David Larrabeiti Piotr Pacyna Universidad Carlos III de...

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David David LarrabeitiLarrabeitiPiotrPiotr PacynaPacyna

Dpto. de IngenierDpto. de Ingenieríía Telema TelemááticaticaUniversidad Carlos III de MadridUniversidad Carlos III de Madrid

PacketPacket--SwitchingSwitching

RearrangableRearrangable networksnetworks

Switching, Chapter 10 2

Programe authors and contributorsLecturers (Spanish course):

David Larrabeiti LópezJosé Félix Kukielka

Teaching material :

David Larrabeiti, Ricardo Romeral,Jorg Diederich, Huw OliverPiotr Pacyna (English course)

2


Objectives

Introduce basic concepts and termof packet switching

Introduce important network types

Present examples of networks

Talk about properties of networks

Comment on application areas


Part I. Basic information

3


Background info

Origins of the solutionsconnecting networks, inteconnection networkscentralised or distributed control

Connection capability of networksaccessibility

blocking


Primary discriminator(functional property)

Non-blocking

Blocking„internal congestion”

4


Non-blocking property

Strict-sense non-blocking

Wide sense non-blocking

Rearrangable non-blocking

cost index: the number of cross-points (traditionally)


Reference net.: crossbar network

N

M

5


Properties of crossbar network

C = N x M

strictly non-blocking

can set arbitrary permutationof inlets to outlets

< = > permutation set P = n!(fully accessible)

C = N 2


General model for multi-stage nets.

N M

1 2 s

1

2

r1

1

2

r2 rs

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General model for Multi-stage nets.

N Mni mi


Inter-stage inter-connection patterns

mi*ri = n i+1 * r i+1

i i+1„full connection”

7


Network parameters

N x M N: # inlets to network,M: # outlets from network

s: # stages

ri: # crossbar matrices at stage i,(i=1 ... s)

ni , mi - are the „dimensions” of crossbars at stage i(all are identical at stage i)


Network parameters (2)

ri * mi => total number of outlets fromstage stage i

N / n1 => number of crossbars at 1-st stage

8


Interconnection types

interconnection patterns are importantelement that define the network

EGS is an important classmany sub-classes

( ji , ki ) ( li+1 , mi+1 )

e.g. „full connection”

outlet crossbar inlet crossbar


Interconnections:Extented Generalized Shuffle

ii+1

( )

+−=

++

11

1inti

iiii r

jkmj

( ji , ki ) ( ji+1 , ki+1 )

( ) 11

mod1

11 +

+

+−=

++

irrjkmk

i

iiii

9


Network graphsNetworks vs. network graphs

Matrices verticesConnections edges

Isomorphic graphs

Isomorphic networks


Equivalence between networks

Topological equivalence

Isomorphism

Functional equivalence

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Equivalence between networks

Source: A. Pattavina, „Switching Theory, architecture and performance ....”


Isomorphic networks

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Isomorphic networks


Channel graph of multistage net.

channel = I/O path in a graph

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Part II. ( Full-connection ) multi-stage networks


General model of full-connectionmulti-stage net.

and

ni x mi switching element is a crossbar

r i-1 = n i m i-1 = r i

13


Multi-stage network types

Full-connection multistage networks

Partial-connection multistage networksPartial self-routing networks

Horizontal extensions

Vertical replication

Fully self-routing networks


Two-stage network

Two-stage FC networks:

not rearrangeable since only one path exists between any inlet/outlet pair

n x r2

#1

#r1

n x r2

#1

#r2

r1 x m

r1 x m

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Full-Connection Multistage Networks

At least three stages needed for FC multi-stage to be rearrangeable !


Example: Three-Stage Switch

NxM three stage switch:

r1 = N/nmatrices

r2 matricesN

ninputs

mM

nN×

kn ×

mM

nN×

kn ×

...

...

...

moutputs

......

...... m

outputsn

inputs

r3 = M/mmatrices

mk ×

mk ×

M

...

k internal paths

possible

1st stage 3rd stage

2nd stage

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Condition for rearrangability

Minimum requirement : Given a NxM three stage switch with r1 matrices in the 1st stage, r2 matrices in the 2nd stage, and r3matrices in the 3rd stage

Such a switch is rearrangeable if and only if the second stage has more than max(n, m) matrices of the size r1xr3. (n = N/r1 , m = M/r3)


Part III. Partial-connection multi-stage networks

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General properties

Matrices in intermediate stage are connecteed only to a subset of the matrices in the adjacent stages

Significant size !

High-performance in switching required

High degree of parallel processing required


Design principles

Built as multi-stage

Switching elements (SE) small : 2k x 2k k = 1 ... 5

SE - autonomous processing

self-routingblocking :=(

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Partial-Connection Multistage Rearrangeable Networks

Rearrangeable (i.e. non-blocking) networks can be built from banyan (i.e. blocking) networks Two classes of rearrangeable networks:

1. Partially self-routing networksSelf routing only in parts of the networkRequires processing to build up the required network permutation being partially distributed (for the self-routing parts of the network) and partially centralized (to determine the switching element states at the other network stages)

1. Horizontal extension

At least one stage added to basic banyan network2. Vertical replication

Whole banyan network replicated several times2. Fully self-routing networks

Self routing applied at all stages


Typical network permutations (1)

h-shuffle:

F h (an-1 .... a0) = an-1 ... a h+1 ah-1 .... a0ah

( 0 < = h <= n-1 )

h-unshuffle:F h

-1 (an-1 .... a0) = an-1 ... a h+1 a0ah-1 .... a1ah

( 0 < = h <= n-1 )h= n-1 = perfect suffle / unshuffle

18



butterfly permutation:

$ h (an-1 .... a0) = an-1 ... a h+1 a0ah-1 .... a1ah

( 0 < = h <= n-1 )

Identity permutation:j (an-1 .... a0) = an-1 ... a0

( 0 < = h <= n-1 )




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Other network permutations

Bit-switch:

* (an-1 .... a0) = a1 a2 ....... an-1a0

bit reversal:D (an-1 .... a0) = a0 a1 ... an-2 an-1




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Banyan NetworksMulti-stage

Simple SE, size: b x b

Only one path between any inlet / outlet I/O

Different I/O paths can share interstage links

# stages: n = log b N

each stage has: N / b SE


Horizontal Extension: Extended Banyan Networks

Objective: provide multiple paths through the network between one inlet / outlet pairResult: Extended Banyan Network (EBN)Construction procedure: Mirror imaging

A NxN EBN network with n+m stages (m ≤ n – 1) is obtained by attaching m stages to the 1st stage of a banyan network

Connection pattern: mirror image of the permutations in the last m stage of the original banyan network

Note: Adding m stages means making available 2m paths between an inlet and an outletSelf-routing: Only in the last n = log2N stagesCentralized control necessary for the first m stages

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Horizontal Extension: ExampleN = 16, m = 1

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Baseline banyan networkm=1



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Baseline banyan networkm=2

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m=3


Horizontal Extension: Rearrangeability

Now: Multiple paths in EBNsProblem: Are EBNs rearrangeable?The following statement can be shown:

An EBN is rearrangeable if m = n – 1Proof built on the Ofman theorem (for SW-banyan)

Ofman theorem:It is always possible to split an arbitrary permutation of size N into two sub-permutations of size N/2 such that, if the permutation is to be set up by the NxN network shown on the next slide, then the two sub-permutations are set up by the two non-blocking N/2xN/2 central subnetworks and no conflicts occur at the first and the last switching stage of the overall network.

23


Recursive Network Construction(Ofman Theorem)

N/4 x N/4

N/4 x N/4

••••••

•••

•••

N/4 x N/4

N/4 x N/4•••

•••

•••

•••

•••

•••

•••

•••

••••••

N/2 x N/2

•••

•••

N inlets N outlets

N/2 x N/2


Ofman Theorem: Properties

Iterate Ofman property to split each permutation of size N/2 into two of size N/4Finally: Rearrangeable network after n-1 steps when the central subnetwork is 2x2

N/2 permutations of size 2Number of 2x2 stages: 2(n-1) + 1 = 2n-1

(n-1): n-1 iterations

“2”: Each iteration adds two stages (symmetric design)

‘+1’: central 2x2 matrices

If an EBN with m = n-1 has n+m = 2n-1 stages, then a rearrangeable network can be constructed following the theorem of Ofman

24


Benes Networks

Construct EBNs recursively from a reverse baseline banyan network

Reverse baseline banyan network: Inverts the permutations of a baseline banyan network

Example (B16):B2

B4

B8

Reverse baseline banyan network

m = 3


Benes Networks: Characteristics

Number of matrices:Number of stages: s = 2 log2N-1 = 2n-1Number of matrices per stage: N/2Number of matrices: S = N/2 * s = Nlog2N –N/2

Costs:C2x2 = 4 (for each 2x2 matrix)Total costs for the network:

C = 4 * S = 4Nlog2N – 2N

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Setting Up Connections in Matrices

Set of connections / paths:Requested paths to be found (between the inlets and the outlets of the switch)

Complete set of connections:If N connections in an NxN network needed

Incomplete set of connections:If less than N connections required in an NxN network

Define: Size of the connection setNumber of connections to be set up

To establish a connection set of an arbitrary size: Necessary to determine the state of all matrices crossed by the connection

Use a recursive algorithm which can be applied to a recursive network of the Ofman type (e.g., a Benes network): The looping algorithm


The Looping Algorithm I

Starting point: An NxN network of 3 stages The 1st and last stage having N/2 matrices Two middle networks of size N/2xN/2

Called upper (U) and lower (L) subnetworksTerminology:

A network termination (i.e. inlet or outlet) is said to be busy (idle) if a connection has (has not) been requested

Busy inlets: Those where packets will arrive which have to be delivered to a certain outlet

Idle inlets: Inlets where no packets will arriveThree steps:

1. Loop start2. Forward connection3. Backward connection

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The Looping Algorithm II

1. Loop start: At the 1st stage: a) Select the unconnected busy inlet of an already connected elementb) Else: select a busy inlet of an unconnected elementc) Else: Stop the algorithm (no more paths to be built up)

2. Forward connectiona) Connected the selected inlet to the requested outlet

I. through the only accessible subnetwork if the element is already connected to the other subnetwork

II.Or through a randomly selected subnetwork if the element is not yet connectedb) If the other outlet of the element just reached is busy, select it step 3; c) Otherwise step 1

3. Backward connectiona) Connect the selected outlet to the requested inlet through the subnetwork

not used in the forward connection;b) If the other inlet of the element just reached is busy and not yet connected,

select it step 2; c) Otherwise step 1


Example: Looping Algorithm I

Connection set: 0 5, 1 3, 4 7, 5 2, 6 1, 7 4 Incomplete set

Busy inlets: 0, 1, 4, 5, 6, 7; Busy outlets: 1, 2, 3, 4, 5, 7Loop start: 0 5Case b) Select upper inlet 0 and go to step 2

Forward connection: 0 5Case a) II) Connect (randomly) to subnetwork L set inlet matrix

0 to “cross” set outlet matrix 2 to “straight”Remember: L must connect L’s inlet 0 L’s outlet 2

Case b) Outlet 4 busy select outlet 4 and go to step 3

B2

B4

01234567

01234567

Inlets OutletsU

L

U:

L: 0 2

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Example: Looping Algorithm II

Backward connection: 7 4Case a) Connect: 7 4 via the upper subnetwork

set inlet matrix 3 to “cross”Remember: Subnetwork U has to connect U’s inlet 3 U’s outlet 2

Case b) Inlet 6: busy go to step 2

B2

B4

01234567

01234567

Inlets OutletsU

L

U: 3 2

L: 0 2


Example: Looping Algorithm III

Forward connection: 6 1Case a) I) Connect: 6 1 via the lower subnet

set outlet matrix 0 to “straight”Remember: Subnetwork L has to connect L’s inlet 3 with L’s outlet 0

Case c) Outlet 0 idle go to step 1

B2

B4

01234567

01234567

Inlets OutletsU

L

U: 3 2

L: 0 2, 3 0

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Example: Looping Algorithm IV

Loop start:Connection set: (0 5), 1 3, 4 7, 5 2, (6 1), (7 4)

a) 1 3 not connected yet go to step 2Forward connection: 1 3Case a) I): Connect 1 3 via subnetwork U set outlet

matrix 1 to “cross”Remember: U must connect U’s inlet 0 U’s outlet 1

Case b) Outlet 2 busy go to step 3

B2

B4

01234567

01234567

Inlets OutletsU

L

U: 3 2, 0 1

L: 0 2, 3 0


Example: Looping Algorithm V

Backward connection: 5 2Case a) Connect: 5 2 via the lower subnetwork

set inlet matrix 2 to “straight”Remember: Subnetwork L has to connect L’s inlet 2 L’s outlet 1

Case b) Inlet 4: busy go to step 2

B2

B4

01234567

01234567

Inlets OutletsU

L

U: 3 2, 0 1

L: 0 2, 3 0,2 1

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Example: Looping Algorithm VI

Forward connection: 4 7Case a) Connect: 4 7 via the upper subnetwork set

outlet matrix 3 to “cross”Remember: Subnetwork U has to connect U’s inlet 2 U’s outlet 3

Case c) Outlet 6: idle, but connected go to step 1

Loop start: No busy unconnected inlets left: STOP

B2

B4

01234567

01234567

Inlets OutletsU

L

U: 3 2, 0 1,2 3

L: 0 2, 3 0,2 1


Example: Looping Algorithm VII

First recursion doneMust repeat the same algorithm for subnetwork U with connection set 3 2, 0 1, 2 3 and subnetwork L with connection set 0 2, 3 0, 2 1

B2

B4

01234567

01234567

Inlets OutletsU

L

U: 3 2, 0 1,2 3

L: 0 2, 3 0,2 1

30


Example: Looping Algorithm VIII

Results:

Overall Benes network at the end of the looping algorithm:

UU: 3 2, 0 1,

2 3

L: 0 2, 3 0,2 1

L

0123

0123

0123

0123

01234567

01234567

Inlets Outlets


Vertical Replication: Replicated Banyan Networks

Use N splitter of type 1xK, K banyan networks of type NxN, N combiners Kx1

Create a Replicated Banyan Network (RBN)

Objective: Find a K that guarantees rearrangeability

Utilization factor

0

1

K-1

01

N-1

01

N-1

1xK NxN Kx1

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Vertical Replication: Utilization Factor

Definition: The utilization factor uk of a generic link in stage k is the maximum number of paths between inlets and outlets which share this linkOr: It is the minimum from

The number of inlets that can reach this link

The number of outlets that can be reached from

this link


Utilization Factor in Banyan Networks

For a Banyan network topology:Easy to show that all links in the same interstage pattern have the same utilization factorExamples:

N=16N=32N=64N=128N=256

Maximum utilization factor:

NxN replicated banyan networkNumber of stages: n = log2NMaximum: At the central stage(s)

2 242 44 22 84 4 22 84 4 28

2 84 4 28 16

= 2max 2

n

u

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Rearrangeability in RBN

Theorem:A Replicated Banyan Network (RBN) of size NxN is rearrangeable if and only if:

K ≥ umax

“Proof”: umax paths share the central stage

must be routed onto umax distinct banyan

networks as each banyan network provides a single

path per inlet / outlet pair


RBN: Characteristics

Cost:

Splitters and combinersThus:

Horizontal extension more cost-effective than vertical replication when building a rearrangeable NxN network

Horizontal extension: Cost grows with Nlog2N

Vertical replication: Cost grows with N3/2log2N

)1(log2222log2

24 222

22 +=⋅+⋅=

NNNNNCnnn

33


Part IV. Fully self-routing networks


Fully Self-Routing Networks

Problem until now:Either: Self-routing, but blocking possible (Banyan networks)Or: Non-blocking, but some centralized control necessary (Horizontal extension, vertical replication)

Batcher-Banyan networksSelf-routing and rearrangeable non-blockingBasic concept:

Use a banyan network

Modify the configuration of input packets such that

blocking cannot occurCyclically compact and monotone sequence

34


Some Definitions I

Sequence of N elements:X = x0, ..., xN-1

xi = e: empty element or xi ∈[0, N-1]Each non-empty xi occurs only once in the sequenceUsually: Sequence determines the requested outlets in a switch

Example: X = (1, 3, e, 2)On inlet 0: packet destined to outlet 1On inlet 1: packet destined to outlet 3On inlet 2: no packet currentlyOn inlet 3: packet destined for outlet 2

Size of a sequence k:Number of non-empty elements of a sequence


CCM by example

Sequence: X1 = (5, 6, e, e, e, 0, 1, 3):N = 8 (number of elements = inlets/outlets)Size k = 5 (number of non-empty elements)Base m = 5 (has to be determined by looking at the sequence sharply)j = 0: x5+0 mod 8 = x5 : “0”j = 1: x5+1 mod 8 = x6 : “1”j = 2: x5+2 mod 8 = x7 : “3”j = 3: x5+3 mod 8 = x0 : “5”j = 4 = k-1: x5+4 mod 8 = x1 : “6”“0”, “1”, “3”, “5”, “6”: monotonically increasing CCM

35


Definitions I

Cyclically compact and monotone sequence (CCM):

If and only if there is a base m such that

x(m+j) mod N is non-empty for each 0 ≤ j ≤ k-1

x(m+i) mod N < x(m+j) mod N (increasing CCM) or

x(m+i) mod N > x(m+j) mod N (decreasing CCM)

for i, j ∈[0, k-1] and i < j


Definition II

Compact and monotone sequences (CM)Subset of CCMSize k and base m as for CCMCondition:

m ≤ (m + k – 1) mod N

36


Further Examples: CCM and CM

Sequence: X2 = (6, 5, 4, 3, 2, 1, 0, 7):N = 8 (number of elements = inlets/outlets)Size k = 8 (number of non-empty elements)Base m = 7 (has to be determined by looking at the sequence sharply; element with highest value here)

“7”, “6”, “5”, “4”, “3”, “2”, “1”, “0”: Monotonically decreasing CCMSequence: X3 = (e, e, e, e, e, 5, 4, 1):

N = 8, k = 3, m = 5Monotonically decreasing CM (not cyclically)


Analysis of Banyan Networks

Main question: For what input sequences is a Banyan Network non-blocking?

Theorem:An Omega banyan network is rearrangeable non-blocking for each set of input packets represented by a CCM sequence of arbitrary size m ≤ N.this means: If the input to the banyan network is monotonically

sorted (increasing or decreasing), then there will be no internal blocking

Assumption: No external blockingI.e.: There are no two elements in the input sequence with the same value (i.e., the same destination outlet)

Note: This theorem holds also for n-cube banyan networksFunctionally equivalent to Omega banyan networks

Use pre-sorting of the input sequence to make a banyan network non-blocking!

37


Batcher-Banyan Networks

Sorting networks: Can sort up to N packets received in an arbitrary configurationOutput: a CM sequence

Cascade sorting network and banyan network: Batcher-Banyan Network

Rearrangeable non-blocking (if no external blocking)

Completely self-routingc0a0

Batcher Bitonic Sorting Network

Banyan Routing Network

(Omega orn-cube)

c1

cN-1

•••

•••

•••

a1

aN-1 n(n+1)/2 n


Discussion: Batcher-Banyan Networks

Number of stages:

Batcher BanyanNumber of matrices:

N/2 per stage

Cost:C2x2 = 4;

NNNNNs 222222 log

23log

21log)1(loglog

21

+=++=

[ ]NNNNNNNS 2222

22 log3log

4log

43log

4+=+=

[ ]NNNC 222 log3log +=

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Demonstration of the Theorem I

Theorem:An Omega banyan network is rearrangeable non-blocking for each set of input packets represented by a CCM sequence of arbitrary size m ≤ N

“Proof” (better: demonstration)Omega banyan networks allow the identification of the used matrices at each stage and the outlet

Assumption:Inlet shall be X = xn-1 .... x0

Outlet shall be Y = yn-1 .... y0

Because of the Omega network topology:First matrix is xn-2...x0


Demonstration of the Theorem II

At the first stage:Use the first bit yn-1 of the outlet identifier Y for self-routingOutlet address of link at stage 1: xn-2...x0yn-1

Second stage:Use the second bit yn-2 of Y for self routingOutlet address of link at stage 2: xn-3...x0yn-1yn-2

Stage k:Use the kth bit yn-k of Y for self routingOutlet address at stage k: xn-k-1...x0yn-1...yn-k

Stage n (after n iterations):Outlet address: Y = yn-1...y0 As expected for the last outlet = the outlet of the network

Comparable to shifting a window of size n over the string xn-1...x0yn-1...y0, starting from bit k+1

39


Example (Omega Network)

Inlet: 0101; Destination: 0111xn-1...x0yn-1...y0 = 01010111

110

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010

011

000

001

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111

100

101

010

011

000

001

0000000100100011010001010110011110001001101010111100110111101111

0000000100100011010001010110011110001001101010111100110111101111

01010111 01010111 01010111 01010111 01010111


Theorem Proof I

Proof by contradiction: (For a CM sequence first)Given two inlet/outlet pairs a—b and a’—b’ from a CM sequence such that a’ > a and b’ > b

a = an-1...a0 ; a’ = a’n-1...a’0 ; b = bn-1...b0 ; b’ = b’n-1...b’0

ai, a’i, bi, b’i ∈[0,1] for i = 0...n-1Assume these two connections share the same output link at stage k (internal blocking):

an-k-1...a0bn-1...bn-k = a’n-k-1...a’0b’n-1...b’n-k

Previous example (N = 16, n = 4 stages): a—b = an-1...a0 bn-1...b0 = 01010111a’—b’ = a’n-1...a’0 b’n-1...b’0 = 01110111k = 3; n-k-1 = 0; n-1 = 3; n-k = 1an-k-1...a0bn-1...bn-k = a0b3b2b1 = 1011 outlet at stage 3a’n-k-1...a’0b’n-1...b’n-k = a’0b’3b’2b’1 = 1011 outlet at stage 3

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Example (Omega Network)

an-1...a0bn-1...b0 = 01010111a’n-1...a’0b’n-1...b’0 = 01110111k = 3 shared outlet at stage 3: 1011

110

111

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0000000100100011010001010110011110001001101010111100110111101111

0000000100100011010001010110011110001001101010111100110111101111


Theorem Proof II

Then:a’ – a = a’n-1...a’0 – an-1...a0

= a’n-1... a’n-k a’n-k-1 ...a’0 – an-1... an-k an-k-1 ...a0= a’n-1... a’n-k a’n-k-1 ...a’0 – an-1... an-k a’n-k-1 ...a’0= 2n-k (a’n-1...a’n-k – an-1...an-k) ≥ 2n-k

Example (decimal notation): 515 – 315 = 102 (5-3) ≥ 100Similarly:

b’ – b = b’n-1...b’0 – bn-1...b0 = b’n-1... b’n-k b’n-k-1 ...b’0 – bn-1... bn-k bn-k-1 ...b0= b’n-1... b’n-k b’n-k-1 ...b’0 – b’n-1... b’n-k bn-k-1 ...b0= 2n-k (b’n-k-1...b’0 – bn--k-1...b0) ≤ 2n-k -1

Example (decimal notation): 84352 – 84342 = 52 – 42 ≤ 102 -1Because of monotone and compact features:

a’ – a ≤ b’ – b 2n-k ≤ a’ – a ≤ b’ – b ≤ 2n-k -1 2n-k ≤ 2n-k -1 Contradiction!a — b and a’ — b’ from a CM sequence cannot share a path!

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Theorem Proof III

Proof by contradiction: (For a CCM sequence)Choose a’ and a such that a’ < a

Then:(a’ – a) mod N = N + a’ – a

= 1a’n-1...a’0 – 0 an-1...a0 (N = 2n = 100...0) = 2n-k (1a’n-1...a’n-k – 0 an-1...an-k) ≥ 2n-k

Example: (2-7) mod 8 = 8 + (2 – 7) = 3

From previous CM sequence proof:(b’ – b) ≤ 2n-k – 1

Because of cyclic compactness:(a’ – a) mod N ≤ b’ – b2n-k ≤ (a’ – a) mod N ≤ b’ – b ≤ 2n-k – 1Contradiction!a — b and a’ — b’ from a CCM sequence cannot share a path!


Exam: June 2003 I

The shown figure is a 8x8 Batcher network

a) Determine which of the switching elements (2x2 matrices) are sorting in increasing order (min on outlet 0 and max on outlet 1) and which are sorting in decreasing order.

b) Determine the state of the switching elements (“straight” or “cross”) of the complete network if the following inlet/outlet connections are required:

0 5, 1 2, 2 6, 3 1, 4 4, 5 7, 6 3, 7 0

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Exam June 2003 I: Solution

a)

b)


Exam June 2003 II

c) Add to the output of the Batcher sorting network a Benes network

Draw clearly the network topology, indicating the number of stages, the number of switching elements and the costs of the network.

d) Determine the state of the matrices in this EBN if all inlets shall be connected to the outlet with the same number (0 0, 1 1,...). Use the looping algorithm, showing all the details of the different steps (e.g., why a certain path / middle stage is chosen). Take into account that the inlets to the EBN are the outlets of the Batcher network.

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Exam June 2003 II: Solution

c) + d)

Number of stages: 5Number of switching elements: 20Cost of the network: 80

01234567

01234567

Inlets Outlets


References

A. Pattavina, “Switching theory”, Wiley 1998 (Available in the library).

Chapter 3 without 3.2.1.3, 3.2.1.4, 3.2.3, 3.3

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Thank you for your attention

David Larrabeiti Piotr Pacyna Universidad Carlos III de...

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