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courses:rg:2013:convolution-kernels [2013/02/26 10:03]
dusek vytvořeno
courses:rg:2013:convolution-kernels [2013/03/12 11:27] (current)
popel <latex>x</latex> was not rendered
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   - Find an error in one of the formulae in the paper.   - Find an error in one of the formulae in the paper.
 +==== Answers ====
 +  - 
 +    * **Generative models** use a two-step setup. They learn class-conditional (likelihood) <latex>P(x|y)</latex>, prior <latex>P(y)</latex> and use the Bayes rule to obtain the posterior.
 +      * they learn the joint distributions:  marginalize P(y), condition P(y|x) = P(x,y) / P(x)
 +      * They learn more than is actually needed, but are not prone to partially missing input data.
 +      * They are able to "generate" fake inputs, but this feat is not used very often.
 +      * Examples: Naive Bayes, Mixtures of Gaussians, HMM, Bayesian Networks, Markov Random Fields
 +    * **Discriminative models** do everything in one-step -- they learn the posterior <latex>P(y|x)</latex> as a function of some features of <latex> x</latex>.
 +      * They are simpler and can use many more features, but are prone to missing inputs.
 +      * Examples: SVM, Logistic Regression, Neural network, k-NN, Conditional Random Fields
 +  - Each CFG rule generates just one level of the derivation tree. Therefore, using "standard" nonterminals, it is not possible to generate e.g. this sentence:
 +    * ''(S (NP (PRP He)) (VP (VBD saw)(NP (PRP himself))))''
 +      * It could be modelled with an augmentation of the nonterminal labels.
 +    * CFGs can't generate non-projective sentences.
 +      * But they can be modelled using traces.
 +  - The derivation is actually quite simple:
 +    - <latex>h(T_a)\cdot h(T_b) = \sum_i h_i(T_a) \cdot h_i(T_b)</latex> -- (definition of the dot product)
 +    - <latex>= \sum_i \left(\sum_{n_a \in N_a} I_i(n_a)\right) \left(\sum_{n_b \in N_b} I_i(n_b)\right)</latex> (from the definition of <latex>I</latex> in the paragraph above the formula)
 +    - <latex>= \sum_i\sum_{n_a \in N_a}\sum_{n_b \in N_b} I_i(n_b)\cdot I_i(n_a)</latex> (since <latex>(a+b)(c+d) = ac+ad+bc+bd</latex>)
 +    - <latex>= \sum_{n_a \in N_a}\sum_{n_b \in N_b}\sum_i I_i(n_b)\cdot I_i(n_a)</latex> (change summation order)
 +    - <latex>= \sum_{n_a \in N_a}\sum_{n_b \in N_b}C(n_a, n_b)</latex> (definition of <latex> C </latex>)
 +  - Convolution is defined like this: <latex>(f*g)_k = \sum_i f_i g_{k-i}</latex>, so it measures the presence of structures that //complement// each other. Here, we have a measure of structures that are //similar//. So it is something different. But the main idea is the same -- we can combine smaller structures (kernels) into more complex ones.
 +  - There is a (tiny) error in the last formula of Section 3. You cannot actually multiply tree parses, so it should read: <latex>\bar{w}^{*} \cdot h(\mathbf{x}) = \dots</latex>
 +==== Report ====
 +We discussed the answers to the questions most of the time. Other issues raised in the discussion were:
 +  * **Usability** -- the approach is only usable for //reranking// the output of some other parser.
 +  * **Scalability** -- they only use 800 sentences and 20 candidates per sentence for training. We believe that for large data (milions of examples) this will become too complex.
 +  * **Evaluation** -- it looks as if they used a non-standard evaluation metric to get "better" results. The standard here would be F1-score.

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