Model Counting of Query Expressions: Limitations of Propositional Methods Paul Beame 1 Jerry Li 2...

Model Counting of Query Expressions:Limitations of Propositional Methods

Paul Beame1 Jerry Li2 Sudeepa Roy1 Dan Suciu1

1University of Washington2 MIT

11

Probabilistic Databases AsthmaPatien

t

Ann

Bob

Friend

Ann Joe

Ann Tom

Bob Tom

Smoker

Joe

Tom

Boolean query Q: x y AsthmaPatient(x) Friend (x, y) Smoker(y)

• Tuples are probabilistic (and independent)▫ “Ann” is present with probability 0.3

• Lineage FQ,D = (x1y1z1) (x1y2z2) (x2y3z2)▫ Q is true on D FQ,D is true

• What is the probability that Q is true on D?• Two main evaluation techniques: lifted vs. grounded

inference

x1

x2

z1

z2

y1

y2

y3

0.30.1

0.51.0

0.90.5

0.7

Pr(x1) = 0.3

2

3

Lifted Inference

Q: x y AsthmaPatient(x) Friend (x, y) Smoker(y)

Work with explicit query structure, i.e. the first order logic

Dichotomy Theorem [Dalvi, Suciu 12] For any UCQ, evaluating it is either

▫#P-hard▫Polynomial time computable using lifted inference▫and there is a simple condition to tell which case holds

4

Grounded Inference

FQ,D = (x1y1z1) (x1y2z2) (x2y3z2)

Work with the boolean formula

Folklore sentiment: Lifted inference is strictly stronger than grounded inference

We give the first clear proof of this

5

Outline

•Background: Model Counting, DPLL algorithms▫Extensions (Caching & Component Analysis)▫Knowledge Compilation (FBDDs & Decision-DNNF)

•Our Contributions▫Statement of separation▫Sketch of FBDD lower bound

•Conclusions

6

Model Counting• Probability Computation Problem:

Given F, and independent Pr(x), Pr(y), Pr(z), …, compute Pr(F)

• Model Counting Problem:

Given a Boolean formula F, compute #F = #Models (satisfying assignments) of F

e.g. F = (x y) (x u w) (x u w z) #Assignments on x, y, u, z, w which make F = true

7

•CDP•Relsat•Cachet•SharpSAT•c2d•Dsharp•…

Known Model Counting Algorithms

Search-based/DPLL-based(explore the assignment-space and count the satisfying ones)

Knowledge Compilation-based(compile F into a “computation-friendly” form)

[Survey by Gomes et. al. ’09]

Both techniques explicitly or implicitly • use DPLL-based algorithms • produce FBDD or Decision-DNNF compiled forms (output or trace)

[Huang-Darwiche’05, ’07]

[Birnbaum et. al.’99]

[Bayardo Jr. et. al. ’97, ’00]

[Sang et. al. ’05]

[Thurley ’06]

[Darwiche ’04]

[Muise et. al. ’12]

DPLL Algorithms

Davis, Putnam, Logemann, Loveland [Davis et. al. ’60, ’62]

8

x

z

0

y

1

u 01

1

0

w

1

0

0

1

1 0

u11

1

0

w

1

0

0

1

1 0

1 0 1 0

01

11

F: (xy) (xuw) (xuwz)

uwz

uw

w

uw

½

¾ ¾

y(uw)3/87/8

5/8

w½

Assume uniform distribution for simplicity

// basic DPLL:Function Pr(F):

if F = false then return 0if F = true then return 1select a variable x, return

½ Pr(FX=0) + ½ Pr(FX=1)

DPLL Algorithms

9

x

z

0

y

1

u 01

1

0

w

1

0

0

1

1 0

u11

1

0

w

1

0

0

1

1 0

1 0 1 0

01

11


uwz

uw

w

uw

½

¾ ¾

y(uw)3/87/8

5/8

w½

The trace is a Decision-Tree for F

10

Extensions to DPLL

• Caching Subformulas

• Component Analysis

• Conflict Directed Clause Learning▫ Affects the efficiency of the algorithm, but not the final “form” of the trace

Extensions to DPLL: Caching

11



½ Pr(FX=0) + ½ Pr(FX=1)

x

z

0

y

1

u 01

1

0

w

1

0

0

1

1 0

u11

1

0

w

1

0

0

1

1 0


uwz

uw

w

uw

y(uw)

w

// DPLL with caching:Cache F and Pr(F);look it up before computing

Caching & FBDDs

12

x

z

0

y

1

0

1 0

u11

1

0

w

1

0

0

1

1 0


uwz

uw

w

y(uw)The trace is a decision-DAG for F

FBDD (Free Binary Decision Diagram)or

ROBP (Read Once Branching Program)

• Every variable is tested at most once on any path

Extensions to DPLL: Component Analysis

13

x

z

0

y

1

0

1 0

u11

1

0

w

1

0

0

1

1 0


uwz

uw

w

y (uw)



½ Pr(FX=0) + ½ Pr(FX=1)

// DPLL with component analysis (and caching):

if F = G Hwhere G and H have disjoint sets of variablesPr(F) = Pr(G) × Pr(H)

Components & Decision-DNNF

14

x

z

1u1

1

1

0

w

1

0

0

1

1 0

uwz

w

y (uw)

0

y

1

0


The trace is a Decision-DNNF [Huang-Darwiche ’05, ’07]

FBDD + “Decomposable” AND-nodes

(Two sub-DAGs do not share variables)

y

01AND Node

uw

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How much power does component analysis add?

Theorem [BLRS]: decision-DNNF for F of size N FBDD for F of size Nlog N + 1 [UAI ’13]

Conversion works even when we allow negation and arbitrary decomposable binary gates. [ICDT ’14]

Corollary: Exponential lower bound for FBDD(F) exponential lower bound for decision-DNNF(F)

16

Implications for Lower Bounds?

•All real world exact model counters compile into FBDDs or decision-DNNFs

•By conversion, an exponential size lower bound for FBDDs implies an exponential lower bound for decision-DNNFs

•Thus suffices to consider FBDDs

17

Outline

•Background: Model Counting, DPLL algorithms▫Extensions (Caching & Component Analysis)▫Knowledge Compilation (FBDDs & Decision-DNNF)

•Our Contributions▫Statement of separation▫Sketch of FBDD lower bound

•Conclusions

18

An important class of queries

H1 =R(x)S(x,y) S(x,y)T(y)

Hk =R(x)S1(x,y) ... Si(x,y)Si+1(x,y) ... Sk(x,y)T(y)

▫ [Dalvi, Suciu 12]: Hk is #P-hard to evaluate

▫Known to “capture” hardness for probabilistic DB queries▫But, some functions of the hki are poly-time computable

using lifted inference, e.g. (h30 h32) (h30 h33) (h31 h33)

hk0 hki hkk

19

New Lower Bounds

Theorem: For all k, FBDD(Hk) = 2(𝑛), which implies

Decision-DNNF(Hk) = 2(√n)

Theorem: Any Boolean function f of hk0,...,hkk that depends on all of them requires

FBDD(f) = 2(𝑛)

which implies Decision-DNNF(f) = 2(√n)

Corollary: Grounded inference requires 2(√𝑛) time even on probabilistic DB instances with poly(n) time algorithms using lifted inference.

Implies separation between grounded and lifted inference

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Proof for H1

H1 = R(x)S(x,y) S(x,y)T(y)

Over the complete database of size n,

H1 = ∨n R(i) S(ij) ∨n S(ij) T(j)

Q: why is H1 hard for FBDDs?

21

Matrix view

T(1) T(2) T(3) T(4) T(5)

R(1) S(1,1) S(1,2) S(1,3) S(1,4) S(1,5)

R(2) S(2,1) S(2,2) S(2,3) S(2,4) S(2,5)

R(3) S(3,1) S(3,2) S(3,3) S(3,4) S(3,5)

R(4) S(4,1) S(4,2) S(4,3) S(4,4) S(4,5)

R(5) S(5,1) S(5,2) S(5,3) S(5,4) S(5,5)


22

Matrix view

T(1) T(2) T(3) T(4) T(5)

R(1) S(1,1)

S(1,2)

S(1,3)

S(1,4)

S(1,5)

R(2) S(2,1)

S(2,2)

S(2,3)

S(2,4)

S(2,5)

R(3) S(3,1)

S(3,2)

S(3,3)

S(3,4)

S(3,5)

R(4) S(4,1)

S(4,2)

S(4,3)

S(4,4)

S(4,5)

R(5) S(5,1)

S(5,2)

S(5,3)

S(5,4)

S(5,5)


R(1)

23

Matrix view

T(1) T(2) T(3) T(4) T(5)

R(1) S(1,1)

S(1,2)

S(1,3)

S(1,4)

S(1,5)

R(2) S(2,1)

S(2,2)

S(2,3)

S(2,4)

S(2,5)

R(3) S(3,1)

S(3,2)

S(3,3)

S(3,4)

S(3,5)

R(4) S(4,1)

S(4,2)

S(4,3)

S(4,4)

S(4,5)

R(5) S(5,1)

S(5,2)

S(5,3)

S(5,4)

S(5,5)


R(1)0 1

24

Matrix view

T(1) T(2) T(3) T(4) T(5)

R(1) S(1,1)

S(1,2)

S(1,3)

S(1,4)

S(1,5)

R(2) S(2,1)

S(2,2)

S(2,3)

S(2,4)

S(2,5)

R(3) S(3,1)

S(3,2)

S(3,3)

S(3,4)

S(3,5)

R(4) S(4,1)

S(4,2)

S(4,3)

S(4,4)

S(4,5)

R(5) S(5,1)

S(5,2)

S(5,3)

S(5,4)

S(5,5)


R(1)0 1

25

Matrix view

T(1) T(2) T(3) T(4) T(5)

R(1) S(1,1)

S(1,2)

S(1,3)

S(1,4)

S(1,5)

R(2) S(2,1)

S(2,2)

S(2,3)

S(2,4)

S(2,5)

R(3) S(3,1)

S(3,2)

S(3,3)

S(3,4)

S(3,5)

R(4) S(4,1)

S(4,2)

S(4,3)

S(4,4)

S(4,5)

R(5) S(5,1)

S(5,2)

S(5,3)

S(5,4)

S(5,5)


Can’t Cache!

R(1)

0

1

S(1,1)

S(1,5)

…

0

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A “unit rule” for FBDDs

Variable x in a formula Φ is a unit if Φ = x v G

A FBDD follows the unit rule if each node tests a unit variable whenever possible

Can we assume that FBDDs follow the unit rule?

G

x

1

1 0

Unit Node

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A “unit rule” for FBDDs

Lemma: Given an FBDD for a monotone DNF formula Φ of size N, there exists an FBDD for Φ that follows the unit rule of size at most |var (Φ)| N.

Proof: Alter FBDD to test units whenever possible, then restore read-once property

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Bound for H1

• Idea: specify a set of “admissible” partial paths A so that:1. None of them cache2. Each takes n – 1 degrees of freedom to specify

Given this set A:▫ Each partial path in A must end at a unique node (they

don’t cache)▫ There are 2𝑛-1 such paths

(n – 1 degrees of freedom) Implies A has at least 2𝑛-1 nodes

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Admissible Paths

Let A be the set of partial paths P which1. Don’t end at a leaf node2. Touch n – 1 rows and/or columns, but not more3. Never set R(i) = S(ij) = T(j) = 0, for any i, j

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Bound for H1

Proposition: If P, Q are paths in A which end at the same node v, then they test the same set of R and T variables, and assign them the same value.

Proof:Suppose P sets R(i), Q does not The subformula at v cannot contain any term R(i)S(ij) Q sets every S(ij) = 0 or every T(j) = 1 (unit rule) #Col(Q) = n, contradiction

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Paths don’t cache

Intuition: given that R(i), T(j) are set, S(ij) is determined

Since two paths that end at the same node v set the same R, T variables, they set the same S variables

R(i) S(ij) T(j)

0 1 0

1 0 0

0 0 1

1 0 1

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n – 1 degrees of freedomLet P be an admissible path

w.l.o.g. |Row(P)| = n - 1

At each node where we first visit a row, could’ve chosen either edge and still been admissible!

n – 1 degrees of freedom

2𝑛-1 distinct admissible paths

FBDD(H1) = 2(𝑛)

R(2)

S(1, 4)

S(5, 4)

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Proof for Hk

Same basic structure. We only need to change definition of admissible path.Let A be the set of partial paths P which

1. Don’t end at a leaf node2. Touch n – 1 rows and/or columns, but not more3. Always set i,j consistent with the following table:

R(i) S1(ij)

S2(ij) S3(ij) T(j)

0 1 0 1 0

1 0 1 0 1

0 0 1 0 1

1 0 1 0 1

34

New Lower Bounds

Theorem: For all k, FBDD(Hk) = 2(𝑛), which implies

Decision-DNNF(Hk) = 2(√n)

Theorem: Any Boolean function f of hk0,...,hkk that depends on all of them requires

FBDD(f) = 2(𝑛)

which implies Decision-DNNF(f) = 2(√n)

Corollary: Grounded inference requires 2(√𝑛) time even on probabilistic DB instances with poly(n) time algorithms using lifted inference.

Implies separation between grounded and lifted inference

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Boolean combinations of hk0,...,hkk

f is a Boolean function that depends on all its inputs

Ψ = f(hk0, hk1,…, hkk)

We give a reduction from any FBDD for Ψ into an FBDD for Hk

Intuitively: to compute Ψ using an FBDD, you must compute the hk0, hk1, … , hkk, so that FBDD can also compute Hk

36

Summary

• FBDDs and decision-DNNFs bound the power of known model counting algorithms

• Exponential lower bounds on FBDDs & decision-DNNFs

• Which implies a separation between lifted and grounded inference

37

Open Problems

• A polynomial conversion of decision-DNNFs to FBDDs?

• General Dichotomy theorem for grounded inference?

• Approximate model counting?

38

Thank You

Questions?

Model Counting of Query Expressions: Limitations of Propositional Methods Paul Beame 1 Jerry Li 2...

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