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Practical Dependence Test

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Practical Dependence Test

Gina Goff, Ken Kennedy, Chau-Wen Tseng

PLDI ’91

presented by

Chong Liang Ooi

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

- 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

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

2. Separability

Separable if indices do not occur in other subscript

Otherwise, coupled

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

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

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

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

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

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)

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

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

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

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

Subscript-by-subscript test may yield false dep.

Delta Test Algorithm

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>

If the intersect of all constraints is empty set

no dependence

The End

Questions?