Matching. Slides designed by Kevin Wayne.
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1 Maching Maching. Inpu: undireced graph G = (V, E). M E i a maching if each node appear in a mo edge in M. Max maching: find a max cardinaliy maching. Slide deigned by Kevin Wayne.
2 Biparie Maching Biparie maching. Inpu: undireced, biparie graph G = (L R, E). M E i a maching if each node appear in a mo edge in M. Max maching: find a max cardinaliy maching. ' ' ' maching -', -', -' ' L ' R
3 Biparie Maching Biparie maching. Inpu: undireced, biparie graph G = (L R, E). M E i a maching if each node appear in a mo edge in M. Max maching: find a max cardinaliy maching. ' ' ' max maching -', -', -' -' ' L ' R
4 Biparie Maching Max flow formulaion. Creae digraph G' = (L R {, }, E' ). Direc all edge from L o R, and aign infinie (or uni) capaciy. Add ource, and uni capaciy edge from o each node in L. Add ink, and uni capaciy edge from each node in R o. G' ' ' ' ' L ' R
5 Biparie Maching: Proof of Correcne Theorem. Max cardinaliy maching in G = value of max flow in G'. Pf. Given max maching M of cardinaliy k. Conider flow f ha end uni along each of k pah. f i a flow, and ha cardinaliy k. ' ' ' ' ' ' ' ' G ' ' G'
6 Biparie Maching: Proof of Correcne Theorem. Max cardinaliy maching in G = value of max flow in G'. Pf. Le f be a max flow in G' of value k. Inegraliy heorem k i inegral and can aume f i 0-. Conider M = e of edge from L o R wih f(e) =. each node in L and R paricipae in a mo one edge in M M = k: conider cu (L, R ) ' ' ' ' ' ' ' ' G' ' ' G 6
7 Perfec Maching Def. A maching M E i perfec if each node appear in exacly one edge in M. Q. When doe a biparie graph have a perfec maching? Srucure of biparie graph wih perfec maching. Clearly we mu have L = R. Wha oher condiion are neceary? Wha condiion are ufficien? 7
8 Perfec Maching Noaion. Le S be a ube of node, and le N(S) be he e of node adjacen o node in S. Obervaion. If a biparie graph G = (L R, E), ha a perfec maching, hen N(S) S for all ube S L. Pf. Each node in S ha o be mached o a differen node in N(S). ' ' ' No perfec maching: S = {,, } N(S) = { ', ' }. ' L ' R 8
9 Marriage Theorem Marriage Theorem. [Frobeniu 97, Hall 9] Le G = (L R, E) be a biparie graph wih L = R. Then, G ha a perfec maching iff N(S) S for all ube S L. Pf. Thi wa he previou obervaion. ' ' ' No perfec maching: S = {,, } N(S) = { ', ' }. ' L ' R 9
10 Ford-Fulkeron nework flow algorihm: Augmen along any pah from o a each ep. Wih ineger flow value, he run-ime i O(m * f) where f i he maximum flow in he nework. Edmond-Karp ha beer performance depending only on n and m and no on he maximum flow amoun. I ake ime O(n m ). Each ep uing an O(m) BFS o find an augmening pah. A lea one edge e i auraed (ha flow equal o i capaciy). Thi edge e in he aux. graph increae i diance from he ource. The graph ha m edge whoe diance can range from 0 o n- from he ource. Thu here i a mo n*m ieraion. 0
11 The max. flow f i he ize of a maximum maching o f n. I ake O(m) wor cae ime o find each augmening pah.
12
13
14 7.6 Dijoin Pah
15 Edge Dijoin Pah Dijoin pah problem. Given a digraph G = (V, E) and wo node and, find he max number of edge-dijoin - pah. Def. Two pah are edge-dijoin if hey have no edge in common. Ex: communicaion nework. 6 7
16 Edge Dijoin Pah Dijoin pah problem. Given a digraph G = (V, E) and wo node and, find he max number of edge-dijoin - pah. Def. Two pah are edge-dijoin if hey have no edge in common. Ex: communicaion nework
17 Edge Dijoin Pah Max flow formulaion: aign uni capaciy o every edge. Theorem. Max number edge-dijoin - pah equal max flow value. Pf. Suppoe here are k edge-dijoin pah P,..., P k. Se f(e) = if e paricipae in ome pah P i ; ele e f(e) = 0. Since pah are edge-dijoin, f i a flow of value k. 7
18 Edge Dijoin Pah Max flow formulaion: aign uni capaciy o every edge. Theorem. Max number edge-dijoin - pah equal max flow value. Pf. Suppoe max flow value i k. Inegraliy heorem here exi 0- flow f of value k. Conider edge (, u) wih f(, u) =. by conervaion, here exi an edge (u, v) wih f(u, v) = coninue unil reach, alway chooing a new edge Produce k (no necearily imple) edge-dijoin pah. Or ue BFS. can eliminae cycle o ge imple pah if deired 8
19 Nework Conneciviy Nework conneciviy. Given a digraph G = (V, E) and wo node and, find min number of edge whoe removal diconnec from. Def. A e of edge F E diconnec from if every - pah ue a lea one edge in F
20 Edge Dijoin Pah and Nework Conneciviy Theorem. [Menger 97] The max number of edge-dijoin - pah i equal o he min number of edge whoe removal diconnec from. Pf. Suppoe he removal of F E diconnec from, and F = k. Every - pah ue a lea one edge in F. Hence, he number of edge-dijoin pah i a mo k
21 Dijoin Pah and Nework Conneciviy Theorem. [Menger 97] The max number of edge-dijoin - pah i equal o he min number of edge whoe removal diconnec from. Pf. Suppoe max number of edge-dijoin pah i k. Then max flow value i k. Max-flow min-cu cu (A, B) of capaciy k. Le F be e of edge going from A o B. F = k and diconnec from. A
22 k-regular Biparie Graph Dancing problem. Excluive Ivy league pary aended by n men and n women. Each man know exacly k women; each woman know exacly k men. Acquainance are muual. I i poible o arrange a dance o ha each woman dance wih a differen man ha he know? Mahemaical reformulaion. Doe every k-regular biparie graph have a perfec maching? ' ' Ex. Boolean hypercube. ' ' women ' men
23 k-regular Biparie Graph Have Perfec Maching Theorem. [König 96, Frobeniu 97] Every k-regular biparie graph ha a perfec maching. Pf. Size of max maching = value of max flow in G'. Conider flow: /k if (u, v) E f (u, v) if u or v 0 oherwie f i a flow and i value = n perfec maching. /k ' flow f ' G' ' ' '
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