Clearly

Bertinoro, 5/5/08 1/57

Clearly

But this is a complicated way to write this.Hereby we suggest some improvement.

First lecture in Mathematics

1 + 1 = 2


StepStep 11

and

and


which is clearly more intuitive

So can be simplifid as 1 + 1 = 2


and

StepStep 22


So

can be further simplifid as

1 + 1 = 2


StepStep 22

and

hence


So eventually gets its final form: 1 + 1 = 2

Now, decide for yourself:

1. which of the forms is the simplest.

2. which of the forms will make your career faster?

3. which of the forms will mostly impress your

boss/boyfriend/girlfriend/mother?


We are given lightpathswe want to get a coloring SS such that cost(S)cost(S) is minimal.What is cost(S)cost(S) ?

1.21.2. . Cost Cost functionsfunctions


1. number of wavelengths

#colors = 4

1.21.2. . Cost Cost functionsfunctions

Bertinoro, 5/5/08 10/57

ADM

OADM

#ADMs + #OADMs

2. Switching cost

Bertinoro, 5/5/08 11/57

ADM

OADM

#ADMs + #OADMs = 12 + 8

Bertinoro, 5/5/08 12/57

but number of OADMs is fixed, so …

Bertinoro, 5/5/08 13/57

2. number of ADMs

#ADMs = 12

Bertinoro, 5/5/08 14/57

#ADMs=12 #ADMs=9

Bertinoro, 5/5/08 15/57

#ADMs=12

#colors=4

#ADMs=9

#colors=3

Trade-off between #colors and #ADMs

Bertinoro, 5/5/08 16/57

#ADMs=8

#colors=2

#ADMs=7

#colors=3

Bertinoro, 5/5/08 17/57

g=2

#ADMs=9

#ADMs=8

With grooming

Bertinoro, 5/5/08 18/57

Minimize the number of ADMs Minimize the number of ADMs with and without groomingwith and without grooming ComplexityComplexity special networks, general networksspecial networks, general networks Approximation algorithmsApproximation algorithms on-lineon-line

1.31.3. P. Problems in this roblems in this talktalk

Bertinoro, 5/5/08 19/57

RingRingFlammini, Moscardeli, Gianpierro, Shalom, Z.

2005/6/7

ln g

Traffic Traffic GroomingGrooming

Shalom, Wong, Zaks, 2007

On-line On-line

74

Bertinoro, 5/5/08 20/57

ALG 2N N OPT ALG 2 x OPT

N: # of lightpathsALG: #ADMs used by the algorithmOPT: #ADMs used by an optimal solution

2.1 approximation 2.1 approximation ratioratio

Bertinoro, 5/5/08 21/57

w/out grooming: N ALG 2N N OPT 2N

ALG 2 x OPT

R: # of lightpathsALG: #ADMs used by the algorithmOPT: #ADMs used by an optimal solution

w/ grooming: N/g ALG

2N N/g OPT

2N ALG 2g x OPT

Bertinoro, 5/5/08 22/57

#ADMs = N + #chains

N lightpaths

cycles

chains

2.2 Basic 2.2 Basic relationrelation

Cycles are good, chains are bad

Bertinoro, 5/5/08 23/57

In the approximation algorithms there are two common techniques for saving ADMs:

Eliminate cycles of lightpaths Find matchings of lightpaths

#ADMs = N + #chains

Bertinoro, 5/5/08 24/57

cost(S) = N + chains=13+6=19 Every path costscosts 1 ADM

cost(S) = 2N-savings=26-7=19 Every connection savessaves 1 ADM

N lightpaths

2.3 2.3 NoteNote

N=13

Min ADM problem: (cost=#ADMs)(cost=#ADMs)

Connections are good, chains are bad

Bertinoro, 5/5/08 25/57

Assume that an optimal solution S* saves x ADMs,

and a solution S saves y ADMs

³

£

cost(S*) if y then

k1

cost(S) (2- )cost(S*)k

Lemma:

2.4 a basic 2.4 a basic lemmalemma

³

£

cost(S*) if y then

23

cost(S

f o

)

r example:

cost2

(S*)

Bertinoro, 5/5/08 26/57

Optimal solution S* saves x ADMsa solution S saves y ADMs

³ ³

= £ £ = =

cost(S*) Ny

k k

cost(S) = 2N - y

cost(S*) =2N - x

N N N2N - 2N - 2N -2N - ycost(S) 1k k k 2-

cost(S*) 2N - x 2N - x 2N -N N k

Bertinoro, 5/5/08 27/57

D0(S) – lightpath not sharing any ADM

D1(S) – lightpath sharing ONE ADM

D2(S) – lightpath sharing BOTH ADMs

d0(S) = 2

d1(S) = 4

d2(S) = 19

d0(S) + d1(S) + d2(S) = 25 = N

N=25 lightpaths

2.5 2.5 notaton

Bertinoro, 5/5/08 28/57

S – ALG, S* - OPT

0 1 2d (S)+d(S)+d (S) =N

0 110

2d (S)+d (S)-2#chains(S*)d (S)=d (S)+ - #chains(S*) =

2 2

0 2N +d (S)- d (S)-2#chains(S*)=

2

Solution S=chains + cycles

#ADMs = N + #chains

cost(S)- cost(S*) =#chains(S)- #chains(S*) =

0 2d (S)- d (S)-2#chains(S*)1N(1+ )

2 N

0 1 0 22d (S)+d(S) =N +d (S)- d (S)

2.6 a basic 2.6 a basic tooltool

Bertinoro, 5/5/08 29/57

We define:

and get

= 0 2d (S)- d (S)-2 #chains(S*)ε(S)

N

1cost(S) =cost(S*)+ N(1+ε(S))

2

0 2d (S)- d (S)-2#chains(S*)1N(1+ )

2 Ncost(S)- cost(S*) =

Bertinoro, 5/5/08 30/57

D0(S) – lightpath not sharing any ADM

D1(S) – lightpath sharing ONE ADM

D2(S) – lightpath sharing BOTH ADMs

d0(S) = 2

d1(S) = 4

d2(S) = 19

d0(S) + d1(S) + d2(S) = 25 = N

N=25 lightpaths

2.7 2.7 exampleexample

#ADMs=29

Bertinoro, 5/5/08 31/57

= =

=-

0 2d (S)- d (S)-2 #chains(S*)ε(S)

2-19- 4 21N

25 25

d0(S) = 2

d2(S) = 19

N=25

suppose #chains(S*)=2, cost(S*)=25+2=27

1 25 21cost(S) =cost(S*)+ N(1+ε(S)) =cost(S*)+ (1- ) =

2 2 25cost(S*)+2=29

Bertinoro, 5/5/08 32/57

path of length 2

3.3 minADM with grooming is NP-complete 3.3 minADM with grooming is NP-complete

for a starfor a star

path of length 1

Bertinoro, 5/5/08 33/57

path of length 2

path of length 1

Star, g=1( trivial )

Bertinoro, 5/5/08 34/57

The number of used ADM is exactly equal to the lower bound of needed ADM:

é ùê ú

é ùê úê úê úê úê ú ê úê ú

ê úê ú

åå

n

i ni ii=1

i=1

yx +y

+2 2node

0

nodes 1,…,n

01

n

2

ixi paths of length

2

yi paths of length

1

Star, g=2

Bertinoro, 5/5/08 35/57

Sketch for g=3:

3-Exact Cover

Edge Partition into 3-regular graphs

Star grooming, g=3

Star, g≥3 - NP-complete

Bertinoro, 5/5/08 36/57

3-Exact Cover:

Input:Input: set A of size 3n, and a collection S of subsets of A of size 3 each.

Output:Output: are there n subsets in the collection S that cover A?

Bertinoro, 5/5/08 37/57

Edge Partition into 3-regular graphs:

Input:Input: undirected graph G = (V,E).

Output:Output: can E be partitioned into subsets E1,…,Em , each inducing a 3-regular subgraph G=(Vt,Et), t=1,…,m ?

Bertinoro, 5/5/08 38/57

3-Exact Cover


Star grooming, g=3

Bertinoro, 5/5/08 39/57

sets

elements

Bertinoro, 5/5/08 40/57

3-Exact Cover


Star grooming, g=3

Bertinoro, 5/5/08 41/57

Claim: there exists a solution using at most 2|E|/3 ADMs iff the edges of G can be partitioned into 3-regular graphs.

0

1

2

1

2

4

3

3

4

G S

Bertinoro, 5/5/08 42/57

2

1

5 4

3

5 4

3

2

10

g=2=2

Bertinoro, 5/5/08 43/57

Preprocessing: While there is a cycle C of length ≤

do: Remove (the lightpaths of) C from the

instance

Processing: Designate each lightpath as a chain Do

Build the matching graph M of the chains Find a maximum matching MM of M Combine chains according to MM

Until M has no edges

4.1 basic algorithm4.1 basic algorithm

eliminate short cycles, then find matchings

( )PI M l

Bertinoro, 5/5/08 44/57

The running time of the algorithm is exponential in due to the preprocessing phase

By removing the preprocessing phase (=1) we obtain algorithm PIM(1)

Recall:

= 0 2d (S)- d (S)-2 #chains(S*)ε(S)

N


2

Bertinoro, 5/5/08 45/57

algorithm PIM()

( ) (1 )2

NPIM OPT l

10

2

l

Bertinoro, 5/5/08 46/57

1

2= £

+0 2d (S)- d (S)-2 #chains(S*)

ε(S)N l

2£ £

+0 2 0 2

Nd (S)- d (S)-2 #chains(S*) d (S)- d (S)- #chains(S*)

l

Need to prove:

We will show:

Bertinoro, 5/5/08 47/57

2£

+0 2

Nd (S)- d (S)- #chains(S*)

l

Take S*:

in the example :

38 lightpaths, 5 chains, 3 cycles

Bertinoro, 5/5/08 48/57

2£

+0 2


l

preprocessing stage of S: eliminate cycles of

The lightpaths that we used are colored red:

£size l . . .

in the example : 10 lightpaths are used to form cycles in S

Bertinoro, 5/5/08 49/57

2£

+0 2


l

So, after preprocessing stage of S: in S* we have the following lightpaths:

Bertinoro, 5/5/08 50/57

2£

+0 2


l

Now the algorithm is doing MM,MM,MM,…

We show that already after the first MM we are ok.

Bertinoro, 5/5/08 51/57

2£

+0 2


l

We show that in the remaining lightpaths there is a

MM’ ≤ MM that is ok.

Bertinoro, 5/5/08 52/57

in the example : 0d (S) =8

0d (S) - number of isolated lightpaths

1 for each odd path and for each odd cycle

2£

+0 2


l

Bertinoro, 5/5/08 53/57

2£

+0 2


l

0 0 0d (S) = d (odd cycles)+d (odd chains)

2£

+0

Nd (odd cycles)

l

£0 2d (odd chains) d (S)+ #chains(S*)

Which implies

Show:

2£

+0 2


l

Bertinoro, 5/5/08 54/57

2£

+0 2


l

2£

+0

Nd (odd cycles)

l

Since there are at most N lightpath left, and each odd cycle is of size at least 2+l

Bertinoro, 5/5/08 55/57

2£

+0 2


l

1. Original chain of S* that was untouched


1 here is matched with 1 here

Bertinoro, 5/5/08 56/57

2£

+0 2


l

2. A a chain of S* that was partitioned into t parts


t-1 here are matched with t-1 here

1 here is matched with 1 here

Bertinoro, 5/5/08 57/57

2£

+0 2


l

3. A a cycle S* that was partitioned into t parts


Each 1 here is matched with at least 1 here

Bertinoro, 5/5/08 58/57

Preprocessing: While there is a cycle C of length ≤ do:

Remove (the lightpaths of) C from the instance

Processing: Designate each lightpath as a chain Do

Build the matching graph M of the chains Find a maximum matching MM of M Combine chains according to MM

Until M has no edges

4.2 basic algorithm without 4.2 basic algorithm without preprocessingpreprocessing

PIM(l)

Bertinoro, 5/5/08 59/57

This is optimal.

£1

ε5

£ +3

PI M(1) OPT N5


2

Without preprocessing:

Bertinoro, 5/5/08 60/57

Upper bound

Recall: and

to show that

we prove that (S) ≤ 1/5

0 2d (S)- d (S)-2#chains(S*)ε(S) =

N1

cost(S) =cost(S*)+ N(1+ε(S))2

£ +3

PI M(1) OPT N5

Bertinoro, 5/5/08 61/57

Orient the chains and cycles of S*.

Bertinoro, 5/5/08 62/57

Let LAST be the set of nodes which are last elements of the chains according to this orientation.

Bertinoro, 5/5/08 63/57

By mapping a path in D0(S) to either

a path of D2(S)or to a chainor to a cycle of size ≥ 5

£0 2d (S)- d (S)-2#chains(S*) 1ε(S) =

N 5

£

£

0 2

0 2

show:

Nd (S)- d (S)-2#chains(S*)

5or

Nd (S) d (S)+#chains(S*)+

5

Bertinoro, 5/5/08 64/57

The Mapping

0 ( )p D SÎ

q0=p

If q0 is the last node of a path of S* then:

p’=q0

map p to p’

return

Otherwise, q1 is the next node in q0’s path/cycle in S*

q1

q1 can not be in D0(S), otherwise the algorithm would add the edge (q0,q1) to the matching.If q1 is in D2(S) then:

p”=q1

map p to p”

return

q2

Otherwise q1 has exactly one neighbor q2 in GS. Obviously q2 is not in D0(S).

If q2 is in D2(S) then:

p”=q2

map p to p”

return

If q2 is the last node of a path of S* then:

p’=q2

map p to p’

return

Otherwise, q3 is the next node in q2’s path/cycle in S*

q3

q3 can not be in D0(S), otherwise the above path would be an augmenting path of the maximum matching found by the algorithm.

As the graph is finite, this process continues until p is mapped, or we re-encounter a node. In this case:

C = {qi}

Map p to C

return

It is easy to see that |C| is odd. We also show that |C| > 3.

Bertinoro, 5/5/08 65/57

Algorithm

¬ ÈS S {A}

Input: Graph G, set of lightpaths P, g > 0

Step 1: Choose a parameter k = k(g).

Step 2: Consider all subsets of P of size

If a subset A is 1-colorable (i.e., any edge is used at most g times) then

weight[A]=endpoints(A)

£ ×k g

5.1 5.1 algorithm algorithm

Bertinoro, 5/5/08 66/57

Step 3: COVER an approximation to the Minimum Weight Set Cover of S

Step 4: Convert COVER to a PARTITION

Output: the coloring induced by PARTITION

S – collection of all legal sets of at most kg lightpaths, each with its switching cost.

Bertinoro, 5/5/08 67/57

Legal coloring

For any fixed g, the number of subsets constructed in the first phase is g kO n

Bertinoro, 5/5/08 68/57

Legal coloring

, B is 1-colorable

A is 1-colorable ( correctness).

(and cost(A) cost(B).)

A B

5.2 analysis: ln(g)-approximation for a 5.2 analysis: ln(g)-approximation for a ringring

Bertinoro, 5/5/08 69/57

k g

cost(PARTI TI ON)

weight(COVER)

H weight(MI NCOVER)

(1+ln(k g))weight

ALG=

(SC)

for every set cover SC.

Bertinoro, 5/5/08 70/57

Lemma: There is a set cover SC, s.t.: 2g

weight(SC) 1+ Ok

PT

=cost(PARTI TI ON)

(1+ln(k g)) weigh

AL

C

G

t(S )

Bertinoro, 5/5/08 71/57

k g

cost(PARTI TI ON)

weight(COVER)

H weight(MI NCOVE

ALG=

SC

R)

(1+ln(k g))weight( )

2g(1+ln(k g)) 1+

kOPT

Conclusion:

For k = g ln g : 2lng+ALGOP

)T

o(lng

Bertinoro, 5/5/08 72/57

Lemma: There is a set cover SC, s.t.:

2gweight(SC) 1+ O

kPT

5.3 proof of 5.3 proof of lemma lemma

Bertinoro, 5/5/08 73/57

Consider OPT x - a color of OPT. Px - paths colored x. endpoints(Px) - the set of ADMs operating at wavelength x. (assume |endpoints(Px)|= ) Partition endpoints(Px) into m sets of k consecutive nodes in the example: k=5, m=4

m k

Use OPT to build SC

Bertinoro, 5/5/08 74/57

k k k k

iweight[S ] k+g k g

S1 S2 Sm

M=4 k=5

{paths starting at S1}, {paths starting at S2}, …,

{paths starting at Sm}

Each of these sets was in S !

All these sets, for all colors, cover A

Bertinoro, 5/5/08 75/57

i

m

ii=1

m

i xi=1

weight[S ] k+g

weight[S ] m(k+g)

( OPT =m k)

gweight[S ] OPT 1+

k

x

w/o the assumption we have: m

i xi=1

2gweight[S ] OPT 1+

k

1

m

ii

2gweight[S ] OPT 1+

k

x

Bertinoro, 5/5/08 76/57

undirected:

ALG 2ln( g) OPT

directed:

ALG 2lng OPT

5.4 trees 5.4 trees

Bertinoro, 5/5/08 77/57

On-line problem Input arrives one at a time, and a decision is

made (and cannot be changed). In the minADM problem: lightpaths arrive one at a time, and need to be colored.

Competitive analysis An on-line algorithm A is c-competitive if A(I) c OPT(I)for any input sequence I.(A(I) and OPT(I) are #ADMs used by A and by an optimal offline algorithm OPT.)

6.1 on-line6.1 on-line

Bertinoro, 5/5/08 78/57

When a new path arrives:1. if closes a unicolor cycle2. if any endpoint colored3. else (no side colored)

- assign same color

- assign same color- assign a new color

#ADMs=7

6.2 algorithm ALG6.2 algorithm ALG

Bertinoro, 5/5/08 79/57

3

2

32

21

When a new path arrives:

1. if closes a unicolor cycle2. if any endpoint colored3. else (no side colored)

- assign same color- assign same color

- assign a new color

32

Bertinoro, 5/5/08 80/57

Theorem: ALG is 7/4-competitive on any topology. This is optimal even for a ring.

Theorem: ALG is 3/2-competitive on a path. This is optimal.

6.3 6.3 ResultsResults

Bertinoro, 5/5/08 81/57

ALG: ADM=7

OPT: ADM=4

ALG ≥ 7/4

6.4 ALG 6.4 ALG ≥ 7/4 even for a ring

Bertinoro, 5/5/08 82/57

k paths

k-1 spaces:

x between same colork-1-x between different colors

k=12

x=6

6.5 any algorithm for a path 6.5 any algorithm for a path ≥ 3/2

Bertinoro, 5/5/08 83/57

So far: any algorithm uses 2k ADMs

now – a short path at each gap of diffferent colors

(k-1-x such gaps)

Any algorithm uses at least one more ADM for each (ALG uses exactly one)

So: any algorithm ≥ 2k + (k-1-x) ADMs

k=12, x=6, 12-1-6=5

Bertinoro, 5/5/08 84/57

now – two long paths at each of the k gap of same color

So far: use ≥ 2k + (k-1-x) ADMs

Any algorithm must use 2 ADMs for each

So: any algorithm ≥ 2k + (k-1-x) + 4x =

= 3k+3x-1 ADMs

Bertinoro, 5/5/08 85/57

OPT: the short paths ≤ 2k ADMsfor the long paths 2x ADMs

any algorithm/OPT 3/2 – 1/(2k)

We showed: any algorithm uses ≥ 3k+3x-1 ADMs

OPT 2k + 2x

Bertinoro, 5/5/08 86/57

Optimal solution S* saves x ADMsOur solution S saves y ADMs

) ) )

)

£ = £ £

Þ £

cost(S)- cost(S*) =(2N - y)- (2N - x) =

x 1 1 1=x- y x - (1

Proo

- (1- (1-k k k k

1c

x N

ost(S) cost(S*

cost(

)(2

S )

f

-

:

k

*

³ Þ £x 1

y cost(S) (2- )cLem ost(S*)k k

ma: ≤ -> ≤

≤ ≤ ≤

≤£ £ x N costNote: (S*)

6.6 ALG for a path 6.6 ALG for a path ≤ 3/2

Bertinoro, 5/5/08 87/57

³

£

x if y then

k1

cost(S) (2- )co

Lemma

k

:

st(S*)

³x

y2

We show that

Þ £3

cost(S) cost(S*)2

≤

≤

≤

≤

Bertinoro, 5/5/08 88/57

³x

y2

Claim:

optimal S* – max matching at each point

≤

Bertinoro, 5/5/08 89/57

For the proof choose a specific S*

savings of S* (x)savings of S (y)

Bertinoro, 5/5/08 90/57

a b

a came before bc

c

map

1-1

Bertinoro, 5/5/08 91/57

savings of S (y)

³ 2

³x

y2

Savings of S* (x)

≤

≤

Bertinoro, 5/5/08 93/57

Lemma 1 d0(S) d2(S) + |chains(S*)| + N/2orient S*

for any u D0(S)

1. if u is last in some chain of S*, map u to this chain

2.else

i. u’ D0(S) contradiction

ii. u’ D1(S) map u to {u, u’}

iii.u’ D2(S) map u to u’

u u’S*

Bertinoro, 5/5/08 94/57

Case a:

7/4=1.75

6.8 6.8 lower bound of lower bound of 7/4 , even for a ring

Bertinoro, 5/5/08 95/57

Case b: Case b1:

6/3 = 2

Case b2: 5/3 = 1.67

any algorithm ≥ 1.67 any algorithm ≥ 1.75-

Exercise:

Bertinoro, 5/5/08 96/57

K M

C

HBDG

EFG

A

D

F

E

G

B

6.9. 6.9. a simpler lower bound of 7/a simpler lower bound of 7/4 (not for a ring)

so: BDG

Bertinoro, 5/5/08 97/57

K M

C

HBDG

EFG

A

D

F

E

G

B

Bertinoro, 5/5/08 98/57

K M

C

HBDG

EABDG

GFEAB

EFG

A

D

F

E

G

B

#ADMS=7

#OPT=4

Competitive Ratio: 7/4Competitive Ratio: 7/4

Bertinoro, 5/5/08 99/57

K M

C

HBDG

EFG

A

D

F

E

G

B

#ADMS=6

#OPT=3

Competitive Ratio: 6/3 > 7/4Competitive Ratio: 6/3 > 7/4

BAE

so: BAE

Bertinoro, 5/5/08 100/57

K M

C

HBDG

EFG

A

D

F

E

G

B

BAE

EFKMHG

so: EFKMHG

Bertinoro, 5/5/08 101/57

K M

C

HBDG

EFG

A

D

F

E

G

B

BAE

EFKMHG

Bertinoro, 5/5/08 102/57

K M

C

HBDG

EFG

A

D

F

E

G

B

BAE

EFKMHG

EABDCHG#ADMS=9

#OPT=5

Competitive Ratio: 9/5 > 7/4Competitive Ratio: 9/5 > 7/4

Bertinoro, 5/5/08 103/57

K M

C

HBDG

EFG

A

D

F

E

G

B

BAE

EFKMHG

Hw: finish this case

Bertinoro, 5/5/08 104/57

K M

C

HBDG

EFG

A

D

F

E

G

B

BAE

Hw: finish this case

Bertinoro, 5/5/08 105/57

#colors=2, #ADMs=8

#colors=3, #ADMs=7

1st open problem: what can be said about the trade-off between #colors and #ADMs=8 ?

Bertinoro, 5/5/08 106/57


ALG, path ≤ 3/2

³

£

cost(S*) if y then

k1

cost(S) (2- )cost(S*)k

Lemma:

Bertinoro, 5/5/08 107/57


³ ³

= £ £ = =

cost(S*) Ny

k k

cost(S) = 2N - y

cost(S*) =2N - x

N N N2N - 2N - 2N -2N - ycost(S) 1k k k 2-

cost(S*) 2N - x 2N - x 2N -N N k

Bertinoro, 5/5/08 108/57


,³ £

= £ = =

xy x N

k

cost(S) = 2N - y

cost(S*) =2N - x

x x2N - 2N -2N - ycost(S) k k

cost(S*) 2N - x 2N -N N

x=2-

Nk

Bertinoro, 5/5/08 109/57

On-line algorithms

when a request arrives: if no endpoint common with others then assign a new color if one endpoint in common with other(s)

then assign same color if two endpoints in common with others

then assign one of the colors

Bertinoro, 5/5/08 110/57

Recall Algorithm ALG When a new path arrives:

if closes a unicolor cycle if closes a unicolor path if any endpoint colored

else (no side colored)

- assign same color

- assign same color

- assign same color


ADM=7

Bertinoro, 5/5/08 111/57

ALG-TRIANGLEWhen a lightpath arrives1. if p is length-2

if closes unicolor cycle, assign same color else assign new color

2. if p is length-1 if closes unicolor cycle containing length-2

lightpath p’, assign same color if there are two unmarked length-1

lightpaths p’ & p’’ with different color, assign to color of either p’ or p’’ and mark p, p’, p’’

else assign new color

Bertinoro, 5/5/08 112/57

When a new path arrives:if closes a unicolor cycleif closes a unicolor pathif any endpoint colored

else (no endpoint colored)

- assign same color

- assign same color

- assign same color


Bertinoro, 5/5/08 113/57


,³ £

= £ = =

xy x N

k

cost(S) = 2N - y

cost(S*) =2N - x

x x2N - 2N -2N - ycost(S) k k

cost(S*) 2N - x 2N -N N

x=2-

Nk

Bertinoro, 5/5/08 114/57

ONLINE COLORING – INPUT 2

w(BDG)=2

w(BAE)=1

w(EFKMHG)=2

w(EFG)=1

Total ADMS: 0

w(GFEAB)=3

w(EABDG)=4

w(BDGFE)=5

w(EABCDG)=6

24568101214

K M

H

DF

E

G

BA C

Bertinoro, 5/5/08 115/57

OFFLINE COLORING – INPUT 2

w(BDG)=2

w(BAE)=3

w(EFKMHG)=4

w(EFG)=1

Total ADMS:

w(GFEAB)=2

w(EABDG)=1

w(BDGFE)=3

w(EABCDG)=4K M

H

DF

E

G

BA C

Competitive Ratio: 14/8=7/4


8

Bertinoro, 5/5/08 116/57


w(BDG)=2

w(BAE)=1

w(EFKMHG)=3

w(EFG)=1

Total ADMS: 0

w(EABDCHG)=4

24579

K M

H

DF

E

G

BA C

Bertinoro, 5/5/08 117/57


w(BDG)=2

w(BAE)=1

w(EFKMHG)=3

w(EFG)=1

Total ADMS: 0

w(EABDCHG)=4

5

K M

H

DF

E

G

BA C

Competitive Ratio: 9/5 >7/4

Competitive Ratio: 9/5 >7/4

Bertinoro, 5/5/08 118/57


w(BDG)=2

w(BAE)=2

w(BDCHG)=1

w(EFG)=1

Total ADMS: 0

w(EABDG)=3

24568

K M

H

DF

E

G

C

w(GFEAB)=4

w(GKFEAB)=5

w(EFGDB)=6

101214

BA

Bertinoro, 5/5/08 119/57


w(BDG)= 2

w(BAE)=3

w(BDCHG)=4

w(EFG)=1

Total ADMS: 8

w(EABDG)=1

K M

H

DF

E

G

C

w(GFEAB)=4

w(GKFEAB)=2

w(EFGDB)=3

BA



Bertinoro, 5/5/08 120/57


w(BDG)=2

w(BAE)=2

w(BDCHG)=3

w(EFG)=1

Total ADMS: 0

w(GHMKFEAB)=4

24579

K M

H

DF

E

G

BA C

Bertinoro, 5/5/08 121/57


w(BDG)=2

w(BAE)=1

w(BDCHG)=1

w(EFG)=1

Total ADMS: 5

w(GHMKFEAB)=2

K M

H

DF

E

G

BA C

Competitive Ratio: 9/5>7/4

Competitive Ratio: 9/5>7/4

Clearly

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