Minimizing the number of ADMs with and without traffic grooming

Minimizing the number of ADMs with and without traffic grooming

Minimizing the number of ADMs
with and without traffic grooming:
complexity and approximability
Shmuel Zaks
Bertinoro, 5/5/08
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Minimizing the number of ADMs
with and without traffic grooming:
complexity and approximability
Prudence Wong – Liverpool
Michele Flammini, Gianpiero Monaco,
Luca Moscardelli - L’Aquilla
Mordechai
Shalom, Shmuel Zaks - Technion
Bertinoro, 5/5/08
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1.
Model, problems
2.
Warm up examples
3.
Complexity
4.
ADMs - Approximation
5.
Grooming – appproximation
6.
ADMs - On-line
7.
Open problems
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1.
Model, problems
2.
Warm up examples
3.
Complexity
4.
ADMs - Approximation
5.
Grooming – approximation
6.
ADMs - On-line
7.
Open problems
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1.1. Basics
Optical networks - 1st generation
the fiber serves as a transmission medium
Electronic
switch
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Optic
fiber
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Optical networks - 2nd generation
Routing in the optical domain
Two complementing technologies:
- Wavelength Division Multiplexing (WDM):
Transmission of data simultaneously at
multiple wavelengths over same fiber
- Optical switches: the output port is
determined according to the input port and
the wavelength
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lightpaths
ADM
OADM
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Directed:
Symmetric:
Undirected:
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λ4
λ3
λ2
λ1
Optic Fiber
λ4
λ3
λ2
λ1
Optic Fiber
λ4
λ3
λ2
λ1
Optic Fiber
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OADM (optical add-drop multiplexer)
Optical
switch
lightpath
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ADM (add-drop multiplexer)
Electronic device at the endpoints of
lightpaths
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Where can we save?
an ADM can be shared by two lightpaths
2 ADMs
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1 ADM
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0
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1
2
3
1
2
3
4
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lightpaths
p1
Valid coloring
w( p1 ) ≠ w( p2 )
p2
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Traffic grooming
„
„
„
low capacity requests can be groomed
into high capacity wavelengths
(colors).
colors can be assigned such that at
most g lightpaths with the same color
can share an edge
g is the grooming factor
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lightpaths - with grooming
g=2
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Valid coloring
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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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1.2. Cost functions
1. number of wavelengths
#colors = 4
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2. Switching cost
OADM
ADM
α#ADMs + β#OADMs
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ADM
OADM
α#ADMs + β#OADMs = 12α + 8β
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but number of OADMs is fixed, so …
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2. number of ADMs
#ADMs = 12
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#ADMs=12
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#ADMs=9
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Trade-off between #colors and #ADMs
#ADMs=12
#ADMs=9
#colors=4
#colors=3
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#ADMs=8
#ADMs=7
#colors=2
#colors=3
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With grooming
#ADMs=9
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g=2
#ADMs=8
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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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1.4. Problems not covered
™ Other cost functions (e.g., OADM+ADM)
™
™
™
Design the placement of ADMs in a network
to cope with a family of demands
Given pairs to connect, design a routing and
a coloring
more …
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1.5. References
General references
Minimizing # of ADMs
Gerstel, Lin, Sasaki, 1998
Traffic grooming
Gerstel, Ramaswamy, Sasaki, 1998
Zhu, Mukherjee, 2003
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Complexity
NP-complete of minADM
g=1:
ring – Eilam, Moran, Z., 2002
g>1:
ring – Chiu, Modiano, 2000
fixed g, path, ring - Shalom, Unger, Z. , 2006
star, fixed g>3 ( poly for g=1,2)
Shalom, Flammini, Monaco, Moscardelli, Z., 2007
Inapproximability
OPT+Nα ,α<1 – impossible
Eilam, Moran, Z. 2002
NoPTAS Amini, Perennes, Sau, 2007
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Approximation
General topology
3
OPT + N
5
Eilam, Moran, Z., 2002
Calinescu, Frieder, Wan, 2002
N
1
O PT +
(1 + ε ) ( 0 ≤ ε ≤
)
2
+2
Calinescu, Frieder, Wan, 2002
Ring
3
10 2
+ε
7
10
7
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Calinescu, Wan , 2002
Shalom, Z. , 2004
Epstein, Levin, 2004
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Traffic Grooming
Ring
ln g
Flammini, Moscardeli, Gianpierro, Shalom, Z. 2005/6/7
On-line
7
4
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Shalom, Wong, Zaks, 2007
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Misc
Bermond, Coudert, 2003
Bermond, Braud, Coudert, 2005
Shalom, Z. , 2005
Bermond, Coudert, Munoz, Sau, 2006
Munoz, Sau, 2008
More …
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1.
Model, problems
2.
Warm up examples
3.
Complexity
4.
ADMs - Approximation
5.
Groomin – approximation
6.
ADMs - On-line
7.
Open problems
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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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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
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w/ grooming:
N/g ≤ ALG ≤ 2N
N/g ≤ OPT ≤ 2N
ALG ≤ 2g x OPT
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2.2 Basic relation
Cycles are good, chains are bad
N lightpaths
cycles
chains
#ADMs = N + #chains
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#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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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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2.4 a basic lemma
Assume that an optimal solution S*
saves x ADMs,
and a solution S saves y ADMs
cost(S*)
Lemma : if y ≥
then
k
1
cost(S) ≤ (2 - )cost(S*)
k
cost(S*)
for example : if y ≥
then
2
3
cost(S) ≤ cost(S*)
2
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Optimal solution S* saves x ADMs
a solution S saves y ADMs
y≥
cost(S*) N
≥
k
k
cost(S) = 2N - y
cost(S*) = 2N - x
N
N
N
2N 2N 2N - y
cost(S)
1
k
k
k
=
≤
≤
=
= 2cost(S*) 2N - x 2N - x 2N -N
N
k
2N -
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2.5 notaton
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
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d0(S) + d1(S) + d2(S) = 25 = N
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2.6 a basic tool
Solution S=chains + cycles
d0 (S) + d1 (S) + d2 (S) = N
2d0 (S) + d1 (S) = N + d0 (S) - d2 (S)
#ADMs = N + #chains
S – ALG, S* - OPT
co st(S ) - co st(S * ) = # c h a in s(S ) - # ch a in s(S * ) =
= d0 (S) +
2d (S) + d1 (S) -2#chains(S*)
d1 (S)
- #chains(S*) = 0
2
2
N + d0 (S) - d2 (S) -2#chains(S*)
=
2
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d0 (S) - d2 (S) -2#chains(S*)
1
N(1 +
)
2
N
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cost(S)-cost(S*) =
d (S) - d2 (S) -2#chains(S*)
1
N(1 + 0
)
2
N
We define:
ε(S) =
and get
d0 (S) - d2 (S) -2 #chains(S*)
N
1
cost(S) = cost(S*) + N(1 + ε(S))
2
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2.7 example
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
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d0(S) + d1(S) + d2(S) = 25 = N
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N=25
suppose #chains(S*)=2, cost(S*)=25+2=27
ε(S) =
d0 (S) - d2 (S) -2 #chains(S*)
2 -19 - 4
25
=−
21
25
N
=
d0(S) = 2
d2(S) = 19
1
25
21
cost(S) = cost(S*) + N(1 +ε(S)) = cost(S*) +
(1 - ) =
2
2
25
cost(S*) +2 = 29
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1.
Model, problems
2.
Warm up examples
3.
Complexity
4.
ADMs - Approximation
5.
Grooming – approximation
6.
ADMs - On-line
7.
Open problems
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3.1 line
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Greedy coloring
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1st cost function: Number of colors
1. In any coloring #_of_colors ≥ max_load
2. In the greedy coloring #_of_colors = max_load
3. Hence this is optimal
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max load = 4
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2nd cost function: Number of ADMs (no grooming)
1. At each point any coloring can save ≤ min{left,right} ADMs
2. The greedy coloring saves min{left,right} ADMs
3. Hence it is optimal
left=3, right=2, can save at most min{2,3}=2
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3nd cost function: Number of ADMs (with grooming)
APX – hard any g > 1
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3.2 minADM is NP-complete for a ring
minADM
Input: a graph, a set of lightpaths, t>o.
Output: can the lightpath be colored such that
#ADMs ≤ t ?
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Coloring of an interval graph
Always possible with max load
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k=4
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Coloring of a circular arc graph
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Coloring of a circular arc graph
Not always possible with max load
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Coloring of a circular arc graph
Input: circular arc graph G, k>o.
Output: can the arcs be colored by ≤ k colors?
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Coloring of a circular arc graph
Input: circular arc graph G, k>o.
Output: can the arcs be colored with ≤ k colors?
G
minADM
Input: a graph, a set of lightpaths, t>o.
Output: can the lightpath be colored
such that #ADMs ≤ t ?
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Given an instance of the circular arc graph
problem, construct an instance H of minADM:
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Claim: can color G with ≤ k colors
iff can color H with ≤ k colors
iff can color H with #ADMs ≤ N.
G
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H
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Claim: can color H with ≤ 3 colors iff
can color H with #ADMs ≤ 13
Assume a
coloring with ≤ 3
colors …
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Claim: can color with ≤ 3 colors iff
can color the lightpaths with ≤ 13 ADMs
Assume a
coloring with ≤
13 ADMs …
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Note:
impossible to color in 2 colors
impossible to color with 5 ADMs
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3.3 minADM with grooming is NP-complete
for a star
path of length 1
path of length 2
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Star, g=1
( trivial )
path of length 1
path of length 2
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Star, g=2
The number of used ADM is exactly equal to
the lower bound of needed ADM:
1
n
node 0
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2
0
i
⎡ n
⎤
⎢∑ yi ⎥
⎢ i= 1
⎥
⎢
⎥+
⎢ 2
⎥
⎢
⎥
⎢
⎥
n
∑
i= 1
xi
paths of length 2
yi
paths of length 1
⎡ xi+yi
⎢
⎢⎢ 2
⎤
⎥
⎥⎥
nodes 1,…,n
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Star, g≥3 - NP-complete
Sketch for g=3:
3-Exact Cover ⇒
Edge Partition into 3-regular graphs ⇒
Star grooming, g=3
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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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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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3-Exact Cover ⇒
Edge Partition into 3-regular graphs ⇒
Star grooming, g=3
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sets
elements
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3-Exact Cover ⇒
Edge Partition into 3-regular graphs ⇒
Star grooming, g=3
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4
1
4
G
S
3
1
0
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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4
5
5
3
1
1
4
0
3
2
2
δ=2
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g=2
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1.
Model, problems
2.
Warm up examples
3.
Complexity
4.
ADMs - Approximation
5.
Grooming – approximation
6.
ADMs - On-line
7.
Open problems
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4.1 basic algorithm
PIM( )
eliminate short cycles, then find matchings
„
Preprocessing:
„
While there is a cycle C of length ≤
„
„
do:
Remove (the lightpaths of) C from the instance
Processing:
„
„
„
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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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„
„
„
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:
1
cost(S) = cost(S*) + N(1 +ε(S))
2
d (S) - d2 (S) -2 #chains(S*)
ε(S) = 0
N
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algorithm PIM( )
N
P IM ( ) = O P T +
(1 + ε )
2
0 ≤ ε ≤
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1
+ 2
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Need to prove:
ε(S) =
d0 (S) - d2 (S) -2 #chains(S*)
N
≤
1
+2
We will show:
N
d0 (S) - d2 (S) -2 #chains(S*) ≤ d0 (S) - d2 (S) - #chains(S*) ≤
+2
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N
d0 (S) - d2 (S) - #chains(S*) ≤
+2
Take S*:
in the example :
38 lightpaths, 5 chains, 3 cycles
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d0 (S) - d2 (S) - #chains(S*) ≤
N
+2
preprocessing stage of S: eliminate cycles of size ≤
The lightpaths that we used are colored red:
in the example : 10
lightpaths are used to form
cycles
in S
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d0 (S) - d2 (S) - #chains(S*) ≤
N
+2
So, after preprocessing stage of S: in S* we have the
following lightpaths:
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d0 (S) - d2 (S) - #chains(S*) ≤
N
+2
Now the algorithm is doing MM,MM,MM,…
We show that already after the first MM we are ok.
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N
d0 (S) - d2 (S) - #chains(S*) ≤
+2
We show that in the remaining lightpaths there is a
MM’ ≤ MM that is ok.
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d0 (S) - d2 (S) - #chains(S*) ≤
N
+2
d0 (S) - number of isolated lightpaths
in the example :
d0 (S) = 8
1 for each odd path and
forBertinoro,
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odd cycle
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N
d0 (S) - d2 (S) - #chains(S*) ≤
+2
d0 (S) = d0 (odd cycles) + d0 (odd chains)
Show:
d0 (odd cycles) ≤
N
+2
d0 (odd chains) ≤ d2 (S) + #chains(S*)
Which implies
N
d0 (S) - d2 (S) - #chains(S*) ≤
+2
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N
d0 (S) - d2 (S) - #chains(S*) ≤
+2
N
d0 (odd cycles) ≤
+2
Since there are at most N lightpath
left, and each odd cycle is of size at
least
+2
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d0 (S) - d2 (S) - #chains(S*) ≤
N
+2
d0 (odd chains) ≤ d2 (S) + #chains(S*)
1. Original chain of S* that was untouched
1 here is matched with 1 here
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d0 (S) - d2 (S) - #chains(S*) ≤
N
+2
d0 (odd chains) ≤ d2 (S) + #chains(S*)
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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d0 (S) - d2 (S) - #chains(S*) ≤
N
+2
d0 (odd chains) ≤ d2 (S) + #chains(S*)
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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4.2 basic algorithm without preprocessing
PIM(l)
„
Preprocessing:
„
„
While there is a cycle C of length ≤ do:
„ Remove (the lightpaths of) C from the instance
Processing:
„
„
„
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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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Without preprocessing:
1
ε≤
5
1
cost(S) = cost(S*) + N(1 +ε(S))
2
3
PIM(1) ≤ OPT + N
5
This is optimal.
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x
w
a
d
b
e
c
u
v
x
a
x
d
w
v
x
e
x
b
u
c
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A solution
„
A Partition of P into feasible chains and
cycles
„ Feasible Æ 1-colorable
x
d
u
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a
b
e
c
v
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PIM(3)
x
w
a
d
b
u
x
a
e
c
v
x
d
w
v
x
e
x
b
u
c
Cost=5
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PIM(1)
x
d
u
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w
a
b
e
c
v
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x
w
a
d
e
b
c
u
v
x
a
x
d
w
Cost = 8
v
x
e
x
b
u
c
= OPT + 3 =
= OPT + 0.6x5
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Lower Bound
The performance of PIM(1) can be as
3
bad as
PIM(1) ≤ OPT + N
5
...
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Upper bound
d0 (S) - d2 (S) -2#chains(S*)
„ Recall: ε(S) =
N
and
1
cost(S) = cost(S*) + N(1 + ε(S))
2
„ to show that
3
PIM(1) ≤ OPT + N
5
we prove that ε(S) ≤ 1/5
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Orient the chains and cycles of S*.
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Let LAST be the set of nodes which are last
elements of the chains according to this
orientation.
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ε(S) =
d0 (S) - d2 (S) -2#chains(S*) 1
≤
N
5
show :
N
d0 (S) - d2 (S) -2#chains(S*) ≤
5
or
N
d0 (S) ≤ d2 (S) + #chains(S*) +
5
By mapping a path in D0(S) to
either
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a path of D2(S)
or to a chain
or to a cycle of size ≥ 5
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The Mapping
As the p
graph
∈ Dis0 (finite,
S ) this process continues until p is
mapped, or we re-encounter a node. In this case:
q1
q2
q3
q0=p
9C = {qi}
9Map p to C
q
If
If
q
q
is
is
not
in
the
D
be
last
(S)
in
then:
node
Dthe
(S),
of
otherwise
a
path
of
the
S*
algorithm
then:
would
Otherwise,
Otherwise,
q
is
is
the
next
next
node
node
in
in
q
q
’s
’s
path/cycle
path/cycle
in
in S*
S*
If
1 can
q
20
is
the
2q
last
node
0exactly
of
a
path
of
S*
then:
3
1
2
0
If
Otherwise
q
2
is
in
D
q
(S)
has
then:
one
neighbor
q
in
G
.
q
can
not
be
in
D
(S),
otherwise
the
above
path
would
1
2
1
2
S
9return
3
0 ) to the matching.
add
the
edge
(q
,q
0
Obviously
q2 is not1path
in Dof
9p”=q
9p’=q
be
an
augmenting
the maximum matching
0(S).
9p’=q
2
0
9p”=q
2
1
Itfound
is easy
tothe
see
that |C| is odd. We also show that |C|
by
algorithm.
9map
to
p’
> 3. 9map
9map p
p to
to p”
p’
9map
pp
to
p”
9return
9return
9return
9return
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4.3 extension of analysis
Improved analysis of Algorithm PIM ( )
N
P IM ( ) = O P T +
(1 + ε )
2
1
1
≤ ε ≤
.
2 + 3
1 .5 + 3
( p r e v io u s ly : 0 ≤ ε ≤
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1
)
+ 2
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∈ GS
Lower bound
∉ GS
+2 nodes
+1 nodes
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All the possible edges
between the two cycles
are in the conflict graph
There are no feasible
cycles of length <= .
The matching leaves
only one isolated node,
therefore maximum.
The matching can not
be extended to a bigger
solution, because of the
conflict edges.
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+2 nodes
+1 nodes
„
„
„
„
d0(S)=1
d2(S)=0
chains(S*)=0
N=2 +3
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1
ε=
2 +3
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Upper bound
ε(S) =
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d0 (S) - d0 (S) +2#chains(S*)
1
≤
.
1.5 + 3
N
110/169
Comb.
Matching
Lemma
of -odd
-one
even
cycles
cycles
even
Comb.
cycles
lemma
with
odd
color
cycles
The Matching
I1
I
Max ind.
odd
cycles
I2
ID
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D1
D
D2
Other
odd
cycles
ED
E
E2
Even
cycles
111/169
The
Evenmaximality
cycles, one
ofcolor:
I implies:
|ID|≤|E
|D1D|≤|I
| 2|
At least 2 cycles per isolated node, therefore at least
2 +3 (≥ 3/2 ( +2)) nodes per isolated node.
The Matching - Analysis
I
I1
D1
D
D2
I2
ID
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ED
E
E2
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≥ ( +2)/3 “purple” nodes (Lemma 1)
≥ (3/2)( +2) nodes
≥ (1/2)( +2)/3) white nodes (Lemma 2)
The Matching - Analysis
I
I1
D
≥ +2
nodes
D2
E
E2
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1.
Model, problems
2.
Warm up examples
3.
Complexity
4.
ADMs - Approximation
5.
Grooming – approximation
6.
ADMs - On-line
7.
Open problems
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5.1 algorithm
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 ≤ k ⋅ g
If a subset A is 1-colorable (i.e., any edge is used at
most g times) then
weight[A]=endpoints(A)
S ← S ∪ {A}
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S – collection of all legal sets of at most kg
lightpaths, each with its switching cost.
Step 3: COVERÅ 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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Legal coloring
For any fixed g, the number of subsets
constructed in the first phase is O ng ik
( )
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5.2 analysis: ln(g)-approximation for a ring
Legal coloring
A ⊂ B , B is 1-colorable ⇒
A is 1-colorable (⇒ correctness).
(and cost(A) ≤ cost(B).)
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ALG = cost(PARTITION)
≤ weight(COVER) ≤
≤H
k ⋅g
weight(MINCOVER)
≤ (1 +ln(k ⋅ g))weight(SC)
for every set cover SC.
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A L G = cost(PA R T IT IO N ) ≤
≤ (1 + ln(k ⋅ g)) w eigh t(S C )
Lemma: There is a set cover SC, s.t.:
⎛ 2g ⎞
weight(SC) ≤ ⎜ 1 +
⎟ O PT
k ⎠
⎝
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Conclusion:
ALG = cost(PARTITION)
≤ weight(COVER) ≤
≤H
k ⋅g
weight(MINCOVE R)
≤ (1 + ln(k ⋅ g))weight(SC ) ≤
⎛ 2g ⎞
(1 + ln(k ⋅ g)) ⋅ ⎜ 1 +
⋅ OPT
⎟
k ⎠
⎝
ALG
For k = g ln g :
≤ 2lng + o(lng)
OPT
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5.3 proof of lemma
Lemma: There is a set cover SC, s.t.:
⎛ 2g ⎞
weight(SC) ≤ ⎜ 1 +
⎟ O PT
k ⎠
⎝
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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)|= m ⋅ k)
Partition endpoints(Px) into m
sets of k consecutive nodes
in the example: k=5, m=4
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M=4
k=5
S1
k
S2
k
k
k
Sm
weight[S]
i ≤ k + g ≤ k ⋅g
{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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weight[S]
i ≤k+g
m
∑ weight[S] ≤ m(k + g)
i=1
i
( OPTx = m ⋅ k)
⎛ g⎞
weight[S]
∑
i ≤ OPTx ⎜ 1 +
⎟
⎝ k⎠
i=1
m
w/o the assumption we have:
⎛ 2g ⎞
weight[S]
∑
i ≤ OPTx ⎜ 1 +
⎟
k ⎠
⎝
i=1
m
⎛ 2g ⎞
weight[S]
∑∑
i ≤ OPT ⎜ 1 +
⎟
k ⎠
⎝
x i =1
m
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5.4 trees
undirected:
ALG ≤ 2ln(δ g) OPT
directed:
ALG ≤ 2lng OPT
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1.
Model, problems
2.
Warm up examples
3.
Complexity
4.
ADMs - Approximation
5.
Grooming – approximation
6.
ADMs - On-line
7.
Open problems
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6.1 on-line
ƒ On-line problem
ƒ Input arrives one at a time, and a decision is
made (and cannot be changed).
ƒ In the minADM problem: lightpaths arrive
one at a time, and need to be colored.
ƒ Competitive analysis
ƒ 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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6.2 algorithm ALG
When a new path arrives:
1. if closes a unicolor cycle- assign same color
2. if any endpoint colored - assign same color
3. else (no side colored) - assign a new color
#ADMs=7
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When a new path arrives:
1. if closes a unicolor cycle - assign same color
2. if any endpoint colored - assign same color
- assign a new color
3. else (no side colored)
2
1
2
2
3
3
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3
2
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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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6.4 ALG ≥ 7/4 even for a ring
ALG: ADM=7
OPT: ADM=4
ALG ≥ 7/4
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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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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, 12-1-6=5
Any algorithm uses at least one more ADM
for each (ALG uses exactly one)
So: any algorithm
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≥ 2k + (k-1-x) ADMs
134/169
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
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ADMs
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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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6.6 ALG for a path ≤ 3/2
Optimal solution S* saves x ADMs
Our solution S saves y ADMs
Lemma :
x
y≥
≥
k
1
⇒ cost(S) ≤ (2 - )cost(S*)
->
k
Proo f : cost(S) - cost(S*) = (2N - y) - (2N - x) =
=x-y≤
≤x-
⇒
x
1
1
1
N
cost(
S
*
)
= x (1 - ) ≤
)
≤
)
(1
(1
≤
≤
k
k
k
k
1
c ost(S) ≤
cost(S*
)(2
)
≤
k
Bertinoro, 5/5/08
Note :
x ≤ N ≤ cost(S*)
137/169
≥
Lemma : if y ≥
x
then
k
1
cost(S) ≤
≤ (2 - )cost(S*)
k
We show that
⇒
Bertinoro, 5/5/08
x
y≥
≥2
3
cost(S) ≤
≤ cost(S*)
2
138/169
Claim:
x
y≥
≥2
optimal S* – max matching at each point
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For the proof choose a specific S*
savings of S
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(y)
savings of S*
(x)
140/169
a
b
c
a came before b
c
map
1-1
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Savings of S*
(x)
savings of S (y)
≥≥
2
x
y≥
≥
2
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6.7 ALG is 7/4-competitve for any topology
1. cost(S) - cost(S*) =
= N/2 + (d0(S)-d2(S)-2chains(S*))/2
2. d0(S) ≤ d2(S) + chains(S*) + N/2
Combining, we have
3
cost(S) - cost(S*) ≤ N ≤ cost(S*)
4
7
cost(S) ≤ cost(S*)
4
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d0(S) ≤ d2(S) + chains(S*) + N/2
orient S*
for u ∈ D0(S)
1. if u is last in some chain of S*, map u to this
chain
2. else
S*
u
u’
i. u’ ∈ D0(S)
contradiction
ii. u’ ∈ D1(S)
map u to {u, u’}
iii. u’ ∈ D2(S)
map u to u’
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6.8 lower bound of 7/4 , even for a ring
Case a:
7/4=1.75
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Case b:
Case b1:
6/3 = 2
Case b2:
5/3 = 1.67
any algorithm ≥ 1.67
Exercise:
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any algorithm ≥ 1.75- ε
146/169
6.9. a simpler lower bound of 7/4
7/ (not for a ring)
A
B
E
D
C
EFG
F
G
H
K
M
BDG
so: BDG
Bertinoro, 5/5/08
147/169
A
B
E
D
C
EFG
F
Bertinoro, 5/5/08
G
H
K
M
BDG
148/169
A
B
E
D
C
EFG
F
G
H
BDG
EABDG
GFEAB
#ADMS=7
#OPT=4
K
M
Competitive Ratio: 7/4
Bertinoro, 5/5/08
149/169
A
B
E
D
C
EFG
F
G
H
K
M
BDG
BAE
#ADMS=6
#OPT=3
Competitive Ratio: 6/3 > 7/4
Bertinoro, 5/5/08
so: BAE
150/169
A
B
E
D
C
EFG
F
G
H
BDG
BAE
EFKMHG
K
M
so: EFKMHG
Bertinoro, 5/5/08
151/169
A
B
E
D
C
EFG
F
G
H
BDG
BAE
EFKMHG
K
Bertinoro, 5/5/08
M
152/169
A
B
E
D
C
EFG
F
G
H
BDG
BAE
EFKMHG
EABDCHG
#ADMS=9
#OPT=5
K
M
Competitive Ratio: 9/5 > 7/4
Bertinoro, 5/5/08
153/169
A
B
E
D
C
EFG
F
G
H
BDG
BAE
EFKMHG
K
M
Hw: finish this case
Bertinoro, 5/5/08
154/169
A
B
E
D
C
EFG
F
G
H
K
M
BDG
BAE
Hw: finish this case
Bertinoro, 5/5/08
155/169
1.
Model, problems
2.
Warm up examples
3.
Complexity
4.
ADMs - Approximation
5.
ADMs - On-line
6. Grooming - approximation
7. Open problems
Bertinoro, 5/5/08
156/169
Open Problems
Closing gaps between lower and upper bounds
Extension to other topologies (e.g., on-line,
ring of size n)
Extension and traffic patterns
Trade-offs between w,#ADM,g
Complexity, inapproximability
Bertinoro, 5/5/08
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Placement of the ADMs
Routing and coloring
Other traffic grooming models
More cost functions (OADMs) and criteria
(average cost, #savings)
Wavelength conversion
On-line and dynamic networks
Game theory
...
Bertinoro, 5/5/08
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Open Problems
Closing gaps between lower and upper bounds,
Extension to other topologies (e.g., on-line, ring
of size n), Extension and traffic patterns,
Trade-offs between w,#ADM,g, Complexity,
inapproximability, Placement of the ADMs,
Routing and coloring, Other traffic grooming
models, More cost functions (OADMs) and
criteria,(average cost, #savings), Wavelength
conversion, On-line and dynamic networks, Game
theory, . . .
Bertinoro, 5/5/08
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Open Problems
Closing gaps between lower and upper
bounds, Extension to other topologies (e.g.,
on-line, ring of size n), Extension and
traffic patterns, Trade-offs between
w,#ADM,g, Complexity, inapproximability,
Placement of the ADMs, Routing and
coloring, Other traffic grooming models,
More cost functions (OADMs) and criteria
(average cost, #savings), Wavelength
conversion, On-line and dynamic networks,
Game theory, …
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Questions ?
Bertinoro, 5/5/08
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Bertinoro, 5/5/08
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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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Step 1
1 = ln e
and
1 = sin α + cos α
2
2
and
2 =
∞
∑
n=0
Bertinoro, 5/5/08
⎛ 1 ⎞
⎜ ⎟
⎝ 2⎠
n
164/169
So
1+1=2
can be simplifid as
∞
⎛1⎞
ln e + sin α + cos α = ∑ ⎜ ⎟
n =0 ⎝ 2 ⎠
2
2
n
which is clearly more intuitive
Bertinoro, 5/5/08
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Step 2
1 = cosh( q ) * 1 − tanh ( q )
2
and
⎛ 1⎞
e = lim ⎜ 1 + ⎟
z →∞
z⎠
⎝
Bertinoro, 5/5/08
z
166/169
So
1+1=2
can be further simplifid as
⎛ ⎛ 1⎞
ln⎜lim⎜1+ ⎟
⎜ z→∞⎝ z ⎠
⎝
z
Bertinoro, 5/5/08
2
∞
⎞
cosh(q)* 1− tanh (q)
2
2
⎟⎟ +sin α + cos α = ∑
n
2
n=0
⎠
167/169
Step 3
0!= 1
and
(X ) − (X )
T −1
hence
( ) − (X )
⎛
⎜ X
⎝
Bertinoro, 5/5/08
−1 T
T −1
=0
−1 T
⎞
⎟!= 1
⎠
168/169
So eventually
⎛ ⎛⎛ T
ln⎜lim⎜⎜ X
⎜ z→∞⎝⎝
⎝
1+1=2
( ) −( X )
−1
−1 T
2
⎞ 1⎞
⎟!+ z ⎟
⎠ ⎠
gets its final form:
2
∞
⎞
−
cosh(
q
)*
1
tanh
(q)
2
2
⎟ +sin α +cos α = ∑
n
⎟
2
n=0
⎠
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?
Bertinoro, 5/5/08
169/169