Question

Consider a weighted, undirected, graph G. Let e(u,v) be the weight of the edge between nodes u and v, where e(u,u)=0 and e(u,v)=[infinity] if u and v is disconnected. Assume the graph is a connected component, i.e., there exists a path between every two nodes. Suppose the path length, d(u,v), is defined as follows:

d(u,v) = f(x) = 0, if u=v;
e(u,v), if there is an edge between u and v
​min w=u=v d(u,w)+d(w,v), otherwise;
​ if there is an edge between u and vIs d(u,v) a metric? State your reasons clearly. (Check whether the positivity, symmetry, and triangle inequality properties are preserved).

Answer 1

Yes, d(u,v) is a metric because it satisfies the positivity, symmetry, and triangle inequality properties.

Step-by-step explanation:

The distance metric, d(u,v), is indeed a metric.

To check for the positivity property, we need to verify that d(u,v) is always greater than or equal to zero. In this case, if there is a direct edge between nodes u and v, then e(u,v) will be the weight of that edge, which would be a non-negative value. If there is no edge between u and v, then e(u,v) is defined as infinity, which is greater than or equal to zero. When evaluating d(u,w)+d(w,v), we take the minimum of the previous distances, which means d(u,w)+d(w,v) will be greater than or equal to zero. Therefore, d(u,v) is always greater than or equal to zero.

To check for the symmetry property, we need to verify that d(u,v) is equal to d(v,u) for all u and v. In this case, if u=v, then d(u,v)=f(x)=0 and d(v,u)=f(x)=0, so they are equal. If there is a direct edge between u and v, then e(u,v) will be the weight of that edge and e(v,u) will also be the weight of the same edge, so they are equal. When evaluating d(u,w)+d(w,v), we are taking the minimum of the previous distances, so d(u,w)+d(w,v) is the same as d(v,w)+d(w,u). Therefore, d(u,v) is equal to d(v,u) for all u and v.

To check for the triangle inequality property, we need to verify that d(u,v) is less than or equal to d(u,w)+d(w,v) for all u, v, and w. In this case, if u=v, then both sides of the inequality will be zero. If there is a direct edge between u and v, then both sides of the inequality will be the weight of that edge, which means they are equal. When evaluating d(u,w)+d(w,v), we are taking the minimum of the previous distances, so d(u,w)+d(w,v) will be greater than or equal to the actual distance between u and v. Therefore, d(u,v) is less than or equal to d(u,w)+d(w,v) for all u, v, and w.