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Active Forums => Programs => Topic started by: FellippeHeitor on August 01, 2020, 09:11:11 pm

Title: Estimating Pi
Post by: FellippeHeitor on August 01, 2020, 09:11:11 pm
Of course we already have _PI, this is just a cool way to visualize it.

Code: QB64: [Select]
  1. SCREEN _NEWIMAGE(600, 600, 32)
  2.  
  3. DIM inCircle AS _UNSIGNED LONG, total AS _UNSIGNED LONG
  5.  
  7.     px = RND * _WIDTH
  8.     py = RND * _HEIGHT
  9.  
  10.     total = total + 1
  11.     IF dist(_WIDTH / 2, _HEIGHT / 2, px, py) <= 300 THEN
  12.         inCircle = inCircle + 1
  13.         PSET (px, py), _RGB32(100)
  14.     ELSE
  15.         PSET (px, py), _RGB32(70)
  16.     END IF
  17.     pi = (inCircle * 4) / total
  18.  
  19.     LOCATE 1, 1
  20.     PRINT pi
  21.  
  22.     CIRCLE (_WIDTH / 2, _HEIGHT / 2), 299, _RGB32(216, 144, 22)
  23.     _DISPLAY
  25.  
  26. FUNCTION dist! (x1!, y1!, x2!, y2!)
  27.     dist! = _HYPOT((x2! - x1!), (y2! - y1!))

As seen on:
Title: Re: Estimating Pi
Post by: _vince on August 02, 2020, 12:45:23 am
It is a great simulation! Another one is called "buffon's needles", I wanted to write a program that demonstrates the experiment graphically and then numerically computes the results to iteratively approximated pi as more needles are dropped, open challenge I guess

ie
(https://en.wikipedia.org/wiki/Buffon%27s_needle_problem#/media/File:Buffon_needle_experiment_compressed.gif)
Title: Re: Estimating Pi
Post by: SierraKen on August 02, 2020, 12:55:12 am
Really awesome Felippe! I have no idea how you and others find Pi, was just ages ago since school. But I remembered a year ago (which turned out to be exactly a year ago), I found code to output as many digits as you want. I posted it in another thread if anyone is interested. Just funny that it's been one year, around the Sun, using Pi LOL. Oh, I also updated it tonight to make it show how many digits it has left to calculate before it shows a Notepad of it.
Title: Re: Estimating Pi
Post by: FellippeHeitor on August 02, 2020, 01:04:52 am
Well, I guess that only proves that life is a circle, @SierraKen 😂

@_vince looking forward to your project!
Title: Re: Estimating Pi
Post by: SierraKen on August 02, 2020, 01:05:25 am
LOL true.
Title: Re: Estimating Pi
Post by: bplus on August 02, 2020, 12:49:32 pm
Quote from: _vince on August 02, 2020, 12:45:23 am
It is a great simulation! Another one is called "buffon's needles", I wanted to write a program that demonstrates the experiment graphically and then numerically computes the results to iteratively approximated pi as more needles are dropped, open challenge I guess

ie
(https://en.wikipedia.org/wiki/Buffon%27s_needle_problem#/media/File:Buffon_needle_experiment_compressed.gif)

A very slow road to Pi:
Code: QB64: [Select]
  1. _TITLE "Buffon Pi Calc " 'b+ 2020-08-2
  2. '2020-08-02 try this with Buffon Pi Calc  pi ~ 2 * nTrials / nCross  from
  3. ' pi ~ 2*L*N/(T*H) L=length of needle, T=spacing between lines N = Number of trials (or pins) H = number that cross line
  6.  
  7. CONST xmax = 800, ymax = 600
  8. SCREEN _NEWIMAGE(xmax, ymax, 32)
  9. _DELAY .25
  12. L = 100 ' both pin length and line spacing
  13. N = 0 '   number of trials
  14. C = 0 '   number of pins that cross the line
  15. SS = 0 '  pin head and point lay in Same Space
  16.  
  18.     CLS
  19.     'redraw lines
  20.     FOR y = 0 TO _HEIGHT - 1 STEP L
  21.         LINE (0, y)-(_WIDTH - 1, y), &HFFFFFFFF
  22.     NEXT
  23.  
  24.     'drop a random pin
  25.     px1 = RND * (_WIDTH - 200) + 100
  26.     py1 = RND * (_HEIGHT - 200) + 100
  27.     ra = RND * _PI(2)
  28.     px2 = px1 + L * COS(ra)
  29.     py2 = py1 + L * SIN(ra)
  30.     LINE (px1, py1)-(px2, py2), _RGB32(50 + RND * 150)
  31.     N = N + 1 '                             another trial pin
  32.  
  33.     'did our pin cross a line
  34.     FOR y = 0 TO _HEIGHT - 1 STEP L
  35.         IF py1 >= y AND py1 < y + L THEN ' one side of pin is between y and y + L
  36.             IF py2 >= y AND py2 < y + L THEN ' the pin did NOT cross a line both the head and point lay in same space
  37.                 SS = SS + 1
  38.             END IF
  39.         END IF
  40.     NEXT
  41.     C = N - SS
  42.     'Print results of trials
  43.     LOCATE 1, 1: PRINT "Trials:"; N
  44.     LOCATE 2, 1: PRINT "Crossings:"; C
  45.     LOCATE 3, 1: PRINT "Pi est:"; 2 * N / C
  46.     _DISPLAY
  47.     IF N = 1 THEN
  48.         _DELAY 2
  49.     ELSEIF N < 5 THEN
  50.         _DELAY 1
  51.     ELSEIF N < 10 THEN
  52.         _DELAY .5
  53.     ELSEIF N < 50 THEN
  54.         _DELAY .25
  55.     ELSEIF N < 1000 THEN
  56.         _DELAY .05
  57.     ELSEIF N <= 10000 THEN
  58.         IF N MOD 1000 = 0 THEN _DELAY .25
  59.     ELSE
  60.         IF N MOD 10000 = 0 THEN _DELAY .05
  61.     END IF
  63.  
  64.  
  65.  
Title: Re: Estimating Pi
Post by: _vince on August 02, 2020, 01:10:45 pm
Yes, that is it. I've uncommented CLS to watch the needles pile up. Nice work, bplus
Title: Re: Estimating Pi
Post by: bplus on August 02, 2020, 01:25:36 pm
Thanks @_vince  this might be slightly more efficient for check needle crossing the line, skipping a FOR loop and maybe an extra IF THEN check:
Code: QB64: [Select]
  1. _TITLE "Buffon Pi Calc " 'b+ 2020-08-2
  2. '2020-08-02 try this with Buffon Pi Calc  pi ~ 2 * nTrials / nCross  from
  3. ' pi ~ 2*L*N/(T*H) L=length of needle, T=spacing between lines N = Number of trials (or pins) H = number that cross line
  6.  
  7. CONST xmax = 800, ymax = 600
  8. SCREEN _NEWIMAGE(xmax, ymax, 32)
  9. _DELAY .25
  12. L = 100 ' both pin length and line spacing
  13. N = 0 '   number of trials
  14. C = 0 '   number of pins that cross the line
  15. 'SS = 0 '  pin head and point lay in Same Space
  16.  
  18.     CLS
  19.     'redraw lines
  20.     FOR y = 0 TO _HEIGHT - 1 STEP L
  21.         LINE (0, y)-(_WIDTH - 1, y), &HFFFFFFFF
  22.     NEXT
  23.  
  24.     'drop a random pin
  25.     px1 = RND * (_WIDTH - 200) + 100
  26.     py1 = RND * (_HEIGHT - 200) + 100
  27.     ra = RND * _PI(2)
  28.     px2 = px1 + L * COS(ra)
  29.     py2 = py1 + L * SIN(ra)
  30.     LINE (px1, py1)-(px2, py2), _RGB32(50 + RND * 150)
  31.     N = N + 1 '                             another trial pin
  32.  
  33.     'did our pin cross a line
  34.     IF py1 < py2 THEN
  35.         IF py2 >= 100 * INT(py1 / 100) + L THEN C = C + 1
  36.     ELSE
  37.         IF py1 >= 100 * INT(py2 / 100) + L THEN C = C + 1
  38.     END IF
  39.     'FOR y = 0 TO _HEIGHT - 1 STEP L
  40.     '    IF py1 >= y AND py1 < y + L THEN ' one side of pin is between y and y + L
  41.     '        IF py2 >= y AND py2 < y + L THEN ' the pin did NOT cross a line both the head and point lay in same space
  42.     '            SS = SS + 1
  43.     '            EXIT FOR ' speed up a little
  44.     '        END IF
  45.     '    END IF
  46.     'NEXT
  47.     'C = N - SS
  48.     'Print results of trials
  49.     LOCATE 1, 1: PRINT "Trials:"; N
  50.     LOCATE 2, 1: PRINT "Crossings:"; C
  51.     LOCATE 3, 1: PRINT "Pi est:"; 2 * N / C
  52.     _DISPLAY
  53.     IF N = 1 THEN
  54.         _DELAY 2
  55.     ELSEIF N < 5 THEN
  56.         _DELAY 2
  57.     ELSEIF N < 10 THEN
  58.         _DELAY 2
  59.     ELSEIF N < 50 THEN
  60.         _DELAY .25
  61.     ELSEIF N < 1000 THEN
  62.         _DELAY .05
  63.     ELSEIF N <= 10000 THEN
  64.         IF N MOD 1000 = 0 THEN _DELAY .25
  65.     ELSE
  66.         IF N MOD 10000 = 0 THEN _DELAY .05
  67.     END IF
  69.  
  70.  
  71.  
Title: Re: Estimating Pi
Post by: bplus on August 02, 2020, 02:21:40 pm
There might be a problem with the code or the Random Number Generator as I have a definite pattern of "random" needle color "waves" along the bottom and right borders of the screen.

Anyone have an idea of the error, blunder or glitch?

 


Both versions produce this pattern.
Title: Re: Estimating Pi
Post by: _vince on August 02, 2020, 02:33:57 pm
looks like the bounds of the needle coordinates are too far inside the screen
Title: Re: Estimating Pi
Post by: bplus on August 02, 2020, 04:01:50 pm
Quote from: _vince on August 02, 2020, 02:33:57 pm
looks like the bounds of the needle coordinates are too far inside the screen

No I don't think it was that, here is the fix of the "pattern" and even speedier calc's seems 3.140 is limit of precision.

Code: QB64: [Select]
  1. _TITLE "Buffon Pi Calc " 'b+ 2020-08-2
  2. '2020-08-02 try this with Buffon Pi Calc  pi ~ 2 * nTrials / nCross  from
  3. ' pi ~ 2*L*N/(T*H) L=length of needle, T=spacing between lines N = Number of trials (or pins) H = number that cross line
  6.  
  7. CONST xmax = 800, ymax = 600
  8. SCREEN _NEWIMAGE(xmax, ymax, 32)
  9. _DELAY .25
  12.  
  13. L = 100 ' both pin length and line spacing
  14. N = 0 '   number of trials
  15. C = 0 '   number of pins that cross the line
  17.     'CLS
  18.     'redraw lines
  19.     'drop a random pin
  20.     px1 = RND * (_WIDTH - 201) + 100
  21.     py1 = RND * (_HEIGHT - 201) + 100
  22.     ra = RND * _PI(2)
  23.     IF RND < .5 THEN dir = -1 ELSE dir = 1 ' << this fixed the pattern developing along right and bottom border of screen
  24.     px2 = px1 + dir * L * COS(ra)
  25.     py2 = py1 + dir * L * SIN(ra)
  26.  
  27.     N = N + 1 '                             another trial pin
  28.  
  29.     'did our pin cross a line
  30.     IF py1 < py2 THEN
  31.         IF py2 >= 100 * INT(py1 / 100) + L THEN C = C + 1
  32.     ELSE
  33.         IF py1 >= 100 * INT(py2 / 100) + L THEN C = C + 1
  34.     END IF
  35.     IF N < 100000 THEN 'display
  36.         FOR y = 0 TO _HEIGHT - 1 STEP L
  37.             LINE (0, y)-(_WIDTH - 1, y), &HFFFFFFFF
  38.         NEXT
  39.         LINE (px1, py1)-(px2, py2), _RGB32(50 + RND * 150)
  40.  
  41.         'Print results of trials
  42.         LOCATE 1, 1: PRINT "Trials:"; N
  43.         LOCATE 2, 1: PRINT "Crossings:"; C
  44.         LOCATE 3, 1: PRINT "Pi est:"; 2 * N / C
  45.         _DISPLAY
  46.         IF N = 1 THEN
  47.             _DELAY 2
  48.         ELSEIF N < 5 THEN
  49.             _DELAY 2
  50.         ELSEIF N < 10 THEN
  51.             _DELAY 2
  52.         ELSEIF N < 50 THEN
  53.             _DELAY .25
  54.         ELSEIF N < 1000 THEN
  55.             _DELAY .05
  56.         ELSEIF N <= 10000 THEN
  57.             IF N MOD 1000 = 0 THEN _DELAY .25
  58.         ELSE
  59.             IF N MOD 10000 = 0 THEN _DELAY .05
  60.         END IF
  61.     ELSE
  62.         IF N MOD 1000000 = 0 THEN
  63.             LOCATE 1, 1: PRINT "Trials:"; N
  64.             LOCATE 2, 1: PRINT "Crossings:"; C
  65.             LOCATE 3, 1: PRINT "Pi est:"; 2 * N / C
  66.             _DISPLAY
  67.         END IF
  68.     END IF
  70.  

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