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## PowerPoint Slideshow about ' Practical Dependence Test' - stian

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Contribution

Efficient and precise dependence test is essential

General tests (Banerjee and GCD) are unnecessary

Most array refs in scientific Fortran program are simple

Based on these simple cases, this paper proposed

Partition-Based Algorithm

Classification of Subscripts

- Complexity

No of unique array indices a subscript has

- ZIV– Zero Index Variable
- SIV – Single Index Variable
- MIV – Multiple Index Variable

DO 10 i

DO 10 j

DO 10 k

10 A(5, i+1, j) = A(N, i, k) + c

Classification of Subscripts

2. Separability

Separable if indices do not occur in other subscript

Otherwise, coupled

A(i, j, j) = A(i, j, k) + c

Partition-Based Algorithm

Partition subscripts separable & minimal coupled group

Label each subscript as ZIV, SIV or MIV

Apply Single Subscript Test based on complexity

Apply Multiple Subscript Test to coupled group

If any test yields independence, no dependence exist

Otherwise, merge all direction vectors into single set

Single Subscript Test – ZIV

ZIV takes 2 loop invariant expressions

Proves 2 expressions cannot be equal

Can be extended for symbolic expressions

If differences is non-zero constant independence

SS Test – Strong SIV

Ref pair of form: <ai+c1, ai’+c2> for a Є [1,10]

Dependence Distance, d = i’-i = (c1-c2)/a

Dependence exist, if |d| <= U - L

Dependence Direction =

< if d > 0

= if d = 0

> if d < 0

Exact & efficient

Extendable to Symbolic Expr

by eval d symbolically

SS Test – Weak-Zero SIV

Ref pair of form: <a1i+c1, a2i’+c2> for a1 != a2

a1i+c1=a2i’+c2 is a line in 2D space of i vs i’

Check whether line intersect with any integer points

Weak-zero SIV

For a1=0 or a2=0

Let a2=0 i=(c2-c1 )/a1

Check iЄI and |i| < U - L

SS Test – Weak-Zero SIV

Usually, i=0 or last iteration

Loop peeling transformation can help

DO 10 i=1, N

10 Y(i, N)=Y(1,N)+Y(N,N)

Y(1,N)=Y(1,N)+Y(N,N)

DO 10 i=2, N-1

- Y(i, N)=Y(1,N)+Y(N,N)

Y(N,N)=Y(1,N)+Y(N,N)

SS Test – Weak-Crossing SIV

Weak-Crossing SIV

For a2 = -a1

Let i=i’ i=(c2-c1 )/2a1

Check |i| < U – L and iЄI or 1/2

Typically in Choleskey decomposition

SS Test – Weak-Crossing SIV

Loop splitting transformation can help

DO 10 i=1, N

10 A(i)=A(N-i+1)+C

DO 10 i=1, (N+1)/2

10 A(i)=A(N-i+1)+C

DO 20 i=(N+1)/2+1, N

20 A(i)=A(N-i+1)+C

SS Test – Restricted Double Index Var

Ref pair of form: <a1i+c1, a2j+c2>

SIV Tests can be used with 2 loop bounds for i & j

Delta Test Constraints

Assertions on indices derived from subscripts

<a1i+c1, a2i’+c2> a1i - a2i’= c2 - c1

Constraint vector, C=(del1, del2, …)

one constraint for each index in the coupled group

Del can be

dependence line: <ax+by=c>

dependence distance: <d>

dependence point: <x,y>

Questions?

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