Post on 25-Jun-2020
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
The Push-Relabel AlgorithmA Better Solution to the Max-Flow Problem
Hao Lu Yu Shi Boyuan Kong Jiefeng Chen
Department of Computer ScienceShanghai Jiao Tong University
2015–11–24
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Outline
1 A Consice Review of the Max-Flow ProblemMax-Flow Problem and Ford-Fulkerson MethodAn example where FF never terminatesImprovements by Specifying the Order of Path Choice
2 Generic Push-Relabel AlgorithmConcepts and Basic OperationsAn Intuitive Example
3 Push-Relabel Algorithm AnalysisCorrectness of Push-Relabel AlgorithmRunning Time Analysis of Push-Relabel Algorithm
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Outline
1 A Consice Review of the Max-Flow ProblemMax-Flow Problem and Ford-Fulkerson MethodAn example where FF never terminatesImprovements by Specifying the Order of Path Choice
2 Generic Push-Relabel AlgorithmConcepts and Basic OperationsAn Intuitive Example
3 Push-Relabel Algorithm AnalysisCorrectness of Push-Relabel AlgorithmRunning Time Analysis of Push-Relabel Algorithm
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Outline
1 A Consice Review of the Max-Flow ProblemMax-Flow Problem and Ford-Fulkerson MethodAn example where FF never terminatesImprovements by Specifying the Order of Path Choice
2 Generic Push-Relabel AlgorithmConcepts and Basic OperationsAn Intuitive Example
3 Push-Relabel Algorithm AnalysisCorrectness of Push-Relabel AlgorithmRunning Time Analysis of Push-Relabel Algorithm
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Basics
Max-Flow Problem and Ford-Fulkerson Method
• G = (V ,E , c), c : V × V 7→ R+
• Find a flow that maximize val(f ) =∑
v∈V f (s, v)
• Fold-Fulkerson Method
Ford-Fulkerson-Method(G , s, t)
1 f ← 02 while there exists an augmenting path p
from s to t in Gf
3 f ← f + fp4 return f
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Basics
Max-Flow Problem and Ford-Fulkerson Method
• G = (V ,E , c), c : V × V 7→ R+
• Find a flow that maximize val(f ) =∑
v∈V f (s, v)
• Fold-Fulkerson Method
Ford-Fulkerson-Method(G , s, t)
1 f ← 02 while there exists an augmenting path p
from s to t in Gf
3 f ← f + fp4 return f
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Non-termination of FF
Initialization
v2v1
s
v3 v4
t
MM
M
e1, r0 e3, r
0 e2, r1
MM
M
• r =√5−12 , r2 + r1 = r0, rn+2 + rn+1 = rn
• M ≥ 1 + r
• p ′ = s → v2 → v3 → t
• p0 = s → v4 → v3 → v2 → v1 → t
• p1 = s → v2 → v3 → v4 → t
• p2 = s → v1 → v2 → v3 → t
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Non-termination of FF
Step 0–4
v2v1
s
v3 v4
t
MM
M
r0 r1r0
MM
M
• p ′ = s → v2 → v3 → t, send flow: r0
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r1
• p1 = s → v2 → v3 → v4 → t, send flow: r1
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r2
• p2 = s → v1 → v2 → v3 → t, send flow: r2
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Non-termination of FF
Step 0–4
v2v1
s
v3 v4
t
MM
M
r0 r1
r0
MM
M
• p ′ = s → v2 → v3 → t, send flow: r0
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r1
• p1 = s → v2 → v3 → v4 → t, send flow: r1
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r2
• p2 = s → v1 → v2 → v3 → t, send flow: r2
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Non-termination of FF
Step 0–4
v2v1
s
v3 v4
t
MM
M
r0 r1
r0
MM
M
• p ′ = s → v2 → v3 → t, send flow: r0
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r1
• p1 = s → v2 → v3 → v4 → t, send flow: r1
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r2
• p2 = s → v1 → v2 → v3 → t, send flow: r2
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Non-termination of FF
Step 0–4
v2v1
s
v3 v4
t
MM
M
r0 − r1
r1 r1
r1
r0 − r1
MM
M
• p ′ = s → v2 → v3 → t, send flow: r0
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r1
• p1 = s → v2 → v3 → v4 → t, send flow: r1
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r2
• p2 = s → v1 → v2 → v3 → t, send flow: r2
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Non-termination of FF
Step 0–4
v2v1
s
v3 v4
t
MM
M
r0 − r1
r1 r1
r1
r0 − r1
MM
M
• p ′ = s → v2 → v3 → t, send flow: r0
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r1
• p1 = s → v2 → v3 → v4 → t, send flow: r1
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r2
• p2 = s → v1 → v2 → v3 → t, send flow: r2
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Non-termination of FF
Step 0–4
v2v1
s
v3 v4
t
MM
M
r0 − r1
r1
r1
r0
MM
M
• p ′ = s → v2 → v3 → t, send flow: r0
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r1
• p1 = s → v2 → v3 → v4 → t, send flow: r1
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r2
• p2 = s → v1 → v2 → v3 → t, send flow: r2
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Non-termination of FF
Step 0–4
v2v1
s
v3 v4
t
MM
M
r2
r1
r1
r0
MM
M
• p ′ = s → v2 → v3 → t, send flow: r0
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r1
• p1 = s → v2 → v3 → v4 → t, send flow: r1
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r2
• p2 = s → v1 → v2 → v3 → t, send flow: r2
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Non-termination of FF
Step 0–4
v2v1
s
v3 v4
t
MM
M
r0
r3
r1 − r3
r2
r0 − r2
MM
M
• p ′ = s → v2 → v3 → t, send flow: r0
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r1
• p1 = s → v2 → v3 → v4 → t, send flow: r1
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r2
• p2 = s → v1 → v2 → v3 → t, send flow: r2
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Non-termination of FF
Step 0–4
v2v1
s
v3 v4
t
MM
M
r0
r3
r1 − r3
r2
r0 − r2
MM
M
• p ′ = s → v2 → v3 → t, send flow: r0
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r1
• p1 = s → v2 → v3 → v4 → t, send flow: r1
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r2
• p2 = s → v1 → v2 → v3 → t, send flow: r2
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Non-termination of FF
Step 0–4
v2v1
s
v3 v4
t
MM
M
r2
r0 − r2
r3
r1 − r3r0
MM
M
• p ′ = s → v2 → v3 → t, send flow: r0
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r1
• p1 = s → v2 → v3 → v4 → t, send flow: r1
• p0 = s → v4 → v3 → v2 → v1 → t, send flow: r2
• p2 = s → v1 → v2 → v3 → t, send flow: r2
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Non-termination of FF
Iteration
v2v1
s
v3 v4
t
MM
M
r0 r1
r0
MM
M
v2v1
s
v3 v4
t
MM
M
r2
r0−r2
r3
r1−r3r0
MM
M
• After Step 4n, e1 : r2n, e2 : r
2n+1, e3 : r0
• Send flow: r0 + 2∑∞
i=1 ri = 1 + 2 · r
1−r = 3 + 2r
• Max flow: 2M + 1
• If M > 1 + r , it even not converges to the max-flow’s value!
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Non-termination of FF
Iteration
v2v1
s
v3 v4
t
MM
M
r0 r1
r0
MM
M
v2v1
s
v3 v4
t
MM
M
r2
r0−r2
r3
r1−r3r0
MM
M
• After Step 4n, e1 : r2n, e2 : r
2n+1, e3 : r0
• Send flow: r0 + 2∑∞
i=1 ri = 1 + 2 · r
1−r = 3 + 2r
• Max flow: 2M + 1
• If M > 1 + r , it even not converges to the max-flow’s value!
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Non-termination of FF
Iteration
v2v1
s
v3 v4
t
MM
M
r0 r1
r0
MM
M
v2v1
s
v3 v4
t
MM
M
r2
r0−r2
r3
r1−r3r0
MM
M
• After Step 4n, e1 : r2n, e2 : r
2n+1, e3 : r0
• Send flow: r0 + 2∑∞
i=1 ri = 1 + 2 · r
1−r = 3 + 2r
• Max flow: 2M + 1
• If M > 1 + r , it even not converges to the max-flow’s value!
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Non-termination of FF
Iteration
v2v1
s
v3 v4
t
MM
M
r0 r1
r0
MM
M
v2v1
s
v3 v4
t
MM
M
r2
r0−r2
r3
r1−r3r0
MM
M
• After Step 4n, e1 : r2n, e2 : r
2n+1, e3 : r0
• Send flow: r0 + 2∑∞
i=1 ri = 1 + 2 · r
1−r = 3 + 2r
• Max flow: 2M + 1
• If M > 1 + r , it even not converges to the max-flow’s value!
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Non-termination of FF
Iteration
v2v1
s
v3 v4
t
MM
M
r0 r1
r0
MM
M
v2v1
s
v3 v4
t
MM
M
r2
r0−r2
r3
r1−r3r0
MM
M
• After Step 4n, e1 : r2n, e2 : r
2n+1, e3 : r0
• Send flow: r0 + 2∑∞
i=1 ri = 1 + 2 · r
1−r = 3 + 2r
• Max flow: 2M + 1
• If M > 1 + r , it even not converges to the max-flow’s value!
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Non-termination of FF
Iteration
v2v1
s
v3 v4
t
MM
M
r0 r1
r0
MM
M
v2v1
s
v3 v4
t
MM
M
r2
r0−r2
r3
r1−r3r0
MM
M
• After Step 4n, e1 : r2n, e2 : r
2n+1, e3 : r0
• Send flow: r0 + 2∑∞
i=1 ri = 1 + 2 · r
1−r = 3 + 2r
• Max flow: 2M + 1
• If M > 1 + r , it even not converges to the max-flow’s value!
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Improvements
Several Improvement Ideas
• Some thought: Select one path with the largest capacity
• Edmonds Karp: Select one shortest path
• · · ·
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Basics
Preflow
s
u
v
t
1/2
1/1
1/1
1/3
• It’s a flow!
• It’s not a flow, but apreflow!
• e(u) = 2 − 1 = 1
• Overflowing: e(u) > 0
• A preflow is a flow where:• inflow(u) > outflow(u) is allowed• Flow conservation is relaxed
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Basics
Preflow
s
u
v
t
2/2
1/1
1/1
1/3
• It’s a flow!
• It’s not a flow, but apreflow!
• e(u) = 2 − 1 = 1
• Overflowing: e(u) > 0
• A preflow is a flow where:• inflow(u) > outflow(u) is allowed• Flow conservation is relaxed
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Basics
Preflow
s
u
v
t
2/2
1/1
1/1
1/3
• It’s a flow!
• It’s not a flow, but apreflow!
• e(u) = 2 − 1 = 1
• Overflowing: e(u) > 0
• A preflow is a flow where:• inflow(u) > outflow(u) is allowed• Flow conservation is relaxed
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Basics
Preflow
s
u
v
t
3/2
1/1
1/1
2/3
• It’s a flow!
• It’s not a flow, but apreflow!
• e(u) = 2 − 1 = 1
• Overflowing: e(u) > 0
• A preflow is a flow where:• inflow(u) > outflow(u) is allowed• Flow conservation is relaxed
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Basics
Preflow
s
u
v
t
3/2
1/1
1/1
2/3
• It’s a flow!
• It’s not a flow, but apreflow!
• e(u) = 2 − 1 = 1
• Overflowing: e(u) > 0
• A preflow is a flow where:• inflow(u) > outflow(u) is allowed• Flow conservation is relaxed
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Basics
Height Function
0
1
2
3
u
v1
v2
v3
v4
× A height function h : V 7→ N• h(t) = 0, h(s) = |V |• h(v) ≤ h(u) + 1,∀ 〈v , u〉 ∈ Ef
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Basics
Height Function
0
1
2
3
u
v1
v2
v3
v4
× A height function h : V 7→ N• h(t) = 0, h(s) = |V |• h(v) ≤ h(u) + 1,∀ 〈v , u〉 ∈ Ef
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Basics
The Push Operation
0
1
2
3
u
v1
v2 v
3/4
2/2
3/4
• u is overflowing
• cf (u, v) > 0
• h(u) = h(v) + 1
• Height function maintains
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Basics
The Push Operation
0
1
2
3
u
v1
v2 v
A saturating push
3/4
2/2
4/4
• u is overflowing
• cf (u, v) > 0
• h(u) = h(v) + 1
• Height function maintains
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Basics
The Push Operation
0
1
2
3
u
v1
v2 v
A saturating push
3/4
2/2
4/4
• u is overflowing
• cf (u, v) > 0
• h(u) = h(v) + 1
• Height function maintains
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Basics
The Push Operation
0
1
2
3
u
v1
v2 v
3/4
0/2
0/4
• u is overflowing
• cf (u, v) > 0
• h(u) = h(v) + 1
• Height function maintains
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Basics
The Push Operation
0
1
2
3
A non-saturating push
u
v1
v2 v
3/4
0/2
3/4
• u is overflowing
• cf (u, v) > 0
• h(u) = h(v) + 1
• Height function maintains
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Basics
The Relabel Operation
0
1
2
3
u
v1
v2
v3
• u is overflowing
• h(u) ≤ h(v),∀ 〈u, v〉 ∈ Ef
• Height function maintains
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Basics
The Relabel Operation
0
1
2
3
u
v1
v2
v3
• u is overflowing
• h(u) ≤ h(v),∀ 〈u, v〉 ∈ Ef
• Height function maintains
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Basics
The Relabel Operation
0
1
2
3
u
v1
v2
v3
• u is overflowing
• h(u) ≤ h(v),∀ 〈u, v〉 ∈ Ef
• Height function maintains
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Basics
The Relabel Operation
0
1
2
3
u v1
v2
v3
• u is overflowing
• h(u) ≤ h(v),∀ 〈u, v〉 ∈ Ef
• Height function maintains
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Basics
Generic Algorithm
Push-Relabel(G )
1 h(u) = 0 except h(s) = |V |2 f (u, v) = 0 except f (s, v) = c(s, v),∀v connected to s3 while there exists an applicable push or relabel operation4 do it5 return f
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Example
Example
s+0 0
v2+0 0
v1+0 0
v3+0 0
t+0 0
12
14
5
16
8
7
10
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Example
Example
s−26 5
v2+14 0
v1+12 0
v3+0 0
t+0 0
12
14
5
16
8
7
10
• Initial the preflow
• Relabel v1
• Push 12 from v1 to t
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Example
Example
s−26 5
v2+14 0
v1+12 1
v3+0 0
t+0 0
12
14
5
16
8
7
10
• Initial the preflow
• Relabel v1
• Push 12 from v1 to t
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Example
Example
s−26 5
v1+0 1
v2+14 0
v3+0 0
t+12 0
12
14
5
4
12
8
7
10
• Initial the preflow
• Relabel v1
• Push 12 from v1 to t
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Example
Example
s−26 5
v1+0 1
v2+14 1
v3+0 0
t+12 0
12
14
5
4
12
8
7
10
• Relabel v2
• Push 8 from v2 to v3
• Relabel v3
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Example
Example
s−26 5
v1+0 1
v2+6 1
v3+8 0
t+12 0
12
14
5
4
12
8
7
10
• Relabel v2
• Push 8 from v2 to v3
• Relabel v3
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Example
Example
s−26 5
v1+0 1
v2+6 1
v3+8 1
t+12 0
12
14
5
4
12
8
7
10
• Relabel v2
• Push 8 from v2 to v3
• Relabel v3
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Example
Example
s−26 5
v1+0 1
v2+6 1
v3+0 1
t+20 0
12
14
5
4
12
8
7
2
8
• Push 8 from v3 to t
• Relabel v2
• Push 6 from v2 to s
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Example
Example
s−26 5
v1+0 1
v2+6 6
v3+0 1
t+20 0
12
14
5
4
12
8
7
2
8
• Push 8 from v3 to t
• Relabel v2
• Push 6 from v2 to s
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Example
Example
s−20 5
v1+0 1
v2+0 6
v3+0 1
t+20 0
12
6
8
5
4
12
8
7
2
8
• Push 8 from v3 to t
• Relabel v2
• Push 6 from v2 to s
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Correctness
Termination with a Flow
Lemma (Termination with a flow)
G , f , h, u is overflowing. Then either a push or a relabel applies to it.
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Correctness
Termination with a Flow
Lemma (Termination with a flow)
G , f , h, u is overflowing. Then either a push or a relabel applies to it.
0
1
2
3
u v1
v2
v3
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Correctness
Termination with a Flow
Lemma (Termination with a flow)
G , f , h, u is overflowing. Then either a push or a relabel applies to it.
0
1
2
3
u
v1
v2
v3
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Correctness
Non-Existence of s ; t Path
Lemma
G , f , h. There is no path from s to t in Gf .
|V |
≥|V |−1
≥|V |−2 · · ·
0
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Correctness
Non-Existence of s ; t Path
Lemma
G , f , h. There is no path from s to t in Gf .
|V |
≥|V |−1
≥|V |−2 · · ·
0
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Correctness
Correctness
Theorem
Push-Relabel Algorithm generates a max-flow.
Proof.
• Termination with a flow.
• Non-existence of s ; t path.
• Termination with a max-flow.
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Correctness
Correctness
Theorem
Push-Relabel Algorithm generates a max-flow.
Proof.
• Termination with a flow.
• Non-existence of s ; t path.
• Termination with a max-flow.
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Running Time
Excess Flow Can Return
Lemma
G , f . For any overflowing vertex w , there is a simple path w ; s in Gf .
Proof.
Yes, or WHERE did the flow come from!
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Running Time
Excess Flow Can Return
Lemma
G , f . For any overflowing vertex w , there is a simple path w ; s in Gf .
Proof.
Yes, or WHERE did the flow come from!
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Running Time
Bound on Heights
Lemma (Bound on heights)
∀v ∈ V : h(v) ≤ 2|V |− 1
Proof.
• Heights only increase during relabel operations.
• Relabel operations only applies to overflowing vertices.
• Overflowing vertices are connected to s.
• h(v) − h(s) ≤ |V |− 1
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Running Time
Bound on Heights
Lemma (Bound on heights)
∀v ∈ V : h(v) ≤ 2|V |− 1
Proof.
• Heights only increase during relabel operations.
• Relabel operations only applies to overflowing vertices.
• Overflowing vertices are connected to s.
• h(v) − h(s) ≤ |V |− 1
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Running Time
Bound on Relabel Operations
Lemma (Bound on relabel operations)
# Relabel operations = O(|V |2)
Proof.
• A relabel operation increases h(u) by at least 1.
• h(u) ≤ 2|V |− 1
• (|V |− 1) · (2|V |− 1) = O(|V |2)
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Running Time
Bound on Relabel Operations
Lemma (Bound on relabel operations)
# Relabel operations = O(|V |2)
Proof.
• A relabel operation increases h(u) by at least 1.
• h(u) ≤ 2|V |− 1
• (|V |− 1) · (2|V |− 1) = O(|V |2)
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Running Time
Bound on Saturating Pushes
Lemma (Bound on saturating pushes)
# Saturating pushes = O(|V ||E |)
Proof.•• (|E |) · (2|V |− 1) = O(|V ||E |)
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Running Time
Bound on Saturating Pushes
Lemma (Bound on saturating pushes)
# Saturating pushes = O(|V ||E |)
Proof.
•0
1
2
u
v1
• (|E |) · (2|V |− 1) = O(|V ||E |)
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Running Time
Bound on Saturating Pushes
Lemma (Bound on saturating pushes)
# Saturating pushes = O(|V ||E |)
Proof.
•0
1
2
u
v1
• (|E |) · (2|V |− 1) = O(|V ||E |)
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Running Time
Bound on Saturating Pushes
Lemma (Bound on saturating pushes)
# Saturating pushes = O(|V ||E |)
Proof.
• 0
1
2
u
v1
• (|E |) · (2|V |− 1) = O(|V ||E |)
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Running Time
Bound on Non-saturating Pushes
Lemma (Bound on non-saturating pushes)
# Non-saturating pushes = O(|V |2|E |)
Proof.
Define Φ =∑
v : e(v)>0 h(v)
• A relabel operation increases Φ by at most 2|V |− 1.
• A saturating push operation increases Φ by at most 2|V |− 1.
• A non-saturating push operation decreases Φ by at least 1: u will nolonger be overflowing, v may become overflowing, h(u) = h(v) + 1
• (2|V |− 1) · (O(|V |2) + O(|V ||E |)) = O(|V |2|E |)
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Running Time
Bound on Non-saturating Pushes
Lemma (Bound on non-saturating pushes)
# Non-saturating pushes = O(|V |2|E |)
Proof.
Define Φ =∑
v : e(v)>0 h(v)
• A relabel operation increases Φ by at most 2|V |− 1.
• A saturating push operation increases Φ by at most 2|V |− 1.
• A non-saturating push operation decreases Φ by at least 1: u will nolonger be overflowing, v may become overflowing, h(u) = h(v) + 1
• (2|V |− 1) · (O(|V |2) + O(|V ||E |)) = O(|V |2|E |)
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Running Time
Bound on Non-saturating Pushes
Lemma (Bound on non-saturating pushes)
# Non-saturating pushes = O(|V |2|E |)
Proof.
Define Φ =∑
v : e(v)>0 h(v)
• A relabel operation increases Φ by at most 2|V |− 1.
• A saturating push operation increases Φ by at most 2|V |− 1.
• A non-saturating push operation decreases Φ by at least 1: u will nolonger be overflowing, v may become overflowing, h(u) = h(v) + 1
• (2|V |− 1) · (O(|V |2) + O(|V ||E |)) = O(|V |2|E |)
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Running Time
Bound on Non-saturating Pushes
Lemma (Bound on non-saturating pushes)
# Non-saturating pushes = O(|V |2|E |)
Proof.
Define Φ =∑
v : e(v)>0 h(v)
• A relabel operation increases Φ by at most 2|V |− 1.
• A saturating push operation increases Φ by at most 2|V |− 1.
• A non-saturating push operation decreases Φ by at least 1: u will nolonger be overflowing, v may become overflowing, h(u) = h(v) + 1
• (2|V |− 1) · (O(|V |2) + O(|V ||E |)) = O(|V |2|E |)
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Running Time
Running Time
Theorem (Running time)
There is an implementation of Push-Relabel Algorithm whose runningtime is O(|V |2|E |).
Proof.
There is an implementation which satisfies:
• Every push takes O(1) time, totalling O(|V |2|E |).• Every relabel takes O(|V |) time, totalling O(|V |3).? Actually, relabel operations take O(|V ||E |).
There is also an implementation, called Relabel-To-FrontAlgorithm, whose running time is O(|V |3).
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Running Time
Running Time
Theorem (Running time)
There is an implementation of Push-Relabel Algorithm whose runningtime is O(|V |2|E |).
Proof.
There is an implementation which satisfies:
• Every push takes O(1) time, totalling O(|V |2|E |).• Every relabel takes O(|V |) time, totalling O(|V |3).? Actually, relabel operations take O(|V ||E |).
There is also an implementation, called Relabel-To-FrontAlgorithm, whose running time is O(|V |3).
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Running Time
Running Time
Theorem (Running time)
There is an implementation of Push-Relabel Algorithm whose runningtime is O(|V |2|E |).
Proof.
There is an implementation which satisfies:
• Every push takes O(1) time, totalling O(|V |2|E |).• Every relabel takes O(|V |) time, totalling O(|V |3).? Actually, relabel operations take O(|V ||E |).
There is also an implementation, called Relabel-To-FrontAlgorithm, whose running time is O(|V |3).
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Summary
• Augmenting paths: O(|V ||E |2)• Preflows: O(|V |3), O(|V ||E | log|E |/(|V | lg |V |)|V |)• Blocking flows: O(min(|V |2/3, |E |1/2)|E | lg(|V |2/|E |+ 2) lgC )
• A 2013 algorithm by Orlin achieves O(|V ||E |)
Slides by Hao Lu
Max-Flow Review Push-Relabel Algorithm Analysis Summary
Bibliography
Advanced Algorithms, Spring 11, Course Website.
Thomas H. Cormen, Charles E Leiserson, Ronald L Rivest, andClifford Stein.Introduction to algorithms.MIT Press, Cambridge, Mass., 3rd ed edition, 2009.
Andrew V. Goldberg and Robert Endre Tarjan.Efficient maximum flow algorithms.Commun. ACM, 57(8):82–89, 2014.
Slides by Hao Lu
Algorithm Analysis
Layering
Definition
L0 L1 L2
· · ·
Lk
• A layering is a partition V =⋃k
i=0 Li s.t.• s ∈ L0
• u ∈ Li , v ∈ Lj , 〈u, v〉 ∈ E ⇒ j ≤ i + 1
• dist(s, u) ≥ i , u ∈ Li• Canonical Layering
• Li = {v |distG (s, v) = i }• L∞ = {v | @s ; vpath}
Slides by Hao Lu
Algorithm Analysis
Basics
Preflow and Height Function
• A preflow is a function f : V × V 7→ R s.t.• Capacity Constraints: f ≤ c• Skew Symmetry: f (u, v) = −f (v , u)• Relaxation of Flow Conservation:
∑v∈V f (v , u) ≥ 0,∀u ∈ V \ {s}
• The excess flow into u is defined as e(u) =∑
v∈V f (v , u).
• A vertex v ∈ V \ {s, t} is overflowing if e(u) > 0.
• G = (V ,E ), f is a preflow; a height function h : V 7→ N is definedas:
• h(s) = |V |, h(t) = 0• h(u) ≤ h(v) + 1, ∀ 〈u, v〉 ∈ Ef
Lemma
If h(u) > h(v) + 1, then 〈u, v〉 /∈ Ef .
Slides by Hao Lu
Algorithm Analysis
Basics
The Push Operation
• Applies when:• u is overflowing• cf (u, v) > 0• h(u) = h(v) + 1
• Push min(e(u), cf (u, v)) units of flow from u to v
• Non-saturating Push, saturating push
• After a saturating push, u is no longer overflowing.
Slides by Hao Lu
Algorithm Analysis
Basics
The Relabel Operation
• Applies when:• u is overflowing• h(u) ≤ h(v),∀ 〈u, v〉 ∈ Ef
• Increase h(u) to 1 + min {h(v) | 〈u, v〉 ∈ Ef }
• Feasibility: {h(v) | 〈u, v〉 ∈ Ef } 6= ∅ since u is overflowing.
Slides by Hao Lu
Algorithm Analysis
Basics
Generic Algorithm
Push-Relabel(G )
1 h(u) = 0 except h(s) = |V |2 f (u, v) = 0 except f (s, v) = c(s, v),∀v connected to s3 while there exists an applicable push or relabel operation4 do it5 return f
Slides by Hao Lu
Algorithm Analysis
Correctness
Termination with a Flow
Lemma (Termination with a flow)
G , f , h, u is overflowing. Then either a push or a relabel applies to it.
Proof.
∀ 〈u, v〉 ∈ Ef : h(u) ≤ h(v) + 1If ∃ 〈u, v〉 ∈ Ef : h(u) = h(v) + 1, a push applies.Else, ∀ 〈u, v〉 ∈ Ef : h(u) < h(v)+1, or h(u) ≤ h(v), a relabel applies.
Slides by Hao Lu
Algorithm Analysis
Correctness
Heights Never Decrease
Lemma (Heights never decrease)
During the execution of the Push-Relabel algorithm on G , for everyu ∈ V , h(u) never decreases. Moreover, a relabel operation increasesh(u) by at least 1.
Proof.
Push operations don’t involve height changing.Before a relabel operation, ∀ 〈u, v〉 ∈ Ef : h(u) ≤ h(v). After it,h(u) = min {h(v) | 〈u, v〉 ∈ Ef }+ 1.
Slides by Hao Lu
Algorithm Analysis
Correctness
Height Function Maintains
Lemma (Height function maintains)
During the execution of the Push-Relabel algorithm on G , h willalways be a height function.
Proof.
• Relabel Operations: For 〈u, v〉 ∈ Ef ,h(u) = min {h(v) | 〈u, v〉 ∈ Ef }+ 1 ensures h(u) ≤ h(v) + 1. For〈w , u〉 ∈ Ef , before the relabel operation h(w) ≤ h(u) + 1.According to, this property maintains.
• Push Operations: During a push operation on 〈u, v〉, a new edge〈v , u〉 may appear. h(u) = h(v)+1 guarantees that h(v) ≤ h(u)−1.With a saturating push, 〈u, v〉 will disappear, but it doesn’t hurt.
Slides by Hao Lu
Algorithm Analysis
Correctness
Non-Existence of s ; t Path
Lemma
G , f , h. There is no path from s to t in Gf .
Proof.
For the sake of contradiction, Gf has some s ; t paths. Let the shortestone be p = 〈v0, v1, . . . , vk〉, where v0 = s, vk = t, k < |V |.
|V | = h(s) ≤ h(v1) + 1 ≤ · · · ≤ h(t) + k = k
which leads to a contradiction.
Slides by Hao Lu
Algorithm Analysis
Correctness
Correctness
Theorem
Push-Relabel Algorithm generates a max-flow.
Proof.
• Preflow maintains.
• Height function maintains.
• Termination with a flow.
• Non-existence of s ; t path.
• Termination with a max-flow.
Slides by Hao Lu
Algorithm Analysis
Running Time
Excess Flow Can Return
Lemma
G , f . For any overflowing vertex w , there is a simple path w ; s in Gf .
Proof.
For any overflowing vertex w , let U = {v |∃w ; v in Gf }.Claim: s ∈ U∑
u∈U
e(u) =∑u∈U
∑v∈V
f (v , u) =∑u∈U
∑v∈U
f (v , u) +∑u∈U
∑v∈U
f (v , u)
=∑u∈U
∑v∈U
f (v , u)
If s /∈ U, the sum is greater than 0, whichmeans∃u ′ ∈ U, v ′ ∈ U : f (v ′, u ′) > 0, thus ∃w ; u ′ → v ′, v ′ ∈ U.
Slides by Hao Lu
Algorithm Analysis
Running Time
Bound on Heights
Lemma (Bound on heights)
∀v ∈ V : h(v) ≤ 2|V |− 1
Proof.
• Heights only increase during relabel operations.
• Relabel operations only applies to overflowing vertices.
• Overflowing vertices are connected to s.
• h(v) − h(s) ≤ |V |− 1
Slides by Hao Lu
Algorithm Analysis
Running Time
Bound on Relabel Operations
Lemma (Bound on relabel operations)
# Relabel operations = O(|V |2)
Proof.
• A relabel operation increases h(u) by at least 1.
• h(u) ≤ 2|V |− 1
• (|V |− 1) · (2|V |− 1) = O(|V |2)
Slides by Hao Lu
Algorithm Analysis
Running Time
Bound on Saturating Pushes
Lemma (Bound on saturating pushes)
# Saturating pushes = O(|V ||E |)
Proof.
• A saturating push operation changes the direction e = 〈u, v〉.• Between two such pushes on e, h(u) + h(v) increases by at least 2.
• For each e, there are at most 2|V |− 1 saturating pushes.
• (|E |) · (2|V |− 1) = O(|V ||E |)
Slides by Hao Lu
Algorithm Analysis
Running Time
Bound on Non-saturating Pushes
Lemma (Bound on non-saturating pushes)
# Non-saturating pushes = O(|V |2|E |)
Proof.
Define Φ =∑
v : e(v)>0 h(v)
• A relabel operation increases Φ by at most 2|V |− 1.
• A saturating push operation increases Φ by at most 2|V |− 1.
• A non-saturating push operation decreases Φ by at least 1:u will no longer be overflowing, v may become overflowing,h(u) = h(v) + 1
• (2|V |− 1) · (O(|V |2) + O(|V ||E |)) = O(|V |2|E |)
Slides by Hao Lu
Algorithm Analysis
Running Time
Running Time
Theorem (Running time)
There is an implementation of Push-Relabel Algorithm whoserunning time is O(|V |2|E |).
Proof.
There is an implementation which satisfies:
• Every push takes O(1) time, totalling O(|V |2|E |).• Every relabel takes O(|V |) time, totalling O(|V |3).? Actually, relabel operations take O(|V ||E |).