Generating a pattern from user-defined rules

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

I want to generate a list of $L$ and $S$ according to the following rule:
upon each iteration, replace

• $S rightarrow L$


• $L rightarrow LS$

Such that, starting with $L$:

$ L rightarrow LS rightarrow LSL rightarrow LSLLS rightarrow LSLLSLSL rightarrow LSLLSLSLLSLLS rightarrow dots$

How can I automate this with Mathematica? I just want to specify the starting letter and the number of iteration and I'd want it to spit out the final result.

list-manipulation pattern-matching string-manipulation

share|improve this question

asked 57 mins ago

SuperCiocia

376210

add a comment | 

up vote
1
down vote

favorite

I want to generate a list of $L$ and $S$ according to the following rule:
upon each iteration, replace

• $S rightarrow L$


• $L rightarrow LS$

Such that, starting with $L$:

$ L rightarrow LS rightarrow LSL rightarrow LSLLS rightarrow LSLLSLSL rightarrow LSLLSLSLLSLLS rightarrow dots$

How can I automate this with Mathematica? I just want to specify the starting letter and the number of iteration and I'd want it to spit out the final result.

list-manipulation pattern-matching string-manipulation

share|improve this question

asked 57 mins ago

SuperCiocia

376210

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I want to generate a list of $L$ and $S$ according to the following rule:
upon each iteration, replace

• $S rightarrow L$


• $L rightarrow LS$

Such that, starting with $L$:

$ L rightarrow LS rightarrow LSL rightarrow LSLLS rightarrow LSLLSLSL rightarrow LSLLSLSLLSLLS rightarrow dots$

How can I automate this with Mathematica? I just want to specify the starting letter and the number of iteration and I'd want it to spit out the final result.

list-manipulation pattern-matching string-manipulation

share|improve this question

asked 57 mins ago

SuperCiocia

376210

I want to generate a list of $L$ and $S$ according to the following rule:
upon each iteration, replace

• $S rightarrow L$


• $L rightarrow LS$

Such that, starting with $L$:

$ L rightarrow LS rightarrow LSL rightarrow LSLLS rightarrow LSLLSLSL rightarrow LSLLSLSLLSLLS rightarrow dots$

How can I automate this with Mathematica? I just want to specify the starting letter and the number of iteration and I'd want it to spit out the final result.

list-manipulation pattern-matching string-manipulation

list-manipulation pattern-matching string-manipulation

share|improve this question

asked 57 mins ago

SuperCiocia

376210

share|improve this question

asked 57 mins ago

SuperCiocia

376210

share|improve this question

share|improve this question

asked 57 mins ago

SuperCiocia

376210

asked 57 mins ago

SuperCiocia

376210

asked 57 mins ago

SuperCiocia

376210

376210

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

For the 5th iterate:

Nest[
 Replace[#, L -> Sequence[L, S], S -> Sequence[L], 1] &,
 L,
 5
 ]

L, S, L, L, S, L, S, L, L, S, L, L, S

And for the list of first 5 iterates:

NestList[
 Replace[#, L -> Sequence[L, S], S -> Sequence[L], 1] &,
 L,
 5
 ]

L, L, S, L, S, L, L, S, L, L, S, L, S, L, L, S, L, S,
L, L, S, L, L, S, L, S, L, L, S, L, L, S

Edit:

SubstitutionSystem operates on strings instead. It is a bit faster because it is specifically designed for Lindenmayer systems:

L = "L";
S = "S";
a = StringJoin /@ NestList[
 Replace[#, L -> Sequence[L, S], S -> Sequence[L], 1] &,
 L,
 20
 ]; // AbsoluteTiming // First
b = SubstitutionSystem["L" -> "LS", "S" -> "L", "L", 20]; // AbsoluteTiming // First

a == b

0.004506

0.001265

True

share|improve this answer

edited 18 mins ago

answered 42 mins ago

Henrik Schumacher

38.1k250109

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What if I just wanted a random sequence of $L$ and $S$?
  – SuperCiocia
  36 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then you can use RandomChoice: RandomChoice[L, S, 10000] will generate a list of 10000 random letter at once (uniformly distributed). You can also prescribe probabilities for each letter, e.g. RandomChoice[0.1, 0.9 -> L, S, 10000].
  – Henrik Schumacher
  34 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Great, thank you.
  – SuperCiocia
  32 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're welcome.
  – Henrik Schumacher
  29 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @SuperCiocia Please forget what I said about SequenceReplace. Better use Replace with level specification 1. That does also not suffer from using symbols.
  – Henrik Schumacher
  17 mins ago
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

You may use ReplaceRepeatedwith its MaxIterationsoption.

ReplaceRepeated[L, L -> Sequence[L, S], S -> Sequence[L], 
 MaxIterations -> 5]

L, S, L, L, S, L, S, L, L, S, L, L, S

Hope this helps.

share

answered 7 mins ago

Edmund

24.4k32995

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

For the 5th iterate:

Nest[
 Replace[#, L -> Sequence[L, S], S -> Sequence[L], 1] &,
 L,
 5
 ]

L, S, L, L, S, L, S, L, L, S, L, L, S

And for the list of first 5 iterates:

NestList[
 Replace[#, L -> Sequence[L, S], S -> Sequence[L], 1] &,
 L,
 5
 ]

L, L, S, L, S, L, L, S, L, L, S, L, S, L, L, S, L, S,
L, L, S, L, L, S, L, S, L, L, S, L, L, S

Edit:

SubstitutionSystem operates on strings instead. It is a bit faster because it is specifically designed for Lindenmayer systems:

L = "L";
S = "S";
a = StringJoin /@ NestList[
 Replace[#, L -> Sequence[L, S], S -> Sequence[L], 1] &,
 L,
 20
 ]; // AbsoluteTiming // First
b = SubstitutionSystem["L" -> "LS", "S" -> "L", "L", 20]; // AbsoluteTiming // First

a == b

0.004506

0.001265

True

share|improve this answer

edited 18 mins ago

answered 42 mins ago

Henrik Schumacher

38.1k250109

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What if I just wanted a random sequence of $L$ and $S$?
  – SuperCiocia
  36 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then you can use RandomChoice: RandomChoice[L, S, 10000] will generate a list of 10000 random letter at once (uniformly distributed). You can also prescribe probabilities for each letter, e.g. RandomChoice[0.1, 0.9 -> L, S, 10000].
  – Henrik Schumacher
  34 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Great, thank you.
  – SuperCiocia
  32 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're welcome.
  – Henrik Schumacher
  29 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @SuperCiocia Please forget what I said about SequenceReplace. Better use Replace with level specification 1. That does also not suffer from using symbols.
  – Henrik Schumacher
  17 mins ago
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

For the 5th iterate:

Nest[
 Replace[#, L -> Sequence[L, S], S -> Sequence[L], 1] &,
 L,
 5
 ]

L, S, L, L, S, L, S, L, L, S, L, L, S

And for the list of first 5 iterates:

NestList[
 Replace[#, L -> Sequence[L, S], S -> Sequence[L], 1] &,
 L,
 5
 ]

L, L, S, L, S, L, L, S, L, L, S, L, S, L, L, S, L, S,
L, L, S, L, L, S, L, S, L, L, S, L, L, S

Edit:

SubstitutionSystem operates on strings instead. It is a bit faster because it is specifically designed for Lindenmayer systems:

L = "L";
S = "S";
a = StringJoin /@ NestList[
 Replace[#, L -> Sequence[L, S], S -> Sequence[L], 1] &,
 L,
 20
 ]; // AbsoluteTiming // First
b = SubstitutionSystem["L" -> "LS", "S" -> "L", "L", 20]; // AbsoluteTiming // First

a == b

0.004506

0.001265

True

share|improve this answer

edited 18 mins ago

answered 42 mins ago

Henrik Schumacher

38.1k250109

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What if I just wanted a random sequence of $L$ and $S$?
  – SuperCiocia
  36 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then you can use RandomChoice: RandomChoice[L, S, 10000] will generate a list of 10000 random letter at once (uniformly distributed). You can also prescribe probabilities for each letter, e.g. RandomChoice[0.1, 0.9 -> L, S, 10000].
  – Henrik Schumacher
  34 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Great, thank you.
  – SuperCiocia
  32 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're welcome.
  – Henrik Schumacher
  29 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @SuperCiocia Please forget what I said about SequenceReplace. Better use Replace with level specification 1. That does also not suffer from using symbols.
  – Henrik Schumacher
  17 mins ago
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

For the 5th iterate:

Nest[
 Replace[#, L -> Sequence[L, S], S -> Sequence[L], 1] &,
 L,
 5
 ]

L, S, L, L, S, L, S, L, L, S, L, L, S

And for the list of first 5 iterates:

NestList[
 Replace[#, L -> Sequence[L, S], S -> Sequence[L], 1] &,
 L,
 5
 ]

L, L, S, L, S, L, L, S, L, L, S, L, S, L, L, S, L, S,
L, L, S, L, L, S, L, S, L, L, S, L, L, S

Edit:

SubstitutionSystem operates on strings instead. It is a bit faster because it is specifically designed for Lindenmayer systems:

L = "L";
S = "S";
a = StringJoin /@ NestList[
 Replace[#, L -> Sequence[L, S], S -> Sequence[L], 1] &,
 L,
 20
 ]; // AbsoluteTiming // First
b = SubstitutionSystem["L" -> "LS", "S" -> "L", "L", 20]; // AbsoluteTiming // First

a == b

0.004506

0.001265

True

share|improve this answer

edited 18 mins ago

answered 42 mins ago

Henrik Schumacher

38.1k250109

For the 5th iterate:

Nest[
 Replace[#, L -> Sequence[L, S], S -> Sequence[L], 1] &,
 L,
 5
 ]

L, S, L, L, S, L, S, L, L, S, L, L, S

And for the list of first 5 iterates:

NestList[
 Replace[#, L -> Sequence[L, S], S -> Sequence[L], 1] &,
 L,
 5
 ]

L, L, S, L, S, L, L, S, L, L, S, L, S, L, L, S, L, S,
L, L, S, L, L, S, L, S, L, L, S, L, L, S

Edit:

SubstitutionSystem operates on strings instead. It is a bit faster because it is specifically designed for Lindenmayer systems:

L = "L";
S = "S";
a = StringJoin /@ NestList[
 Replace[#, L -> Sequence[L, S], S -> Sequence[L], 1] &,
 L,
 20
 ]; // AbsoluteTiming // First
b = SubstitutionSystem["L" -> "LS", "S" -> "L", "L", 20]; // AbsoluteTiming // First

a == b

0.004506

0.001265

True

share|improve this answer

edited 18 mins ago

answered 42 mins ago

Henrik Schumacher

38.1k250109

share|improve this answer

share|improve this answer

edited 18 mins ago

edited 18 mins ago

edited 18 mins ago

answered 42 mins ago

Henrik Schumacher

38.1k250109

answered 42 mins ago

Henrik Schumacher

38.1k250109

answered 42 mins ago

Henrik Schumacher

38.1k250109

38.1k250109

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What if I just wanted a random sequence of $L$ and $S$?
  – SuperCiocia
  36 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then you can use RandomChoice: RandomChoice[L, S, 10000] will generate a list of 10000 random letter at once (uniformly distributed). You can also prescribe probabilities for each letter, e.g. RandomChoice[0.1, 0.9 -> L, S, 10000].
  – Henrik Schumacher
  34 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Great, thank you.
  – SuperCiocia
  32 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're welcome.
  – Henrik Schumacher
  29 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @SuperCiocia Please forget what I said about SequenceReplace. Better use Replace with level specification 1. That does also not suffer from using symbols.
  – Henrik Schumacher
  17 mins ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What if I just wanted a random sequence of $L$ and $S$?
  – SuperCiocia
  36 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then you can use RandomChoice: RandomChoice[L, S, 10000] will generate a list of 10000 random letter at once (uniformly distributed). You can also prescribe probabilities for each letter, e.g. RandomChoice[0.1, 0.9 -> L, S, 10000].
  – Henrik Schumacher
  34 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Great, thank you.
  – SuperCiocia
  32 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You're welcome.
  – Henrik Schumacher
  29 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @SuperCiocia Please forget what I said about SequenceReplace. Better use Replace with level specification 1. That does also not suffer from using symbols.
  – Henrik Schumacher
  17 mins ago
  
  

  
  
  
  

What if I just wanted a random sequence of $L$ and $S$?
– SuperCiocia
36 mins ago

What if I just wanted a random sequence of $L$ and $S$?
– SuperCiocia
36 mins ago

Then you can use RandomChoice: RandomChoice[L, S, 10000] will generate a list of 10000 random letter at once (uniformly distributed). You can also prescribe probabilities for each letter, e.g. RandomChoice[0.1, 0.9 -> L, S, 10000].
– Henrik Schumacher
34 mins ago

Then you can use RandomChoice: RandomChoice[L, S, 10000] will generate a list of 10000 random letter at once (uniformly distributed). You can also prescribe probabilities for each letter, e.g. RandomChoice[0.1, 0.9 -> L, S, 10000].
– Henrik Schumacher
34 mins ago

Great, thank you.
– SuperCiocia
32 mins ago

Great, thank you.
– SuperCiocia
32 mins ago

You're welcome.
– Henrik Schumacher
29 mins ago

You're welcome.
– Henrik Schumacher
29 mins ago

@SuperCiocia Please forget what I said about SequenceReplace. Better use Replace with level specification 1. That does also not suffer from using symbols.
– Henrik Schumacher
17 mins ago

@SuperCiocia Please forget what I said about SequenceReplace. Better use Replace with level specification 1. That does also not suffer from using symbols.
– Henrik Schumacher
17 mins ago

add a comment | 

up vote
1
down vote

You may use ReplaceRepeatedwith its MaxIterationsoption.

ReplaceRepeated[L, L -> Sequence[L, S], S -> Sequence[L], 
 MaxIterations -> 5]

L, S, L, L, S, L, S, L, L, S, L, L, S

Hope this helps.

share

answered 7 mins ago

Edmund

24.4k32995

add a comment | 

up vote
1
down vote

You may use ReplaceRepeatedwith its MaxIterationsoption.

ReplaceRepeated[L, L -> Sequence[L, S], S -> Sequence[L], 
 MaxIterations -> 5]

L, S, L, L, S, L, S, L, L, S, L, L, S

Hope this helps.

share

answered 7 mins ago

Edmund

24.4k32995

add a comment | 

up vote
1
down vote

up vote
1
down vote

You may use ReplaceRepeatedwith its MaxIterationsoption.

ReplaceRepeated[L, L -> Sequence[L, S], S -> Sequence[L], 
 MaxIterations -> 5]

L, S, L, L, S, L, S, L, L, S, L, L, S

Hope this helps.

share

answered 7 mins ago

Edmund

24.4k32995

You may use ReplaceRepeatedwith its MaxIterationsoption.

ReplaceRepeated[L, L -> Sequence[L, S], S -> Sequence[L], 
 MaxIterations -> 5]

L, S, L, L, S, L, S, L, L, S, L, L, S

Hope this helps.

share

answered 7 mins ago

Edmund

24.4k32995

share

share

answered 7 mins ago

Edmund

24.4k32995

answered 7 mins ago

Edmund

24.4k32995

answered 7 mins ago

Edmund

24.4k32995

24.4k32995

add a comment | 

add a comment | 

 

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