2008 Generalized SEN

Optimal All-To-All PersonalizedExchange Algorithms in Generalized

Shuffle-Exchange Networks

Student: YuChieh Chiu Advisor: Chiuyuan Chen

Department of Applied Mathematics

National Chiao Tung University

July 31, 2008

Shuffle

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When N = 2n, the shuffle operation is:

π(in−1in−2 · · · i1i0) = in−2 · · · i1i0in−1

.

Binary Switch

A binary switch is a 2× 2 Switch Element (SE)

2x2Switch

Legitimate States = 4

Permutation Connections = 2

Binary Switch

Control bit

0 for straight and 1 for exchange (cross)

the 2 broadcast states are not used in this paper

Straight Exchange

Upper-broadcast Lower-broadcast

The different setting of the 2X2 SE

Multistage Interconnection Network(MIN)

three typical MINs

8× 8 baseline network, shuffle exchange network, and indirectbinary n-cube network

MIN Implementation

Control (X)

Source (S) Destination (D)

N ×N Shuffle Exchange Networks

N ×N Shuffle Exchange Networks = N ×N SENs

N = ♯ of nodes, n = ♯ of stages

N = 2n

Figure: 4× 4, and 8× 8 SENs

N ×N Generalized Shuffle Exchange Networks

N ×N Generalized Shuffle Exchange Networks = N ×N

GSENs

N = ♯ of nodes, n + 1 = ♯ of stages

2n < N ≤ 2n+1

SENs are contained in GSENs

Figure: 4× 4, 6× 6, and 8× 8 GSENs

NOT unique-path

A MIN is unique-path if there is a unique path between eachpair of input and output.

SENs are unique-path

GSENs are NOT unique-path

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stage 0 stage 1 stage 2 stage 3

Communications among processors

one-to-one

one-to-many

all-to-all

all-to-all broadcastall-to-all personalized exchange (ATAPE) ←−we focus on here

Definition

In ATAPE, each processor sends a specific message to every otherprocessor.

Why ATAPE?

ATAPE occurs in many applications:

matrix transposition

fast Fourier transform (FFT)

Compare MIN with other networks

network model scalability communication delay

hypercubes poor shortermeshes better highertori better higherMINs better shorter

is ATAPE easy?

Figure: 6× 6 GSEN

Stage control

Stage Control (SC): all the SEs at the same stage are set tothe same state.

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stage 0 stage 1 stage 2 stage 3

Figure: SC in 10× 10 GSEN

Configuration

Network configuration: the states of switches of the network

in Matrix form

1 0 0 11 0 0 11 0 0 11 0 0 11 0 0 1

under stage control, configuration represented as n + 1-tuple

(1 0 0 1)

or an integer C

(1001)2 = 9

Previous Results

Yang & Wang (2000),IEEE TRANSACTIONS ON PARALLEL AND DISTRIBUTED

SYSTEMS,Optimal all-to-all personalized exchange in self-routablemultistage networks.

Yang & Wang propose an optimal all-to-all personalizedexchange algorithm with stage control technique in which theLatin square method is used.

Yang & Wang’s algorithm requires constructing a Latin squarein advance and allocating memory for storing the Latin square.

MINs in this paper must be unique path and self-routable.

Previous Results

Massini (2003),DISCRETE APPLIED MATHEMATICS,All-to-all personalized communication on multistageinterconnection networks.

Massini’ algorithm does not require precomputation for Latinsquare and extra memory space.

MINs in this paper must be unique path and self-routable.

Motivation

The purpose of this thesis

To our knowledge, no one has studied ATAPE algorithms inGSENs.

The purpose of this thesis is to propose ATAPE algorithms forGSENs.

We propose two algorithms

Algorithm 1 uses the stage control technique and works for alleven N .

On the contrary, Algorithm 2 works for all N ≡ 2 (mod 4)without stage control.

Both are optimal.

R(N) and Rsc(N)

Definition

Let R(N) denote the minimum number of network configurationsrequired to realize ATAPE in an N ×N GSEN.Also, let Rsc(N) denote the minimum number of networkconfigurations required to realize ATAPE in an N ×N GSEN whenthe stage control technique is assumed.

Lemma 1

N ≤ R(N) ≤ Rsc(N) ≤ 2n+1.

Main results

We propose two optimal ATAPE algorithms for N ×N GSENs.

Algorithm 1

with stage control

need 2n+1 configurations

need to construct a destination matrix in advance

Algorithm 2

without stage control

need N configurations

compute destinations directly

Algorithm 1

Preprocessing-Phase Destination matrix constructing phase

for each processor i (0 ≤ i < N) do in parallel

for each time k (0 ≤ k < 2n+1) do in sequential

prepare a null message;equip the message with configuration k and send it out;when an output (say, j) receives the message, set sj,k = i

for each j (0 ≤ j < N) do

for each k (0 ≤ k < 2n+1) do

if sj,k = i then set di,k = j;for each i (0 ≤ i < N) do

for each j (0 ≤ j < N) do

set mark[j] = 0;

Algorithm 1

Preprocessing-Phase (cont.)

for each k (0 ≤ k < 2n+1) do

if mark[di,k] = 0 then set mark[di,k] = 1;else set di,k = −1;

S =

0 4 8 5 7 6 1 3 5 9 3 0 2 1 6 84 0 5 8 6 7 3 1 9 5 0 3 1 2 8 68 3 0 2 1 5 7 9 3 8 5 7 6 0 2 43 8 2 0 5 1 9 7 8 3 7 5 0 6 4 27 2 6 3 0 9 4 5 2 7 1 8 5 4 9 02 7 3 6 9 0 5 4 7 2 8 1 4 5 0 96 1 7 9 4 8 0 2 1 6 2 4 9 3 5 71 6 9 7 8 4 2 0 6 1 4 2 3 9 7 55 9 4 1 3 2 6 8 0 4 9 6 8 7 1 39 5 1 4 2 3 8 6 4 0 6 9 7 8 3 1

, D =

0 1 2 3 4 5 6 7 8 9 − − − − − −7 6 9 8 2 3 0 1 − − 4 5 − − − −5 4 3 2 9 8 7 6 − − − − 0 1 − −3 2 5 4 8 9 1 0 − − − − 7 6 − −1 0 8 9 6 7 4 5 − − − − − − 3 28 9 1 0 3 2 5 4 − − − − − − 6 76 7 4 5 1 0 8 9 − − − − 2 3 − −4 5 6 7 0 1 2 3 − − − − 9 8 − −2 3 0 1 7 6 9 8 − − 5 4 − − − −9 8 7 6 5 4 3 2 1 0 − − − − − −

Algorithm 1

Phase 1: The message preparing phase



if di,k 6= −1 then prepare a personalized message to di,k

else prepare a null message;equip the message with configuration k

insert the message into the message queue of i;

Phase 2: The message sending phase



do send a message in the message queue of i;

Correct and Optimal

Correctness

Easy part. Since we use every configuration k, 0 ≤ k < 2n+1.

Optimality

Hard part. We have to claim every configuration contains a

unique-path.

Definition

stage 0 stage 1 stage n

nb

nf 1n

b 1n

f

0b

0fi

j

Forward control tag F = fn2n + · · · + f121 + f02

0

Backward control tag B = bn2n + · · ·+ b121 + b02

0

Observation

sketch of proof

C B F !Lem.82 2 1n n "# Lem.9

2 2 1n n " ! !Lem.10

Lem.11 no 2 holes

12ni FB

N

"$ "% & % &' (

Lem.7

Thm FBCT

Thm.21( ) 2nN ")

"#$

Without SC

Can we do better, if we abandon the SC technique?

N ≤ R(N), at least N configurations

Alternating stage control (ASC)

A variation of stage control, which means the states of theswitches of a stage alternate between straight and cross.

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Observation

To choose N configurations

We reordered the configurations AF by the control tag F of inputprocess 0, then we found A0 = AN , A1 = AN+1, . . .

Algorithm 2



calculate mi by the formula:

mi =

{

(i · 2n+1) mod N, if i is even;((i + 1) · 2n+1 − 1) mod N, if i is odd;

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Algorithm 2


for each time k (0 ≤ k < N) do in sequential

prepare a personalized message for destination processor{

(mi + k) mod N, if i is even;(mi − k) mod N, if i is odd;

equip the message with configuration Ak = k ⊕⌊

k2

⌋

;insert the message into the message queue of i;

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9876543210i\k

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85032149677

58301294766

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36189072544

41698705233

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9876543210i\k

Algorithm 2

Phase 2: The message sending phase


for each time k (0 ≤ k < N) do in sequential

send a message in the message queue of i;

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Correct and Optimal

Optimality

Easy part. Since we use only N configurations Ak, 0 ≤ k < N .N ≤ R(N)

Correctness

Hard part. We proved that the link from any input i to any output

j exists in our algorithm, and the message sent by input i would

reach exactly the output j calculated in our algorithm.

Why N ≡ 2 (mod 4)?

In the proof of correctness, property (∗) only holds when N ≡ 2(mod 4).Thus Algorithm 2 can work when N ≡ 2 (mod 4) only.

Property (∗)

Property (∗)

1 If the alternating control bit is 0, E0−→ E, O

1−→ O;

2 Else the control bit is 1, E1−→ O, O

0−→ E.

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(a) (b)

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(c) (d)

Property (∗)

Proof of property (∗)

N2

is odd because N ≡ 2 (mod 4).

x0 = y and x1 = y + N2.

Thus one of the input port of switch y

is even while the other input port isodd.

z0 is even and z1 is odd.

Now set the control bit 0 or 1,andswitch y even or odd; totally 4 cases.

By examine the states of all switches,we have done.

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N-1

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N-2

N-1

0x

1x

1z

0zy

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N

1, 2 1n

k kF k F k

!" " # #Lem.17

1( 2 )modnj i T N

" $ Lem.6

F F! "Lem.16

Thm.15 relation between

0 1,E E O O%%& %%&

Property (*)

(i)

(ii)

Thm.19 Algorithm is correct

Thm.18 0 1 1, ,...,

NA A A

# fulfill ATAPE

, ,F F A!

1 0,E O O E%%& %%&

Concluding remarks

In this thesis

We have proposed two optimal ATAPE algorithms for GSENs

We obtained Rsc(N) = 2n+1

We obtained N = R(N), if N ≡ 2 (mod 4)

Open question

To determine R(N) for N ≡ 0 (mod 4).

Generalize to d× d switch elements.

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Initial GSEN

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Round 1

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Round 2

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Round 3

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Round 4

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Round 5

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Round 6

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Round 7

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Round 8

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Round 9

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Round 10

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Round 11

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Round 12

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Round 13

0725436189

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6381092745

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4169870523

1496785032

2947658301

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2008 Generalized SEN

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