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Compiling Array Assignments

Compiling Array Assignments. Allen and Kennedy, Chapter 13. Fortran 90. Fortran 90: successor to Fortran 77 Slow to gain acceptance: Need better/smarter compiler techniques to achieve same level of performance as Fortran 77 compilers

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Compiling Array Assignments

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  1. Compiling Array Assignments Allen and Kennedy, Chapter 13

  2. Fortran 90 • Fortran 90: successor to Fortran 77 • Slow to gain acceptance: • Need better/smarter compiler techniques to achieve same level of performance as Fortran 77 compilers • This chapter focuses on a single new feature - the array assignment statement: A[1:100] = 2.0 • Intended to provide direct mechanism to specify parallel/vector execution • This statement must be implemented for the specific available hardware. In an uniprocessor, the statement must be converted to a scalar loop: Scalarization

  3. Fortran 90 • Range of a vector operation in Fortran 90 denoted by a triplet: <lower bound: upper bound: increment> A[1:100:2] = B[2:51] + 3.0 • Semantics of Fortran 90 require that for vector statements, all inputs to the statement are fetched before any results are stored

  4. Outline • Simple scalarization • Safe scalarization • Techniques to improve on safe scalarization • Loop reversal • Input prefetching • Loop splitting • Multidimensional scalarization • A framework for analyzing multidimensional scalarization

  5. Scalarization • Replace each array assignment by a corresponding DO loop • Is it really that easy? • Two key issues: • Wish to avoid generating large array temporaries • Wish to optimize loops to exhibit good memory hierarchy • performance

  6. Simple Scalarization • Consider the vector statement: A(1:200) = 2.0 * A(1:200) • A scalar implementation: S1 DO I = 1, 200 S2 A(I) = 2.0 * A(I) ENDDO • However, some statements cause problems: A(2:201) = 2.0 * A(1:200) • If we naively scalarize: DO i = 1, 200 A(i+1) = 2.0 * A(i) ENDDO

  7. Simple Scalarization • Meaning of statements changed by the scalarization process: scalarization faults • Naive algorithm which ignores such scalarization faults: procedure SimpleScalarize(S) let V0 = (L0: U0: D0 ) be the vector iteration specifier on left side of S; // Generate the scalarizing loop let I be a new loop index variable; generate the statement DO I = L0 ,U0 ,D ; for each vector specifier V = (L: U: D) in S do replace V with (I+L-L0 ); generate an ENDDO statement; end SimpleScalarize

  8. Scalarization Faults • Why do scalarization faults occur? • Vector operation semantics: All values from the RHS of the assignment should be fetched before storing into the result • If a scalar operation stores into a location fetched by a later operation, we get a scalarization fault • Principle 13.1: A vector assignment generates a scalarization fault if and only if the scalarized loop carries a true dependence. • These dependences are known as scalarization dependences • To preserve correctness, compiler should never produce a scalarization dependence

  9. Safe Scalarization • Naive algorithm for safe scalarization: Use temporary storage to make sure scalarization dependences are not created • Consider: A(2:201) = 2.0 * A(1:200) • can be split up into: T(1:200) = 2.0 * A(1:200) A(2:201) = T(1:200) • Then scalarize using SimpleScalarize DO I = 1, 200 T(I) = 2.0 * A(I) ENDDO DO I = 2, 201 A(I) = T(I-1) ENDDO

  10. Safe Scalarization • Procedure SafeScalarize implements this method of scalarization • Good news: • Scalarization always possible by using temporaries • Bad News: • Substantial increase in memory use due to temporaries • More memory operations per array element • We shall look at a number of techniques to reduce the effects of these disadvantages

  11. Loop Reversal A(2:256) = A(1:255) + 1.0 • SimpleScalarize will produce a scalarization fault • Solution: Loop reversal DO I = 256, 2, -1 A(I) = A(I+1) + 1.0 ENDDO

  12. Loop Reversal • When can we use loop reversal? • Loop reversal maps dependences into antidependences • But also maps antidependences into dependences A(2:257) = ( A(1:256) + A(3:258) ) / 2.0 • After scalarization: DO I = 2, 257 A(I) = ( A(I-1) + A(I+1) ) / 2.0 ENDDO • Loop Reversal gets us: DO I = 257, 2 A(I) = ( A(I-1) + A(I+1) ) / 2.0 ENDDO • Thus, cannot use loop reversal in presence of antidependences

  13. Input Prefetching A(2:257) = ( A(1:256) + A(3:258) ) / 2.0 • Causes a scalarization fault when naively scalarized to: DO I = 2, 257 A(I) = ( A(I-1) + A(I+1) ) / 2.0 ENDDO • Problem: Stores into first element of the LHS in the previous iteration • Input prefetching: Use scalar temporaries to store elements of input and output arrays

  14. Input Prefetching • A first-cut at using temporaries: DO I = 2, 257 T1 = A(I-1) T2 = ( T1 + A(I+1) ) / 2.0 A(I) = T2 ENDDO • T1holds element of input array, T2 holds element of output array • But this faces the same problem. Can correct by moving assignment to T1 into previous iteration...

  15. Input Prefetching T1 = A(1) DO I = 2, 256 T2 = ( T1 + A(I+1) ) / 2.0 T1 = A(I) A(I) = T2 ENDDO T2 = ( T1 + A(257) ) / 2.0 A(I) = T2 • Note: We are using scalar replacement, but the motivation for doing so is different than in Chapter 8

  16. Already seen in Chapter 8, we need as many temporaries as the dependence threshold + 1. Example: DO I = 2, 257 A(I+2) = A(I) + 1.0 ENDDO Can be changed to: T1 = A(1) T2 = A(2) DO I = 2, 255 T3 = T1 + 1.0 T1 = T2 T2 = A(I+2) A(I+2) = T3 ENDDO T3 = T1 + 1.0 T1 = T2 A(258) = T3 T3 = T1 + 1.0 A(259) = T3 Input Prefetching

  17. Input Prefetching • Can also unroll the loop and eliminate register to register copies • Principle 13.2: Any scalarization dependence with a threshold known at compile time can be corrected by input prefetching.

  18. Input Prefetching • Sometimes, even when a scalarization dependence does not have a constant threshold, input prefetching can be used effectively A(1:N) = A(1:N) / A(1) • which can be naively scalarized as: DO i = 1, N A(i) = A(i) / A(1) ENDDO • true dependence from first iteration to every other iteration • antidependence from first iteration to itself • Via input prefetching, we get: tA1 = A(1) DO i = 1, N A(i) = A(i) / tA1 ENDDO

  19. Loop Splitting • Problem with using input prefetching with thresholds > 1: Temporaries must be saved for each iteration of the loop up to the threshold • Can potentially use loop splitting to solve this problem A(3:6) = ( A(1:4) + A(5:8) ) / 2.0 • Scalarization loop: DO I = 3, 6 A(I) = (A(I-2) + A(I+2) ) / 2.0 ENDDO • True dependence and antidependence with threshold of 2 • Split up into two independent loops which do not interact with each other...

  20. Loop Splitting DO I = 3, 6 A(I) = (A(I-2) + A(I+2) ) / 2.0 ENDDO • Both true and anti dependence has threshold of 2 • With loop splitting becomes: DO I = 3, 5, 2 A(I) = (A(I-2) + A(I+2) ) / 2.0 ENDDO DO I = 4, 6, 2 A(I) = (A(I-2) + A(I+2) ) / 2.0 ENDDO • Note that the threshold becomes 1 for each loop. Could have produced incorrect results if threshold of antidependence was not divisible by 2

  21. Loop Splitting • Can write the splitting as a nested pair of loops: DO i1 = 3, 4 DO i2 = i1, 6, 2 A(i2) = (A(i2-2) + A(i2+2) ) / 2.0 ENDDO ENDDO • The inner loop carries a scalarization dependence with threshold 1 and the outer loop carries no dependence. Can apply input prefetching: DO i1 = 3, 4 T1 = A(i1-2) DO i2 = i1, 6, 2 T2 = (T1 + A(i2 + 2)) / 2.0 T1 = A(i2) A(i2) = T2 ENDDO ENDDO

  22. Loop Splitting • Principle 13.3: Any scalarization loop in which all true dependences have the same constant threshold T and all antidependences have a threshold that is divisible by T can be transformed, using input prefetching and loop splitting, so that all scalarization dependences are eliminated.

  23. Scalarization Algorithm • Revised scalarization Algorithm: procedureFullScalarize for each vector statement Sdo begin compute the dependences of S on itself as though S had been scalarized; ifS has no scalarization dependences upon itself thenSimpleScalarize(S); elseifS has scalarization dependences, but no self antidependences then begin SimpleScalarize(S); reverse the scalarization loop; end

  24. Scalarization Algorithm else if all scalarization dependences have a threshold of 1 then begin SimpleScalarize(S); InputPrefetch(S); end else if all scalarization dependences for S have the same constant threshold T and all antidependences have thresholds that are divisible by T thenSplitLoop(S); else if all antidependences for S have the same constant threshold T and all true dependences have thresholds that are divisible by Tthen begin reverse the loop; SplitLoop(S); end elseSafeScalarize(S, SL); end endFullScalarize

  25. Multidimensional Scalarization • Vector statements in Fortran 90 in more than 1 dimension: A(1:100, 1:100) = B(1:100, 1, 1:100) • corresponds to: DO J = 1, 100 A(1:100, J) = B(1:100, 1, J) ENDDO • Scalarization in multiple dimensions: A(1:100, 1:100) = 2.0 * A(1:100, 1:100) • Obvious Strategy: convert each vector iterator into a loop: DO J = 1, 100, 1 DO I = 1, 100 A(I,J) = 2.0 * A(I,J) ENDDO ENDDO

  26. Multidimensional Scalarization • What should the order of the loops be after scalarization? • Familiar question: We dealt with this issue in Loop Selection/Interchange in Chapter 5 • Profitability of a particular configuration depends on target architecture • For simplicity, we shall assume shorter strides through memory are better • Thus, optimal choice for innermost loop is the leftmost vector iterator

  27. Multidimensional Scalarization • Extending previous results to multiple dimensions: • Each vector iterator is scalarized separately, starting from the leftmost vector iterator in the innermost loop and the rest of the iterators from left to right • Once the ordering is available: 1. Test to see if the loop carries a scalarization dependence. If not, then proceed to the next loop. • If the scalarization loop carries only true dependences, reverse the loop and proceed to the next loop. • Apply input prefetching, with loop splitting where appropriate, to eliminate dependences to which it applies. Observe, however, that in outer loops, prefetching is done for a single submatrix (the remaining dimensions). 4. Otherwise, the loop carries a scalarization fault that requires temporary storage. Generate a scalarization that utilizes temporary storage and terminate the scalarization test for this loop, since temporary storage will eliminate all scalarization faults.

  28. Outer Loop Prefetching A(1:N, 1:N) = (A(0:N-1, 2:N+1) + A(2:N+1, 0:N-1)) / 2.0 • If we try to scalarize this (keeping the column iterator in the innermost loop) we get a true scalarization dependence (<, >) involving the second input and an antidependence (>, <) involving the first input • Cannot use loop reversal...

  29. Outer Loop Prefetching A(1:N, 1:N) = (A(0:N-1, 2:N+1) + A(2:N+1, 0:N-1)) / 2.0 • We can use input prefetching on the outer loop. The temporaries will be arrays: T0(1:N) = A(2:N+1, 0) DO j = 1, N-1 T1(1:N)=( A(0:N-1, j+1) + T0(1:N) ) / 2.0 T0(1:N) = A(2:N+1, j) A(1:N, j) = T1(1:N) ENDDO T1(1:N) = ( A(0:N-1, N) + T0(1:N) ) / 2.0 A(1:N, N) = T1(1:N) • Total temporary space required = 2 rows of original matrix • Better than storage required for copy of the result matrix

  30. Loop Interchange • Sometimes, there is a tradeoff between scalarization and optimal memory hierarchy usage A(2:100, 3:101) = A(3:101, 1:201:2) • If we scalarize this using the prescribed order: DO I = 3, 101 DO 100 J = 2, 100 A(J,I) = A(J+1,2*I-5) ENDDO ENDDO • Dependences (<, >) (I = 3, 4) and (>, >) (I = 6, 7) • Cannot use loop reversal, input prefetching • Can use temporaries

  31. Loop Interchange • However, we can use loop interchange to get: DO J = 2, 100 DO I = 3, 101 A(J,I) = A(J+1,2*I-5) ENDDO ENDDO • Not optimal memory hierarchy usage, but reduction of temporary storage • Loop interchange is useful to reduce size of temporaries • It can also eliminate scalarization dependences

  32. General Multidimensional Scalarization • Goal: To vectorize a single statement which has m vector dimensions • Given an ideal order of scalarization (l1, l2, ..., lm) • (d1, d2, ..., dn) be direction vectors for all true and antidependences of the statement upon itself • The scalarization matrix is a n  m matrix of these direction vectors • For instance: A(1:N, 1:N, 1:N) = A(0:N-1, 1:N, 2:N+1) + A(1:N, 2:N+1, 0:N-1) > = < < > =

  33. General Multidimensional Scalarization • If we examine any column of the direction matrix, we can immediately see if the corresponding loop can be safely scalarized as the outermost loop of the nest: • If all entries of the column are = or >, it can be safely scalarized as the outermost loop without loop reversal. • If all entries are = or <, it can be safely scalarized with loop reversal. • If it contains a mixture of < and >, it cannot be scalarized by simple means.

  34. General Multidimensional Scalarization • Once a loop has been selected for scalarization, the dependences carried by that loop, any dependence whose direction vector does not contain a = in the position corresponding to the selected loop may be eliminated from further consideration. • In our example, if we move the second column to the outside, we get: > = < = > < < > = > < = • Scalarization in this way will reduce the matrix to: < >

  35. A Complete Scalarization Algorithm procedureCompleteScalarize(S, loop_list) let M be the scalarization direction matrix resulting from scalarization S to loop_list; while there are more loops to be scalarized do begin let l be the first loop in loop list that can be simply scalarized with or without loop reversal (determine by examining the columns of M from left to right); if there is no such l then begin let l be the first loop on loop_list; section l by input prefetching if the previous step fails then section S using the naive temporary method and exit; end

  36. A Complete Scalarization Algorithm else // make l the outermost loop section l directly or with loop reversal, depending on the entries in the column of M corresponding to l (if l is the last loop, use hardware section length); remove l from loop_list; let M' be M with the column corresponding to l and the rows corresponding to non “=“ entries in that column eliminated; M := M'; end //while endCompleteScalarize • Time Complexity = O(m2 n) where • m = number of loops • n = number of dependences

  37. A Complete Scalarization Algorithm • Correctness follows from the definition of the scalarization matrix and the method to remove dependences from the matrix • For a given statement S and loop list, Scalarize produces a correct scalarization with the following properties: • input prefetching is applied to the innermost loop possible • the order of scalarization loops is the closest possible to the order specified on input among scalarizations with property (1).

  38. Scalarization Example DO J = 2, N-1 A(2:N-1,J) = A(1:N-2,J) + A(3:N,J) + & A(2:N-1,J-1) + A(2:N-1,J+1)/4. ENDDO • Loop carried true dependence, antidependence • Naive compiler could generate: DO J = 2, N-1 DO i = 2, N-1 T(i-1) = (A(i-1,J) + A(i+1,J) + & A(i,J-1) + A(i,J+1) )/4 ENDDO DO i = 2, N-1 A(i,J) = T(i-1) ENDDO ENDDO • 2  (N-2)2accesses to memory due to array T

  39. Scalarization Example • However, can use input prefetching to get: DO J = 2, N-1 tA0 = A(1, J) DO i = 2, N-2 tA1 = (tA0+A(i+1,J)+A(i,J-1)+A(i,J+1))/4 tA0 = A(i-1, J) A(i,J) = tA1 ENDDO tA1 = (tA0+A(N,J)+A(N-1,J-1)+A(N-1,J+1))/4 A(N-1,J) = tA1 ENDDO • If temporaries are allocated to registers, no more memory accesses than original Fortran 90 program

  40. Post Scalarization Issues • Issues due to scalarization: • Generates many individual loops • These loops carry no dependences. So reuse of quantities in registers is not common • Solution: Use loop interchange, loop fusion, unroll-and-jam, and scalar replacement

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