/// 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);
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)