s1 i X s1 {X i} px i (terms that involve X si ) Reasoning Under Uncertainty: Variable Elimination CPSC 322 Uncertainty 7, Slide 9

13 Summing out a variable efficiently To sum out a variable X sj from a product f 1,..., f k of factors: Partition the factors into those that don t contain X sj, say f 1,..., f i, those that contain X sj, say f i+1,..., f k We know: f 1 f k = (f 1 f i ) f i+1 f k. X sj Xsj ( ) f Xsj i+1 f k is a new factor; let s call it f. Now we have X sj f 1 f k = f 1 f i f. Store f explicitly, and discard f i+1,..., f k. Now we ve summed out X sj. Reasoning Under Uncertainty: Variable Elimination CPSC 322 Uncertainty 7, Slide 10

14 Variable elimination algorithm To compute P (X q X o1 = v 1... X oj = v j ): Construct a factor for each conditional probability. Set the observed variables to their observed values. For each of the other variables X si {X s1,..., X sk }, sum out X si Multiply the remaining factors. Normalize by dividing the resulting factor f(x q ) by X q f(x q ). Reasoning Under Uncertainty: Variable Elimination CPSC 322 Uncertainty 7, Slide 11

15 Lecture Overview 1 Recap 2 Variable Elimination 3 Variable Elimination Example Reasoning Under Uncertainty: Variable Elimination CPSC 322 Uncertainty 7, Slide 12

16 Variable elimination example A Compute P (G H = h 1 ). Elimination order: A, C, E, H, I, B, D, F P (G, H) = A,B,C,D,E,F,I P (A, B, C, D, E, F, G, H, I) 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 CPSC 322 Uncertainty 7, Slide 13

17 Variable elimination example Compute P (G H = h 1 ). Elimination order: A, C, E, H, I, B, D, F 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) 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 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 CPSC 322 Uncertainty 7, Slide 13

18 Variable elimination example Compute P (G H = 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) = 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 f 1(B) := a dom(a) 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 CPSC 322 Uncertainty 7, Slide 13

19 Variable elimination example Compute P (G H = h 1 ). Elimination order: A, C, E, H, I, B, D, F P (G, H) = B,D,E,F,I f1(b) f2(b, D, E) P (F D) P (G F, E) P (H G) P (I G) Eliminate E: 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 f 1(B) := a dom(a) 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) 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 CPSC 322 Uncertainty 7, Slide 13

20 Variable elimination example Compute P (G H = h 1 ). Elimination order: A, C, E, H, I, B, D, F 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 (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 f 1(B) := a dom(a) 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) 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 CPSC 322 Uncertainty 7, Slide 13

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) = B,D,F,I f1(b) f3(b, D, F, G) P (F D) f4(g) P (I G) Eliminate I: 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 f 1(B) := a dom(a) 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) 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) f 5(G) := i dom(i) P (I = i G) G H I Reasoning Under Uncertainty: Variable Elimination CPSC 322 Uncertainty 7, Slide 13

22 Variable elimination example Compute P (G H = 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 (G, H = h 1) = D,F f6(d, F, G) P (F D) f4(g) f5(g) A B D F C E f 1(B) := a dom(a) 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) 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) f 5(G) := i dom(i) 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 CPSC 322 Uncertainty 7, Slide 13

23 Variable elimination example Compute P (G H = h 1 ). Elimination order: A, C, E, H, I, B, D, F P (G, H = h 1) = D,F f6(d, F, 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 f 1(B) := a dom(a) 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) 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) f 5(G) := i dom(i) P (I = i G) f 6(D, F, G) := 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 CPSC 322 Uncertainty 7, Slide 13

24 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 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 f 1(B) := a dom(a) 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) 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) f 5(G) := i dom(i) P (I = i G) f 6(D, F, G) := 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) f 8(G) := f dom(f ) f7(f = f, G) Reasoning Under Uncertainty: Variable Elimination CPSC 322 Uncertainty 7, Slide 13

25 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 ) g dom(g) P (G,H = h 1) A B D F H G C E I f 1(B) := a dom(a) 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) 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) f 5(G) := i dom(i) P (I = i G) f 6(D, F, G) := 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) f 8(G) := f dom(f ) f7(f = f, G) Reasoning Under Uncertainty: Variable Elimination CPSC 322 Uncertainty 7, Slide 13

26 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 CPSC 322 Uncertainty 7, Slide 14

27 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 CPSC 322 Uncertainty 7, Slide 14

28 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 CPSC 322 Uncertainty 7, Slide 14

29 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 CPSC 322 Uncertainty 7, Slide 14

30 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 CPSC 322 Uncertainty 7, Slide 15

31 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 CPSC 322 Uncertainty 7, Slide 15

32 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 CPSC 322 Uncertainty 7, Slide 15