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Grid/lib/stencil/Lebesgue.cc

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#include <Grid.h>
namespace Grid {
int LebesgueOrder::UseLebesgueOrder;
LebesgueOrder::IndexInteger LebesgueOrder::alignup(IndexInteger n){
n--; // 1000 0011 --> 1000 0010
n |= n >> 1; // 1000 0010 | 0100 0001 = 1100 0011
n |= n >> 2; // 1100 0011 | 0011 0000 = 1111 0011
n |= n >> 4; // 1111 0011 | 0000 1111 = 1111 1111
n |= n >> 8; // ... (At this point all bits are 1, so further bitwise-or
n |= n >> 16; // operations produce no effect.)
n++; // 1111 1111 --> 1 0000 0000
return n;
};
LebesgueOrder::LebesgueOrder(GridBase *grid)
{
_LebesgueReorder.resize(0);
// Align up dimensions to power of two.
const IndexInteger one=1;
IndexInteger ND = grid->_ndimension;
std::vector<IndexInteger> dims(ND);
std::vector<IndexInteger> adims(ND);
std::vector<std::vector<IndexInteger> > bitlist(ND);
for(IndexInteger mu=0;mu<ND;mu++){
dims[mu] = grid->_rdimensions[mu];
assert ( dims[mu] != 0 );
adims[mu] = alignup(dims[mu]);
}
// List which bits of padded volume coordinate contribute; this strategy
// i) avoids recursion
// ii) has loop lengths at most the width of a 32 bit word.
int sitebit=0;
int split=2;
for(int mu=0;mu<ND;mu++){ // mu 0 takes bit 0; mu 1 takes bit 1 etc...
for(int bit=0;bit<split;bit++){
IndexInteger mask = one<<bit;
if ( mask&(adims[mu]-1) ){
bitlist[mu].push_back(sitebit);
sitebit++;
}
}
}
for(int bit=split;bit<32;bit++){
IndexInteger mask = one<<bit;
for(int mu=0;mu<ND;mu++){ // mu 0 takes bit 0; mu 1 takes bit 1 etc...
if ( mask&(adims[mu]-1) ){
bitlist[mu].push_back(sitebit);
sitebit++;
}
}
}
// Work out padded and unpadded volumes
IndexInteger avol = 1;
for(int mu=0;mu<ND;mu++) avol = avol * adims[mu];
IndexInteger vol = 1;
for(int mu=0;mu<ND;mu++) vol = vol * dims[mu];
// Loop over padded volume, following Lebesgue curve
// We interleave the bits from sequential "mu".
std::vector<IndexInteger> ax(ND);
for(IndexInteger asite=0;asite<avol;asite++){
// Start with zero and collect bits
for(int mu=0;mu<ND;mu++) ax[mu] = 0;
int contained = 1;
for(int mu=0;mu<ND;mu++){
// Build the coordinate on the aligned volume
for(int bit=0;bit<bitlist[mu].size();bit++){
int sbit=bitlist[mu][bit];
if(asite&(one<<sbit)){
ax[mu]|=one<<bit;
}
}
// Is it contained in original box
if ( ax[mu]>dims[mu]-1 ) contained = 0;
}
if ( contained ) {
int site = ax[0]
+ dims[0]*ax[1]
+dims[0]*dims[1]*ax[2]
+dims[0]*dims[1]*dims[2]*ax[3];
_LebesgueReorder.push_back(site);
}
}
assert( _LebesgueReorder.size() == vol );
}
}