Lecture 02-08-17

Discuss take home exam 2

1. Was late to class
2. Sorting adjacency lists
   • Go through adjacency lists in order and always adding in order
   • Don’t need extra space?
   • Twist: if directed:
     • First create lists of edges into the vertex
     • Reverse edges
   • Standard assumption:
     • Vertices are numbered 1…n
3. Identify in O(n+m) all vertices in path from x-y
   • Make subgraph of biconnected components (biconnected component tree)
   • Account for case where starting point is articulation point
   • Also can’t DFS on the original graph without keeping track of whether it is leaving a biconnected component or not
4. Single vertex has path to all other vertices in digraph
   • Break into strongly connected components
   • True if exactly 1 vertex has in degree 0 in the strongly connected component graph
   • Derived from harder problem:
     • Notion of connectivity (uni-connected)
       • Path of at least x-y or y-x

Union-Find

• Create a data structure where both UNION and FIND operations are “fast”
• Based on trees (inverted trees)
• Inverted tree:
  • Every node keeps track of parent
  • Not the same as the MST
  • Will keep the property: the vertices it’s connected to is a single set
• Union
  • O(1)
  • Make one tree’s root the child of the other’s root
• Find
  • Requires that the name of the set it belongs to is at the top of the tree
  • Real implementation is a array representation of a tree
    • Necessary to instantly access node
  • Runtime dependent on the height of the tree: O(n)
    • Height based union
    • Generalized: make root of “smaller”, child of “bigger”
    • Definition of smaller:
      • Height vs vertices
  • Strategies:
    • Height based union
    • Size based union
  • Note: addition is by convention O(1) but is actually more complex
  • Not immediately obvious why find is based on height
    • See that it is not easy to keep track of the height of the tree
    • Will partially collapse this tree