- 108 Views
- Uploaded on
- Presentation posted in: General

Algorithmic Transformations

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Algorithmic Transformations

- The goal: Get the DSP algorithm in an amenable form before heading off to synthesize the design on the selected platform (FPGA or PDSP)
- No changes to the actual algorithms, just changes to the way the algorithms are prepared for implementation.
- This will require understanding aspects of
- timing,
- pipelining,
- parallelism

(C)2002-2004 Yu Hen Hu

- Algorithm Representations and Iteration Bound
- Parallelism and Pipelining
- Retiming
- Unfolding
- Folding

(C)2002-2004 Yu Hen Hu

(C)2002-2004 Yu Hen Hu

(C)2002-2004 Yu Hen Hu

(C)2002-2004 Yu Hen Hu

Node:

Computation

Associated with a computing time.

Direct edge:

data path and delay

Delay: iteration count

Example

y(n) = a*y(n-1) + b*u(n)

The delay of 1 u.t. indicates that to compute y(n+1) in the next iteration depends on result y(n) of the present iteration.

Delay labeled with D or positive integer on edges

(C)2002-2004 Yu Hen Hu

Intra-iteration dependency

A direct edge without any delay

Inter-iteration dependency

Direct edge with 1 or more delays

Node computing delay labeled with parenthesis.

Critical path: longest path between registers

Example: critical path delay = 4+2+2 = 8 t.u.

Recursive DFG: contains loops. Must have at least one delay element along any loop. Otherwise, the algorithm is NON-computable!

x(n)

D

D

M1

M2

(4)

(4)

M0

(4)

y(n)

A1

A0

(2)

(2)

(C)2002-2004 Yu Hen Hu

T{A-B-A} = (2+4)/2 = 3 t.u.

T = max{(2+4)/2, (2+4+5)/1}

= max{3, 11} = 11

D

(2)

(5)

(4)

A

B

C

2D

(2)

(4)

A

B

2D

(C)2002-2004 Yu Hen Hu

(C)2002-2004 Yu Hen Hu

(C)2002-2004 Yu Hen Hu

- To achieve high-speed, the length of the critical path can be reduced by pipelining and parallel processing

(C)2002-2004 Yu Hen Hu

- Algorithm Representations and Iteration Bound
- Parallelism and Pipelining
- Retiming
- Unfolding
- Folding

(C)2002-2004 Yu Hen Hu

Parallel processing

Pipelined processing

time

time

P1

P2

P3

P4

P1

P2

P3

P4

a1

a2

a3

a4

a1

b1

c1

d1

b1

b2

b3

b4

a2

b2

c2

d2

c1

c2

c3

c4

a3

b3

c3

d3

d1

d2

d3

d4

a4

b4

c4

d4

Less inter-processor communication

Complicated processor hardware

More inter-processor communication

Simpler processor hardware

Colors: different types of operations performed

a, b, c, d: different data streams processed

(C)2002-2004 Yu Hen Hu

Parallel processing requires NO data dependence between processors

Pipelined processing will involve inter-processor communication

P1

P2

P3

P4

P1

P2

P3

P4

time

time

(C)2002-2004 Yu Hen Hu

By inserting latches or registers between combinational logic circuits, the critical path can be shortened.

Consequence:

reduce clock cycle time,

increase clock frequency.

Suitable for DSP applications that have (infinity) long data stream.

Method to incorporate pipelining: Cut-set retiming

Cut set:

A cut set is a set of edges of a graph. If these edges are removed from the original graph, the remaining graph will become two separate graphs.

Retiming:

The timing of an algorithm is re-adjusted while keeping the partial ordering of execution unchanged so that the results correct

(C)2002-2004 Yu Hen Hu

(C)2002-2004 Yu Hen Hu

(C)2002-2004 Yu Hen Hu

(C)2002-2004 Yu Hen Hu

To further reduce TM.

Critical Path = Max {TM1, TM2, TA}

(C)2002-2004 Yu Hen Hu

x[n]

z-1

z-1

h[0]

h[1]

y[n]

h[2]

?

=

- The transfer function of a signal flow graph remain unchanged if
- The directions of each arc is reversed
- The input and output labels are switched.

u[n]

y[n]

z-1

z-1

h[2]

h[0]

h[1]

x[n]

(C)2002-2004 Yu Hen Hu

Algorithm transform may lead to pipelined structure without adding additional delays.

Given a FIR filter SFG

Critical path TM+2TA

Use graph transposition theorem:

Reverse all arcs

Reverse input/output

We obtain

Critical path Max(TM, TA)

No additional delay added!

(C)2002-2004 Yu Hen Hu

One form of vectorized parallel processing of DSP algorithms. (Not the parallel processing in most general sense)

Block vector: [x(3k) x(3k+1) x(3k+2)]

Clock cycle: can be 3 times longer

Original (FIR filter):

Rewrite 3 equations at a time:

Define block vector

Block formulation:

(C)2002-2004 Yu Hen Hu

(C)2002-2004 Yu Hen Hu

(C)2002-2004 Yu Hen Hu

(C)2002-2004 Yu Hen Hu

x(1)

x(2)

x(3)

x(4)

MAC

1

2

3

4

y(1)

y(2)

y(3)

y(4)

- Pipelining
- Block processing

x(1)

x(2)

x(3)

x(4)

x(5)

x(6)

x(7)

x(7)

Add

1

2

3

4

5

6

7

8

y(1)

y(2)

y(3)

y(4)

y(5)

y(6)

y(7)

y(7)

a y(1)

Mul

1

2

3

4

5

6

7

8

x(2)

x(4)

x(6)

x(8)

2

2

4

4

6

6

8

8

x(1)

x(3)

x(5)

x(7)

1

1

3

3

5

5

7

7

(C)2002-2004 Yu Hen Hu

- Algorithm Representations and Iteration Bound
- Parallelism and Pipelining
- Retiming
- Unfolding
- Folding

(C)2002-2004 Yu Hen Hu

- Retiming
Retiming is a mapping from a given DFG, G to a retimed DFT, Gr such that the corresponding transfer function of G and Gr differ by a pure delay z-L.

- Purposes
- To facilitate pipelining to reduce clock cycle time
- To reduce number of registers needed.

(C)2002-2004 Yu Hen Hu

(C)2002-2004 Yu Hen Hu

(C)2002-2004 Yu Hen Hu

(C)2002-2004 Yu Hen Hu

Feed-forward cut-set:

Feed-back cut-set

Delay transfer theorem

Adding arbitrary non-negative number of delays to each edge of a feed-forward cut-set of a DFG will not alter its output, except the output timing will be delayed.

Transfer the same amount of delays from edges of the same direction across a feed-back cut set of a DFG to all edges of opposing edges across the same cut set will not alter the output, but its timing.

(C)2002-2004 Yu Hen Hu

Consider the FIR digital filter and its DFG:

y(n) = b0x(n) + b1x(n-1)

Critical path length = TM+TA

Select a cut set

Insert a delay each to each edge in the cut set.

Retiming:

ynew(n) = b0x(n-1) + b1x(n-2)

ynew(n) = y(n-1)

Critical path = Max(TM, TA)

D

x(n)

x(n-1)

X

b0

X

b1

D

x(n)

x(n-1)

+

y(n)

X

b0

X

b1

D

D

+

y(n)

(C)2002-2004 Yu Hen Hu

Consider an IIR digital filter

y(n) = a·y(n-2) + x(n)

loop bound = (TM+TA)/2

clock cycle = TM+TA

Shift 1 delay to the other edge across a feed-back cut set

Filter remains unchanged.

loop bound = (TM+TA)/2

clock cycle = Max(TM ,TA)

x(n)

y(n)

x(n)

y(n)

+

+

2D

D

D

a

a

(C)2002-2004 Yu Hen Hu

Consider an IIR digital filter

y(n) = ay(n-1) + x(n)

loop bound = (TM+TA)

throughput = 1/(TM+TA)

x(2k-1)=x(k)

x(2k) = 0

Clock period = (TM+TA)

Throughput = 1/[2(TM+TA)]

x(n)

y(n)

+

x(m)

y(m)

+

D

2D

a

a

(C)2002-2004 Yu Hen Hu

(C)2002-2004 Yu Hen Hu

(C)2002-2004 Yu Hen Hu

(C)2002-2004 Yu Hen Hu

Start with

y(n) = a y(n-1) + x(n)

clock cycle = Max(TM ,TA)

Throughput = 1/[2max(TM,TA)]

Start with

y(n) = a y(n-2) + x(n)

loop bound = (TM+TA)/2

clock cycle = Max(TM ,TA)

throughput = 1/ Max(TM ,TA)

x(n)

y(n)

x(m)

y(m)

+

+

D

D

D

D

a

a

(C)2002-2004 Yu Hen Hu

(C)2002-2004 Yu Hen Hu

Node delay = 1 t.u.

Before retiming:

Critical path: a3 a4 a5 a6

Clock cycle time = 4

2 delay units

After cut-set retiming

Critical path: a3 a5, a4 a6

Clock cycle time = 2

6 delay units

After additional retiming

Critical path: none

Clock cycle time = 1

11 delay units

D

a4

a2

D

a6

D

a1

D

D

D

a3

a5

D

a4

a2

a6

a1

D

a5

a3

2D

a4

a2

D

D

a6

2D

a1

D

D

D

2D

a3

a5

(C)2002-2004 Yu Hen Hu

Transfer delay through a node in DFG:

r(v) = # of delays transferred from out-going edges to incoming edges of node v w(e) = # of delays on edge e

wr(e) = # of delays on edge e after retiming

Retiming equation:

subject to wr(e) 0.

Let p be a path from v0 to vk

then

…

e0

e1

ek

v0

v1

vk

e

v

u

D

3D

2D

r(v) = 2

v

v

2D

3D

D

p

(C)2002-2004 Yu Hen Hu

- Retiming does NOT change the total number of delays for each cycle.
- Retiming does not change loop bound or iteration bound of the DFG
- If the retiming values of every node v in a DFG G are added to a constant integer j, the retimed graph Gr will not be affected. That is, the weights (# of delays) of the retimed graph will remain the same.

(C)2002-2004 Yu Hen Hu

r(2) = 1

(C)2002-2004 Yu Hen Hu

T = max. {(1+2+1)/2, (1+2+1)/3} = 2

Cr. Path Delay = max{2,2,1+1} = 2 t.u

T = max. {(1+2+1)/2, (1+2+1)/3} = 2

Cr. Path delay = 2+1 = 3 t.u

(C)2002-2004 Yu Hen Hu

Note that retiming will NOT alter iteration bound T.

Iteration bound is the theoretical minimum clock period to execute the algorithm.

Let edge e connect node u to node v. If the node computing time t(u) + t(v) > T, then clock period T > T. For such an edge, we require that

To generalize, for any path from v0 to vk, we have

In other words, for any possible critical path in the DFG that is larger than T, we require wr(e) 1.

(C)2002-2004 Yu Hen Hu

wr(e21) 0, since t(2)+t(1) = 2 = T.

wr(e13) 1, since t(1)+t(3) = 3 > T.

wr(e14) 1, since t(1)+t(4) = 3 > T.

wr(e32) 1, since t(3)+t(2) = 3 > T.

wr(e42) 1, since t(4)+t(2) = 3 > T.

Use eq. wr(euv) = w(e) + r(v) – r(u),

w(e21) + r(1) – r(2) = 1 + r(1) – r(2) 0

w(e13) + r(3) – r(1) = 1 + r(3) – r(1) 1

w(e14) + r(4) – r(1) = 2 + r(4) – r(1) 1

w(e32) + r(2) – r(3) = 0 + r(2) – r(3) 1

w(e42) + r(2) – r(4) = 0 + r(2) – r(4) 1

(C)2002-2004 Yu Hen Hu

Since the retimed graph Gr remain the same if all node retiming values are added by the same constant. We thus can set r(1) = 0.

The inequalities become

1 – r(2) 0 or r(2) 1

1 + r(3) 1 or r(3) 0

2 + r(4) 1 or r(4) –1

r(2) – r(3) 1 or r(3) r(2) - 1

r(2) – r(4) 1 or r(2) r(4) + 1

Since

one must have r(2) = +1.

This implies r(3) 0. But we also have r(3) 0. Hence r(3)=0.

These leave –1 r(4) 0.

Hence the two sets of solutions are:

r(3) = 0, r(2) = +1, and r(4) = 0 or -1.

(C)2002-2004 Yu Hen Hu

Given a systems of inequalities:

r(i) – r(j) k; 1 i,j N

Construct a constraint graph:

Map each r(i) to node i. Add a node N+1.

For each inequality

r(i) – r(j) k,

draw an edge eji

such that w(eji) = k.

Draw N edges eN+1,i = 0.

The system of inequalities has a solution if and only if the constraint graph contains no negative cycles

If a solution exists, one solution is where ri is the minimum length path from the node N+1 to the node i.

Shortest path algorithms: Bellman-Ford algorithm

Floyd-Warshall algorithm

(C)2002-2004 Yu Hen Hu

- Algorithm Representations and Iteration Bound
- Parallelism and Pipelining
- Retiming
- Unfolding
- Folding

(C)2002-2004 Yu Hen Hu

Unfolding is the process of unfolding a loop so that several iterations are unrolled into the same iteration.

Also known as

Loop unrolling (in compilers for parallel programs)

Block processing

Applications

Reducing sampling period to achieve iteration bound (desired throughput rate) T.

Parallel (block processing) to execute several iterations concurrently.

Digit-serial or bit-serial processing

(C)2002-2004 Yu Hen Hu

Block processing formulation

J = 3, 9/J = 3 (an integer)

X(k) = [x(3k) x(3k+1) x(3k+2)]T

Y(k) = [y(3k) y(3k+1) y(3k+2)]T

Y(k) = a*Y(k- 3 ) + X(k)

J = 2, 9/J = ? (not an integer)

X(k) = [x(2k) x(2k+1)]T

Y(k) = [y(2k) y(2k+1)]T

Y(k) = a*Y(k- ? ) + X(k)

Before unfolding:

For n = 0 to N-1,

y(n)=a*y(n-9)+x(n)

end

Unfolding once (J = 2)

For k = 0 to N/2-1,

y(2k)=a*y(2k-9)+x(2k)

y(2k+1)=a*y(2k-8)+x(2k+1)

end

Unfolding twice (J = 3)

For k = 0 to N/3-1,

y(3k)=a*y(3k-9)+x(3k)

y(3k+1)=a*y(3k-8)+x(3k+1)

y(3k+2)=a*y(3k-7)+x(3k+2)

end

(C)2002-2004 Yu Hen Hu

Rewrite the algorithm formulation:

y(2k)=a*y(2k-9)+x(2k)

y(2k+1)=a*y(2k-8)+x(2k+1)

y(2k)=a*y(2(k-5)+1)+x(2k)

y(2k+1)=a*y(2(k-4))+x(2k+1)

After J-folded unfolding, the clock period T = J Ts, where Ts is the data sampling period.

T=Ts

T=J Ts

(C)2002-2004 Yu Hen Hu

Define

Step 1. For each node U in original DFG, draw J nodes {Ui; 0 iJ-1} in the unfolded DFG

Step 2. For each edge from U to V with w delays, draw J edges from Ui to V(i+w)%J with (i+w)/J delays

(C)2002-2004 Yu Hen Hu

J=2

S0

Q0

T0

S

R0

Q

T

3D

2D

S1

R

Q1

T1

T=3

R1

Step 1. Duplicate J copies of each node

(C)2002-2004 Yu Hen Hu

J=2

S0

Q0

T0

S

R0

Q

T

3D

2D

S1

R

Q1

T1

T=3

R1

Step 2. Add all edges with 0 delay on them.

(C)2002-2004 Yu Hen Hu

J=2

S0

Q0

T0

S

D

R0

Q

T

2D

D

3D

2D

S1

R

Q1

T1

T=3

D

R1

Step 3. Use table on the left to figure out edges with delays.

T=6

(C)2002-2004 Yu Hen Hu

Unfolding preserves the number of registers (delays) in a DFG

For a loop with w delays in a DFG that has been unfolded J times, it leads to

g.c.d.(w, J) loops in the unfolded DFG, with each of these loops containing

w/(g.c.d.(w,J)) delays and

J/(g.c.d.(w,J)) copies of each node that appear in the original loop.

Unfolding a DFG with iteration bound T results in a J-folded DFG with iteration bound JT.

A path with w (< J) delays in a DFG will lead to J-w paths with no delays, and w paths with 1 delay each in the J-unfolded DFG.

Any path in the original DFT containing J or more delays leads to J paths 2ith 1 or more delay in each path. Therefore, it can not create a critical path in the J-unfolded DFT

Any clock period that can be achieved by retiming a J-unfolded DFG can be achieved by retiming the original DFG and followed by J-unfolding.

(C)2002-2004 Yu Hen Hu