-- Sep 27 In-Class Exercise
In the first division step we divide the points into two groups:
G1 ={(0,0),(1,12),(1,−12)}
G2 = {(2,5),(2,−5),(3,0).}
Subsequently we divide G1 into further groups until we reach:
G3 = {(0,0)}
G4 = {(1,1/2)}
G5 = {(1,-1/2)}
G2 into two groups
G6 = {(2,5)}
G7 = {(2,-5)}
G8 = {(3,0)}
G3, G4, G5 merge (two at a time) to form the left triangle
G6, G7 and G8 merge (again two at a time) to form the right triangle
Then we find the point with the maximum coordinates in one of the dimensions. In our case lets take y. These points turn out to be (2,5) and (2,-5).
Now we try to figure out the points form the left that triangle that these should connect to. Consider the vector from (2,5) to any of the points, lets say (1,1/2). As we are connecting from right to left and downwards (since it has max y), when we try to connect to the next point if the vector rotates clockwise we consider this as the new connection point . Hence we consider (2,5) to (0,0) as the new edge and discard (2,5) connection to (1,-1/2). Similarly we create the edge between (2,-5) to (0,0) but in this case we have to consider the rotation counterclockwise as we are y min.
We draw a vector from (2,5) to
In the first division step we divide the points into two groups:
G1 ={(0,0),(1,12),(1,−12)}
G2 = {(2,5),(2,−5),(3,0).}
Subsequently we divide G1 into further groups until we reach:
G3 = {(0,0)}
G4 = {(1,1/2)}
G5 = {(1,-1/2)}
G2 into two groups
G6 = {(2,5)}
G7 = {(2,-5)}
G8 = {(3,0)}
G3, G4, G5 merge (two at a time) to form the left triangle
G6, G7 and G8 merge (again two at a time) to form the right triangle
Then we find the point with the maximum coordinates in one of the dimensions. In our case lets take y. These points turn out to be (2,5) and (2,-5).
Now we try to figure out the points form the left that triangle that these should connect to. Consider the vector from (2,5) to any of the points, lets say (1,1/2). As we are connecting from right to left and downwards (since it has max y), when we try to connect to the next point if the vector rotates clockwise we consider this as the new connection point . Hence we consider (2,5) to (0,0) as the new edge and discard (2,5) connection to (1,-1/2). Similarly we create the edge between (2,-5) to (0,0) but in this case we have to consider the rotation counterclockwise as we are y min.
We draw a vector from (2,5) to