MARKOV LOGIC NASSLLI 2010 Mathias Niepert

MARKOV LOGIC: INTUITION A logical KB is a set of hard constraints on the set of possible worlds  Let’s make some of them soft constraints: When a world violates a formula, it becomes less probable, not impossible  Give each formula a weight (Higher weight  Stronger constraint) 

P(world)  exp  weights of formulas it satisfies

MARKOV LOGIC: DEFINITION 

A Markov Logic Network (MLN) is a set of pairs where

(Fi, wi)

Fi is a formula in first-order logic  wi is a real-valued weight 



Together with a finite set of constants C, it defines a Markov network with One binary node for each grounding of each predicate in the MLN. The value of the node is 1 if the ground atom is true, and 0 otherwise.  One feature for each grounding of each formula F in the MLN, with the corresponding weight wi 

LOG-LINEAR MODELS 

A distribution is a log-linear model over a Markov network H if it is associated with 



A set of features F = {f1(D1),…,fk(Dk)}, where each Di is a complete subgraph (clique) of H, A set of weights w1 ,…,wk such that

P (X1; :::Xn) =

1 Z

exp

hP k

i i=1 wi fi (Di )

ASSUMPTIONS 1.

2.

3.

Unique names. Different constants refer to different objects. (Genesereth & Nilsson, 1987) Domain closure. The only objects in the domain are those representable using the constant and function symbols (Genesereth & Nilsson, 1987) Known functions. For each function, the value of that function applied to every possible tuple of arguments is known, and is an element of C

EXAMPLE: FRIENDS & SMOKERS Smoking causes cancer. Friends have similar smoking habits.

EXAMPLE: FRIENDS & SMOKERS Smoking causes cancer. Friends have similar smoking habits.

x Smokes( x)  Cancer ( x) x, y Friends ( x, y )  Smokes( x)  Smokes( y) 

EXAMPLE: FRIENDS & SMOKERS Smoking causes cancer. Friends have similar smoking habits.

1 .5 1 .1

x Smokes( x)  Cancer ( x) x, y Friends ( x, y )  Smokes( x)  Smokes( y) 

EXAMPLE: FRIENDS & SMOKERS Smoking causes cancer. Friends have similar smoking habits.

1.5 x Smokes( x)  Cancer ( x) 1.1 x, y Friends ( x, y )  Smokes( x)  Smokes( y) 

Two constants: Anna (A) and Bob (B)

EXAMPLE: FRIENDS & SMOKERS 1.5 x Smokes( x )  Cancer ( x ) 1.1 x, y Friends ( x, y )  Smokes( x )  Smokes( y ) 

Two constants: Anna (A) and Bob (B)

Smokes(A)

Cancer(A)

Smokes(B)

Cancer(B)

EXAMPLE: FRIENDS & SMOKERS 1.5 x Smokes( x )  Cancer ( x ) 1.1 x, y Friends ( x, y )  Smokes( x )  Smokes( y ) 

Two constants: Anna (A) and Bob (B) Friends(A,B)

Friends(A,A)

Smokes(A)

Smokes(B)

Cancer(A)

Friends(B,B)

Cancer(B) Friends(B,A)

EXAMPLE: FRIENDS & SMOKERS 1.5 x Smokes( x)  Cancer ( x) 1.1 x, y Friends ( x, y )  Smokes( x)  Smokes( y) 

Two constants: Anna (A) and Bob (B) Friends(A,B)

Friends(A,A)

Smokes(A)

Smokes(B)

Cancer(A)

Friends(B,B)

Cancer(B) Friends(B,A)

EXAMPLE: FRIENDS & SMOKERS 1.5 x Smokes( x )  Cancer ( x ) 1.1 x, y Friends ( x, y )  Smokes( x )  Smokes( y ) 

Two constants: Anna (A) and Bob (B) Friends(A,B)

Friends(A,A)

Smokes(A)

Smokes(B)

Cancer(A)

Friends(B,B)

Cancer(B) Friends(B,A)

MARKOV LOGIC NETWORKS MLN is template for ground Markov networks  Probability of a world x: 

1   P( x)  exp   wi ni ( x)  Z  i  Weight of formula i

No. of true groundings of formula i in x

Typed variables and constants greatly reduce size of ground Markov net  Functions, existential quantifiers, etc.  Infinite and continuous domains are possible 

EXAMPLE: FRIENDS & SMOKERS 1.5 x Smokes( x)  Cancer ( x) 1.1 x, y Friends ( x, y )  Smokes( x)  Smokes( y) 

Two constants: Anna (A) and Bob (B) 1.5 : Smokes( A)  Cancer ( A) 1.5 : Smokes( B)  Cancer ( B)

1.1 : Friends ( A, A)  Smokes( A)  Smokes( A)  1.1 : Friends ( A, B)  Smokes( A)  Smokes( B)  1.1 : Friends ( B, A)  Smokes( B)  Smokes( A)  1.1 : Friends ( B, B)  Smokes( B)  Smokes( B) 

P( S ( A)  T , S ( B)  F , F ( A, A)  T , F ( A, B)  T , F ( B, A)  F , F ( B, B)  T , C ( A)  F , C ( B)  T ) 

1   1 exp   wi ni ( x)   exp 1.5 *1  1.1* 3 Z  i  Z

RELATION TO FIRST-ORDER LOGIC Infinite weights  First-order logic  Satisfiable KB, positive weights  Satisfying assignments = Modes of distribution  Markov logic allows inconsistencies (contradictions between formulas) 

MAP INFERENCE IN MARKOV LOGIC NETWORKS 

Problem: Find most likely state of world y given evidence e

arg max P( y | e) y

Query

Evidence

MAP INFERENCE 

Problem: Find most likely state of world y given evidence e

1   arg max exp   wi ni ( y, e)  Ze y  i  ni is the feature corresponding to formula Fi  wi is the weight corresponding to formula Fi 

MAP INFERENCE 

Problem: Find most likely state of world y given evidence e

  arg max   wi ni ( y, e)  y  i  ni is the feature corresponding to formula Fi  wi is the weight corresponding to formula Fi 

MAP INFERENCE 1.5 x Smokes( x )  Cancer ( x ) 1.1 x, y Friends ( x, y )  Smokes( x )  Smokes( y ) 

Two constants: Anna (A) and Bob (B) Evidence: Friends(A,B), Friends(B,A), Smokes(B) true Friends(A,B) ? Friends(A,A)

Cancer(A)

? Smokes(A)

true Smokes(B)

? Friends(B,A)

true

? Friends(B,B)

Cancer(B)

?

MAP INFERENCE 1.5 x Smokes( x )  Cancer ( x ) 1.1 x, y Friends ( x, y )  Smokes( x )  Smokes( y ) 

Two constants: Anna (A) and Bob (B) Evidence: Friends(A,B), Friends(B,A), Smokes(B) true Friends(A,B) false Friends(A,A)

Cancer(A)

true Smokes(A)

true Smokes(B)

true Friends(B,A)

true

false Friends(B,B)

Cancer(B)

true

MAP INFERENCE 

Problem: Find most likely state of world given evidence e

arg max y

 w n ( y, e) i i

i

This is the weighted MAX-SAT problem  Use weighted MAX-SAT solver (e.g. MaxWalkSAT)  Better: Integer Linear Programming 

MAP INFERENCE 1.5 x Smokes( x )  Cancer ( x ) 1.1 x, y Friends ( x, y )  Smokes( x )  Smokes( y ) 

Two constants: Anna (A) and Bob (B) Evidence: Friends(A,B), Friends(B,A), Smokes(B) :Smokes(A) _ Cancer(A) 1.5 :Smokes(B) _ Cancer(B) 1.5 :Smokes(A) _ Cancer(B) 1.5 :Smokes(B) _ Cancer(A) 1.5 :Friends(A,B) _ :Smokes(A) _ Smokes(B) 0.55 :Friends(A,B) _ :Smokes(B) _ Smokes(A) 0.55 …

RELATION TO STATISTICAL MODELS  Special       

cases:

Markov networks Bayesian networks Log-linear models Exponential models Max. entropy models Hidden Markov models …

 Obtained

by making all predicates zeroarity

Every probability distribution over discrete or finiteprecision numeric variables can be represented as a Markov logic network.

MARKOV LOGIC 

Declarative language, several challenges Inference  Weight Learning  Structure Learning 



Many ways to perform probabilistic inference Conditional probability query  MAP query 

There’s a large body of work on probabilistic inference in graphical models  We’ll talk about some of these methods and how they can be put to work in Markov logic networks 

INFERENCE IN GRAPHICAL MODELS 

Conditional probability queries P(Y|E=e) where Y µX and E µX and Y and E disjoint

P (YjE = e) =



Let W = X – Y – E be the variables that are neither query nor evidence, then

P(y; e) = 

P (Y;e) P (e)

P

w P(y; e; w)

P(e) can be computed reusing the previous computation P

P (e) =

y P (y; e)

COMPLEXITY OF INFERENCE The process of “summing out” the joint not satisfactory as it leads to the  Exponential blowup that the graphical model representation was supposed to avoid  Problem: Exponential blow-up in the worst case is unavoidable   Worse: Approximate inference is also NP-hard  But: We really care about the cases that we encounter in practice not the worst-case 

COMPLEXITY OF INFERENCE Theoretical analysis can focus on Bayesian networks as they can be converted to Markov networks with no increase in representation size  First question: How do we encode a BN? 

 

Decision Problem BN-Pr-DP: 



DAG structure + worst-case representation of each CPD as a table of size |Val({X_i} [ PaXi )| Given a BN B over X, a variable X 2 X, and a value x 2 Val(X), decide whether PB (X=x) > 0

BN-Pr-DP is NP-complete

COMPLEXITY OF INFERENCE 

BN-Pr-DP is in NP: We guess an assignment x to the network variables  We check whether X=x holds in x and whether P(x)>0  The latter can be accomplished in linear time using the chain rule of BNs 



BN-Pr-DP is NP-hard: Reduction from 3-SAT  Given any 3-SAT formula f we can create a Bayesian network B with some distinguished binary variable X such that f is satisfiable if and only if PB(X=x1)>0  The BN has to be constructible in polynomial time 

COMPLEXITY OF INFERENCE

For each prop. variable qi one root node Qi with P(qi1)=0.5  For each clause ci one node Ci with edge from Qi to Cj if qi or ¬qi occurs in the clause cj 

COMPLEXITY OF INFERENCE

c1 = q1 _ :q3

c10

c11

q10 , q30

0

1

q11 , q30

0

1

q10 , q31

1

0

q11 , q31

0

1

COMPLEXITY OF INFERENCE

We cannot connect the Ci’s (i=1,…,m) directly to the variable X as the CPD for X would be exponentially large  We introduce m-2 AND nodes 

COMPLEXITY OF INFERENCE

Now, X has value 1 if and only if all of the clauses are satisfied  All nodes have at most three parents and, therefore, the size of the BN is polynomial in the size of f 

COMPLEXITY OF INFERENCE

Prior probability of each possible assignment is 1/2n  P(X=x1) = number of satisfying assignments to f divided by 2n  f has a satisfying assignment iff P(x1) > 0 

COMPLEXITY OF INFERENCE Probabilistic inference is a numerical problem not a decision problem  We can use a similar construction to show the following problem is #P-complete 



Given a BN B over X, a variable X 2 X, and a value x 2 Val(X), compute PB (X=x)

We have to do a weighted count of instantiations  #P is the set of the counting problems associated with the decision problems in the set NP  #P problem must be at least as hard as the corresponding NP problem 

COMPLEXITY OF APPROXIMATE INFERENCE Goal is to compute P(Y|e)  An estimate r has relative error e for P(y|e) if: 

r 1 e 

 P( y | e)  r (1  e )

We can use a similar construction again to show the following problem is NP-hard 

Given a BN B over X, a variable X 2 X, and a value x 2 Val(X), find a number r that has relative error e for PB (X=x)

COMPLEXITY OF APPROXIMATE INFERENCE Goal is to compute P(Y|e)  An estimate r has absolute error e for P(y|e) if: 

| P( y | e)  r | e Computing P(X=x) up to some absolute error r has a randomized polynomial time algorithm  However, when evidence is introduced, we’re back to NP-hardness  Following problem is NP-hard for any e 2 (0,1/2) 



Given a BN B over X, a variable X 2 X, a value x 2 Val(X), and an observation E=e, find a number r that has absolute error e for PB (X=x|e)

MONTE CARLO PRINCIPLE ? 







Consider the game of solitaire: What’s the probability of winning a game? Hard to compute analytically because winning or losing depends on a complex procedure of reorganizing cards Let’s play a few hands, and see empirically how many do in fact win Idea: Approximate a probability distribution using only samples from that distribution

Lose Lose Lose Win Lose Chance of winning is 1 in 5!

SAMPLING FROM A BAYESIAN NETWORK 

Generate samples (particles) from a Bayesian network using a random number generator i0 normal i1 high

d0

easy d1 difficult

g1 A g2 B g3 C

1 1

l0 weak l1 strong 3

s0 low s1 high

SAMPLING FROM A BAYESIAN NETWORK 

Generate samples (particles) from a Bayesian network using a random number generator i0 normal i1 high

d0

easy d1 difficult

g1 A g2 B g3 C

1 1

l0 weak l1 strong 3

s0 low s1 high

SAMPLING FROM A BAYESIAN NETWORK 

Generate samples (particles) from a Bayesian network using a random number generator i0 normal i1 high

d0

easy d1 difficult

g1 A g2 B g3 C

1 1

l0 weak l1 strong 3

s0 low s1 high

SAMPLING FROM A BAYESIAN NETWORK 

Generate samples (particles) from a Bayesian network using a random number generator i0 normal i1 high

d0

easy d1 difficult

g1 A g2 B g3 C

1 1

l0 weak l1 strong 3

s0 low s1 high

SAMPLING FROM A BAYESIAN NETWORK 

Generate samples (particles) from a Bayesian network using a random number generator i0 normal i1 high

d0

easy d1 difficult

g1 A g2 B g3 C

1 1

l0 weak l1 strong 3

s0 low s1 high

SAMPLING One sample can be computed in linear time  Sampling process generates a set of particles D = {x[1],…,x[M]}  When computing P(y), the estimate is simply the fraction of particles in which y “was seen” 

P^D =

1 M

PM

m=1 1fy[m]

= yg

with 1 the indicator function and y[m] the assignment to Y in particle x[m]

EXAMPLE: BAYESIAN NETWORK INFERENCE  





Suppose we have a Bayesian network with variables X Our state space is the set of all possible assignments of values to variables We can draw a sample in time that is linear in the size of the network Draw N samples, use them to approximate the joint

1st sample: D=d0,I=i1,G=g2,S=s0, L=l1 2nd sample: D=d1,I=i1,G=g1,S=s1, L=l1 …

REJECTION SAMPLING 







Suppose we have a Bayesian network with variables X

We wish to condition on some evidence E=e and compute the posterior over Y=X-E

Draw samples and reject them when not compatible with evidence e Inefficient if the evidence is itself improbable  we must reject a large number of samples

1st sample: D=d0,I=i1,G=g2,S=s0, L=l1 2nd sample: D=d1,I=i1,G=g1,S=s1, L=l1 …

reject accept

SAMPLING IN MARKOV LOGIC NETWORKS Sampling is performed on the ground Markov logic network  Alchemy uses a variant of the MCMC (Markov Chain Monte Carlo) method  Can answer arbitrary queries of the form P(Fi|MLNC,L)  Example: P(Cancer(Alice)|MLNC,L) 

MAP INFERENCE IN GRAPHICAL MODELS 

The following problem is NP-complete: 



Given a BN B over X and a number t, decide whether there exists an assignment x to X such that P(x) > t.

There exist several algorithms for MAP inference with reasonable performance on most practical problems

MAP INFERENCE IN MARKOV LOGIC NETWORKS 

Problem: Find most likely state of world y given evidence e

  arg max   wi ni ( y, e)  y  i  ni is the feature corresponding to formula Fi  wi is the weight corresponding to formula Fi 

MAP INFERENCE IN MARKOV LOGIC NETWORKS 

Problem: Find most likely state of world given evidence e

arg max y

 w n ( y, e) i i

i

This is the weighted MAX-SAT problem  Use weighted MAX-SAT solver (e.g. MaxWalkSAT)  Better: Integer Linear Programming 

THE MAXWALKSAT ALGORITHM for i ← 1 to max-tries do solution = random truth assignment for j ← 1 to max-flips do if ∑ weights(sat. clauses) > threshold then return solution c ← random unsatisfied clause with probability p flip a random variable in c else flip variable in c that maximizes ∑ weights(sat. clauses) return failure, best solution found

MAP INFERENCE IN MARKOV LOGIC NETWORKS We’ve tried Alchemy (and MaxWalkSAT) and the results were poor  Better results with integer linear programming (ILP)  ILP performs exact inference  Works very well on the problems we are concerned with  Originated in the field of operations research 

LINEAR PROGRAMMING A linear programming problem is the problem of maximizing (or minimizing) a linear function subject to a finite number of linear constraints  Standard form of linear programming: 

n

maximize

c x j

j 1 n

subject to

a x j 1

ij

xj

j

j

 bi

(i



( j  1, 2, ...,

0

 1, 2, ..., m) n)

INTEGER LINEAR PROGRAMMING An integer linear programming problem is the problem of maximizing (or minimizing) a linear function subject to a finite number of linear constraints  Difference to LP: Variables only allowed to have integer values 

n

maximize

c x j

j 1

subject to

n

a x j 1

ij

xj

j

j

 bi

(i



( j  1, 2, ...,

0

 1, 2, ..., m) n)

x j  {...,1,0,1,...} 54

MAP INFERENCE 1.5 x Smokes( x )  Cancer ( x ) 1.1 x, y Friends ( x, y )  Smokes( x )  Smokes( y ) 

Two constants: Anna (A) and Bob (B) Evidence: Friends(A,B), Friends(B,A), Smokes(B) :Smokes(A) _ Cancer(A) 1.5 :Smokes(B) _ Cancer(B) 1.5 :Smokes(A) _ Cancer(B) 1.5 :Smokes(B) _ Cancer(A) 1.5 :Friends(A,B) _ :Smokes(A) _ Smokes(B) 0.55 :Friends(A,B) _ :Smokes(B) _ Smokes(A) 0.55 …

MAP INFERENCE - EXAMPLE :Smokes(A) _ Cancer(A) 1.5

Introduce new variable for each ground atom: sa , ca Introduce new variable for each formula: xj Add the following three constraints:  sa + xj ¸ 1  -ca + xj ¸ 0  -xj - sa + ca ¸ -1 Add 1,5xj to the objective function

n

maximize

c x j 1

subject to

j

j

j

 bi

n

a x j 1

ij

x j  {0,1}

(i  1, 2, ..., m)

TOMORROW Ontology Matching with Markov Logic  Weight Learning  Experiments 