HPL_plindx0 - Man Page

#include "hpl.h"
 void HPL_plindx0( HPL_T_panel * PANEL, const int K, int * IPID, int * LINDXA, int * LINDXAU, int * LLEN );

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.

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.

HPL_pdlaswp00N (3), HPL_pdlaswp00T (3), HPL_pdlaswp01N (3), HPL_pdlaswp01T (3).

HPL_pdlaswp00N(3), HPL_pdlaswp00T(3).