1. Dinitz's algorithm for finding a maximum flow in a network.
Presented by:
Ilan Kadar and Sivan Albagli
April 18, 2019

2. Content Introduction Ford and Fulkerson Algorithm
Original Dinitz’ algorithm 2 2 DA
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3. Introduction The max flow problem definition
Capacitated directed graph G=(V,E,c,s,t).
The capacities are non-negative. 1 G s t 2
source
sink 2 1 3 3 DA
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4. Introduction The max flow problem definition
A flow f, is defined as a function on the directed edges satisfying the following:
Capacity constraint: eE : 0f(e)c(e)
Flow conservation: vV\{s,t}: (v,u)Ef(v,u) - (u,v)Ef(u,v)=0 4 4 DA
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5. Q: How do we know that the flow is maximum?
Introduction The max flow problem definition
The goal is to find the maximum flow from the source to the sink.
1/1 G s t
0/2
source
sink
0/1
Q: How do we know that the flow is maximum? 5 5 DA
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6. Introduction Motivation
Finding the maximal way to ship goods from a set of factories to a set of stores.
Bipartite matching 6 6 6 DA
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7. Introduction Naïve algorithm2 2 S 2 t 4 3 B 7 7 7 DA
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8. Introduction Naïve algorithm2 2 S 2 t 4 3 B 8 8 8 DA
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9. Introduction Naïve algorithm
2/2
0/2 S
2/2 t
2/4
0/3 B 9 9 9 DA
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10. Introduction Naïve algorithm
2/2
0/2 S
2/2 t
2/4
0/3 B 10 10 10 DA
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11. There is no path where one can increase the flow
Introduction Naïve algorithm
There is no path where one can increase the flow A B S t
2/2
0/2
4/4
2/3
Therefore, the algorithm terminates 11 11 11 DA
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12. Introduction Naïve algorithm
Is it the maximum flow? A
2/2
0/2
(Flow value = 4) S
2/2 t
4/4
2/3 B 12 12 12 DA
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13. Introduction Naïve algorithmB S t
2/2
1/2
4/4
3/3
Maximum flow = 5 13 13 13 DA
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14. Introduction Equivalent flow function
For each edge add ,
if doesn’t exists, with capacity and flow zero. 5 1 8
source s t
sink 1 2 2 G 14 14 14 DA
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15. Introduction Equivalent flow function
Define for each edge: 15 15 15 DA
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16. f Introduction Equivalent flow function 3/5 2/2 1/5 -1/2 v u v =3-2
=2-3 v DA
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17. Introduction Equivalent flow function
satisfy the following constraints:
Capacity constraint v u
3/5
2/2
1/5
-1/2
f(v,u)<=c therefore, f(v,u)-(positive number)<=c 17 17 17 DA
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18. Introduction Equivalent flow function
satisfy the following constraints:
2. Skew symmetry v u
3/5
2/2
1/5
-1/2
Fnew(v,u)=f(v,u)-f(u,v) = -(f(v,u)-f(u,v))=f(u,v)-f(v,u)=-Fnew(u,v) 18 18 18 DA
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19. Introduction Equivalent flow function
satisfy the following constraints:
3. Flow conservation
Flow conservation 19 19 19 DA
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20. Introduction Equivalent flow function
The net flow from v is defined as:
The flow value is defined as the net flow from s: 20 20 20 DA
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21. Introduction Equivalent flow function
Edge is called
saturated if
Residual capacity
(possibility to add flow)
1st representation
2nd representation
Another equvalent is formed by considering a flow as a skew-symmetric function on the pairs of vertices connected by at least one edge 21 21 21 DA
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22. Edge is called saturated if:
Introduction Equivalent flow function
Edge is called saturated if:
saturated v u
5/5
0/2
-5/2
saturated DA
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23. Introduction Equivalent flow function
Unsaturated edges:
unsaturated v u
5/5
1/2
4/5
-4/2
unsaturated DA
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24. Introduction Equivalent flow function
Residual capacity (possibility to add flow):
cf = 5-3+2=4 v u
3/5
2/2
1/5
-1/2
=cf
cf = 5-1=4
=cf DA
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25. Introduction Equivalent flow function
We will use the second representation, denoting it by f DA
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26. Ford and Fulkerson Algorithm
Define Gf=(V,Ef) – the residual network.
Ef - the unsaturated edges with capacity cf(u,v)=c(u,v)-f(u,v)>0. 2 1
-1/0
-2/0
1/4 s t
1/1
1/2
2/2
0/5 G 3 5 1 Gf
The idea is to make iterations of finding a flow augmenting path p from s to t in the residual network Gf, and updating Gf along p, until it such a path cannot be found. 1
-1/0 1
-1/0
0/0 DA
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27. Ford and Fulkerson Algorithm
A path is an augmenting path if it contains only unsaturated edges.
for example: A B S t
1/2
3/4
3/3
0/0
Wikipedia claimed 1956. DA
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28. Ford and Fulkerson Algorithm
Ford&Fulkerson(G,s,t)
1. initialize flow f(v,u) to 0
2. while there exists an augmenting path P do:
augment flow f along P
4. return f
Published in 1956
When no augmenting path exists, the current flow is maximum (Ford & Fulkerson Theorem).
Wikipedia claimed 1956. DA
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29. Ford and Fulkerson Algorithm
Ford&Fulkerson(G,s,t)
1. for each edge (u,v) Ef do:
f(u,v)0
f(v,u)0
4. while there exists a path P from s to t in the Gf do: cf(P) = min ePcf(e)>0
for each edge (u,v) in P do:
f(u,v)  f(u,v) + cf(P)
f(v,u)  f(u,v) - cf(P)
לכל קשת ולקשת ההופכית שלה נאתחל את הזרעמה ל0 DA
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30. Ford and Fulkerson Algorithm
Comments:
Notice that it is not certain that E = Ef,
as sending flow on (u,v) might close (u,v) (it is saturated),
and may open a new edge (v,u) in the residual network.
אחרי כל איטרציה, השינוי ברשת השירוית הוא: הוא "מאבד" את כל הקשתות שנהיו רוויות במסלול הנבחר להעביר זרימה, ומקבל קשתות חדשות - בכיוון הפוך מהקשתות שנהיו רוויות במסלול הנ"ל. DA
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31. Ford and Fulkerson Example
0/1 s t G 1 Gf DA
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32. Ford and Fulkerson Example
0/1 s t G 1 s t Gf
First iteration DA
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33. First iteration – continue
Ford and Fulkerson Example
0/1 s t
1/1 G 1 s t Gf
First iteration – continue DA
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34. First iteration – continue
Ford and Fulkerson Example
0/1 s t
1/1 G 1 s t Gf
First iteration – continue DA
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35. Ford and Fulkerson Example
0/1 s t
1/1 G 1 s t Gf
Second iteration DA
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36. Second iteration-continue
Ford and Fulkerson Example
1/1 s t
0/1 G 1 s t Gf
Second iteration-continue DA
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37. There is no path at the residual network Gf
Ford and Fulkerson Example
1/1 s t
0/1 G 1 s t Gf
There is no path at the residual network Gf
The flow is maximal DA
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38. Ford and Fulkerson Time analysis
With rational capacities, the algorithm will always terminate.
With irrational capacities, the algorithm may run forever.
Runtime is bounded by O(E*|f*|) when the capacities are integers. DA
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39. Ford and Fulkerson Problem 1 (Integer case)
The algorithm runs in pseudo polynomial time:
How many iterations?? DA
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40. Ford and Fulkerson Problem 1 (Integer case)
The algorithm runs in pseudo polynomial time:
How many iterations??
2,000,000 DA
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41. Ford and Fulkerson Problem 2 (General case)
Convergence isn’t guaranteed : DA
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42. Ford and Fulkerson Open question?
The question of the existence of a polynomial, or finite algorithm, for the general case, remained open, for almost 10 years.
This was settled by Edmonds and Karp and by Y. Dinitz independently, at the late 60s.
Both algorithms are modifications of F&F.
Maybe to change 10 to ___ DA
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43. Ford and Fulkerson Edmonds-Karp(1972)
A small fix to the Ford-Fulkerson algorithm makes it work in polynomial time.
Compute the augmenting path using BFS on the residual network.
Run in time O(VE2)
Maybe to change 10 to ___ DA
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44. Dinitz’ Algorithm A Historical Remark
The DA was invented in response to a pre-class-exercise in Adel’son-Vel’sky’s Algorithm class.
At that time, the author was not aware the basic facts regarding FF DA
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45. Dinitz’ Algorithm Motivation
All parts of an iteration of FF except of finding an augmenting path P cost O(|P|).
Finding an augmenting path is the bottleneck of an iteration– O(|E|).
Improving FF by using a smart data structure. DA
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46. Dinitz’ Algorithm Motivation
Example BFS Tree: s t 1 5 2 4 3 DA
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47. Dinitz’ Algorithm Motivation
Example BFS Tree: s t 1 5 2 4 3 DA
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48. Dinitz’ Algorithm Motivation
saturated s t 1 5 2 4 3 DA
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49. Dinitz’ Algorithm Motivations t 5 2 1 3 4
There is no easy means to connect them again using unsaturated edges.
s and t become disconnected in the BFS tree! DA
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50. Dinitz’ Algorithm Extended BFS
Dist = dist from s
Dist=2
Dist=3
Dist=4 s t
Dist=1
Dist=3
Dist=2
Dist=1
Dist=3
Dist=2
Extended BFS includes the first edge found, leading to each vertex from the previous layer. DA
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51. Dinitz’ Algorithm Layered Network – L(s)V0 V1 V2 V3 V4 s t
Vi is the set of all nodes with distance i from s
Ei is the set of all edges going from Vi-1 to Vi DA
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52. Dinitz’ Algorithm Layered Network – L(s)V0 V1 V2 V3 V4 s t
L(s) = (Vi , Ei) – the union of all shortest paths from s in the graph. DA
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53. Running time of regular BFS == Running time of extended BFS
Dinitz’ Algorithm Layered Network: Definitions
Vi – The i layer: dist(v)=i  vVi
Ei – Edges from Vi -1 to Vi
L(s) = (Vi , Ei) – the union of all shortest paths from s in the graph.
Running time of regular BFS == Running time of extended BFS DA
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54. Dinitz’ Algorithm Layered Networks t
BFS from s in G
L(s) – extended BFS
from s in G DA
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55. Dinitz’ Algorithm Layered Network: next step
Goal: we want to maintain our data structure
as the union of all the shortest paths from s to t. s t
How can it be done?
Prune into
Run extended BFS from t on L(s), in the opposite edge direction DA
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56. Dinitz’ Algorithm Layered Network: next step
How can it be done?
Prune into s t
L(s) in the
opposite edge direction
Then run BFS from t on L(s) DA
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57. Dinitz’ Algorithm Layered Network: next step
How can it be done?
Prune into s t s t DA
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58. Dinitz’ Algorithm Layered Network
Claim:
is the union of all the shortest paths from s to t(Invariant).
Property of :
Doesn't have vertices with no any incoming edge, except s.
Doesn’t have vertices with no any outgoing edge, except t. DA
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59. Dinitz’ Algorithm Augmenting path finding
Just “walk” from s over After l steps t is reached. 1 s t 1 2 2
We could walk from t to s( usually that’s what we do) 1 1 1 DA
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60. Dinitz’ Algorithm Augmenting path finding
Remove the saturated edges. 1 s t 1 2 2
Just walk from s over After l steps t is reached.
This is the current augmenting path. 1 1 1 DA
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61. Dinitz’ Algorithm Augmenting path finding
Just “walk” from s over After l steps t is reached 1 s t 1 1 1
Just walk from s over After l steps t is reached.
This is the current augmenting path. 1 1 DA
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62. Dinitz’ Algorithm Cleaning dead ends
We might have vertices without outgoing edges called the dead-ends vertices.
We wish to remove them, so the next augmenting path finding from s won’t get stuck and will work in O(l) time.
- After an edge removing we might have vertexes with no out-coming edges.
This kind of vertexes is called dead-ends vertex.
- We wish to remove them, so the next augmenting path finding from s
won’t got stuck and will work in short time – O(l) DA
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63. Dinitz’ Algorithm Cleaning dead ends
Initialize two queues Ql and Qr by the list of saturated edges sat that were removed.
Define two procedures:
Left pass
Right Pass
- After an edge removing we might have vertexes with no out-coming edges.
This kind of vertexes is called dead-ends vertex.
- We wish to remove them, so the next augmenting path finding from s
won’t got stuck and will work in short time – O(l) DA
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64. Dinitz’ Algorithm Cleaning dead ends – Left Pass1 a d c s t b 1 2 2
- After an edge removing we might have vertexes with no out-coming edges.
This kind of vertexes is called dead-ends vertex.
- We wish to remove them, so the next augmenting path finding from s
won’t got stuck and will work in short time – O(l) 1 1 1
(d,t) Ql DA
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65. Dinitz’ Algorithm Cleaning dead ends – Left Pass1 a d c s t b 1 2 2
- After an edge removing we might have vertexes with no out-coming edges.
This kind of vertexes is called dead-ends vertex.
- We wish to remove them, so the next augmenting path finding from s
won’t got stuck and will work in short time – O(l) 1
(c,d) 1
(a,d) Ql DA
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66. Dinitz’ Algorithm Cleaning dead ends – Left Pass1 a c s t b 1 2
- After an edge removing we might have vertexes with no out-coming edges.
This kind of vertexes is called dead-ends vertex.
- We wish to remove them, so the next augmenting path finding from s
won’t got stuck and will work in short time – O(l) 1
(s,c) Ql DA
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67. Dinitz’ Algorithm Cleaning dead ends – Left Pass1 a s t b 1 2
- After an edge removing we might have vertexes with no out-coming edges.
This kind of vertexes is called dead-ends vertex.
- We wish to remove them, so the next augmenting path finding from s
won’t got stuck and will work in short time – O(l)
(s,c) Ql DA
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68. Dinitz’ Algorithm Cleaning dead ends
After applying LeftPass all the vertices(except t) have outgoing edge.
After applying RightPass all the vertices(except s) have incoming edge.
Why we need both procedures??? DA
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69. Dinitz’ Algorithm Cleaning dead ends
No path in the layered network from s to t could be removed.
Next augmenting path of length l can be found in O(l) time incase the cleaned layer is not vanished. DA
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70. Dinitz’ Algorithm The original algorithm
Phase
Until when, Phases?
Iteration DA
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71. Dinitz’ Algorithm The original algorithm
Constructing of the layered network using two BFS-es, at the beginning of a phase costs O(|E|).
Phase
Iteration DA
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72. Dinitz’ Algorithm The original algorithm
Phase
Augmenting path finding using a walk from s over the layered network costs O(l)=O(|V|) every time thanks to the cleaning maintenance .
Iteration DA
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73. Dinitz’ Algorithm The original algorithm
Phase
Flow updating costs O(l)=O(|V|) as well.
Iteration DA
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74. Dinitz’ Algorithm The original algorithm
Phase
Cleaning dead ends vertices
Iteration DA
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75. Dinitz’ Algorithm The original algorithm
Phase
In practice a single iteration might cost Ω(|E|+|V|) because we should remove every edge and vertex from the graph.
Iteration DA
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76. E Dinitz’ Algorithm Time analysis V How many Iterations?
Cost of the algorithm
O(#iterations * |V|+ #Phases* |E|) E V
How many Iterations?
How many Phases? DA
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77. Dinitz’ Algorithm Time analysis
Intuition: After any iteration, there is no augmenting path of length less then l. s t s t
Augmenting path
New path DA
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78. Dinitz’ Algorithm Time analysis
If there are paths of length l they are contained in s t s t -
Augmenting path
New path DA
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79. Dinitz’ Algorithm Time analysis
In the end of the phase(when the layer vanish), doesn’t contain paths of length l.
“We” can prove that the residual network doesn’t contain paths of length l.
Conclusion: The length of the layered network grows from phase to phase. DA
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80. Dinitz’ Algorithm Time analysis
- In every iteration at least one edge is being removed (min edge of path augmenting).
- At most |E| edges are being removed during a phase 
There are at most |E| iterations
during a phase.
Phase
Iteration DA
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81. Dinitz’ Algorithm Time analysis
Constructing of the layered network = O(|E|)
Path finding and capacity updating =O(l)=O(|V|).
It is done at most |E| times during a phase.
Maintenance of the layered network = O(|E|).
So, a phase costs O(|E|+|E||V|+|E|)=O(|E||V|).
Phase
Phase
Iteration DA
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82. Dinitz’ Algorithm Time analysis
- Distance(s,t)|V|-1 
There are at most |V|-1 phases in DA
Phase
Iteration DA
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83. Dinitz’ Algorithm Time analysis
DA running time is
O(|E||V||V|) = O(|V|2|E|)
Phase
Iteration DA
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84. Further Progress Time Method Discover Year O(|V| * |E|2) Shortest Path
Edmonds and Karp
1972
O(|V|3)
Preflow-push
Karzanov’s Algorithm
1974
Dynamic trees
Sleator-Tarjan
1983
FIFO preflow-push
Goldberg-Tarjan
1986
Length function
Goldberg-Rao
1997