Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revisionPrevious revision
Next revision
Previous revision
courses:cs211:winter2012:journals:paul:home [2012/04/06 01:54] – [Chapter 7] nguyenpcourses:cs211:winter2012:journals:paul:home [2012/04/06 02:17] (current) – [Chapter 7] nguyenp
Line 776: Line 776:
               * 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)|<|A|. A perfect matching or an appropriate subset A can be found in O(mn) time.                * 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)|<|A|. A perfect matching or an appropriate subset A can be found in O(mn) time. 
   * Section 7.7 - Extensions to Max-Flow Problem   * 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
courses/cs211/winter2012/journals/paul/home.txt · Last modified: 2012/04/06 02:17 by nguyenp
CC Attribution-Noncommercial-Share Alike 4.0 International
Driven by DokuWiki Recent changes RSS feed Valid CSS Valid XHTML 1.0