Max Flows and Minimum Cuts Carlo Mannino (from Geir Dahl notes) - PowerPoint PPT Presentation

Max Flows and Minimum Cuts Carlo Mannino (from Geir Dahl notes) University of Oslo, INF-MAT5360 - Autumn 2011 (Mathematical optimization)

st-flow Given a directed graph D = ( V,E ) ( x , c ) 2 distinct vertices s,t  V (1,1) 1 4 (2,2) (2,3) s source: no edges entering s (1,1) (1,1) s t t sink : no edges outgoing from t (2,3) (2,2) 2 3 Edge capacity function c : E  R + (3,3) An st-flow is a function x : E  R, satisfying     flow conservation constraints x ( e ) x ( e ) ( v V \ { s , t })       e ( v ) e ( v )  x  non-negativity and capacity constraints 0 c Obs: div x ( v ) = 0 for all v  s,t .

Flow value e    val ( x ) x ( e ) div ( s ) VALUE of an st-flow = flow outgoing from s x    ( s ) Theorem (net st-flow through an st-cut) Let  + ( S ) be an st- cut of D = ( V,E) and x be an st -flow. Then   Hajhsas   x ( e ) x ( e ) val ( x ) da       e ( S ) e ( S )    Since s  S and t  S we have div ( v ) div ( s ) val ( x ) x x  v S      x ( e ) x ( e ) div ( v ) The result follows from        v S e ( S ) e ( S ) Corollary (the flow sent by s equals the flow into t ) Let  + ( S ) be a st- cut of D = ( V,E) and x be an st -flow. Then   Hajhsas   val ( x ) x ( e ) div ( t ) da x    e ( t )

The maximum st -flow problem Maximum st-flow problem Given a directed graph D = (V,E) with edge capacity c : E  R + , Hajhsas source s and sink t , find an st-flow of maximum value. da  maximize flow value max x ( e )      e ( s )   flow conservation constraints x ( e ) x ( e ) ( v V \ { s , t })       e ( v ) e ( v )  x  non-negativity and capacity constraints 0 c LP- program: optimum exists. We can apply simplex method. More effective combinatorial algorithms.

Minimum st-cut K 1 Let K =  + ( S ) be an st-cut ( s  S, t  S ) 1 4 2 3 s t CAPACITY of K: cap c ( K ) =  1 1 ( ) c e 3 2  K 2 3 e 3 cap c ( K ) = 6 Minimum st-cut problem Given a directed graph D = ( V , E ) with edge capacity c : E  R + , Hajhsas source s and sink t , find an st-cut with minimum capacity. da Combinatorial optimization problem with linear objective function.

Minimum st-cut Lemma (weak max-flow min-cut property ) The following inequality holds: Hajhsas max{val( x ): x is st-flow }  min{cap c ( K ): K is st-cut } da Let K =  + ( S ) be a (minimum capacity) st-cut ( s  S, t  S )          val ( x ) x ( e ) x ( e ) x ( e ) c ( e ) cap ( K ) c             e ( S ) e ( S ) e ( S ) e ( S )

The max-flow min-cut theorem Theorem (max-flow / min-cut ) The value of a maximum st -flow equals the minimum st-cut capacity: Hajhsas max{val( x ): x is st-flow } = min{cap c ( K ): K is st-cut } da ( l , u ) Let M = min{cap c ( K ): K is st-cut } (0, c e ) D ’ The flow value is no larger than M s t Build graph D’ from D adding edge ( t,s ) Let l ( t,s ) = u ( t,s ) = M ; l ( e ) = 0, u ( e ) = c ( e ) for e  E ( M , M ) We show that D’ , l and u satisfy Hoffman’s conditions     l ( e ) u ( e ) ( S V )       e ( S ) e ( S ) and D’ admits a circulation x’.

Minimum st-cut   Hoffman’s   (0, c e ) l ( e ) u ( e ) ( S V ) D ’ condition       e ( S ) e ( S ) s t       1. If ( t,s )   - ( S ) l(e) 0 c(e) u(e)       δ δ δ e (S) e (S) e (S) e  ( M , M )  l ( e ) M 2. If ( t,s )   - ( S )    ( S ) s  S , t  S K =  + ( S ) is an st -cut      satisfying Hoffman’s condition ( ) ( ) cap ( ) u e c e K M c       e ( S ) e ( S ) There exists a circulation x in D’ with l  x  u  x(t,s) = M      x ( e ) x ( e ) x ( t , s ) M x circulation       e ( s ) e ( s ) The restriction of x to the edges of D provides an st-flow with value M

Augmenting paths D = ( V,E ), capacity c  R E + , st-flow x  R E (1,4) 1 4 (1,2) (2,3) Auxiliary Graph D x = ( V,E x ) (1,1) (0,1) s t = { e  E: x e < u e }  { e -1 : e  E, 0 < x e } E x (1,3) (2,2) 2 3 (2,2) x -AUGMENTING PATH: st-path P in D x Forward edges P + = {e  E : e  E x } 1 4 P Backward edges P - = {e  E : e -1  E x } s t 2 3 D x

The augmenting path theorem Theorem (augmenting path theorem ) An st -flow x is maximum if and only if D x contains no x-augmenting Hajhsas path. da Proof . ( Necessity ) By contradiction. 1 4 Suppose x is maximum and P st-path of D x P Let  + = min{ c (e)- x ( e ): e  P + } > 0 s t Let  - = min{ x ( e ): e  P - } > 0 D x 2 3  = min{  - ,  + } > 0 (1,2) 1 4 (1,2) (2,3) Send  units of flow on P. Obtain flow x’ (1,1) (0,1) s t x ( e ) if e  P (1,3) (2,2) x ( e )+  if e  P + 2 3 x’ ( e ) = (2,2) x ( e )-  if e  P -

The augmenting path theorem Proof . ( Necessity ) (1,4) 1 4 (1,2) (2,3) Send  units of flow on P. Obtain flow x’ (1,1) (0,1) s t x ( e ) if e  P (1,3) (2,2) x ( e )+  if e  P + 2 3 x’ ( e ) = (2,2) x ( e )-  if e  P - 2 2 (1,2) x ' is a feasible st-flow (show it!) 1 4 (1,2) (2,3) (1,1) (0,1) s t Since P is an st-path, we have: 0 (1,3) (2,2) 2 3 2    (2,2)       val ( x ' ) x ' ( e ) x ( e ) x ( e ) val ( x )          e ( s ) e ( s ) e ( s ) and x is not maximum, a contradiction .

The augmenting path theorem Proof . ( Sufficiency ) (2,4) 1 4 (2,2) (2,3) Suppose no st-path in D x (0,1) (0,1) s t Then there is an empty st -cut K =  + ( S ) in D x (2,3) (2,2) 2 3 (i.e. S = { w  V : there is a sw-path in D x } (2,2) If D = ( V,E ) contains an edge ( w,z ) with w  S to z  V\S    x wz = c wz (otherwise ( w,z )  E x ) x ( e ) c ( e )       e ( S ) e ( S ) D D If D = ( V,E ) contains an edge ( z,w ) with w  S to z  V/S   x zw = 0 (otherwise ( z,w ) -1 =( w,z )  E x ) x ( e ) 0    e ( S ) D  + ( S )        val ( x ) x ( e ) x ( e ) c ( e ) cap ( K ) 1 4 c          e ( S ) e ( S ) e ( S ) P s t x is a maximum flow and K is a minimum cut D x 2 3 S ={ s ,1,2,4}

Ford-Fulkerson algorithm 1. Start with the zero flow x = 0 2. Look for an x-augmenting path P in D x 3. if P exists, then find the maximum possible increase  along P and augment x. Goto 2 . 4. if no such P exists then x is a maximum flow. A minimum st -cut is  + (S(x)) where S(x)={ w  V : there is a sw-path in D x }. The algorthm can take C iterations, where C is the maximum capacity of an edge. This is not polynomial time (why?). There are more efficient versions (polynomial in | V | and | E |).

Using LP duality to prove the same things

The dual of max st -flow problem    max -div x ( t ) max x x vt tv       vt ( t ) tv ( t )      x x 0 ( v V \ { s , t }) y v wv wv PRIMAL       wv ( v ) vw ( v ) (MF)   z vw x c ( vw E ) vw vw  0 x  min c vw z vw  E vw       y y z 0 ( vw E , v , w V \ { s , t }) v w vw    y z 0 ( sw E ) w sw     y z 0 ( vs E) v vs DUAL max flow     y z 1 ( tw E) w tw ( DMF )     y z 1 ( vt E) v vt   z e 0 ( e E)

The dual of max st -flow problem  min c vw z vw  E vw       y y z 0 ( vw E , v , w V \ { s , t }) v w vw    y z 0 ( sw E ) w sw     y z 0 ( vs E) v vs ( DMF )     y z 1 ( tw E) w tw     1 ( y z vt E) v vt   z e 0 ( e E) By introducing y s = 0 and y t = -1 we make it less ugly  min c vw z vw  E vw    ( DMF ) y 0 , y 1 s t      y y z 0 ( vw E ) v w vw   0 ( z e e E)

The dual of max st -flow problem  min c vw z vw ( DMF )  E vw    y 0 , y 1 s t      0 ( ) y y z vw E v w vw   z e 0 ( e E) ( y * ,z * ) is an optimal solution to dual if and only if ( y * + 1 ,z * ) is optimal to:  min c vw z vw ( DMF )  E vw   y 1 , y 0 s t      y y z 0 ( vw E ) v w vw   z e 0 ( e E)