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ChrisKFlag for United States of America

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VB5 best route from point A to point B

Not exactly sure how to describe this but I need a formula I believe on how to plot a location of an obect and get it from point A to point B.
For example, I've got a black dot at 0,0...and I want to move him to 46,36 taking the most direct route (a line).  If it were straight lines, it would be easy...but how do I do it with diagonals?  I know there is some sort of formula for this used in games...I need it :)
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deighton
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If you want a standard linear equation in the form

y = ax + b

then for a line passing thru (x1,y1) and (x2,y2) its

a = (y2 - y1)/(x2 - x1)

b = (y1 * (x2 - x1) - x1 * (y2 - y1))/(x2 - x1)
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ASKER

excellent...but what about x on your comment?
I'll try and explain my y = ax + b formula a bit more.

Given 'cartesian coordinates' that is with an x and y axis, you can express a formula for a straight line using the formula

y = ax + b

so for any x co-ordinate you can find the y co-ordinate which corresponds to the line.

In your case you have a line with b = 0 and a = 18/23

so at x = 46 you have y = 36

at x = 23 you have y = 18 so (23,18) is another point on your line.

I don't know if you'd want to use this in your program, one problem is that for a vertical line, a which is the gradient of your line becomes infinite.  I just put it in as an afterthought in case it was what you were trying to work out.
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Why wouldn't I want to use this?  To me it looks like I can draw an imaginary line from point A to point B and have a dot travel the correct pixel path to those points, which is exactly what I need.
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ASKER

If what you say only works for diagonals, that's fine too because then for straight lines I simply check the previous X or Y and make sure one of those matches the new X or Y signifying a straight line.  If not, then use the formula you posted.  That sounds right don't ya think?
Yes thats true, I was making things too complcated.  
The formula works fine for horizontals where a = 0, as you say when x1 = x2 you've got a vertical straight line.
the angle to the horizontal is atn((y2 - y1) / (x2 - x1))

By the way deighton, I can't get the right answer using this bit.
You might me getting an answer in radians, to get back to degrees multiply by 360/(2 * pi)
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ASKER

You guys must be mathimaticians :)  Could you help out a mathimatically challenged soul by supplying a little example source :)  nothing big, just something like here's point A, here's point B, and move a pixel along the path.
Private Sub Command1_Click()
    Me.DrawWidth = 4
    Call fLine(799, 0, 950, 5000, Me, vbBlack, vbWhite)
End Sub

Private Function fLine(x1 As Long, y1 As Long, x2 As Long, y2 As Long, vObject As Object, lLineColor As Long, lPixelColor As Long)

    Dim a As Double
    Dim b As Double
    Dim dTemp As Double
    Dim bPlot As Boolean
   
    Dim x As Double
   
    Dim dLen As Double
   
    Dim xO As Long, yO As Long

    'vObject.Line (x1, y1)-(x2, y2), lLineColor
   
    dLen = Sqr((x1 - x2) ^ 2 + (y1 - y2) ^ 2)
   
    a = (y2 - y1) / (x2 - x1)
   
    b = (y1 * (x2 - x1) - x1 * (y2 - y1)) / (x2 - x1)
   
    For x = x1 To x2 Step (x2 - x1) / dLen
   
        y = Int(a * x + b + 0.5)
        If Not bPlot Or x <> xO Or y <> yO Then
       
            vObject.PSet (x, y), lLineColor
            bPlot = True
            xO = x
            yO = y
           
        End If
       
    Next
   
    bPlot = False
   
   
    For x = x1 To x2 Step (x2 - x1) / dLen
   
        y = Int(a * x + b + 0.5)
       
        If x <> xO And y <> yO Or Not bPlot Then
       
            If bPlot Then vObject.PSet (xO, yO), lLineColor
           
            vObject.PSet (x, y), lPixelColor
           
           
               
            bPlot = True
               
            xO = x
            yO = y
               
        End If
       
    Next
           
                               
       
       
       
       
   
   
End Function


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ASKER

Thanks...now I kinda understand it.