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courses:cs211:winter2014:journals:haley:chapter7 [2014/04/02 04:50] archermcclellanhcourses:cs211:winter2014:journals:haley:chapter7 [2014/04/02 05:20] (current) archermcclellanh
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     * If all capacities, this algorithm (the Ford-Fulkerson algorithm) terminates in C ( = sum(c_e) for all out edges of s) iterations of the while-loop     * If all capacities, this algorithm (the Ford-Fulkerson algorithm) terminates in C ( = sum(c_e) for all out edges of s) iterations of the while-loop
        * And it can be implemented in O(mC) time        * And it can be implemented in O(mC) time
 +    * Wordy but interesting, 7/10
  
 ===== 7.2: Network Flows & Minimum Cuts ===== ===== 7.2: Network Flows & Minimum Cuts =====
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    * If f is an s-t flow so that the residual graph has no s-t path, there is an s-t cut (_A,_B) in G so that f(v) = c(_A,_B). Then, v(f) is the maximum of all flows in G and (_A,_B) has the minimum capacity of any s-t cut.    * If f is an s-t flow so that the residual graph has no s-t path, there is an s-t cut (_A,_B) in G so that f(v) = c(_A,_B). Then, v(f) is the maximum of all flows in G and (_A,_B) has the minimum capacity of any s-t cut.
    * the Ford-Fulkerson flow is a maximum flow    * the Ford-Fulkerson flow is a maximum flow
 +   * Reminded me of real analysis, straightforward, 7/10
  
 ===== 7.5: Bipartite Matching ===== ===== 7.5: Bipartite Matching =====
  
-   * We wish to +   * We wish to match edges to a subset such that each node appears in max one edge of the edge subset, such that the matching has the largest possible size. 
 +   * We solve this as follows: 
 +       * Look at G, constructing a flow network by directing all edges in G from X to Y, then adding a node s and an (s,x) edge from s to each node in X. We add a node t and an edge (y,t) from every edge in Y to t. We give every edge in the flow network a capacity of 1. Then we compute the maximum s-t flow. 
 +    * The size of the maximum matching in G equals the max flow in the flow network G' and edges carrying flow from X to Y in G' are in the matching in G. 
 +    * We can do this in O(mn) time. 
 +    * What if the graph has no perfect matching? 
 +        * If this //isn't// the case, for all A in side X of G, we have more nodes in A's adjacent nodes than in the A component. 
 +        * Either we can find a perfect matching or an appropriate subset _G in O(mn) time. 
 +    * 4/10
  
-===== 7.7: Maximum Flow & Ford-Fulkerson =====+===== 7.7: Maximum Flow Extensions ===== 
 + 
 +    * What if we have a set of sources and a set of sinks? 
 +    * Consider the case where sources have fixed supply and sinks have fixed demand, like rainway lines shipping from factories to retail outlets. 
 +    * Now the circulation fulfills 
 +        * for each edge e, 0<=f(e)<=c_e 
 +        * for each vertex v, the in minus out flows equal the demand of v. 
 +    * We can also do this with circulations that have lower bounds for demand. 
 +    * Boring and wordy, 2/10
courses/cs211/winter2014/journals/haley/chapter7.1396414248.txt.gz · Last modified: 2014/04/02 04:50 by archermcclellanh
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