Algorithms

25-Jul-07 9:15

Hiya guys.. Using C++ does anyone know how i can tell if a line is linear or not? I want the user to enter 5 readings which i can plot on a chart. However i would like a function to find out if the plotted line is linear.

Any help would be greatly appreciated!

Dan Neely25-Jul-07 9:35

Dan Neely

25-Jul-07 9:35

EDIT: Not sure if this is needed or not: if the slope between x_1,y_1 and x_5,y_5 is greater than 1, swap the x and y values for each coordinate pair. This swap is needed to keep the delta value in the next step smaller than 1.

Create a function for the line y = f(x) using points 1 and 5. Then of each of the 3 points in the middle calculate f(x_n), if ABS(f(x_n) - y_n) < 1, then x_n,y_n is on on the line (give or take the rounding error from forcing integral values).

--
You have to explain to them [VB coders] what you mean by "typed". their first response is likely to be something like, "Of course my code is typed. Do you think i magically project it onto the screen with the power of my mind?" --- John Simmons / outlaw programmer

Luc Pattyn25-Jul-07 11:14

25-Jul-07 11:14

Hi Dan,

Same idea, but without any math showing, the lazy approach:

- Create a GraphicsPath
- add the line connecting the end points
- widen with a pen (I guess width=3, more gives more tolerance)
- check all points fall within (that's GraphicsPath.IsVisible)

Smile | :)

Luc Pattyn

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flippydeflippydebop25-Jul-07 20:30

25-Jul-07 20:30

Hiya Luc. Thanks also for your reply. This is an interesting approach here, however GraphicsPath is a .net class while i am using native C++/MFC.

flippydeflippydebop25-Jul-07 20:26

25-Jul-07 20:26

Cheers Dan. This is great..

Im just a little confused about the functions f(x) and f(x_n). What is the body of these functions?

Dan Neely26-Jul-07 2:19

Dan Neely

26-Jul-07 2:19

The same function, f(x) is the generic function, x_n is x_1, x_2,... (the x points of your test coordinates). Probably best done using the slope-intercept form. Google if you need help getting the values from two points.

flippydeflippydebop26-Jul-07 2:25

26-Jul-07 2:25

ahh-ahhh..... the penny has dropped. Thanks again.

CPallini25-Jul-07 11:12

CPallini

25-Jul-07 11:12

Try to apply Linear Regression to your points.
Smile | :)

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.

Luc Pattyn25-Jul-07 11:16

25-Jul-07 11:16

Hi CPallini,

it depends on what the points represent:
- for ideal (x,y) points, yes
- for graphic approximation (such as Bresenham), not so good

Greetings

Luc Pattyn

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CPallini25-Jul-07 11:24

CPallini

25-Jul-07 11:24

It appears to me like a mathematical problem rather than a (digital) graphics one. Anyway, only the OP can clarify the point (I hope!)
Smile | :)

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.

Habib Ahmed Bhutto30-Jul-07 13:24

Habib Ahmed Bhutto

30-Jul-07 13:24

hi flippydeflippydebop
D'Oh! | :doh:

"entering 5 readings" u mean input 5 values then in each reading u will get a point's value mean(x,y)
it means
int x[5], y[5]
for( i=0; i<5; i++ )
read x[i], y[i]

resultX = x[0] - x[1] - x[2] - x[3] - x[4]
resultY = y[0] - y[1] - y[2] - y[3] - y[4]

if( ( resultX == 0 ) and ( resultY == 0 ) )
print "line is linear"
else
print "line is not linear"

i think this is the solution if it is true plz inform me if it is wrong then also inform me plz..

in case it is true u can further enhance and can be maked easy and powerful then this.

plz inform me about its correctness

habib bhutto

Habib Ahmed Bhutto30-Jul-07 13:32

Habib Ahmed Bhutto

30-Jul-07 13:32

hi flippydeflippydebop
D'Oh! | :doh:

habib bhutto

Dan Neely31-Jul-07 2:09

Dan Neely

31-Jul-07 2:09

won't work.

points 10,10 11,11 12,12 13,13 14,14. Obviously linear, but resultx and resulty are both -50.

Russell^'3-Aug-07 4:19

Russell^'

3-Aug-07 4:19

To understand if a line is 'approzimately' a line you can use the Hough transform.
Hope helps

Russell

Russell^'3-Aug-07 4:20

Russell^'

3-Aug-07 4:20

here is a link to the algorithm.

http://en.wikipedia.org/wiki/Hough_transform[^]

Big Grin | :-D

Russell

Finding a point on a circle [modified]

stevepqr7-Aug-07 6:02

stevepqr

7-Aug-07 6:02

You could use the equation of a line y=mx+c, we don't know the value of c (and we can't calculate it because the line might not be straight) but we can force it to be zero by shifting all the x,y coordinates by the same amount such that the first point is 0,0 now c=0 and we can calculate m for each point, m=y/x (except for the first point since this will result in divide by zero!)

If the line is straight then the value for m will be the same for each point. Should be relatively easy and quick to code... Smile | :)

Apathy Rules - I suppose...

images

zainab ibrahim24-Jul-07 22:47

zainab ibrahim

24-Jul-07 22:47

hi..I'd like to ask if someone here know how can I convert 2dim.image to 3dim. image i.e.2d information to 3d information & how can I get patches of information for extraction images.In my project I use matlab.Thank you.

zz

Re: images

shiraztk26-Jul-07 9:24

shiraztk

26-Jul-07 9:24

Hi
Is it possible at all to convert 2d image data to 3d image data? In 2d image dat a you wont have the depth information, but for creating 3d image data you need depth information.

Regards

The Best Religion is Science.
Once you understand it, you will know God.

max2929720-Jul-07 12:36

max29297

20-Jul-07 12:36

This one has me frustrated... What I need is a function that will give me the x and y coordinates of a point on a circle (defined by it's x and y coordinates and its width and height) when given a number in degrees. For example...

public Point GetPointOnCircle(int circleX, int circleY, int circleWidth, int circleHeight, int degrees) {
	return new Point(x, y);
}

I've tried to figure it out on my own but it's proven to be a bit more complicated than I thought it would be. Could anyone help me out?

P.S. I'm using C#

-----
*** Never give me an answer having anything to do with Visual Studio. I don't have this because I have two computers, one being my dad's mac, which is connected to the internet, the other being my pc, which is, sadly, not connected to the internet. The setup for the Visual C# program I think is called a "bootstrap" program, and it needs to connect to the internet to install the program. THEREFORE I cannot install this program onto my pc.***

Re: Finding a point on a circle [modified]

Luc Pattyn20-Jul-07 16:23

Re: Finding a point on a circle

20-Jul-07 16:23

Hi Max,

for a real circle with origin (0,0) and radius 1 the coordinates of the point at angle alpha
are given by (cos(alpha), sin(alpha))

Sanity checks: alpha=0 ==> (1,0) alpha=90 ==> (0,1) alpha=180 ==> (-1,0)

multiply by r to enlarge the circle, and add (cx,cy) to translate it.

BTW I expect Math.Sin/Cos to accept radians, not degrees; so you need to convert that
probably based on Math.Pi

for an ellipse it is similar but slightly more complex, ~~dont know by heart~~.

it is (cx+rx*cos(alpha), cy+ry*sin(alpha))
where rx and ry are the horizontal and vertical radii (that's for an orthogonal ellipse,
one that has its major axes parallel with the coordinate system's axes)

Sanity check: x must and will be in range (cx-rx,cx+rx) and y in range (cy-ry,cy+ry)

Smile | :)

-- modified at 0:30 Saturday 21st July, 2007

Luc Pattyn

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max2929720-Jul-07 20:09

max29297

20-Jul-07 20:09

Thanks a lot! That's pretty much exactly what I needed.

Re: Finding a point on a circle

Luc Pattyn21-Jul-07 0:49