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--- courses:rg:2013:dep-tree-kernels [2013/03/04 22:28]
popel pdf link
+++ courses:rg:2013:dep-tree-kernels [2013/03/12 00:14]
kosao7am
@@ Line 16: / Line 16: @@
     * "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 contignous and sparse kernels will effecively 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 contignous (no missing indices) it can be shown (eg. by induction) that the equation holds, unless some of the indices is repeated. Strictly speaking (1,1,1) is a valid sequence.
+  - 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)=0 (or not defined)''
+  - 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)+\lambda^2K_0(t_2,u_1)+\lambda^2K_0(t_2,u_2)+\lambda^2K_0(t_3,u_1)+\lambda^2K_0(t_3,u_2)+\lambda^4K_0(\{t_1,t_2\},\{u_1,u_2\})+\lambda^4K_0(\{t_2,t_3\},\{u_1,u_2\})+\lambda^5K_0(\{t_1,t_3\},\{u_1,u_2\})= 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)+\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 (very likely) positive
+  - It was argued that we'll probably end up with different relation-args, 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

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