Category Archives: Fractals

Altars of Mathness

by in Album Art, Arithmetic, Base Behavior, Fibonacci Sequence, Fractals, Integer Sequences, Mathematics, Polygonal numbers, Powers, Prime numbers, Triangular Numbers and tagged , , , , , , , , | Leave a comment

What could be duller than digits? They just sit there on the page or screen, mindlessly marking mathematics:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100…

But perhaps they become more interesting as images. Let’s display the final digit of the integers, or counting numbers, on a graph. Running left-right and up-down, the graph represents the final or rightmost digit of 1, 2, 3, … 10, 11, 12, 13, … 100, 101, 102, 103, …, 1000, 1001, 1002, 1003, …:

Rightmost single digit of the integers (click for larger)

No, that’s still dull: the graph just generates endlessly repeating triangles. After all, the final digits fall into a cycle: 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3… So do the final two digits: 1, 2, 3, 4, 5, […] 94, 95, 96, 97, 98, 99, 00, 01, 02, 03… Here they are as a graph:

Rightmost two digits of the integers

Now the triangles look like waves sweeping to shore. That’s a bit more interesting, but not much. So let’s try something different. The trailing digits of the integers generate triangles, so let’s see what the triangular numbers generate. The triangular numbers — 0, 1, 3, 6, 10, 15, 21… — are very simple to form. You just sum the integers: 1, 3 = 1 + 2, 6 = 1 + 2 + 3, 10 = 1 + 2 + 3 + 4, 15 = 1 + 2 + 3 + 4 + 5, 21 = 1 + 2 + 3 + 4 + 5 + 6, 28 = 1 + 2 + 3 + 4 + 5 + 6 + 7, 36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8, 45 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9, 55 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10… Here are the final digits of the triangulars — 1, 3, 6, 0, 5, 1, 8, 6… — as a graph:

Final digit of triangular numbers in base 10 (click for larger)

Now something interesting has appeared. The final digits form a repeated palindromic pattern (counting 0 as the zero-th triangular number):

0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, …

An Altar of Mathness created by the final digit of triangular numbers in base 10

And those palindromic digits create symmetric shapes that remind me of little altars — let’s call them “altars of mathness” in tribute to Morbid Angel’s genre-defining album Altars of Madness (1989). And what about the final two digits of the triangular numbers? Here’s the graph (adjusted so that 99 fits into the same space as 9):

Final two digits of triangulars in b10

Final two triangular digits in b10 (horizontal scale compressed)

The final two digits form palindromes too. And this time we don’t get just triangles, but curves too. But that’s in base 10. What happens with the trailing triangular digits in other bases? Well, here’s the final triangular digit creating more altars of mathness in different bases (note that the altars are more elaborate in even bases):

Final triangular digit in base 4

Final triangular digit in b5

Final triangular digit in b6

Final triangular digit in b7

Final triangular digit in b8

Final triangular digit in b9

Final triangular digit in b14

And here’s the graph for the final triangular digit in base 100:

Final triangular digit in b100

The graph for final single digit in b100 should look familiar, because it’s identical to the graph for final double triangular digits in b10:

Final two digits of triangulars in b10

That’s because two digits in b10 are equivalent in one digit in b100, four digits in b10 are equivalent to two digits in b100, and so on. But b100 can’t capture three digits in b10 (the graph is again adjusted so that 999 fits into the same space as 9 and 99 above):

Final three triangular digits in b10

If you compress the x-axis for that graph, you can see how long the symmetries are:

Final three triangular digits in b10 (x-axis / 2)

Final three triangular digits in b10 (x-axis / 4)

The final four digits of the triangulars in b10 create even longer symmetries:

Final quadruple triangular digits in b10

Final quadruple triangular digits in b10 (x-axis / 2)

Final quadruple triangular digits in b10 (x-axis / 8)

Note how, as the length of the final digits rises, you need to compress the x-axis more and more to see the symmetries. But integer sequences obviously don’t end with the counting numbers and triangulars. What about squares and powers of n? What about primes and Fibonacci numbers? Here’s the final two digits of the squares — 1, 4, 9, 16, 25, 49, 64, 81, 100, 121, 144, 169… — in b10:

Final two digits of the squares in b10

It’s reminiscent of the triangular numbers (so are the final-digit graphs for other polygonal numbers). So what about the powers of 2? That’s 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024… Here’s the graph for final single digits of 2^p in b10:

Final single digits of 2^p in b10

This time there’s repetition, but not symmetry. Here’s the graph for final double digits, or 2-digits, of 2^p in b10:

Final 2-dig of 2^p in b10

Now the graph looks a little like a range of eroded mountains. Now try dig-4, the final four digits of 2^p in b10:

Final 4-dig of 2^p in b10

The patterns are similar to those of dig-2 and don’t need compressing in the x-axis. This similarity and lack of need for compression are true of any number of final digits in 2^p. The final 10 digits look like this:

Final 10-dig of 2^p in b10

And the final 20 and 30 digits like this:

Final 20-dig of 2^p in b10

Final 30-dig of 2^p in b10

Powers don’t behave like polygonals: the finals are fractals. That is, the final digits create similar patterns at all scales: 1-dig, 2-dig, 10-dig, 100-dig, 1000-dig and so on. That’s true in other bases:

Final 5-dig of 3^p in b2

But a glimpse of b2 is all you’re going to get of other bases. There are other fish to fry — Fibonacci fish. The Fibonacci sequence, whose terms are equal to the sum of the previous two numbers (after seeding with “1, 1”), starts like this: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811… And what about the graphs for final fib-digits? As you’ll see, final Fib-digits are fractal too. Indeed, Fibonacci final-graphs look like 2-power final-graphs (in a way, Fibonacci numbers are powers of φ = 1.6180339887498948482…). The patterns are similar at all scales. And they remind me of the skyline of a ruined city in an Oriental tale, with collapsed domes and crumbling minarets:

Final 1-dig of Fibonacci numbers in b10

Final 2-fibdig in b10

Final 3-fibdig in b10

Final 4-fibdig in b10

Final 5-fibdig in b10

Final 10-fibdig in b10

Final 15-fibdig in b10

Final 20-fibdig in b10

Final 25-fibdig in b10

So final fibdigs are fractal. But final prime digits aren’t:

Final 1-digit of primes in b10

Final 1-digit of primes in b5

Final 2-digit of primes in b10

Primes aren’t final-digitally fractal like Fibonaccis and powers of 2. But there’s occasional symmetry in the prime fin-digs. I’ve marked some palindromic patterns in red and green:

Palindromic patterns in final 1-digits of the primes in b10 (click for larger)

The palindromic patterns, or pal-pats, in the primes look like the altars of mathness in the triangulars. They’re created by digital palindromes like these:

19, 23, 29 (c=3)
347, 349, 353, 359, 367 (c=5)
937, 941, 947, 953, 967, 971, 977 (c=7)
1951, 1973, 1979, 1987, 1993, 1997, 1999, 2003, 2011 (c=9)
26423, 26431, 26437, 26449, 26459, 26479, 26489, 26497, 26501, 26513 (c=10)

Here are the first few pal-pats in the primes (note that 157, 163, 167 and 163, 167, 173 overlap):

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607…

And are there palindromes among the final 2-digits, 3-digits and higher n-digits of the primes in different bases? Yes, you can easily find some. But I haven’t put them on a graph yet:

Base 10 (2-dig)

58789, 58831, 58889 (c=3)
286873, 286927, 286973 (c=3)
360649, 360653, 360749 (c=3)
404851, 404941, 404951 (c=3)
590437, 590489, 590537 (c=3)
623071, 623107, 623171 (c=3)
651517, 651587, 651617 (c=3)

Base 6 (2-dig)

300335, 300401, 300441, 300501, 300535 (c=5) (23459 to 23531 in base 10)
1030255, 1030331, 1030351, 1030431, 1030455 (c=5) (50651 to 50723 in b10)
1140451, 1140501, 1140521, 1141001, 1141051 (c=5) (59791 to 59863 in b10)
1402451, 1402545, 1403031, 1403045, 1403051 (c=5) (78367 to 78439 in b10)
1435431, 1435451, 1435505, 1435551, 1440031 (c=5) (82891 to 82963)
2400505, 2401001, 2401015, 2401101, 2401105 (c=5) (124601 to 124673)
2442235, 2442311, 2442351, 2442411, 2442435 (c=5) (130127 to 130199)
2444215, 2444225, 2444311, 2444325, 2444415 (c=5) (130547 to 130619)
2533105, 2533121, 2533215, 2533221, 2533305 (c=5) (136769 to 136841)

Base 4 (3-dig)

20013013, 20013133, 20020013 (c=3) (33223 to 33287 in base 10)
21031111, 21031303, 21032111 (c=3) (37717 to 37781)
22310011, 22310333, 22311011 (c=3) (44293 to 44357)
33030121, 33031001, 33031121 (c=3) (62233 to 62297)
102031333, 102032131, 102032333 (c=3) (74623 to 74687)
110013121, 110013311, 110020121 (c=3) (82393 to 82457)

Base 3 (3-dig)

112121020012, 112121021211, 112121022021, 112121100211, 112121101012 (c=5) (287393 to 287501 in base 10)
202002212002, 202002212101, 202002220001, 202002221101, 202010000002 (c=5) (395741 to 395849)
1001012111212, 1001012112202, 1001012121022, 1001012121202, 1001012122212 (c=5) (555143 to 555251)
1010112112012, 1010112112201, 1010112120222, 1010112121201, 1010112200012 (c=5) (601079 to 601187)
1011202211211, 1011202212212, 1011202220022, 1011202221212, 1011202222211 (c=5) (625369 to 625477)

Base 2 (5-dig)

101110001111101, 101110010000111, 101110010001001, 101110010100111, 101110010111101 (c=5) (23677 to 23741 in base 10)
10000001000111101, 10000001001011001, 10000001001110001, 10000001001111001, 10000001001111101 (c=5) (66109 to 66173)
10111100110011011, 10111100110011111, 10111100110111001, 10111100110111111, 10111100111011011 (c=5) (96667 to 96731)
11010000111110001, 11010001000001101, 11010001000011001, 11010001000101101, 11010001000110001 (c=5) (106993 to 107057)

And I conjecture that you’ll get palindromes for any number of final digits in all bases. And can these palindromes be of arbitrary length? Again, I conjecture so. There are infinitely many primes and very rare patterns can occur infinitely often in an infinite set of numbers.

Post-Performative Post-Scriptum

Here’s Dan Seagrave’s classic cover for Morbid Angel’s Altars of Madness (1989):

• Morbid Angel — official website
• Dan Seagrave — official website

Elsewhere Other-Accessible…

• Formulas Focal to the Flesh — a pre-previous post paronomasizing the title of a Morbid-Angel album…

The Hex Crystals

by in Fractals, Geometry, Hexagons, Mathematics, Pentagons and tagged , , , , , , , , | Leave a comment

To coin a phrase: Never Mind the Bollocks — Here’s the Hex Crystals! And what is a hex crystal? It’s what I call a shape that’s created algorithmo inside a hexagon and looks like a crystal:

A hex crystal

Here are some more hex-crystals:

I came across hex-crystals when I was looking at an interesting little geometrical question. How does sum(vd), the sum of distances to the vertices of a square, vary from different points, (x,y), inside the square? Say the square is created inside a circle of radius = 500 units and centered on (x,y) = (0,0). When the point is at (0,0), the center of the square, sum(vd) is obviously 2000, because the four vertices all fall on the perimeter of the circle at 500 units from the center and 4 * 500 = 2000:
0

sum(vd) = 2000 = sum of distances to vertices from (0,0)

When is sum(vd) at a maximum? When the point is on one or another of the vertices, which are at (+/-354,+/-354) units in relation to the center at (0,0):

sum(vd) = 2414 = sum of distances to vertices from (354,-354)

More precisely, the sum is 2414.213562373… = 1000 * (√2 + 1) units and the vertices are at (+/-353.55339…, +/-353.55339…) units, as simple geometry dictates for a square inside a circle of radius 500. Accordingly, sum(vd) varies between exactly 2000 and 2414.213562373… as the point moves inside the square:

sum(vd) = 2165 from (132,256)

sum(vd) = 2182 from (-135,271)

sum(vd) = 2069 from (177,51)

I wondered what shapes appeared as one traced the route of a point jumping, say, 1/2 towards the vertices according to tests on sum(vd). For example, if the point starts at (0,0) at time t0) and sum(vd) at time ti has to be alternately greater and less than sum(vd) at ti-1 for successive jumps, you get this shape:

jump = 1/2, test = sum(vd,ti) >,< sum(vd,ti-1)

You can use the binary number 10bin to represent the test on sum(vd) at ti-1 and ti-1, i.e. the test at jump 1 is sum(vd,ti) > sum(vd,ti-1), at step 2 is sum(vd,ti) < sum(vd,ti-1), and so on. Using the same test and a jump of 1/3, you get this shape:

jump = 1/3, test = sum(vd,ti,10bin)

Now the shape is clearly a fractal. So are some of the other shapes I found by applying the same kind of tests to a point jumping inside a pentagon:

vertex = 5, jump = 55/144 = fib(10) / fib(12), test on sum(vd) = 10bin

v = 5, j = 55/144, test = 10010bin

v = 5, j = 55/144, test = 11000bin

When test = 10010bin, you read the binary number left-to-right and check for s1><s0,s2<s1,s3<s2,s4>s3,s5<s4. Then you apply the same tests to subsequent jumps, i.e., you return to the beginning of the binary number and read it left-to-right again. Now let’s apply similar tests to hexagons and create some hex-crystals:

v = 6, j = 1/2, test = 10bin

Various hex-crystals (animated gif courtesy EZgif)

I searched an array to calculate the possible routes, so the same test yielded different results depending on dp, the depth of the search. This is because tl, the length of the test, fits more or less well into dp by dp modulo tl, that is, by whether tl is a factor of dp. For example, when the test is 110 and tl = 3, you get this with dp = 9:

v = 6, j = 1/2, test = 110, dp = 9

And you get this when dp = 10 (i.e., dp = 9+1):

v = 6, j = 1/2, test = 110bin, dp = 10dec

Here are some more hex-crystals:

test = 1100bin

test = 1110bin

test = 10010bin

test = 11010bin

test = 11100bin

test = 101000, dp = 12

test = 101100bin

test = 111100bin

test = 111100, dp = 11

test = 1110010bin

test = 1111100bin

test = 10010110bin

test = 10011110bin

test = 11000110bin

test = 11001110bin

test = 11010110bin

test = 11100110bin

test = 11101000bin

test = 11110010bin

test = 100101000bin

test = 100111110bin

test = 110011110bin

test = 110111000bin

test = 1001101010bin

test = 1001111000bin

test = 1001111010bin

test = 1010011110bin

test = 1011101110bin

test = 1101010000bin

test = 1110001110bin

test = 1110101000bin

test = 1110101010bin

test = 1111100010bin

j = 1/3, test = 1 (i.e., for all jumps sum(vd) at ti > sum(vd) at ti-1, center point

j = 2/3, test = 11100bin

j = 2/5, test = 10010bin

Finally, here are some hex-crystals based on a test of sorted distances from (x,y), i.e. how the vertices rank by distance from (x,y):

Worms in Terms of Perms

by in Arithmetic, Continued Fractions, Fractals, Mathematics and tagged , , , , , , , , , | Leave a comment

If you go back far enough, we’re all worms. All us animals, that is. But in a subtler sense, all life is vermiform — animals, plants, fungi, bacteria. DNA is a kind of worm, a string of chemicals encoding the recipe for an animal, plant, fungus or bacterium. And the worms of DNA can be turned into numbers, just as some numbers can be turned into worms:

3/7 = 0·0.428571428571428571428571…
154/183 = 0.841530054644808743169398907…
√2 = 1.414213562373095048801688…
π = 3.1415926535897932384626433…

Those are decimals, but there’s another kind of worm for such numbers. It’s called a continued fraction:

contfrac(3/7) = [0,2,3]
contfrac(154/183) = [0,1,5,3,4,2]
contfrac(√2) = [1,2,2,2,2,2…]
contfrac(π) = [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, 1, 84, 2, 1, 1, 15…]

Extracting and enacting continued fractions is very simple. Here’s the extracting:

3/7 → 1/(3/7) = 7/3 = 2+1/3 – 2 = 1/3 → 1(1/3) = 3, ∴ contfrac(3/7) = [0,2,3]
154/183 → 1/(154/183) = 183/154 = 1 + 29/154 – 1 = 29/154 → 1/(29/154) = 154/29 = 5 + 9/29 – 5 = 9/29 → 1/(9/29) = 29/9 = 3 + 2/9 – 3 = 2/9 → 1/(2/9) = 9/2 = 4 + 1/2 – 4 = 1/2 → 1/(1/2) = 2 – 2 = 0, ∴ contfrac(154/183) = [0,1,5,3,4,2]

And here’s the enacting:

[0,2,3] → 3 → 1/3 → 1/3 + 2 = 7/3 → 1/(7/3) = 3/7
[0,1,5,3,4,2] → 2 → 1/2 → 1/2 + 4 = 9/2 → 2/9 + 3 = 29/9 → 9/29 + 5 = 154/29 → 29/154 + 1 = 183/154 → 1/(183/154) = 154/183

Once you’ve got the worm of a continued fraction, you can perm the worm, as it were, generating different fractions like this (I’m dropping the initial [0,…] of the contfracs):

[2,3,4] = contfrac(13/30)
[2,4,3] → 13/29
[3,2,4] → 09/31
[3,4,2] → 09/29
[4,2,3] → 07/31
[4,3,2] → 07/30

Reversing a continued fraction is a kind of permutation, so the fractal below represents one kind of worms in terms of perms:

Variant of a limestone fractal or gryke fractal

I call that graph a fract-L, because it’s shaped like an L and the x axis represents the simplified fractions 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5…, while the y axis represents the fractions you get by reversing the continued fractions of 1/2, 1/3, 2/3…:

contfrac(1/2) = [2] → 1/2
contfrac(1/3) = [3] → 1/3
contfrac(2/3) = [1,2] → 1/3
contfrac(1/4) = [4] → 1/4
contfrac(3/4) = [1,3] → 1/4
contfrac(1/5) = [5] → 1/5
contfrac(2/5) = [2,2] → 2/5
contfrac(3/5) = [1,1,2] → 2/5
contfrac(4/5) = [1,4] → 1/5
contfrac(1/6) = [6] → 1/6
contfrac(5/6) = [1,5] → 1/6
contfrac(1/7) = [7] → 1/7
contfrac(2/7) = [3,2] → 3/7
contfrac(3/7) = [2,3] → 2/7
contfrac(4/7) = [1,1,3] → 2/7
contfrac(5/7) = [1,2,2] → 3/7
contfrac(6/7) = [1,6] → 1/7
contfrac(1/8) = [8] → 1/8
contfrac(3/8) = [2,1,2] → 3/8
contfrac(5/8) = [1,1,1,2] → 3/8
contfrac(7/8) = [1,7] → 1/8
contfrac(1/9) = [9] → 1/9
contfrac(2/9) = [4,2] → 4/9
contfrac(4/9) = [2,4] → 2/9
contfrac(5/9) = [1,1,4] → 2/9
contfrac(7/9) = [1,3,2] → 4/9
contfrac(8/9) = [1,8] → 1/9
[…]

If you perm the worm in other ways, you get other shapes on the fract-L. I looked at continued fractions of fixed length, 4, 5 and 6, and permed them using one of the permutations of [1,2,3,4], [1,2,3,4,5] and [1,2,3,4,5,6]. Here’s a graph for fractions, a/b, and permed fractions, perm(a/b), where length(contfrac(a/b)) = 4:

x = a/b when length(contfrac(a/b)) = 4, y = fraction from contfrac(a/b) permed with [1,3,2,4]

The x axis represents simplified fractions, a/b, when len(cf(a/b)) = 4. The y axis represents the fractions found by applying the perm [1,3,2,4] to contfrac(a/b). That is, the first number of the contfrac stays where it is, the third number moves to position 2, the second number moves to position 3 and the fourth number stays where it is. In short, you simply swap the middle two numbers of contfrac(a/b). Here’s an example:

contfrac(9/43) = [4,1,3,2] → [4,3,1,2] → 11/47, because contfrac(11/47) = [4,3,1,2]

Here are more fract-Ls representing worms in terms of perms:

fract-L for contfrac(a/b) permed by [2,1,3,4]

fract-L for contfrac(a/b) permed by [3,2,1,4]

fract-L for contfrac(a/b) permed by [1,4,2,3,5] (i.e. a/b where len(contfrac(a/b)) = 5)

fract-L for contfrac(a/b) permed by [1,5,3,4,2]

fract-L for contfrac(a/b) permed by [2,1,4,3,5]

fract-L for contfrac(a/b) permed by [3,4,1,2,5]

fract-L for contfrac(a/b) permed by [4,2,3,1,5]

fract-L for contfrac(a/b) permed by [4,2,5,3,1]

fract-L for contfrac(a/b) permed by [4,3,2,1,5]

fract-L for contfrac(a/b) permed by [5,3,4,2,1]

fract-L for contfrac(a/b) permed by [2,1,4,3,5,6] (i.e. a/b where len(contfrac(a/b)) = 6)

fract-L for contfrac(a/b) permed by [2,1,5,4,3,6]

fract-L for contfrac(a/b) permed by [3,2,1,4,5,6]

fract-L for contfrac(a/b) permed by [3,2,1,5,4,6]

fract-L for contfrac(a/b) permed by [3,5,1,4,2,6]

fract-L for contfrac(a/b) permed by [4,2,5,1,3,6]

fract-L for contfrac(a/b) permed by [4,3,2,1,5,6]

fract-L for contfrac(a/b) permed by [4,5,2,3,1,6]

fract-L for contfrac(a/b) permed by [1,3,2,6,5,4,7] (i.e. a/b where len(contfrac(a/b)) = 7)

fract-L for contfrac(a/b) permed by [1,5,2,6,3,4,7]

fract-L for contfrac(a/b) permed by [5,6,3,7,4,1,2]

fract-L for contfrac(a/b) permed by [6,2,3,5,4,1,7]

fract-L for contfrac(a/b) permed by [6,2,5,4,7,3,1]

Post-Performative Post-Scriptum

Much as I hate the phrase “in terms of”, I was happy to use it in the title of this post. After all, it isn’t ugly but assonant there. And it began life in mathematics, where it still has its proper meaning rather than being pretentious and prolix:

How did this complex preposition come into being? The OED [Oxford English Dictionary] reveals that it has been in use since the mid-18c. as a mathematical expression “said of a series…stated in terms involving some particular (my emphasis) quantity”, and illustrates this technical usage by citing examples from the work of Herbert Spencer (1862), J. F. W. Herschel (1866), and other writers. From this technical use came at first a trickle and, after the 1940s, a flood of imitative uses by non-mathematicians. — “Terminal Trinity”

Elsewhere Other-Engageable

• A Fracteasel on a Fract-L — an earlier look at continued fractions and fractal fract-Ls

A FracTeasel on a Fract-L

by in Arithmetic, Continued Fractions, Fractals, Mathematics and tagged , , , , , , , , , , , , | Leave a comment

Here are two new fractals, both of which remind me of the seedheads of the wildflower known as a teasel, Dipsacus fullonum:

A FracTeasel fractal

Dried seedheads of teasel, Dipsacus fullonum (Wikipedia)

Another FracTeasel fractal (embedded in the first)

Flowering seedhead of teasel, Dipsacus fullonum (Wikipedia)

How do you create the two FracTeasels? Let’s look first at the fractal they’re inspired by. In “Back to Frac’” I talked about this fractional fractal, a variant of what I call the limestone fractal:

Variant of a limestone fractal or gryke fractal

It’s a fractal on a fract-L, that is, the x and y co-ordinates of the red L represent pairs of fractions generating decimals between 0 and 1. The x represents the fractions a1/b1 = 1/n to (n-1)/n in simplest form: 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 5/6, 1/7, 2/7, 3/7, 4/7, 5/7, 6/7, 1/8, 3/8, 5/8, 7/8,…

And what about the y? It represents the fraction found by taking the continued fraction of a1/b1, reversing it, and generating a new fraction, a2/b2, from the reversal. For example, here’s the continued fraction of a1/b1 = 3/23 = 0.1304347826…:

contfrac(3/23) = 7,1,2

The continued fraction of a1/b1 = 3/23 is used like this to reconstruct a1/b1:

7,1,2

0 → 1 / (0 + 2) = 1/2 → 1 / (1/2 + 1) = 2/3 → 1 / (7 + 2/3) = 3/23

Now reverse the continued fraction, 7,1,2 → 2,1,7, and generate a2/b2:

2,1,7

0 → 1 / (0 + 7) = 1/7 → 1 / (1/7 + 1) = 7/8 → 1 / (2 + 7/8) = 8/23 = 0.3478260869565…

The limestone fractal above appears when a1/b1 → a2/b2 for a1/b1 = 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 5/6, 1/7, 2/7, 3/7, 4/7, 5/7, 6/7, 1/8, 3/8, 5/8, 7/8,… But you can do other things to contfrac(a1/b1) beside just reversing it. What about the permutations of contfrac(a1/b1), for example? If length(contfrac(a1/b1)) = n, the permutations can generate up to n! (factorial n) new a2/b2 for the y co-ordinate (if all the numbers of contfrac(a1/b1) are different, you’ll get n! permutations). The resultant fractal is the first of the FracTeasels above (note that a2/b2 isn’t multiplied by two):

FracTeasel #1 from fract-L for y = perm(contfrac(a1/b1))

If you think about it, you’ll see that the fractal from permed contfrac(a1/b1) contains the fractal from reversed contfrac(a1/b1). It also contains the second FracTeasel:

FracTeasel #2

How so? Because the second FracTeasel — let’s call it the stemmed FracTeasel — is created by shifting some numbers in contfrac(a1/b1) and leaving others alone. For example:

contfrac(940/1089) = 1, 6, 3, 4, 5, 2 → 1, 4, 3, 2, 5, 6 = contfrac(1008/1243)

So the function is finding one particular permutation of contfrac(a1/b1) to generate a2/b2, not all permutations. And so the function creates the stemmed FracTeasel, which carries an infinite number of seedheads on the same stem. To show that, here’s an animated gif zooming in on the bend of the fract-L for the stemmed FracTeasel:

Zooming the FracTeasel (animated at ezGif)

Elsewhere Other-Accessible…

• I Like Gryke — a first look at the limestone fractal
• Lime Time — more on the limestone fractal

Back to Frac’

by in Arithmetic, Continued Fractions, Fractals, Mathematics and tagged , , , , , , , , , | Leave a comment

Here’s a second serendipitous fractal:

A serendipitous fractal on a fract-L

It looks like (and is related to) the limestone fractal and I found it similarly serendipitously. This time I was looking at continued fractions, a simple yet subtle and seductive way of representing non-integer numbers like 2/3 and 7/9 (or √2 and π). To generate a continued fraction from a/b < 1, you divide a/b into 1 and take away the integer part. Then you repeat with the remainder until nothing is left (or, as with irrationals like 1/√2 and 1/π, you've calculated long enough for your needs). The integers at each stage are the numbers of the continued fraction. Here is the working for contfrac(2/3), the continued fraction of 2/3:

int(1/(2/3)) = int(3/2) = int(1.5) = 1
3/2 – 1 = 1/2
int(1/(1/2)) = int(2) = 2
2 – 2 = 0

contfrac(2/3) = 1, 2

By working backwards with (1, 2), you can use the continued fraction to reconstruct the original number a/b. Start with a/b = 0/1:

1 / (0/1 + 2) = 1 / ((0+2*1)/2) = 1 / (2/1) = 1/2
1 / (1/2 + 1) = 1 / ((1+2*1)/2) = 1 / (3/2) = 2/3

And here’s the working for contfrac(7/9), the continued fraction of 7/9:

int(1/(7/9)) = int(9/7) = int(1.285714…) = 1
9/7 – 1 = 2/7
int(1/(2/7)) = int(7/2) = int(3.5) = 3
7/2 – 3 = 1/2
int(1/(1/2)) = int(2) = 2
2 – 2 = 0

contfrac(7/9) = 1, 3, 2

And here’s the reconstruction of 7/9 from its continued fraction, starting again with a/b = 0/1:

1 / (0/1 + 2) = 1 / ((0+2*1)/2) = 1 / (2/1) = 1/2
1 / (1/2 + 3) = 1 / ((1+2*3)/2) = 1 / (7/2) = 2/7
1 / (2/7 + 1) = 1 / ((2+7*1)/7) = 1 / (9/7) = 7/9

From that simple algorithm arise subtle and seductive things. Look at some continued fractions, cf(a/b), for a/b in simplest form (giving only the first few reciprocals, 1/b, because cf(1/b) = b). Interesting patterns appear, e.g. when a/b uses adjacent or nearly adjacent Fibonacci numbers:

cf(1/3) = 3 = cf(0.333333333…)
cf(2/3) = 1,2 = cf(0.666666666…)
cf(1/4) = 4 = cf(0.25)
cf(3/4) = 1,3 = cf(0.75)
cf(1/5) = 5 = cf(0.2)
cf(2/5) = 2,2 = cf(0.4)
cf(3/5) = 1,1,2 = cf(0.6)
cf(4/5) = 1,4 = cf(0.8)
cf(5/6) = 1,5 = cf(0.833333333…)
cf(2/7) = 3,2 = cf(0.285714285…)
cf(3/7) = 2,3 = cf(0.428571428…)
cf(4/7) = 1,1,3 = cf(0.571428571…)
cf(5/7) = 1,2,2 = cf(0.714285714…)
cf(6/7) = 1,6 = cf(0.857142857…)
cf(3/8) = 2,1,2 = cf(0.375)
cf(5/8) = 1,1,1,2 = cf(0.625)
cf(7/8) = 1,7 = cf(0.875)
cf(2/9) = 4,2 = cf(0.222222222…)
cf(4/9) = 2,4 = cf(0.444444444…)
cf(5/9) = 1,1,4 = cf(0.555555555…)
cf(7/9) = 1,3,2 = cf(0.777777777…)
cf(8/9) = 1,8 = cf(0.888888888…)
cf(3/10) = 3,3 = cf(0.3)
cf(7/10) = 1,2,3 = cf(0.7)
cf(9/10) = 1,9 = cf(0.9)
cf(2/11) = 5,2 = cf(0.181818181…)
cf(3/11) = 3,1,2 = cf(0.272727272…)
cf(4/11) = 2,1,3 = cf(0.363636363…)
cf(5/11) = 2,5 = cf(0.454545454…)
cf(6/11) = 1,1,5 = cf(0.545454545…)
cf(7/11) = 1,1,1,3 = cf(0.636363636…)
cf(8/11) = 1,2,1,2 = cf(0.727272727…)
cf(9/11) = 1,4,2 = cf(0.818181818…)
cf(10/11) = 1,10 = cf(0.909090909…)
cf(5/12) = 2,2,2 = cf(0.416666666…)
cf(7/12) = 1,1,2,2 = cf(0.583333333…)
cf(11/12) = 1,11 = cf(0.916666666…)
cf(2/13) = 6,2 = cf(0.153846153…)
cf(3/13) = 4,3 = cf(0.230769230…)
cf(4/13) = 3,4 = cf(0.307692307…)
cf(5/13) = 2,1,1,2 = cf(0.384615384…)
cf(6/13) = 2,6 = cf(0.461538461…)
cf(7/13) = 1,1,6 = cf(0.538461538…)
cf(8/13) = 1,1,1,1,2 = cf(0.615384615…)
cf(9/13) = 1,2,4 = cf(0.692307692…)
cf(10/13) = 1,3,3 = cf(0.769230769…)
cf(11/13) = 1,5,2 = cf(0.846153846…)
cf(12/13) = 1,12 = cf(0.923076923…)
cf(3/14) = 4,1,2 = cf(0.214285714…)
cf(5/14) = 2,1,4 = cf(0.357142857…)
cf(9/14) = 1,1,1,4 = cf(0.642857142…)
cf(11/14) = 1,3,1,2 = cf(0.785714285…)
cf(13/14) = 1,13 = cf(0.928571428…)
cf(2/15) = 7,2 = cf(0.133333333…)
cf(4/15) = 3,1,3 = cf(0.266666666…)
cf(7/15) = 2,7 = cf(0.466666666…)
cf(8/15) = 1,1,7 = cf(0.533333333…)
cf(11/15) = 1,2,1,3 = cf(0.733333333…)
cf(13/15) = 1,6,2 = cf(0.866666666…)
cf(14/15) = 1,14 = cf(0.933333333…)
cf(3/16) = 5,3 = cf(0.1875)
cf(5/16) = 3,5 = cf(0.3125)
cf(7/16) = 2,3,2 = cf(0.4375)

After investigating some of those patterns, I wondered what happened when you reversed the continued fraction cf(a/b) and used those reversed numbers backward (that is, used the numbers of cf(a/b) forward) to generate another and different a/b. And a/b will always be different unless cf(a/b) is a palindrome, like cf(5/12) = 2,2,2 or cf(5/13) = 2,1,1,2 or cf(4/15) = 3,1,3. Note that a continued fraction never ends in 1, so that when reversing, say, cf(5/8) = (1, 1, 1, 2), you need an adjustment from (2, 1, 1, 1) to (2, 1, 1+1) = (2, 1, 2). Here’s a little of what happens when you reverse cf(a1/b1) to generate a2/b2:

cf(1/2) = 2 → 2 = cf(1/2)
1/2 = 0.5 : 0.5 = 1/2
cf(1/3) = 3 → 3 = cf(1/3)
1/3 = 0.333333333 : 0.333333333 = 1/3
cf(2/3) = 1, 2 → 2, 1 → 3 = cf(1/3)
2/3 = 0.666666666 : 0.333333333 = 1/3
cf(3/4) = 1, 3 → 3, 1 → 4 = cf(1/4)
3/4 = 0.75 : 0.25 = 1/4
cf(2/5) = 2, 2 → 2, 2 = cf(2/5)
2/5 = 0.4 : 0.4 = 2/5
cf(3/5) = 1, 1, 2 → 2, 1, 1 → 2, 2 = cf(2/5)
3/5 = 0.6 : 0.4 = 2/5
cf(4/5) = 1, 4 → 4, 1 → 5 = cf(1/5)
4/5 = 0.8 : 0.2 = 1/5
cf(5/6) = 1, 5 → 5, 1 → 6 = cf(1/6)
5/6 = 0.833333333 : 0.166666666 = 1/6
cf(2/7) = 3, 2 → 2, 3 = cf(3/7)
2/7 = 0.285714286 : 0.428571428 = 3/7
cf(3/7) = 2, 3 → 3, 2 = cf(2/7)
3/7 = 0.428571429 : 0.285714286 = 2/7
cf(4/7) = 1, 1, 3 → 3, 1, 1 → 3, 2 = cf(2/7)
4/7 = 0.571428571 : 0.285714286 = 2/7
cf(5/7) = 1, 2, 2 → 2, 2, 1 → 2, 3 = cf(3/7)
5/7 = 0.714285714 : 0.428571429 = 3/7
cf(6/7) = 1, 6 → 6, 1 → 7 = cf(1/7)
6/7 = 0.857142857 : 0.142857143 = 1/7
cf(3/8) = 2, 1, 2 → 2, 1, 2 = cf(3/8)
0.375 : 0.375
cf(5/8) = 1, 1, 1, 2 → 2, 1, 1, 1 → 2, 1, 2 = cf(3/8)
0.625 : 0.375
cf(7/8) = 1, 7 → 7, 1 → 8 = cf(1/8)
0.875 : 0.125
cf(2/9) = 4, 2 → 2, 4 = cf(4/9)
0.222222222 : 0.444444444
cf(4/9) = 2, 4 → 4, 2 = cf(2/9)
0.444444444 : 0.222222222

And if you plot x = a1/b1 and y = (a2/b2 * 2) on a fract-L, that is, a graph whose horizontal and vertical arms represent 0 to 1, you get the fractal right at the beginning:

Fract-L for x = a1/b1 and y = (a2/b2 * 2), where a2/b2 is generated from reversed(cf(a1/b1))

You need to use (a2/b2 * 2) because a2/b2 from reversed(cf(a1/b1)) is always <= 0.5, so using raw a2/b2 generates this graph:

Fract-L for x = a1/b1 and y = a2/b2 (i.e. a2/b2 is unadjusted)

Why is it always true that a2/b2 <= 0.5? For two reasons. First, a/b > 0.5 always generate continued fractions that start with 1, like cf(2/3) = 1, 2 or cf(3/4) = 1, 3 or cf(3/5) = 1, 1, 2. Second, as previously mentioned, no continued fraction ends with 1. Therefore a reversed cf(a1/b1), where the final number, n > 1, moves to the beginning, will never begin with 1 and the a2/b2 generated from reversed(cf(a1/b1)) will always be less than 0.5 (or equal to it in the solitary case of cf(1/2) = 2).

Now let's look at the development of the fractal as a1/b1 uses larger and larger denominators:

Fract-L for x = a1/b1 and y = (a2/b2 * 2) for a1/b1 <= 6/7

Fract-L for for a1/b1 <= 14/15

Fract-L for a1/b1 <= 30/31

Fract-L for a1/b1 <= 62/63

Fract-L for a1/b1 <= 126/127

Fract-L for a1/b1 <= 254/255

Fract-L for a1/b1 <= 357/358

Fract-L for a1/b1 <= 467/468

Animated fract-L for x = a1/b1 and y = (a2/b2 * 2) (animated at ezGif)

The fractal changes subtly when you restrict the b1 of a1/b1 in some way, say using multiples of 2, 3, 4, 5…:

Fract-L for x = a1/b1 and y = (a2/b2 * 2) for b1 = n = 2, 3, 4, 5, 6, 7, 8…

Fract-L for b1 = 2n = 2, 4, 6, 8, 10…

Fract-L for b1 = 3n = 3, 6, 9, 12, 15…

Fract-L for b1 = 4n

Fract-L for b1 = 5n

Fract-L for b1 = 6n

Animated fract-L for b1 = 1n..12n (animated at ezGif)

Finally, here are fract-Ls when b1 is a triangular, square, hexagonal or octagonal number:

Fract-L for x = a1/b1 and y = (a2/b2 * 2) for triangular(b1) = 3, 6, 10, 15, 21, 28,…

Fract-L for square(b1) = 4, 9, 16, 25, 36, 49,…

Fract-L for hexagonal(b1) = 6, 15, 28, 45, 66, 91,…

Fract-L for octagonal(b1) = 8, 21, 40, 65, 96, 133,…

Elsewhere Other-Accessible…

• Back to Drac’ — a parallel pun for a pre-previous fractal
• I Like Gryke — a first look at the limestone fractal
• Lime Time — more on the limestone fractal

Fractional Fractal Fract-Ls

by in Fractals, Mathematics and tagged , , , , , , , , , , , , , , , , | Leave a comment

This is the surpassingly special Stern-Brocot sequence:

0, 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4, 1, 5, 4, 7, 3, 8, 5, 7, 2, 7, 5, 8, 3, 7, 4, 5, 1, 6, 5, 9, 4, 11, 7, 10, 3, 11, 8, 13, 5, 12, 7, 9, 2, 9, 7, 12, 5, 13, 8, 11, 3, 10, 7, 11, 4, 9, 5, 6, 1, 7, 6, 11, 5, 14, 9, 13, 4, 15, 11, 18, 7, 17, 10, 13, 3, 14, 11, 19, 8, 21, 13, 18, 5, 17, 12, 19, … (A002487 at the Online Encyclopedia of Integer Sequences)

And why is the sequence special? Because if you take successive pairs of the apparently arbitrarily varying numbers, you get every rational fraction in its simplest form exactly once. So 1/2, 2/3, 6/11 and 502/787 appear once and then never again. And so do 2/1, 3/2, 11/6 and 787/502. Et cetera, ad infinitum. If you map the Stern-Brocot sequence against the related Calkin-Wilk sequence, which has the same “all-simplest-fractions-exactly-once” properties, you can create this fractal, which I call a limestone fractal or gryke fractal:

Gryke fractal by mapping Stern-Brocot sequence against Calkin-Wilf sequence

The graph is what I call a Fract-L, because the lines for the x,y coordinates create an L. Each coordinate runs from 0 to 1, with the x set by the fraction from the Stern-Brocot sequence and the y set by the fraction from the Calkin-Wilf sequence (if a > b in a/b, use the conversion 1/(a/b) = b/a). But you can also find interesting patterns by mapping the Stern-Brocot sequence against itself. That is, you use two Stern-Brocot sequences that start in different places. Now, there are complicated ways to create the Stern-Brocot sequence using mathematical trees and sequential algorithms and so on. But there’s also an astonishingly simple way, a formula created by the Israeli mathematician Moshe Newman. If (a,b) is one pair of successive numbers in the sequence, the next pair (a,b) is found like this:

c = b
b = (2 * int(a/b) + 1) * b – a
a = c

This means that you can seed a Stern-Brocot sequence with any (correctly simplified) a/b and it will continue in the right way. If the two SB-sequences for x and y are both seeded with (0,1), you get this 45° line, because each successive a/b for (x,y) is identical:

Stern-Brocot pairs seeded with x ← (0,1) and y ← (0,1)

The further you extend the sequences, the less broken the 45° line will appear, because the points determined by a/b for x and y will get closer and closer together (but the line will never be solid, because any two rationals are separated by an infinity of irrationals). Now try offsetting the SB-sequences for x,y by using different seeds. Different fractal patterns appear, which all appear to be subsets (or fractions) of the limestone fractal above (see animated gif below):

Stern-Brocot pairs seeded with x ← (0,1) and y ← (1,1)

x ← (0,1) and y ← (1,2)

x ← (0,1) and y ← (1,3)

x ← (0,1) and y ← (2,3)

x ← (0,1) and y ← (3,4)

x ← (0,1) and y ← (6,7)

x ← (1,2) and y ← (1,9)

x ← (1,4) and y ← (1,6)

x ← (1,7) and y ← (1,8)

x ← (2,3) and y ← (4,5) — apparently identical to x ← (1,4) and y ← (1,6) above

x ← (26,25) and y ← (1,10)

Gryke fractal compared with Stern-Brocot-pair patterns (animated at ezGif)

And here’s what happens when the seed-fractions for x run from 1/3 to 12/13, while the seed-fraction for y is held constant at 1/23:

x ← (1,13) and y ← (1,23)

x ← (2,13) and y ← (1,23)

x ← (3,13) and y ← (1,23)

x ← (4,13) and y ← (1,23)

x ← (5,13) and y ← (1,23)

x ← (6,13) and y ← (1,23)

x ← (7,13) and y ← (1,23)

x ← (8,13) and y ← (1,23)

x ← (9,13) and y ← (1,23)

x ← (10,13) and y ← (1,23)

x ← (11,13) and y ← (1,23)

x ← (12,13) and y ← (1,23)

Animated gif for x ← (n,13) and y ← (1,23) (animated at ezGif)

Previously Pre-Posted

• I Like Gryke — a first look at the limestone fractal
• Lime Time — more on the fractal

Fractional Fractal

by in Arithmetic, Fractals, Geometry, Mathematics | Leave a comment

Serendipity in some simplification statistics. That’s what I encountered the other day. I was looking at the ways to simplify this set of fractions:

1/2, 1/3, 2/3, 1/4, 2/4, 3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 2/6, 3/6, 4/6, 5/6, 1/7, 2/7, 3/7, 4/7, 5/7, 6/7, 1/8, 2/8, 3/8, 4/8, 5/8, 6/8, 7/8, 1/9, 2/9, 3/9, 4/9, 5/9, 6/9, 7/9, 8/9, 1/10, 2/10, 3/10, 4/10, 5/10, 6/10, 7/10, 8/10, 9/10, 1/11, 2/11, 3/11, …

The underlined fractions are not in their simplest possible form. For example, 2/4 simplifies to 1/2, 2/6 to 1/3, 3/6 to 1/2, and so on:

1/2, 1/3, 2/3, 1/4, 2/4 → 1/2, 3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 2/6 → 1/3, 3/6 → 1/2, 4/6 → 2/3, 5/6, 1/7, 2/7, 3/7, 4/7, 5/7, 6/7, 1/8, 2/8 → 1/4, 3/8, 4/8 → 1/2, 5/8, 6/8 → 3/4, 7/8, 1/9, 2/9, 3/9 → 1/3, 4/9, 5/9, 6/9 → 2/3, 7/9, 8/9, 1/10, 2/10 → 1/5, 3/10, 4/10 → 2/5, 5/10 → 1/2, 6/10 → 3/5, 7/10, 8/10 → 4/5, 9/10, 1/11, 2/11, 3/11, …

I counted the number of times the simplest possible fractions occurred when simplifying all fractions a/b for (b = 2..n, a = 1..b-1), then displayed the stats as a graph running from 0/1 to 1/1. And there was serendipity in these simplification statistics, because a fractal had appeared:

Graph for count of simplest possible fractions from a/b for (b = 2..n, a = 1..b-1)

It’s interesting to work out what fractions appear where. For example, the peak in the middle is 1/2, but what are the next-highest peaks on either side? The answers are more obvious in the colored version of the graph:

Colored graph for count of simplest possible fractions
Key: n/2, n/3, n/4, n/5, n/6, n/7

The line for 1/2 is red, the lines for 1/3 and 2/3 in light green, the lines for 1/4 and 3/4 in yellow, and so on. But those graphs appear in an equilateral triangle, as it were. The fractals get easier to see in the full-sized versions of these widened graphs:

Widened graph for count of simplest possible fractions from a/b for (b = 2..n, a = 1..b-1)

(click for full size)

Widened and colored graph for count of simplest possible fractions

(click for full size)

The Wyrm Ferns

by in Aesthetics, Dragon Curves, Fractals, Geometry, Mathematics and tagged , , , , , , | Leave a comment

A fern is a fractal, a shape that contains copies of itself at smaller and smaller scales. That is, part of a fern looks like the fern as a whole:

Fern as fractal (source)

Millions of years after Mother Nature, man got in on the fract, as it were:

The Sierpiński triangle, a 2d fractal

The Sierpiński triangle is a fractal created in two dimensions by a point jumping halfway towards one or another of the three vertices of a triangle. And here is a fractal created in one dimension by a point jumping halfway towards one or another of the two ends of a line:

A 1d fractal

In one dimension, the fractality of the fractal isn’t obvious. But you can try draggin’ out (or dragon out) the fractality of the fractal by ferning the wyrm, as it were. Suppose that after the point jumps halfway towards one or another of the two points, it’s rotated by some angle around the midpoint of the two original points. When you do that, the fractal becomes more and more obvious. In fact, it becomes what’s called a dragon curve (in Old English, “dragon” was wyrm or worm):

Fractal with angle = 5°

Fractal 10°

Fractal 15°

Fractal 20°

Fractal 25°

Fractal 30°

Fractal 35°

Fractal 40°

Fractal 45°

Fractal 50°

Fractal 55°

Fractal 60°

Fractal 0° to 60° (animated at ezGif)

But as the angle gets bigger, an interesting aesthetic question arises. When is the ferned wyrm, the dragon curve, at its most attractive? I’d say it’s when angle ≈ 55°:

Fractal 50°

Fractal 51°

Fractal 52°

Fractal 53°

Fractal 54°

Fractal 55°

Fractal 56°

Fractal 57°

Fractal 58°

Fractal 59°

Fractal 60°

Fractal 50° to 60° (animated)

At angle >= 57°, I think the dragon curve starts to look like some species of bristleworm, which are interesting but unattractive marine worms:

A bristleworm, Nereis virens (see polychaete at Wikipedia)

Finally, here’s what the ferned wyrm looks like in black-and-white and when it’s rotating:

Fractal 0° to 60° (b&w, animated)

Fractal 56° (rotating)

Fractal 56° (b&w, rotating)

Double fractal 56° (b&w, rotating)

Previously Pre-Posted (Please Peruse)…

• Curvous Energy — a first look at dragon curves
• Back to Drac’ — another look at dragon curves