Jaehyeon
Heo
(Teddy)

Software Engineer
Tutorial 6 question 1 prompt

Initial Residual Network GfG_f for initial ff (0 assignment)

98534965497sbadct
Forward residual capacities equal the original capacities; no backward edges yet.
IterationAugmenting PathBottleneck Δ\DeltaFlow Value
1

sacts \rightarrow a \rightarrow c \rightarrow t

6666
2

sbats \rightarrow b \rightarrow a \rightarrow t

3399
3

sbcts \rightarrow b \rightarrow c \rightarrow t

331212
4

sbdts \rightarrow b \rightarrow d \rightarrow t

221414
5sats \rightarrow a \rightarrow t221616
6sdts \rightarrow d \rightarrow t552121

Interactive Walkthrough

Iteration 1 of 6

Iteration 1

Autoplay paused
36063685349547sbadct
  • Path: s -> a -> c -> t
  • Bottleneck capacity: min(9, 6, 9) = 6
  • Flow value after augmentation: 6

We can now notice that all the edges going into vertex tt has capacity 00, so we know for sure that no more simple path exists from ss to tt.