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FellippeHeitor

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Visualizing _PI challenge
« on: March 15, 2019, 10:37:15 am »
Anyone would be willing to take on the challenge to recreate this or even reinvent it? https://www.reddit.com/r/dataisbeautiful/comments/b13h1d/adding_to_the_pi_visualizations_oc/
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Offline bplus

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Re: Visualizing _PI challenge
« Reply #1 on: March 15, 2019, 11:02:41 am »
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    • Hi Fellippe,

    Is there some sort of calculation of Pi going on or is this just bouncing from digit to digit of a Pi expansion?

    Repeating digits could use a little loop, internal or external.
    « Last Edit: March 15, 2019, 11:04:26 am by bplus »
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    Offline Qwerkey

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    Re: Visualizing _PI challenge
    « Reply #2 on: March 15, 2019, 11:37:32 am »
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    • ... And when you have taken up this challenge, unless you produce pi to more than 31 trillion digits, we shall think that you have hardly tried at all.  (To any mathematicians out there, why do mathematicians do that sort of thing and who knows if they're correct?).
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    Offline Petr

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    Re: Visualizing _PI challenge
    « Reply #3 on: March 15, 2019, 12:28:29 pm »
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    • It looks interesting. To calculate pi, we could use the formula π = 2 x (2 x 2 x 4 x 4 x 6 x 6 ...) / (1 x 3 x 3 x 5 x 5 x 7 ...)
    but a sample example uses only a few decimal places to visualize ...

    Code: QB64: [Select]
    1. SCREEN _NEWIMAGE(800, 600, 32)
    2.  
    3. TYPE nrs
    4.     x AS INTEGER
    5.     y AS INTEGER
    7. DIM N(10) AS nrs
    8.  
    9. CONST CX = 400
    10. CONST CY = 300
    11. pi$ = "3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273"
    12.  
    13. FOR p = -60 TO 264 STEP 36
    14.     q = _D2R(p)
    15.     N(w).x = CX + COS(q) * 150
    16.     N(w).y = CY + SIN(q) * 150
    17.     w = w + 1
    18.     CIRCLE (CX + COS(q) * 150, CY + SIN(q) * 150), 3
    19.     _PRINTSTRING (CX + COS(q) * 170, CY + SIN(q) * 170), _TRIM$(STR$(num))
    20.     num = num + 1
    22.  
    23. CIRCLE (400, 300), 150
    24.  
    25. FOR sh = 1 TO LEN(pi$)
    26.     IF ASC(pi$, sh) >= 48 AND ASC(pi$, sh) <= 57 THEN
    27.         v$ = MID$(pi$, sh, 1)
    28.         LINE (ox, oy)-(ox2, oy2), _RGB32(127, 127, 127)
    29.          i = ASC(pi$, sh) - 48
    30.         LOCATE 1, 1: PRINT MID$(pi$, 1, sh)
    31.         ox = N(oldN).x: oy = N(oldN).y: ox2 = N(i).x: oy2 = N(i).y
    32.         LINE (N(oldN).x, N(oldN).y)-(N(i).x, N(i).y), _RGB32(255, 255, 255): oldN = i
    33.         oldN = i
    34.         _DELAY .2
    35.     END IF
    36. NEXT sh
    37.  
    « Last Edit: March 15, 2019, 12:34:06 pm by Petr »
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    Offline Petr

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    Re: Visualizing _PI challenge
    « Reply #4 on: March 15, 2019, 12:46:30 pm »
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    • There is one more way - the infinite division using MOD, but the fraction 22/7 is inaccurate, because there is a wrong result in the third decimal place. Another fraction is 355/113, but there will also be an error in the seventh decimal place.
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    FellippeHeitor

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    Re: Visualizing _PI challenge
    « Reply #5 on: March 15, 2019, 12:53:51 pm »
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    • Wow, that was fast, Petr. Good job!

    @bplus
    I don't know if they're calculating as they go or if it's just fetching from a precalculated list of digits. I just liked the visualization very much when I fist came across it.

    @Qwerkey
    It's indeed silly to try to break any supercomputer's record in findind the next digits of pi... Maybe mathematicians just wanna have fun in the end too (and measure dick).

    I'm looking to take part in the challenge myself. I'll keep this thread as a post-it reminder for me to do it eventually.
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    Offline Petr

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    Re: Visualizing _PI challenge
    « Reply #6 on: March 15, 2019, 12:58:30 pm »
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    • Thank you Fellippe, it was a welcome change, such a rebound from the 3D maze editor that I am writing.
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    Offline jack

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    Re: Visualizing _PI challenge
    « Reply #7 on: March 15, 2019, 01:08:36 pm »
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    • I think the Spigot algorithm would be perfect for this https://rosettacode.org/wiki/Pi
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    Offline Pete

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    Re: Visualizing _PI challenge
    « Reply #8 on: March 15, 2019, 01:15:32 pm »
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    • I haven't figured out how far my string math routines would go, I just know if it was accurate and infinite, it would take about 40 years to surpass the record pi calculation to date, of 31.4 trillion digits. Anyone know where I can find a computer that will last 40 years, without a %^&^$# Windows update?

    Pete
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    Want to learn how to write code on cave walls? https://www.tapatalk.com/groups/qbasic/qbasic-f1/

    Offline bplus

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    Re: Visualizing _PI challenge
    « Reply #9 on: March 15, 2019, 02:14:22 pm »
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    • Hey Petr, nice and fast.

    I had basic scheme worked out when I saw you had already posted. So I better have some better features!

    I do. :)

    This draws an outer loop around digit when digit repeats sequentially (you have to wait awhile for 7), plus it has the pointer moving between points...
    Code: QB64: [Select]
    1. _TITLE "Visualizing Pi" 'B+ 2019-03-15
    2.  
    3. 'Featuring a circle loop around the a repeated digit.
    4. ' 3 first, 8, 9, 4, 1, 6, 2, 5, 0 (300+ digits), 7 (560 digits) EDIT
    5.  
    6. 'ref  http://www.math.com/tables/constants/pi.htm  First 100 digits
    7. CONST pi100 = "3141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067"
    8. 'ref same as above 1000 digits, see if can circle all digits pi shortened to 560 digits
    9. CONST pi = "314159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196442881097566593344612847564823378678316527120190914564856692346034861045432664821339360726024914127372458700660631558817488152092096282925409171536436789259036001133053054882046652138414695194151160943305727036575959195309218611738193261179310511854807446237996274956735188575272489122793818301194912983367336244065664308602139494639522473719070217986094370277"
    10. 'PRINT INSTR(pi, "77") ' last number to get circled for repeating at 560
    11.  
    12. CONST xmax = 800
    13. CONST ymax = 720
    14. CONST x0 = 400
    15. CONST y0 = 420
    16. CONST radius = 240
    17. CONST r2 = 20
    18.  
    19. TYPE vectorType
    20.     x AS SINGLE
    21.     y AS SINGLE
    23.  
    24. SCREEN _NEWIMAGE(xmax, ymax, 32)
    25. _SCREENMOVE 300, 20
    26. s2& = _NEWIMAGE(xmax, ymax, 32) 'to update drawing lines in isolation
    27. t = TIMER(.001)
    28. ca = _PI(2 / 10)
    29. DIM vdigit(0 TO 9) AS vectorType
    30. DIM outer(0 TO 9) AS vectorType
    31. CIRCLE (x0, y0), radius
    32. FOR i = 0 TO 9
    33.     vdigit(i).x = x0 + radius * COS(ca * i - 2 * ca)
    34.     vdigit(i).y = y0 + radius * SIN(ca * i - 2 * ca)
    35.     CIRCLE (vdigit(i).x, vdigit(i).y), 2
    36.     x = x0 + (radius + 18) * COS(ca * i - 2 * ca) - 2
    37.     y = y0 + (radius + 18) * SIN(ca * i - 2 * ca) - 6
    38.     _PRINTSTRING (x, y), LTRIM$(STR$(i))
    39.     outer(i).x = x0 + (radius + r2) * COS(ca * i - 2 * ca)
    40.     outer(i).y = y0 + (radius + r2) * SIN(ca * i - 2 * ca)
    42. _PUTIMAGE , 0, s2&
    43.  
    44. digitPointer = 1
    45. cx = x0: cy = y0: speed = 10
    46. WHILE digitPointer <= LEN(pi)
    47.     d = VAL(MID$(pi, digitPointer, 1))
    48.     targetx = vdigit(d).x: targety = vdigit(d).y
    49.     targetangle = _ATAN2(targety - cy, targetx - cx)
    50.     rc~& = _RGB32(255 * RND, 255 * RND, 255 * RND)
    51.     IF lastd = d THEN
    52.         _DEST s2&
    53.         CIRCLE (outer(d).x, outer(d).y), r2, rc~&
    54.         _DEST 0
    55.         CLS
    56.         _PUTIMAGE , s2&, 0
    57.         CIRCLE (cx, cy), 2, &HFFFFFFFF, BF
    58.         LOCATE 1, 1: PRINT MID$(pi, 1, digitPointer - 1)
    59.         _DISPLAY
    60.     ELSE
    61.         DO
    62.             dist = SQR((targetx - cx) ^ 2 + (targety - cy) ^ 2)
    63.             IF dist < speed THEN
    64.                 newX = targetx
    65.                 newY = targety
    66.             ELSE
    67.                 newX = cx + speed * COS(targetangle)
    68.                 newY = cy + speed * SIN(targetangle)
    69.             END IF
    70.             _DEST s2&
    71.             LINE (cx, cy)-(newX, newY), rc~&
    72.             _DEST 0
    73.             CLS
    74.             _PUTIMAGE , s2&, 0
    75.             CIRCLE (newX, newY), 2, &HFFFFFFFF, BF
    76.             cx = newX: cy = newY
    77.             LOCATE 1, 1: PRINT MID$(pi, 1, digitPointer - 1)
    78.             _DISPLAY
    79.             _LIMIT 60
    80.         LOOP UNTIL cx = targetx AND cy = targety
    81.     END IF
    82.     digitPointer = digitPointer + 1
    83.     lastd = d
    85. LOCATE 1, 1: PRINT MID$(pi, 1, digitPointer)
    86. PRINT TIMER(.001) - t; " secs"
    88.  


    « Last Edit: March 15, 2019, 02:25:15 pm by bplus »
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    Offline Petr

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    Re: Visualizing _PI challenge
    « Reply #10 on: March 15, 2019, 02:31:17 pm »
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    • Hi BPlus!  It's very nice. I even discovered the BF parameter for Circle with your program, I haven't seen it yet. Motion in circle is also very interesting.
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    Offline bplus

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    Re: Visualizing _PI challenge
    « Reply #11 on: March 15, 2019, 02:34:44 pm »
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    • Quote from: Petr on March 15, 2019, 02:31:17 pm
    Hi BPlus!  It's very nice. I even discovered the BF parameter for Circle with your program, I haven't seen it yet. Motion in circle is also very interesting.

    HA! Twice!
    Code: QB64: [Select]
    1. CIRCLE (cx, cy), 2, &HFFFFFFFF, BF  
    2.  

    I started with a line for a solid pointer, forgot to erase that BF!
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    Offline Petr

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    Re: Visualizing _PI challenge
    « Reply #12 on: March 15, 2019, 02:37:40 pm »
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    • Quote
    I started with a line for a solid pointer, forgot to erase that BF!

    Really? I thought I discovered something new! :-D
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    Offline jack

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    Re: Visualizing _PI challenge
    « Reply #13 on: March 15, 2019, 04:36:30 pm »
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    • hello bplus
    here's spigot sub for you
    Code: QB64: [Select]
    1. DIM pi(100) AS _BYTE
    3. spigot pi()
    4.  
    5. PRINT LTRIM$(STR$(pi(1))); ".";
    6. FOR i = 2 TO UBOUND(pi)
    7.     PRINT LTRIM$(STR$(pi(i)));
    11.  
    12. SUB spigot (pi() AS _BYTE)
    13.     DIM N AS LONG
    14.     N = UBOUND(pi)
    15.     DIM ln AS LONG
    16.     ln = INT((10 * N) / 3) + 1
    17.     DIM A(0 TO ln - 1) AS LONG
    18.     DIM npi AS LONG
    19.     DIM i AS LONG, j AS LONG, k AS LONG
    20.     npi = 0
    21.     FOR i = 0 TO ln - 1
    22.         A(i) = 2
    23.     NEXT
    24.  
    25.     DIM nines AS LONG
    26.     DIM predigit AS LONG
    27.     nines = 0
    28.     predigit = 0
    29.     FOR j = 1 TO N
    30.         DIM q AS LONG
    31.         q = 0
    32.         FOR i = ln TO 1 STEP -1
    33.             DIM x AS LONG
    34.             x = (10 * A((i - 1))) + (q * i)
    35.             A((i - 1)) = x MOD ((2 * i) - 1)
    36.             q = x \ ((2 * i) - 1)
    37.         NEXT
    38.         A(0) = q MOD 10
    39.         q = q \ 10
    40.  
    41.         IF 9 = q THEN
    42.             nines = nines + 1
    43.         ELSEIF 10 = q THEN
    44.             pi(npi) = predigit + 1: npi = npi + 1
    45.             FOR k = 0 TO nines - 1
    46.                 pi(npi) = 0: npi = npi + 1
    47.             NEXT
    48.             predigit = 0
    49.             nines = 0
    50.         ELSE
    51.             pi(npi) = predigit: npi = npi + 1
    52.             predigit = q
    53.  
    54.             IF 0 <> nines THEN
    55.                 FOR k = 0 TO nines - 1
    56.                     pi(npi) = 9: npi = npi + 1
    57.                 NEXT
    58.                 nines = 0
    59.             END IF
    60.         END IF
    61.     NEXT
    62.     pi(npi) = predigit
    64.  
    « Last Edit: March 17, 2019, 03:25:47 am by jack »
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    Offline bplus

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    Re: Visualizing _PI challenge
    « Reply #14 on: March 15, 2019, 05:53:27 pm »
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    • Thanks Jack I will check it out.

    Update: That compares very well with the reference Pi I was using. I used the following code as a check:
    Code: QB64: [Select]
    1. 'from Jack 2019-03-15 QB64 forum "Visualizing Pi" B+ mod to check Pi with reference Pi.
    2. _TITLE "Compare Spigot to Pi Ref digit by digit"
    3. SCREEN _NEWIMAGE(600, 720, 32)
    4. _SCREENMOVE 300, 20
    5. CONST piRef = "314159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196442881097566593344612847564823378678316527120190914564856692346034861045432664821339360726024914127372458700660631558817488152092096282925409171536436789259036001133053054882046652138414695194151160943305727036575959195309218611738193261179310511854807446237996274956735188575272489122793818301194912983367336244065664308602139494639522473719070217986094370277"
    6. DIM pi%(561)
    7. spigot pi%()
    8.  
    9. FOR i = 1 TO UBOUND(pi%)
    10.     PRINT i; " digit spigot vrs ref "; pi%(i); " vrs "; MID$(piRef, i, 1)
    11.     IF i MOD 30 = 29 THEN
    12.         INPUT "OK, press enter "; wate$
    13.         CLS
    14.     END IF
    17.  
    18. SUB spigot (pi%())
    19.     DIM N%
    20.     N% = UBOUND(pi%)
    21.     DIM ln AS LONG
    22.     ln = INT((10 * N%) / 3) + 1
    23.     DIM A(0 TO ln - 1) AS LONG
    24.     DIM npi AS LONG
    25.     DIM i AS INTEGER, j AS INTEGER, k AS INTEGER
    26.     npi = 0
    27.     FOR i = 0 TO ln - 1
    28.         A(i) = 2
    29.     NEXT
    30.  
    31.     DIM nines AS LONG
    32.     DIM predigit AS LONG
    33.     nines = 0
    34.     predigit = 0
    35.     FOR j = 1 TO N%
    36.         DIM q AS LONG
    37.         q = 0
    38.         FOR i = ln TO 1 STEP -1
    39.             DIM x AS LONG
    40.             x = (10 * A((i - 1))) + (q * i)
    41.             A((i - 1)) = x MOD ((2 * i) - 1)
    42.             q = x \ ((2 * i) - 1)
    43.         NEXT
    44.         A(0) = q MOD 10
    45.         q = q \ 10
    46.  
    47.         IF 9 = q THEN
    48.             nines = nines + 1
    49.         ELSEIF 10 = q THEN
    50.             pi%(npi) = predigit + 1: npi = npi + 1
    51.             FOR k = 0 TO nines - 1
    52.                 pi%(npi) = 0: npi = npi + 1
    53.             NEXT
    54.             predigit = 0
    55.             nines = 0
    56.         ELSE
    57.             pi%(npi) = predigit: npi = npi + 1
    58.             predigit = q
    59.  
    60.             IF 0 <> nines THEN
    61.                 FOR k = 0 TO nines - 1
    62.                     pi%(npi) = 9: npi = npi + 1
    63.                 NEXT
    64.                 nines = 0
    65.             END IF
    66.         END IF
    67.     NEXT
    68.     pi%(npi) = predigit
    70.  
    71.  
    « Last Edit: March 15, 2019, 07:14:11 pm by bplus »
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