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Draw a curve that touches 3 points (1 starting point, 2 medium, 3 final point)

Do not use functions of a library, implement the curve() function yourself
coordinates:(x,y) starting point (10,10) medium point (100,200) final point (200,10)

Go

There are, of course, an infinity of curves which can be fitted to 3 points. The most obvious solution is to fit a quadratic curve (using Lagrange interpolation) and so that's what we do here.

As we're not allowed to use library functions to draw the curve, we instead divide the x-axis of the curve between successive points into equal segments and then join the resulting points with straight lines.

The resulting 'curve' is then saved to a .png file where it can be viewed with a utility such as EOG. <lang go>package main

import "github.com/fogleman/gg"

var p = [3]gg.Point{{10, 10}, {100, 200}, {200, 10}}

func lagrange(x float64) float64 {

   return (x-p[1].X)*(x-p[2].X)/(p[0].X-p[1].X)/(p[0].X-p[2].X)*p[0].Y +
       (x-p[0].X)*(x-p[2].X)/(p[1].X-p[0].X)/(p[1].X-p[2].X)*p[1].Y +
       (x-p[0].X)*(x-p[1].X)/(p[2].X-p[0].X)/(p[2].X-p[1].X)*p[2].Y

}

func getPoints(n int) []gg.Point {

   pts := make([]gg.Point, 2*n+1)
   dx := (p[1].X - p[0].X) / float64(n)
   for i := 0; i < n; i++ {
       x := p[0].X + dx*float64(i)
       pts[i] = gg.Point{x, lagrange(x)}
   }
   dx = (p[2].X - p[1].X) / float64(n)
   for i := n; i < 2*n+1; i++ {
       x := p[1].X + dx*float64(i-n)
       pts[i] = gg.Point{x, lagrange(x)}
   }
   return pts

}

func main() {

   const n = 50 // more than enough for this
   dc := gg.NewContext(210, 210)
   dc.SetRGB(1, 1, 1) // White background
   dc.Clear()
   for _, pt := range getPoints(n) {
       dc.LineTo(pt.X, pt.Y)
   }
   dc.SetRGB(0, 0, 0) // Black curve
   dc.SetLineWidth(1)
   dc.Stroke()
   dc.SavePNG("quadratic_curve.png")

}</lang>

J

   NB. coordinates:(x,y) starting point (10,10) medium point (100,200) final point (200,10)

   X=: 10 100 200
   Y=: 10 200 10

   NB. matrix division computes polynomial coefficients
   NB. %. implements singular value decomposition
   NB. in other words, we can also get best fit polynomials of lower order.

   polynomial=: (Y %. (^/ ([: i. #)) X)&p.


   assert 10 200 10 -: polynomial X  NB. test



   Filter=: (#~`)(`:6)

   Round=: adverb def '<.@:(1r2&+)&.:(%&m)'
   assert 100 120 -: 100 8 Round 123  NB. test, round 123 to nearest multiple of 100 and of 8



   NB. libraries not permitted, character cell graphics are used.


   GRAPH=: 50 50 $ ' '  NB. is an array of spaces

   NB. place the axes
   GRAPH=: '-' [`(([:<0; i.@:#)@:])`]} GRAPH
   GRAPH=: '|' [`(([:<0;~i.@:#)@:])`]} GRAPH
   GRAPH=: '+' [`((<0;0)"_)`]} GRAPH           NB. origin


   NB. clip the domain.
   EXES=: ((<:&(>./X) *. (<./X)&<:))Filter 5 * i. 200
   WHYS=: polynomial EXES


   NB. draw the curve
   1j1 #"1 |. 'X' [`((<"1 WHYS ;&>&:([: 1 Round %&5) EXES)"_)`]} GRAPH


   NB. were we to use a library:
   load'plot'
   'title 3 point fit' plot (j. polynomial) i.201

Julia

To make things more specific, find the circle determined by the points. The curve is then the arc between the 3 points. <lang julia>using Makie

struct Point; x::Float64; y::Float64; end

Find a circle passing through the 3 points

const p1 = Point(10, 10) const p2 = Point(100, 200) const p3 = Point(200, 10) const allp = [p1, p2, p3]

set up problem matrix and solve.
if (x - a)^2 + (y - b)^2 = r^2 then for some D, E, F, x^2 + y^2 + Dx + Ey + F = 0
therefore Dx + Ey + F = -x^2 - y^2

v = zeros(Int, 3) m = zeros(Int, 3, 3) for row in 1:3

   m[row, 1:3] .= [allp[row].x, allp[row].y, 1]
   v[row] = -(allp[row].x)^2 - (allp[row].y)^2

end q = (m \ v) # [-210.0, -162.632, 3526.32] a, b, r = -q[1] / 2, -q[2] / 2, sqrt((q[1]^2/4) + q[2]^2/4 - q[3])

println("The circle with center at x = $a, y = $b and radius $r.")

x = a-r:0.25:a+r y0 = sqrt.(r^2 .- (x .- a).^2) scene = lines(x, y0 .+ b, color = :red) lines!(scene, x, b .- y0, color = :red) scatter!(scene, [p.x for p in allp], [p.y for p in allp], markersize = r / 10)

</lang>

Output:

The circle with center at x = 105.0, y = 81.31578947368422 and radius 118.78948534384199.

Perl

Hilbert curve task code repeated here, with the addition that the 3 task-required points are marked. Mostly satisfies the letter-of-the-law of task specification while (all in good fun) subverting the spirit of the thing. <lang perl>use SVG; use List::Util qw(max min);

use constant pi => 2 * atan2(1, 0);

Compute the curve with a Lindemayer-system

%rules = (

   A => '-BF+AFA+FB-',
   B => '+AF-BFB-FA+'

); $hilbert = 'A'; $hilbert =~ s/([AB])/$rules{$1}/eg for 1..6;

Draw the curve in SVG

($x, $y) = (0, 0); $theta = pi/2; $r = 5;

for (split //, $hilbert) {

   if (/F/) {
       push @X, sprintf "%.0f", $x;
       push @Y, sprintf "%.0f", $y;
       $x += $r * cos($theta);
       $y += $r * sin($theta);
   }
   elsif (/\+/) { $theta += pi/2; }
   elsif (/\-/) { $theta -= pi/2; }

}

$max = max(@X,@Y); $xt = -min(@X)+10; $yt = -min(@Y)+10; $svg = SVG->new(width=>$max+20, height=>$max+20); $points = $svg->get_path(x=>\@X, y=>\@Y, -type=>'polyline'); $svg->rect(width=>"100%", height=>"100%", style=>{'fill'=>'black'}); $svg->polyline(%$points, style=>{'stroke'=>'orange', 'stroke-width'=>1}, transform=>"translate($xt,$yt)"); my $task = $svg->group( id => 'task-points', style => { stroke => 'red', fill => 'red' },); $task->circle( cx => 10, cy => 10, r => 1, id => 'point1' ); $task->circle( cx => 100, cy => 200, r => 1, id => 'point2' ); $task->circle( cx => 200, cy => 10, r => 1, id => 'point3' );

open $fh, '>', 'curve-3-points.svg'; print $fh $svg->xmlify(-namespace=>'svg'); close $fh;</lang> Hilbert curve passing through 3 defined points (offsite image)

Phix

Translation of: zkl

<lang Phix>include pGUI.e

Ihandle dlg, canvas cdCanvas cddbuffer, cdcanvas

enum X, Y constant p = {{10,10},{100,200},{200,10}}

function lagrange(atom x)

  return (x - p[2][X])*(x - p[3][X])/(p[1][X] - p[2][X])/(p[1][X] - p[3][X])*p[1][Y] +
         (x - p[1][X])*(x - p[3][X])/(p[2][X] - p[1][X])/(p[2][X] - p[3][X])*p[2][Y] +
         (x - p[1][X])*(x - p[2][X])/(p[3][X] - p[1][X])/(p[3][X] - p[2][X])*p[3][Y]

end function

function getPoints(integer n)

   sequence pts = {}
   atom {dx,pt,cnt} := {(p[2][X] - p[1][X])/n, p[1][X], n}
   for j=1 to 2 do
       for i=0 to cnt do
           atom x := pt + dx*i;
           pts = append(pts,{x,lagrange(x)});
       end for
       {dx,pt,cnt} = {(p[3][X] - p[2][X])/n, p[2][X], n+1};
   end for
   return pts

end function

procedure draw_cross(sequence xy)

   integer {x,y} = xy
   cdCanvasLine(cddbuffer, x-3, y, x+3, y) 
   cdCanvasLine(cddbuffer, x, y-3, x, y+3)

end procedure

function redraw_cb(Ihandle /*ih*/, integer /*posx*/, integer /*posy*/)

   cdCanvasActivate(cddbuffer)
   cdCanvasSetForeground(cddbuffer, CD_BLUE)
   cdCanvasBegin(cddbuffer,CD_OPEN_LINES)
   atom {x,y} = {p[1][X], p[1][Y]}; -- curve starting point
   cdCanvasVertex(cddbuffer, x, y)
   sequence pts = getPoints(50)
   for i=1 to length(pts) do
       {x,y} = pts[i]
       cdCanvasVertex(cddbuffer, x, y)
   end for
   cdCanvasEnd(cddbuffer)
   cdCanvasSetForeground(cddbuffer, CD_RED)
   for i=1 to length(p) do draw_cross(p[i]) end for
   cdCanvasFlush(cddbuffer)
   return IUP_DEFAULT

end function

function map_cb(Ihandle ih)

   cdcanvas = cdCreateCanvas(CD_IUP, ih)
   cddbuffer = cdCreateCanvas(CD_DBUFFER, cdcanvas)
   cdCanvasSetBackground(cddbuffer, CD_WHITE)
   return IUP_DEFAULT

end function

procedure main()

   IupOpen()
   
   canvas = IupCanvas(NULL)
   IupSetAttribute(canvas, "RASTERSIZE", "220x220")
   IupSetCallback(canvas, "MAP_CB", Icallback("map_cb"))

   dlg = IupDialog(canvas,"DIALOGFRAME=YES")
   IupSetAttribute(dlg, "TITLE", "Quadratic curve")
   IupSetCallback(canvas, "ACTION", Icallback("redraw_cb"))
   IupCloseOnEscape(dlg)

   IupMap(dlg)
   IupSetAttribute(canvas, "RASTERSIZE", NULL)
   IupShowXY(dlg,IUP_CENTER,IUP_CENTER)
   IupMainLoop()
   IupClose()

end procedure main()</lang>

Raku

(formerly Perl 6)

Works with: Rakudo version 2018.10

Kind of bogus. There are an infinite number of curves that pass through those three points. I'll assume a quadratic curve. Lots of bits and pieces borrowed from other tasks to avoid relying on library functions.

Saved as a png for wide viewing support. Note that png coordinate systems have 0,0 in the upper left corner.

<lang perl6>use Image::PNG::Portable;

Solve for a quadratic line that passes through those points

my (\a, \b, \c) =

 rref([[10², 10, 1, 10],[100², 100, 1, 200],[200², 200, 1, 10]])[*;*-1];

General case quadratic line equation

sub f (\x) { a*x² + b*x + c }

Scale it up a bit for display

my $scale = 2;

my ($w, $h) = (500, 500); my $png = Image::PNG::Portable.new: :width($w), :height($h);

my ($lastx, $lasty) = 8, f(8).round; (9 .. 202).map: -> $x {

   my $f = f($x).round;
   line($lastx, $lasty, $x, $f, $png, [0,255,127]);
   ($lastx, $lasty) = $x, $f;

}

Highlight the 3 defining points

dot(|$_, $png, 2) for (10,10,[255,0,0]), (100,200,[255,0,0]), (200,10,[255,0,0]);

$png.write: 'Curve-3-points-perl6.png';

Assorted helper routines

sub rref (@m) {

   return unless @m;
   my ($lead, $rows, $cols) = 0, +@m, +@m[0];
   for ^$rows -> $r {
       $lead < $cols or return @m;
       my $i = $r;
       until @m[$i;$lead] {
           ++$i == $rows or next;
           $i = $r;
           ++$lead == $cols and return @m;
       }
       @m[$i, $r] = @m[$r, $i] if $r != $i;
       my $lv = @m[$r;$lead];
       @m[$r] »/=» $lv;
       for ^$rows -> $n {
           next if $n == $r;
           @m[$n] »-=» @m[$r] »*» (@m[$n;$lead] // 0);
       }
       ++$lead;
   }
   @m

}

sub line($x0 is copy, $y0 is copy, $x1 is copy, $y1 is copy, $png, @rgb) {

   my $steep = abs($y1 - $y0) > abs($x1 - $x0);
   ($x0,$y0,$x1,$y1) »*=» $scale;
   if $steep {
       ($x0, $y0) = ($y0, $x0);
       ($x1, $y1) = ($y1, $x1);
   }
   if $x0 > $x1 {
       ($x0, $x1) = ($x1, $x0);
       ($y0, $y1) = ($y1, $y0);
   }
   my $Δx = $x1 - $x0;
   my $Δy = abs($y1 - $y0);
   my $error = 0;
   my $Δerror = $Δy / $Δx;
   my $y-step = $y0 < $y1 ?? 1 !! -1;
   my $y = $y0;
   next if $y < 0;
   for $x0 .. $x1 -> $x {
       next if $x < 0;
       if $steep {
           $png.set($y, $x, |@rgb);
       } else {
           $png.set($x, $y, |@rgb);
       }
       $error += $Δerror;
       if $error >= 0.5 {
           $y += $y-step;
           $error -= 1.0;
       }
   }

}

sub dot ($X is copy, $Y is copy, @rgb, $png, $radius = 3) {

   ($X, $Y) »*=» $scale;
   for ($X X+ -$radius .. $radius) X ($Y X+ -$radius .. $radius) -> ($x, $y) {
       $png.set($x, $y, |@rgb) if ( $X - $x + ($Y - $y) * i ).abs <= $radius;
   }

}</lang> See Curve-3-points-perl6.png (offsite .png image)

zkl

Translation of: Go

Uses Image Magick and the PPM class from http://rosettacode.org/wiki/Bitmap/Bresenham%27s_line_algorithm#zkl <lang zkl>const X=0, Y=1; // p.X == p[X] var p=L(L(10.0, 10.0), L(100.0, 200.0), L(200.0, 10.0)); // (x,y)

fcn lagrange(x){ // float-->float

  (x - p[1][X])*(x - p[2][X])/(p[0][X] - p[1][X])/(p[0][X] - p[2][X])*p[0][Y] +
  (x - p[0][X])*(x - p[2][X])/(p[1][X] - p[0][X])/(p[1][X] - p[2][X])*p[1][Y] +
  (x - p[0][X])*(x - p[1][X])/(p[2][X] - p[0][X])/(p[2][X] - p[1][X])*p[2][Y]

}

fcn getPoints(n){ // int-->( (x,y) ..)

 pts:=List.createLong(2*n+1);
 dx,pt,cnt := (p[1][X] - p[0][X])/n, p[0][X], n;
 do(2){
    foreach i in (cnt){

x:=pt + dx*i; pts.append(L(x,lagrange(x)));

    }
    dx,pt,cnt = (p[2][X] - p[1][X])/n, p[1][X], n+1;
 }
 pts

}

fcn main{

  var [const] n=50; // more than enough for this
  img,color := PPM(210,210,0xffffff), 0;     // white background, black curve
  foreach x,y in (p){ img.cross(x.toInt(),y.toInt(), 0xff0000) } // mark 3 pts

  a,b := p[0][X].toInt(), p[0][Y].toInt(); // curve starting point
  foreach x,y in (getPoints(n)){
     x,y = x.toInt(),y.toInt();
     img.line(a,b, x,y, color);	 // can only deal with ints
     a,b = x,y;
  }
  img.writeJPGFile("quadraticCurve.zkl.jpg");

}();</lang>

Output:

Image at quadratic curve