Update 2_dtw.md

docs/2_method/2_dtw.md

@@ -39,7 +39,7 @@ $$
 
 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 $y$. Below is an example of this cost matrix $C$ and the warping path through it.
 
-![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.png)
+![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/](https://github.com/Battery-Intelligence-Lab/dtw-cpp/blob/main/media/Merged_document.png)
 
 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 $c_{i-1,j-1}$, $c_{i-1,j}$, $c_{i,j-1}$).