Polymorphous Pursuit

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

Spijit

The only two digits found in all standard bases are 1 and 0. But they behave quite differently. Suppose you take the integers 1 to 100 and compare the number of 1s and 0s in the representation of each integer, n, in bases 2 to n-1. For example, 10 would look like this:

1010 in base 2
101 in base 3
22 in base 4
20 in base 5
14 in base 6
13 in base 7
12 in base 8
11 in base 9

So there are nine 1s and four 0s. If you check 1 to 100 using this all-base function, the count of 1s goes like this:

1, 1, 2, 3, 5, 5, 8, 5, 9, 9, 11, 10, 15, 12, 14, 13, 15, 12, 17, 14, 20, 19, 20, 15, 23, 19, 22, 22, 25, 24, 31, 21, 25, 24, 24, 27, 33, 27, 31, 29, 34, 29, 36, 30, 34, 35, 34, 30, 40, 33, 36, 35, 38, 34, 42, 37, 43, 40, 41, 37, 48, 39, 42, 42, 44, 43, 48, 43, 47, 46, 51, 42, 53, 44, 48, 50, 51, 50, 55, 48, 59, 55, 55, 54, 64, 57, 57, 55, 60, 57, 68, 60, 64, 63, 64, 59, 68, 58, 61, 63.

And the count of 0s goes like this:

0, 1, 0, 2, 1, 2, 0, 4, 4, 4, 2, 5, 1, 2, 2, 7, 4, 8, 4, 7, 4, 3, 1, 8, 4, 4, 6, 8, 4, 7, 1, 10, 8, 7, 7, 12, 5, 6, 5, 10, 4, 8, 2, 6, 7, 4, 2, 12, 6, 9, 7, 8, 4, 11, 6, 10, 5, 4, 2, 12, 2, 3, 5, 14, 11, 13, 7, 10, 8, 11, 5, 17, 7, 8, 10, 10, 8, 10, 4, 13, 12, 10, 8, 16, 8, 7, 7, 12, 6, 14, 6, 8, 5, 4, 4, 16, 6, 10, 11, 15.

The bigger the numbers get, the bigger the discrepancies get. Sometimes the discrepancy is dramatic. For example, suppose you represented the prime 1014719 in bases 2 to 1014718. How 0s would there be? And how many 1s? There are exactly nine zeroes:

1014719 = 11110111101110111111 in base 2 = 1220112221012 in base 3 = 40B27B in base 12 = 1509CE in base 15 = 10[670] in base 1007.

But there are 507723 ones. The same procedure applied to the next integer, 1014720, yields 126 zeroes and 507713 ones. However, there is a way to see that 1s and 0s in the all-base representation are behaving in a similar way. To do this, imagine listing the individual digits of n in bases 2 to n-1 (or just base 2, if n <= 3). When the digits aren’t individual they look like this:

1 = 1 in base 2
2 = 10 in base 2
3 = 11 in base 2
4 = 100 in base 2; 11 in base 3
5 = 101 in base 2; 12 in base 3; 11 in base 4
6 = 110 in base 2; 20 in base 3; 12 in base 4; 11 in base 5
7 = 111 in base 2; 21 in base 3; 13 in base 4; 12 in base 5; 11 in base 6
8 = 1000 in base 2; 22 in base 3; 20 in base 4; 13 in base 5; 12 in base 6; 11 in base 7
9 = 1001 in base 2; 100 in base 3; 21 in base 4; 14 in base 5; 13 in base 6; 12 in base 7; 11 in base 8
10 = 1010 in base 2; 101 in base 3; 22 in base 4; 20 in base 5; 14 in base 6; 13 in base 7; 12 in base 8; 11 in base 9

So the list would look like this:

1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 2, 1, 1, 1, 1, 0, 2, 0, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 2, 1, 1, 1, 0, 0, 0, 2, 2, 2, 0, 1, 3, 1, 2, 1, 1, 1, 0, 0, 1, 1, 0, 0, 2, 1, 1, 4, 1, 3, 1, 2, 1, 1, 1, 0, 1, 0, 1, 0, 1, 2, 2, 2, 0, 1, 4, 1, 3, 1, 2, 1, 1

Suppose that these digits are compared against the squares of a counter-clockwise spiral on a rectangular grid. If the spiral digit is equal to 1, the square is filled in; if the spijit is not equal to 1, the square is left blank. The 1-spiral looks like this:
1spiral
Now try zero. If the spijit is equal to 0, the square is filled in; if not, the square is left blank. The 0-spiral looks like this:
0spiral
And here’s an animated gif of the n-spiral for n = 0..9:
animspiral

Persecution Complex

Imagine four mice sitting on the corners of a square. Each mouse begins to run towards its clockwise neighbour. What happens? This:

Four mice chasing each other

Four mice chasing each other


The mice spiral to the centre and meet, creating what are called pursuit curves. Now imagine eight mice on a square, four sitting on the corners, four sitting on the midpoints of the sides. Each mouse begins to run towards its clockwise neighbour. Now what happens? This:

Eight mice chasing each other

Eight mice chasing each other


But what happens if each of the eight mice begins to run towards its neighbour-but-one? Or its neighbour-but-two? And so on. The curves begin to get more complex:

square+midpoint+2


(Please open the following image in a new window if it fails to animate.)

square+midpoint+3


You can also make the mice run at different speeds or towards neighbours displaced by different amounts. As these variables change, so do the patterns traced by the mice:

• Continue reading Persecution Complexified