maurizio patrignani seminar on the paper on the single-source unsplittable flow problem authored by...

Maurizio Patrignani

seminar on the paper

on the single-sourceunsplittable flow problem

authored by

Yefim Dinitz Naveen Garg

Michel X. Goemans

FOCS ‘98

the model – capacitated graph

• directed graph G(V,E)• capacity on edges c : E R+

4.1

2.3

1.62.9

2.6 1.7

1.0 1.5

2.2

2.1

2.6

8.5

3.4

3.2

4.1

2.3

1.62.9

2.6 1.7

1.0 1.5

2.2

2.1

2.6

8.5

3.4

3.2

• a single source s and k terminals ti with demands di R+

• a vertex may contain an arbitrary number of terminals

the model – source and terminals

s

t11.2

t2t3

0.9

3.1

t43.1

t51.5

t62.1

4.1

2.3

1.62.9

2.6 1.7

1.0 1.5

2.2

2.1

2.6

8.5

3.4

3.2

s

t11.2

t2t3

0.9

3.1

t43.1

t51.5

t62.1

• each commodity flows along a single path from s to ti (unsplittable)

• the flow on each edge is a real number (fractional) f : E R+ 0

the model – unsplittable, fractional flows

• if flows were allowed to split• a (splittable) single source multicommodity flow can be obtained

from a single source network flow• consider a flow of the simple network flow problem and iterate:

• find a maximal route (blue) from s to ti on the flow network (red)• add such route to the splittable flow of ti

• update flow and demand di

the model – relationships with other models

3.0

0.5

1.0 1.0

0.5

7.0 3.0

2.0s

4.5 2.0

0.5

1.0

1.0

0.5

6.0 3.0

2.0s

3.5

simple network flow is an easy problemsimple network flow problem (and, hence, splittable single source

multicommodity flow problem) admits a solution

for any vertex set S (sS) the total demand of terminals in S is at most the total capacity of the edges entering S

4.1

2.3

1.62.9

2.6 1.7

1.0 1.5

2.2

2.1

2.68.5

3.4

3.2

s

1.20.9

3.11.5

2.1cut condition

four possible questions

feasibility: can one route all commodities unsplittably?

minimum congestion: find the smallest 1, such that the problem is feasible if all capacities are multiplied by ?

maximum routable demands: max(iT di) over all subsets T of demands D

minimum number of rounds: if we partition the demands into rounds, how many rounds are needed in order to have all feasible rounds?

2C

Cthe obtained instance is feasible

the partition instance admits a

solution

reduction from partition

feasibility is NP-hard

C

Cs

1.20.9

3.13.1

1.52.1

2C

Cthe minimum congestion =C

the partition instance admits a

solution

reduction from partition

minimum congestion is NP-hard

1

1s

1.20.9

3.13.1

1.52.1

C

2C

C2 rounds are needed to

unsplittably route all demands

the partition instance admits a solution

reduction from partition

minimum number of rounds is NP-hard

s

1.20.9

3.13.1

1.52.1

C

2C

CC demands can be unsplittably

satisfied

the partition instance admits a solution

reduction from partition

maximum routable demand is NP-hard

s

1.20.9

3.13.1

1.52.1

example of applicationscheduling on a parallel machine

machines jobs

T

T

T

T

p1

p2

p3

p4

p5

pn

notations and assumptions

notations:• dmax = maximum demand• cmin = minimum capacity

assumptions:• flow is acyclic (thus, graph G is acyclic)• cmin dmax, that is, any commodity can be routed on any edge

results of the paper

if the cut condition is verified:

1. all demands can be unsplittably routed violating the capacities by at most the maximum demand dmax

• since cmin dmax then congestion is at most two (best possible)

2. all demands can be unsplittably routed in at most 5 rounds3. 22.6% of demands can be routed unsplittably

we will focus on result 1.

basic idea of the paper

start from a single-source, splittable, fractional flow satisfying all demands

start from a single-source, splittable, fractional flow satisfying all demands

transform it into an unsplittable flow that, in the worst case, violates capacities of dmax

transform it into an unsplittable flow that, in the worst case, violates capacities of dmax

preliminary (simplification) phase

f

if f d

f-d

• remove each edge e with f(e)=0• remove each ti that is on the same vertex as s• remove each vertex that has no incident edge

stop when all terminals are regular

a terminal is regular i fi < d

d d

f1

f2

d

algorithm architecture

preliminary simplification

search for alternating cycle

augment flow

move terminals

s is the only vertex?

end

yes

no

invariant condition

invariant condition: each vertex v containing terminals has at least a regular terminal (it follows that v has at least two incoming edges)

rather than asking for all regular terminals, a relaxed invariant condition will be pursued

f1

f2

d2d1

search for alternating cycle

as soon as it is possible start a new forward path

arbitrarily chosen outgoing edge

go forward as long as possible (but the grafo is acyclic!)

forward path

arbitrarily chosen incoming edge

no outgoing edges

backward path

go backward till you find vertices with only one outgoing edge

arbitrarily chosen vertex

search for alternating cyclestop when you reach an already visited vertex

it may happen that• a forward path intersects a forward path (represented in the figure above)• a forward path intersects a backward path• a backward path intersects a forward path

surely a terminal here

surely a terminal here

augment flowdecrease the flow on forward paths and increase on backward paths of the same quantity

the balance of each vertex is granted

augment flowaugment/decrement flow until one of these two conditions holds: condition one

a forward edge disappears

condition twoa non-movable terminal becomes movable

an edge e = (u,v) is called singular if the directed subgraph reachable from e is a path

singular edges

the edges of a backward path are all singular

singular edges

once an edge becomes singular it stays singular until it is removed

move terminal t along edge e if:

1) e is singular and f(e) = d

move terminals

2) e is not singular and f(e) d

with a preference for (1)

in other words:1) you move preferentially through singular edges, only if the flow is matched

(and the singular edge is then removed)2) you move on non-singular edges if no other move can be done

singular edges

non singular edge

exampleconsider this instanceall terminals are regular

4

7

u

v

ws

86

2

3 2

examplesearch for an alternating cycle

4

7

u

v

ws

86

2

3 2

exampleaugment/decrease flow

4

7

u

v

ws

86

2

3 2

-2 -2

+2+2

exampleaugment/decrease flow

4

7

u

v

ws

86

3 2

6

54

20

examplemove terminals

4

7

u

v

ws

66

5 4

examplemove terminals

7

u

v

s

66

5 4

• it is not possible to move terminal 4 to s since (s,v) is singular and f(s,v)>4

• it is not possible to move terminal 4 to u since (u,v) is singular and f(u,v)>4

end of the first iteration

examplesearch for an alternating cycle (iteration 2)

7

u

v

s

66

5 4

exampleaugment/decrease flow (iteration 2)

7

u

v

s

66

5 4

-2

-2

+2

exampleaugment/decrease flow (iteration 2)

7

u

v

s

66

5 4

4

7

4

examplemove terminals (iteration 2)

7

u

v

s

44

7 4

• it is not possible to move terminal 4 to s since (s,v) is singular and f(s,v) > 4

examplemove terminals (iteration 2)

u

v

s

44

4

done

7

examplemove terminals (iteration 2)

u

s

4

4

done

7

examplemove terminals (iteration 2)

s

47

done

examplefinal solution of the unsplittable flow problem

4

7

u

v

ws

86

2

3 2

47

main result

the algorithm finds an unsplittable flow such that the total flow on each edge e violates its capacity (actually, violates the initial flow on e) of a quantity that is less than the maximum demand

• before edge e becomes singular• the value of all terminals moved through e is at most the initial

flow f(e)

• after edge e becomes singular• at most one terminal may be moved through e (and this deletes e)

the sum of the flows of all commodities, with the exception of at most one, on edge e is less than the initial flow f(e)

corollary

tightness

M

M

M

M

M

M

M

M

• an unsplittable flow necessarily violates the capacity of M-M/q on some edge

• growing q, M/q goes to zero and the unsplittable flow violates the capacity of M, the maximum demand

s

M

M-M/q

M-M/q

M-M/q

M-M/q

consider this example

preliminary simplification

search for alternating cycle

augment flow

move terminals

s is the only vertex?

end

yes

no

running time

O(n)

O(m) times (removes at

least one edge)

O(k)

O(km) total

O(nm + km)

m = |E|n = |V|k = no. terminals

O(km)

correctness

lemma 3.2: if, at the beginning of an iteration, v contains an irregular terminal, then

1. the out-degree of v is zero2. v contains no other irregular terminal3. v contains another (regular) terminal

lemma 3.2: if, at the beginning of an iteration, v contains an irregular terminal, then

1. the out-degree of v is zero2. v contains no other irregular terminal3. v contains another (regular) terminal

lemma 3.3: at the beginning of any iteration, the in-degree of any vertex containing one or more terminals is at least 2

lemma 3.3: at the beginning of any iteration, the in-degree of any vertex containing one or more terminals is at least 2

lemma 3.4: as long as all terminals have not reached the source, the algorithm always finds an alternating cycle

lemma 3.4: as long as all terminals have not reached the source, the algorithm always finds an alternating cycle

lemma 3.1: whenever an edge becomes singular, it remains so until it is removed from the graph

lemma 3.1: whenever an edge becomes singular, it remains so until it is removed from the graph

lemma 3.1

• when e becomes singular, the statement obviously true

• no edge is added during the process (edges are removed only)

• by removing an edge the property can not be violated

euv e’

lemma 3.1: whenever an edge becomes singular, it remains so until it is removed from the graph

lemma 3.1: whenever an edge becomes singular, it remains so until it is removed from the graph

lemma 3.2

v

lemma 3.2: if, at the beginning of an iteration, v contains an irregular terminal tj, then

1. the out-degree of v is zero2. v contains no other irregular terminal3. v contains another (regular) terminal

lemma 3.2: if, at the beginning of an iteration, v contains an irregular terminal tj, then

1. the out-degree of v is zero2. v contains no other irregular terminal3. v contains another (regular) terminal

79

8

8

18

• terminal with demand 7 is not regular

• terminal with demand 18 is regular

lemma 3.2

v

79

8

8

18

• since tj is irregular, at least one edge e entering v must have f(e) > dj

• consider the iteration i when you moved terminal tj to v

• since tj was not moved further backwards, edge e was singular

• f(e) can not be increased from below tj to above tj in a following iteration

• hence edge (v,w) was singular and out-degree of v is zero after

iteration i

we

lemma 3.2

v

79

8

8

18

• at most one irregular terminal can be moved to a vertex v

• this proves that v contains no other irregular terminals

• since f(e) > tj and out-degree of v is zero, the conservation of the

flow implies that there is at least another (regular) terminal on v

e

lemma 3.3

if a vertex has irregular terminals it has also (at least) one regular terminal

vertices with regular terminals have in-degree at least two by definition

lemma 3.3: at the beginning of any iteration, the in-degree of any vertex containing one or more terminals is at least 2

lemma 3.3: at the beginning of any iteration, the in-degree of any vertex containing one or more terminals is at least 2

lemma 3.4lemma 3.4: as long as all terminals have not reached the source, the algorithm always finds an alternating cycle

lemma 3.4: as long as all terminals have not reached the source, the algorithm always finds an alternating cycle

• when constructing a forward path you stop at a vertex v with no outgoing edges

• flow conservation implies that v has terminals• by lemma 3.3 vertex v has in-degree at least 2

• when constructing a backward path you may be stuck at a vertex v with no incoming edges

• necessarily v=s• contradiction

s

how can this branch be connected to s?

the end