CSC 173
Tues. Nov 19, 2002
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Project 5 due Monday Dec 9
  Project 6 will be out before then! (a written assignment, ~one week)
Quiz on Thursday
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TSP - Traveling salesman Problem
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Hamiltonian Cycle of shortest length
  Simple cycle that contains each vertex

Finding an exact solution, naive solution:
  list all permutations of vertices
  find the cost of each, select the minimum
  Not tractable!  running time ~ 2^n

Triangle inequality - 
  often, it is always cheapest to go directly from a place a to "a" place
  "c" rather than going through an intermediate stop "b"

  Going through "b" CANNOT be less expensive
  Cutting out an intermediate stop never increases the cost

  always true for graphs where distance is euclidian

Approx TSP
  1. form a MST 
  2. Let L be the list of vertices visited in a preorder walk of T
  3. Let the TSP path be the vertices in the order L

Assuming the graph satisfies triangle inequality, this approximation is never
more than twice the cost of an optimal tour.

Proof:  Let H' be an optimal tour
  Let H be the tour returned by the above alg.
  We need to show that cost(H) <= 2*cost(H')

  Since we can obtain a spanning tree by deleting any edge from a tour, if T
  is a MST, then cost(T)<=cost(H')

  A full walk of T lists the vertices when they are first visited and also
  whenever they are returned to after a visit to a subtree.  Call this walk W.

  This walk traverses every edge exactly twice, so
    cost(W) = 2*cost(T)
 
  because cost(T)<=cost(H') and cost(W) = 2*cost(T), it is implied that
    cost(W) <= 2*cost(H')

  so the cost of W is within a factor of 2 of the cost of optimal

  But W is not a tour, but by triangle inequality, we can delete a visit to
  any vertex and the cost will not increase.

  The resulting order after repeatedly removing the edges results in the same
  order as a preorder walk of the tree.  Let H be the cycle corresponding to
  this preorder walk.  
    H is the same cycle computed by our alg, so
    cost(H) <= cost(W)

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Food for thought: 
  if we drop the assumption of triangle inequality,
    good approximate tours cannot be found in polynomial time

  If we assume the vertices are points in the plane and cost of an edge is
  eudlicean distance, an optimal tour never crosses itself!

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Maximum flow
  liquid running through pipes
  parts going through assembly lines
  information going through communication networks

  directed graphs, each edge has a capacity (max rate)
    e.g., 200 gallons per hour, 100 MB/s
  vertices, no collection (no buffering)
    sum of inputs must be equal to sum of outputs

  max flow - greatest rate at which material can be shipped from the source to
  the sink without violating any capacity constraints

  ex. cities, trucks shipping crates, edges are roads, labels on edges are
  number of crates per day along a road.
 
    do not care about how long it takes to get a crate to the destination

    find the max crates per day that can be shipped

  can convert multiple source, multiple sink to single source, single sink:
    add new source node s'.  Create edges from s' to all original sources.
    Capacity on the link is infinite!  Create a new sink node t'.  Creat edges
    from all original sinks to t'.  Capacity on the link is infinite!

Ford-Fulkerson
  1. start with 0 flow
  2. find an augmenting path along which we can push flow
  3. incrase flow along this path.
  4. iterate until no more augmenting paths can be found

This yields the maximum flow!

residual networks
  given a flow network and a flow, the residual network consists of edges that
  can admit more net flow

Augmenting path
  simple path from source to sink in the residual network
  residual capacity is the minimum edge weight

Cuts of flow networks
  cut is a partition of vertices, put s and t in different partitions
  capacity of the cut is the difference between the sum of the forward edges
    and the sum of the backward edges
  value of a flow in a network is the net flow across any cut of the network
  
Running time is 

  use BFS or DFS to find augmenting path O(E)
  at must you must do this f times, where f is the resulting flow of the
    graph. Overall:  O(Ef)
  good when f is small

  Show example for what happens if f is large and capacity of one link is very
  low.  BFS will perform better than DFS

  if we use bfs to find the shortest path, the result is known as the
  Edmonds-Karp algorithm, and it runs in O(VE^2)