Ch9: Exact Inference: Variable Elimination. Shimi Salant, Barak Sternberg

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1 Ch9: Exact Inference: Variable Elimination Shimi Salant Barak Sternberg

2 Part 1

3 Reminder introduction (1/3) We saw two ways to represent (finite discrete) distributions via graphical data structures: Bayesian Network is a DAG =Π factorizes over for each = Markov Network means: : an edge is an undirected graph = Φ Φ factorizes over means: each is a clique in Φ Φ Φ D I G S L = ϕ ϕ ϕ

4 Reminder introduction (2/3) We reviewed a concept of separation in graphs and saw we can query the graph itself in order to get independence assertions all of which apply in the represented distribution. Bayesian Network ={ B d-sepg Markov Network ; } ={ Φ Φ seph ; } * we also mentioned that these are solid representations in the sense that are "infinitely many more" 's for which = than those for which.

5 Reminder introduction (3/3) We will now use these graphical data structure in order to: answer probabilistic queries such as = =? show that properties of the graphs determine upper bounds for computational cost of answering query i.e. these properties provide a way to gauge/reduce costs. show algorithms that take these properties into consideration.

6 Definition of the inference task (1/2) Some general context: types of tasks we may wish to carry out with PGMs: inference learning structure learning given graph and factors/cpds find probabilities e.g. =? given graph and data find factors/cpds (namely: their parameters) given variables and data find graph structure and factors/cpds * extra characteristic of learning tasks: data might be partially observed i.e. we're not given values for all variables.

7 Definition of the inference task (2/2) The exact inference task is defined as: Given a fully parameterized BN or MN over variables = and : Compute = are the query variables are the evidence/observed variables is the evidence itself. can be empty in which case we're after. * note that we're after a distribution size of answer is which - to begin with - is exponential in.

8 Hardness of inference task (1/12) We first consider only BNs: = =? For any specific valuation we can compute the joint.5.5 = = = :

9 Hardness of inference task.5.5 = =

10 Hardness of inference task.5.5 = =

11 Hardness of inference task.5.5 = =

12 Hardness of inference task.5.5 = =

13 Hardness of inference task.5.5 = = =. 8

14 Hardness of inference task (2/12) Task: compute =. Naive solution: sum the joint. Denote Σ =. For each Then compute Σ Then for each / = This would entail = = compute:.. compute. = work for each.

15 Hardness of inference task (3/12) Task: compute =. Any solution: we show it's NP-hard. Reminder: The 3-SAT decision problem: Given a formula over binary variables = where every = ± ± ± is a disjunction of 3 literals [example: = - decide whether there is a satisfying assignment to. 3-SAT is in NP: i.e. there is an polynomial-time verifier algorithm that for a given problem instance and an assignment checks if proves that that is satisfiable. 3-SAT is NP-hard: i.e. every other problem in NP can be reduced to 3-SAT. ]

16 Hardness of inference task (4/12) Consider the following tasks each a sub-case of the next: (1) (2) (3) (4) = =? = = =? = =? = >? ( can be empty) ( can be empty is a single variable) (a decision problem) If (4) is NP-hard the original inference task (1) certainly is. The BN-Positive problem: Given a BN a variable and value decide whether = >.

17 Hardness of inference task (5/12) The BN-Positive problem: Given a BN a variable and value decide whether = >. BN-Positive is NP-hard: We next show a reduction from 3-SAT to BN-Positive. (also: BN-Positive is in NP: Given (a candidate proof) values to all variables where = in we can compute = ξ and check for positivity. i.e. BN-Positive is NP-Complete.)

18 Hardness of inference task (6/12) Given a 3-SAT formula = we construct a BN: (in polynomial time) for each formula variable = =. over binary variables { : a binary variable with for each formula clause : a deterministic binary variable with ( = ) = iff is satisfied by values of a layer of deterministic binary AND variables { = = iff all are valued. } } s.t.

19 Hardness of inference task (7/12) is not satisfiable no assignment to { } can yield is satisfiable an assignment to { } can yield = = = hence = hence = = > we can solve 3-SAT. i.e. if we can answer whether = BN-Positive problem (4) is NP-hard. = =? our initial inference problem (1) is NP-hard for BNs. =. >.

20 Hardness of inference task (8/12) For MNs: we can easily translate a BN to an MN (hence MNs are NP-hard too): Given a BN where =Π we can construct an MN Φ modeling the same distribution by: using CPDs as factors: ( )= and Φ =. translate directed to undirected (also need to add edges between all parents of a node cost of addition is polynomial). Φ = D I G S L = Φ ϕ

21 Hardness of inference task (9/12) * extra note: BN-Positive is a decision problem i.e. of the for are there a y solutio s (in BN above: = >? i.e. is there an event in the event space - all possible valuations to all variables - in which = and event has a positive probability). The general i fere e task is of the for how a y solutio s are there (in BN above: knowing how many such = > events there are we would have divided this amount by size of event space to obtain the probability = ). It turns out the inference task belongs to an even harder class of problems #P-hard whi h o sists of su h ou ti g pro le s.

22 Hardness of inference task (10/12) Approximate inference: obtain an approximation for. We consider measuring accuracy of = rather than that of def: def: has absolute error if. might be too coarse of a criterion e.g. = has relative error if + + is between. e.g. true answer for. and.. 8.

23 Hardness of inference task (11/12) def: def: has absolute error if. might be too coarse of a criterion e.g. = has relative error if + + is between. e.g. true answer If we can get such a relative error approximation > > and we can solve BN-Positive which is NP-hard relative error inference is NP-hard. for. and :. 8.

24 Hardness of inference task (12/12) def: def: has absolute error if. might be too coarse of a criterion e.g. = has relative error if + + is between. e.g. true answer for. and For absolute error inference the situation is more specific: For the case where no evidence is present i.e. there exists a randomized polynomial-time algorithm. For the case where evidence is present i.e. and. - absolute error inference is NP-hard.. Namely - any useful absolute error inference is NP-hard when evidence is present. 8.

25 End of part 1

26

27 VE algorithm is + is number of initial where is number of variables factors in factor set Φ and is the size (number of entries) of the largest intermediate factor created throughout run.

28

29 def 9.5: the induced graph for factor set Φ (over ) and an ordering (over eliminated variables ) we define the induced graph Φ as an undirected graph over with edges where and appear in the scope of some intermediate factor when Sum-Product-VE(Φ ) is executed. theorem 9.6: For Φ and their induced graph Φ : The scope of every is a clique in Φ. Every maximal clique in Φ is the scope of a certain. 1st statement means: size of largest clique in Φ bounds 2st statement means: bound is tight it is encountered..

30

31 Part 3

32 Finding elimination orderings (1/12) We reached the problem of finding an optimal ordering for VE. Considering a network Φ we saw each ordering yields an induced graph Φ whose largest clique size is an exponent in run-time bound. We now consider the ordering problem in purely graphical terms. We can observe the induced graph K of the network's graph (instead of its factor set Φ) since the induced graph depends only on how variables appear together in factors i.e. depends only on structure of network's graph.

33 Finding elimination orderings Compute = : Sum-Product-VE(Φ = { } = Φ= { )) }

34 Finding elimination orderings Compute : Sum-Product-VE(Φ = { } = Φ= { Eliminate )) = (factor size: 2) = Σ = Σ an added edge to induced-graph (was already present in original graph) }

35 Finding elimination orderings Compute : Sum-Product-VE(Φ = { } = Φ= { Eliminate )) } = (factor size: 3) =Σ=Σ an added fill-edge to induced-graph (wasn't present in original graph)

36 Finding elimination orderings Compute : Sum-Product-VE(Φ = { } = Φ= { Eliminate )) } = (factor size: 2) =Σ =Σ

37 Finding elimination orderings Compute : Sum-Product-VE(Φ = { } = Φ= { Eliminate )) } = (factor size: 2) = Σ = ΣS

38 Finding elimination orderings Compute : Sum-Product-VE(Φ = { } = Φ= { } = )) =

39 Finding elimination orderings (2/12) def: def: the induced-width of is the size of largest clique in (the -1 is for having zero width for an which is a tree) the tree-width of is its minimal induced-width = min (i.e. size of the smallest clique we can hope for in an induced graph of minus 1) The ordering problem is now: find = rgmin - 1.

40 Finding elimination orderings (3/12) (1) = rgmin =? (2) = min =? (3) ℕ? (find optimal ordering) (find min. induced width i.e. tree-width) (decide whether tree-width ) A theorem from graph theory: (3) is NP-complete. (2) is NP-hard (1) is NP-hard and we have now shown that finding optimal ordering is NP-hard. We will now translate the ordering problem to a different graphical problem thereby yielding an ordering algorithms for a certain type of graphs.

41 Finding elimination orderings (4/12) def: A graph (directed or undirected) is chordal if every loop of length greater than 3 has a chord. Such a graph is tria gulated all polygons are divided to triangles.

42 Finding elimination orderings (5/12) We next show that the class of induced graphs is the class of chordal graphs. * remark: this is also the class of graphs for which there are prefect I-maps i.e. for a graph in this class = for a distribution that factorize over graph.

43 Finding elimination orderings (6/12) is induced is chordal: Assume is not chordal: it has a loop Assume WLOG was eliminated first. After line 3 of Sum-Product-Eliminate-Var(Φ no more edges are added to. ) has the edges : when line 3 runs: appears in some factor with appears in some factor with will appear in of line 3 has the edge Since.

44 Finding elimination orderings (7/12) is chordal is induced: def: For an undirected graph over a tree is a clique tree for if: every node in the tree is a clique in the graph. every maximal clique in the graph is a node in the tree. for each an edge in : separates nodes in : those appearing on 's side of the tree from those appearing on 's side.

45 Finding elimination orderings (8/12) def: For an undirected graph over a tree is a clique tree for if: every node in the tree is a clique in the graph. every maximal clique in the graph is a node in the tree. for each an edge in : separates nodes in : those appearing on 's side of the tree from those appearing on 's side.

46 Finding elimination orderings (9/12) Not every graph has a clique tree but: Every chordal graph has a clique tree. A property of clique trees: Every leaf in the tree has a node present only there.

47 Finding elimination orderings (10/12) is chordal is induced: We show that if is chordal then = for some ordering. Claim: all variables of a chordal graph over variables can be eliminated without introducing fill-edges. Induction on : is chordal it has a clique tree. is a clique tree it has a leaf node present only in. and in that leaf node a variable can be eliminated from without introducing fill-edges: present only in all of 's neighbors are in. is a clique in all neighbors already connected to each other. Removing from we are left with a chordal graph of variables.

48 Finding elimination orderings Example: graph = 's clique tree induced graph

49 Finding elimination orderings graph = 's clique tree induced graph

50 Finding elimination orderings graph = 's clique tree induced graph

51 Finding elimination orderings graph = 's clique tree induced graph

52 Finding elimination orderings graph = 's clique tree induced graph

53 Finding elimination orderings graph = 's clique tree induced graph

54 Finding elimination orderings graph = 's clique tree induced graph

55 Finding elimination orderings graph = 's clique tree induced graph

56 Finding elimination orderings (11/12) For a chordal graph we have an algorithm that does not produce fill-edges: Max-Cardinality produces an ordering consistent with always choosing a node which is only present in a leaf of the (current) graph's clique-tree and only in that leaf. Meaning: Max-Cardinality solves the ordering problem for chordal graphs.

57 Finding elimination orderings (12/12) The general ordering problem therefore reduces to a well-known problem from graph-theory: Minimal-Triangulation: Given a graph find a chordal graph containing such that 's largest clique is as small as possible. There are different graph-theoretic algorithms addressing Minimal-Triangulation offering different levels of performance guarantees.

58 Greedy search for elimination ordering (1/3) We saw that the inference problem = =? is NP-hard. Moreover even finding an optimal ordering for variable elimination for running Sum-Product-VE(Φ ) is NP-hard. A practical approach is therefore to search the ordering space for a sub-optimal yet as-good-as-possible ordering. We can search this space greedily each time choosing a variable whose elimination will result in least cost (for some definition of cost) given current state of induced graph. We do not have formal complexity guarantees for this approach yet it works surprisingly well in practice.

59 Greedy search for elimination ordering (2/3) * note that is the induced graph. Possible evaluation metrics: min-neigh min-neigh-weight min-fill min-fill-weight = number of neighbors of = product of cardinalities of neighbors of = number of added fill-edges if is eliminated = sum of weights of fill-edges where the weight of a fill-edge is the product of cardinalities of edge's nodes

60 Greedy search for elimination ordering (3/3) Neither of these is universally better than others. A variant of the greedy approach is to introduce stochasticity into it by not always deterministically selecting the that minimizes. For example: upon each elimination round select out of a random half of remaining variables. This serves to introduce exploration into the ordering search (rather than have it be fully exploitative). Note that the greedy algorithm runs in polynomial time and it can compute the number of operations the VE algorithm itself will execute. A suggested practice for large networks where such pre-computation time is negligible is to execute the greedy algorithm multiple times and use the best ordering obtained.

61

62 *extra material

63 Conditioning (1/6) Consider MNs: Φ = Φ Φ = Φ Φ = Φ Sum-Product-VE calculated Sum-Product-VE(Φ ) ( = ) returned such that: =Σ =Σ =Σ = Π Φ Φ = Φ :... and if Φ is needed - renormalize.

64 Conditioning (2/6) We used the ability to perform a sum-product calculation for getting conditional probabilities: Cond-Prob-VE calculated Φ and Φ : It returned s.t. (now = ) and a scalar:... and if Φ =Σ =Σ =Σ =Σ =Σ is needed: Φ Π Φ[=] [ = ] [ = ] Φ = Φ Φ = Φ / Φ = Φ

65 Conditioning (3/6) Calculating conditional probabilities = can reduce sizes of encountered intermediate factors - since evidence variables can render other variables independent.

66 Conditioning (4/6) Example: same network as before with evidence = (using Cond-Prob-VE) = Compute = for Φ = { } = )) Φ= { } Φ Φ[I = i ] = { } = { [ ] [ ] }

67 Conditioning Compute = for Φ = { } = Φ= { Eliminate )) [ ] [ ] = (factor size: 2) = Σ = Σ }

68 Conditioning Compute = for Φ = { } = Φ= { )) [ ] Eliminate [ ] } = [ ] (factor size: 1) [ ] = Σ =Σ [ ] [ ] ℝ a scalar dependent on.

69 Conditioning Compute = for Φ = { } = )) Φ= { [ ] ℝ [ ] Eliminate = [ ] [ ] = Σ = Σ [ ] } (factor size: 2)

70 Conditioning Compute = for Φ = { } = )) Φ= { [ ] ℝ [ ] } = [ ] [ ] =

71 Conditioning (5/6) So computing Φ can introduce smaller intermediate factors compared to computation of Φ. We can do the following (where we build upon Cond-Prob-VE - our ability to calculate Φ and Φ ): For some for every Φ = Φ then compute Φ = Φ Σ and have: / Φ= Φ Φ Φ Computing Φ Φ compute and Φ = and Σ as such is called conditioning. Φ Φ = Φ

72 Conditioning (6/6) Conditioning does no less work than ordinary VE we've seen so far it generally does more by running VE many times but the maximal intermediate factor ever encountered over VE runs can be smaller -- which is a necessity if we cannot hold very large intermediate factors in memory.

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