Reduction in Strength CS 480

Our sample calculation for i := 1 to n for j := 1 to m c [i, j] := 0 for k := 1 to p c[i, j] := c[i, j] + a[i, k] * b[k, j]

Graph rep of our program Node 1 t1 <- n i <- 1 (implicit goto node 2) Node 2 If i > t1 goto node 10 Node 3 t2 <- m j <- 1 Node 4 if j > t2 goto node 9 Node 5 (c + ((i*4-4) * m + (j*4-4))) <- 0 t3 <- p k <- 1 Node 6 If k > t3 goto node 8 Node 7 t4 <- i*4-4 t5 <- j*4-4 t7 <- t4 * m + t5 t8 <- k*4-4 (c + t7) + (t4 * p + t8)) + (t8 * m + t5)) k <- k + 1 goto node 6 Node 8 j <- j + 1 goto Node 4 Node 9 i <- i + 1 goto Node 2 Node 10 (procedure exit)

Innermost loop 7 *, 12 + Node 5 (c + ((i*4-4) * m + (j*4-4))) <- 0 t3 <- p k <- 1 Node 6 If k > t3 goto node 8 Node 7 t4 <- i*4-4 t5 <- j*4-4 t7 <- t4 * m + t5 t8 <- k*4-4 (c + t7) + (t4 * p + t8)) + (t8 * m + t5)) k <- k + 1 goto node 6

Loop Invariant Code Removal Last time we looked for expressions that did not change within a loop, and moved them out Called Loop Invariant Code removal We were able to reduce the operations to 8 additions and 3 multiplications

After loop invariant code removal Node 5 t4 <- i * 4 – 4 t5 <- j * 4 – 4 t7 <- t4 * m + t5 t6 <- t4 * p (c + t7) <- 0 t3 <- p k <- 1 Node 6 If k > t3 goto node 8 Node 7 t8 <- k*4-4 (c + t7) + (t6 + t8)) + (8 * m + t5)) k <- k + 1 goto node 6

So, are we done? Originally had +/-: 12 and */%: 7 in innermost loop Now have +/-: 8 and */%: 3 Are we done? Can we do better? Sure we can.. Next optimization

Reduction in Strength Reduce a costly (“strong”) operation to a less costly (“weak”) operation Typically replace multiplications by additions, which can be executed much faster

Reduction in Strength algorithm Step 1 within a loop, look for variables that are changing by the addition or subtraction of a constant Called Induction Variables Step 2 look for expressions of the form (iv + c) * c + c (these come up a lot in subscript calculations) Step 3. Replace these with a temporary variable

Justification Think about the variable k in the innermost loop How is it changing? 1,2,3,……. Then how does the expression k * 4 – 4 change?

Finding induction variables Same type of analysis as before, only now we need to know not only is a variable being changed, but NOW it is being changed Look for iv <- iv + c Most commonly from for statements, but can actually come from anyplace (or even multiple assignments)

Finding expressions Look for expressions of the form (iv + c) * c + c (again, these are common in subscripts) Create a temporary that tracks these expressions Every time our induction variable changes, change the temporary (by adding a constant)

See what the constant is Suppose we have iv = iv + c 1 Suppose we have (iv + c 2 ) * c 3 + c 4 New expression is (iv + c 1 + c 2 ) * c 3 + c 4 Which is (iv + c 2 ) * c 3 + c 4 + c 1 * c 2 Which is old expression + c 1 * c 2 So new expression is just a constant change from old expression

Lets try it Node 5 t4 <- i * 4 – 4 t5 <- j * 4 – 4 t7 <- t4 * m + t5 t6 <- t4 * p (c + t7) <- 0 t3 <- p k <- 1 t8 <- 0 (this is k * 4 – 4 at this point) Node 6 If k > t3 goto node 8 Node 7 t8 <- k*4-4 (NO LONGER NEEDED) (c + t7) + (t6 + t8)) + (t8 * m + t5)) k <- k + 1 t8 <- t8 + 4 goto node 6

Are we done yet? Nope. As aways, after we have done one optimization, we have a different program Need to look once more since there might now be further optimizations that were not possible previously

New induction variable T8 is now an induction variable It changes only by a constant amount (namely, m) each time through the loop Can replace with another temporary

Lets try it Node 5 t4 <- i * 4 – 4 t5 <- j * 4 – 4 t7 <- t4 * m + t5 t6 <- t4 * p (c + t7) <- 0 t3 <- p k <- 1 t8 <- 0 t11 <- 0 (this is t8* m at this point) Node 6 If k > t3 goto node 8 Node 7 (c + t7) + t6 + t8) + t11+ t5) k <- k + 1 t8 <- t8 + 4 t11 <- t11 + 4m goto node 6

Operation counts Original 7 * 12 + After loop invariant code removal 3 * 8 + After reduction in strength 1 * 9 + Still other types of optizations that could be done, but this is pretty good