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Nope! I had it right the first time. Revert the operative part of r139135 and 

add more showing of my work.

llvm-svn: 139136
This commit is contained in:
Nick Lewycky 2011-09-06 06:39:54 +00:00
parent 6f86e001d6
commit 702cf1eccc
1 changed files with 8 additions and 5 deletions
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13
llvm/lib/Analysis/ScalarEvolution.cpp
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@ -652,7 +652,7 @@ static void GroupByComplexity(SmallVectorImpl<const SCEV *> &Ops,
/// Assume, K > 0.
static const SCEV *BinomialCoefficient(const SCEV *It, unsigned K,
ScalarEvolution &SE,
Type* ResultTy) {
Type *ResultTy) {
// Handle the simplest case efficiently.
if (K == 1)
return SE.getTruncateOrZeroExtend(It, ResultTy);
@ -1976,12 +1976,15 @@ const SCEV *ScalarEvolution::getMulExpr(SmallVectorImpl<const SCEV *> &Ops,
OtherIdx < Ops.size() && isa<SCEVAddRecExpr>(Ops[OtherIdx]);
++OtherIdx)
if (AddRecLoop == cast<SCEVAddRecExpr>(Ops[OtherIdx])->getLoop()) {
// {A,+,B}<L> * {C,+,D}<L> --> {A*C,+,A*D + B*C - B*D,+,2*B*D}<L>
// {A,+,B}<L> * {C,+,D}<L> --> {A*C,+,A*D + B*C + B*D,+,2*B*D}<L>
//
// For reference, given that {X,+,Y,+,Z} = x + y*It + z*It^2 then
// X = x, Y = y-z, Z = 2z.
// {A,+,B} * {C,+,D} = A+It*B * C+It*D = A*C + (A*D + B*C)*It + B*D*It^2
// Given an equation of the form x + y*It + z*It^2 (above), we want to
// express it in terms of {X,+,Y,+,Z}.
// {X,+,Y,+,Z} = X + Y*It + Z*(It^2 - It)/2.
// Rearranging, X = x, Y = x+y, Z = 2z.
//
// x = A*C, y = (A*D + B*C), z = B*D
// x = A*C, y = (A*D + B*C), z = B*D.
// Therefore X = A*C, Y = (A*D + B*C) - B*D and Z = 2*B*D.
for (; OtherIdx != Ops.size() && isa<SCEVAddRecExpr>(Ops[OtherIdx]);
++OtherIdx)