# Combinatory analysis by Percy Alexander MacMahon

By Percy Alexander MacMahon

Initially released in 1915-16. This quantity from the Cornell collage Library's print collections used to be scanned on an APT BookScan and switched over to JPG 2000 layout by means of Kirtas applied sciences. All titles scanned disguise to hide and pages may perhaps comprise marks notations and different marginalia found in the unique quantity.

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Extra info for Combinatory analysis

Example text

Therefore, no augmenting path exists. 2. Initializing the Algorithm. The push-relabel algorithm requires a pre ow and a distance labeling before it can get started. To generate a pre ow ll all the arcs leaving s to capacity and set the ows in all the other arcs to zero. 13. An initial distance labeling is given by d(s) = n and d(v ) = 0 for all other nodes v . Notice that any arc from s to a node v is at its capacity, so it is not a residual arc, and therefore we do not require d(s) d(v ) + 1. Also, for any residual arc (v w), d(v ) = 0 0 + 1 = d(w) + 1 so that d is indeed a distance labeling.

2. E cient practical implementation. There are two re nements that may speed up the algorithm by more quickly showing that nodes with positive excess have no paths to the sink. We will describe these two heuristic improvements in the following: (1) Global Relabeling: We can periodically bring the distance labels up to date by performing a breadth- rst-search backward from the sink. Applying this initially and every time the algorithm does cm work, for some constant c > 1 will not a ect the worst-case running time of the algorithm.

De ne the interior of a path to be the path without its end nodes. Consider interiors of paths from nodes with excess to the sink. The interior of such a path has length at least p n ; 2. After the excess ow at v 2 X Y (if any) is moved from v to s or to t along the corresponding paths of f , in the residual graph either in- or out-degree of v is one. From this, it is not hard to see that the path interiors are node-disjoint. Therefore the number p of such paths to t (and hence the residual ow value) is at most nn 2 = O( n).