1. First lecture in Mathematics
Clearly
1 + 1 = 2
But this is a complicated way to write this.
Hereby we suggest some improvement.
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2. Step 1
and
and
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3. 1 + 1 = 2 So can be simplifid as which is clearly more intuitive
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4. Step 2
and
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5. So
1 + 1 = 2
can be further simplifid as
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6. Step 2
and
hence
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7. 1 + 1 = 2 So eventually gets its final form: 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?
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8. We are given lightpaths
1.2. Cost functions
We are given lightpaths
we want to get a coloring S such that cost(S) is minimal.
What is cost(S) ?
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9. 1. number of wavelengths 1.2. Cost functions #colors = 4
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10. 2. Switching cost
OADM
ADM
#ADMs + #OADMs
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11. ADM
OADM
#ADMs + #OADMs = 12 + 8
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12. but number of OADMs is fixed, so …
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13. 2. number of ADMs
#ADMs = 12
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14. #ADMs=12
#ADMs=9
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15. Trade-off between #colors and #ADMs
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16. #ADMs=8
#colors=2
#ADMs=7
#colors=3
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17. g=2
With grooming
#ADMs=9
#ADMs=8
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18. 1.3. Problems in this talk Minimize the number of ADMs
with and without grooming
Complexity
special networks, general networks
Approximation algorithms
on-line
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19. Traffic Grooming Ring On-line
Flammini, Moscardeli, Gianpierro, Shalom, Z /6/7
On-line
Shalom, Wong, Zaks, 2007
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20. ALG: #ADMs used by the algorithm
2.1 approximation ratio
N: # of lightpaths
ALG: #ADMs used by the algorithm
OPT: #ADMs used by an optimal solution
ALG  2N
N  OPT
ALG  2 x OPT
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21. w/out grooming: N  ALG  2N N  OPT  2N ALG  2 x OPT w/ grooming:
R: # of lightpaths
ALG: #ADMs used by the algorithm
OPT: #ADMs used by an optimal solution
w/out grooming:
N  ALG  2N
N  OPT  2N
ALG  2 x OPT
w/ grooming:
N/g  ALG  2N
N/g  OPT  2N
ALG  2g x OPT
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22. 2.2 Basic relation Cycles are good, chains are bad N lightpaths cycles
#ADMs = N + #chains
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23. Eliminate cycles of lightpaths Find matchings of lightpaths
#ADMs = N + #chains
In the approximation algorithms there are two common techniques for saving ADMs:
Eliminate cycles of lightpaths
Find matchings of lightpaths
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24. cost(S) = 2N-savings=26-7=19 Every connection saves 1 ADM
2.3 Note
Min ADM problem: (cost=#ADMs)
Connections are good, chains are bad
N lightpaths
N=13
cost(S) = N + chains=13+6=19
Every path costs 1 ADM
cost(S) = 2N-savings=26-7=19
Every connection saves 1 ADM
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25. Assume that an optimal solution S* saves x ADMs,
2.4 a basic lemma
Assume that an optimal solution S* saves x ADMs,
and a solution S saves y ADMs
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26. Optimal solution S* saves x ADMs a solution S saves y ADMs
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27. 2.5 notaton N=25 lightpaths D0(S) – lightpath not sharing any ADM
D1(S) – lightpath sharing ONE ADM
D2(S) – lightpath sharing BOTH ADMs
N=25 lightpaths
d0(S) = 2
d1(S) = 4
d2(S) = 19
d0(S) + d1(S) + d2(S) = 25 = N
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28. 2.6 a basic tool Solution S=chains + cycles #ADMs = N + #chains
S – ALG, S* - OPT
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29. We define:
and get
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30. 2.7 example N=25 lightpaths D0(S) – lightpath not sharing any ADM
D1(S) – lightpath sharing ONE ADM
D2(S) – lightpath sharing BOTH ADMs
N=25 lightpaths
d0(S) = 2
d1(S) = 4
#ADMs=29
d2(S) = 19
d0(S) + d1(S) + d2(S) = 25 = N
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31. suppose #chains(S*)=2, cost(S*)=25+2=27
d0(S) = 2
d2(S) = 19
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32. 3.3 minADM with grooming is NP-complete for a star
path of length 1
path of length 2
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33. path of length 1 path of length 2 ( trivial ) Star, g=1
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34. xi paths of length 2 yi paths of length 1
Star, g=2
The number of used ADM is exactly equal to the lower bound of needed ADM: 2 1
xi paths of length 2
yi paths of length 1 i n
nodes 1,…,n
node 0
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35. Edge Partition into 3-regular graphs  Star grooming, g=3
Star, g≥ NP-complete
Sketch for g=3:
3-Exact Cover 
Edge Partition into 3-regular graphs 
Star grooming, g=3
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36. 3-Exact Cover:
Input: set A of size 3n, and a collection S of subsets of A of size 3 each.
Output: are there n subsets in the collection S that cover A?
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37. Edge Partition into 3-regular graphs:
Input: undirected graph G = (V,E).
Output: can E be partitioned into subsets E1,…,Em , each inducing a 3-regular subgraph G=(Vt,Et), t=1,…,m ?
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38. Edge Partition into 3-regular graphs  Star grooming, g=3
3-Exact Cover 
Edge Partition into 3-regular graphs 
Star grooming, g=3
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39. sets
elements
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40. Edge Partition into 3-regular graphs  Star grooming, g=3
3-Exact Cover 
Edge Partition into 3-regular graphs 
Star grooming, g=3
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41. 4 1 4 S G 3 1 3 2 2
Claim: there exists a solution using at most 2|E|/3 ADMs iff the edges of G can be partitioned into 3-regular graphs.
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42. 4 5 5 4 3 1 1 3 2 2
=2
g=2
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43. eliminate short cycles, then find matchings
4.1 basic algorithm
eliminate short cycles, then find matchings
Preprocessing:
While there is a cycle C of length ≤ l 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
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44. By removing the preprocessing phase (l=1) we obtain algorithm PIM(1)
The running time of the algorithm is exponential in l due to the preprocessing phase
By removing the preprocessing phase (l=1) we obtain algorithm PIM(1)
Recall:
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45. algorithm PIM(l)
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46. Need to prove:
We will show:
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47. 38 lightpaths, 5 chains, 3 cycles
Take S*:
in the example :
38 lightpaths, 5 chains, 3 cycles
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48. in the example : 10 lightpaths are used to form cycles in S
preprocessing stage of S: eliminate cycles of
The lightpaths that we used are colored red:
in the example : 10 lightpaths are used to form cycles in S
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49. So, after preprocessing stage of S: in S
So, after preprocessing stage of S: in S* we have the following lightpaths:
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50. Now the algorithm is doing MM,MM,MM,…
We show that already after the first MM we are ok.
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51. We show that in the remaining lightpaths there is a
MM’ ≤ MM that is ok.
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52. in the example : 1 for each odd path and for each odd cycle
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53. Show:
Which implies
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54. Since there are at most N lightpath left, and each odd cycle is of size at least
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55. 1. Original chain of S* that was untouched
1 here is matched with 1 here
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56. 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
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57. 3. A a cycle S* that was partitioned into t parts
Each 1 here is matched with at least 1 here
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58. 4.2 basic algorithm without preprocessing
PIM(l)
Preprocessing:
While there is a cycle C of length ≤ l 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
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59. Without preprocessing:
This is optimal.
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60. Upper bound Recall: and to show that we prove that e(S) ≤ 1/5
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61. Orient the chains and cycles of S*.
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62. Let LAST be the set of nodes which are last elements of the chains according to this orientation.
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63. By mapping a path in D0(S) to either
a path of D2(S)
or to a chain
or to a cycle of size ≥ 5
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64. The Mapping
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. q1 q2 q3
q0=p
q1 can not be in D0(S), otherwise the algorithm would add the edge (q0,q1) to the matching.
If q2 is in D2(S) then:
p”=q2
map p to p”
return
If q0 is the last node of a path of S* then:
p’=q0
map p to p’
return
Otherwise, q3 is the next node in q2’s path/cycle in S*
Otherwise, q1 is the next node in q0’s path/cycle in S*
If q2 is the last node of a path of S* then:
p’=q2
map p to p’
return
If q1 is in D2(S) then:
p”=q1
map p to p”
return
Otherwise q1 has exactly one neighbor q2 in GS. Obviously q2 is not in D0(S).
q3 can not be in D0(S), otherwise the above path would be an augmenting path of the maximum matching found by the algorithm.
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65. Algorithm 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)
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66. Step 3: COVER an approximation to the Minimum Weight Set Cover of S
S – collection of all legal sets of at most kg lightpaths, each with its switching cost.
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
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67. Legal coloring
For any fixed g, the number of subsets constructed in the first phase is
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68. 5.2 analysis: ln(g)-approximation for a ring
Legal coloring
, B is 1-colorable 
A is 1-colorable ( correctness).
(and cost(A)  cost(B).)
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69. for every set cover SC.
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70. Lemma: There is a set cover SC, s.t.:
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71. Conclusion:
For k = g ln g :
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72. Lemma: There is a set cover SC, s.t.:
5.3 proof of lemma
Lemma: There is a set cover SC, s.t.:
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73. endpoints(Px) - the set of ADMs operating at wavelength x.
Use OPT to build SC
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
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74. M=4 k=5 k k k k
S S Sm
{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
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75. w/o the assumption we have:
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76. 5.4 trees
undirected:
directed:
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77. On-line problem Competitive analysis 6.1 on-line
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.)
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78. When a new path arrives:
6.2 algorithm ALG
When a new path arrives:
if closes a unicolor cycle
if any endpoint colored
else (no side colored)
- assign same color
- assign same color
- assign a new color
#ADMs=7
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79. When a new path arrives:
1. if closes a unicolor cycle
2. if any endpoint colored
3. else (no side colored)
- assign same color
- assign same color
- assign a new color 1 2 2 2 3 3 3 2
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80. Theorem: ALG is 3/2-competitive on a path. This is optimal.
6.3 Results
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.
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81. ALG ≥ 7/4 6.4 ALG ≥ 7/4 even for a ring ALG: ADM=7 OPT: ADM=4
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82. 6.5 any algorithm for a path ≥ 3/2
k paths
k=12
x=6
k-1 spaces:
x between same color
k-1-x between different colors
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83. So far: any algorithm uses 2k ADMs
now – a short path at each gap of diffferent colors
(k-1-x such gaps)
k=12, x=6, =5
Any algorithm uses at least one more ADM for each (ALG uses exactly one)
So: any algorithm ≥ 2k + (k-1-x) ADMs
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84. So far: use ≥ 2k + (k-1-x) ADMs
now – two long paths at each of the k gap of same color
Any algorithm must use 2 ADMs for each
So: any algorithm ≥ 2k + (k-1-x) + 4x =
= 3k+3x-1 ADMs
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85. any algorithm/OPT  3/2 – 1/(2k)
We showed: any algorithm uses ≥ 3k+3x-1 ADMs
OPT: the short paths ≤ 2k ADMs
for the long paths 2x ADMs
OPT  2k + 2x
any algorithm/OPT  3/2 – 1/(2k)
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86. 6.6 ALG for a path ≤ 3/2 ≤ -> ≤ ≤ ≤ ≤ ≤
Optimal solution S* saves x ADMs
Our solution S saves y ADMs ≤
-> ≤ ≤ ≤ ≤ ≤
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87. ≤ ≤
We show that ≤ ≤
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88. optimal S* – max matching at each point
Claim: ≤
optimal S* – max matching at each point
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89. For the proof choose a specific S*
savings of S (y)
savings of S* (x)
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90. b a c
a came before b c
map
1-1
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91. Savings of S* (x)
savings of S (y) ≤ ≤
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92. Theorem: cost(S)  7cost(S*)/4
6.7 ALG is 7/4-competitve for any topology
Lemmas
1. cost(S) - cost(S*) = = N/2 + (d0(S)-d2(S)-2|chains(S*)|)/2
2. d0(S)  d2(S) + |chains(S*)| + N/2
Combining, we have
|cost(S)| - |cost(S*)|  3N/4 3 |cost(S*)|/4
previous slide
next slide
Theorem: cost(S)  7cost(S*)/4
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93. Lemma 1 d0(S)  d2(S) + |chains(S*)| + N/2 orient S* for any u  D0(S)
if u is last in some chain of S*, map u to this chain
else
u’  D0(S) contradiction
u’  D1(S) map u to {u, u’}
u’  D2(S) map u to u’ S* u u’
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94. 6.8 lower bound of 7/4 , even for a ring
Case a:
7/4=1.75
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95. any algorithm ≥ 1.67 Exercise: any algorithm ≥ 1.75- Case b: Case b1:
6/3 = 2
Case b2:
5/3 = 1.67
any algorithm ≥ 1.67
Exercise:
any algorithm ≥ 1.75-
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96. 6.9. a simpler lower bound of 7/4 (not for a ring)C E D
EFG F
BDG G H K M
so: BDG
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97. A B C E D
EFG F
BDG G H K M
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98. A B C E D EFG F BDG G H EABDG GFEAB #ADMS=7 K M #OPT=4
Competitive Ratio: 7/4
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99. Competitive Ratio: 6/3 > 7/4B C E D
EFG F
BDG G H
BAE
#ADMS=6 K M
#OPT=3
so: BAE
Competitive Ratio: 6/3 > 7/4
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100. so: EFKMHG A B C E D EFG F BDG G H BAE EFKMHG K M Bertinoro, 5/5/08
100

101. A B C E D
EFG F
BDG G H
BAE
EFKMHG K M
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101

102. Competitive Ratio: 9/5 > 7/4B C E D
EFG F
BDG G H
BAE
EFKMHG
EABDCHG
#ADMS=9 K M
#OPT=5
Competitive Ratio: 9/5 > 7/4
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102

103. Hw: finish this case A B C E D EFG F BDG G H BAE EFKMHG K M
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103

104. Hw: finish this case A B C E D EFG F BDG G H BAE K M Bertinoro, 5/5/08
104

105. 1st open problem: what can be said about the trade-off between #colors and #ADMs=8 ?
#colors=2, #ADMs=8
#colors=3, #ADMs=7
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106. ALG, path ≤ 3/2 Optimal solution S* saves x ADMs
Our solution S saves y ADMs
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107. Optimal solution S* saves x ADMs Our solution S saves y ADMs
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108. Optimal solution S* saves x ADMs Our solution S saves y ADMs
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109. 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
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110. When a new path arrives:
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
- assign a new color
ADM=7
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111. ALG-TRIANGLE When a lightpath arrives if p is length-2
if closes unicolor cycle, assign same color
else assign new color
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’’
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112. When a new path arrives: if closes a unicolor cycle
if closes a unicolor path
if any endpoint colored
else (no endpoint colored)
- assign same color
- assign same color
- assign same color
- assign a new color
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113. Optimal solution S* saves x ADMs Our solution S saves y ADMs
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114. ONLINE COLORING – INPUT 2A B C
Total ADMS: 10 12 14 4 5 8 2 6 E
w(EFG)=1 D
w(BDG)=2 F
w(BAE)=1 G H
w(EFKMHG)=2
w(GFEAB)=3
w(EABDG)=4
w(BDGFE)=5 K M
w(EABCDG)=6
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114

115. OFFLINE COLORING – INPUT 2A B C
Total ADMS: 8 E
w(EFG)=1 D
w(BDG)=2 F
w(BAE)=3 G H
w(EFKMHG)=4
w(GFEAB)=2
w(EABDG)=1
w(BDGFE)=3 K M
w(EABCDG)=4
Competitive Ratio: 14/8=7/4
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115

116. ONLINE COLORING – INPUT 3A B C
Total ADMS: 4 5 9 2 7 E
w(EFG)=1 D
w(BDG)=2 F
w(BAE)=1 G H
w(EFKMHG)=3
w(EABDCHG)=4 K M
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116

117. OFFLINE COLORING – INPUT 3A B C
Total ADMS: 5 E
w(EFG)=1 D
w(BDG)=2 F
w(BAE)=1 G H
w(EFKMHG)=3
w(EABDCHG)=4 K M
Competitive Ratio: 9/5 >7/4
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117

118. ONLINE COLORING – INPUT 4A B C
Total ADMS: 14 12 10 6 5 8 4 2 E
w(EFG)=1 D
w(BDG)=2 F
w(BAE)=2 G H
w(BDCHG)=1
w(EABDG)=3
w(GFEAB)=4
w(GKFEAB)=5 K M
w(EFGDB)=6
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118

119. OFFLINE COLORING – INPUT 4A B C
Total ADMS: 8 E
w(EFG)=1 D
w(BDG)= 2 F
w(BAE)=3 G H
w(BDCHG)=4
w(EABDG)=1
w(GFEAB)=4
w(GKFEAB)=2 K M
w(EFGDB)=3
Competitive Ratio: 14/8=7/4
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119

120. ONLINE COLORING – INPUT 5A B C
Total ADMS: 4 5 9 2 7 E
w(EFG)=1 D
w(BDG)=2 F
w(BAE)=2 G H
w(BDCHG)=3
w(GHMKFEAB)=4 K M
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120

121. OFFLINE COLORING – INPUT 5A B C
Total ADMS: 5 E
w(EFG)=1 D
w(BDG)=2 F
w(BAE)=1 G H
w(BDCHG)=1
w(GHMKFEAB)=2 K M
Competitive Ratio: 9/5>7/4
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121