Name
HPL_plindx0 Compute local swapping index arrays.
Synopsis
#include <hpl.h>
void
HPL_plindx0(
HPL_T_panel *
PANEL,
const int
K,
int *
IPID,
int *
LINDXA,
int *
LINDXAU,
int *
LLEN
);
Description
HPL_plindx0
computes two local arrays LINDXA and LINDXAU containing
the local source and final destination position resulting from the
application of row interchanges.
On entry, the array IPID of length K is such that the row of global
index IPID(i) should be mapped onto row of global index IPID(i+1).
Let IA be the global index of the first row to be swapped. For k in
[0..K/2), the row of global index IPID(2*k) should be mapped onto the
row of global index IPID(2*k+1). The question then, is to determine
which rows should ultimately be part of U.
First, some rows of the process ICURROW may be swapped locally. One
of this row belongs to U, the other one belongs to my local piece of
A. The other rows of the current block are swapped with remote rows
and are thus not part of U. These rows however should be sent along,
and grabbed by the other processes as we progress in the exchange
phase.
So, assume that I am ICURROW and consider a row of index IPID(2*i)
that I own. If I own IPID(2*i+1) as well and IPID(2*i+1) - IA is less
than N, this row is locally swapped and should be copied into U at
the position IPID(2*i+1) - IA. No row will be exchanged for this one.
If IPID(2*i+1)-IA is greater than N, then the row IPID(2*i) should be
locally copied into my local piece of A at the position corresponding
to the row of global index IPID(2*i+1).
If the process ICURROW does not own IPID(2*i+1), then row IPID(2*i)
is to be swapped away and strictly speaking does not belong to U, but
to A remotely. Since this process will however send this array U,
this row is copied into U, exactly where the row IPID(2*i+1) should
go. For this, we search IPID for k1, such that IPID(2*k1) is equal to
IPID(2*i+1); and row IPID(2*i) is to be copied in U at the position
IPID(2*k1+1)-IA.
It is thus important to put the rows that go into U, i.e., such that
IPID(2*i+1) - IA is less than N at the begining of the array IPID. By
doing so, U is formed, and the local copy is performed in just one
sweep.
Two lists LINDXA and LINDXAU are built. LINDXA contains the local
index of the rows I have that should be copied. LINDXAU contains the
local destination information: if LINDXAU(k) >= 0, row LINDXA(k) of A
is to be copied in U at position LINDXAU(k). Otherwise, row LINDXA(k)
of A should be locally copied into A(-LINDXAU(k),:). In the process
ICURROW, the initial packing algorithm proceeds as follows.
for all entries in IPID,
if IPID(2*i) is in ICURROW,
if IPID(2*i+1) is in ICURROW,
if( IPID(2*i+1) - IA < N )
save corresponding local position
of this row (LINDXA);
save local position (LINDXAU) in U
where this row goes;
[copy row IPID(2*i) in U at position
IPID(2*i+1)-IA; ];
else
save corresponding local position of
this row (LINDXA);
save local position (-LINDXAU) in A
where this row goes;
[copy row IPID(2*i) in my piece of A
at IPID(2*i+1);]
end if
else
find k1 such that IPID(2*k1) = IPID(2*i+1);
copy row IPID(2*i) in U at position
IPID(2*k1+1)-IA;
save corresponding local position of this
row (LINDXA);
save local position (LINDXAU) in U where
this row goes;
end if
end if
end for
Second, if I am not the current row process ICURROW, all source rows
in IPID that I own are part of U. Indeed, they are swapped with one
row of the current block of rows, and the main factorization
algorithm proceeds one row after each other. The processes different
from ICURROW, should exchange and accumulate those rows until they
receive some data previously owned by the process ICURROW.
In processes different from ICURROW, the initial packing algorithm
proceeds as follows. Consider a row of global index IPID(2*i) that I
own. When I will be receiving data previously owned by ICURROW, i.e.,
U, row IPID(2*i) should replace the row in U at pos. IPID(2*i+1)-IA,
and this particular row of U should be first copied into my piece of
A, at A(il,:), where il is the local row index corresponding to
IPID(2*i). Now,initially, this row will be packed into workspace, say
as the kth row of that work array. The following algorithm sets
LINDXAU[k] to IPID(2*i+1)-IA, that is the position in U where the row
should be copied. LINDXA(k) stores the local index in A where this
row of U should be copied, i.e il.
for all entries in IPID,
if IPID(2*i) is not in ICURROW,
copy row IPID(2*i) in work array;
save corresponding local position
of this row (LINDXA);
save position (LINDXAU) in U where
this row should be copied;
end if
end for
Since we are at it, we also globally figure out how many rows every
process has. That is necessary, because it would rather be cumbersome
to figure it on the fly during the bi-directional exchange phase.
This information is kept in the array LLEN of size NPROW. Also note
that the arrays LINDXA and LINDXAU are of max length equal to 2*N.
Arguments
PANEL (local input/output) HPL_T_panel *
On entry, PANEL points to the data structure containing the
panel information.
K (global input) const int
On entry, K specifies the number of entries in IPID. K is at
least 2*N, and at most 4*N.
IPID (global input) int *
On entry, IPID is an array of length K. The first K entries
of that array contain the src and final destination resulting
from the application of the interchanges.
LINDXA (local output) int *
On entry, LINDXA is an array of dimension 2*N. On exit, this
array contains the local indexes of the rows of A I have that
should be copied into U.
LINDXAU (local output) int *
On exit, LINDXAU is an array of dimension 2*N. On exit, this
array contains the local destination information encoded as
follows. If LINDXAU(k) >= 0, row LINDXA(k) of A is to be
copied in U at position LINDXAU(k). Otherwise, row LINDXA(k)
of A should be locally copied into A(-LINDXAU(k),:).
LLEN (global output) int *
On entry, LLEN is an array of length NPROW. On exit, it
contains how many rows every process has.
See Also
HPL_pdlaswp00N,
HPL_pdlaswp00T,
HPL_pdlaswp01N,
HPL_pdlaswp01T