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Combinations calculations

I have, for my own amusement, written the following code. However
I am in grave doubts as to the accuracy of the code in the function
"variations", having spent 50 odd years forgetting everything about
these calculations. Criticism and suggestions welcomed.

Note that the solutions presented to variations are all ordered,
but may have trailing zeroes.

/* find integral solutions to x*x + y*y + z*z ... = r*r */
/* Public domain, by C.B. Falconer, 2005-03-25 */

#include <stdio.h>
#include <limits.h>
#include <stdlib.h>

/* 181 allows operation with 16 bit ints */
#define MAXDIMS 10

#endif

struct soln {
int dims;
int id;
int sv[MAXDIMS+1]; /* solution vector */
};

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

static void help(void)
{
" with radius <= %d, dimensions in 1..%d\n",
);
exit(0);
} /* help */

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

/* Problem - how to compute the independant solutions
generated by negating points. 0 points can't be
negated, and two identical points are trivially
interchangeable, all of which reduce the soln count.
The negations generate reflections about some axis.
I don't remember any combinatorial maths.
*/
/* NULL input returns (and resets) accumulated sum */
static long int variations(struct soln *s)
{
long int vars;
static long int vartotal;
int d, f;

vars = vartotal;
if (!s) vartotal = 0;
else {
/* How many variations of s are available by reflections */
/* useful test group for this are (rad, dims) =
9, 3; 25, 3; 45, 3; 10, 4; 3, 9; 3, 10 */
vars = 1;
s->sv = 0;
for (d = 1, f = s->dims; d <= s->dims; d++, f--) {
/* errors here on duplicated entries ?? */
if (s->sv[d]) {
/* Two possible values here, negatable */
if (s->sv[d] != s->sv[d-1]) vars *= 2;
vars *= f;
}
}
vartotal += vars;
}
return vars;
} /* variations */

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

static void dumpsolution(struct soln *s)
{
int d;

if (s) { /* else just accumulated total vars */
s->id++;
printf("%5d:", s->id);
for (d = 1; d <= s->dims; d++) {
printf(" %3d", s->sv[d]);
}
}
printf(" [%ld]\n", variations(s));
} /* dumpsolution */

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

/* brute force exhaustive search */
/* I have rediscovered the backtracking algorithm! :-) */
static void solve(int r, int dims)
{
int d; /* dimension index */
int needed[MAXDIMS+1]; /* solution criterion */
struct soln s; /* a solution develops here */

s.dims = dims; s.radius = r; s.id = 0;
for (d = 1; d <= dims; d++) s.sv[d] = 0;
needed = squares[r];
s.sv = 1; /* just to start the sequence */

for (d = 1; ; d++) {
for (s.sv[d] = s.sv[d-1]; s.sv[d] <= r; s.sv[d]++) {
needed[d] = needed[d-1] - squares[s.sv[d]];
if (needed[d] <= 0) { /* backtrack */
if (0 == needed[d]) dumpsolution(&s);
s.sv[d] = 0;
d--; /* use prev. dim */
if (d) continue;
return; /* all done */
}
else if (d < dims) break; /* check next dimension */
}
}
} /* solve */

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

int main(int argc, char **argv)
{
size_t temp;
char *errp;
unsigned int r, dims;

r = 0; dims = 2;
if ((argc < 2) || (argc > 3)) help();
if (argc > 1) {
temp = strtoul(argv, &errp, 10);
if ((temp > MAXRADIUS) || (errp == argv)
|| (temp < 1)) help();
else r = temp;
}
if (argc > 2) {
temp = strtoul(argv, &errp, 10);
if ((temp > MAXDIMS) || (errp == argv)
|| (temp < 1)) help();
else dims = temp;
}
for (temp = 0; temp <= MAXRADIUS; temp++) {
squares[temp] = temp * temp;
}
printf("solve %d %d\n", r, dims);
solve(r, dims);
dumpsolution(NULL); /* the accumulated total */
return 0;
} /* diophan main */

--
"If you want to post a followup via groups.google.com, don't use
the broken "Reply" link at the bottom of the article. Click on
"show options" at the top of the article, then click on the

Nov 14 '05 #1
1 1316 [A warning about follow-ups would have been nice. ;)
Sent to clc as was original, F'up to comp.programming.]

CBFalconer wrote:
I have, for my own amusement, written the following code.
However I am in grave doubts as to the accuracy of the code
in the function "variations", having spent 50 odd years
forgetting everything about these calculations. Criticism
and suggestions welcomed.

Note that the solutions presented to variations are all ordered,
but may have trailing zeroes.

/* find integral solutions to x*x + y*y + z*z ... = r*r */
/* Public domain, by C.B. Falconer, 2005-03-25 */
...
It's arguable whether 0 ordinates should count as a solution,

I don't like the way you've used hidden static variables, but
again, I'll put that aside.

You _are_ over-reporting the number of variations.
...
/* Problem - how to compute the independant solutions
generated by negating points. 0 points can't be
negated, and two identical points are trivially
interchangeable, all of which reduce the soln count.
The negations generate reflections about some axis.
I don't remember any combinatorial maths.
*/
/* NULL input returns (and resets) accumulated sum */
static long int variations(struct soln *s)
{
long int vars;
static long int vartotal;
int d, f;

vars = vartotal;
if (!s) vartotal = 0;
else {
/* How many variations of s are available by reflections */
/* useful test group for this are (rad, dims) =
9, 3; 25, 3; 45, 3; 10, 4; 3, 9; 3, 10 */
vars = 1;
s->sv = 0;
for (d = 1, f = s->dims; d <= s->dims; d++, f--) {
/* errors here on duplicated entries ?? */
if (s->sv[d]) {
/* Two possible values here, negatable */
if (s->sv[d] != s->sv[d-1]) vars *= 2;
vars *= f;
}
}
vartotal += vars;
}
return vars;
} /* variations */

% diophan 3 10
solve 3 10
1: 1 1 1 1 1 1 1 1 1 0 
2: 1 1 1 1 1 2 0 0 0 0 
3: 1 2 2 0 0 0 0 0 0 0 
4: 3 0 0 0 0 0 0 0 0 0 


The variations need to be a factored by 2 for each non-zero
dimension (+ or -). To calculate the combinations, you need
to group and count the number of similar ordinates.

For example, the variations for the solutions above should be...

1: 9x1 1x0 [10!/9!1! * 2^9]
2: 5x1 1x2 4x0 [10!/5!1!4! * 2^6]
3: 1x1 2x2 7x0 [10!/1!2!7! * 2^3]
4: 1x3 9x0 [10!/1!9! * 2^1]

Try...

static long int variations(struct soln *s)
{
long int vars;
static long int vartotal;
int d, f;

vars = vartotal;
if (!s) vartotal = 0;
else {
for (vars = 1, f = 0, d = s->dims; d; d--) {
if (s->sv[d]) vars *= 2;
if (d != s->dims && s->sv[d] != s->sv[d+1]) f = 0;
vars = vars * d / ++f;
}
vartotal += vars;
}
return vars;
}

The calculation of combinations (for 2: above) is done as...

2: 1 1 1 1 1 2 0 0 0 0

10! - 1 2 3 4 5 6 7 8 9 10
5!1!4! - 5 4 3 2 1 1 4 3 2 1

For no particular reason, I start from the right and work left.
I used f to count the number of ordinate repeats, reseting it
when a different ordinate is encountered.

With the change, I get...

% diophan 3 10
solve 3 10
1: 1 1 1 1 1 1 1 1 1 0 
2: 1 1 1 1 1 2 0 0 0 0 
3: 1 2 2 0 0 0 0 0 0 0 
4: 3 0 0 0 0 0 0 0 0 0 


--
Peter

Nov 14 '05 #2

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