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Dynamic Wavelength Allocation and Wavelength Conversion

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Wavelength Converters A wavelength converter is modeled by a bipartite graph. For any two adjacent edges x,y, we define the corresponding conversion graph G xy =(V x,V y,E xy ) where: V x ={x 0,x 1,…x W-1 } V y ={y 0,y 1,…y W-1 } if and only if wavelength i on endpoint x can be converted to wavelength j on endpoint y (i is compatible with j on x-y ).

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Wavelength Converters A wavelength converter is symmetric if: A full wavelength converter corresponds to a complete bipartite graph. The degree of a wavelength converter is the maximum degree of a node in the bipartite graph.

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Full Wavelength Converters Any instance can be colored using W=L wavelengths. (In fact we need only d>=L) Proof: –Given a path, we can color its edges independently of each other, because the full conversion capability. – Consider any edge : There are at most L-1 paths other than p traversing this edge. They use at most L-1=W-1 colors. We can use any one from the remaining colors for p.

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Expander Graphs Definition: Given any, we define the neighborhood of S, namely: Definition: A bipartite graph (V 1,V 2, E) is an ( , , d)-expander if: –each node has degree at most d. –0 < < ½ – > 1 –for any,

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Expander Graphs Lemma: There is a triple ( , ,d), such that: for every sufficiently large n, there is an ( , ,d)-expander with n nodes.

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Limited Wavelength Converters (Any Graph) Theorem: There exists two constants k>1 and d>1, such that every instance can be colored with –W=kL colors –Using wavelength converters with degree d. Proof: Between each two adjacent edges we use the converter which correspond to the ( , ,d)- expander whose existence is guaranteed by the previous lemma. Let k=1/min{ ( -1),1- } We will prove that as long as L <= W min{ ( - 1),1- }, any path can be colored.

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Limited Wavelength Converters (Any Graph) Assume L <= W ( -1) and L<=W(1- ) Consider a path p=(e 1, e 2, …, e l ) to be colored. For any edge e 1, e 2, …, e l a color is said to be busy if it is used by another path. For any edge e i, (i>1) a color c is said to be busy also if all the colors compatible with c are busy in e i-1

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Limited Wavelength Converters (Any Graph) Claim:There are W colors which are not busy (idle) in e i. (By induction on i) –i=1:L<=W(1- ), therefore there are W colors idle in e 1. –i > 1: In edge e i+1 there are at least W colors compatible with the idle colors of e i At most L

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Limited Wavelength Converters (Rings) Theorem: Any instance of ring graph can be colored with –W=L log L + 4 L colors (independent of N !!) –using converters of degree 2. Proof: –Divide the ring into segments of length at least L, but less than 2L. W line (N,L)<=L log N (prove) W line (2L,L)<=L log N + L We can color the intra-segment paths with L logN + L colors with no wavelength conversion.

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Limited Wavelength Converters (Rings) –Use the following graph to color inter-segment paths: An edge of the graph is a color. A vertex joins compatible colors. First segment Intermediate segment Last segment u1u1 u2u2 uLuL v1v1 v2v2 vLvL

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Incremental WLA in Rings Claim: Any instance in Ring graphs can be colored using W <= max{L, 2L-d}colors. Algorithm: –Initialization: M = max {0, L-d} for i=0 to M do POOL(0)={1,…,min{L, d}} w=d for i=1 to M do POOL(i)={++w, ++w}

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Incremental WLA in Rings Notation: l(e/S) -Load induced on edge e by paths in S, namely: Note that l(e)=l(e/P). F i the set of paths received before path i, namely: F i ={p 1,p 2,…,p i-1 }

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Incremental WLA in Rings Algorithm (path p) –i=0; –While L(p/S i ) >= d+i do i++ – –Color the edges of p using wavelengths from SHELF(i)

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Incremental WLA in Rings Lemma: Let then Proof: Assume and, therefore contradicting to the fact that.

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Incremental WLA in Rings Lemma: Let then Proof: w.l.o.g. x

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Incremental WLA in Rings Assume, therefore The algorithm would place p y in SHELF(j) for some j>i.

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Incremental WLA in Rings The maximum load induced by the paths of SHELF(0) is d. By code inspection. The maximum load induced by the paths of SHELF(i) (i>0) is 2. –Assume otherwise. There is an edge with three paths traversing it. By previous lemma, none of them contains the other. W.l.o.g assume they are sorted by their starting points:

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Incremental WLA in Rings By the above picture, for any set S of paths: By the first lemma: Load is non decreasing: Combining, we get: The algorithm would place p y in SHELF(j) for some j

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