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Unformatted text preview: ' n "i . hmgto Unhemts mSt Lows Engineering Max Flow Problem
Din/C’s Algorithm Jon Turner
Computer Science & Engineering
Washington University www.ar1.wustl.edu/~jst $Wash1ngton UanCFSIthStlDUIS Engineering Dinic’s Algorithm  Shortest augmenting path algorithm often explores
same “deadends” multiple times >> can avoid some redundant effort by ignoring edges that cannot
possibly appear in augmenting paths of the “current length" I Let f be flow function on G and let R be residual graph >> /eve/(u)=number of edges in a shortest path from s to u in R
>> only edges (u,v) for which Ievel(v)=level(u)+1 can possibly be
on a shortest augmenting path
I Dinic’s algorithm restricts path search to those edges
(u,v) for which res(u,v)>0 and /eve/(v)= /eve/(u)+1
>> this can be done by explicitly constructing a subgraph of R with
only these edges >> or, it can be done by efficiently ignoring other edges during
path searches ;. Washmgton Unwersm 1n Stloum Engineering
I Dinic's algorithm begins by setting f(u,v) to zero for all
edges in G, computes leve/(u) for all vertices u and then repeats the following step as long as /eve/(t)<n >> while there is an augmenting path using edges (u,v) with level
(v)=/eve/(u)+1, select such a path and saturate it >> recalculate [eve/(u) for all u
.The subroutine that finds augmenting paths uses depthfirst search and maintains certain information
across searches >> for each vertex u, maintain a pointer nextedge(u) into
adjacency list for u >> outgoing searches for u always start at nextedge(u) >> pointer is advanced past edges used for unsuccessful searches
from u >> nextedge(u) is reset to the start of the adjacency list every
time level is recalculated :1, Mshmgton UniversuymStloms Engineering Review Questions 1. Draw a flow graph with at least 6 vertices and 12 directed edges. Assign capacities
to each edge and add flow to two source/sink paths. Now, draw the residual graph
for this flow and for each vertex u, write the value of leve/(u) next to u in the
graph. Draw an x through each edge (u,v) for which /eve/(v)=#/eve/(u)+1. The
subgraph of the residual graph defined by the edges that have not been crossed out
is called the level graph. . For each vertex u, assign an ordering to the outgoing edges from u in the level
graph. You can use any ordering you want, but a simple way to do this is to use the
“clockwise” ordering in the drawing of the level graph (that is start with the edge
that is closest to the “12 o’clock” position and proceed in clockwise fashion around
the vertex). For each vertex u, define nextedge(u) to be the first vertex in the
clockwise ordering. . Try to find a path from s to t by following the edges defined by the nextedge
values. That is, starting at 5, follow nextedge(s), and continue until you reach t or a
deadend. If you reach a deadend, then backup to the previous vertex x and
advance nextedge(x) to the next edge in the list of outgoing edges from x, then
resume your search. Continue in this fashion, backing up and adjusting the
nextedge values until you find a path from s to t. Write down the final nextedge
values. :1, Mshmgton UnlverSItymStlouls Engineering Implementing Dinic’s Algorithm class dinic : public augPath {
public:
dinic(Flograph&,int&);
~dinic();
private:
int* nextEdge; // ignore edges before nextEdge[u]
int* level; // level[u]=# of edges in path from source bool findPath(vertex); // find augmenting path
newPhase(); // prepare for a new phase }; dinic::dinic(Flograph& fgl, ints floVal) : augPath(fgl,floVal) {
level  new int[fg—>n()+1];
nextEdge = new edge[fg>n()+1]; floVal = 0; while (newPhase()) {
while (findPath(fg—>src())) floVal += augment(); at nggitpraieps $Msh1ngton UanCFSIthStlDUIS Engineering bool dinic::newphase() ( // Prepare for new phase. Return true if there is a source/sink path.
vertex u,v; edge e;
UiList q(G>n()); for (u = 1; u <= fg—>n(); u++) { level[u] = fg—>n(); nextEdge[u] = fg—>first(u); breadthfirst search
} 0(m) time
q.addLast(fg—>src()); 1evel[fg—>src()] 0; ;;;Z:////F————————_J
while (lq.empty()) { u = q.first(); q.removefirst();
for (e = fg>firstAt(u); e != 0; e = fg>nextAt(u,e)) (
v = fg—>mate(u,e);
if (fg>res(u,e) > 0 && level[v] == fg>n()) {
level[v] = level[u] + 1;
if (v == fg—>snk()) return true; q.addLast(v);
) M } early return prevents
return false; useless exploration of
vertices that are further from source than sink ‘IIIIIIIIIIIIIIIIIIIIII
Engineering E Washington University in St. Louis bool dinic::findpath(vertex u) ( diig;ggst // Find a shortest path with unused residual capacity.
vertex v; edge e; for (e = nextEdge[u]; e != 0; e fg>next(u,e)) {
v = fg>mate(u,e);
if (fg—>res(u,e) ==  1eve1[v] != 1eve1[u] 1) continue;
if (v == fg—>snk()  findpath(v)) {
pEdge[v] = e; nextEdge[u] = e; return true; } ignore ineligible
} edges
nextEdge[u] = 0; return false; return path in pEdge and
update nextEdge on failure, make
nextEdge null to
block future searches :1, Mshmgton UniverSItymStlouis Engineering Analysis of Dinic’s Algorithm l Define a phase of Dinic’s algorithm to be the period
from one calculation of level function to the next  Theorem 8.4. Dinic’s algorithm executes sn—l phases
proofsketch. Level(t) increases by 21 after each phase;
since no path is longer than n—1, there can be at most n—l phases
l Theorem 8.7. Dinic's algorithm spends at most 0(mn)
time per phase and finds a max flow in O(mn2) time Proof. Time per phase is O(mn) excluding time for path searches.
Let N, be number of edges traced by ith top level call to findpath
in a phase. Time for all calls to findpath is proportional to 215,5”, N, A nextedge pointer is advanced for every edge traced in ith top
level call that is not part of path returned by findpath. So pointers
are advanced at least (N,—n) times during the i—th call. Thus, 2lsism(Ni_n) 5 2m and 215/5", N,‘ is So, 0(mn) time per phase and O(mn2) time overall I shi' n nir'rsiIi .1" ' '
\Va ngto U W [l HSthUb Engineering Performance on Hard Cases I For each path length, Dinic does k2 path searches,
spending 0(k) time per search, so 0(k4) altogether I Because augmenting path algorithm uses breadthfirst
search, it reexamines “deadends” on every search :1, Washmgton UniverSItymStlouls Engineering Dinic’s Algorithm on Unit Networks I In a unit network all edges have capacity one and
every vertex other than 5 or t has either a single
entering edge or a single outgoing edge  Theorem 8.5. On unit network, Dinic's algorithm takes
0(m) time/phase and finds max flow in O(mn1/2) time Proof. Time for ith top level call to findpath in a phase is
proportional to number of edges N, that are traced A nextedge pointer is advanced for every edge traced in ith call
that is not part of the path returned by findpath So, if k, is length of ith augmenting path, nextedge pointers are
advanced at least (N,—k,) times during findpath and all k, edges on
path are saturated when flow is added Consequently, 2,k, sm and 2,(N,— k,)52m, so 2,N,s 2m+2,k,s 3m and
thus, number of edges examined in all calls to ﬁndpath during a
phase is 0(m) In :1, Mshington UniversitymStlouis Engineering
Now show there are at most 2[(n2)1/2] phases; suppose the we’ve
completed k phases, and let f be current flow, f* be a max flow
and R be the residual graph for f. The number of remaining phases is at most f*f, so the total
number of phases is at most k+f*f. We can complete the
analysis by finding an upper bound on this expression. Note that f*f is a flow on R, R is a unit network and f*fis zero
or one on every edge. We can partition edges on which f* f is one into a collection of
If *f paths from s to t and possibly some cycles . Since R is unit, any vertex other than s or t can be on at most one
of these paths; hence, there is some augmenting path in R with at
most (n—2)/(f*—f)+1 edges Since we’ve completed k phases, the next augmenting path will
have at least k+1 edges, so k+1 s (n—2)/(f*—f) +1 or f*—f s (n—2)/k
Hence, the total number of phases is sk+(n—2)/k, no matter what
value k has. Choosing k=[(n—2)1/2] yields the desired result. I II \X/ashington University in St. Louis Exercises 1‘ The diagram below shows an intermediate state in the
execution of Dinic's algorithm. In the diagram, the numbers
by the vertices represent the value of the level function The
arcs around each vertex indicate the order of the edges in the
adjacency list, and the heavy dot on each arc shows the
position of the next edge pointer 50, for example, the next
edge at vertex e is (he) Which augmenting path is selected next? see/{ft Show the state of the algorithm after flow is added to the
next augmenting path, Be sure to update the next edge
pointers as appropriate. Engineering The new shite appears below. 0.
. 3,3 2 3
5,3 6,6
' 0
4,4 2'0 ,3 1,1 4'3
0&33ﬁj @H‘QD»;
1 1,0
3,2 2'0 4 1,0 3,2
5,2 2,213 1 2 5
i Explain how Dinic's algorithm can be used to compute a
maximum size matching in a bipartite graph in Omit“)
time We have seen earlier how a maximum size iiiatchiiigﬂwr a bipartite
graph can be computed by solving a maximumﬂow problem. The
flow problem used ta salve the matching problem is a unit network,
Tliereﬂu‘e, [free solve it usingY Diiiie’s algorithm, the rmmiii‘s‘Y time
is 0(mtll/3) by the analysis (in pages 9 and 10. 1’) E \X/ashington University in St. Louis 3, The lefthand diagram below shows a residual graph R for
some ﬂow. The number next to each vertex is its level value
(that is, the number of edges in a shortest path from the
source). The subgraph shown at right excludes the edges tat
are ignored by Dinic’s algorithm and is sometimes called the
low] graph, We can view each phase of Dinic's algorithm as finding a
lrlm‘kiugﬂow in the level graph. A flow is called a blocking
flow, if every sourcetosink path contains some saturated
edge. Find a blocking flow with a total value of 2 in the level
graph, Show that this is not a maximum flow. Iwa mid our uni! ofﬂow to the pull: surf and one unit ofﬂin to
the path sluit, we get 17 liloi‘kingﬂow. Tngct n maximumﬂnw, add one unit oft/law to mull oft/Iv Hm’v
paths sndf, sin" and slu‘t, —
Engineering I? ...
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 Spring '12
 JonTurner

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