Compile-time Loop Splitting for Distributed Memory ... - Stanford AI Lab

Original Generalization
for(I step,0,100,10)� OriginalLoop
i=I step; A[i] =(i+1)/10 + (i-2)%10; � body;�
i=I step+1; A[i] =(i+1)/10 + (i-2)%10;
I div 10 = I step/10; IntervalLoop �
I rem 10 = I step%10; HeadLoop
for(i,I step+2,I step+9,1) � body;�
� A[i] =Idiv 10 + I rem 10; BodyLoop
I rem 10++;� � optimized body;�
i=I step+9; A[i] =(i+1)/10 + (i-2)%10;� TailLoop
� body;� �
Table 3.8: The previous example with correct loop splitting, showing peeled iterations; the
general loop splitting transformation.
iteration (� aH) uses the value 0 in place of the modulo function, where 0P7IHaV is the
correct value. Because the range of values for the modulo in the interval is from 8 to 9 and
then from 0 to 7, the restriction of Section 3.3.2 (a modulo cannot flip to zero in a loop
splitting interval) has been violated. Clearly, the loop splitting interval cannot account for
the offsets correctly.
One solution is to optimize only expressions without offsets. When many such division
and modulo expressions exist, this option may be appealing, especially if optimizing the
expressions with offsets reintroduces redundant computations by reducing the interval
length. Of course, if all expressions have offsets, none are optimized.
A better, more general solution is to divide the interval into segments of continuous
iterations with common values. In the example of Table 3.7, the ten-iteration interval is
divided into three segments – iterations 1 and 2, iterations 3 through 9, and iteration 10.
The first group of peeled iterations accounts for the modulo’s negative offset, and the third
group accounts for the division’s positive offset. Table 3.8 shows the same loop with the
first, second, and last iterations peeled away. All iterations now have the correct value,
though only the middle segment has been optimized.
In general, many different offset values may exist, leading to the appearance of many
small segments at both ends of the original interval. This thesis groups these little
segments into two segments, one representing the maximum positive offset, and the other
representing the minimum negative offset. Because they contain different sets of values,
36