Day 20, Mat385

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• Announcements
  • The quiz:
    • The Key
      • About cardinality:

        We use the word cardinality even when sets are infinite in size. If you want a fun discussion of infinite sets, try Hilbert's paradox of the Grand Hotel.

      • The induction proof requested for problem three starts with a base case of a set containing two elements, which has a single subset containing two elements (itself).

        I find the induction step easiest if one assumes the result for \(k\), then considers a set of \(k+1\) elements, then removes one element \(a\) and considers cases: subsets of two elements containing \(a\), and those which don't.

        I assigned problem 26, which asked about subsets of three elements, and relied on this exercise (#25).

        Here's my argument.

  • This week's quiz will cover graphs and trees (but not decision trees).

  • Next week, Wednesday, we have our second midterm exam. It will cover everything we've covered from 4.1 through 7.3 (or topic for Wednesday).

• Last time:
  • We wrapped up trees.

  • Then we began section 6.3, decision trees.

    We arrived at a lower bound on the number of comparisons \(C\) for searching a list of \(m\) elements, based on the depth of the binary decision tree for searching: \[ C = \lfloor \log_2(m) \rfloor + 1 \]

    Upshot:

    • if you want the best worst-case scenario, make/use a bushy binary tree, the bushiest of which (full binary trees) contain \(2^d\) nodes at depth \(d\).

      By the time you've gotten depth \(d\) filled, you've got \(2^{d+1}-1\) internal nodes. If you're lucky and \(m= 2^{d+1}-1\), and you filled your binary tree very wisely, \[ m= 2^{d+1}-1 \iff d=\log_2(m+1) - 1 \]

      The number of comparisons is one more than the depth of the search tree -- it's the depth of the decision tree, which contains the leaves which report the results. So \[ C = \lfloor log_2(m) \rfloor + 1 \]

    • If for some perverse reason you want the worst worst-case scenario for searching make a chain by inserting elements into a binary search tree in order (alphabetic or reverse alphabetic order).

    • We could take another approach to get to essentially the same result (and which we'll take with the sort problem): the number of leaves of the search (see our author's example) is \(m+1\). If we can manage to put them all at one depth, then that depth (the total number of comparisons \(C\)) will satisfy \[ 2^C \ge m+1 \] or \[ C \ge \lceil \log(m+1) \rceil \ge \lfloor \log(m) \rfloor + 1 \]

    • Now we'll take exactly the same approach to searching....

  • Marked up
    • highlights from 6.2
    • highlights from 6.3

• Today:
  • We'll wrap up Section 6.3: Decision Trees.
  • Then we'll hit Section 7.2 (Euler Paths and Hamiltonian Circuits).

• Some Fun, Motivating Examples from Propositional and Predicate Calculus