Day 32 mat385

• Announcements:
  • Homework for 5.3
    • #12 - Don't reproduce nodes (e.g. the). Once it's in, it's in. Think "clients" -- you wouldn't want to create two separate accounts for the same client....

      If you treated the tree as binary, and permitted a tree to have a right child without a left (in contrast to our author's algorithm), then the inorder traversal produced a sorted list.

      If you follow the author's convention that single children are left children, then the result would be "Beloved Best Elephant far had and high In no O off the trunk times."

    • #17 - You were doing some unusual counting here: if you've got 8 leaves, then I should see 8 leaves in the final decision tree. And if you've got 8 leaves, but are using only binary trees, then you're going to need to go to depth 3 (not 2) to provide for them.

      We should have only 4 leaves, however: we only care which coin is counterfeit, not whether it is heavy or light.

      It doesn't pay to compare one pair to another: we know that they'll be unbalanced, but we won't be able to tell from that which pair has the counterfeit coin in it!

      In this problem one trick was that you didn't need to know whether the coin was heavy or light. That cuts the number of leaves in half....

  • Homework for 6.2 due next time
  • Homework for 6.4: pp. 521--, #1, 2, 4, 9, 10, 11, 14, 16, 17, 20 (due Mon, 4/13)

• Questions on things?

• We'll wrap up section 6.3: Shortest Path and Minimal Spanning Tree

• then start section 6.4: Traversal Algorithms

Links:

• Details of the workings of Dijkstra's algorithm, Bellman-Ford, and Floyd's algorithm.
• An animation of Dijkstra's algorithm (without stated destination node)