Tag Archives: recreational maths

Tri-Way to L

by in Fractals, Geometry, Mathematics, Rep-Tiles and tagged , , , , , , , , , , , , , , , , , , | Leave a comment

The name is more complicated than the shape: L-triomino. The shape is simply three squares forming an L. And it’s a rep-tile — it can be divided into four smaller copies of itself.

l-triomino

An L-triomino — three squares forming an L

l-triomino_anim

L-triomino as rep-tile

That means it can also be turned into a fractal, as I’ve shown in Rep-Tiles Revisited and Get Your Prox Off #2. First you divide an L-triomino into four sub-copies, then discard one sub-copy, then repeat. Here are the standard L-triomino fractals produced by this technique:

l-triomino_123_134

Fractal from L-triomino — divide and discard

l-triomino_234

l-triomino_124

l-triomino_124_upright

l-triomino_124_upright_static1

(Static image)

l-triomino_124_upright_static2

(Static image)

But those fractals don’t exhaust the possibilities of this very simple shape. The standard L-triomino doesn’t have true chirality. That is, it doesn’t come in left- and right-handed forms related by mirror-reflection. But if you number its corners for the purposes of sub-division, you can treat it as though it comes in two distinct orientations. And when the orientations are different in the different sub-copies, new fractals appear. You can also delay the stage at which you discard the first sub-copy. For example, you can divide the L-triomino into four sub-copies, then divide each sub-copy into four more sub-copies, and only then begin discarding.

Here are the new fractals that appear when you apply these techniques:

l-triomino_124_exp

Delay before discarding

l-triomino_124_exp_static

(Static image)

l-triomino_124_tst2_static1

(Static image)

l-triomino_124_tst2_static2

(Static image)

l-triomino_124_tst1

l-triomino_124_tst1_static1

(Static image)

l-triomino_124_tst1_static2

(Static image)

l-triomino_134_adj1

Adjust orientation

l-triomino_134_adj2

l-triomino_134_adj3

l-triomino_134_adj3_tst3

(Static image)

l-triomino_134_adj4

l-triomino_134_exp_static

(Static image)

l-triomino_234_exp

Can You Dij It? #2

by in Base Behavior, Binary numbers, Integer Sequences, Mathematics and tagged , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

It’s very simple, but I’m fascinated by it. I’m talking about something I call the digit-line, or the stream of digits you get when you split numbers in a particular base into individual digits. For example, here are the numbers one to ten in bases 2 and 3:

Base = 2: 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010…
Base = 3: 1, 2, 10, 11, 12, 20, 21, 22, 100, 101…

If you turn them into digit-lines, they look like this:

Base = 2: 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0… (A030190 in the Online Encyclopedia of Integer Sequences)
Base = 3: 1, 2, 1, 0, 1, 1, 1, 2, 2, 0, 2, 1, 2, 2, 1, 0, 0, 1, 0, 1… (A003137 in the OEIS)

At the tenth digit of the two digit-lines, both digits equal zero for the first time:

Base = 2: 1, 1, 0, 1, 1, 1, 0, 0, 1, 0…
Base = 3: 1, 2, 1, 0, 1, 1, 1, 2, 2, 0…

When the binary and ternary digits are represented together, the digit-lines look like this:

(1,1), (1,2), (0,1), (1,0), (1,1), (1,1), (0,1), (0,2), (1,2), (0,0)…

But in base 4, the tenth digit of the digit-line is 1. So when do all the digits of the digit-line first equal zero for bases 2 to 4? Here the early integers in those bases:

Base 2: 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101…

Base 3: 1, 2, 10, 11, 12, 20, 21, 22, 100, 101, 102, 110, 111, 112, 120, 121, 122, 200, 201, 202, 210, 211, 212, 220, 221, 222, 1000, 1001, 1002…

Base 4: 1, 2, 3, 10, 11, 12, 13, 20, 21, 22, 23, 30, 31, 32, 33, 100, 101, 102, 103, 110, 111, 112, 113, 120, 121, 122, 123, 130, 131, 132, 133, 200…

And here are the digits of the digit-line in bases 2 to 4 represented together:

(1,1,1), (1,2,2), (0,1,3), (1,0,1), (1,1,0), (1,1,1), (0,1,1), (0,2,1), (1,2,2), (0,0,1), (1,2,3), (1,1,2), (1,2,0), (0,2,2), (1,1,1), (1,0,2), (1,0,2), (1,1,2), (0,0,3), (0,1,3), (0,1,0), (1,0,3), (0,2,1), (0,1,3), (1,1,2), (1,0,3), (0,1,3), (1,1,1), (0,1,0), (1,1,0), (0,1,1), (1,2,0), (1,1,1), (1,2,1), (1,0,0), (0,1,2), (0,2,1), (1,1,0), (1,1,3), (0,2,1), (1,2,1), (1,2,0), (1,0,1), (1,0,1), (0,2,1), (1,0,1), (1,1,1), (1,2,2), (1,0,1), (1,2,1), (0,2,3), (0,1,1), (0,0,2), (0,2,0), (1,1,1), (0,1,2), (0,2,1), (0,1,1), (1,2,2), (1,2,2), (0,2,1), (0,0,2), (1,2,3), (0,2,1), (1,1,3), (0,2,0), (0,2,1), (1,2,3), (1,1,1), (1,0,1), (0,0,3), (1,0,2), (0,1,1), (0,0,3), (1,0,3), (0,1,2), (1,1,0), (0,0,0)…

At the 78th digit, all three digits equal zero. But the 78th digit of the digit-line in base 5 is 1. So when are the digits first equal to zero in bases 2 to 5? It’s not difficult to find out, but the difficulty of the search increases fast as the bases get bigger. Here are the results up to base 13 (note that bases 11 and 12 both have zeroes at digit 103721663):

dig=0 in bases 2 to 3 at the 10th digit of the digit-line
dig=0 in bases 2 to 4 at the 78th digit of the digit-line
dig=0 in bases 2 to 5 at the 182nd digit of the digit-line
dig=0 in bases 2 to 6 at the 302nd digit of the digit-line
dig=0 in bases 2 to 7 at the 12149th digit of the digit-line
dig=0 in bases 2 to 8 at the 45243rd digit of the digit-line
dig=0 in bases 2 to 9 at the 255261st digit of the digit-line
dig=0 in bases 2 to 10 at the 8850623rd digit of the digit-line
dig=0 in bases 2 to 12 at the 103721663rd digit of the digit-line
dig=0 in bases 2 to 13 at the 807778264th digit of the digit-line

I assume that, for any base b > 2, you can find some point in the digit-line at which d = 0 for all bases 2 to b. Indeed, I assume that this happens infinitely often. But I don’t know any short-cut for finding the first digit at which this occurs.

Previously pre-posted:

• Can You Dij It? #1

Dice in the Witch House

by in Mathematics, Probability, Randomness and tagged , , , , , , , , , , , , , , , , , , , , , | Leave a comment

“Who could associate mathematics with horror?”

John Buchan answered that question in “Space” (1911), long before H.P. Lovecraft wrote masterpieces like “The Call of Cthulhu” (1926) and “Dreams in the Witchhouse” (1933). But Lovecraft’s use of mathematics is central to his genius. So is his recognition of both the importance and the strangeness of mathematics. Weird fiction and maths go together very well.

But weird fiction is about the intrusion or eruption of the Other into the everyday. Maths can teach you that the everyday is already Other. In short, reality is weird — the World is a Witch House. Let’s start with a situation that isn’t obviously weird. Suppose you had three six-sided dice, A, B and C, each with different set of numbers, like this:

Die A = (1, 2, 3, 6, 6, 6)
Die B = (1, 2, 3, 4, 6, 6)
Die C = (1, 2, 3, 4, 5, 6)

If the dice are fair, i.e. each face has an equal chance of appearing, then it’s clear that, on average, die A will beat both die B and die C, while die B will beat die C. The reasoning is simple: if die A beats die B and die B beats die C, then surely die A will beat die C. It’s a transitive relationship: If Jack is taller than Jim and Jim is taller than John, then Jack is taller than John.

Now try another set of dice with different arrangements of digits:

Die A = (1, 2, 2, 5, 6, 6)
Die B = (1, 1, 4, 5, 5, 5)
Die C = (3, 3, 3, 3, 4, 6)

If you roll the dice, on average die A beats die B and die B beats die C. Clearly, then, die A will also beat die C. Or will it? In fact, it doesn’t: the dice are what is called non-transitive. Die A beats die B and die B beats die C, but die C beats die A.

But how does that work? To see a simpler example of non-transitivity, try a simpler set of random-number generators. Suppose you have a triangle with a short rod passing through its centre at right angles to the plane of the triangle. Now imagine numbering the edges of the triangles (1, 2, 3) and throwing it repeatedly so that it spins in the air before landing on a flat surface. It should be obvious that it will come to rest with one edge facing downward and that each edge has a 1/3 chance of landing like that.

In other words, you could use such a spiked triangle as a random-number generator — you could call it a “trie”, plural “trice”. Examine the set of three trice below. You’ll find that they have the same paradoxical property as the second set of six-sided dice above. Trie A beats trie B, trie B beats trie C, but trie C beats trie A:

Trie A = (1, 5, 8)
Trie B = (3, 4, 7)
Trie C = (2, 3, 9)

When you throw two of the trice, there are nine possible outcomes, because each of three edges on one trie can be matched with three possible edges on the other. The results look like this:

Trie A beats Trie B 5/9ths of the time.
Trie B beats Trie C 5/9ths of the time.
Trie C beats Trie A 5/9ths of the time.

To see how this works, here are the results throw-by-throw:

Trie A = (1, 5, 8)
Trie B = (3, 4, 7)

When Trie A rolls 1…

…and Trie B rolls 3, Trie B wins (Trie A has won 0 out of 1)
…and Trie B rolls 4, Trie B wins (0 out of 2)
…and Trie B rolls 7, Trie B wins (0 out of 3)

When Trie A rolls 5…

…and Trie B rolls 3, Trie A wins (1/4)
…and Trie B rolls 4, Trie A wins (2/5)
…and Trie B rolls 7, Trie B wins (2/6)

When Trie A rolls 8…

…and Trie B rolls 3, Trie A wins (3/7)
…and Trie B rolls 4, Trie A wins (4/8)
…and Trie B rolls 7, Trie A wins (5/9)

Trie B = (3, 4, 7)
Trie C = (2, 3, 9)

When Trie B rolls 3…

…and Trie C rolls 2, Trie B wins (Trie B has won 1 out of 1)
…and Trie C rolls 3, it’s a draw (1 out of 2)
…and Trie C rolls 9, Trie C wins (1 out of 3)

When Trie B rolls 4…

…and Trie C rolls 2, Trie B wins (2/4)
…and Trie C rolls 3, Trie B wins (3/5)
…and Trie C rolls 9, Trie C wins (3/6)

When Trie B rolls 7…

…and Trie C rolls 2, Trie B wins (4/7)
…and Trie C rolls 3, Trie B wins (5/8)
…and Trie C rolls 9, Trie C wins (5/9)

Trie C = (2, 3, 9)
Trie A = (1, 5, 8)

When Trie C rolls 2…

…and Trie A rolls 1, Trie C wins (Trie C has won 1 out of 1)
…and Trie A rolls 5, Trie A wins (1 out of 2)
…and Trie A rolls 8, Trie A wins (1 out of 3)

When Trie C rolls 3…

…and Trie A rolls 1, Trie C wins (2/4)
…and Trie A rolls 5, Trie A wins (2/5)
…and Trie A rolls 8, Trie A wins (2/6)

When Trie C rolls 9…

…and Trie A rolls 1, Trie C wins (3/7)
…and Trie A rolls 5, Trie C wins (4/8)
…and Trie A rolls 8, Trie C wins (5/9)

The same reasoning can be applied to the six-sided non-transitive dice, but there are 36 possible outcomes when two of the dice are thrown against each other, so I won’t list them.

Die A = (1, 2, 2, 5, 6, 6)
Die B = (1, 1, 4, 5, 5, 5)
Die C = (3, 3, 3, 3, 4, 6)

Elsewhere other-posted:

• At the Mountains of Mathness
• Simpson’s Paradox — a simple situation with a very weird outcome

De Pluribus Unum

by in Mathematics, Natural History, Puzzles and tagged , , , , , , , , , , , , | Leave a comment

A beautifully subtle puzzle:

Scrambled Box Tops

Imagine you have three boxes, one containing two black marbles, one containing two white marbles, and the third, one black and one white marble. The boxes are labelled according to their contents — BB, WW, and BW — but someone has switched the labels so that every box is now incorrectly labelled. You are allowed to take one marble at a time out of any box, without looking inside, and by this process of sampling you are to determine the contents of all three boxes. What is the smallest number of drawings needed to do this? — Martin Gardner, Mathematical Puzzles and Diversions (1959), chapter 3, “Nine Problems”, #5.

Bald eagle, Haliaeetus leucocephalus (Linnaeus 1776)

Bald eagle, Haliaeetus leucocephalus (Linnaeus 1776)

Answer: You can learn the contents of all three boxes by drawing just one marble. The key to the solution is your knowledge that the labels on all three boxes are incorrect. You must draw a marble from the box labelled “black-white”. Assume that the marble drawn is black. You know then that the other marble in the box must be black also, otherwise the label on the box would be correct. Since you have now identified the box containing two black marbles, you can tell at once the contents of the box labelled “white-white”: you know it cannot contain two white marbles, because its label has to be wrong; it cannot contain two black marbles, because you have identified that box; therefore it must contain one black and one white marble. The third box, of course, must then be the one containing two white marbles. You can solve the puzzle by the same reasoning if the marble you draw from the “black-white” box happens to be white instead of black.

For Revver and Fevver

by in Base Behavior, Fractals, Geometry, Mathematics and tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

This shape reminds me of the feathers on an exotic bird:

feathers

(click or open in new window for full size)

feathers_anim

(animated version)

The shape is created by reversing the digits of a number, so you could say it involves revvers and fevvers. I discovered it when I was looking at the Halton sequence. It’s a sequence of fractions created according to a simple but interesting rule. The rule works like this: take n in base b, reverse it, and divide reverse(n) by the first power of b that is greater than n.

For example, suppose n = 6 and b = 2. In base 2, 6 = 110 and reverse(110) = 011 = 11 = 3. The first power of 2 that is greater than 6 is 2^3 or 8. Therefore, halton(6) in base 2 equals 3/8. Here is the same procedure applied to n = 1..20:

1: halton(1) = 1/10[2] → 1/2
2: halton(10) = 01/100[2] → 1/4
3: halton(11) = 11/100[2] → 3/4
4: halton(100) = 001/1000[2] → 1/8
5: halton(101) = 101/1000[2] → 5/8
6: halton(110) = 011/1000 → 3/8
7: halton(111) = 111/1000 → 7/8
8: halton(1000) = 0001/10000 → 1/16
9: halton(1001) = 1001/10000 → 9/16
10: halton(1010) = 0101/10000 → 5/16
11: halton(1011) = 1101/10000 → 13/16
12: halton(1100) = 0011/10000 → 3/16
13: halton(1101) = 1011/10000 → 11/16
14: halton(1110) = 0111/10000 → 7/16
15: halton(1111) = 1111/10000 → 15/16
16: halton(10000) = 00001/100000 → 1/32
17: halton(10001) = 10001/100000 → 17/32
18: halton(10010) = 01001/100000 → 9/32
19: halton(10011) = 11001/100000 → 25/32
20: halton(10100) = 00101/100000 → 5/32…

Note that the sequence always produces reduced fractions, i.e. fractions in their lowest possible terms. Once 1/2 has appeared, there is no 2/4, 4/8, 8/16…; once 3/4 has appeared, there is no 6/8, 12/16, 24/32…; and so on. If the fractions are represented as points in the interval [0,1], they look like this:

line1_1_2

point = 1/2

line2_1_4

point = 1/4

line3_3_4

point = 3/4

line4_1_8

point = 1/8

line5_5_8

point = 5/8

line6_3_8

point = 3/8

line7_7_8

point = 7/8

line_b2_anim

(animated line for base = 2, n = 1..63)

It’s apparent that Halton points in base 2 will evenly fill the interval [0,1]. Now compare a Halton sequence in base 3:

1: halton(1) = 1/10[3] → 1/3
2: halton(2) = 2/10[3] → 2/3
3: halton(10) = 01/100[3] → 1/9
4: halton(11) = 11/100[3] → 4/9
5: halton(12) = 21/100[3] → 7/9
6: halton(20) = 02/100 → 2/9
7: halton(21) = 12/100 → 5/9
8: halton(22) = 22/100 → 8/9
9: halton(100) = 001/1000 → 1/27
10: halton(101) = 101/1000 → 10/27
11: halton(102) = 201/1000 → 19/27
12: halton(110) = 011/1000 → 4/27
13: halton(111) = 111/1000 → 13/27
14: halton(112) = 211/1000 → 22/27
15: halton(120) = 021/1000 → 7/27
16: halton(121) = 121/1000 → 16/27
17: halton(122) = 221/1000 → 25/27
18: halton(200) = 002/1000 → 2/27
19: halton(201) = 102/1000 → 11/27
20: halton(202) = 202/1000 → 20/27
21: halton(210) = 012/1000 → 5/27
22: halton(211) = 112/1000 → 14/27
23: halton(212) = 212/1000 → 23/27
24: halton(220) = 022/1000 → 8/27
25: halton(221) = 122/1000 → 17/27
26: halton(222) = 222/1000 → 26/27
27: halton(1000) = 0001/10000 → 1/81
28: halton(1001) = 1001/10000 → 28/81
29: halton(1002) = 2001/10000 → 55/81
30: halton(1010) = 0101/10000 → 10/81

And here is an animated gif representing the Halton sequence in base 3 as points in the interval [0,1]:

line_b3_anim

Halton points in base 3 also evenly fill the interval [0,1]. What happens if you apply the Halton sequence to a two-dimensional square rather a one-dimensional line? Suppose the bottom left-hand corner of the square has the co-ordinates (0,0) and the top right-hand corner has the co-ordinates (1,1). Find points (x,y) inside the square, with x supplied by the Halton sequence in base 2 and y supplied by the Halton sequence in base 3. The square will gradually fill like this:

square1

x = 1/2, y = 1/3

square2

x = 1/4, y = 2/3

square3

x = 3/4, y = 1/9

square4

x = 1/8, y = 4/9

square5

x = 5/8, y = 7/9

square6

x = 3/8, y = 2/9

square7

x = 7/8, y = 5/9

square8

x = 1/16, y = 8/9

square9

x = 9/16, y = 1/27…

square_anim

animated square

Read full page: For Revver and Fevver…

Pigmental Paradox

by in Logic, Mathematics, Quotations and tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

From Raymond Smullyan’s Logical Labyrinths (2009):

We now visit another knight/knave island on which, like on the first one, all knights tell the truth and all knaves lie. But now there is another complication! For some reason, the natives refuse to speak to strangers, but they are willing to answer yes/no questions using a secret sign language that works like this:

Each native carries two cards on his person; one is red and the other is black. One of them means yes and the other means no, but you are not told which color means what. If you ask a yes/no question, the native will flash one of the two cards, but unfortunately, you will not know whether the card means yes or no!

Problem 3.1. Abercrombie, who knew the rules of this island, decided to pay it a visit. He met a native and asked him: “Does a red card signify yes?” The native then showed him a red card.

From this, is it possible to deduce what a red card signifies? Is it possible to deduce whether the native was a knight or a knave?

Problem 3.2. Suppose one wishes to find out whether it is a red card or a black card that signifies yes. What simple yes/no question should one ask?

Rep-tilian Rites

by in Fractals, Geometry, Mathematics, Rep-Tiles and tagged , , , , , , , , , , , , , , , , , , | Leave a comment

A pentomino is one of the shapes created by laying five squares edge-to-edge. There are twelve of them (not counting reflections) and this is the P-pentomino:

p_pentomino

But it’s not just a pentomino, it’s also a rep-tile, or a shape that can divided into smaller copies of itself. There are two ways of doing this (I’ve rotated the pentomino 90° to make the images look better):

p_pentomino_a

p_pentomino_b

Once you’ve divided the shape into four copies, you can divide the copies, then the copies of the copies, and the copies of the copies of the copies, and so on for ever:

p_pentomino_a_anim

p_pentomino_a_anim

And if you’ve got a reptile, you can turn it into a fractal. Simply divide the shape, discard one or more copies, and continue:

p_pentomino_a_124_1

Pentomino-based fractal stage 1

p_pentomino_a_124_2

Pentomino-based fractal stage 2

p_pentomino_a_124_3

Pentomino-based fractal stage 3

p_pentomino_a_124_4

Stage 4

p_pentomino_a_124_5

Stage 5

p_pentomino_a_124_6

Stage 6

p_pentomino_a_124_7

Stage 7

p_pentomino_a_124_8

Stage 8

p_pentomino_a_124_9

Stage 9

p_pentomino_a_124_10

Stage 10

Here are some more fractals created using the same divide-and-discard process:

p_pentomino_b_234

p_pentomino_b_234anim

Animated version

p_pentomino_b_134

p_pentomino_b_134anim

Animated version

p_pentomino_b_124

p_pentomino_b_124anim

p_pentomino_b_123

p_pentomino_b_123anim

p_pentomino_a_134anim

p_pentomino_a_134

p_pentomino_a_234anim

p_pentomino_a_234

p_pentomino_a_124

p_pentomino_a_124anim

p_pentomino_a_123

p_pentomino_a_123anim

You can also use variants on a standard rep-tile dissection, like rotating the copies or trying different patterns of dissection at different levels to see what new shapes appear:

p_pentomino_adj_13

p_pentomino_adj_anim13

p_pentomino_adj_6

p_pentomino_adj_anim6

p_pentomino_adj_anim5

p_pentomino_adj_5

p_pentomino_adj_3

p_pentomino_adj_anim3

p_pentomino_adj_2

p_pentomino_adj_anim2

p_pentomino_adj_1

p_pentomino_adj_anim1

p_pentomino_adj_17

p_pentomino_adj_anim17

p_pentomino_adj_15

p_pentomino_adj_anim15

p_pentomino_adj_16

p_pentomino_adj_anim16

p_pentomino_adj_8

p_pentomino_adj_anim8

p_pentomino_adj_10

p_pentomino_adj_anim10

p_pentomino_adj_11

p_pentomino_adj_anim11

p_pentomino_adj_14

p_pentomino_adj_anim14

p_pentomino_adj_anim4

p_pentomino_adj_anim12

p_pentomino_adj_anim9

p_pentomino_adj_anim7

Rock’n’Roll Suislide

by in Art, Geometry, Mathematics, Puzzles and tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

Q. Each face of a convex polyhedron can serve as a base when the solid is placed on a horizontal plane. The center of gravity of a regular polyhedron is at the center, therefore it is stable on any face. Irregular polyhedrons are easily constructed that are unstable on certain faces; that is, when placed on a table with an unstable face as the base, they topple over. Is it possible to make a model of an irregular convex polyhedron that is unstable on every face?

Portrait of Luca Pacioli (1495)

Portrait of Luca Pacioli (1495)

A. No. If a convex polyhedron were unstable on every face, a perpetual motion machine could be built. Each time the solid toppled over onto a new base it would be unstable and would topple over again.

 — From “Ridiculous Questions” in Martin Gardner’s Mathematical Magical Show (1965), chapter 10.

Fragic Carpet

by in Fractals, Geometry, Mathematics, Rep-Tiles and tagged , , , , , , , , , , , , , , , , , | Leave a comment

Maths is like a jungle: rich, teeming and full of surprises. A waterfall here, a glade of butterflies there, a bank of orchids yonder. There is always something new to see and a different route to try. But sometimes a different route will take you to the same place. I’ve already found two ways to reach this fractal (see Fingering the Frigit and Performativizing the Polygonic):

carpet2x2

Fractal Carpet

Now I’ve found a third way. You could call it the rep-tile route. Divide a square into four smaller squares:

square2x2

Add an extra square over the centre:

square2x2_1

Then keep dividing the squares in the same way:

carpet2x2_anim_1

Animated carpet (with coloured blocks)

carpet2x2_anim_2

Animated carpet (with empty blocks)

The colours of the fractal appear when the same pixel is covered repeatedly: first it’s red, then green, yellow, blue, purple, and so on. Because the colours and their order are arbitrary, you can use different colour schemes:

carpet2x2_col1

Colour scheme #1

carpet2x2_col2

Colour scheme #2

carpet2x2_col3

Colour scheme #3

Here are more colour-schemes in an animated gif:

carpet2x2_col

Various colour-schemes

Now try dividing the square into nine and sixteen, with an extra square over the centre:

carpet3x3

3×3 square + central square

carpet3x3_anim

3×3 square + central square (animated)

carpet4x4

4×4 square + central square

carpet4x4_anim

4×4 square + central square (animated)

You can also adjust the size of the square added to the 2×2 subdivision:

carpet2x2_1_2

2×2 square + 1/2-sized central square

carpet2x2_3_4

2×2 square + 3/4-sized central square

Elsewhere Other-Posted:

• Fingering the Frigit
• Performativizing the Polygonic

Polymorphous Pursuit

by in Geometry, Mathematics, Pursuit Curves and tagged , , , , , , , , , , , , , , , , , , , , , | 2 Comments

Suppose four mice are standing on the corners of a large square. Each mouse begins running at the same speed towards the mouse one place away, reckoning clockwise. The mice will meet at the centre of the square and the path taken by each mouse will be what is known as a pursuit curve:

v4_mi1

vertices = 4, mouse-increment = 1

v4_mi1_animated

v = 4, mi = 1 (animated)

As I showed in “Persecution Complex”, it’s easy to find variants on the basic pursuit curve. If mi = 2, i.e. each mouse runs towards the mouse two places away, the mice will run in straight lines direct to the centre of the square:

v4_mi2

v = 4, mi = 2

v4_mi2_animated

v = 4, mi = 2 (animated)

That variant is trivial, but suppose there are eight mice, four starting on the corners of the square and four starting on the midpoints of the sides. Mice starting on the corners will run different pursuit curves to those starting on the midpoints, because the corners are further from the centre than the midpoints are:

v4_si1_mi1

v = 4, si = 1, mi = 1

v4_si1_mi1_extra

If mi = 3, the pursuit curves look like this:

v4_si1_mi3

v = 4, si = 1, mi = 3

v4_si1_mi3_animated

v = 4, si = 1, mi = 3 (animated)

Suppose there are twelve mice, four on each corner and two more on each side. If each mouse runs towards the mouse four places away, then the pursuit curves don’t all meet in the centre of the square. Instead, they meet in groups of three at four points equidistant from the centre, like this:

v4_si2_mi4

v4_si2_mi4_curves

v = 4, si = 2, mi = 4

v4_si2_mi4_animated

v = 4, si = 2, mi = 4 (animated)

v4_si4_mi4_animated

v = 4, si = 4, mi = 4 (animated)

v4_si4_mi4_large

v = 4, si = 4, mi = 4 (zoom)

Now suppose each mouse become sophisticated and runs toward the combined positions of two other mice, one two places away, the other three places away, like this:

v4_si1_mi2_3

v = 4, si = 1, mi = (2, 3)

v4_si1_mi2_3_animated

v = 4, si = 1, mi = (2, 3) (animated)

These polypursuits, as they could be called, can have complicated central regions:

v4_si2_mi1_4

v = 4, si = 2, mi = (1, 4)

v4_si2_mi1_4_animated

v = 4, si = 2, mi = (1, 4) (animated)

v4_si_va_mi_va

v = 4, si = various, mi = various

And what if you have two teams of mice, running towards one or more mice on the other team? For example, suppose two mice, one from each team, start on each corner of a square. Each mouse on team 1 runs towards the mouse on team 2 that is one place away, while each mouse on team 2 runs towards the mouse on team 1 that is two places away. If the pursuits curves of team 1 are represented in white and the pursuit curves of team 2 in green, the curves look like this:

2v4_mi1_mi2

v = 4 * 2, vmi = 1, vmi = 2

2v4_mi1_mi2_green

v = 4 * 2, vmi = 1, vmi = 2

2v4_mi1_mi2_animated

v = 4 * 2, vmi = 1, vmi = 2 (animated)

Now suppose the four mice of team 1 start on the corners while the mice of team 2 start at the centre of the square.

v4_c4_vmi1_cmi2_white

v = 4, centre = 4, vmi = 1, cmi = 2 (white team)

v4_c4_vmi1_cmi2_green

v = 4, centre = 4, vmi = 1, cmi = 2 (green team)

v4_c4_vmi1_cmi2_both

v = 4, centre = 4, vmi = 1, cmi = 2 (both teams)

v4_c4_vmi1_cmi2_animated

v = 4, centre = 4, vmi = 1, cmi = 2 (animated)

Here are more variants on pursuit curves formed by two teams of mice, one starting on the corners, one at the centre:

v4_c4_vmi0_1_cmi0

v = 4, centre = 4, vmi = (0, 1), cmi = 0

v4_c4_vmi0_2_cmi0

v = 4, centre = 4, vmi = (0, 2), cmi = 0

v4_c4_vmi0_3_cmi0

v = 4, centre = 4, vmi = (0, 3), cmi = 0

2v4_mi1_mi2_both