{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# hw4sol - 10708 Graphical Models Homework 4 Solutions 1...

This preview shows pages 1–4. Sign up to view the full content.

10708 Graphical Models: Homework 4 Solutions 1 Markov Network Representations 1. We can convert the triangle graph on ( A, B, C ) with potential Ψ( A, B, C ) into a pair- wise Markov Random Field by introducing a variable X and connecting the nodes as shown in figure 1. X will take on 8 possible values, with each value corresponding to a joint setting of ( A, B, C ). The edge potentials are given in the following tables. (The leftmost column of each table gives the 8 possible values of X and each table belongs to the edge potential between X and one of A , B or C ) A 0 1 000 1 0 001 1 0 010 1 0 011 1 0 100 0 1 101 0 1 110 0 1 111 0 1 B 0 1 000 1 0 001 1 0 010 0 1 011 0 1 100 1 0 101 1 0 110 0 1 111 0 1 C 0 1 000 1 0 001 0 1 010 1 0 011 0 1 100 1 0 101 0 1 110 1 0 111 0 1 The node potential of X is just set to mirror Ψ: Φ( X ) 000 15 001 5 010 1 011 7 100 12 101 14 110 22 111 8 2. The original graph ( A, B, C, D ) is chordal, but not after converting to a pairwise Markov Random Field. 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
A X B C Figure 1: Pairwise Markov Random Field for part (a) A X B Y C D Figure 2: Pairwise Markov Random Field for part (b) 2 Hammersley-Clifford Assume for the sake of contradiction that P factors over the graph H : P = ψ AB · ψ BC · ψ CD · ψ DA Now observe that P (0010) = 0, which implies that ψ AB (0 , 0) · ψ BC (0 , 1) · ψ CD (1 , 0) · ψ DA (0 , 0) = 0 This means that at least one of the factors in this product, ψ AB (0 , 0), ψ BC (0 , 1), ψ CD (1 , 0), or ψ DA (0 , 0) is zero. But on the other hand, we have that P (0000) > 0 which implies that ψ AB (0 , 0) is nonzero, and P (0011) > 0 which implies that ψ BC (0 , 1) is nonzero, and P (1110) > 0 which implies that ψ CD (1 , 0) is nonzero, and P (0000) > 0 which implies that ψ DA (0 , 0) is also nonzero. This gives us the desired contradiction, and so P cannot factor over H . 2
3 Importance Sampling Since the question changed after its initial release, we were lenient in grading. (a) Computing the probability of a complete instantiation of the variables in a Markov random field is computationally intractable because one has to compute the partition function, which involves a sum or integral over the exponentially many possible values of all the variables.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

hw4sol - 10708 Graphical Models Homework 4 Solutions 1...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online