Update 2_dtw.md

docs/2_method/2_dtw.md

@@ -19,29 +19,28 @@ $$
 y=(y_1, y_2, ..., y_m).
 $$
 
-The DTW distance is the sum of the Euclidean distance between each point and its matched point(s) in the other vector, as shown in \autoref{fig:warping_signals}. The following constraints must be met: 
+The DTW distance is the sum of the Euclidean distance between each point and its matched point(s) in the other vector. The following constraints must be met: 
 
 1. The first and last elements of each series must be matched.
 2. Only unidirectional forward movement through relative time is allowed, i.e., if $x_1$ is mapped to $y_2$ then $x_2$ may not be mapped to
     $y_1$ (monotonicity). 
 3. Each point is mapped to at least one other point, i.e., there are no jumps in time (continuity).
 
-![Two time series with DTW pairwise alignment between each element, showing one-to-many mapping properties of DTW (left). Cost matrix $C$ for the two time series, showing the warping path and final DTW cost at $C_{14,13}$ (right). \label{fig:warping_signals}](../media/Merged_document.pdf)
+![Two time series with DTW pairwise alignment between each element, showing one-to-many mapping properties of DTW (left). Cost matrix $C$ for the two time series, showing the warping path and final DTW cost at $C_{14,13}$ (right).](../../media/Merged_document.pdf)
 
 Finding the optimal warping arrangement is an optimisation problem that can be solved using dynamic programming, which splits the problem into easier sub-problems and solves them recursively, storing intermediate solutions until the final solution is reached. To understand the memory-efficient method used in ''DTW-C++``, it is useful to first examine the full-cost matrix solution, as follows. For each pairwise comparison, an ($n$) by ($m$) matrix $C^{n\times m}$ is calculated, where each element represents the cumulative cost between series up to the points $x_i$ and $y_j$:
 
-\begin{equation}
-    \label{c}
-    c_{i,j} = (x_i-y_j)^2+\min\begin{cases}
-    c_{i-1,j-1}\\
-    c_{i-1,j}\\
-    c_{i,j-1}
-    \end{cases}
-\end{equation}
+\[
+c_{i,j} = (x_i-y_j)^2+\min \left\{
+\begin{array}{ccc}
+c_{i-1,j-1} & c_{i-1,j} & c_{i,j-1}
+\end{array}
+\right\}
+\]
 
 The final element $c_{n,m}$ is then the total cost, $C_{x,y}$, which provides the comparison metric between the two series $x$ and $y$. \autoref{fig:warping_signals} shows an example of this cost matrix $C$ and the warping path through it.
 
-For the clustering problem, only this final cost for each pairwise comparison is required; the actual warping path (or mapping of each point in one time series to the other) is superfluous for k-medoids clustering. The memory complexity of the cost matrix $C$ is $O(nm)$, so as the length of the time series increases, the memory required increases greatly. Therefore, significant reductions in memory can be made by not storing the entire $C$ matrix. When the warping path is not required, only a vector containing the previous row for the current step of the dynamic programming sub-problem is required (i.e., the previous three values $c_{i-1,j-1}$, $c_{i-1,j}$, $c_{i,j-1}$), as indicated in \autoref{c}.
+For the clustering problem, only this final cost for each pairwise comparison is required; the actual warping path (or mapping of each point in one time series to the other) is superfluous for k-medoids clustering. The memory complexity of the cost matrix $C$ is $O(nm)$, so as the length of the time series increases, the memory required increases greatly. Therefore, significant reductions in memory can be made by not storing the entire $C$ matrix. When the warping path is not required, only a vector containing the previous row for the current step of the dynamic programming sub-problem is required (i.e., the previous three values $c_{i-1,j-1}$, $c_{i-1,j}$, $c_{i,j-1}$).
 
 The DTW distance $C_{x,y}$ is found for each pairwise comparison. Pairwise distances are then stored in a separate symmetric matrix, $D^{p\times p}$, where ($p$) is the total number of time series in the clustering exercise. In other words, the element $d_{i,j}$ gives the distance between time series ($i$) and ($j$).