Five Lectures on CA 1. Signals and Markings Thomas Worsch - PowerPoint PPT Presentation

Five Lectures on CA 1. Signals and Markings Thomas Worsch Department of Informatics Karlsruhe Institute of Technology http://liinwww.ira.uka.de/~thw/vl-hiroshima/ at Hiroshima University, January 2012

Outline Simple Signals Markings

Simple Signals Markings

Example: signal with speed 1 Q = { , > } 3 4 5 6 7 t = 10 > t = 11 > empty places: t = 12 quiescent state > t = 13 > t = 14 >

Example: signal with speed 1 Q = { , > } 3 4 5 6 7 t = 10 > t = 11 > empty places: t = 12 quiescent state > t = 13 > t = 14 > ℓ ( − 1) ℓ (0) ℓ (1) f ( ℓ ) > > > otherwise f ( ℓ ) = ℓ (0)

Example: signal with speed 1 Q = { , > } 3 4 5 6 7 t = 10 > t = 11 > empty places: t = 12 quiescent state > t = 13 > t = 14 > ℓ ( − 1) ℓ (0) ℓ (1) f ( ℓ ) > > > otherwise f ( ℓ ) = ℓ (0) a signal s : { 10 , 11 , 12 , 13 , 14 } → R

Example: signal with speed 1 Q = { , > } 3 4 5 6 7 t = 10 > t = 11 > empty places: t = 12 quiescent state > t = 13 > t = 14 > ℓ ( − 1) ℓ (0) ℓ (1) f ( ℓ ) > > > otherwise f ( ℓ ) = ℓ (0) a signal s : { 10 , 11 , 12 , 13 , 14 } → R speed: 1 cell / 1 step = 1

Example: signal gets reflected Q = { , > , | , < } t = 0 > | t = 1 > | t = 2 > | t = 3 < | t = 4 < | t = 5 < |

Example: signal gets reflected Q = { , > , | , < } ℓ ( − 1) ℓ (0) ℓ (1) f ( ℓ ) t = 0 > | > t = 1 > | > > t = 2 > | > | < < | t = 3 < | < t = 4 < | < < t = 5 < | otherwise f ( ℓ ) = ℓ (0)

Example: signal gets reflected Q = { , > , | , < } ℓ ( − 1) ℓ (0) ℓ (1) f ( ℓ ) t = 0 > | > t = 1 > | > > t = 2 > | > | < < | t = 3 < | < t = 4 < | < < t = 5 < | otherwise f ( ℓ ) = ℓ (0)

Example: another type of reflection Q = { , > , | , <| , < } t = 0 > | t = 1 > | t = 2 > | t = 3 > | t = 4 <| t = 5 < | t = 6 < |

Example: another type of reflection Q = { , > , | , <| , < } ℓ ( − 1) ℓ (0) ℓ (1) f ( ℓ ) t = 0 > | > t = 1 > | > > t = 2 > | > | > > | <| t = 3 > | <| | t = 4 <| <| < t = 5 < | < t = 6 < | < < otherwise f ( ℓ ) = ℓ (0)

Example: yet another type of reflection Q = { , > , | , >| , <| , < } t = 0 > | t = 1 > | t = 2 > | t = 3 > | t = 4 >| t = 5 <| t = 6 < | t = 7 < |

Definition A signal is ◮ a function s : T → R where ◮ T = { a , . . . , b } or ◮ T = { a , . . . } is a (finite of infinite) set of points in time.

Example: signals with constant speed t = 0 1> t = 1 2> t = 2 3> t = 3 4> t = 4 5> t = 5 1> t = 6 2> t = 7 3> t = 8 4> t = 9 5>

Example: signals with constant speed ℓ ( − 1) ℓ (0) ℓ (1) f ( ℓ ) t = 0 1> 1> 2> t = 1 2> 2> 3> t = 2 3> 3> t = 3 4> 3> 4> t = 4 5> 4> 5> t = 5 1> 5> t = 6 2> 5> 1> t = 7 3> otherwise t = 8 4> . . . whatever is useful . . . t = 9 5>

Example: signals with constant speed ℓ ( − 1) ℓ (0) ℓ (1) f ( ℓ ) t = 0 1> 1> 2> t = 1 2> 2> 3> t = 2 3> 3> t = 3 4> 3> 4> t = 4 5> 4> 5> t = 5 1> 5> t = 6 2> 5> 1> t = 7 3> otherwise t = 8 4> . . . whatever is useful . . . t = 9 5> speed: ◮ above: 2 / 5 ◮ in general: average number of cells per time step

Which signal speeds are constructible?

Which signal speeds are constructible? ◮ depends on neighborhood N

Which signal speeds are constructible? ◮ depends on neighborhood N ◮ if N = {− 1 , 0 , 1 }

Which signal speeds are constructible? ◮ depends on neighborhood N ◮ if N = {− 1 , 0 , 1 } : ◮ any rational speed − 1 ≤ r ≤ 1 ◮ negative r means move to the left

Which signal speeds are constructible? ◮ depends on neighborhood N ◮ if N = {− 1 , 0 , 1 } : ◮ any rational speed − 1 ≤ r ≤ 1 ◮ negative r means move to the left ◮ no speed > 1

Which signal speeds are constructible? ◮ depends on neighborhood N ◮ if N = {− 1 , 0 , 1 } : ◮ any rational speed − 1 ≤ r ≤ 1 ◮ negative r means move to the left ◮ no speed > 1 ◮ more difficult (Korec, 1993): some signals with “irrational speed” are constructible

Algorithm: zigzag signals in a continuous world it would look like this: ✆ ✝ � ✁ � ✄ ✆ ✞ � ✂ � ☎ ✆ ✟

v A = 1 / 2 t = 1 a t = 2 b t = 3 a t = 4 b t = 5 a t = 6 b t = 7 a t = 8 b t = 9 a t = 10 b t = 11 a t = 12 b t = 13 a t = 14 b t = 15 a t = 16 b t = 17 a t = 18 b t = 19 t = 20 t = 21 t = 22 t = 23

v A = 1 / 2 v B = 1 / 5 t = 1 1a t = 2 2b t = 3 3 a t = 4 4 b t = 5 5 a t = 6 1 b t = 7 2 a t = 8 3 b t = 9 4 a t = 10 5 b t = 11 1 a t = 12 2 b t = 13 3 a t = 14 4 b t = 15 5 a t = 16 1 b t = 17 2 a t = 18 3 b t = 19 4 t = 20 5 t = 21 1 t = 22 2 t = 23 3

v A = 1 / 2 v B = 1 / 5 v R = 1 v L = − 1 t = 1 1a t = 2 2b t = 3 3> a t = 4 4 <b t = 5 5> a t = 6 1> b t = 7 2 > a t = 8 3 <b t = 9 4 < a t = 10 5> b t = 11 1> a t = 12 2 > b t = 13 3 > a t = 14 4 > b t = 15 5 > a t = 16 1 <b t = 17 2 < a t = 18 3 < b t = 19 4 < t = 20 5> t = 21 1> t = 22 2 > t = 23 3 >

Algorithm: discretized parabola How can one construct this signal? t = 0 | t = 1 | t = 2 | t = 3 | t = 4 | t = 5 | t = 6 | t = 7 | t = 8 | t = 9 | t = 10 | t = 11 | t = 12 | t = 13 | t = 14 | t = 15 | t = 16 | t = 17 | t = 18 | t = 19 | t = 20 | t = 21 | t = 22 | t = 23 | t = 24 | t = 25 |

Algorithm: discretized parabola Hint: t = 0 ( | t = 1 ( | t = 2 ( | t = 3 ( | t = 4 ( | t = 5 ( | t = 6 ( | t = 7 ( | t = 8 ( | t = 9 ( | t = 10 ( | t = 11 ( | t = 12 ( | t = 13 ( | t = 14 ( | t = 15 ( | t = 16 ( | t = 17 ( | t = 18 ( | t = 19 ( | t = 20 ( | t = 21 ( | t = 22 ( | t = 23 ( | t = 24 ( | t = 25 ( |

Algorithm: discretized parabola 1 2 3 4 5 t = 0 (<| t = 1 (>| t = 2 ( <| t = 3 (< | t = 4 (> | t = 5 ( >| t = 6 ( <| t = 7 ( < | t = 8 (< | t = 9 (> | t = 10 ( > | t = 11 ( >| t = 12 ( <| t = 13 ( < | t = 14 ( < | t = 15 (< | t = 16 (> | t = 17 ( > | t = 18 ( > | t = 19 ( >| t = 20 ( <| t = 21 ( < | t = 22 ( < | t = 23 ( < | t = 24 (< | t = 25 (> |

Simple Signals Markings

Algorithm: mark cell n / 2 Task: ◮ in a segment of n cells 1 2 3 4 5 6 7 8 9 10 11 12 13 ( ) ◮ mark cell n / 2 1 2 3 4 5 6 7 8 9 10 11 12 13 ( * )

Algorithm: mark cell n / 2 Task: ◮ in a segment of n cells 1 2 3 4 5 6 7 8 9 10 11 12 13 ( ) ◮ mark cell n / 2 1 2 3 4 5 6 7 8 9 10 11 12 13 ( * ) different possibilities; for example: ◮ 2 signals with speeds r < 1 and 1 starting at the left end

Algorithm: mark cell n / 2 Task: ◮ in a segment of n cells 1 2 3 4 5 6 7 8 9 10 11 12 13 ( ) ◮ mark cell n / 2 1 2 3 4 5 6 7 8 9 10 11 12 13 ( * ) different possibilities; for example: ◮ 2 signals with speeds r < 1 and 1 starting at the left end ◮ reflect the faster signal at the right end ◮ signals meet in the middle

Algorithm: mark cell n / 2 Task: ◮ in a segment of n cells 1 2 3 4 5 6 7 8 9 10 11 12 13 ( ) ◮ mark cell n / 2 1 2 3 4 5 6 7 8 9 10 11 12 13 ( * ) different possibilities; for example: ◮ 2 signals with speeds r < 1 and 1 starting at the left end ◮ reflect the faster signal at the right end ◮ signals meet in the middle if r = 1 / 3

Algorithm: marking cells with exponentially increasing distances How could one achieve that?

1 2 4 8 t = 1 1a t = 2 2b t = 3 3> a t = 4 4 <b t = 5 5> a t = 6 1> b t = 7 2 > a t = 8 3 <b t = 9 4 < a t = 10 5> b t = 11 1> a t = 12 2 > b t = 13 3 > a t = 14 4 > b t = 15 5 > a t = 16 1 <b t = 17 2 < a t = 18 3 < b t = 19 4 < t = 20 5> t = 21 1> t = 22 2 > t = 23 3 >

zigzag signal: a computation for points P i = ( x i , t i ) in the space time diagram: = x 1 + ( t 2 − t 1 ) v L t 2 v B x 1 − x 1 t 2 ( v B − v L ) = v A v L v A − v L t 2 = x 1 v A ( v B − v L ) ✆ ✝ � ✁ v B ( v A − v L ) x 2 = x 1 � ✄ ✆ ✞ � ✂ v A ( v B − v L ) t 3 v A = x 2 + ( t 3 − t 2 ) v R � ☎ v B ( v A − v L ) ( v A − v L ) v R t 3 ( v A − v R ) = x 1 v A ( v B − v L ) − x 1 v A ( v B − v L ) ✆ ✟ ( v B − v R )( v A − v L ) t 3 = x 1 v A ( v B − v L )( v A − v R ) ( v B − v R )( v A − v L ) x 3 = x 1 ( v B − v L )( v A − v R )