Differences

This shows you the differences between two versions of the page.

--- courses:rg:2013:dep-tree-kernels [2013/03/04 22:28]
popel pdf link
+++ courses:rg:2013:dep-tree-kernels [2013/03/12 11:14] (current)
popel
@@ 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 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

[ Back to the navigation ] [ Back to the content ]

Institute of Formal and Applied Linguistics Wiki

Differences