By Utpal Banerjee
Dependence Analysis can be thought of to be the second one variation of the author's 1988 ebook, Dependence research for Supercomputing. it's, even though, a totally new paintings that subsumes the cloth of the 1988 book. This booklet is the 3rd quantity in the sequence Loop Transformations for Restructuring Compilers. This sequence has been designed to supply an entire mathematical concept of ameliorations that may be used to immediately swap a sequential software containing FORTRAN-like do loops into an an identical parallel shape.
In Dependence Analysis, the writer extends the version to a application which include do loops and project statements, the place the loops don't need to be sequentially nested and are allowed to have arbitrary strides. within the context of any such software, the writer experiences, intimately, dependence among statements of this system because of software variables which are components of arrays.
Dependence Analysis is directed towards graduate and undergraduate scholars, writers of restructuring compilers. The prerequisite for the ebook includes a few wisdom of programming languages, and familiarity with calculus and graph thought. No wisdom of linear programming is needed.
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6) b e c o m e s 4~ - 6~ = 2. 7) b e c o m e 0 < [ < 49 and 0 < ~ < 49. 8) s u c h t h a t ~ a n d ~ are non_negative i n t e g e r s n o t larger t h a n 49. One s u c h s o l u t i o n is clearly (2, 1), a n d t h e r e are m a n y o t h e r s . However, a n y s u c h s o l u t i o n m u s t n e c e s s a r i l y s a t i s f y ~ > ~ (Exercise 3). 7) t h a t T 6 S h o l d s a n d t h a t S 6 T d o e s not. W h e t h e r T 6 S is t r u e w o u l d d e p e n d o n t h e p a r t o f t h e loop b o d y n o t s h o w n .
Since c = f(x,y) for s o m e p o i n t ( x , y ) E P, it follows that re_in f ( x , y ) <_ c <_ m a x f ( x , y ) . (x,y)EP (x,y)EP By definition, g ( a , b, p, q, ~) a n d v ( a , b, p, q, ~) are the m i n i m u m and m a x i m u m values, respectively, of the f u n c t i o n ( x , y ) ~ a x + b y in the triangle b o u n d e d by t h e lines: x = p , y = q , a n d x < y - ~. Hence, we have rn~n f ( x , y ) = u ( a , - b , O, ~, O) (x,y)~P max f(x,y) (x,y)~P = v(a,-b,O,~,O), so that the following inequalities hold: g(a,-b,0,~,0) < c < v(a,-b,O,~,O).
I~o - - ~. 5. Since a ~: b, go to Step 9. ] 9 "- 1, (~o,,3o) -- (1, 2). Since c m o d 9 = 0, continue. 10. ( ~ 2 1 , ~ 1 , ¢ 1) "-- (alg, blg, c/9) = (7,3,-5). 5. SOLUTION TO THE LINEAR PROBLEM 11. Go to Case (bl > 0). Vl -- [ - c l [ o / b l ] = 2, T 2 '-- [ ( 3 Cl[o)/bl] = 14. Go to Case (al > 0). T 1 '-- m a x (T1, [ - - C 1 , ~ 0 / / 2 1 ] ) = 2, T 2 "-- m i n (T2, [(t~ - C l ) 0 ) / ~ / 1 ] ) = Since T1 --< T 2 , continue. 12. ~ ~ Cl([0 - 3o)/(al - bl) 43 6. = 5/4. ] 13. T3 "-- [ ~ - - 11 = 1, ,-- + I J = 2, T5 "" m i n ( T 2 , T 3 ) -- max = 1, ('rl, V4) = 2.