ASCII Sierpinski Triangle

[Ashish]

Hi everyone!
Here is Sierpinski Triangle in SCREEN 0!
I can't make it more bigger because QB64 gives error for 21 factorial (I mean it is out of range for QB64)
Anyway, enjoy this!

Code: QB64: [Select]

1. '############################################
2. '# ASCII Sierpinski Triangle By Ashish      #
3. '#  Fractal is generated with the help of   #
4. '#         Pascal  Triangle.                #
5. '#          23 Oct, 2019                    #
6. '############################################
7.  
8. _DEFINE A-Z AS _UNSIGNED _INTEGER64
9.  
10. _TITLE "ASCII Sierpinski Triangle"
11. x = _WIDTH / 2 'start from the centre of the screen
12.  
13. FOR i = 0 TO 15
14.     d$ = ""
15.     LOCATE , x
16.     FOR j = 0 TO i
17.         c$ = _TRIM$(STR$(nCr(i, j))) 'calculate the number value of the triangle
18.         'c$ = "*"
19.         IF nCr(i, j) MOD 2 = 1 THEN c$ = "*" ELSE c$ = " "
20.  
21.         d$ = d$ + c$ + " "
22.     NEXT
23.     PRINT d$;
24.     PRINT
25.     x = x - 1 'shift each step towards left as the triangle row move down.
26. NEXT
27.  
28. FUNCTION nCr (n, r) '(n!)/r!*(n-r)!
29.     IF r > n THEN EXIT FUNCTION
30.     nCr = fact(n) / (fact(r) * fact(n - r))
31. END FUNCTION
32.  
33. FUNCTION fact (n)
34.     fact = 1
35.     IF n = 0 THEN EXIT FUNCTION
36.     FOR i = 1 TO n
37.         fact = fact * i
38.     NEXT
39. END FUNCTION
40.  

[bplus]

Hi Ashish,

Very interesting! I did not know Sierpinski could be pulled out of Pascal Triangle.

I did work out a way to get around the huge number problem for nCr (Combination) calcs. I did it for number of 5 card hands you could be dealt from 52, might be able to code generalized solution... I will look into it. But! didn't you post code for finding large factorials? I think I saved it somewhere, guess I will look that up too :)

Update: Found it, it was how many hands of 13 from a deck of 52

Code: QB64: [Select]

1. _TITLE "How many hands of 13 from a deck" ' B+ started 2019-03-17
2.  
3. '   How many hands of 13 from a deck?
4.  
5. ' EDIT: formula for Combination of N items taken R at a time = N!/(R! * (N-R)!) so we need
6. ' 52! / (39! * 13!) =    52 * 51 * 50 * 49 * ... * 40
7. '                      / 13 * 12 * 11 * ...      * 1
8.  
9. DIM a52_39(1 TO 13) AS INTEGER, a13(1 TO 13) AS INTEGER
10. FOR i = 1 TO 13
11.     a52_39(i) = 39 + i ' = 40, 41, 42, 43, ... 52 numerator multipliers
12.     a13(i) = i '             1, 2, 3, 4, ... 13   denominator multipliers
13. NEXT
14.  
15. ' To keep the numerator from getting too big reduce numerator terms by demoinator terms that divide evenly
16.  
17. 'reduce terms in a52_39 by terms in denominator mult 2 to 13
18. FOR i = 13 TO 2 STEP -1
19.     found = 0
20.     FOR j = 1 TO 13 'is the array item divisible by i
21.         IF a52_39(j) MOD i = 0 THEN a52_39(j) = a52_39(j) \ i: found = 1: EXIT FOR
22.     NEXT
23.     IF found = 1 THEN a13(i) = 1 ELSE a13(i) = i
24. NEXT
25.  
26. ' multiply whats left in numerator and denomiator
27. f~&& = 1: g = 1
28. FOR i = 1 TO 13
29.     f~&& = a52_39(i) * f~&& 'multiple numerators left
30.     g = a13(i) * g 'multiply denominators left
31. NEXT
32.  
33. PRINT "The number of hands of 13 in a deck is ";
34. PRINT USING "###,###,###,###,###.###"; f~&& \ g
35. PRINT "                                compare to 635,013,559,600"
36.  

I think the method can be generalized to any combination up to a limit.


And I found your's, FPM: https://www.qb64.org/forum/index.php?topic=1468.msg106680#msg106680
But have to do commands from command line,  :p

Maybe through SHELL can do this with QB64 program?

[johnno56]

Nice Triangle... Oh. A possible use... Vary the colour of each 'star' and overlay it onto an Xmas tree...

[Petr]

Nice work, Ashish.

[bplus]

Yes, an alternate way to calculate nCr buys us a bigger triangle, here I have reached the height limit of my screen:

Code: QB64: [Select]

1. '############################################
2. '# ASCII Sierpinski Triangle By Ashish      #
3. '#  Fractal is generated with the help of   #
4. '#           Pascal  Triangle.              #
5. '#             23 Oct, 2019                 #
6. '#  B+ mod of nCr function allows for more. #
7. '#        Oh, hey try Johnno's idea!        #
8. '############################################
9.  
10. _DEFINE A-Z AS _UNSIGNED _INTEGER64
11. CONST xmax = 800, ymax = 800
12. SCREEN _NEWIMAGE(xmax, ymax, 32)
13. _SCREENMOVE 400, 0
14. _TITLE "ASCII Sierpinski Triangle"
15. COLOR 0, &HFF004400
16. x = (_WIDTH / 8) / 2 'start from the centre of the screen
17. FOR i = 0 TO 47
18.     d$ = " "
19.     LOCATE i + 1, x
20.     FOR j = 0 TO i
21.         c$ = _TRIM$(STR$(nCr(i, j))) 'calculate the number value of the triangle
22.         IF nCr(i, j) MOD 2 = 1 THEN c$ = "*" ELSE c$ = " "
23.         d$ = d$ + c$ + " "
24.     NEXT
25.     FOR j = 1 TO LEN(d$)
26.         c$ = MID$(d$, j, 1)
27.         IF c$ <> " " THEN
28.             COLOR _RGB32(RND * 255, RND * 255, RND * 255)
29.         ELSE
30.             COLOR , _RGB32(RND * 25, RND * 15 + 20, RND * 25)
31.         END IF
32.         PRINT c$;
33.     NEXT
34.     'PRINT
35.     x = x - 1 'shift each step towards left as the triangle row move down.
36. NEXT
37. SLEEP
38.  
39. ' B+ alt way to calculate nCr, generalized 2019-10-23
40. FUNCTION nCr~&& (n AS INTEGER, r AS INTEGER)
41.     'Note: I am using notes from a successful calc to generalize this method.
42.     ' How many hands of 13 from a deck?
43.     ' Formula for Combination of N items taken R at a time = N!/(R! * (N-R)!) so we need
44.     ' 52! / (39! * 13!) =    52 * 51 * 50 * 49 * ... * 40  / 13 * 12 * 11 * ...      * 1
45.  
46.     DIM a AS INTEGER, i AS INTEGER, j AS INTEGER, found AS INTEGER, numerator AS _UNSIGNED _INTEGER64, denomenator AS _UNSIGNED LONG
47.     a = n - r
48.     IF a < r THEN r = a 'results are symmteric for r and n-r so use smaller of 2
49.     a = n - r
50.  
51.     DIM numer(1 TO r) AS INTEGER, denom(1 TO r) AS INTEGER
52.     ' DIM a52_39(1 TO 13) AS INTEGER, a13(1 TO 13) AS INTEGER
53.     FOR i = 1 TO r
54.         numer(i) = a + i ' = 40, 41, 42, 43, ... 52 numerator multipliers
55.         denom(i) = i '     =  1,  2,  3,  4, ... 13 denominator multipliers
56.     NEXT
57.  
58.     ' To keep the numerator from getting too big reduce numerator terms by demoinator terms that divide evenly
59.     FOR i = r TO 2 STEP -1
60.         found = 0
61.         FOR j = 1 TO r 'is the array item divisible by i
62.             IF numer(j) MOD i = 0 THEN numer(j) = numer(j) \ i: found = 1: EXIT FOR
63.         NEXT
64.         IF found = 1 THEN denom(i) = 1
65.     NEXT
66.  
67.     ' multiply whats left in numerator and denomiator
68.     numerator = 1: denomenator = 1
69.     FOR i = 1 TO r
70.         numerator = numerator * numer(i) 'multiple numerators left
71.         denomenator = denomenator * denom(i) 'multiply denominators left
72.     NEXT
73.     nCr~&& = numerator \ denomenator
74. END FUNCTION
75.  
76.  

[Ashish]

Cool Bplus! Very nice MOD.
I could use FPM, but it is string based. Also, it can give me factorial of any +ve number but the problem comes when you have to divide the quantities.

[bplus]

Cool Bplus! Very nice MOD.
I could use FPM, but it is string based. Also, it can give me factorial of any +ve number but the problem comes when you have to divide the quantities.

Ah! the division, yes that would be a problem.

[TempodiBasic]

Hey Bplus and Ashish...
cool!

PS:
how many slices of math have you at breakfast every day? :-)

[Ashish]

Hey Bplus and Ashish...
cool!

PS:
how many slices of math have you at breakfast every day? :-)

Not everyday. It's just what taught in school, I make a simple program on that. Factorials, Permutations, Combination, Binomial Theorem,etc are in the syllabus this year. (Also The Calculus Ghost).

[bplus]

I am wondering how someone "saw" Sierpinsky in the odd combinations of Pascal's Triangle, though it's not exactly the Sierpinsky I would draw with lines or dots, still it's dang close!

Ashish, did you "see" that or did someone show you the trick?

[Ashish]

I am wondering how someone "saw" Sierpinsky in the odd combinations of Pascal's Triangle, though it's not exactly the Sierpinsky I would draw with lines or dots, still it's dang close!

Ashish, did you "see" that or did someone show you the trick?

I'm not that intelligent. I have known this fact from numberphile's video on Pascal Triangle.

There are also many other amazing properties which I have known from that video.

[bplus]

Oh excellent link, thanks Ashish!