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-## K-means Codebook
+# K-means Codebook
 
 A common technique for codebook generation involves utilising K-means clustering on a sample of the
 image descriptors. In this way descriptors may be mapped to *visual* words which lend themselves to
@@ -14,7 +14,7 @@ The number of clusters or the number of centroids determines the vocabulary size
 
 ## Bag-of-words histogram quantisation of descriptor vectors
 
-An example histograms for training and testing images is shown on figure \ref{fig:histo_tr}, computed with a vocubulary size of 100. The histograms of the same class appear to have comparable magnitudes for their respective keywords, demonstrating they had a similar number of descriptors which mapped to each of the clusters. The effect of the vocubalary size (as determined by the number of K-means centroids) on the classificaiton accuracy is shown in figure \ref{fig:km_vocsize}. A small vocabulary size tends to misrepresent the information contained in the different patches, resulting in poor classification accuracy. Conversly a large vocabulary size (many K-mean centroids), may display overfitting. In our tests, we observe a plateau after a cluster count of 60 on figure \ref{fig:km_vocsize}. This proccess of partitioning the input space into K distinct clusters is a form of **vector quantisation**.
+An example of histograms for training and testing images is shown on figure \ref{fig:histo_tr}, computed with a vocubulary size of 100. The histograms of the same class appear to have comparable magnitudes for their respective keywords, demonstrating they had a similar number of descriptors which mapped to each of the clusters. The effect of the vocubalary size (as determined by the number of K-means centroids) on the classificaiton accuracy is shown in figure \ref{fig:km_vocsize}. A small vocabulary size tends to misrepresent the information contained in the different patches, resulting in poor classification accuracy. Conversly a large vocabulary size (many K-mean centroids), may display overfitting. In our tests, we begin to observe a plateau after a cluster count of 60 on figure \ref{fig:km_vocsize}. This proccess of partitioning the input space into K distinct clusters is a form of **vector quantisation**.
 
 \begin{figure}
 \begin{center}
@@ -26,10 +26,10 @@ An example histograms for training and testing images is shown on figure \ref{fi
 \end{figure}
 
 
-The time complexity of quantisation with a K-means codebooks is $O(DNK)$, where N is the number of entities to be clustered (descriptors), D is the dimension (of the descriptors) and K is the cluster count [@km-complexity]. As the computation time is high, the tests we use a subsample of descriptors to compute the centroids (a random selection of 100 thousand descriptors). An alternative method we tried is applying PCA to the descriptors vectors to improve time performance. However, the descriptor dimension of 128 is relatiely small and as such we found PCA to be unnecessary.
+The time complexity of quantisation with a K-means codebooks is $O(DNK)$, where N is the number of entities to be clustered (descriptors), D is the dimension (of the descriptors) and K is the cluster count [@km-complexity]. As the computation time is high, the tests use a subsample of descriptors to compute the centroids (a random selection of 100 thousand descriptors). An alternative method we tried is applying PCA to the descriptors vectors to improve time performance. However, the descriptor dimension of 128 is relatiely small and as such we found PCA to be unnecessary.
 
 K-means is a process that converges to local optima and heavily depends on the initialization values of the centroids.
-Initializing K-means is an expensive process, based on sequential attempts of centroids placement. Running for multiple instances significantly affects the computation process, leading to a linear increase in execution time. We did not observe increase in accuracy with K-means estimator size larger than one, and therefore present results accuracy and execution time results with a single K-Mean estimator.
+Initializing K-means is an expensive process, based on sequential attempts of centroids placement. Running for multiple instances significantly affects the computation process, leading to a linear increase in execution time. We did not observe increase in accuracy with K-means estimator size larger than one, and therefore present results for accuracy and execution time with a single K-Mean estimator.
 
 \begin{figure}
 \begin{center}
@@ -59,7 +59,7 @@ We expect a large tree depth to lead into overfitting. However for the data anal
 \end{center}
 \end{figure}
 
-Random forests will select a random number of features on which to apply a weak learner (such as axis aligned split) and then chose the best feature of the sampled ones to perform the split on, based on a given criteria (our results use the *Gini index*). The fewer features that are compared for each split the quicker the trees are built and the more random they are. Therefore the randomness parameter can be considered the number of features used when making splits. We evaluate accuracy given different randomness when using a K-means vocabulary in figure \ref{fig:kmeanrandom}. The results in the figure \ref{fig:kmeanrandom} use a forest size of 100 as we infered that this is the estimatator count for which performance gains tend to plateau (when selecting $\sqrt{n}$ random features).
+Random forests will select a random number of features on which to apply a weak learner (such as axis aligned split) and then choose the best feature of the sampled ones to perform the split on, based on a given criteria (our results use the *Gini index*). The fewer features that are compared for each split the quicker the trees are built and the more random they are. Therefore the randomness parameter can be considered as the number of features used when making splits. We evaluate accuracy given different randomness when using a K-means vocabulary of size 100 in figure \ref{fig:kmeanrandom}. The results in the figure \ref{fig:kmeanrandom} use a forest size of 100 as we infered that this is the estimatator count for which performance gains tend to plateau (when selecting $\sqrt{n}$ random features).
 This parameter also affects correlation between trees. We expect in fact trees to be more correlated when using a large number of features for splits.
 
 \begin{figure}