Searching for the best prescaler (closest integer multiple) - math

Question

Let's say we have a timer with a counter (cnt) and a prescaler (psc), which triggers clk_freq / div times per second (Hz), where div = cnt * psc and clk_freq is a clock frequency, which is irrelevant. Value range for both counter and prescaler is [1;65536].
Which is the best (quickest) method of picking cnt and psc so that cnt * psc would be as close as possible to some desired_div value?
Example:
desired_div = 3849586
psc = 251, cnt = 15337, psc * cnt = 3849587, distance = 1
psc = 313, cnt = 12299, distance = 1
psc = 1757, cnt = 2191, distance = 1
psc = 2191, cnt = 1757, distance = 1
psc = 12299, cnt = 313, distance = 1
psc = 15337, cnt = 251, distance = 1
The code used to acquire these values (not effective enough):
int main(int argc, char *argv[])
{
uint32_t val, cnt, psc;
if (argc < 2) {
printf("Usage: %s <u32_value>\n", argv[0]);
return 1;
}
sscanf(argv[1], "%u", &val);
printf("Searching for %u\n", val);
uint32_t hlimit = min(65536, val);
uint32_t llimit = max(1, val / 65536);
uint32_t minpsc = 0, minredif = 0xFFFFFFFF;
for (psc = llimit; psc <= hlimit; ++psc) {
cnt = min(65536, val / psc);
uint32_t redif = abs(val - cnt * psc);
if (cnt < 65536) {
redif = min(redif, abs((cnt + 1) * psc - val));
}
if (redif < minredif) {
minpsc = psc;
minredif = redif;
}
printf("%u %u\n", psc, redif);
}
printf("Optimal psc: %u, difference: %u\n", minpsc, minredif);
return 0;
}

Answer 1

not sure how much faster or slower, not calling any libraries so that helps (well other than printf).
#include <stdio.h>
void fun ( unsigned int x )
{
unsigned int ra;
unsigned int rb;
unsigned int rc;
unsigned int rd;
unsigned int min;
min=0; min--;
for(ra=1;ra<32768;ra++)
{
rb=x/ra;
if(rb>65536) continue;
if(1)
{
rc=rb*ra;
if(rc>x) rd=rc-x;
else rd=x-rc;
if(rd<=min)
{
printf("%u %u %u\n",ra,rb,rd);
min=rd;
}
}
if(1)
{
rc=(rb+1)*ra;
if(rc>x) rd=rc-x;
else rd=x-rc;
if(rd<=min)
{
printf("%u %u %u\n",ra,rb+1,rd);
min=rd;
}
}
}
}
for your value the +1 gave the closest results.
59 65247 13
61 63108 2
122 31554 2
183 21036 2
244 15777 2
251 15337 1
313 12299 1
1757 2191 1
2191 1757 1
12299 313 1
15337 251 1
but dont know off hand if that will always be the case (expect not).
You shouldnt need to go past the halfway point as you are just testing the same pairs twice. ab = ba