Max-Flow: Non-terminating example with irrational capatcities

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1 Max-Flow: Non-terminating example with irrational capatcities
4/22/2019 Max-Flow: Non-terminating example with irrational capatcities Copyright 2000, Kevin Wayne 1

2 Max-Flow Instance G: Φ=(√5-1)⁄2=0.618….. s 2 5 3 4 t 4/22/2019 4 4 4 Φ
Copyright 2000, Kevin Wayne 2

3 Step 1. G: Gf: X X X Flow value = 1 s s 2 2 5 5 3 3 4 4 t t 4/22/2019
4 X 1 4 4 4 4 2 2 Gf: G: 5 5 Φ 1 1 1 1 Φ X 1 1 1 3 3 4 4 4 4 4 4 X 1 4 4 t t Flow value = 1 Copyright 2000, Kevin Wayne 3

4 Step 2. G: Gf: X X X X Φ^2 X Flow value = 1+Φ s s 2 2 5 5 3 3 4 4 t t
4/22/2019 Step 2. s s 2 1 3 2 X Φ 1 4 4 4 2 2 Gf: G: 5 5 Φ 1 1 X X Φ 1 Φ 1 1 X Φ Φ^2 1 3 3 4 4 4 2 2 1 4 3 1 Φ X 2 t t Flow value = 1+Φ Copyright 2000, Kevin Wayne 4

5 Step 3. G: Gf: X 1+Φ Φ X Φ^2 X X Φ Flow value = 1+Φ+Φ s s 2 2 5 5 3 3
4/22/2019 Step 3. s s 2 1 Φ 1 Φ X 2 1 2-Φ 1+Φ 2 2 2 2 Gf: G: 5 5 Φ Φ X Φ 1-Φ Φ 1-Φ Φ 1 Φ^2 1 X Φ 1 3 3 4 4 2 X Φ 2 2 1 1 1 2 Φ Φ 2-Φ t t Flow value = 1+Φ+Φ Copyright 2000, Kevin Wayne 5

6 Step 4. G: Gf: X 1+Φ Φ^2 X 1 X X X Φ X Flow value = 1+Φ+Φ+Φ^2 s s 2 2
4/22/2019 Step 4. s s X 1 2 Φ Φ 1+Φ 2 1-Φ 2-Φ 1+Φ 2 2 2 2 Gf: G: 5 5 Φ^2 Φ Φ X Φ 1 1-Φ Φ 1 1 1 X Φ X 1 3 3 4 2-Φ 4 2 Φ 2 1 1 1 Φ Φ X X 2 Φ 1 2-Φ t t Flow value = 1+Φ+Φ+Φ^2 Copyright 2000, Kevin Wayne 6

7 Flow value = 1+Φ+Φ+Φ^2+Φ^2
4/22/2019 Step 5. s s 1 2 1+Φ 1 2 1+Φ 1-Φ 1 Φ^2 2 X 2 2 2 Gf: G: 5 5 Φ^2 1 Φ 1 1-Φ Φ 2Φ-1 1 Φ X 1 Φ 1-Φ 1 X 3 3 4 2-Φ 4 2 Φ 1+Φ^2 2 1 1 X 1 Φ 1 2 1 1 t t Flow value = 1+Φ+Φ+Φ^2+Φ^2 Copyright 2000, Kevin Wayne 7

8 Flow value = 1+Φ+Φ+Φ^2+Φ^2+Φ^3
4/22/2019 Step 6. s s 1+Φ^3 1 X Φ^2 2 1 1+Φ 2 1-Φ 1 Φ^2 1+Φ 2 2-Φ^2 2 2 Gf: G: 5 Φ 5 Φ^2 X 1 X Φ^3 Φ 1-Φ Φ 1 1 1 1-Φ^3 X Φ 1-Φ 1-Φ^4 3 3 4 2-Φ 4 2 Φ 1+Φ^2 Φ 2-Φ 2 Φ 2 1 1 X 1 1+Φ^3 t t Flow value = 1+Φ+Φ+Φ^2+Φ^2+Φ^3 Copyright 2000, Kevin Wayne 8

9 Flow value = 1+Φ+Φ+Φ^2+Φ^2+Φ^3
4/22/2019 Step 7. s Φ^2 1+Φ^3 1+Φ 1-Φ 1-Φ^3 2-Φ^2 2 Gf: 5 1-Φ^4 1-Φ^3 Φ^4 Φ Φ^3 3 4 2-Φ Φ 2-Φ Φ 1-Φ^3 1+Φ^3 Flow value = 1+Φ+Φ+Φ^2+Φ^2+Φ^3 t Zwick, Uri (21 August 1995). "The smallest networks on which the Ford-Fulkerson maximum flow procedure may fail to terminate". Theoretical Computer Science 148 (1): 165–170. Copyright 2000, Kevin Wayne 9

10 Step N. If Flow(2,4)==1 do Choose Step 5 as the path.
4/22/2019 Step N. If Flow(2,4)==1 do Choose Step 5 as the path. Else if Flow(5,3)==Φ do Choose Step 3 as the path. Else if Flow(2,3)==1 do Choose Step 4(6) as the path. End If Copyright 2000, Kevin Wayne