Lecture Notes, 8 Dec 2010

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Index to all lecture notes
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§9.6, esp. pp. 653–655
§9.8, esp. p. 668;
§4.3(!), esp. p. 302;
§10.1, esp. p. 685–686:

Graphs (cont'd) and Rooted Trees

1. The Traveling Salesman Problem (TSP) is "NP":
   1. ∃ no known algorithm for solving it in Polynomial time
      by a deterministic Turing Machine (e.g., your laptop)
      • A problem is computable in P-time
        iff
        in the worst case, the time it takes to compute a solution for it
        on a deterministic Turing Machine
        is a polynomial function of the size of its input n.
        • i.e.) time is no worse than a_kn^k + a_k–1n^k–1 + … + a₁n + a₀,
          for some integer k and real constants a_i

        • e.g.) exponential time (order of 2²) is not in P

   2. It can only be computed in Non-deterministic Polynomial time
      • i.e.) solvable in polynomial time by a non-deterministic TM
        • A non-deterministic TM
          is
          a TM that does all possible "next steps" simultaneously until it finds a solution.

      • hence, ≡ SAT
        • SAT = finding an assignment of truth values that "sat"isfies a proposition of propositional logic
          • See "Propositional Logic & NP-Completeness"
          • i.e.) computing a truth table with n atomic propositions

   3. P = NP?
      • The greatest unsolved problem of CS
      • All functions divide into:
        • those not computable even in principle
        • those that are computable
          (by a Turing Machine, or by a recursive function),
          which divide into:
          • those computable in practice
            (i.e., polynomial time, or "P")

          • those not computable in practice
            (i.e., non-deterministic polynomial time, or "NP")

      • Read Goldman's Theorem
        • (humorous novel about P=NP written by a guy I went to elementary school with :-)

   4. and TSP is NP-hard:
      • at least as hard as—and possibly harder than—any other NP problem

   5. and TSP is NP-complete:
      • if can solve it in P-time,
        then can solve all other problems in P-time

   ---

2. The 4-Color Theorem:
   1. Thm (4CT): No map needs >4 colors
      • Convention: Two "countries" that share a linear boundary must be colored differently.

   2. How many colors does this map need?
      
      • It can be colored with 5 colors,
        but does it need 5?
        
        uses colors 1,…,5

      • No:
        
        only needs colors 1,…,4

   3. Can represent a map as a graph:
      • vertices represent countries;
        edges represent boundaries
        

   4. Then can use graph theory + a computer! to prove 4CT
      • Appel & Haken (1976) had to consider 2000 special cases
      • Univ. of Illinois postmark: "4 Colors Suffice!"

      • Does this change the nature of mathematical proof?
        (Do we need to prove that the computer program that proved 4CT was correct?)

   ---

3. Trees & Rooted Trees:
   1. Def:
      Let T = (V,E) be a graph.
      Then T is a tree
      =def
      T is a connected graph with no simple circuits
      1. E.g.)
         

      2. But this is not a tree (has a circuit):
         

      3. Nor is this (it's not connected;
         it's a "forest": a graph consisting of two (or more) trees)
         

      4. This "rooted", directed tree (with a "start" vertex, or "root") is a tree:
         
         (think of each edge as being directed "downwards")

   2. Recursive Def of "Rooted Tree":
      Base Case:
      Let r be a vertex

      Then T = ({r}, ∅)

      • (where V = {r} & E = ∅)

      is_def a rooted tree with root r

      • (∴ the base case of a rooted tree, T, is a graph with 1 vertex & no edges!)

      Recursive Case:

      Let T₁, … , T_n be (a forest of) disjoint rooted trees with roots r₁, … , r_n
      • ("disjoint" means that their vertex sets are disjoint)

      Let r be a new vertex ≠ each r_i

      Then the directed graph consisting of r with an edge from r to each r_i

      is_def
      a rooted tree with root r

   3. Set-Theoretical Recursive Def of Rooted Tree:
      1. Just as a test of your mathematical reading skills ;-)

      2. Base Case:
         Let r be a vertex.
         Then T = ( {r}, ∅ ) is a tree rooted at r

      3. Recursive Case:
         ∀ i ∈ {1, … , n}:
         Let r_i be a vertex.
         Let V_i be a set of vertices.
         Let E_i be a set of edges.
         Let T_i = (V_i, E_i) be a tree rooted at r_i.

         Let T₁, … , T_n be a forest of disjoint trees (rooted at r₁, … , r_n).

         Let r be a vertex ≠ each r_i.

         Then:
         T_n+1 = ( V_n+1 = ∪_i∈{1,…n}V_i ∪ {r},

                   E_n+1 = ∪_i∈{1,…n}E_i ∪ ∪_i∈{1,…n}{(r,r_i)} )

         is a tree rooted at r.

   4. E.g.)
      T₁ = • r₁

      T₂ = • r₂

      T₃ = • r₃

      are rooted trees with roots r_i by the base case.

      T₄ =
      
      is a rooted tree with root r₄ by the recursive case
      (and T₁, T₂, T₃ are subtrees of T₄)

      and if T₅ =
      
      is a rooted tree with root r₅,
      then T₇ =
      
      is a rooted tree with root r₇

   5. Terminology for Rooted Trees:
      1. Recall that a rooted tree is a digraph!
      2. Consider this rooted tree:
         

         Then here is the "feminist"/"male chauvinist"/neutral terminology:

      3. "a" is the mother/father/parent of "b" and "c"
      4. "b" is a daughter/son/child of "a"
      5. "b" and "c" are sister/brother/sibling vertices
      6. "b" and "d" are descendants of "a"
      7. "a" is an ancestor of "b",…,"e"

         and, switching metaphors:

      8. "a" is the root
      9. "d", "e", and "c" are leaves
      10. "b" is the root of a subtree with leaves "d" and "e"

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