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courses:rg:2013:dep-tree-kernels [2013/03/12 00:11]
kosao7am |
courses:rg:2013:dep-tree-kernels [2013/03/12 11:14] (current)
popel |
| You want to check the relation between entities "US" and "Baghdad". Compute (estimate) <latex>K_1</latex> and <latex>K_2</latex>. | You want to check the relation between entities "US" and "Baghdad". Compute (estimate) <latex>K_1</latex> and <latex>K_2</latex>. |
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| ====== Answers ====== | ====== 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. |
| - 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 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. |
| - 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). | - 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'' | * ''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)'' | * ''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> | - 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> | * <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 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 | - 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, thus there will be no match | - 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 |
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| ====== Misc ====== | ====== Misc ====== |
| - There was some discussion what are the features for bag-of-words kernel (just presence of a word in sentence?) | - 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 | - Feature selection, mainly the relation-args feature |
| - "Two level" classification, why it might be a good idea | - "Two level" classification, why it might be a good idea |
