1. Class Plan 2-1．Minimum Spanning Tree 最小木
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
3-0. Shortest path 最短路
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　

2. Network Flows Variants in flows classical , traditional (?)
generalized flow
multicommodity flow
dynamic flow
maximum flow
minimum cost flow
Problem Formulation
Optimality condition
(Algorithm)

3. Maximum Flow Problem (最大流問題)
Given : directed graph G=(V,A)
total out flow from s
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

4. Are the following flows feasible/maximum ?
Example
capacity
instance 3 4 5 2 s 3 4 t 出口 入口 4 3 2
Are the following flows feasible/maximum ? 3 1 4 2 s t 3 4 2 1 s t 2) 3) 3 4 2 1 s t 1) 4) 5) 3 3 4 4 4 3 2 2 s 1 1 1 t s 2 t 4 4 4 3 3 2

5. For evaluate a bottleneck of flow value …
s-t cut :
capacity of an s-t cut 3 3 4 4 5 5 2 2 s 3 4 t s 3 4 t 4 3 4 3 2 2
total out flow from s :
v≠s

6. Feasible flow and s-t cut
total out flow from s :

7. Maximum-flow Minimum-cut Theorem
Total out flow ∂φ（ｓ）　≦　capacity of δ+(X)
Maximum-flow Minimum-cut Theorem
(Corollary 4.2)
(s-t cut with minimum capacity)
Ford-Fulkerson(1956)

8. Augmenting path method(偽)
step0:
Augmenting path
step1:
instance
capacity s 2
step2: ε 4 2 1 2 3 3 ε t 1 3
flow: s s 2 s 2 1 2 2 t t t 2 3 s 2 2 1 2 ?
There is no s-t directed path. 1 t 3

9. cap ☆ residual network :reverse arc of a2 4 s 2 +1 s 2 1 2 1 2 3 3 2 1 2 1 2 -1 1 1 3 t 1 +1 2 t t 3 -1 2 s 2 1 2 1 1 1
☆ residual network 2 t 1 3
:reverse arc of a
arcs represent the possibility of
pushing flow back on arcs s 2 4 2 2 1 2 1 1 2 1 3 t

10. 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.
←　augmenting path ε
cap s 2 4 2 1 2 3 3 1 3 t s 2 s 2 1 2 1 2 1 2 1 1 1 2 t t 3 1 3 s 2 4 s 2 3 1
no augmenting path 2 1 2 2 1 2 1 2 2 1 1 1 2 1 1 1 3 t 1 3 t

11. When no augmenting path exists …
: feasible flow
There is no s-t directed path (augmenting path ) in s s t t X
X : set of vertices reachable from s
→　s-t cut
total out flow from s
“Maximum “

12. Maximum-flow Minimum-cut Theorem
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)

13. Max … min… duality Max flow problem Min cut problem
Subject to
Show the dual problem
Show the complementary slackness condition

14. 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 ≦ 4x1+2x2+4x3≦26
2nd constraint ×2
Obj func. = 4x1+x2+3x3 ≦ 5x1+x2+3x3≦27
2nd constraint ×2 + 1st constraint

15. 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 ≦ y1+13y2
We need y2 ≧０
y1+2y2 ≧ 4
-y1+y2 ≧1
-y1+2y2 ≧3 If
coefficient of each xi is at least as big as the corresponding coefficient in obj. function.
then obj. function ≦ y1+13y2

16. Review of LP (dual problem)
primal
The dual of the dual problem is always the primal problem

17. Review of LP (duality theorem)
: primal opt. sol
⇒ There exists
: dual opt. sol
s.t.

18. Review of LP (complementary slackness)
: primal feasible sol
: dual feasible sol
: opt. solutions
(proof)
Form

19. 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. ε

20. Instance 1 M M s 1 t M-1 M M M-1 1 1 s 1 t 1 1 residual network M-1
capacity M M s 1 t
M-1 M M
M-1 1 1 s 1 t 1 1
residual network
M-1
M-1 M
M-1 M M 1 s 1 s 1 t t 1 M
M-1 M M
Number of iterations : ?

21. Instance 2 M 1 3 1 M M r 1 t s M M 2 4 M P1=(s,2,1,3,4,t)
Find maximum flow
If the algorithm chooses augmenting path as P1, P2, P1, P3, P1, P2, P1, P3, ,,, how many iterations the algorithm performs?

22. Rules for selecting augmenting paths
Maximum capacity augmenting path
… Find an augmenting path maximizing α ε
Max capacity augmenting path algo.
Capacity scaling algo.
MA ordering algo.
Minimum length augmenting path
… Find an augmenting path with least edges
Min length augmenting path algo.
Dinitz’s blocking flow algo.

23. 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.
An s-t directed path maximize
min{uφ(a)|a∈A(P)} ε

24. Max capacity augmenting path algo.(complexity)
※　integer capacity
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
U=max{u(a)|a∈A}

25. Capacity scaling algorithm
※　integer capacity
Bit scaling
Capacity of kth scaling phase
φ ←２φ
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 s 5 5 s
101
101 3 1
011 2 2
001
101 3
010 5
010
011 t 4 3 t
100
011
K=3 (log25=2.32…)

26. Capacity scaling algorithm
101
101
uk(a)=2uk-1(a) or 2uk-1(a)+1
011
001
101
010
010
011 t
100
011 u1 u2 u3 s 1 1 s 10 10 s
101
101 1 01
011 00 10
001
101 01
010 01
010 01
011 t 1 t t 10 01
100
011 ２ ２
Max flow ×２ 1 1 ２ 1 t
Max flow t

27. Capacity scaling algorithm(complexity)
Step0: φ:=0, k:=1
Step1: [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.
uk(a)=2uk-1(a) or 2uk-1(a)+1 ２ ２
Max flow ×２ 1 1 ２
Min cut value in the auxiliary network: ≦m
(the number of cut edges) 1 t
Max flow t

28. Minimum length augmenting path

29. Other algorithms Push-relabel algorithm preflow flow bound constraints
Network simplex algorithm
preflow
flow bound constraints
No augmenting path
mass balance constraints

30. Max flow on planar graph
s-t planar graph … the graph added the arc (t, s) is still planar 2 1 3 4 s t
capacity
s-t cut = s*-t* path
min cut = shortest path

31. Class Plan 2-1．Minimum Spanning Tree 最小木
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　

32. 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 → R
Find : flow :A →　R+ minimizing total cost
which satisfies
・flow bound constraints
and
・mass balance constraints
feasible flow
minimum cost flow

33. Example (cap,cost) instance (3,-1) (4,3) (3,0) (2,2) (1,-1) (1,-3)
(2,1)
(4,2) 3 1 2 1 1 1 2 1 1 2 1 2
total cost = 0
total cost = -2
total cost = 12

34. residual network w.r.t. a flow
:reverse arc of a
arcs represent the possibility of
pushing flow back on arcs

35. optimality condition (negative cycle…)
Let be a feasible flow
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
directed cycle C with

36. (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

37. negative cycle canceling algorithm
step0 : = 0
step1 : 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
(cap,cost)
(3,-1)
(4,3)
(3,0)
(2,2)
(1,-1)
(1,-3) 1
(2,2)
(2,1)
(4,2) 1 1
residual network
optimal
(2,-1)
(3,-1)
(1,1)
(3,0)
(4,3)
(3,0)
(4,3)
(2,2)
(1,2)
(1,-2)
(1,-1)
(1,-1)
(1,-3)
(1,3)
(2,2)
(2,2)
(2,1)
(2,1)
(4,2)
(4,2)

38. LP form Minimize Subject to Show the dual problem
Show the complementary slackness condition
Compare to optimality conditions

39. Optimality condition (reduced cost)
potential
Kilter diagram
u(a)

40. Modified network w.r.t. p
G=(V, A)
Modified capacity

41. 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
Positive cut : a cut with

42. 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
(cap,cost)
(3,-1)
(4,3)
(3,0)
(2,2)
(1,-1)
(1,-3) 1
(2,2)
(2,1)
(4,2) 1
(0, 3)0
(0, 3)0
(3, 3)-1
(0, 0)3
(0, 3)0 1
(0, 0)3
(0, 3)0
(0, 0)2
(0, 1)0 1
(0, 3)0
(0, 0)3
(1, 1)-4
(0, 0)2
(1, 1)-1
(0, 2)0
(1, 1)-3
(0, 0)2
(0, 2)0
(0, 1)0 1
(0, 0)1
(1, 1)-2
(0, 0)2
(0, 0)2
(0, 2)0
(0, 2)0
(0, 0)2 2
modified network
(Ip,up)cp
(0, 0)4

43. Other algorithms Primal dual algorithm Network simplex algorithm
+Scaling technique