1. CS314 – Section 5 Recitation 13
Long Zhao
SIV, ZIV
Vectorization
Slides available at

2. Dependence Distance Vector
Definition
Suppose that there is a dependence from S1 on the i-th level iteration x to S2 on iteration y; then the i-th element of the dependence distance vector d is defined as d = y - x
Example:
DO I = 1, DO J = 1, I S1 A(I+1,J) = A(I,J) ENDDO ENDDO
True dependence between S1 and itself on i = (1,1) and j = (2,1): d(i, j) = (1,0) i = (2,1) and j = (3,1): d(i, j) = (1,0) i = (2,2) and j = (3,2): d(i, j) = (1,0)

3. Dependence Direction Vector
Definition
Suppose that there is a dependence from S1 on iteration i and S2 on iteration j; then the dependence direction vector D(i, j) is defined as
“<” if d(i, j)k > 0 “=” if d(i, j)k = 0 “>” if d(i, j)k < 0
D(i, j)k =

4. Example 4 DO I = 1, 4 DO J = 1, 4 S1 A(I,J+1) = A(I-1,J) ENDDO ENDDO 32 1
Distance vector is (1,1)
S1 (<,<) S1 I 1 2 3 4

5. ZIV Test e1, e2 are constants or loop invariant symbols
DO j = 1, 100
S A(e1) = A(e2) + B(j)
ENDDO
e1, e2 are constants or loop invariant symbols
If (e1-e2)!=0 No Dependence exists

6. Strong SIV Test Strong SIV subscripts are of the form
For example the following are strong SIV subscripts

7. Strong SIV Test Strong SIV test when
DO i1 = L1, U1
S1 A(f(i1,...,in)) = ...
S = A(g(i1,...,in))
ENDDO
Strong SIV test when
f(...) = ai+c1 and g(…) = ai’+c2
Plug in α,β and solve for dependence:
β-α = (c1 – c2)/a
A dependence exists from S1 to S2 if:
β-α is an integer
|β-α| ≤ U1- L1

8. Strong SIV Test Example
DO k = 1, 100
DO j = 1, 100
S1 A(j+1,k) = ...
S = A(j,k) + 32
ENDDO

9. Weak SIV Tests Weak SIV subscripts are of the form
For example the following are weak SIV subscripts

10. Weak-zero SIV Test Weak-Zero SIV test when
DO i1 = L1, U1
S1 A(f(i1,...,in)) = ...
S = A(g(i1,...,in))
ENDDO
Weak-Zero SIV test when
f(...) = ai+c1 and g(…) = c2
Plug in α and solve for dependence:
α = (c2 – c1)/ a
A dependence exists from S1 to S2 if:
α is an integer
L1 ≤ α ≤ U1

11. Weak-crossing SIV Test
DO i1 = L1, U1
S1 A(f(i1,...,in)) = ...
S = A(g(i1,...,in))
ENDDO
Weak-Crossing SIV test when
f(...) = ai+c1 and g(…) = -ai+c2
To find crossing point, set α = β and solve:
α = (c2 – c1) / 2a
A dependence exists from S1 to S2 if:
2α is an integer
L1 ≤ α ≤ U1

12. Vectorization Simple vectorizer assumptions: singly-nested loops
constant upper and lower bounds, step is always 1
body is sequence of assignment statements to array variables
simple array index expressions of induction variable (i +/- c or c); can use ZIV or SIV test
no function calls

13. Vectorization (1)
A single-statement loop that carries no dependence can be vectorized
DO I = 1, 4 S1 X(I) = X(I) + C ENDDO
vectorize
S1 X(1:4) = X(1:4) + C
X(1)
X(2)
X(3)
X(4) +
Fortran 90 array statement C
(anti), d = 0 S1
X(1)
X(2)
X(3)
X(4)
Vector operation

14. Vectorization (2)
This example has a loop- carried dependence (true dependence), and is therefore invalid
S1 (<) S1
DO I = 1, N S1 X(I+1) = X(I) + C ENDDO
(true), d = 1
S1 X(2:N+1) = X(1:N) + C S1
Invalid Fortran 90 statement

15. Vectorization (2)
This example has a loop- carried dependence (anti- dependence), and is valiad
S1 (<) S1
DO I = 2, N S1 X(I-1) = X(I) + C ENDDO
(anti), d = 1
S1 X(1:N-1) = X(2:N) + C S1

16. S1
Vectorization (3)
(true), d = 1 S2
DO I = 1, N S1 A(I+1) = B(I) + C S2 D(I) = A(I) + E ENDDO
Because only single loop statements can be vectorized, loops with multiple statements must be transformed using the loop distribution transformation
Loop has no loop-carried dependence or has forward flow dependences
loop distribution
DO I = 1, N S1 A(I+1) = B(I) + C ENDDO DO I = 1, N S2 D(I) = A(I) + E ENDDO
vectorize
S1 A(2:N+1) = B(1:N) + C S2 D(1:N) = A(1:N) + E

17. S1
Vectorization (4)
(true), d = 1 S2
DO I = 1, N S1 D(I) = A(I) + E S2 A(I+1) = B(I) + C ENDDO
When a loop has backward flow dependences and no loop-independent dependences, interchange the statements to enable loop distribution
loop distribution
DO I = 1, N S2 A(I+1) = B(I) + C ENDDO DO I = 1, N S1 D(I) = A(I) + E ENDDO
vectorize
S2 A(2:N+1) = B(1:N) + C S1 D(1:N) = A(1:N) + E

18. Vectorization (5) S1 S2
DO I = 2, 100 S1 A(I) = D(I-1) + B(I-1) S2 B(I) = A(I-1) * 2 S3 C(I) = C(I-1) + 3 S4 D(I) = C(I) ENDDO S3 S4

19. Vectorization (5) S1 S2 S3 S4 B: (true), d = 1 A: (true), d = 1
DO I = 2, 100 S1 A(I) = D(I-1) + B(I-1) S2 B(I) = A(I-1) * 2 S3 C(I) = C(I-1) + 3 S4 D(I) = C(I) ENDDO
D: (true), d = 1
(true), d = 1 S3
(true), d = 0 S4

20. Vectorization (5) S3 S1 S4 S2 S1 S3 S2 S4 (true), d = 1
B: (true), d = 1
A: (true), d = 1 S4 S2
D: (true), d = 1
D: (true), d = 1
(true), d = 1 S1 S3
A: (true), d = 1
B: (true), d = 1
(true), d = 0 S2 S4

21. Vectorization (5) S3 S4 S1 S2 (true), d = 1
DO I = 2, 100 S1 A(I) = D(I-1) + B(I-1) S2 B(I) = A(I-1) * 2 S3 C(I) = C(I-1) + 3 S4 D(I) = C(I) ENDDO S3
(true), d = 0 S4
D: (true), d = 1
DO I = 2, 100 S3 C(I) = C(I-1) + 3 S4 D(I) = C(I) S1 A(I) = D(I-1) + B(I-1) S2 B(I) = A(I-1) * 2 ENDDO S1
A: (true), d = 1
B: (true), d = 1 S2

22. Vectorization (5) S3 S4 S1 S2 (true), d = 1
DO I = 2, 100 S3 C(I) = C(I-1) + 3 S4 D(I) = C(I) S1 A(I) = D(I-1) + B(I-1) S2 B(I) = A(I-1) * 2 ENDDO S3
(true), d = 0 S4
D: (true), d = 1
DO I = 2, 100 S3 C(I) = C(I-1) ENDDO
DO I = 2, 100 S4 D(I) = C(I) S1 A(I) = D(I-1) + B(I-1) S2 B(I) = A(I-1) * 2 ENDDO S1
A: (true), d = 1
B: (true), d = 1 S2

23. Vectorization (5) S3 S4 S1 S2 DO I = 2, 100 S3 C(I) = C(I-1) + 3 ENDDO
DO I = 2, 100 S4 D(I) = C(I) S1 A(I) = D(I-1) + B(I-1) S2 B(I) = A(I-1) * 2 ENDDO
(true), d = 1 S3
(true), d = 0 S4
D: (true), d = 1
DO I = 2, 100 S3 C(I) = C(I-1) ENDDO
S4 D(2:100) = C(2:100)
DO I = 2, 100 S1 A(I) = D(I-1) + B(I-1) S2 B(I) = A(I-1) * 2 ENDDO S1
A: (true), d = 1
B: (true), d = 1 S2

24. Vectorization (5) S3 S4 S1 S2 DO I = 2, 100 S3 C(I) = C(I-1) + 3 ENDDO
S4 D(2:100) = C(2:100)
DO I = 2, 100 S1 A(I) = D(I-1) + B(I-1) S2 B(I) = A(I-1) * 2 ENDDO
(true), d = 1 S3
(true), d = 0 S4
D: (true), d = 1 S1
A: (true), d = 1
B: (true), d = 1 S2