e Some cient solutions to the a ne scheduling problem Part II Multidimensional time
Paul Feautrier Laboratoire MASI, Institut Blaise Pascal Universite de Versailles Saint-Quentin 45 Avenue des Etats-Unis 78035 VERSAILLES CEDEX FRANCE e-mail :
[e-mail]
October 7, 1992
This paper extends the algorithms which were given in Part I to cases in which there is no a ne schedule, i.e. to problems whose parallel complexity is polynomial but not linear. The natural generalization is to multidimensional schedules with lexicographic ordering as temporal succession. Multidimensional a ne schedules, are, in a sense, equivalent to polynomial schedules, and are much easier to handle automatically. Furthermore, there is a strong connexion between multidimensional schedules and loop nests, which allows one to prove that a static control program always has a multidimensional schedule. Roughly, a larger dimension indicates less parallelism. In the algorithm which is presented here, this dimension is computed dynamically, and is just su cient for scheduling the source program. The algorithm lends itself to a \divide and conquer" strategy. The paper gives some experimental evidence for the applicability, performances and limitations of the algorithm.
Abstract
1
2
Resume
Dans cet article, les algorithmes qui ont ete propose dans la premiere partie sont etendus au cas ou le programme source n'a pas de base de temps a ne, c'est-a-dire a des algorithmes dont la complexite parallele est polynomiale mais non lineaire. La solution naturelle est l'emploi de bases de temps a plusieurs dimensions, l'ordre de succession temporelle etant l'ordre lexicographique. Les bases de temps multidimensionnelles sont, en un certain sens, equivalentes a des bases de temps polynomiales, et sont beaucoup plus faciles a manipuler algorithmiquement. De plus, il y a une connexion forte entre bases de temps multidimensionnelles et nids de boucles, ce qui permet de demontrer qu'un programme a contr^le statique a toujours une o base de temps multidimensionnelle. En gros, plus grande est la dimension et moins il y a de parallelisme. Dans l'algorithme ici presente, cette dimension est determinee dynamiquement; elle est juste su sante pour permettre l'ordonnancement du programme source. En n, cet algorithme se pr^te a e l'application de la strategie \diviser pour regner". On presente en conclusion quelques resultats experimentaux permettant de juger du domaine d'application, des performances et des limitations de l'algorithme.
3
do i = 0,n do j = 0,i s = s + a(i,j) end do end do
1
Figure 1: A simple program with no linear schedule Instruction 1 Edge 1 2 Domain 0 i n 0 j i Source Destination Condition (1; i; j ? 1) (1; i; j ) j 1 (1; i ? 1; i ? 1) (1; i; j ) j < 1 ^ i 1 Table 1: The DFG of program 1
1 Introduction
In the rst part of this paper 1], I have presented a new algorithm for computing a ne and piecewise a ne schedules for Generalized Dependence Graphs and a ne systems of recurrence equations. The algorithm is simple and e cient. However, there are programs and systems which do not have such a schedule. This is equivalent to the observation that there are programs which cannot be executed in linear time on a paracomputer, and should not come as a surprise.
Consider the program in gure 1. For de niteness, suppose that + in this computation stands for some operator with no special algebraic property. As a consequence, the computation must be executed as written; there is no possibility of sharing the work between processors by taking advantage, e.g. of associativity1 . The DFG of program 1 is given in table 1. The only ow of data in this program is throught the successive values of variable s. Each iteration of instruction 1 uses a value of s which has been produced by the immediately preceding iteration, which is h1; i; j ? 1i if j 1 and h1; i ? 1; i ? 1i if not. To construct an a ne schedule, let us apply the Farkas algorithm. The prototype schedule is: (i; j ) =
1
0
+
1
i+
2
(n ? i) +
3
j+
4
(i ? j ):
This remark will stand for all examples in this paper. Computing schedules for operators with nontrivial algebraic properties is a largely open problem.
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Edge 1 is a uniform dependence, which gives simply:
3
?
4
1:
(1)
Edge 2 is nonuniform; we must use Farkas lemma. Note that the constraint i 0 is redundant in the presence of i 1. The result is: + 1 i + 2 (n ? i) + 3 j + 4 (i ? j ) ? ( 0 + 1(i ? 1) + 2(n ? i + 1) + 3(i ? 1)) ? 1 0 + 1 (i ? 1) + 2(n ? i) + 3j + 4(i ? j ) + 5(1 ? j ):
0
This is equivalent to:
1
? + ?1 = ? = ? =
2 3 4 3 3 4
0 1 3 2
0 =
1
? + ? + ? ?
1 2 4
5 4 5
; ; ;
...
FeautrierSchedulingMultiD