====== Questions ======
Aron Culotta, Jeffrey Sorensen: [[http://www.newdesign.aclweb.org/anthology-new/P/P04/P04-1054.pdf|Dependency Tree Kernels for Relation Extraction]], ACL 2004.


  - Given Figure 1, what is the smallest common subtree that includes both t1 (Troops) and t2 (near)?
  - Section 5: "Therefore, d(a)=l(a)." When is this true and why? (Assume this holds for the following questions.)
  - Let <latex>\phi_m</latex> = {general-pos-tag, entity-type, relation-arguments} (in accordance with the paper). Let <latex>\phi_s = \phi_m</latex> (unlike in the paper). Based on Figure 2 and Section 5, compute the following matching functions and similarity functions:
    * ''m(t0,u0)=?     m(t1,u1)=?      m(t2,u2)=?''
    * ''s(t0,u0)=?     s(t1,u1)=?      s(t2,u2)=?''
  - Let <latex>\lambda=0.5</latex>. Compute the contiguous kernel for the two trees in Figure 2: <latex>K_1(T,U)=?</latex>. Provide the final number and some counts along the way, so its clear how you got the number. Optionally, compute also the sparse kernel <latex>K_0(T,U)=?</latex>.
  - Let DT be a function that assigns the correct augmented dependency tree to a sentence. Compute (estimate) contiguous kernel and bag-of-words kernel for the following sentences:
    * <latex>K_1</latex>(DT("Peter sleeps"), DT("Bob runs"))=?
    * <latex>K_2</latex>(DT("Peter sleeps"), DT("Bob runs"))=?
  - Lets have a pair of sentences:   
    * "Bob saw US troops that moved towards Baghdad"
    * "US troops that moved towards Baghdad were seen by Bob"
   You want to check the relation between entities "US" and "Baghdad". Compute (estimate) <latex>K_1</latex> and <latex>K_2</latex>.
   
====== Answers ====== 
  - Depends on the exact definition of smallest common subtree, but keep in mind you need at least some non-trivial "context". The definition should be such that contiguous and sparse kernels will effectively be different things. The whole subtree is probably the right answer here.
  - d(a) is defined as the last member of the sequence - the first member + 1. If the sequence is contiguous (no missing indices) it can be shown (eg. by induction) that the equation holds, unless some of the indices is repeated. Note that e.g. a sequence (1,1,1) is valid according to the definition of sequence <latex>a</latex> in the paper.
  - Depends on how you treat "N/A" values, by the definition you should sum values that are "the same/compatible" (disregarding the "type" of the feature).
    * ''m(t0,u0)=1     m(t1,u1)=1      m(t2,u2)=0''
    * ''s(t0,u0)=5     s(t1,u1)=3      s(t2,u2)=4''
  - First this depends on the previous one (the "N/A" values) and second the paper doesn't say how to compute <latex>K_0(t_i[A],t_j[B])</latex>, where A,B are sequences with more than one member. One proposed solution was to use <latex>K_0(t_i[A],t_j[B])=\sum_{s=0..l(A)}K_0(t_i[a_s],t_j[b_s])</latex>
    * <latex>K_0(T,U)=s(t_0,u_0)+\lambda^2K_0(t_1,u_1)+\lambda^2K_0(t_1,u_2)+</latex><latex>\lambda^2K_0(t_2,u_1)+\lambda^2K_0(t_2,u_2)+\lambda^2K_0(t_3,u_1)+</latex><latex>\lambda^2K_0(t_3,u_2)+\lambda^4K_0(\{t_1,t_2\},\{u_1,u_2\})+</latex><latex>\lambda^4K_0(\{t_2,t_3\},\{u_1,u_2\})+\lambda^5K_0(\{t_1,t_3\},\{u_1,u_2\})</latex><latex>=s(t_0,u_0)+\lambda^2s(t_1,u_1)+\lambda^2s(t_3,u_2)+\lambda^4s(t_4,u_3)+\lambda^4s(t_1,u_1)+</latex><latex>\lambda^4s(t_3,u_2)+\lambda^6s(t_4,u_3)+\lambda^5s(t_1,u_1)+\lambda^5s(t_3,u_2)+\lambda^7s(t_4,u_3)</latex>
    * When counting K_1 you leave out the <latex>(\{t_1,t_3\},\{u_1,u_2\})</latex> part
  - When you regard bag-of-words kernel as number of matching forms then K_2 is zero whereas K_1 is positive
  - It was argued that we'll probably end up with different relation-args (//troops// being ARG_B in the first sentence, but ARG_A in the second sentence), thus there will be no match
  

====== Misc ====== 
  - There was some discussion what are the features for bag-of-words kernel (just presence of a word in sentence?)
  - Feature selection, mainly the relation-args feature
  - "Two level" classification, why it might be a good idea