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| courses:cs211:winter2012:journals:paul:home [2012/04/04 17:00] – [Chapter 6] nguyenp | courses:cs211:winter2012:journals:paul:home [2012/04/06 02:17] (current) – [Chapter 7] nguyenp | ||
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| * Locally optimal, but not globally optimal (it's very easy to think of a counter example, counter example is on page 339 in book) | * Locally optimal, but not globally optimal (it's very easy to think of a counter example, counter example is on page 339 in book) | ||
| * Awesome Solution: Residual Graphs | * Awesome Solution: Residual Graphs | ||
| + | * Edges have a capacity, but also a flow | ||
| + | * We don't have to use all of it at once | ||
| + | * Residual edge is one where you have some going one way and some going the other | ||
| + | * The Ford-Fulkerson Algorithm uses residual edges and works awesomely (optimal) | ||
| + | * page 342 - 344 | ||
| + | * Informal explanation | ||
| + | * graphs | ||
| + | * keep updating the residual edges using the following formula: | ||
| + | * If edge is the same as it was originally, set the forward rate to the bottle neck + original rate (which is the edge with the smallest flow rate) | ||
| + | * Otherwise, set the backwards residual rate to what it originally was minus the bottle neck | ||
| + | * Section 7.2 - Maximum Flows and Minimmum Cuts in a Network | ||
| + | * Cut: s-t cut is a partition (A,B) where s is in A and t is in B | ||
| + | * Problem: Find an s-t cut of minimum capacity | ||
| + | * The capacity of a cut is the sum of all the possible stuff that can go through the edge at once | ||
| + | * Flow Value Lemma: Let f be any flow, and let (A, B) be any s-t cut. Then, the value of the flow is = f_out(A) – f_in(A). | ||
| + | * Weak Duality: let f be any flow and let (A, B) be any s-t cut. Then the value of the flow is at most the cut’s capacity. | ||
| + | * Corollary: Let f be any flow, and let (A, B) be any cut. If v(f) = cap(A, B), then f is a max flow and (A, B) is a min cut | ||
| + | * Augmenting path theorem: Flow f is a max flow iff there are no augmenting paths | ||
| + | * Max-flow min-cut theorem: The value of the max flow is equal to the value of the min cut. | ||
| + | * More explanation on page 350 | ||
| + | * Boom and we are done. Just use the same stuff as in the past section and we have all we need. | ||
| + | * Section 7.5 - A First Application: | ||
| + | * Given an undirected bipartite graph | ||
| + | * We must match so each node appears at most once on each edge | ||
| + | * How do we find one with the max number of nodes? This is our problem. | ||
| + | * Here's how we do it. We got a bunch of red and blue nodes. Add nodes s and t, which will be the start and end nodes. Have the start node s have a unit edge (edge of size 1) connect it to all the red nodes. Have all the blue nodes have a unit edge connect them to the end node t. Still have same edges in the middle. | ||
| + | * Now, use same algorithm as in the first section. BOOOM!!! Right? Yep. It's pretty intuitive, these later sections are more like practical uses rather than actual algorithms. I guess this is kind of good since pretty much all computer science problems go back to graph theory. | ||
| + | * Extensions: The structore of bipartite graphs with no perfect matching | ||
| + | * Dicusses how to tell if it is impossible to find a perfect matching in a bipartite graph. | ||
| + | * Can easily find an algorithm based on the following theorem: | ||
| + | * Assume that the bipartite graph G=(V,E) has two sides X and Y such that |X| and |Y|. Then the graph G either has a perfect matching or these is a subset A such that |\G(A)|< | ||
| + | * Section 7.7 - Extensions to Max-Flow Problem | ||
| + | * Circulations with Demands | ||
| + | * We have a directed graph | ||
| + | * edges have capacities | ||
| + | * nodes have supply and demands | ||
| + | * A circulation is a function that satisfies the following | ||
| + | * the flows for a particular edge are always less than the capacity (makes sense and is very intuitive) | ||
| + | * the demand for a node is equal to the stuff coming in minnus the stuff going out (the the demand that his node has? I guess that makes enough sense...) | ||
| + | * To keep the world form exploding, the sum of supplies == sum of demands | ||
| + | * Algorithm: more details on pg 381 and 382 | ||
| + | * Add a new source s and sink t | ||
| + | * For each v with d(v) < 0, add edge (s, v) with capacity -d(v) | ||
| + | * For each v with d(v) > 0, add edge (v, t) with capacity d(v) | ||
| + | * MAGIC: G has circulation iff G' has max flow of value D | ||
| + | * MORE MAGIC : Given (V, E, c, d), there does not exist a circulation iff there exists a node partition (A, B) such that the sum of all the demands is greater than cap(A,B) | ||
| + | * These two pretty much solve the problem for us. more details are on page 382 if you get confused rereading this sometime in the future | ||
| + | * Circulation with Demands and Lower Bounds | ||
| + | * Same givens as last problem | ||
| + | * except!!!! we have lower bounds on each edge | ||
| + | * algorithm on page 384 | ||
| + | * Here's how we do | ||
| + | * we set teh flow along an edge to be the lower bound and then we update the demand on both nodes attached to it. | ||
| + | * the rest is self explanatory | ||
