1 The basic circulation problem

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1 2WO08: Graphs and Algorihms Lecure 4 Dae: 26/2/2012 Insrucor: Nikhil Bansal The Circulaion Problem Scribe: Tom Slenders 1 The basic circulaion problem We will consider he max-flow problem again, bu his ime here will be no source and sink. Each verex v has a demand d(v), which is more or less he flow ha should ener ha verex. If he demand is negaive ha amoun should leave he verex. A bi more concree we can wrie i like his: Definiion 1 d(v) := e N (v) f(e) e N + (v) f(e) We can say ha a verex v is a semi-sink if d(v) < 0 and a semi-source if d(v) > 0. Noe ha we can rewrie he original max-flow problem as a circulaion problem as follows: d(s) = D (his is our original source) d() = D (his is our original sink) d(v) = 0 v s, Theorem 2 Solving if a circulaion problem of a graph G is feasible and is he same as solving a max-flow problem for a graph G. Proof: We noe ha a necessary condiion for a flow o exis is: v:d(v)<0 d(v) = v:d(v)>0 d(v) =: D (1) Now we add wo exra verices o he graph G: s and. s is a source ha has an ougoing edge o every verex v ha has d(v) < 0, also he edge has a capaciy of d(v). is a sink ha has an incoming edge from every verex v ha has d(v) > 0, also he edge has a capaciy of d(v). For an example see Figure 1. Claim 3 A circulaion problem is feasible if and only if he max-flow value is D We see ha here is a cu separaing s from all oher verices, his cu has capaciy D. If all hese edges are using full capaciy, hen also all edges going o mus have full capaciy. Therefor all oher verices have heir demand me. This also concludes he proof of Theorem 2. We also wan o solve he minimal cos for a circulaion problem if edges have cos, also known as he minimum cos circulaion problem. Theorem 4 If (1) is me and here is no cycle wih infinie capaciy edges s.. w(cycle) 0 hen here exiss a polynomial algorihm ha finds a soluion for he minimum cos circulaion problem. The proof was omied during he lecure. For a proof see [2] 1

2 c=9 d=-9 s c=4 d=-4 d=1 c=7 d=7 d=5 c=5 Figure 1: By adding he verices s and we can rewrie a circulaion problem ino a max-flow problem. 2 lower bounds We will now add some exra resricions o he circulaion problem. For every edge we now inroduce a lower bound on he circulaion flow. For every edge e we have a lower bound l(e) and he following mus hold for every edge: l(e) f(e) c(e). Noe ha we included he resricion for capaciy as well. Theorem 5 Solving he circulaion problem wih lower bounds is feasible and is he same as solving he basic circulaion problem. Proof: For every edge e wih a lower bound going from verex u o verex v we do he following: We add a verex u and an edge going from u o u, also d(u ) = l(e). We add a verex v and an edge going from v o v, also d(v ) = l(e). We change he capaciy of edge e o c(e) l(e). This is illusraed in figure 2. Now we have reduced every edge o an edge wihou lower bound and if only if he demands in u and v can be me, he lower bound circulaion problem can be solved. 2.1 Example: min-cos perfec maching We wan o creae a perfec maching wih minimal coss. We can reduce his o a circulaion problem. We add a source s and sink as in figure 3. The newly creaed edges all have capaciy and lower bound 1. The edges in he original graph have capaciy 1 and no lower bound. 2.2 Example: Rounding in a marix Consider a marix A wih real values. We denoe he sum of each row o be R i := j (a ij) and he sum of each column o be C j := i (a ij). We wan o round all values in he marix, for each 2

3 d(v )=-l(e) v u c(e)-l(e) v u d(u )=l(e) Figure 2: The change o every edge wih a lower bound as in proof 5. s Figure 3: A represenaion of how o solve a min-cos perfec maching by solving a max-flow. 3

4 Rows Columns c = R i, l = R i c = C j, l = C j s c = (a ij ), l = (a ij ) Figure 4: The graph creaed from he rounding in a marix problem. value we can choose o round i up or down. We denoe hese new values in he marix B, so (b ij ) = (a ij ) or (b ij ) = (a ij ). Also we wan o round all C j and R i up or down, such ha for every j: i (b ij) = C j or i (b ij) = C j and also for every i: j (b ij) = R i or j (b ij) = R i. We wan o know if such a marix B exiss and also how o find one. We creae a graph as in figure 4. For every row we creae a verex, for every column we creae a verex and we creae a source verex and a sink verex. Now we draw edges from he source o he row verices, hese edges have a capaciy of R i and a lower bound of R i. We also draw edges from he column verices o he sink, hese edges have a capaciy of C j and a lower bound of C j. Lasly we draw a verex from every row verex i o every column verex j, hese edges have a capaciy of (a ij ) and a lower bound of (a ij ). If we can solve his circulaion problem, we have found a soluion o our problem, since if we solve a circulaion problem wih only ineger values, our circulaion algorihm will reurn an ineger only soluion. We know ha a non-ineger soluion exiss, because we can simply fill in he original values of he (a ij ). This means we will always find an ineger soluion o he saed circulaion problem. 3 Chinese Posman Tour For he Chinese Posman Tour (CPT) we wan o visi all edges of a parially direced graph Γ a leas once by doing a our. Every edge has a cos d and we wan o minimize he oal cos of all edges visied. 4

5 Figure 5: A represenaion of he edges for a dig sie. The op layer mus be dug firs before he lower pars can be reached. We firs noe ha any soluion is a leas (i,j) E d i,j, since we have o visi every edge a leas once. If he graph is Eulerian (his means v : d + (v) = d (v) or in words: The in-degree is he same as he ou-degree), we can simply find an Euler roue and visi every edge exacly once. Finding his Euler roue can be done by saring wih some cycle and adding cycles of unvisied edges. Solving his problem wih boh undireced and direced edges is known o be NP-complee, hough wih only undireced eges i can be solved in polynomial ime, see [3][1]. 4 Projec managemen In projec managemen i is ofen necessary o disribue he way money or ime is spen. To gain money or ime one ofen has o inves o learn cerain skills before one ges (hopefully more) money in reurn by doing a job ha requires hose skills. We can model his problem wih graphs as follows. Every verex v represens a projec ha can be done and has an associaed revenue p(v), possibly negaive. Every (direced) edge represens he prerequisies of a verex, an incoming arrow means ha verex can only be done if he verex from which he edge comes is also done. For example we can look a a mining sie where one can only dig for he lowes pars of he mine if he pars on op are dug firs. The corresponding edges are shown in Figure 5. Some of he lower pars migh be very valuable, bu also need a lo of digging before hey can be reached. We wonder how we can maximize he profi for he Projec managemen problem. Noe ha we may assume ha here are no cycles in our graph G, since a cycle means ha all verices in he cycle mus all be done or none. We could simply represen his as one verex. We creae he following graph G : We use he same verices as in G and every prerequisies edge e is added wih c(e) = inf. We add wo verices: a source s wih ougoing edges o all verices v for which p(v) 0, he capaciy for each edge is p(v), a sink wih incoming edges from all verices v for which p(v) < 0, he capaciy for each edge is p(v). Now we search for a min-cu, he verices ha are on he same side of he cu as s are he projecs ha should be done o maximize profi, we shall call his se S, he oher verices will be called T. See Figure 6. 5

6 s Figure 6: An example of a min-cu o find he lowes projec coss. Noe ha he blue and red arrows are arrows wih infinie capaciy, bu ha only he blue arrows are allowed. The verices in he ligh blue are are in se S, he ohers are in T. We now denoe P (S) he posiive profi ha lies in S, P (T ) he posiive profi ha lies in T, N(S) he negaive profi in S and N(T ) he negaive profi ha lies in T. Now we see we wan o solve max(p (S) N(S)). This also means we wan o solve max(p (S) N + N(T ) wih N = N(T )+N(S) a consan. This means we wan o solve max P (S)+N(T ) = P P (T )+N N(S) wih P = P (S) + P (T ) a consan. Now we wan o solve min(p (T ) + N(S)) and we can see ha his is acually he same as finding a min-cu for our graph G. Hence finding a min-cu in G resuls in an opimal choice S o maximize he oal profi. References [1] Jack Edmonds & Ellis L. Johnson. Maching, euler ours and he chinese posman. Mahemaical Programming, 5:88 124, [2] Éva Tardos. A srongly polynomial minimum cos circulaion algorihm. Combinaorica, 5: , [3] NIST. Chinese posman problem. hp://xlinux.nis.gov/dads//html/chineseposman. hml, May

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