-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- The three rules of Ruby Quiz: 1. Please do not post any solutions or spoiler discussion for this quiz until 48 hours have elapsed from the time this message was sent. 2. Support Ruby Quiz by submitting ideas and responses as often as you can. 3. Enjoy! Suggestion: A [QUIZ] in the subject of emails about the problem helps everyone on Ruby Talk follow the discussion. Please reply to the original quiz message, if you can. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - RSS Feed: http://rubyquiz.strd6.com/quizzes.rss Suggestions?: http://rubyquiz.strd6.com/suggestions -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- ## Random Points within a Circle (#234) Greetings Rubyists, Generating random numbers is a useful feature and Ruby provides us with `Kernel.rand` to generate numbers uniformly distributed between zero and one. If we want to generate random numbers with other distributions, then we'll need to do a little work. The quiz this week is to generate random points uniformly distributed within a circle of a given radius and position. This quiz is relatively simple, great if you are new to Ruby or strapped for time. Remember, it is contributions from people like *you* that make Ruby Quiz succeed! Have fun!

on 2010-06-19 20:06

on 2010-06-21 13:13

Hi ! On 19 June 2010 20:05, Daniel Moore <yahivin@gmail.com> wrote: > > This quiz is relatively simple, great if you are new to Ruby or > strapped for time. Remember, it is contributions from people like > *you* that make Ruby Quiz succeed! > > Have fun! > > -Daniel So here is my solution in a hundred lines, with an ImageMagick visualisation: http://gist.github.com/446707 I had an intuition doing some Math.sqrt about the distance, and it revealed to be exact :) Enjoy, Benoit Daloze

on 2010-06-22 00:32

It's not as advanced as Daloze's solution, and it definitely can (should?) be enhanced, but my solution is available at http://gist.github.com/447554. Any feedback is welcomed. In a nutshell, Point.random implements a random algorithm that basically repositions the head of an initial null vector v until it falls within a circle of the given radius that is centered around the origin. It then translates a copy of that head to the correct position with respect to the given center. BTW, this is the first time I participate in the Ruby Quiz. I'm looking forward to future quizzes :) Regards, Yaser Sulaiman

on 2010-06-22 00:46

On 6/19/10, Daniel Moore <yahivin@gmail.com> wrote: > The quiz this week > is to generate random points uniformly distributed within a circle of > a given radius and position. def random_point_in_a_circle(x,y,r) angle=rand*2*Math::PI r*=rand x+=r*Math.sin(angle) y+=r*Math.cos(angle) return x,y end 7 lines, 5 minutes, 0 tests or visualizations. Super-easy.

on 2010-06-22 01:13

On Jun 21, 2010, at 15:45 , Caleb Clausen wrote: > > return x,y > end > > 7 lines, 5 minutes, 0 tests or visualizations. Super-easy. And wrong, unfortunately. You're selecting a random angle, and then a random distance from the center. This will result in way too many points at the center of the circle.

on 2010-06-22 03:19

On Mon, Jun 21, 2010 at 7:12 PM, Dave Howell <groups.2009a@grandfenwick.net> wrote: >> y+=r*Math.cos(angle) >> >> return x,y >> end >> >> 7 lines, 5 minutes, 0 tests or visualizations. Super-easy. > > And wrong, unfortunately. You're selecting a random angle, and then a random distance from the center. This will result in way too many points at the center of the circle. I whipped up a quick little visualization: def random_point_in_a_circle(x,y,r) angle=rand*2*Math::PI r*=rand x+=r*Math.sin(angle) y+=r*Math.cos(angle) return x,y end require 'java' JFrame = javax.swing.JFrame JPanel = javax.swing.JPanel frame = JFrame.new("Random Points within a Circle") frame.default_close_operation = JFrame::EXIT_ON_CLOSE frame.set_size(400, 400) frame.show class RPwiaC < JPanel def paintComponent(graphics) super(graphics) 10000.times do x,y = random_point_in_a_circle(200,200,150) graphics.draw_line(x-1,y-1,x+1,y+1) graphics.draw_line(x-1,y+1,x+1,y-1) end end end panel = RPwiaC.new frame.add(panel) panel.repaint panel.revalidate

on 2010-06-22 04:20

On Mon, Jun 21, 2010 at 9:12 PM, <brabuhr@gmail.com> wrote: >>> r*=rand > I whipped up a quick little visualization: Quick hack to animate the above method side by side with an alternate method: require 'java' JFrame = javax.swing.JFrame JPanel = javax.swing.JPanel frame = JFrame.new("Random Points within a Circle") frame.default_close_operation = JFrame::EXIT_ON_CLOSE frame.set_size(700, 400) frame.show class RPwiaC < JPanel def initialize times, *procs super() @times = times @procs = procs end def paintComponent(graphics) super(graphics) @times.times do @procs.each_with_index do |proc, i| x,y = proc.call(200 + (300 * i),200,150) graphics.draw_line(x-1,y-1,x+1,y+1) graphics.draw_line(x-1,y+1,x+1,y-1) end end end end panel = RPwiaC.new( 5000, lambda{|x,y,r| angle=rand*2*Math::PI r*=rand x+=r*Math.sin(angle) y+=r*Math.cos(angle) return x,y }, lambda{|x,y,r| loop { a = x + rand(2*r) - r b = y + rand(2*r) - r d = Math.sqrt((x - a) ** 2 + (y - b) ** 2) return a,b if d < r } } ) frame.add(panel) panel.revalidate loop { panel.repaint }

on 2010-06-22 04:43

On 6/21/10, Dave Howell <groups.2009a@grandfenwick.net> wrote: > > On Jun 21, 2010, at 15:45 , Caleb Clausen wrote: >> 7 lines, 5 minutes, 0 tests or visualizations. Super-easy. > > And wrong, unfortunately. You're selecting a random angle, and then a random > distance from the center. This will result in way too many points at the > center of the circle. I guess this is why Benoit had that sqrt in there. I don't quite get why it's necessary. I kinda like Yaser's solution.

on 2010-06-22 18:16

On 22 June 2010 00:30, Yaser Sulaiman <yaserbuntu@gmail.com> wrote: > It's not as advanced as Daloze's solution, and it definitely can (should?) > be enhanced, but my solution is available at http://gist.github.com/447554. > Any feedback is welcomed. I would say it is a good try, and the distinction Vector/Point is interesting, while being a problem with DRY (you repeat the calculation of the distance). A quick tip a friend showed me: Math.sqrt(a**2 + b**2) => Math.hypot(a, b) In the end, you manually add the offset to x and y. As you have Vector/Point, the Point+Vector should be defined and then the 4 last lines would be: "center + v" On 22 June 2010 01:12, Dave Howell <groups.2009a@grandfenwick.net> wrote: >> r*=rand >> x+=r*Math.sin(angle) >> y+=r*Math.cos(angle) >> >> return x,y >> end >> >> 7 lines, 5 minutes, 0 tests or visualizations. Super-easy. > > And wrong, unfortunately. You're selecting a random angle, and then a random distance from the center. This will result in way too many points at the center of the circle. > That was also my initial thought, and I think for most of us. Caleb: > I guess this is why Benoit had that sqrt in there. I don't quite get > why it's necessary. Indeed, please look at my solution and remove the "sqrt", and you will see most of the points will be in the center . The output will look like: min: 1, moy, 1.13, max: 122 # This is way too much max Point: (-2.0,4.0) # which is the center The image would then look like a point in the center. I 'felt' it would result in that because the points near the center will be much closer to each other, as it will be as much points at any distance. Brabuhr's visualization is even *way* better to see it (especially the animated one) ! And finally, why the sqrt? I just felt like I should have as many points in each area, and the area of a circle is ðr^2, so let's reduce this r^2 and becomes r. (Also I wanted the random distance to be likely further from the center, and as rand returns between 0 and 1, we have to do ^(1/2) and not ^2. - Benoit

on 2010-06-22 20:10

2010/6/22 Benoit Daloze <eregontp@gmail.com>: > > The image would then look like a point in the center. > I 'felt' it would result in that because the points near the center > will be much closer to each other, as it will be as much points at any > distance. > > And finally, why the sqrt? I just felt like I should have as many > points in each area, and the area of a circle is ðr^2, so let's reduce > this r^2 and becomes r. (Also I wanted the random distance to be > likely further from the center, and as rand returns between 0 and 1, > we have to do ^(1/2) and not ^2. Found an explanation: http://www.anderswallin.net/2009/05/uniform-random...

on 2010-06-22 21:52

On 6/22/10, brabuhr@gmail.com <brabuhr@gmail.com> wrote: > Found an explanation: > http://www.anderswallin.net/2009/05/uniform-random... Thanks. It took me a while, but it's making sense now.

on 2010-06-22 23:25

This is actually a pretty interesting puzzle. Correct solutions should have points that are distributed as a hump Gaussian-like curve when histogram plotted over the y rand and x range. My first solution had a humped-histogram over the X ranges but was evenly distributed over the Y ranges. Back to the drawing board. The algorithm I ended up using makes points distributed over the Y and X range, but only keeps them if they are within the circle. Used R to plot. Code below: class Circle attr_reader :r,:x,:y def initialize(r,x,y) @r = r @y = y @x = x end #get random x and y from within circle space def internal_rand #to randomly select rise run values rise = rand(0)*@r*rand_sign run = rand(0)*@r*rand_sign #validate that point lies within circle, about 20% fail if Math.sqrt(run**2+rise**2)<@r [@x+run,@y+rise] else internal_rand end end private def rand_sign rand(2) == 0? -1 : 1 end end #test c1 = Circle.new(4,10,10) xs = []; ys = [] 1000.times do point = c1.internal_rand xs << point.first ys << point.last end plotted in R using: x = c(pasted in xs from ruby output) y = c(pasted in ys from ruby output) plot(x,y) library(plotrix) draw.circle(10,10,4) #evenly distribute points within circle hist(x, breaks=20) #Gaussian histogram bar distribution hist(y, breaks=20) #Gaussian histogram bar distribution

on 2010-06-23 01:26

(Apologies for this oldish "re-post": it seems that changing from googlemail to gmail makes emails not recognised by ruby-talk.) On Tue, Jun 22, 2010 at 3:41 AM, Caleb Clausen <vikkous@gmail.com> wrote: > I guess this is why Benoit had that sqrt in there. I don't quite get > why it's necessary. > > I kinda like Yaser's solution. > Yaser's solution is a "Monte Carlo" simulation? My initial reaction was that Yaser's method might not end up with the correct distribution, but on further thought it (I think): 1. Generates points uniformly in a square of side r. 2. Redistributes the points uniformly in a square of side 2*r. (I wondered what the "Flip coin to reflect" bits were doing until I realised that.) 3. Ignores any points outside the circle. Which should leave points uniformly distributed (by area) inside the circle. So Yaser's solution does have the correct distribution of points. And shows that sometimes a way to generate a correct distribution is to use an incorrect distribution and then ignore some values generated from the incorrect distribution. (Strictly speaking, is the test that "v.length > 0" necessary, and should the other test be "v.length <= radius"? It won't make much difference though.) ** The rest of this has now been covered by other posts, but the links might be interesting. The random angle bit in your method is OK. For the square root: the area of an annulus radius r to r+e is pi * ( (r+e)**2 - r**2 ) = pi * 2*r*e + pi * e**2 So, if I treat e as an infinitesimal and neglect the e**2 (and hope that no real mathematicians are looking, or claim I'm using Abraham Robinson's http://en.wikipedia.org/wiki/Abraham_robinson Non-Standard Analysis http://en.wikipedia.org/wiki/Non-standard_analysis and hope that nobody asks me any awkward questions about how that really works) the area of the annulus is (approximately) pi * 2*r*e. (Think of it as a (very) "bent" rectangle of length pi * 2*r with a "width" of e.) If you use a uniform distribution for the radius to the random point then on average you'll be putting the same number of points in an annulus (a0) of width e very close to the centre of the circle as in an annulus (a1) of width e very close to the circumference of the circle. But the annulus (a1) has a much larger area than the annulus (a0), so the density of points will be greater in annulus (a0) than annulus (a1). Which is why Dave Howell made his comment. So if we want to use a method which selects a random angle and a random distance from the centre of the circle, for the random distance from the centre of the circle we need a 0 to 1 distribution which isn't uniform but which has more weight for higher than lower values in 0 to 1. When I saw this quiz I thought of using the random angle and distance method, and after a bit of work guessed that the correct distribution for the distance was probably to take the square root of random numbers generated from a uniform 0 to 1 distribution. BUT: I couldn't immediately see a way to prove that! So I didn't post anything! I suspect Benoit can *prove* that taking the square root gives the correct distribution for the radius! def uniform_random_point_in_ circle( r, x, y ) # circle radius r, centre at x, y r_to_point = Math.sqrt( Kernel.rand ) * r radians_to_point = 2 * Math::PI * Kernel.rand return x + r_to_point * Math.cos( radians_to_point ), y + r_to_point * Math.sin( radians_to_point ) end

on 2010-06-23 03:56

On 6/22/10, Colin Bartlett <colinb2r@googlemail.com> wrote: > (Think of it as a (very) "bent" rectangle of length pi * 2*r with a "width" > of e.) > > If you use a uniform distribution for the radius to the random point then on > average you'll be putting the same number of points in an annulus (a0) of > width e very close to the centre of the circle as in an annulus (a1) of > width e very close to the circumference of the circle. But the annulus (a1) > has a much larger area than the annulus (a0), so the density of points will > be greater in annulus (a0) than annulus (a1). Which is why Dave Howell made > his comment. That was beautiful! The best explanation yet. Thank you. I keep wondering the use of sqrt will cause weird float rounding errors; that is will there be valid [float, float] pairs which are in the circle but can never be be found by an algorithm that includes Math.sqrt(rand)? Maybe someone with a better grip on the math has an answer to this.

on 2010-06-23 07:48

On Sun, 20 Jun 2010 03:05:58 +0900 Daniel Moore <yahivin@gmail.com> wrote: > is to generate random points uniformly distributed within a circle of > a given radius and position. > > This quiz is relatively simple, great if you are new to Ruby or > strapped for time. Remember, it is contributions from people like > *you* that make Ruby Quiz succeed! > > Have fun! Inspired by the other solutions here are two possible algorithms with a little benchmark and Tk visualization (call like ruby circlepoints.rb [n_points=5000] [a || b]). Martin -------------------- # Solution A: Randomly select a point in a square and retry if out of # the circle def mk_points_a(radius, n_points) n_points.times do begin x, y = [-1+2*rand, -1+2*rand] end while Math.sqrt(x**2+y**2).abs > 1 yield x*radius, y*radius end end # Solution B: Random angle and distance def mk_points_b(radius, n_points) n_points.times do angle = rand*2*Math::PI r = Math.sqrt(rand) yield r*Math.sin(angle) * radius , r*Math.cos(angle) * radius end end require 'tk' require 'benchmark' canvas = TkCanvas.new radius = 100 n_points = (ARGV[0] || 5000).to_i # select an algorithm method = ARGV.include?('b') ? :mk_points_b : :mk_points_a bench = Benchmark.measure do send(method, radius, n_points) do |x,y| TkcLine.new(canvas, radius + x, radius + y, radius + x + 1, radius + y + 1, 'fill' => 'black') end end puts "Using: #{method}" puts bench.to_s canvas.pack Tk.mainloop

on 2010-06-24 20:25

2010/6/22 Benoit Daloze <eregontp@gmail.com> > I would say it is a good try, and the distinction Vector/Point is > interesting, while being a problem with DRY (you repeat the > calculation of the distance). > Thanks. I don't know how did I miss it, but it kinda suddenly hit me: I can use Point.distance_to instead of Vector.length. After this realization, I decided to drop the Vector class altogether - although I have to admit that talking about generating random vectors sounds "cooler" :P A quick tip a friend showed me: Math.sqrt(a**2 + b**2) => Math.hypot(a, b) > In the end, you manually add the offset to x and y. As you have > Vector/Point, the Point+Vector should be defined and then the 4 last > lines would be: "center + v" > Because I'm not using Vectors anymore, Point + Point now performs the equivalent 2D translation. Also, Point.distance_to now uses Math.hypot. On Wed, Jun 23, 2010 at 2:25 AM, Colin Bartlett <colinb2r@googlemail.com>wrote: > that sometimes a way to generate a correct distribution is to use an > incorrect distribution and then ignore some values generated from the > incorrect distribution. > I have modified the comments in the code to better reflect the used approach, which you have elaborate on here quite nicely. (Strictly speaking, is the test that "v.length > 0" necessary, and should > the other test be "v.length <= radius"? It won't make much difference > though.) > v.length > 0 was necessary because of the way I was using the until loop (after initializing v to a null vector, v.length is 0 and the second test, v.length < radius, evaluates to true). Anyways, I'm now using the begin-end-until loop with a single test. Regarding using < vs. <=, I think it depends on whether points on the circumference (perimeter) of the circle are considered to be "within" the circle or not. Check the updated code @ http://gist.github.com/447554. Thank you all for your valuable feedback! Regards, Yaser

on 2010-06-25 02:00

Here's a quick solution visualized using RubygameÂ¹: http://www.pastie.org/1018079 I'm sorry about the lack of comments (the code should be mostly self-explanatory, if not just ask.) -- Lars Haugseth [1] http://rubygame.org/

on 2010-06-25 02:05

* Benoit Daloze <eregontp@gmail.com> wrote: > > I had an intuition doing some Math.sqrt about the distance, and it > revealed to be exact :) I bet you didn't do that right from the start, before seeing the first results? At least I didn't. :-)

on 2010-06-25 13:00

On 24 June 2010 21:24, Yaser Sulaiman <yaserbuntu@gmail.com> wrote: > [...] although I have to admit that > talking about generating random vectors sounds "cooler" :P Yes, and a Point + Point is not as meaningful > Because I'm not using Vectors anymore, Point + Point now performs > the equivalent 2D translation. Also, Point.distance_to now uses Math.hypot. Cool, but I believe your implementation of Point#+ is somehow bad, because it returns @y. So you probably want to return self if you accept the Point objects to be mutable, or create a new Point, which is a bit safer, but creates a new object (in both cases you could definitely get rid of this awful "return p" :) ) Also, @@origin should not be modified, and then should be a constant, and be #freeze if you want Point to be mutable. On 25 June 2010 03:05, Lars Haugseth <njus@larshaugseth.com> wrote: > * Benoit Daloze <eregontp@gmail.com> wrote: >> >> I had an intuition doing some Math.sqrt about the distance, and it >> revealed to be exact :) > > I bet you didn't do that right from the start, before seeing the > first results? At least I didn't. :-) > -- > Lars Haugseth I thought while writing it (without sqrt) that it was going wrong, and the next day I thought to sqrt. The night is a good adviser :-) B.D.

on 2010-06-25 15:30

On Fri, Jun 25, 2010 at 1:57 PM, Benoit Daloze <eregontp@gmail.com> wrote: > Cool, but I believe your implementation of Point#+ is somehow bad, > because it returns @y. > So you probably want to return self if you accept the Point objects to > be mutable, > or create a new Point, which is a bit safer, but creates a new object > (in both cases you could definitely get rid of this awful "return p" :) ) > > The _awful_ "return p" is now gone ;) I want Point objects to be mutable, so Point#+ returns self after modifying the x and y coordinates. > Also, @@origin should not be modified, and then should be a constant, > and be #freeze if you want Point to be mutable. Good point. The origin is now a "frozen" class constant.

on 2010-06-27 07:25

A thorough derivation of the distribution frequency transformation for the case of a circle. Behold the power of math! http://excitemike.com/Random_Numbers_and_Probabili...

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