To conclude the process of finding the best path, we now work backwards from the top-right corner, tracing back the cheapest route to that point. This gives us a path through the matrix along which the test and reference signals are most similar to each other:
 

7	B: 7	B: 5	B: 3	BL: 4	BL: 2½	BL: 1½
6	B: 7	B: 5	B: 2	B: 2	BL: 1	BL: 4
5	B: 6	BL: 4	B: 2 BL: 2	BL: 1	L: 2	BL: 6½
4	B: 2	BL: 1 ½	BL: 1	L: 2 BL: 2	L: 2	L: 6
3	B: 1	B: 1 L: 1 BL: 1	BL: 1½	BL: 4	L: 5	L: 6
2	B: 1	L: 1 BL: 1	L: 2	L: 6	L: 7	L: 8
1	Start	L: 2	L: 6	L: 15	L: 19	L: 19
	1	2	3	4	5	6

In this case, it can be seen that there are several equally low-cost paths from (1,1) to (3,4). To remediate this, it is possible to take into account the overall length of the path. This would rule out (1,3), but would still give us a choice of going up and diagonally from the start, or equally well, diagonally first and then up. There is no solution to this problem.

Next: warp paths and alignment