Talk:Constrained random points on a circle

Not 100 points

There are only 89 points in the circle shown in the verilog example output. This is no surprise, because AFAICS the algorithm doesn't make sure that the same point isn't chosen twice. Now given that it's the first example, I guess it's what was meant by the task description, but then the task description probably should be changed to reflect the fact that less points are OK. --Ce 10:55, 3 September 2010 (UTC)

How to check the code

If you increase the number of points produced to 10k, you should get output rather like this (generated with Tcl version; your version may differ). This lets you check that the spread of points produces the expected annulus. –Donal Fellows 11:00, 3 September 2010 (UTC)

               X               
          XXXXXXXXXXX          
        XXXXXXXXXXXXXXX        
      XXXXXXXXXXXXXXXXXXX      
     XXXXXXXXXXXXXXXXXXXXX     
    XXXXXXXXXXXXXXXXXXXXXXX    
   XXXXXXXX         XXXXXXXX   
   XXXXXXX           XXXXXXX   
  XXXXXX               XXXXXX  
  XXXXXX               XXXXXX  
 XXXXXX                 XXXXXX 
 XXXXX                   XXXXX 
 XXXXX                   XXXXX 
 XXXXX                   XXXXX 
 XXXXX                   XXXXX 
XXXXXX                   XXXXXX
 XXXXX                   XXXXX 
 XXXXX                   XXXXX 
 XXXXX                   XXXXX 
 XXXXX                   XXXXX 
 XXXXXX                 XXXXXX 
  XXXXXX               XXXXXX  
  XXXXXX               XXXXXX  
   XXXXXXX           XXXXXXX   
   XXXXXXXX         XXXXXXXX   
    XXXXXXXXXXXXXXXXXXXXXXX    
     XXXXXXXXXXXXXXXXXXXXX     
      XXXXXXXXXXXXXXXXXXX      
        XXXXXXXXXXXXXXX        
          XXXXXXXXXXX          
               X               

Uniform distribution

I'm not very good with stats, but I've seen discussions pop up before about different distributions of points. Is uniform vs normal vs (something?) a significant component of the task? How may it be verified with at most 100 points? --Michael Mol 12:27, 3 September 2010 (UTC)

I guess that if you count the number of points that lie on a line that passes through the center, then the count should not depend on the angle of that line. So it would be wrong to simply pick a uniform-random value of x and then a uniform-random of y at that location, because that would tend to lead to a higher density of points on the left and right sides (look at the maximally dense version above: there are 12 possible values of y at x==0; but only one at x == +/- 15) -- and thus a greater number of points away from the x-axis. A common mistake when generating a set of linked random variables is that the distribution depends on the order in which they are generated.

So, take this Perl code:

my @bitmap = map { " " x 32 } 0 .. 31;
for (1 .. 100) {
    my $x = int rand(31) - 15;


    my $max = sqrt( 225-($x*$x) );
    my $min = 100-($x*$x);
    $min = $min > 0 ? sqrt $min : 0;

    my $y = int rand( $max-$min ) + $min;
    $y = -$y if rand() < .5;


    $x += 15;
    $y += 15;
    substr( $bitmap[$y], $x, 1, "#" );
}

print "$_\n" for @bitmap;


               # #
          #    ##    #
       #   ## ##  ##   #
      #        ##  #   #
         # # # #   #  # #
      #           # #
        ##          #   #
        #
  #                      #
     #                       #
 ## #                    #
  #
 #
                           #
 ## #                         #
  #                        #
    ##                      #

                             #
 #                      #

       #                #  #
   #     #
        ## #      # #
        #     #   #
     #  # #       ##  #
        # #  #      ##
            #       #
             # #

Here you can see that the number of points that lie on the X axis (y==0) is much smaller than the number that lie on the Y axis (x==0).