Match of the Day

Some interesting shapes are mentioned in Derrick Niederman’s Number Freak (2010). Using identical matchsticks, what’s the smallest fully connected shape you can make in which two matches meet at every vertex? That is, what is the smallest 2-regular matchstick graph?

It’s an equilateral triangle:

2match

Now, what is the smallest fully connected shape you can make in which three matches meet at every vertex? That is, what is the smallest 3-regular matchstick graph? It uses twelve identical matches and looks like this:

3match

And here is the smallest known 4-regular matchstick graph, discovered by the German mathematician Heiko Harborth and using 104 identical matches:

4match

But Niederman says that “it’s impossible to create any arrangement in which five or more matchsticks meet at every vertex” (entry for “104”, pg. 230 of the 2012 paperback).

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Radical Sheet

If you take a sheet of standard-sized paper and fold it in half from top to bottom, the folded sheet has the same proportions as the original, namely √2 : 1. In other words, if x = √2 / 2, then 1 / x = √2:

√2 = 1.414213562373…, √2 / 2 = 0.707106781186…, 1 / 0.707106781186… = 1.414213562373…

So you could say that paper has radical sheet (the square or other root of a number is also called its radix and √ is known as the radical sign). When a rectangle has the proportions √2 : 1, it can be tiled with an infinite number of copies of itself, the first copy having ½ the area of the original, the second ¼, the third ⅛, and so on. The radical sheet below is tiled with ten diminishing copies of itself, the final two having the same area:

papersizes

papersizes_static

You can also tile a radical sheet with six copies of itself, two copies having ¼ the area of the original and four having ⅛:

paper_6div_static

paper_6div

This tiling is when you might say the radical turns crucial, because you can create a fractal cross from it by repeatedly dividing and discarding. Suppose you divide a radical sheet into six copies as above, then discard two of the ⅛-sized rectangles, like this:

paper_cross_1

Stage 1


Then repeat with the smaller rectangles:

paper_cross_2

Stage 2


paper_cross_3

Stage 3


paper_cross_4

Stage 4


paper_cross_5

Stage 5


paper_cross

Animated version

paper_cross_static

Fractile cross

The cross is slanted, but it’s easy to rotate the original rectangle and produce an upright cross:

paper_cross_upright

paper_cross_upright_static

Toxic Turntable #4

Currently listening…

• Yott, Baal (1971)
• If, Om-Serpent (1989)
• Sculptrix Sabulorum, Sables Chantantes (2005)
• Binksh, Contempletion (1983)


Previously pre-posted:

Toxic Turntable #1
Toxic Turntable #2
Toxic Turntable #3

Performativizing the Polygonic

Maths is a mountain: you can start climbing in different places and reach the same destination. There are many ways of proving the irrationality of √2 or the infinitude of the primes, for example. But you can also arrive at the same destination by accident. I’ve found that when I use different methods of creating fractals. The same fractals appear, because apparently different algorithms are actually the same underneath.

But different methods can create unique fractals too. I’ve found some new ones by using what might be called point-to-point recursion. For example, there are ten ways to select three vertices from the five vertices of a pentagon: (1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 3, 4), (1, 3, 5), (1, 4, 5), (2, 3, 4), (2, 3, 5), (2, 4, 5), (3, 4, 5). Find the midpoint of the first three-point set, (1, 2, 3). Then select two vertices to go with this midpoint, creating a new three-point set, and find the midpoint again. And so on. The process looks like this, with the midpoints shown for all the three-point sets found at each stage:

v5_p3_stage1

vertices = 5, choose sets of 3 points, find mid-point of each

v5_p3_stage2

v5_p3_stage3


At stage 5, the fractal looks like this:

v5_p3_static

v = 5, p = 3


Note that when pixels are used again, the colour changes. That’s another interesting thing about maths: limits can sometimes produce deeper results. If these fractals were drawn at very high resolution, pixels would only be used once and the colour would never change. As it is, low resolution means that pixels are used again and again. But some are used more than others, which is why interesting colour effects appear.

If the formation of the fractal is animated, it looks like this (with close-ups of even deeper stages):
v5_p3


Here are some more examples:

v4c_p2_static

v = 4 + central point, p = 2 (cf. Fingering the Frigit)

v4c_p2

v = 4c, p = 2 (animated)


v4_p3_static

v = 4, p = 3

v4_p3


v5_p4_static

v = 5, p = 4

v5_p4


v5c_p3_static

v = 5 + central point, p = 3

v5c_p3


v5c_p4

v = 5c, p = 4


v5c_p5

v = 5c, p = 5


v6_1_p6

v = 6 + 1 point between each pair of vertices, p = 6


v6_p2

v = 6, p = 2


v6_p3_static

v = 6, p = 3

v6_p3


v6_p4

v = 6, p = 4


v6c_p2_static

v = 6c, p = 2 (cf. Fingering the Frigit)

v6c_p2


v6c_p3_static

v = 6c, p = 3

v6c_p3


v6c_p4

v = 6c, p = 4


v7_p3

v = 7, p = 3


v7_p4_static

v = 7, p = 4

v7_p4


v7_p5_static

v = ,7 p = 5

v7_p5


v7_p4

v = 7c, p = 4


v3_1_p2

v = 3+1, p = 2


v3_1_p3

v = 3+1, p = 3


v3_1_p4

v = 3+1, p = 4


v3_2_p5

v = 3+2, p = 5


v3c_1_p2

v = 3c+1, p = 2


v3c_1_p4

v = 3c+1, p = 4


v3c_p2

v = 3c, p = 2


v3c_p3

v = 3c, p = 3


v4_1_p3

v = 4+1, p = 3


v4_1_p4

v = 4+1, p = 4


v4_1_p5

v = 4+1, p = 6


v4_1_p6

v = 4+1, p = 2


v4c_1_p4

v = 4c+1, p = 4


v4c_p3_static

v = 4c, p = 3

v4c_p3


v5_1_p4_va

v = 5+1, p = 4 (and more)


v5_p2

v = 5, p = 2


I Have a Threem

And now I have another. My first “threem”, or “three-M”, for this bijou bloguette was the alliterative three-word Latin phrase Mathematica Magistra Mundi, meaning “Mathematics Mistress of the World”. I also use it in the form Mathematica Machina Mundi, which has a variety of translations. In both Latin phrases, the words have five, three and two syllables, respectively. That’s the first three prime numbers in reverse and also part of the Fibonacci sequence in reverse.

You can find the same alliteration in languages derived from Latin, like the French La Mathématique, Maîtresse du Monde, but you don’t get the same syllable-count. So how likely was it that everything – the same alliteration and the same syllable-count – would appear in a language unrelated to Latin? But it does. Here’s a Georgian translation of the threem:

მათემატიკა მსოფლიოს მეფე

Matemat’ik’a Msoplios Mepe

“Mathematics the World’s King”

Msoplios isn’t a typo: Georgian is famous for its exotic consonant clusters and მს- / ms- isn’t a particularly unusual example. But it’s one I particularly like.

He Say, He Sigh, He Sow #36

• “By the time I was twenty-four I had constructed a complete system of philosophy. It rested on two principles: The Relativity of Things and The Circumferentiality of Man. I have since discovered that the first was not a very original discovery. It may be that the other was profound, but though I have racked my brains I cannot for the life of me remember what it was.” — W. Somerset Maugham, The Summing Up (1938), sec. 66.

Digital Rodeo

What a difference a digit makes. Suppose you take all representations of n in bases b <= n. When n = 3, the bases are 2 and 3, so 3 = 11 and 10, respectively. Next, count the occurrences of the digit 1:

digitcount(3, digit=1, n=11, 10) = 3

Add this digit-count to 3:

3 + digitcount(3, digit=1, n=11, 10) = 3 + 3 = 6.

Now apply the same procedure to 6. The bases will be 2 to 6:

6 + digitcount(6, digit=1, n=110, 20, 12, 11, 10) = 6 + 6 = 12

The procedure, n = n + digitcount(n,digit=1,base=2..n), continues like this:

12 + digcount(12,dig=1,n=1100, 110, 30, 22, 20, 15, 14, 13, 12, 11, 10) = 12 + 11 = 23
23 + digcount(23,dig=1,n=10111, 212, 113, 43, 35, 32, 27, 25, 23, 21, 1B, 1A, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 23 + 21 = 44
44 + digcount(44,dig=1,n=101100, 1122, 230, 134, 112, 62, 54, 48, 44, 40, 38, 35, 32, 2E, 2C, 2A, 28, 26, 24, 22, 20, 1L, 1K, 1J, 1I, 1H, 1G, 1F, 1E, 1D, 1C, 1B, 1A, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 44 + 31 = 75

And the sequence develops like this:

3, 6, 12, 23, 44, 75, 124, 202, 319, 503, 780, 1196, 1824, 2766, 4191, 6338, 9546, 14383, 21656, 32562, 48930, 73494, 110361, 165714, 248733, 373303, 560214, 840602, 1261237, 1892269, 2838926, 4258966, 6389157, 9584585, 14377879…

Now try the same procedure using the digit 0: n = n + digcount(n,dig=0,base=2..n). The first step is this:

3 + digcount(3,digit=0,n=11, 10) = 3 + 1 = 4

Next come these:

4 + digcount(4,dig=0,n=100, 11, 10) = 4 + 3 = 7
7 + digcount(7,dig=0,n=111, 21, 13, 12, 11, 10) = 7 + 1 = 8
8 + digcount(8,dig=0,n=1000, 22, 20, 13, 12, 11, 10) = 8 + 5 = 13
13 + digcount(13,dig=0,n=1101, 111, 31, 23, 21, 16, 15, 14, 13, 12, 11, 10) = 13 + 2 = 15
15 + digcount(15,dig=0,n=1111, 120, 33, 30, 23, 21, 17, 16, 15, 14, 13, 12, 11, 10) = 15 + 3 = 18
18 + digcount(18,dig=0,n=10010, 200, 102, 33, 30, 24, 22, 20, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 18 + 9 = 27
27 + digcount(27,dig=0,n=11011, 1000, 123, 102, 43, 36, 33, 30, 27, 25, 23, 21, 1D, 1C, 1B, 1A, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 27 + 7 = 34
34 + digcount(34,dig=0,n=100010, 1021, 202, 114, 54, 46, 42, 37, 34, 31, 2A, 28, 26, 24, 22, 20, 1G, 1F, 1E, 1D, 1C, 1B, 1A, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 34 + 8 = 42
42 + digcount(42,dig=0,n=101010, 1120, 222, 132, 110, 60, 52, 46, 42, 39, 36, 33, 30, 2C, 2A, 28, 26, 24, 22, 20, 1K, 1J, 1I, 1H, 1G, 1F, 1E, 1D, 1C, 1B, 1A, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 42 + 9 = 51

The sequence develops like this:

3, 4, 7, 8, 13, 15, 18, 27, 34, 42, 51, 59, 62, 66, 80, 94, 99, 111, 117, 125, 132, 151, 158, 163, 173, 180, 204, 222, 232, 244, 258, 279, 292, 307, 317, 324, 351, 364, 382, 389, 400, 425, 437, 447, 454, 466, 475, 483, 494, 509, 517, 536, 553, 566, 576, 612, 637, 649, 669, 679, 693, 712, 728, 753, 768, 801, 822, 835, 849, 862, 869, 883, 895, 906, 923, 932, 943, 949, 957, 967, 975, 999, 1011…

If you compare it with the sequence for digit=1, it appears that digcount(n,dig=1,b=2..n) is always larger than digcount(n,dig=0,b=2..n). That is in fact the case, with one exception, when n = 2:

digcount(2,dig=1,n=10) = 1
digcount(2,dig=0,n=10) = 1

When n = 10 (in base ten), there are twice as many ones as zeros:

digcount(10,dig=1,n=1010, 101, 22, 20, 14, 13, 12, 11, 10) = 10
digcount(10,dig=0,n=1010, 101, 22, 20, 14, 13, 12, 11, 10) = 5

As n gets larger, the difference grows dramatically:

digcount(100,dig=1,base=2..n) = 64
digcount(100,dig=0,base=2..n) = 16

digcount(1000,dig=1,base=2..n) = 533
digcount(1000,dig=0,base=2..n) = 25

digcount(10000,dig=1,base=2..n) = 5067
digcount(10000,dig=0,base=2..n) = 49

digcount(100000,dig=1,base=2..n) = 50140
digcount(100000,dig=0,base=2..n) = 73

digcount(1000000,dig=1,base=2..n) = 500408
digcount(1000000,dig=0,base=2..n) = 102

digcount(10000000,dig=1,base=2..n) = 5001032
digcount(10000000,dig=0,base=2..n) = 134

digcount(100000000,dig=1,base=2..n) = 50003137
digcount(100000000,dig=0,base=2..n) = 160

In fact, digcount(n,dig=1,b=2..n) is greater than the digit-count for any other digit: 0, 2, 3, 4, 5… (with the exception n = 2, as shown above). But digit=0 sometimes beats digits >= 2. For example, when n = 18:

digcount(18,dig=0,n=10010, 200, 102, 33, 30, 24, 22, 20, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 9
digcount(18,dig=2,n=10010, 200, 102, 33, 30, 24, 22, 20, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 7
digcount(18,dig=3,n=10010, 200, 102, 33, 30, 24, 22, 20, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 4
digcount(18,dig=4,n=10010, 200, 102, 33, 30, 24, 22, 20, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 2
digcount(18,dig=5,n=10010, 200, 102, 33, 30, 24, 22, 20, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 1

But as n gets larger, digcount(0) will fall permanently behind all these digits. However, digcount(0) will always be greater than some digit d, for the obvious reason that some digits only appear when the base is high enough. For example, the hexadecimal digit A (with the decimal value 10) first appears when n = 21:

digcount(21,dig=A,n=10101, 210, 111, 41, 33, 30, 25, 23, 21, 1A, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 1 digcount(21,dig=0,n=10101, 210, 111, 41, 33, 30, 25, 23, 21, 1A, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10) = 5

There is a general rule for the n at which digit d first appears, n = 2d + 1 (this doesn’t apply when d = 0 or d = 1):

d = 2, n = 5 = 2*2 + 1
digcount(5,dig=2,n=101, 12, 11, 10) = 1

d = 3, n = 7 = 2*3 + 1
digcount(7,dig=3,n=111, 21, 13, 12, 11, 10) = 1

d = 4, n = 9 = 2*4 + 1
digcount(9,dig=4,n=1001, 100, 21, 14, 13, 12, 11, 10) = 1

d = 5, n = 11 = 2*5 + 1
digcount(11,dig=5,n=1011, 102, 23, 21, 15, 14, 13, 12, 11, 10) = 1

It should be apparent, then, that the digit-count for a particular digit starts at 1 and gets gradually higher. The rate at which the digit-count increases is highest for 1 and lowest for 0, with digits 2, 3, 4, 5… in between:

All-Base Graph

Graph for digcount(n,dig=d,b=2..n)


You could think of the graph as a digital rodeo in which these digits compete with each other. 1 is the clear and permanent winner, 0 the gradual loser. Now recall the procedure introduced at the start: n = n + digcount(n,dig=d,b=2..n). When it’s applied to the digits 0 to 5, these are the sequences that appear:

n = n + digcount(n,dig=0,b=2..n)

2, 3, 4, 7, 8, 13, 15, 18, 27, 34, 42, 51, 59, 62, 66, 80, 94, 99, 111, 117, 125, 132, 151, 158, 163, 173, 180, 204, 222, 232, 244, 258, 279, 292, 307, 317, 324, 351, 364, 382, 389, 400, 425, 437, 447, 454, 466, 475, 483, 494, 509, 517, 536, 553, 566, 576, 612, 637, 649, 669, 679, 693, 712, 728, 753, 768, 801, 822, 835, 849, 862, 869, 883, 895, 906, 923, 932, 943, 949, 957, 967, 975, 999, 1011…

n = n + digcount(n,dig=1,b=2..n)

2, 3, 6, 12, 23, 44, 75, 124, 202, 319, 503, 780, 1196, 1824, 2766, 4191, 6338, 9546, 14383, 21656, 32562, 48930, 73494, 110361, 165714, 248733, 373303, 560214, 840602, 1261237, 1892269, 2838926, 4258966, 6389157, 9584585, 14377879…

n = n + digcount(n,dig=2,b=2..n)

5, 6, 8, 12, 16, 22, 31, 37, 48, 60, 76, 94, 115, 138, 173, 213, 257, 311, 374, 454, 542, 664, 790, 935, 1109, 1310, 1552, 1835, 2167, 2548, 2989, 3509, 4120, 4832, 5690, 6687, 7829, 9166, 10727, 12568, 14697, 17182, 20089, 23470, 27425, 32042, 37477, 43768, 51113, 59687, 69705, 81379, 94998, 110910, 129488, 151153, 176429, 205923, 240331, 280490, 327396, 382067, 445858…

n = n + digcount(n,dig=3,b=2..n)

7, 8, 9, 10, 11, 13, 16, 18, 22, 25, 29, 34, 38, 44, 50, 56, 63, 80, 90, 104, 113, 131, 151, 169, 188, 210, 236, 261, 289, 320, 350, 385, 424, 463, 520, 572, 626, 684, 747, 828, 917, 999, 1101, 1210, 1325, 1446, 1577, 1716, 1871, 2040, 2228, 2429, 2642, 2875, 3133, 3413, 3719, 4044, 4402, 4786, 5196, 5645, 6140, 6673, 7257, 7900, 8582, 9315, 10130, 10998, 11942, 12954, 14058…

n = n + digcount(n,dig=4,b=2..n)

9, 10, 11, 12, 13, 14, 16, 18, 20, 23, 25, 28, 34, 41, 44, 52, 61, 67, 74, 85, 92, 102, 113, 121, 134, 148, 170, 184, 208, 229, 253, 269, 287, 306, 324, 356, 386, 410, 439, 469, 501, 531, 565, 604, 662, 703, 742, 794, 845, 895, 953, 1007, 1062, 1127, 1188, 1262, 1336, 1421, 1503, 1585, 1676, 1777, 1876, 2001, 2104, 2249, 2375, 2502, 2636, 2789, 2938, 3102, 3267, 3444, 3644, 3868, 4099…

n = n + digcount(n,dig=5,b=2..n)

11, 12, 13, 14, 15, 16, 17, 19, 21, 23, 26, 28, 29, 33, 37, 41, 48, 50, 55, 60, 64, 67, 72, 75, 83, 91, 96, 102, 107, 118, 123, 129, 137, 151, 159, 171, 180, 192, 202, 211, 224, 233, 251, 268, 280, 296, 310, 324, 338, 355, 380, 401, 430, 455, 488, 511, 536, 562, 584, 607, 638, 664, 692, 718, 748, 778, 807, 838, 874, 911, 951, 993, 1039, 1081, 1124, 1166, 1216, 1264, 1313, 1370, 1432…