Contact
CoCalc Logo Icon
StoreFeaturesDocsShareSupport News AboutSign UpSign In
| Download

Distances between capital cities of New England

Views: 48
print ("Homework 16 (2) 10.2 #2 : Use Kruskal's algorithm to find a minimal spanning tree for the capital cities of New England...") print("") print("ME: Augusta, Maine") # vertex 1 print("MA: Boston, Massachusetts") # vertex 2 print("NH: Concord, New Hampshire") # vertex 3 print("CT: Hartford, Connecticut") # vertex 4 print("VT: Montpelier, Vermont") # vertex 5 print("RI: Providence, Rhode Island") # vertex 6 print("") print ("While not geographically true, the graph below does accurately depict the distances between each city...") edges = [ (1, 2, 165), (1, 3, 148), (1, 4, 266), (1, 5, 190), (1, 6, 208), (2, 3, 75), (2, 4, 103), (2, 5, 192), (2, 6, 43), (3, 4, 142), (3, 5, 117), (3, 6, 109), (4, 5, 204), (4, 6, 70), (5, 6, 223)] G=Graph(edges) G.weighted(True) G.relabel({ 1:'ME' , 2:'MA', 3:'NH', 4:'CT', 5:'VT', 6:'RI'}) G.graphplot(edge_labels=True,save_pos=True).show() print ("Kruskal's algorithm determines the following spanning tree of minimal distances:") G.relabel({'ME':'Augusta' , 'MA':'Boston', 'NH':'Concord', 'CT':'Hartford', 'VT':'Montpelier', 'RI':'Providence'}) from sage.graphs.spanning_tree import kruskal E = kruskal(G, check=True);E T=Graph(E) T.set_pos(G.get_pos()) T.graphplot(edge_labels=True).show()
Homework 16 (2) 10.2 #2 : Use Kruskal's algorithm to find a minimal spanning tree for the capital cities of New England... ME: Augusta, Maine MA: Boston, Massachusetts NH: Concord, New Hampshire CT: Hartford, Connecticut VT: Montpelier, Vermont RI: Providence, Rhode Island While not geographically true, the graph below does accurately depict the distances between each city...
Kruskal's algorithm determines the following spanning tree of minimal distances: [('Augusta', 'Concord', 148), ('Boston', 'Concord', 75), ('Boston', 'Providence', 43), ('Concord', 'Montpelier', 117), ('Hartford', 'Providence', 70)]