/*
* Linpack 100x100 Benchmark In C/C++ For PCs
*
********************************************************************
*
* Original Source from NETLIB
*
* Translated to C by Bonnie Toy 5/88 (modified on 2/25/94 to fix
* a problem with daxpy for unequal increments or equal increments
* not equal to 1. Jack Dongarra)
*
* To obtain rolled source BLAS, add -DROLL to the command lines.
* To obtain unrolled source BLAS, add -DUNROLL to the command lines.
*
* You must specify one of -DSP or -DDP to compile correctly.
*
* You must specify one of -DROLL or -DUNROLL to compile correctly.
*
********************************************************************
*
* Changes in this version
*
* 1. Function prototypes are declared and function headers have
* embedded parameter types to produce code for C and C++
*
* 2. Arrays aa and a are declared as [200*200] and [200*201] to
* allow compilation with prototypes.
*
* 3. Function second changed (compiler dependent).
*
* 4. Timing method changed due to inaccuracy of PC clock (see below).
*
* 5. Additional date function included (compiler dependent).
*
* 6. Additional code used as a standard for a series of benchmarks:-
* Automatic run time calibration rather than fixed parameters
* Initial calibration with display to show linearity
* Results displayed at reasonable rate for viewing (5 seconds)
* Facilities for typing in details of system used etc.
* Compiler details in code in case .exe files used elsewhere
* Results appended to a text file (Linpack.txt)
*
* Roy Longbottom 101323.2241@compuserve.com 14 September 1996
*
************************************************************************
*
* Timing
*
* The PC timer is updated at about 18 times per second or resolution of
* 0.05 to 0.06 seconds which is similar to the time taken by the main
* time consuming function dgefa on a 100 MHz Pentium. Thus there is no
* point in running the dgefa/dges1 combination three times as in the
* original version. Main timing for the latter, in the loop run NTIMES,
* executes matgen/dgefa, summing the time taken by matgen within the
* loop for later deduction from the total time. On a modern PC this sum
* can be based on a random selection of 0 or 0.05/0.06. This version
* executes the single pass once and the main timing loop five times,
* calculating the matgen overhead separately.
*
*************************************************************************
*
* Example of Output
*
* Rolled Double Precision Linpack Benchmark - PC Version in 'C/C++'
*
* Compiler Watcom C/C++ 10.5 Win 386
* Optimisation -zp4 -otexan -fp5 -5r -dDP -dROLL
*
*
* norm resid resid machep x[0]-1 x[n-1]-1
* 0.4 7.41628980e-014 1.00000000e-015 -1.49880108e-014 -1.89848137e-014
*
*
* Times are reported for matrices of order 100
* 1 pass times for array with leading dimension of 201
*
* dgefa dgesl total Mflops unit ratio
* 0.06000 0.00000 0.06000 11.44 0.1748 1.0714
*
*
* Calculating matgen overhead
*
* 10 times 0.11 seconds
* 20 times 0.22 seconds
* 40 times 0.44 seconds
* 80 times 0.87 seconds
* 160 times 1.76 seconds
* 320 times 3.52 seconds
* 640 times 7.03 seconds
*
* Overhead for 1 matgen 0.01098 seconds
*
*
* Calculating matgen/dgefa passes for 5 seconds
*
* 10 times 0.71 seconds
* 20 times 1.38 seconds
* 40 times 2.80 seconds
* 80 times 5.66 seconds
*
* Passes used 70
*
* This is followed by output of the normal data for dgefa, dges1,
* total, Mflops, unit and ratio with five sets of results for each.
*
************************************************************************
*
* Example from output file Linpack.txt
*
* LINPACK BENCHMARK FOR PCs 'C/C++' n @ 100
*
* Month run 9/1996
* PC model Escom
* CPU Pentium
* Clock MHz 100
* Cache 256K
* Options Neptune chipset
* OS/DOS Windows 95
* Compiler Watcom C/C++ 10.5 Win 386
* OptLevel -zp4 -otexan -fp5 -5r -dDP -dROLL
* Run by Roy Longbottom
* From UK
* Mail 101323.2241@compuserve.com
*
* Rolling Rolled
* Precision Double
* norm. resid 0.4
* resid 7.41628980e-014
* machep 1.00000000e-015 (8.88178420e-016 NON OPT)
* x[0]-1 -1.49880108e-014
* x[n-1]-1 -1.89848137e-014
* matgen 1 seconds 0.01051
* matgen 2 seconds 0.01050
* Repetitions 70
* Leading dimension 201
* dgefa dgesl total Mflops
* 1 pass seconds 0.06000 0.00000 0.06000
* Repeat seconds 0.06092 0.00157 0.06249 10.99
* Repeat seconds 0.06077 0.00157 0.06234 11.01
* Repeat seconds 0.06092 0.00157 0.06249 10.99
* Repeat seconds 0.06092 0.00157 0.06249 10.99
* Repeat seconds 0.06092 0.00157 0.06249 10.99
* Average 10.99
* Leading dimension 200
* Repeat seconds 0.05936 0.00157 0.06093 11.27
* Repeat seconds 0.05936 0.00157 0.06093 11.27
* Repeat seconds 0.05864 0.00157 0.06021 11.40
* Repeat seconds 0.05936 0.00157 0.06093 11.27
* Repeat seconds 0.05864 0.00157 0.06021 11.40
* Average 11.32
*
************************************************************************
*
* Examples of Results
*
* Precompiled codes were produced via a Watcom C/C++ 10.5 compiler.
* Versions are available for DOS, Windows 3/95 and NT/Win 95. Both
* non-optimised and optimised programs are available. The latter has
* options as in the above example. Although these options can place
* functions in-line, in this case, daxpy is not in-lined. Optimisation
* reduces 18 instructions in the loop in this function to the following:
*
* L85 fld st(0)
* fmul qword ptr [edx]
* add eax,00000008H
* add edx,00000008H
* fadd qword ptr -8H[eax]
* inc ebx
* fstp qword ptr -8H[eax]
* cmp ebx,esi
* jl L85
*
* Results produced are not consistent between runs but produce similar
* speeds when executing at a particular dimension (see above). An example
* of other results is 11.4/10.5 Mflops. Most typical double precision
* rolled results are:
*
* Opt No Opt Version/
* MHz Cache Mflops Mflops Make/Options Via
*
* AM80386DX 40 128K 0.53 0.36 Clone Win/W95
* 80486DX2 66 128K 2.5 1.9 Escom SIS chipset Win/W95
* 80486DX2 66 128K 2.3 1.9 Escom SIS chipset NT/W95
* 80486DX2 66 128K 2.8 2.0 Escom SIS chipset Dos/Dos
* Pentium 100 256K 11 4.2 Escom Neptune chipset Win/W95
* Pentium 100 256K 11 5.5 Escom Neptune chipset NT/W95
* Pentium 100 256K 12 4.4 Escom Neptune chipset Dos/Dos
* Pentium Pro 200 256K 48 19 Dell XPS Pro200n NT/NT
*
* The results are as produced when compiled as Linpack.cpp. Compiling as
* Linpack.c gives similar speeds but the code is a little different.
*
***************************************************************************
*/
#define ROLL
#define SP
#ifdef SP
#define REAL float
#define ZERO 0.0
#define ONE 1.0
#define PREC "Single "
#endif
#ifdef DP
#define REAL double
#define ZERO 0.0e0
#define ONE 1.0e0
#define PREC "Double "
#endif
#ifdef ROLL
#define ROLLING "Rolled "
#endif
#ifdef UNROLL
#define ROLLING "Unrolled "
#endif
#define NTIMES 10
#include <stdio.h>
#include <math.h>
//#include "conio.h"
#include <stdio.h>
#include <stdlib.h>
static REAL atime[9][15];
static int this_month;
static int this_year;
void print_time (int row);
void matgen (REAL a[], int lda, int n, REAL b[], REAL *norma);
void dgefa (REAL a[], int lda, int n, int ipvt[], int *info);
void dgesl (REAL a[],int lda,int n,int ipvt[],REAL b[],int job);
void dmxpy (int n1, REAL y[], int n2, int ldm, REAL x[], REAL m[]);
void daxpy (int n, REAL da, REAL dx[], int incx, REAL dy[], int incy);
REAL epslon (REAL x);
int idamax (int n, REAL dx[], int incx);
void dscal (int n, REAL da, REAL dx[], int incx);
REAL ddot (int n, REAL dx[], int incx, REAL dy[], int incy);
/* TIME TIME TIME TIME TIME TIME TIME TIME TIME TIME TIME TIME TIME */
#if 1
#include <time.h> /* for following time functions only */
REAL second()
{
REAL secs;
clock_t Time;
Time = clock();
secs = (REAL)Time / (REAL)CLOCKS_PER_SEC;
return secs ;
}
#endif
/* DATE DATE DATE DATE DATE DATE DATE DATE DATE DATE DATE DATE DATE */
#if 1
#include <time.h> /* for following date functions only */
void what_date()
{
time_t *t,timer;
struct tm *tm;
timer = time(t);
tm = localtime(&timer);
this_month = tm->tm_mon;
this_year = tm->tm_year;
return;
}
#endif
main ()
{
static REAL aa[200*200],a[200*201],b[200],x[200];
REAL cray,ops,total,norma,normx;
REAL resid,residn,eps,t1,tm2,epsn,x1,x2;
REAL mflops;
static int ipvt[200],n,i,j,ntimes,info,lda,ldaa;
int Endit, pass, loop;
REAL overhead1, overhead2, time1, time2;
FILE *outfile;
char *compiler, *options, general[9][80] = {" "};
outfile = fopen("Linpack.txt","a+");
if (outfile == NULL)
{
printf ("Cannot open results file \n\n");
printf("Press any key\n");
Endit = getchar();
exit (0);
}
/************************************************************************
* Enter details of compiler and options used *
************************************************************************/
/*----------------- --------- --------- ---------*/
compiler = "INSERT COMPILER NAME HERE";
options = "INSERT OPTIMISATION OPTIONS HERE";
/* Include -dDP or -dSP and -dROLL or -dUNROLL */
lda = 201;
ldaa = 200;
cray = .056;
n = 100;
fprintf(stdout,ROLLING);fprintf(stdout,PREC);
fprintf(stdout,"Precision Linpack Benchmark - PC Version in 'C/C++'\n\n");
fprintf(stdout,"Compiler %s\n",compiler);
fprintf(stdout,"Optimisation %s\n\n",options);
ops = (2.0e0*(n*n*n))/3.0 + 2.0*(n*n);
matgen(a,lda,n,b,&norma);
t1 = second();
dgefa(a,lda,n,ipvt,&info);
atime[0][0] = second() - t1;
t1 = second();
dgesl(a,lda,n,ipvt,b,0);
atime[1][0] = second() - t1;
total = atime[0][0] + atime[1][0];
/* compute a residual to verify results. */
for (i = 0; i < n; i++) {
x[i] = b[i];
}
matgen(a,lda,n,b,&norma);
for (i = 0; i < n; i++) {
b[i] = -b[i];
}
dmxpy(n,b,n,lda,x,a);
resid = 0.0;
normx = 0.0;
for (i = 0; i < n; i++) {
resid = (resid > fabs((double)b[i]))
? resid : fabs((double)b[i]);
normx = (normx > fabs((double)x[i]))
? normx : fabs((double)x[i]);
}
eps = epslon(ONE);
residn = resid/( n*norma*normx*eps );
epsn = eps;
x1 = x[0] - 1;
x2 = x[n-1] - 1;
printf("norm resid resid machep");
printf(" x[0]-1 x[n-1]-1\n");
printf("%6.1f %17.8e%17.8e%17.8e%17.8e\n\n",
(double)residn, (double)resid, (double)epsn,
(double)x1, (double)x2);
fprintf(stderr,"Times are reported for matrices of order %5d\n",n);
fprintf(stderr,"1 pass times for array with leading dimension of%5d\n\n",lda);
fprintf(stderr," dgefa dgesl total Mflops unit");
fprintf(stderr," ratio\n");
atime[2][0] = total;
if (total > 0.0)
{
atime[3][0] = ops/(1.0e6*total);
atime[4][0] = 2.0/atime[3][0];
}
else
{
atime[3][0] = 0.0;
atime[4][0] = 0.0;
}
atime[5][0] = total/cray;
print_time(0);
/************************************************************************
* Calculate overhead of executing matgen procedure *
************************************************************************/
fprintf (stderr,"\nCalculating matgen overhead\n");
pass = -20;
loop = NTIMES;
do
{
time1 = second();
pass = pass + 1;
for ( i = 0 ; i < loop ; i++)
{
matgen(a,lda,n,b,&norma);
}
time2 = second();
overhead1 = (time2 - time1);
fprintf (stderr,"%10d times %6.2f seconds\n", loop, overhead1);
if (overhead1 > 5.0)
{
pass = 0;
}
if (pass < 0)
{
if (overhead1 < 0.1)
{
loop = loop * 10;
}
else
{
loop = loop * 2;
}
}
}
while (pass < 0);
overhead1 = overhead1 / (double)loop;
fprintf (stderr,"Overhead for 1 matgen %12.5f seconds\n\n", overhead1);
/************************************************************************
* Calculate matgen/dgefa passes for 5 seconds *
************************************************************************/
fprintf (stderr,"Calculating matgen/dgefa passes for 5 seconds\n");
pass = -20;
ntimes = NTIMES;
do
{
time1 = second();
pass = pass + 1;
for ( i = 0 ; i < ntimes ; i++)
{
matgen(a,lda,n,b,&norma);
dgefa(a,lda,n,ipvt,&info );
}
time2 = second() - time1;
fprintf (stderr,"%10d times %6.2f seconds\n", ntimes, time2);
if (time2 > 5.0)
{
pass = 0;
}
if (pass < 0)
{
if (time2 < 0.1)
{
ntimes = ntimes * 10;
}
else
{
ntimes = ntimes * 2;
}
}
}
while (pass < 0);
ntimes = 5.0 * (double)ntimes / time2;
if (ntimes == 0) ntimes = 1;
fprintf (stderr,"Passes used %10d \n\n", ntimes);
fprintf(stderr,"Times for array with leading dimension of%4d\n\n",lda);
fprintf(stderr," dgefa dgesl total Mflops unit");
fprintf(stderr," ratio\n");
/************************************************************************
* Execute 5 passes *
************************************************************************/
tm2 = ntimes * overhead1;
atime[3][6] = 0;
for (j=1 ; j<6 ; j++)
{
t1 = second();
for (i = 0; i < ntimes; i++)
{
matgen(a,lda,n,b,&norma);
dgefa(a,lda,n,ipvt,&info );
}
atime[0][j] = (second() - t1 - tm2)/ntimes;
t1 = second();
for (i = 0; i < ntimes; i++)
{
dgesl(a,lda,n,ipvt,b,0);
}
atime[1][j] = (second() - t1)/ntimes;
total = atime[0][j] + atime[1][j];
atime[2][j] = total;
atime[3][j] = ops/(1.0e6*total);
atime[4][j] = 2.0/atime[3][j];
atime[5][j] = total/cray;
atime[3][6] = atime[3][6] + atime[3][j];
print_time(j);
}
atime[3][6] = atime[3][6] / 5.0;
fprintf (stderr,"Average %11.2f\n",
(double)atime[3][6]);
fprintf (stderr,"\nCalculating matgen2 overhead\n");
/************************************************************************
* Calculate overhead of executing matgen procedure *
************************************************************************/
time1 = second();
for ( i = 0 ; i < loop ; i++)
{
matgen(aa,ldaa,n,b,&norma);
}
time2 = second();
overhead2 = (time2 - time1);
overhead2 = overhead2 / (double)loop;
fprintf (stderr,"Overhead for 1 matgen %12.5f seconds\n\n", overhead2);
fprintf(stderr,"Times for array with leading dimension of%4d\n\n",ldaa);
fprintf(stderr," dgefa dgesl total Mflops unit");
fprintf(stderr," ratio\n");
/************************************************************************
* Execute 5 passes *
************************************************************************/
tm2 = ntimes * overhead2;
atime[3][12] = 0;
for (j=7 ; j<12 ; j++)
{
t1 = second();
for (i = 0; i < ntimes; i++)
{
matgen(aa,ldaa,n,b,&norma);
dgefa(aa,ldaa,n,ipvt,&info );
}
atime[0][j] = (second() - t1 - tm2)/ntimes;
t1 = second();
for (i = 0; i < ntimes; i++)
{
dgesl(aa,ldaa,n,ipvt,b,0);
}
atime[1][j] = (second() - t1)/ntimes;
total = atime[0][j] + atime[1][j];
atime[2][j] = total;
atime[3][j] = ops/(1.0e6*total);
atime[4][j] = 2.0/atime[3][j];
atime[5][j] = total/cray;
atime[3][12] = atime[3][12] + atime[3][j];
print_time(j);
}
atime[3][12] = atime[3][12] / 5.0;
fprintf (stderr,"Average %11.2f\n",
(double)atime[3][12]);
/************************************************************************
* Use minimum average as overall Mflops rating *
************************************************************************/
mflops = atime[3][6];
if (atime[3][12] < mflops) mflops = atime[3][12];
fprintf(stderr,"\n");
fprintf(stderr,ROLLING);fprintf(stderr,PREC);
fprintf(stderr," Precision %11.2f Mflops \n\n",mflops);
what_date();
/************************************************************************
* Type details of hardware, software etc. *
************************************************************************/
printf ("Enter the following data which will be "
"appended to file Linpack.txt \n\n");
printf ("PC Supplier/model ?\n ");
scanf ("%[^\n]", general[1]);
fflush (stdin);
printf ("CPU ?\n ");
scanf ("%[^\n]", general[2]);
fflush (stdin);
printf ("Clock MHz ?\n ");
scanf ("%[^\n]", general[3]);
fflush (stdin);
printf ("Cache ?\n ");
scanf ("%[^\n]", general[4]);
fflush (stdin);
printf ("Chipset/options ?\n ");
scanf ("%[^\n]", general[5]);
fflush (stdin);
printf ("OS/DOS version ?\n ");
scanf ("%[^\n]", general[6]);
fflush (stdin);
printf ("Your name ?\n ");
scanf ("%[^\n]", general[7]);
fflush (stdin);
printf ("Where from ?\n ");
scanf ("%[^\n]", general[8]);
fflush (stdin);
printf ("Mail address ?\n ");
scanf ("%[^\n]", general[0]);
fflush (stdin);
/************************************************************************
* Add results to output file LLloops.txt *
************************************************************************/
fprintf (outfile, "----------------- ----------------- --------- "
"--------- ---------\n");
fprintf (outfile, "LINPACK BENCHMARK FOR PCs 'C/C++' n @ 100\n\n");
fprintf (outfile, "Month run %d/%d\n", this_month, this_year);
fprintf (outfile, "PC model %s\n", general[1]);
fprintf (outfile, "CPU %s\n", general[2]);
fprintf (outfile, "Clock MHz %s\n", general[3]);
fprintf (outfile, "Cache %s\n", general[4]);
fprintf (outfile, "Options %s\n", general[5]);
fprintf (outfile, "OS/DOS %s\n", general[6]);
fprintf (outfile, "Compiler %s\n", compiler);
fprintf (outfile, "OptLevel %s\n", options);
fprintf (outfile, "Run by %s\n", general[7]);
fprintf (outfile, "From %s\n", general[8]);
fprintf (outfile, "Mail %s\n\n", general[0]);
fprintf(outfile, "Rolling %s\n",ROLLING);
fprintf(outfile, "Precision %s\n",PREC);
fprintf(outfile, "norm. resid %16.1f\n",(double)residn);
fprintf(outfile, "resid %16.8e\n",(double)resid);
fprintf(outfile, "machep %16.8e\n",(double)epsn);
fprintf(outfile, "x[0]-1 %16.8e\n",(double)x1);
fprintf(outfile, "x[n-1]-1 %16.8e\n",(double)x2);
fprintf(outfile, "matgen 1 seconds %16.5f\n",overhead1);
fprintf(outfile, "matgen 2 seconds %16.5f\n",overhead2);
fprintf(outfile, "Repetitions %16d\n",ntimes);
fprintf(outfile, "Leading dimension %16d\n",lda);
fprintf(outfile, " dgefa dgesl "
" total Mflops\n");
fprintf(outfile, "1 pass seconds %16.5f %9.5f %9.5f\n",
atime[0][0], atime[1][0], atime[2][0]);
for (i=1 ; i<6 ; i++)
{
fprintf(outfile, "Repeat seconds %16.5f %9.5f %9.5f %9.2f\n",
atime[0][i], atime[1][i], atime[2][i], atime[3][i]);
}
fprintf(outfile, "Average %46.2f\n",atime[3][6]);
fprintf(outfile, "Leading dimension %16d\n",ldaa);
for (i=7 ; i<12 ; i++)
{
fprintf(outfile, "Repeat seconds %16.5f %9.5f %9.5f %9.2f\n",
atime[0][i], atime[1][i], atime[2][i], atime[3][i]);
}
fprintf(outfile, "Average %46.2f\n\n",atime[3][12]);
fclose (outfile);
printf("\nPress any key\n");
Endit = getchar();
}
/*----------------------*/
void print_time (int row)
{
fprintf(stderr,"%11.5f%11.5f%11.5f%11.2f%11.4f%11.4f\n", (double)atime[0][row],
(double)atime[1][row], (double)atime[2][row], (double)atime[3][row],
(double)atime[4][row], (double)atime[5][row]);
return;
}
/*----------------------*/
void matgen (REAL a[], int lda, int n, REAL b[], REAL *norma)
/* We would like to declare a[][lda], but c does not allow it. In this
function, references to a[i][j] are written a[lda*i+j]. */
{
int init, i, j;
init = 1325;
*norma = 0.0;
for (j = 0; j < n; j++) {
for (i = 0; i < n; i++) {
init = 3125*init % 65536;
a[lda*j+i] = (init - 32768.0)/16384.0;
*norma = (a[lda*j+i] > *norma) ? a[lda*j+i] : *norma;
/* alternative for some compilers
if (fabs(a[lda*j+i]) > *norma) *norma = fabs(a[lda*j+i]);
*/
}
}
for (i = 0; i < n; i++) {
b[i] = 0.0;
}
for (j = 0; j < n; j++) {
for (i = 0; i < n; i++) {
b[i] = b[i] + a[lda*j+i];
}
}
return;
}
/*----------------------*/
void dgefa(REAL a[], int lda, int n, int ipvt[], int *info)
/* We would like to declare a[][lda], but c does not allow it. In this
function, references to a[i][j] are written a[lda*i+j]. */
/*
dgefa factors a double precision matrix by gaussian elimination.
dgefa is usually called by dgeco, but it can be called
directly with a saving in time if rcond is not needed.
(time for dgeco) = (1 + 9/n)*(time for dgefa) .
on entry
a REAL precision[n][lda]
the matrix to be factored.
lda integer
the leading dimension of the array a .
n integer
the order of the matrix a .
on return
a an upper triangular matrix and the multipliers
which were used to obtain it.
the factorization can be written a = l*u where
l is a product of permutation and unit lower
triangular matrices and u is upper triangular.
ipvt integer[n]
an integer vector of pivot indices.
info integer
= 0 normal value.
= k if u[k][k] .eq. 0.0 . this is not an error
condition for this subroutine, but it does
indicate that dgesl or dgedi will divide by zero
if called. use rcond in dgeco for a reliable
indication of singularity.
linpack. this version dated 08/14/78 .
cleve moler, university of new mexico, argonne national lab.
functions
blas daxpy,dscal,idamax
*/
{
/* internal variables */
REAL t;
int j,k,kp1,l,nm1;
/* gaussian elimination with partial pivoting */
*info = 0;
nm1 = n - 1;
if (nm1 >= 0) {
for (k = 0; k < nm1; k++) {
kp1 = k + 1;
/* find l = pivot index */
l = idamax(n-k,&a[lda*k+k],1) + k;
ipvt[k] = l;
/* zero pivot implies this column already
triangularized */
if (a[lda*k+l] != ZERO) {
/* interchange if necessary */
if (l != k) {
t = a[lda*k+l];
a[lda*k+l] = a[lda*k+k];
a[lda*k+k] = t;
}
/* compute multipliers */
t = -ONE/a[lda*k+k];
dscal(n-(k+1),t,&a[lda*k+k+1],1);
/* row elimination with column indexing */
for (j = kp1; j < n; j++) {
t = a[lda*j+l];
if (l != k) {
a[lda*j+l] = a[lda*j+k];
a[lda*j+k] = t;
}
daxpy(n-(k+1),t,&a[lda*k+k+1],1,
&a[lda*j+k+1],1);
}
}
else {
*info = k;
}
}
}
ipvt[n-1] = n-1;
if (a[lda*(n-1)+(n-1)] == ZERO) *info = n-1;
return;
}
/*----------------------*/
void dgesl(REAL a[],int lda,int n,int ipvt[],REAL b[],int job )
/* We would like to declare a[][lda], but c does not allow it. In this
function, references to a[i][j] are written a[lda*i+j]. */
/*
dgesl solves the double precision system
a * x = b or trans(a) * x = b
using the factors computed by dgeco or dgefa.
on entry
a double precision[n][lda]
the output from dgeco or dgefa.
lda integer
the leading dimension of the array a .
n integer
the order of the matrix a .
ipvt integer[n]
the pivot vector from dgeco or dgefa.
b double precision[n]
the right hand side vector.
job integer
= 0 to solve a*x = b ,
= nonzero to solve trans(a)*x = b where
trans(a) is the transpose.
on return
b the solution vector x .
error condition
a division by zero will occur if the input factor contains a
zero on the diagonal. technically this indicates singularity
but it is often caused by improper arguments or improper
setting of lda . it will not occur if the subroutines are
called correctly and if dgeco has set rcond .gt. 0.0
or dgefa has set info .eq. 0 .
to compute inverse(a) * c where c is a matrix
with p columns
dgeco(a,lda,n,ipvt,rcond,z)
if (!rcond is too small){
for (j=0,j<p,j++)
dgesl(a,lda,n,ipvt,c[j][0],0);
}
linpack. this version dated 08/14/78 .
cleve moler, university of new mexico, argonne national lab.
functions
blas daxpy,ddot
*/
{
/* internal variables */
REAL t;
int k,kb,l,nm1;
nm1 = n - 1;
if (job == 0) {
/* job = 0 , solve a * x = b
first solve l*y = b */
if (nm1 >= 1) {
for (k = 0; k < nm1; k++) {
l = ipvt[k];
t = b[l];
if (l != k){
b[l] = b[k];
b[k] = t;
}
daxpy(n-(k+1),t,&a[lda*k+k+1],1,&b[k+1],1 );
}
}
/* now solve u*x = y */
for (kb = 0; kb < n; kb++) {
k = n - (kb + 1);
b[k] = b[k]/a[lda*k+k];
t = -b[k];
daxpy(k,t,&a[lda*k+0],1,&b[0],1 );
}
}
else {
/* job = nonzero, solve trans(a) * x = b
first solve trans(u)*y = b */
for (k = 0; k < n; k++) {
t = ddot(k,&a[lda*k+0],1,&b[0],1);
b[k] = (b[k] - t)/a[lda*k+k];
}
/* now solve trans(l)*x = y */
if (nm1 >= 1) {
for (kb = 1; kb < nm1; kb++) {
k = n - (kb+1);
b[k] = b[k] + ddot(n-(k+1),&a[lda*k+k+1],1,&b[k+1],1);
l = ipvt[k];
if (l != k) {
t = b[l];
b[l] = b[k];
b[k] = t;
}
}
}
}
return;
}
/*----------------------*/
void daxpy(int n, REAL da, REAL dx[], int incx, REAL dy[], int incy)
/*
constant times a vector plus a vector.
jack dongarra, linpack, 3/11/78.
*/
{
int i,ix,iy,m,mp1;
mp1 = 0;
m = 0;
if(n <= 0) return;
if (da == ZERO) return;
if(incx != 1 || incy != 1) {
/* code for unequal increments or equal increments
not equal to 1 */
ix = 0;
iy = 0;
if(incx < 0) ix = (-n+1)*incx;
if(incy < 0)iy = (-n+1)*incy;
for (i = 0;i < n; i++) {
dy[iy] = dy[iy] + da*dx[ix];
ix = ix + incx;
iy = iy + incy;
}
return;
}
/* code for both increments equal to 1 */
#ifdef ROLL
for (i = 0;i < n; i++) {
dy[i] = dy[i] + da*dx[i];
}
#endif
#ifdef UNROLL
m = n % 4;
if ( m != 0) {
for (i = 0; i < m; i++)
dy[i] = dy[i] + da*dx[i];
if (n < 4) return;
}
for (i = m; i < n; i = i + 4) {
dy[i] = dy[i] + da*dx[i];
dy[i+1] = dy[i+1] + da*dx[i+1];
dy[i+2] = dy[i+2] + da*dx[i+2];
dy[i+3] = dy[i+3] + da*dx[i+3];
}
#endif
return;
}
/*----------------------*/
REAL ddot(int n, REAL dx[], int incx, REAL dy[], int incy)
/*
forms the dot product of two vectors.
jack dongarra, linpack, 3/11/78.
*/
{
REAL dtemp;
int i,ix,iy,m,mp1;
mp1 = 0;
m = 0;
dtemp = ZERO;
if(n <= 0) return(ZERO);
if(incx != 1 || incy != 1) {
/* code for unequal increments or equal increments
not equal to 1 */
ix = 0;
iy = 0;
if (incx < 0) ix = (-n+1)*incx;
if (incy < 0) iy = (-n+1)*incy;
for (i = 0;i < n; i++) {
dtemp = dtemp + dx[ix]*dy[iy];
ix = ix + incx;
iy = iy + incy;
}
return(dtemp);
}
/* code for both increments equal to 1 */
#ifdef ROLL
for (i=0;i < n; i++)
dtemp = dtemp + dx[i]*dy[i];
return(dtemp);
#endif
#ifdef UNROLL
m = n % 5;
if (m != 0) {
for (i = 0; i < m; i++)
dtemp = dtemp + dx[i]*dy[i];
if (n < 5) return(dtemp);
}
for (i = m; i < n; i = i + 5) {
dtemp = dtemp + dx[i]*dy[i] +
dx[i+1]*dy[i+1] + dx[i+2]*dy[i+2] +
dx[i+3]*dy[i+3] + dx[i+4]*dy[i+4];
}
return(dtemp);
#endif
}
/*----------------------*/
void dscal(int n, REAL da, REAL dx[], int incx)
/* scales a vector by a constant.
jack dongarra, linpack, 3/11/78.
*/
{
int i,m,mp1,nincx;
mp1 = 0;
m = 0;
if(n <= 0)return;
if(incx != 1) {
/* code for increment not equal to 1 */
nincx = n*incx;
for (i = 0; i < nincx; i = i + incx)
dx[i] = da*dx[i];
return;
}
/* code for increment equal to 1 */
#ifdef ROLL
for (i = 0; i < n; i++)
dx[i] = da*dx[i];
#endif
#ifdef UNROLL
m = n % 5;
if (m != 0) {
for (i = 0; i < m; i++)
dx[i] = da*dx[i];
if (n < 5) return;
}
for (i = m; i < n; i = i + 5){
dx[i] = da*dx[i];
dx[i+1] = da*dx[i+1];
dx[i+2] = da*dx[i+2];
dx[i+3] = da*dx[i+3];
dx[i+4] = da*dx[i+4];
}
#endif
}
/*----------------------*/
int idamax(int n, REAL dx[], int incx)
/*
finds the index of element having max. absolute value.
jack dongarra, linpack, 3/11/78.
*/
{
REAL dmax;
int i, ix, itemp;
if( n < 1 ) return(-1);
if(n ==1 ) return(0);
if(incx != 1) {
/* code for increment not equal to 1 */
ix = 1;
dmax = fabs((double)dx[0]);
ix = ix + incx;
for (i = 1; i < n; i++) {
if(fabs((double)dx[ix]) > dmax) {
itemp = i;
dmax = fabs((double)dx[ix]);
}
ix = ix + incx;
}
}
else {
/* code for increment equal to 1 */
itemp = 0;
dmax = fabs((double)dx[0]);
for (i = 1; i < n; i++) {
if(fabs((double)dx[i]) > dmax) {
itemp = i;
dmax = fabs((double)dx[i]);
}
}
}
return (itemp);
}
/*----------------------*/
REAL epslon (REAL x)
/*
estimate unit roundoff in quantities of size x.
*/
{
REAL a,b,c,eps;
/*
this program should function properly on all systems
satisfying the following two assumptions,
1. the base used in representing dfloating point
numbers is not a power of three.
2. the quantity a in statement 10 is represented to
the accuracy used in dfloating point variables
that are stored in memory.
the statement number 10 and the go to 10 are intended to
force optimizing compilers to generate code satisfying
assumption 2.
under these assumptions, it should be true that,
a is not exactly equal to four-thirds,
b has a zero for its last bit or digit,
c is not exactly equal to one,
eps measures the separation of 1.0 from
the next larger dfloating point number.
the developers of eispack would appreciate being informed
about any systems where these assumptions do not hold.
*****************************************************************
this routine is one of the auxiliary routines used by eispack iii
to avoid machine dependencies.
*****************************************************************
this version dated 4/6/83.
*/
a = 4.0e0/3.0e0;
eps = ZERO;
while (eps == ZERO) {
b = a - ONE;
c = b + b + b;
eps = fabs((double)(c-ONE));
}
return(eps*fabs((double)x));
}
/*----------------------*/
void dmxpy (int n1, REAL y[], int n2, int ldm, REAL x[], REAL m[])
/* We would like to declare m[][ldm], but c does not allow it. In this
function, references to m[i][j] are written m[ldm*i+j]. */
/*
purpose:
multiply matrix m times vector x and add the result to vector y.
parameters:
n1 integer, number of elements in vector y, and number of rows in
matrix m
y double [n1], vector of length n1 to which is added
the product m*x
n2 integer, number of elements in vector x, and number of columns
in matrix m
ldm integer, leading dimension of array m
x double [n2], vector of length n2
m double [ldm][n2], matrix of n1 rows and n2 columns
----------------------------------------------------------------------
*/
{
int j,i,jmin;
/* cleanup odd vector */
j = n2 % 2;
if (j >= 1) {
j = j - 1;
for (i = 0; i < n1; i++)
y[i] = (y[i]) + x[j]*m[ldm*j+i];
}
/* cleanup odd group of two vectors */
j = n2 % 4;
if (j >= 2) {
j = j - 1;
for (i = 0; i < n1; i++)
y[i] = ( (y[i])
+ x[j-1]*m[ldm*(j-1)+i]) + x[j]*m[ldm*j+i];
}
/* cleanup odd group of four vectors */
j = n2 % 8;
if (j >= 4) {
j = j - 1;
for (i = 0; i < n1; i++)
y[i] = ((( (y[i])
+ x[j-3]*m[ldm*(j-3)+i])
+ x[j-2]*m[ldm*(j-2)+i])
+ x[j-1]*m[ldm*(j-1)+i]) + x[j]*m[ldm*j+i];
}
/* cleanup odd group of eight vectors */
j = n2 % 16;
if (j >= 8) {
j = j - 1;
for (i = 0; i < n1; i++)
y[i] = ((((((( (y[i])
+ x[j-7]*m[ldm*(j-7)+i]) + x[j-6]*m[ldm*(j-6)+i])
+ x[j-5]*m[ldm*(j-5)+i]) + x[j-4]*m[ldm*(j-4)+i])
+ x[j-3]*m[ldm*(j-3)+i]) + x[j-2]*m[ldm*(j-2)+i])
+ x[j-1]*m[ldm*(j-1)+i]) + x[j] *m[ldm*j+i];
}
/* main loop - groups of sixteen vectors */
jmin = (n2%16)+16;
for (j = jmin-1; j < n2; j = j + 16) {
for (i = 0; i < n1; i++)
y[i] = ((((((((((((((( (y[i])
+ x[j-15]*m[ldm*(j-15)+i])
+ x[j-14]*m[ldm*(j-14)+i])
+ x[j-13]*m[ldm*(j-13)+i])
+ x[j-12]*m[ldm*(j-12)+i])
+ x[j-11]*m[ldm*(j-11)+i])
+ x[j-10]*m[ldm*(j-10)+i])
+ x[j- 9]*m[ldm*(j- 9)+i])
+ x[j- 8]*m[ldm*(j- 8)+i])
+ x[j- 7]*m[ldm*(j- 7)+i])
+ x[j- 6]*m[ldm*(j- 6)+i])
+ x[j- 5]*m[ldm*(j- 5)+i])
+ x[j- 4]*m[ldm*(j- 4)+i])
+ x[j- 3]*m[ldm*(j- 3)+i])
+ x[j- 2]*m[ldm*(j- 2)+i])
+ x[j- 1]*m[ldm*(j- 1)+i])
+ x[j] *m[ldm*j+i];
}
return;
}
机器的浮点性能基准测试程序
来源：CSDN
作者：杨幂的咪
链接：https://blog.csdn.net/Hsu_smile/article/details/53004584