From: Phil Garner (garner@signal.dra.hmg.gb)
 Subject: In place matrix transpose
 Newsgroups: sci.math.num-analysis
 Date: 1993-08-05 06:35:06 PST


Someone was talking about matrix transposes earlier on.  It's a
curious subject.  I found that an in-place transpose is about 12 times
slower than the trivial copying method.

Here's somthing I nicked from netlib and translated into C to do the
in-place one for those that are interested: (matrix must be in one
block)


typedef float scalar; /* float -> double for double precision */

/*
 * In Place Matrix Transpose
 * From: Algorithm 380 collected algorithms from ACM.
 * Converted to C by Phil Garner
 *
 * Algorithm appeared in comm. ACM, vol. 13, no. 05,
 * p. 324.
 */
int trans(scalar *a, unsigned m, unsigned n, int *move, int iwrk)
{
  scalar b;
  int i, j, k, i1, i2, ia, ib, ncount, kmi, Max, mn;

  /*
   * a is a one-dimensional array of length mn=m*n, which
   * contains the m by n matrix to be transposed.
   * move is a one-dimensional array of length iwrk
   * used to store information to speed up the process. the
   * value iwrk=(m+n)/2 is recommended. Return val indicates the
   * success or failure of the routine.
   * normal return = 0
   * errors
   * -2, iwrk negative or zero.
   * ret > 0, (should never occur). in this case
   * we set ret equal to the final value of i when the search
   * is completed but some loops have not been moved.
   * check arguments and initialise
   */

  /* Function Body */
  if (n < 2 || m < 2)
    return 0;
  if (iwrk < 1)
    return -2;

  /* If matrix is square, exchange elements a(i,j) and a(j,i). */
  if (n == m)
    {
      for (i = 0; i < m - 1; ++i)
        for (j = i + 1; j < m; ++j)
          {
            i1 = i + j * m;
            i2 = j + i * m;
            b = a[i1];
            a[i1] = a[i2];
            a[i2] = b;
          }   return 0;
    }

  /* Non square matrix */
  ncount = 2;
  for (i = 0; i < iwrk; ++i)
    move[i] = 0;

  if (n > 2)
    /* Count number,ncount, of single points. */
    for (ia = 1; ia < n - 1; ++ia)
      {
        ib = ia * (m - 1) / (n - 1);
        if (ia * (m - 1) != ib * (n - 1))
          continue;
        ++ncount;
        i = ia * m + ib;
        if (i > iwrk)
          continue;
        move[i] = 1;
      }

  /* Set initial values for search. */
  mn = m * n;
  k = mn - 1;
  kmi = k - 1;
  Max = mn;
  i = 1;

  while (1)
    {
      /* Rearrange elements of a loop. */
      /* At least one loop must be re-arranged. */
      i1 = i;
      while (1)
        {
          b = a[i1];
          while (1)
            {
              i2 = n * i1 - k * (i1 / m);
              if (i1 <= iwrk)
                move[i1 - 1] = 2;
              ++ncount;
              if (i2 == i || i2 >= kmi)
                {
                  if (Max == kmi || i2 == i)
                    break;
                  Max = kmi;
                }
              a[i1] = a[i2];
              i1 = i2;
            }

          /* Test for symmetric pair of loops. */
          a[i1] = b;
          if (ncount >= mn)
            return 0;
          if (i2 == Max || Max == kmi)
            break;
          Max = kmi;
          i1 = Max;
        }

      /* Search for loops to be rearranged. */
      while (1)
        {
          Max = k - i;
          ++i;
          kmi = k - i;
          if (i > Max)
            return i;
          if (i <= iwrk)
            {
              if (move[i-1] < 1)
                break;
              continue;
            }
          if (i == n * i - k * (i / m))
            continue;
          i1 = i;
          while (1)
            {
              i2 = n * i1 - k * (i1 / m);
              if (i2 <= i || i2 >= Max)
                break;
              i1 = i2;
            }
          if (i2 == i)
            break;
        }
    } /* End never reached */
}

--
                       ,-----------------------------       ______
 ____                  | Phil Garner.            \___|     |/   \ \   ____
/__/ `--,   _L__L\_    | garner@signal.dra.hmg.gb    |    _|`---'  \_/__/ `--,
`-0---0-'  `-0--0-'    `--OO-------------------O-----'     `---0---' `-0---0-'

 From: Murray Dow (mld900@anusf.anu.edu.au)
 Subject: Re: In place matrix transpose
 Newsgroups: sci.math.num-analysis
 Date: 1993-08-09 19:45:57 PST


In article <23qmp3INN3gl@mentor.dra.hmg.gb>, garner@signal.dra.hmg.gb (Phil Garner) writes:
|> Someone was talking about matrix transposes earlier on.  It's a
|> curious subject.  I found that an in-place transpose is about 12 times
|> slower than the trivial copying method.
|>

Algorithm 380 from CACM is sloweer than ALG 467. Here are my times
from a VP2200 vector computer. Note that the CACM algorithms are scalar.
Times are in seconds, for a 900*904 matrix:

380   NAG   467   disc copy
1.03  1.14  .391  .177

Compare two vector algorithms, one I wrote and the second a matrix
copy:

My Alg    Matrix copy
.0095     .0097

Conclusions: dont use Alg 380 from Netlib. If you have the available memory,
do a matrix copy. If you don't have the memory, I will send you my algorithm
when I have published it.
--
Murray Dow                         GPO Box 4 Canberra ACT 2601 Australia
Supercomputer Facility             Phone: +61 6 2495028
Australian National University     Fax:   +61 6 2473425
mld900@anusf.anu.edu.au

=============================================================================

From: Mark Smotherman (mark@hubcap.clemson.edu)
 Subject: Matrix transpose benchmark [was Re: MIPS R8000 == TFP?]
 Newsgroups: comp.arch, comp.benchmarks, comp.sys.super
 Date: 1994-07-01 06:35:51 PST


mccalpin@perelandra.cms.udel.edu (John D. McCalpin) writes:

>
>Of course, these results are all for the naive algorithm.  I would be
>interested to see what an efficient blocked algorithm looks like.
>Anyone care to offer one?  There is clearly a lot of performance
>to be gained by the effort....

Here is a matrix transpose benchmark generator.  Enter something like

        10d10eij;

and you get a benchmark program with tiles of size 10 for the i and j
inner loops.  Please email code improvements and flames.

Enjoy!


/*---------------------------------------------------------------------------

        Matrix Transpose Generator

        Copyright 1993, Dept. of Computer Science, Clemson University

        Permission to use, copy, modify, and distribute this software and
        its documentation for any purpose and without fee is hereby granted,
        provided that the above copyright notice appears in all copies.

        Clemson University and its Dept. of Computer Science make no
        representations about the suitability of this software for any
        purpose.  It is provided "as is" without express or implied warranty.

        Original author: Mark Smotherman

  -------------------------------------------------------------------------*/


/* tpgen.c version 1.0
 *
 * generate a matrix transpose loop nest, with tiling and unrolling
 * (timing code using getrusage is included in the generated program)
 *
 * mark smotherman
 * mark@cs.clemson.edu
 * clemson university
 * 9 july 1993
 *
 * a loop nest can be described by the order of its loop indices, so
 * this program takes as input a simple language describing these indices:
 *  <number>d  ==> generate tiling loop for index i with step size of <number>
 *  <number>e  ==> generate tiling loop for index j with step size of <number>
 *  <number>i  ==> generate loop for index i with unrolling factor of <number>
 *  <number>j  ==> generate loop for index j with unrolling factor of <number>
 *  ;          ==> input terminator (required)
 * rules are:
 *  i,j tokens must appear
 *  if d appears, it must appear before i
 *  if e appears, it must appear before j
 *  ; must appear
 * matrix size is controlled by #define N in this program.
 *
 * this code was adapted from mmgen.c v1.2 and extended to generate pre-
 * condition loops for unrolling factors that do not evenly divide the
 * matrix size (or the tiling step size for loop nests with a tiling loop).
 * note that this program only provides a preconditioning loop for the
 * innermost loop.  unrolling factors for non-innermost loops that do not
 * evenly divide the matrix size (or step size) are not supported.
 *
 * my interest in this program generator is to hook it to a sentence
 * generator and a minimum execution time finder, that is
 *   while((sentence=sgen())!=NULL){
 *     genprogram=tpgen(sentence);
 *     system("cc -O4 genprogram.c");
 *     system("a.out >> tpresults");
 *   }
 *   findmintime(tpresults);
 * this will find the optimum algorithm for the host system via an
 * exhaustive search.
 *
 * please report bugs and suggestions for enhancements to me.
 */

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#define N 500

#define ALLOC1 temp1=(struct line *)malloc(sizeof(struct line));\
temp1->indentcnt=indentcnt;

#define LINK1 temp1->next=insertbefore;\
insertafter->next=temp1;\
insertafter=temp1;

#define INSERT1 temp1->next=start;\
start=temp1;

#define ALLOC2 temp1=(struct line *)malloc(sizeof(struct line));\
temp2=(struct line *)malloc(sizeof(struct line));\
temp1->indentcnt=indentcnt;\
temp2->indentcnt=indentcnt++;

#define LINK2 temp1->next=temp2;\
temp2->next=insertbefore;\
insertafter->next=temp1;\
insertafter=temp1;\
insertbefore=temp2;

struct line{ int indentcnt; char line[256]; struct line *next; };

int indentcnt;
int iflag,jflag;
int ijflag,jiflag;
int dflag,eflag;
int counter;
int iistep,jjstep;
int iunroll,junroll;
int precond;

char c;
int i,ttp,nt;
char *p0;
char tptype[80];
char number[10];

struct line *start,*head,*insertafter,*insertbefore,*temp1,*temp2;

void processloop();
void processstmt();

main(){

  indentcnt=0;
  iflag=jflag=0;
  ijflag=jiflag=0;
  dflag=eflag=0;
  iunroll=junroll=0;
  counter=1;
  precond=0;
  ttp=0;

  start=NULL;
  ALLOC2
  sprintf(temp1->line,"/* begin */\nt_start=second();\n");
  sprintf(temp2->line,"/* end */\nt_end = second();\n");
  head=temp1; temp1->next=temp2; temp2->next=NULL;
  insertafter=temp1; insertbefore=temp2;

  while((c=getchar())!=';'){
    tptype[ttp++]=c;
    if(isdigit(c)){
      nt=0;
      while(isdigit(c)){
        number[nt++]=c;
        c=getchar();
        if(c==';'){ fprintf(stderr,"unexpected ;!\n"); exit(1); }
        tptype[ttp++]=c;
      }
      number[nt]='\0';
      sscanf(number,"%d",&counter);
    }
    switch(c){
      case 'd':
        if(iflag){ fprintf(stderr,"d cannot appear after i!\n"); exit(1); }
        dflag++;
        ALLOC1
        sprintf(temp1->line,"#define IISTEP %d\n",counter);
        INSERT1
        iistep=counter;
        counter=1;
        ALLOC2
        sprintf(temp1->line,"for(ii=0;ii<%d;ii+=IISTEP){\n",N);
        sprintf(temp2->line,"}\n",N);
        LINK2
        ALLOC1
        sprintf(temp1->line,"it=min(ii+IISTEP,%d);\n",N);
        LINK1
        break;
      case 'e':
        if(jflag){ fprintf(stderr,"e cannot appear after j!\n"); exit(1); }
        eflag++;
        ALLOC1
        sprintf(temp1->line,"#define JJSTEP %d\n",counter);
        INSERT1
        jjstep=counter;
        counter=1;
        ALLOC2
        sprintf(temp1->line,"for(jj=0;jj<%d;jj+=JJSTEP){\n",N);
        sprintf(temp2->line,"}\n",N);
        LINK2
        ALLOC1
        sprintf(temp1->line,"jt=min(jj+JJSTEP,%d);\n",N);
        LINK1
        break;
      case 'i':
        iunroll=counter;
        counter=1;
        iflag++; if(jflag) jiflag++;
        if(dflag) precond=iistep%iunroll; else precond=N%iunroll;
        if(precond&&(jiflag==0)){
          fprintf(stderr,"unrolling factor for outer loop i\n");
          fprintf(stderr,"  does not evenly divide matrix/step size!\n");
          exit(1);
        }
        if(dflag&&(iunroll>1)&&(N%iistep)){
          fprintf(stderr,"with unrolling of i, step size for tiled loop ii\n");
          fprintf(stderr,"  does not evenly divide matrix size!\n");
          exit(1);
        }
        processloop('i',dflag,iunroll,precond,junroll);
        break;
      case 'j':
        junroll=counter;
        counter=1;
        jflag++; if(iflag) ijflag++;
        if(eflag) precond=jjstep%junroll; else precond=N%junroll;
        if(precond&&(ijflag==0)){
          fprintf(stderr,"unrolling factor for outer loop j\n");
          fprintf(stderr,"  does not evenly divide matrix/step size!\n");
          exit(1);
        }
        if(eflag&&(junroll>1)&&(N%jjstep)){
          fprintf(stderr,"with unrolling of j, step size for tiled loop jj\n");
          fprintf(stderr,"  does not evenly divide matrix size!\n");
          exit(1);
        }
        processloop('j',eflag,junroll,precond,iunroll);
        break;
      default: break;
    }
  }
  processstmt();

  tptype[ttp++]=c;

  if((iflag==0)||(jflag==0)){
    fprintf(stderr,
      "one of the loops (i,j) was not specified!\n");
    exit(1);
  }

  temp1=start;
  while(temp1!=NULL){
    printf("%s",temp1->line);
    temp1=temp1->next;
  }
  printf("#include <stdio.h>\n");
  printf("#include <sys/time.h>\n");
  printf("#include <sys/resource.h>\n");
  if(dflag|eflag) printf("#define min(a,b) ((a)<=(b)?(a):(b))\n");
  printf("double second();\n");
  printf("double t_start,t_end,t_total;\n");
  printf("int times;\n");
  printf("\ndouble b[%d][%d],dummy[10000],bt[%d][%d];\n\nmain(){\n"
    ,N,N,N,N);
  if(precond) printf("  int i,j,n;\n"); else printf("  int i,j;\n");
  if(dflag) printf("  int ii,it;\n");
  if(eflag) printf("  int jj,jt;\n");
  printf("/* set coefficients so that result matrix should have \n");
  printf(" * column entries equal to column index\n");
  printf(" */\n");
  printf("  for (i=0;i<%d;i++){\n",N);
  printf("    for (j=0;j<%d;j++){\n",N);
  printf("      b[i][j] = (double) i;\n");
  printf("    }\n");
  printf("  }\n");
  printf("\n  t_total=0.0;\n  for(times=0;times<10;times++){\n\n",N);
  printf("/* try to flush cache */\n");
  printf("  for(i=0;i<10000;i++){\n",N);
  printf("    dummy[i] = 0.0;\n");
  printf("  }\n");
  printf("%s",head->line);
  temp1=head->next;
  while(temp1!=NULL){
    for(i=0;i<temp1->indentcnt;i++) printf("  ");
    while((p0=strstr(temp1->line,"+0"))!=NULL){
      *p0++=' '; *p0=' ';
    }
    printf("%s",temp1->line);
    temp1=temp1->next;
  }
  printf("\n  t_total+=t_end-t_start;\n  }\n");
  printf("/* check result */\n");
  printf("  for (j=0;j<%d;j++){\n",N);
  printf("    for (i=0;i<%d;i++){\n",N);
  printf("      if (bt[i][j]!=((double)j)){\n");
  printf("        fprintf(stderr,\"error in bt[%cd][%cd]",'%','%');
  printf("\\n\",i,j);\n");
  printf("        fprintf(stderr,\" for %s\\n\");\n",tptype);
  printf("        exit(1);\n");
  printf("      }\n");
  printf("    }\n");
  printf("  }\n");
  tptype[ttp]='\0';
  printf("  printf(\"%c10.2f secs\",t_total);\n",'%');
  printf("  printf(\" for 10 runs of %s\\n\");\n",tptype);
  printf("}\n");
  printf("double second(){\n");
  printf("  void getrusage();\n");
  printf("  struct rusage ru;\n");
  printf("  double t;\n");
  printf("  getrusage(RUSAGE_SELF,&ru);\n");
  printf("  t = ((double)ru.ru_utime.tv_sec) +\n");
  printf("    ((double)ru.ru_utime.tv_usec)/1.0e6;\n");
  printf("  return t;\n");
  printf("}\n");

}

void processloop(index,flag,unroll,precond,unroll2)
char index;
int flag,unroll,precond,unroll2;
{
  char build[80],temp[40];
  int n;
  if(precond){
    ALLOC1
    sprintf(temp1->line,"/* preconditioning loop for unrolling factor */\n");
    LINK1
    if(unroll2==1){
      build[0]='\0';
      if(flag){
        if(index='i')
          sprintf(temp,"n=IISTEP%c%d; ",'%',unroll);
        else
          sprintf(temp,"n=JJSTEP%c%d; ",'%',unroll);
        strcat(build,temp);
        sprintf(temp,"for(%c=%c%c;%c<%c%c+n;%c++) ",index,index,index,
          index,index,index,index);
        strcat(build,temp);
      }else{
        sprintf(temp,"n=%d%c%d; ",N,'%',unroll);
        strcat(build,temp);
        sprintf(temp,"for(%c=0;%c<n;%c++) ",index,index,index);
        strcat(build,temp);
      }
      sprintf(temp,"bt[i][j]=b[j][i];\n");
      strcat(build,temp);
      ALLOC1
      sprintf(temp1->line,"%s\n",build);
      LINK1
    }else{
      if(flag){
        ALLOC1
        if(index=='i')
          sprintf(temp1->line,"n=IISTEP%c%d;\n",'%',unroll);
        else
          sprintf(temp1->line,"n=JJSTEP%c%d;\n",'%',unroll);
        LINK1
        ALLOC1
        sprintf(temp1->line,"for(%c=%c%c;%c<%c%c+n;%c++){\n",index,index,index,
          index,index,index,index);
        LINK1
      }else{
        ALLOC1
        sprintf(temp1->line,"n=%d%c%d;\n",N,'%',unroll);
        LINK1
        ALLOC1
        sprintf(temp1->line,"for(%c=0;%c<n;%c++){\n",index,index,index);
        LINK1
      }
      if(index=='i'){
        for(n=0;n<unroll2;n++){
          ALLOC1
          sprintf(temp1->line,"  bt[i][j+%d]=b[j+%d][i];\n",n,n);
          LINK1
        }
      }else{
        for(n=0;n<unroll2;n++){
          ALLOC1
          sprintf(temp1->line,"  bt[i+%d][j]=b[j][i+%d];\n",n,n);
          LINK1
        }
      }
      ALLOC1
      sprintf(temp1->line,"}\n");
      LINK1
    }
    ALLOC2
    if(flag){
      sprintf(temp1->line,"for(%c=%c%c+n;%c<%ct;%c+=%d){\n",index,index,index,
        index,index,index,unroll);
    }else{
      sprintf(temp1->line,"for(%c=n;%c<%d;%c+=%d){\n",index,index,N,index,
        unroll);
    }
    sprintf(temp2->line,"}\n",N);
    LINK2
  }else{
    ALLOC2
    if(unroll==1){
      if(flag){
        sprintf(temp1->line,"for(%c=%c%c;%c<%ct;%c++){\n",index,index,index,
          index,index,index);
      }else{
        sprintf(temp1->line,"for(%c=0;%c<%d;%c++){\n",index,index,N,index);
      }
    }else{
      if(flag){
        sprintf(temp1->line,"for(%c=%c%c;%c<%ct;%c+=%d){\n",index,index,index,
          index,index,index,unroll);
      }else{
        sprintf(temp1->line,"for(%c=0;%c<%d;%c+=%d){\n",index,index,N,index,
          unroll);
      }
    }
    sprintf(temp2->line,"}\n",N);
    LINK2
  }
}

void processstmt()
{
  int i,j;
  for(i=0;i<iunroll;i++){
    for(j=0;j<junroll;j++){
      ALLOC1
      sprintf(temp1->line,"bt[i+%d][j+%d]=b[j+%d][i+%d];\n",i,j,j,i);
      LINK1
    }
  }
}
--
Mark Smotherman, Computer Science Dept., Clemson University, Clemson, SC

=======================================================================
From: has (h.genceli@bre.com)
 Subject: transpose of a nxm matrix stored in a vector !!!
 Newsgroups: sci.math.num-analysis
 Date: 2000/07/25


If I have a matrix nrows x ncols, I can store it in a vector.
so A(i,j) is really a[i*ncols+j]. So really TRANS of A
(say B) is really is also a vector B where

0<=i b[j*nrows+i] <nrows, 0<=j<ncols
b[j*nrows+i] = a[i*ncols+j].

Fine but I want to use only one array a to do this transformation.

i.e a[j*nrows+i] = a[i*ncols+j]. this will itself
erase some elements so each time a swap is necessary in a loop.

temp = a[j*nrows+i]
a[j*nrows+i] = a[i*ncols+j]
a[i*ncols+j] = temp

but still this will lose some info as it is, so indexing
should have more intelligence in it ???? anybody
can give me a lead here, thanks.

Has

 From: wei-choon ng (wng@ux8.cso.uiuc.edu)
 Subject: Re: transpose of a nxm matrix stored in a vector !!!
 Newsgroups: sci.math.num-analysis
 Date: 2000/07/25


has <h.genceli@bre.com> wrote:
> If I have a matrix nrows x ncols, I can store it in a vector.
> so A(i,j) is really a[i*ncols+j]. So really TRANS of A
> (say B) is really is also a vector B where

[snip]

Hey, if you just want to do a transpose-matrix vector multiply, there is
no need to explicitly store the transpose matrix in another array and
doubling the storage!

W.C.
--

 From: Robin Becker (robin@jessikat.fsnet.co.uk)
 Subject: Re: transpose of a nxm matrix stored in a vector !!!
 Newsgroups: sci.math.num-analysis
 Date: 2000/07/25


In article <snr532fo3j1180@corp.supernews.com>, has <h.genceli@bre.com>
writes
>If I have a matrix nrows x ncols, I can store it in a vector.
>so A(i,j) is really a[i*ncols+j]. So really TRANS of A
>(say B) is really is also a vector B where
>
>0<=i b[j*nrows+i] <nrows, 0<=j<ncols
>b[j*nrows+i] = a[i*ncols+j].
>
>Fine but I want to use only one array a to do this transformation.
>
>i.e a[j*nrows+i] = a[i*ncols+j]. this will itself
>erase some elements so each time a swap is necessary in a loop.
>
>temp = a[j*nrows+i]
>a[j*nrows+i] = a[i*ncols+j]
>a[i*ncols+j] = temp
>
>but still this will lose some info as it is, so indexing
>should have more intelligence in it ???? anybody
>can give me a lead here, thanks.
>
>Has
>
>
>

void dmx_transpose(unsigned n, unsigned m, double* a, double* b)
{
        unsigned        size = m*n;
        if(b!=a){
                real    *bmn, *aij, *anm;
                bmn = b + size; /*b+n*m*/
                anm = a + size;
                while(b<bmn) for(aij=a++;aij<anm; aij+=n ) *b++ = *aij;
                }
        else if(size>3){
                unsigned i,row,column,current;
                for(i=1, size -= 2;i<size;i++){
                        current = i;
                        do      {
                                /*current = row+n*column*/
                                column = current/m;
                                row = current%m;
                                current = n*row +  column;
                                } while(current < i);

                        if (current >i) {
                                real temp = a[i];
                                a[i] = a[current];
                                a[current] = temp;
                                }
                        }
                }
}
--
Robin Becker

 From: E. Robert Tisdale (edwin@netwood.net)
 Subject: Re: transpose of a nxm matrix stored in a vector !!!
 Newsgroups: sci.math.num-analysis
 Date: 2000/07/25


Take a look at
The C++ Scalar, Vector, Matrix and Tensor class library

    http://www.netwood.net/~edwin/svmt/

<Type><System>SubVector&
        <Type><System>SubVector::transpose(Extent p, Extent q) {
  <Type><System>SubVector&
                v = *this;
  if (1 < p && 1 < q) {
    // A vector v of extent n = qp is viewed as a q by p matrix U and
    // a p by q matrix V where U_{ij} = v_{p*i+j} and V_{ij} = v_{q*i+j}.
    // The vector v is modified in-place so that V is the transpose of U.
    // The algorithm searches for every sequence k_s of S indices
    // such that a circular shift of elements v_{k_s} <-- v_{k_{s+1}}
    // and v_{k_{S-1}} <-- v_{k_0} effects an in-place transpose.
    Extent      n = q*p;
    Extent      m = 0;                  // count up to n-2
    Offset      l = 0;                  // 1 <= l <= n-2
    while (++l < n-1 && m < n-2) {
      Offset    k = l;
      Offset    j = k;
      while (l < (k = (j%p)*q + j/p)) { // Search backward for k < l.
        j = k;
        }
      // If a sequence of indices beginning with l has any index k < l,
      // it has already been transposed.  The sequence length S = 1
      // and diagonal element v_k is its own transpose if k = j.
      // Skip every index sequence that has already been transposed.
      if (k == l) {                     // a new sequence
        if (k < j) {                    // with 1 < S
          TYPE  x = v[k];               // save v_{k_0}
          do {
            v[k] = v[j];                // v_{k_{s}} <-- v_{k_{s+1}}
            k = j;
            ++m;
            } while (l < (j = (k%q)*p + k/q));
          v[k] = x;                     // v_{k_{S-1}} <-- v_{k_0}
          }
        ++m;
        }
      }
    } return v;
  }



<Type><System>SubVector&

Read the rest of this message... (50 more lines)

 From: Victor Eijkhout (eijkhout@disco.cs.utk.edu)
 Subject: Re: transpose of a nxm matrix stored in a vector !!!
 Newsgroups: sci.math.num-analysis
 Date: 2000/07/25


"Alan Miller" <amiller @ vic.bigpond.net.au> writes:

> The attached routine does an in situ transpose.
> begin 666 Dtip.f90
> M4U5"4D]55$E.12!D=&EP("AA+"!N,2P@;C(L(&YD:6TI#0HA("TM+2TM+2TM

Hm. F90? You're not silently allocating a temporary I hope?

(Why did you have to encode this? Now I have to save, this decode, ...
and all for plain ascii?)

--
Victor Eijkhout
"When I was coming up, [..] we knew exactly who the they were. It was us
versus them, and it was clear who the them was were. Today, we are not
so sure who the they are, but we know they're there." [G.W. Bush]

 From: Alan Miller (amiller_@_vic.bigpond.net.au)
 Subject: Re: transpose of a nxm matrix stored in a vector !!!
 Newsgroups: sci.math.num-analysis
 Date: 2000/07/25


Victor Eijkhout wrote in message ...
>"Alan Miller" <amiller @ vic.bigpond.net.au> writes:
>
>> The attached routine does an in situ transpose.
>> begin 666 Dtip.f90
>> M4U5"4D]55$E.12!D=&EP("AA+"!N,2P@;C(L(&YD:6TI#0HA("TM+2TM+2TM
>
>Hm. F90? You're not silently allocating a temporary I hope?
>
>(Why did you have to encode this? Now I have to save, this decode, ...
>and all for plain ascii?)
>

I know the problem.
I sometimes use a Unix system, and have to use decode64 to read
attachments.   On the other hand, Windows wraps lines around,
formats then and generally makes the code unreadable.

The straight code for dtip (double transpose in place) is attached
this time.

>--
>Victor Eijkhout


--
Alan Miller, Retired Scientist (Statistician)
CSIRO Mathematical & Information Sciences
Alan.Miller -at- vic.cmis.csiro.au
http://www.ozemail.com.au/~milleraj
http://users.bigpond.net.au/amiller/


=================================================================

From: Darran Edmundson (dedmunds@sfu.ca)
 Subject: array reordering algorithm?
 Newsgroups: sci.math.num-analysis
 Date: 1995/04/30


A code I've written refers to a complex array as two separate real arrays.
However, I have a canned subroutine which expects a single array where the
real and imaginary values alternate.  Essentially I have a case of mismatched
data structures, yet for reasons that I'd rather not go into, I'm stuck with them.

Assuming that the two real arrays A and B are sequential in memory, and
that the single array of alternating real/imaginary values C shares the same
space, what I need is a porting subroutine that remaps the data from one format
to the other - using as little space as possible.

I think of the problem as follows.  Imagine an array of dimension 10 containing
the values 1,3,5,7,9,2,4,6,8,10 in this order.

 A(1) /  1   \  C(1)
 A(2) |  3   |  C(2)
 A(3) |  5   |  C(3)
 A(4) |  7   |  C(4)
 A(5) \  9   |  C(5)
             |
 B(1) /  2   |  C(6)
 B(2) |  4   |  C(7)
 B(3) |  6   |  C(8)
 B(4) |  8   |  C(9)
 B(5) \ 10   /  C(10)

Given that I know this initial pattern, I want to sort the array C in-place *without
making comparisons*.  That is, the algorithm can only depend on the initial
knowledge of the pattern.  Do you see what a sort is going to do?  It will
make the A and B arrays alternate, i.e. C(1)=A(1), C(2)=B(1), C(3)=A(2),
C(4)=B(2), etc.  It's not a real sort though because I can't actually refer to the
values above (i.e. no comparisons) because A and B will be holding real data,
not this contrived pattern.  The pattern above exists though - it's the
natural ordering in memory of A and B.

Either pair swapping only or a small amount of workspace can be used.  The
in-place is important - imagine scaling this problem up to an
array of 32 or 64 million double precision values and you can easily see how
duplicating the array is not a feasible solution.

Any ideas?  I've been stumped on this for a day and a half now.

Darran Edmundson
dedmunds@sfu.ca

 From: Roger Critchlow (rec@elf115.elf.org)
 Subject: Re: array reordering algorithm?
 Newsgroups: sci.math.num-analysis
 Date: 1995/04/30


   Any ideas?  I've been stumped on this for a day and a half now.

Here's some code for in situ permutations of arrays that I wrote
a few years ago.  It all started from the in situ transposition
algorithms in the Collected Algorithms of the ACM, the references
for which always get lost during the decryption from fortran.

This is the minimum space algorithm.  All you need to supply is
a function which computes the new order array index from the old
order array index.

If you can spare n*m bits to record the indexes of elements which
have been permuted, then you can speed things up.

-- rec --

------------------------------------------------------------------------
/*
** Arbitrary in situ permutations of an m by n array of base type TYPE.
** Copyright 1995 by Roger E Critchlow Jr, rec@elf.org, San Francisco, CA.
** Fair use permitted, caveat emptor.
*/
typedef int TYPE;

int transposition(int ij, int m, int n)         /* transposition about diagonal from upper left to lower right */
{ return ((ij%m)*n+ (ij/m)); }

int countertrans(int ij, int m, int n)          /* transposition about diagonal from upper right to lower left */
{ return ((m-1-(ij%m))*n+ (n-1-(ij/m))); }

int rotate90cw(int ij, int m, int n)            /* 90 degree clockwise rotation */
{ return ((m-1-(ij%m))*n+ (ij/m)); }

int rotate90ccw(int ij, int m, int n)           /* 90 degree counter clockwise rotation */
{ return ((ij%m)*n+ (n-1-(ij/m))); }

int rotate180(int ij, int m, int n)             /* 180 degree rotation */
{ return ((m-1-(ij/n))*n+ (n-1-(ij%n))); }

int reflecth(int ij, int m, int n)              /* reflection across horizontal plane */
{ return ((m-1-(ij/n))*n+ (ij%n)); }

int reflectv(int ij, int m, int n)              /* reflection across vertical plane */
{ return ((ij/n)*n+ (n-1-(ij%n))); }

int in_situ_permutation(TYPE a[], int m, int n, int (*origination)(int ij, int m, int n))
{
  int ij, oij, dij, n_to_do;
  TYPE b;
  n_to_do = m*n;
  for (ij = 0; ij < m*n && n_to_do > 0; ij += 1) {
    /* Test for previously permuted */
    for (oij = origination(ij,m,n); oij > ij; oij = origination(oij,m,n))
      ;
    if (oij < ij)
      continue;
    /* Chase the cycle */
    dij = ij;
    b = a[ij];
    for (oij = origination(dij,m,n); oij != ij; oij = origination(dij,m,n)) {
      a[dij] = a[oij];
      dij = oij;
      n_to_do -= 1;
    }
    a[dij] = b;
    n_to_do -= 1;
  } return 0;
}

#define TESTING 1
#if TESTING

/* fill a matrix with sequential numbers, row major ordering */
void fill_matrix_rows(a, m, n) TYPE *a; int m, n;
{
  int i, j;
  for (i = 0; i < m; i += 1)
    for (j = 0; j < n; j += 1)
      a[i*n+j] = i*n+j;
}

/* fill a matrix with sequential numbers, column major ordering */
void fill_matrix_cols(a, m, n) TYPE *a; int m, n;
{
  int i, j;
  for (i = 0; i < m; i += 1)
    for (j = 0; j < n; j += 1)
      a[i*n+j] = j*m+i;
}

/* test a matrix for sequential numbers, row major ordering */
int test_matrix_rows(a, m, n) TYPE *a; int m, n;
{
  int i, j, o;
  for (o = i = 0; i < m; i += 1)
    for (j = 0; j < n; j += 1)
      o += a[i*n+j] != i*n+j;
  return o;
}

/* test a matrix for sequential numbers, column major ordering */
int test_matrix_cols(a, m, n) TYPE *a; int m, n;
{
  int i, j, o;
  for (o = i = 0; i < m; i += 1)
    for (j = 0; j < n; j += 1)
      o += a[i*n+j] != j*m+i;
  return o;
}

/* print a matrix */
void print_matrix(a, m, n) TYPE *a; int m, n;
{
  char *format;
  int i, j;
  if (m*n < 10) format = "%2d";
  if (m*n < 100) format = "%3d";
  if (m*n < 1000) format = "%4d";
  if (m*n < 10000) format = "%5d";
  for (i = 0; i < m; i += 1) {
    for (j = 0; j < n; j += 1)
      printf(format, a[i*n+j]);
    printf("\n");
  }
}

#if TEST_TRANSPOSE
#define MAXSIZE 1000

main()
{
  int i, j, m, n, o;
  TYPE a[MAXSIZE];
  for (m = 1; m < sizeof(a)/sizeof(a[0]); m += 1)
    for (n = 1; m*n < sizeof(a)/sizeof(a[0]); n += 1) {
      fill_matrix_rows(a, m, n);                                /* {0 1} {2 3} */
      if (o = transpose(a, m, n))
        printf(">> transpose returned %d for a[%d][%d], row major\n", o, m, n);
      if ((o = test_matrix_cols(a, n, m)) != 0)                 /* {0 2} {1 3} */
        printf(">> transpose made %d mistakes for a[%d][%d], row major\n", o, m, n);
      /* column major */
      fill_matrix_rows(a, m, n);
      if (o = transpose(a, m, n))
        printf(">> transpose returned %d for a[%d][%d], column major\n", o, m, n);
      if ((o = test_matrix_cols(a, n, m)) != 0)
        printf(">> transpose made %d mistakes for a[%d][%d], column major\n", o, m, n);
    } return 0;
}
#endif                                                          /* TEST_TRANSPOSE */


#define TEST_DISPLAY 1
#if TEST_DISPLAY
main(argc, argv) int argc; char *argv[];
{
  TYPE *a;
  int m = 5, n = 5;
  extern void *malloc();
  if (argc > 1) {
    m = atoi(argv[1]);
    if (argc > 2)
      n = atoi(argv[2]);
  }
  a = malloc(m*n*sizeof(TYPE));

  printf("matrix\n");
  fill_matrix_rows(a, m, n);
  print_matrix(a, m, n);
  printf("transposition\n");
  in_situ_permutation(a, m, n, transposition);
  print_matrix(a, n, m);

  printf("counter transposition\n");
  fill_matrix_rows(a, m, n);
  in_situ_permutation(a, m, n, countertrans);
  print_matrix(a, n, m);

  printf("rotate 90 degrees clockwise\n");
  fill_matrix_rows(a, m, n);
  in_situ_permutation(a, m, n, rotate90cw);
  print_matrix(a, n, m);

  printf("rotate 90 degrees counterclockwise\n");
  fill_matrix_rows(a, m, n);
  in_situ_permutation(a, m, n, rotate90ccw);
  print_matrix(a, n, m);

  printf("rotate 180 degrees\n");
  fill_matrix_rows(a, m, n);
  in_situ_permutation(a, m, n, rotate180);
  print_matrix(a, m, n);

  printf("reflect across horizontal\n");
  fill_matrix_rows(a, m, n);
  in_situ_permutation(a, m, n, reflecth);
  print_matrix(a, m, n);

  printf("reflect across vertical\n");
  fill_matrix_rows(a, m, n);
  in_situ_permutation(a, m, n, reflectv);
  print_matrix(a, m, n);

  return 0;
}

#endif
#endif
