his 2.02 to the 0 number that gave node 1 (time 0) in Figure 8.15(a), for a grand total of 2.02. 4. If we started at node 1 (time 0) and moved to node 0 (time 1) using the bottom branch, the first output symbol S1 would be (0, -1). Comparing this to r where the first output symbol is (1.1, 0.1), we say "ds = (1.1-0)2 + (0.1- ( 1)) 2 = 2.42 distance if parent is node 1 (time 0), bottom branch." We add this 2.42 to the 0 number that we gave node 1 (time 0) in Figure 8.15(a), for a grand total of 2.42. Since starting at node 0 (time 0) and moving to node 0 (time 1) along the top branch creates the smallest distance (0.02), we proclaim that the parent node for node 0 (time 1) is node 0 (time 0) and the best branch is the top branch, and that it carries with it the number 0.02 (for total distance with this selection). We repeat this for node 1 (time 1), node 2 (time 1) and node 3 (time 1), and the results are shown in Figure 8.15(c). That is all we do for our first move from left to right through the trellis. At the next time, we do a similar thing. We start again at the top node, this time starting with node 0 (time 2). Looking at Figure 8.16, we can see that this node has two possible parent nodes, which are node 0 (time 1) and node 1 (time 1). Each parent node comes with two possible branches. We decide between these nodes and branches as follows: 1. If we started at node 0 (time 1) and moved to node 0 (time 2) using the top branch, the second output symbol would be (1, 0). Comparing this to r where the second signal is (1.1, 0.1), we say "d1 = (1.1-1 1)2 + (0.1-0)2 = 0.02 distance if parent is node 0 (time 1), top branch." We add this 0.02 to the 0.02 number that we gave node 0 (time 1) in Figure 8.15(b), for a grand total of 0.04. 2. If we started at node 0 (time 1) and moved to node 0 (time 2) using the bottom branch, the second output symbol would be (-1, 0). Comparing this to r where the second signal is (1.1, 0.1), we say "d2 = 4.85 distance if parent is node 0 (time 1), bottom branch." We add this 4.85 to th