Class Plan

• 1-1 ． Graph 　　グラフ• 1-2 ． Graph 　 Search 　グラフ探索• + some graph properties • 1-3 ． Network Flow 　　ネットワーク・フロー• 1-4 ． Algorithms ＆ Complexity 　アルゴリズムと計算量　

•　 2-1 ． Minimum Spanning 　 Tree 　最小木 　　　 ( + data structures and sorting データ構造，整列）•　 2-2 ． Matroid 　マトロイド• 3-1. 　 Maximum Flow 　最大流• 3-2 ．　 Minimum Cost Flow 最小費用流　• ?4-1. 　 Approximation algorithm 　近似解法• ?4-2 ．　 Heuristics ヒューリスティック　

3-0. Shortest path 最短路

Network Flows

classical , traditional (?)

• maximum flow

• minimum cost flow

Variants in flows

• generalized flow

• multicommodity flow

• dynamic flow

• Problem Formulation

• Optimality condition

• (Algorithm)

Maximum Flow Problem ( 最大流問題 )

Given : directed graph G=(V,A)

Find :which satisfies• flow bound constraints : ( 容量制約 )

and• mass balance constraints : ( 流量保存制約 )

=

v

• feasible flow( 可能流 ) : a flow satisfying the above two constraints

• maximum flow( 最大流 ) : a feasible flow maximizing

total out flow from s

3

5

3

4

4

23 4

2

s t

capacity

入口 出口

Are the following flows feasible/maximum ?

Exampleinstance

3

1

1

4

4

02 1

2

s t

3

4

3

4

3

21 1

2

s t

1) 3

3

3

4

3

21 1

2

s t

2) 3)

3

4

4

4

4

21 1

3

s t

4) 3

3

3

4

4

22 1

2

s t

5)

For evaluate a bottleneck of flow value …

• s-t cut :

• capacity of an s-t cut

3

5

3

4

4

23 4

2

s t

3

5

3

4

4

23 4

2

s t

total out flow from s :

v≠s

Feasible flow and s-t cut

total out flow from s :

Maximum-flow Minimum-cut Theorem

(Corollary 4.2)

(s-t cut with minimum capacity)

Ford-Fulkerson(1956)

Total out flow ∂φ （ｓ） ≦ capacity of δ+(X)

Augmenting path method(偽 )

2

2

1

3

4

13

3

s

t

instance capacity

0

0

0

0

0

0 0

0

s

t

flow:

2

0

2

0

0

2

0

0 0

2

s

t

0

2

1

0

2

0

0 0

3

s

t

0

2

2

0

2

0

11

3

s

t

0 ? There is no s-t directed path.

step0:step1:

step2:

Augmenting path

ε

ε

2

2

0

2

0

11

3

s

t

+1

+1-1

-1

0

2

2

0

1

1

12

2

s

t

2

2

1

1

1

12

3

s

t

1

2

2

1

3

4

13

3

s

t

2

☆ residual network

arcs represent the possibility of pushing flow back on arcs

s

t

0

cap

2

1 3

2

2 21

4

121€

Aϕ = AϕF ∪AϕB

AϕF = {a∈ A |ϕ (a) < u(a)}, Aϕ

B = {ar |ϕ (a) > 0} ar :reverse arc of a

€

uϕ (a) =u(a) −ϕ (a) (a∈ AϕF )ϕ (ar) (a∈ AϕB )

⎧ ⎨ ⎩

€

Gϕ = (V ,Aϕ ),uϕ

Augmenting path method (本物 )

step0: step1: construct a residual network step2: Find an s-t directed path P in

If there is no such a path …step3: compute and update flow along P Go to step1.

0

2

2

0

2

0

11

3

s

t

2

2

1

1

1

12

3

s

t

1

s

t

2

1

2

2

2 21 1 1

3

4 s

t

1

1

2

2

1 12 1 2

3

3

1

1no augmenting path

← augmenting path

€

Gϕ = (V ,Aϕ )

ε 2

2

1

3

4

13

3

s

t

2

cap

When no augmenting path exists …

: feasible flowThere is no s-t directed path (augmenting path ) in

t t

ss

X

X : set of vertices reachable from s→ s-t cut

total out flow from s

“Maximum “

€

Gϕ = (V ,Aϕ )

Optimality condition

Theorem4.1 (Augmenting path theorem)

A feasible flow φ is maximum ⇔ there is no augmenting path in

⇒) If there is an augmenting path in …..

　　）　 proved⇒

Maximum-flow Minimum-cut Theorem

(s-t cut with minimum capacity)

Ford-Fulkerson(1956)

€

Gϕ = (V ,Aϕ )

€

Gϕ = (V ,Aϕ )

Max … min…

Max flow problem Min cut problemduality

Subject to

◆Show the dual problem ◆Show the complementary slackness condition

Review of LP (dual problem)Example:

Maximize 4x1+x2+3x3

Subject to x1- x2- x3 ＝ 1 2x1+x2+2x3 13≦ x1, x2, x3≧０

A (quick) estimate of the optimal value (upper bound)

Since x1, x2, x3 are nonnegative,

Obj func. = 4x1+x2+3x3 4x≦ 1+2x2+4x3 26≦

2nd constraint ×2 + 1st constraint

Obj func. = 4x1+x2+3x3 5x≦ 1+x2+3x3 27≦

2nd constraint ×2

Review of LP (dual problem)Example:

Maximize 4x1+x2+3x3

Subject to x1- x2- x3 ＝ 1 2x1+x2+2x3 13≦ x1, x2, x3≧０

Linear combination 1st constraint ×y1 + 2nd constraint ×y2

(y1+2y2)x1+ (-y1+y2)x2 +(-y1+2y2)x3 y≦ 1+13y2

We need y2 ≧０

If

y1+2y2 4≧-y1+y2 1≧-y1+2y2 3≧

coefficient of each xi is at least as big as the corresponding coefficient in obj. function.

then obj. function y≦ 1+13y2

Review of LP (dual problem)

primal

€

maximize c jx jj=1

n

∑

subject to aijx jj=1

n

∑ = bi (i =1,...,l )

aijx jj=1

n

∑ ≤ bi (i = l +1,...,m)

x j ≥ 0 ( j =1,...,n)

€

minimize biyii=1

m

∑

subject to aijyii=1

m

∑ ≥ c j ( j =1,...,n)

yi ≥ 0 (i = l +1,...,m)

dual

€

c jx j ≤ aijyii=1

m

∑ ⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

j=1

n

∑j=1

n

∑ x j ≤ aijx jj=1

n

∑ ⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

i=1

m

∑ yi ≤ biyii=1

m

∑

The dual of the dual problem is always the primal problem

Review of LP (duality theorem)

€

x * : primal opt. sol

⇒ There exists

€

y * : dual opt. sol

s.t.

€

c jx * jj=1

n

∑ = biy *ii=1

m

∑

Review of LP (complementary slackness)

€

x * : primal feasible sol

€

y * : dual feasible sol

€

x *

€

y * : opt. solutions

€

⇔

aijy *i = c j or x * j = 0 ( j =1,...n)i=1

m

∑

aijx * j = bi or y *i = 0 (i =1,...m)j=1

n

∑

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

€

c jx j ≤ aijyii=1

m

∑ ⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

j=1

n

∑j=1

n

∑ x j ≤ aijx jj=1

n

∑ ⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

i=1

m

∑ yi ≤ biyii=1

m

∑

(proof)

Form

Augmenting path method (complexity)

Analyze the worst case time complexity for the augmenting path method

step0: step1: construct a residual network step2: Find an augmenting path P in

If there is no such a path, stop.step3: compute and update flow along P Go to step1.

€

Gϕ = (V ,Aϕ )

ε

Instance 1

s t

M

M

M

M

1

capacity

s t

M

M

M

M

1

residual network

s t

M-1

M

M

M-1

11

1

s t

M-1

M-1

M-1

M-1

11

11

1

Number of iterations : ?

Instance 2

st

1

2

3

4

M

M

M1

1

M

M

M

r€

r = 5 −12

(<1), M > 4

€

rk < rk−1

rk−1 − rk = rk+1

◆Find maximum flow◆If the algorithm chooses augmenting path as P1, P2, P1, P3, P1, P2, P1, P3, ,,, how many iterations the algorithm performs?

P1=(s,2,1,3,4,t)P2=(s,1,2,4,3,t)P3=(s,3,1,2,4,t)

Rules for selecting augmenting paths

Maximum capacity augmenting path

Minimum length augmenting path

… Find an augmenting path maximizing α

… Find an augmenting path with least edges

ε

•Max capacity augmenting path algo.•Capacity scaling algo.•MA ordering algo.

•Min length augmenting path algo.•Dinitz’s blocking flow algo.

Max capacity augmenting path algo.

step0: step1: construct a residual network step2: Find an augmenting path P in

If there is no such a path, stop.step3: compute and update flow along P Go to step1.

€

Gϕ = (V ,Aϕ )

ε An s-t directed path maximize

min{uφ(a)|a A(P)}∈

Max capacity augmenting path algo.(complexity)

€

ϕ* :ϕ :ϕ ':

(1− 1m

)(∂ϕ *(s) −∂ϕ (s)) ≥ (∂ϕ *(s) −∂ϕ '(s))

Opt. flow

A flow

Flow obtained by updating along a max cap. aug. path

The number of iteration O(m log(nU)) × maxmin type shortest path€

1− 1m

⎛ ⎝ ⎜

⎞ ⎠ ⎟m

≤ 12

U=max{u(a)|a A}∈

※ integer capacity

Capacity scaling algorithm

※ integer capacity

Capacity of kth scaling phase

€

uk (a) = u(a)2K−k

⎢ ⎣ ⎢

⎥ ⎦ ⎥

K = logU +1⎣ ⎦

kth scaling phase find a maximum fow φw.r.t uk

(k+1)th scaling phase find a maximum fow φw.r.t uk+1

φ ← ２ φ

Bit scaling

2

5

1

4

2

5

3

5

3

s

t

3 010

101

001

100

010

101

011

s

t

101011

011

K=3 (log25=2.32…)

Capacity scaling algorithm

010

101

001

100

010

101

011

s

t

101011

011

0

1

0

1

0

1

0

s

t

10

001

10

00

10

01

10

01

s

t

1001

01010

101

001

100

010

101

011

s

t

101011

011

u1 u2 u3

uk(a)=2uk-1(a) or 2uk-1(a)+1

Max flow

0 0

0

0

0 t

0

0

1 1

1

0 0

0

0

0 t

0

0

２ ２

２× ２

Max flow

Capacity scaling algorithm(complexity)

uk(a)=2uk-1(a) or 2uk-1(a)+1

Max flow

0 0

0

0

0 t

0

0

1 1

1

0 0

0

0

0 t

0

0

２ ２

２× ２

Max flow

Min cut value in the auxiliary network:

≦m(the number of cut edges)

Step0: φ:=0, k:=1Step1: [k scaling phase] find a maximum flow φ with respect to uk

Step2: If k=K, then stop. Else k:=k+1, φ:=2φ, and go to step 1.

€

K = logU +1⎣ ⎦

€

uk (a) = u(a)2K−k

⎢ ⎣ ⎢

⎥ ⎦ ⎥

Minimum length augmenting path

Other algorithms

Push-relabel algorithm

Network simplex algorithm

flow bound constraints

No augmenting pathmass balance constraints

preflow

Max flow on planar graph

2

2

1

3

4

1

3

3

s

t

capacity

2

s-t planar graph … the graph added the arc (t, s) is still planar

s-t cut = s*-t* path

min cut = shortest path

Class Plan

• 1-1 ． Graph 　　グラフ• 1-2 ． Graph 　 Search 　グラフ探索• + some graph properties • 1-3 ． Network Flow 　　ネットワーク・フロー• 1-4 ． Algorithms ＆ Complexity 　アルゴリズムと計算量　

•　 2-1 ． Minimum Spanning 　 Tree 　最小木 　　　 ( + data structures and sorting データ構造，整列）•　 2-2 ． Matroid 　マトロイド• 3-1. 　 Maximum Flow 　最大流• 3-2 ．　 Minimum Cost Flow 最小費用流　• ?4-1. 　 Approximation algorithm 　近似解法• ?4-2 ．　 Heuristics ヒューリスティック　

minimum circulation problem ( 最小費用循環流問題 )

Given ： directed graph G=(V,A) capacity u: A → R+

cost c: A → RFind : flow :A → R+ minimizing total cost which satisfies ・ flow bound constraints and ・ mass balance constraints

• feasible flow

• minimum cost flow

(3,-1)

(3,0)

(cap,cost)instance

Example

(4,3)

(2,1)

(1,-1)(1,-3)

(2,2)

(4,2)

(2,2)

0

0

0

0

0

00

0

0

3

1

2

2

1

12

2

0

1

1

0

0

1

10

0

0

total cost = 0 total cost = 12 total cost = -2

residual network w.r.t. a flow

☆ residual network

arcs represent the possibility of pushing flow back on arcs

€

Aϕ = AϕF ∪AϕB

AϕF = {a∈ A |ϕ (a) < u(a)}, Aϕ

B = {ar |ϕ (a) > 0} ar :reverse arc of a€

Gϕ = (V ,Aϕ ),uϕ

€

uϕ (a) =u(a) −ϕ (a) (a∈ AϕF )ϕ (ar) (a∈ AϕB )

⎧ ⎨ ⎩

€

c(a) =c(a) (a∈ AϕF )

−c(ar ) (a∈ AϕB )

⎧ ⎨ ⎩

optimality condition (negative cycle…)

The following statements are equivalent :

MCF-(a) : is minimum cost flow

MCF-(b) : the residual network does not have any

negative-cost cycles

MCF-(c) : There exists π ： V → R such that

holds for every

Let be a feasible flow

directed cycle C with

(a) (b)⇒

If the residual network has a negative cost cycle, …

(b) (a)⇒: a flow satisfying (b)

: optimal flow

: feasible flow in

can be decomposed into cycles in

⇒ is optimal

negative cycle canceling algorithm

step0 : = 0step1 : construct the residual network

step2 : Find a negative-cost cycle C in

If there is no such a cycle, stop.

step3 : compute

and update flow along C.

Go to step1.

Shortest path algorithm

0

0

0

0

0

00

0

0

(3,-1)

(3,0) (4,3)

(2,1)

(1,-1)(1,-3)

(2,2)

(4,2)

(2,2)

(cap,cost)

1

0

0

0

0

11

0

0

(3,-1)

(3,0) (4,3)

(2,1)

(1,-1)(1,-3)

(2,2)

(4,2)

(2,2)

(2,-1)

(3,0) (4,3)

(2,1)

(1,-1)(1,3)

(1,2)

(4,2)

(2,2)

(1,-2)

(1,1)

residual network optimal

LP form

◆Show the dual problem ◆Show the complementary slackness condition

Minimize

Subject to

Compare to optimality conditions

Optimality condition (reduced cost)

€

c p (a) = c(a) − p(∂+a) + p(∂−a)

€

p = −π

€

c p (a) > 0⇒ ϕ (a) = 0

c p (a) < 0⇒ ϕ (a) = u(a)

c p (a) = 0⇒ 0 ≤ ϕ (a) ≤ u(a)

Kilter diagram

€

ϕ (a)

€

p(∂+a) − p(∂−a)

u(a)

potential

Modified network w.r.t. p

G=(V, A)

Modified capacity

€

u p (a) =0u(a)u(a)

⎧ ⎨ ⎪

⎩ ⎪

(c p (a) > 0)(c p (a) = 0)(c p (a) < 0)

l p (a) =00u(a)

⎧ ⎨ ⎪

⎩ ⎪

(c p (a) > 0)(c p (a) = 0)(c p (a) < 0)

Optimality condition (positive cut)

A potential p is dual optimal if and only if in the modified network w.r.t. p any cut δ(X) satisfies

€

u p (a)a∈δ + (X )

∑ ≥ l p (a)a∈δ − (X )

∑

Positive cut : a cut with

€

u p (a)a∈δ + (X )

∑ < l p (a)a∈δ − (X )

∑

Positive cut canceling algorithm

step0 : p: any potential

step1 : construct the modified network

step2 : Find a positive cut in modified network

If there is no such a cut, stop.

step3 : compute

and update potential

Go to step1.

Max flow algorithm

0

0 0

0

0 0

(3,-1)

(3,0) (4,3)

(2,1)

(1,-1)(1,-3)

(2,2)

(4,2)

(2,2)

(cap,cost)

1

modified network

(Ip,up)cp

€

α =minmin c p (a) | a∈ δ+(a),c p (a) > 0{ },

min −c p (a) | a∈ δ−(a),c p (a) < 0{ }

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

€

p(v) =p(v) +αp(v)

⎧ ⎨ ⎩

(v ∈ X)(v ∉ X)

(0, 3)0(3, 3)-1

(0, 0)2

(0, 0)3

(0, 0)1

(0, 0)2

(0, 0)2(1, 1)-3

(1, 1)-1

0

0 1

1

0 0

(0, 3)0(0, 3)0

(0, 0)2

(0, 0)3

(0, 2)0

(0, 0)2

(0, 0)2(1, 1)-4

(0, 1)00

0 1

1

2 0

(0, 3)0(0, 3)0

(0, 2)0

(0, 0)3

(0, 2)0

(0, 0)4

(0, 2)0(1, 1)-2

(0, 1)0

Other algorithms

Primal dual algorithmNetwork simplex algorithm

+Scaling technique