[Haskell-cafe] Line drawing algorithm

2009-07-17 Thread CK Kashyap 
Hi All,

I am working on a diagraming utility in Haskell. I started with line drawing.
I am doing the basic stuff using the y = mx + c formula to draw a line between 
(x1,y1) and (x2,y2)

Here's what I need to do - 
if dx  dy where dx = (x2 - x1) and dy = (y2 - y1) then I need to vary x 
between x1 and x2 and find the various y's
however if dy  dx then I need to vary y beteen y1 and y2 and get various x's 

In the code below, I've only taken care of the situation where dx  dy - I was 
thinking if there was a better way to
do it that takes care of the other condition as well without repeating the code.


type Point = (Integer,Integer)

line :: Point - Point - [Point] -- get all the points in the line
line p1@(x1,y1) p2@(x2,y2) = line' start end start slope
  where
(start,end) = reorderPoints p1 p2
slope = ((fromIntegral (y2-y1)) / (fromIntegral (x2-x1)))
reorderPoints (px1,py1) (px2,py2)
| px1  px2 = (p1,p2)
| otherwise = (p2,p1)

line' :: Point - Point - Point - Double - [Point]
line' start@(x1,y1) end@(x2,y2) point@(x3,y3) slope
  | x3 == x2 = [end]
  | otherwise = [point] ++ line' start end (newX,newY) slope
  where
newX = x3 + 1
newY = y1 + round (slope * (fromIntegral (newX - x1)))


hello = line (1,1) (10,10)


Regards,
Kashyap



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Re: [Haskell-cafe] Line drawing algorithm

2009-07-17 Thread Neil Brown 

CK Kashyap wrote:

Hi All,

I am working on a diagraming utility in Haskell. I started with line 
drawing.
I am doing the basic stuff using the y = mx + c formula to draw a line 
between (x1,y1) and (x2,y2)

Hi,

Are you doing this to learn Haskell, learn about drawing lines, or to 
just get it implemented?  If either of the latter two, when drawing a 
straight line you shouldn't need to do floating point operations such as 
this:



newY = y1 + round (slope * (fromIntegral (newX - x1)))


Bresenham's algorithm (or similar variants) allows you to draw a line 
without needing floating point.  A Haskell implementation is here:


http://rosettacode.org/wiki/Bresenham%27s_line_algorithm#Haskell

Although it may not be too understandable!  Wikipedia has an explanation 
of the general algorithm:


http://en.wikipedia.org/wiki/Bresenham%27s_line_algorithm

As to how to cope with the dy  dx case in your code given the dx  dy 
case, you could just swap the x and y coords at the start, then swap 
back the x and y coords of all the output points afterwards.  Odd, but 
effective :-)


Thanks,

Neil.
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Re: [Haskell-cafe] Line drawing algorithm

2009-07-17 Thread CK Kashyap 
Thanks Neil ... 


 Are you doing this to learn Haskell, learn about drawing lines, or to just 
 get it implemented?  If either of the latter two, when drawing a straight 
 line you shouldn't need to do floating point operations such as this:

Actually, my reasons are first and third.

 newY = y1 + round (slope * (fromIntegral (newX - x1)))

 http://rosettacode.org/wiki/Bresenham%27s_line_algorithm#Haskell

Thanks for the link.

 As to how to cope with the dy  dx case in your code given the dx  dy case, 
 you could just swap the x and y coords at the start, then swap back the x and 
 y coords of all the output points afterwards.  Odd, but effective :-)
Slope would differ right for both case right?

Regards,
Kashyap



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