Distance Formula Implecations: Urgent!

KneeLess · 03-18-2004, 07:34 PM

Hello. As we all know, the distance formula is sqrt(sq(x2-x1) + sq(y2 - y1)). sq = square. Now, for some reason, this isn't at all working. Here is my code:

Code:

float distance(float x2, float y2)
{
  float distance = 0;
  static float x1 = 7, y1 = 2;
  distance = sq(x2 - x1) + sq(y2 - y1);
  distance = 4 * sq(distance) + 0.5;
  x1 = x2;
  y1 = y2;
  return distance;
}

Now I don't have the math.h libraries available (working with a special embedded system, using a special compiler) therefore I have no sqrt() availible, so i just squared it a second time. Assume all numbers are correct, because they are. But when I do something like: sqrt(sq(6-7) + sq(1-2)) which in my calculator comes out to 1.41421... but in my program comes out to -46.500000. Also I'm using printf("%f\n", distance) to print. I just don't understance what's wrong, any help would be much appreciated.

XSquared · 03-18-2004, 07:40 PM

...and what's your sq( ) function?

KneeLess · 03-18-2004, 07:42 PM

Nevermind, fixed. Don't you hate that? Worked on it for 6 hours only to fix it in three minutes.

major_blagger · 03-18-2004, 07:49 PM

How did you fix it?

**Bubba** · 03-19-2004, 03:25 AM

You can use a power series to estimate distance or if you know the angle you are looking you can find the distance by

Code:

if (distx>disty)
{
   totaldist=distx/cos(angle);
} else totaldist=disty/sin(angle);

joed · 03-20-2004, 08:00 PM

A little off the subject, but here's a neat way to approximate integer distance, if you don't care about precision. Saw this somewhere and simplified the shifting for my needs:

Code:

int fastdist(int x1, int y1, int x2, int y2)
{
  int dx = abs(x2 - x1);
  int dy = abs(y2 - y1);

  if(dy > dx) {        // XOR swap function
    dx ^= dy;
    dy ^= dx;
    dx ^= dy;
  } 

  return ((dx << 4) + (dx << 2) +
          (dy << 4) - (dy << 3)) >> 4;
}

Making the shifting more complex adds precision, without costing too much, but this was precise enough for the short pixel distances I was measuring. Haven't tried converting to fixed-point.

-Joe

**Bubba** · 03-20-2004, 10:52 PM

Yes you can approximate distance via a power series, but it has an error of like 3 or 4 percent.

03-18-2004, 07:34 PM	#1
KneeLess Registered User Join Date: Aug 2003 Posts: 42	Distance Formula Implecations: Urgent! Hello. As we all know, the distance formula is sqrt(sq(x2-x1) + sq(y2 - y1)). sq = square. Now, for some reason, this isn't at all working. Here is my code: Code: float distance(float x2, float y2) { float distance = 0; static float x1 = 7, y1 = 2; distance = sq(x2 - x1) + sq(y2 - y1); distance = 4 * sq(distance) + 0.5; x1 = x2; y1 = y2; return distance; } Now I don't have the math.h libraries available (working with a special embedded system, using a special compiler) therefore I have no sqrt() availible, so i just squared it a second time. Assume all numbers are correct, because they are. But when I do something like: sqrt(sq(6-7) + sq(1-2)) which in my calculator comes out to 1.41421... but in my program comes out to -46.500000. Also I'm using printf("%f\n", distance) to print. I just don't understance what's wrong, any help would be much appreciated. __________________ Sigh, nothing ever works the first try. Register Linux User #314127
KneeLess is offline

03-18-2004, 07:40 PM	#2
XSquared C++ Developer Join Date: Jun 2002 Location: UWaterloo Posts: 2,718	...and what's your sq( ) function? __________________ Naturally I didn't feel inspired enough to read all the links for you, since I already slaved away for long hours under a blistering sun pressing the search button after typing four whole words! - Quzah You. Fetch me my copy of the Wall Street Journal. You two, fight to the death - Stewie
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03-18-2004, 07:42 PM	#3
KneeLess Registered User Join Date: Aug 2003 Posts: 42	Nevermind, fixed. Don't you hate that? Worked on it for 6 hours only to fix it in three minutes. __________________ Sigh, nothing ever works the first try. Register Linux User #314127
KneeLess is offline

03-18-2004, 07:49 PM	#4
major_blagger Registered User Join Date: Feb 2004 Posts: 72	How did you fix it?
major_blagger is offline

03-19-2004, 03:25 AM	#5
Bubba Super Moderator Join Date: Aug 2001 Posts: 7,817	You can use a power series to estimate distance or if you know the angle you are looking you can find the distance by Code: if (distx>disty) { totaldist=distx/cos(angle); } else totaldist=disty/sin(angle); __________________ If you aim at everything you will hit something but you won't know what it is.
Bubba is offline

03-20-2004, 08:00 PM	#6
joed Registered User Join Date: Mar 2004 Posts: 59	distance A little off the subject, but here's a neat way to approximate integer distance, if you don't care about precision. Saw this somewhere and simplified the shifting for my needs: Code: int fastdist(int x1, int y1, int x2, int y2) { int dx = abs(x2 - x1); int dy = abs(y2 - y1); if(dy > dx) { // XOR swap function dx ^= dy; dy ^= dx; dx ^= dy; } return ((dx << 4) + (dx << 2) + (dy << 4) - (dy << 3)) >> 4; } Making the shifting more complex adds precision, without costing too much, but this was precise enough for the short pixel distances I was measuring. Haven't tried converting to fixed-point. -Joe
joed is offline

03-20-2004, 10:52 PM	#7
Bubba Super Moderator Join Date: Aug 2001 Posts: 7,817	Yes you can approximate distance via a power series, but it has an error of like 3 or 4 percent. __________________ If you aim at everything you will hit something but you won't know what it is.
Bubba is offline

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