Question

Augmenting Paths with the Minimum Number of Backward Arcs. Imagine that, in each iteration, the Ford-Fulkerson algorithm selects the aug. menting path in G, with the fewest number of backward arcs. Prove that: (a) The total number of iterations is (mn). (b) The total running time is 0(mn).

m is the number of arcs and n is the number of vertices

Answer 1

soli Given that Ford - fulkerson (G, 8, 6) for each edge [u, v) e E[G] de flu, v] to F [v,u] O while there exits a path of from s to t in the residual network GF do CF (b) + min [Cç (u, v): (u, v) is in p] For do f[u, v] F[u, v7+ cp (6) P [uiv] + p [Liv] we can know that # Þ is the augmenting Path pooofi a) The total number of iterations is o(mn) let say an edge cuiv) in a residual network ;- Cp is critical on an augmenting path $ -> if the residual capacity of p is the residual capacity of (u, v) ie, if Cp (p) = Cp Cu, v) * After we have angemented flow along an augementing Path any critical edge on the path disappears from the residual network

- At least one edge on any augmenting be exti critical. © we will show that each of the mod m R 1... Iml > it can become critical almost In 1/2-) times. * let use ukv be vertices in in that are connected by an edge in E. → since augementing path we shortest path where u, V is critical for the first time, we have Sp(8,0) = $ (3,4)+1 Se (8, 19) is minimum number of edges between (3, 4) in grap Gp once the flow is augmented The edge is disappear from the residual network. It cannot reappear letter on another augementing path until after the flow from u to v is decreased which occur only if (u, ) appear on the augmenting path TE te e' is the flow in G. when this event occur then we have, Spilf, u)- Sp'(8, 6) +/ Since we know Sp (3,1)< Sp'(8,0)

en then Sp! (3, 4) = S: (8,0) +1 38p! (3,)+1 = 8p (8, 4)+2 consequently, from the time (u, v) becomes critical to the time when it next become critical, the distance of u from the source increases by atteast 2 The distance of u from the source is intially atleast o. The intermediate vertices on a shortest path from 8 tou cannot contain S,uort. Therefore until u becomes un reachable from the source if ever its distance is almost Inl -2. Thus, Curv) can become almost (14! -1) times, since there are o cm) pain of vertices that can have an edge b/w them -> If residual graph, the total number of critical edges during the entine excution of Ocmin) proofi 2) The total running time is 0(m²n) come each interation of fork- fulleerson in plemented cm) time when the augmenting path is found by ist search, hence the total rurning time A no. of interations x ocm) in Olm) time when the breath Rigst search,

-> 0(mn)xo (m); no. of interations = o[mn] - 0 (man) thus, total running time is o(mn).