Simulate Bitwise Cyclic TagSimulate a Minsky Register Machine (I)Simulate a Minsky Register Machine...
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Simulate Bitwise Cyclic Tag
Simulate a Minsky Register Machine (I)Simulate a Minsky Register Machine (II)Simulate any 1D cellular automatonTuring Machine SimulatorInterpret /// (pronounced 'slashes')Simulate a Cyclic Tag SystemOutput “Fit” numbersThe Binary Square Diagonal SequenceHalting Problem for Simplified HexagonySimulate a Cisco Ping
$begingroup$
Challenge
Given two strings in any default I/O format, do the following:
NOTE: The challenge will refer to the first string as the "data" and the second referred to as the "program".
- Change the program to an infinite string which is just the program repeated infinitely (e.g.
10-->1010101010...). The challenge will refer to this as the "infinite program"
While the data is non-empty, do the following while looping over the infinite program:
a. If the current command is "0", delete the left-most bit in the data. If the data is empty, "0" does not do anything.
b. If the current command is "1", append the next character in the program to the data if the left-most bit in the data is a one.
c. If the data is not empty now, output the data.
Test Cases
Data is the left side of the input and the program is the right side.
100, 0 --> 00, 0
1111, 1 --> 11111, 111111, 1111111, ...
10, 011 --> 0, 0, 0
1110, 011 --> 110, 1101, 11010, 1010...
Notes
- The data and program will consist of only 0s and 1s
- For data/programs that do not halt, your program does not need to halt.
- The data and program will not be empty in the input.
- You may have multiple trailing and leading newlines
Standard Loopholes are forbidden- You can use any convenient I/O format
As always with code-golf, shortest code wins!
code-golf interpreter
asked yesterday
MilkyWay90MilkyWay90
740316
$endgroup$
|
show 11 more comments
$begingroup$
Challenge
Given two strings in any default I/O format, do the following:
NOTE: The challenge will refer to the first string as the "data" and the second referred to as the "program".
- Change the program to an infinite string which is just the program repeated infinitely (e.g.
10-->1010101010...). The challenge will refer to this as the "infinite program"
While the data is non-empty, do the following while looping over the infinite program:
a. If the current command is "0", delete the left-most bit in the data. If the data is empty, "0" does not do anything.
b. If the current command is "1", append the next character in the program to the data if the left-most bit in the data is a one.
c. If the data is not empty now, output the data.
Test Cases
Data is the left side of the input and the program is the right side.
100, 0 --> 00, 0
1111, 1 --> 11111, 111111, 1111111, ...
10, 011 --> 0, 0, 0
1110, 011 --> 110, 1101, 11010, 1010...
Notes
- The data and program will consist of only 0s and 1s
- For data/programs that do not halt, your program does not need to halt.
- The data and program will not be empty in the input.
- You may have multiple trailing and leading newlines
Standard Loopholes are forbidden- You can use any convenient I/O format
As always with code-golf, shortest code wins!
code-golf interpreter
asked yesterday
MilkyWay90MilkyWay90
740316
$endgroup$
$begingroup$
This seems a duplicate of /questions/40300/simulate-a-cyclic-tag-system
$endgroup$
– Sanchises
yesterday
$begingroup$
@Sanchises Seems like a borderline duplicate to that, but you have to get the result at a certain generation and that is for any cyclic tag system.
$endgroup$
– MilkyWay90
yesterday
$begingroup$
in the first test case,100goes to10on cmd0, whose definition is "delete the left-most bit in the data." wouldn't the leftmost bit of100be1?
$endgroup$
– Jonah
yesterday
$begingroup$
@Jonah Oh, missed that
$endgroup$
– MilkyWay90
yesterday
$begingroup$
in case (b), if you do the append, does the instruction pointer move right one or two characters?
$endgroup$
– Sparr
yesterday
|
show 11 more comments
$begingroup$
Challenge
Given two strings in any default I/O format, do the following:
NOTE: The challenge will refer to the first string as the "data" and the second referred to as the "program".
- Change the program to an infinite string which is just the program repeated infinitely (e.g.
10-->1010101010...). The challenge will refer to this as the "infinite program"
While the data is non-empty, do the following while looping over the infinite program:
a. If the current command is "0", delete the left-most bit in the data. If the data is empty, "0" does not do anything.
b. If the current command is "1", append the next character in the program to the data if the left-most bit in the data is a one.
c. If the data is not empty now, output the data.
Test Cases
Data is the left side of the input and the program is the right side.
100, 0 --> 00, 0
1111, 1 --> 11111, 111111, 1111111, ...
10, 011 --> 0, 0, 0
1110, 011 --> 110, 1101, 11010, 1010...
Notes
- The data and program will consist of only 0s and 1s
- For data/programs that do not halt, your program does not need to halt.
- The data and program will not be empty in the input.
- You may have multiple trailing and leading newlines
Standard Loopholes are forbidden- You can use any convenient I/O format
As always with code-golf, shortest code wins!
code-golf interpreter
asked yesterday
MilkyWay90MilkyWay90
740316
$endgroup$
Challenge
Given two strings in any default I/O format, do the following:
NOTE: The challenge will refer to the first string as the "data" and the second referred to as the "program".
- Change the program to an infinite string which is just the program repeated infinitely (e.g.
10-->1010101010...). The challenge will refer to this as the "infinite program"
While the data is non-empty, do the following while looping over the infinite program:
a. If the current command is "0", delete the left-most bit in the data. If the data is empty, "0" does not do anything.
b. If the current command is "1", append the next character in the program to the data if the left-most bit in the data is a one.
c. If the data is not empty now, output the data.
Test Cases
Data is the left side of the input and the program is the right side.
100, 0 --> 00, 0
1111, 1 --> 11111, 111111, 1111111, ...
10, 011 --> 0, 0, 0
1110, 011 --> 110, 1101, 11010, 1010...
Notes
- The data and program will consist of only 0s and 1s
- For data/programs that do not halt, your program does not need to halt.
- The data and program will not be empty in the input.
- You may have multiple trailing and leading newlines
Standard Loopholes are forbidden- You can use any convenient I/O format
As always with code-golf, shortest code wins!
code-golf interpreter
code-golf interpreter
asked yesterday
MilkyWay90MilkyWay90
740316
asked yesterday
MilkyWay90MilkyWay90
740316
edited yesterday
MilkyWay90
asked yesterday
MilkyWay90MilkyWay90
740316
asked yesterday
MilkyWay90MilkyWay90
740316
asked yesterday
MilkyWay90MilkyWay90
740316
740316
$begingroup$
This seems a duplicate of /questions/40300/simulate-a-cyclic-tag-system
$endgroup$
– Sanchises
yesterday
$begingroup$
@Sanchises Seems like a borderline duplicate to that, but you have to get the result at a certain generation and that is for any cyclic tag system.
$endgroup$
– MilkyWay90
yesterday
$begingroup$
in the first test case,100goes to10on cmd0, whose definition is "delete the left-most bit in the data." wouldn't the leftmost bit of100be1?
$endgroup$
– Jonah
yesterday
$begingroup$
@Jonah Oh, missed that
$endgroup$
– MilkyWay90
yesterday
$begingroup$
in case (b), if you do the append, does the instruction pointer move right one or two characters?
$endgroup$
– Sparr
yesterday
|
show 11 more comments
$begingroup$
This seems a duplicate of /questions/40300/simulate-a-cyclic-tag-system
$endgroup$
– Sanchises
yesterday
$begingroup$
@Sanchises Seems like a borderline duplicate to that, but you have to get the result at a certain generation and that is for any cyclic tag system.
$endgroup$
– MilkyWay90
yesterday
$begingroup$
in the first test case,100goes to10on cmd0, whose definition is "delete the left-most bit in the data." wouldn't the leftmost bit of100be1?
$endgroup$
– Jonah
yesterday
$begingroup$
@Jonah Oh, missed that
$endgroup$
– MilkyWay90
yesterday
$begingroup$
in case (b), if you do the append, does the instruction pointer move right one or two characters?
$endgroup$
– Sparr
yesterday
$begingroup$
This seems a duplicate of /questions/40300/simulate-a-cyclic-tag-system
$endgroup$
– Sanchises
yesterday
$begingroup$
This seems a duplicate of /questions/40300/simulate-a-cyclic-tag-system
$endgroup$
– Sanchises
yesterday
$begingroup$
@Sanchises Seems like a borderline duplicate to that, but you have to get the result at a certain generation and that is for any cyclic tag system.
$endgroup$
– MilkyWay90
yesterday
$begingroup$
@Sanchises Seems like a borderline duplicate to that, but you have to get the result at a certain generation and that is for any cyclic tag system.
$endgroup$
– MilkyWay90
yesterday
$begingroup$
in the first test case,
100 goes to 10 on cmd 0, whose definition is "delete the left-most bit in the data." wouldn't the leftmost bit of 100 be 1?$endgroup$
– Jonah
yesterday
$begingroup$
in the first test case,
100 goes to 10 on cmd 0, whose definition is "delete the left-most bit in the data." wouldn't the leftmost bit of 100 be 1?$endgroup$
– Jonah
yesterday
$begingroup$
@Jonah Oh, missed that
$endgroup$
– MilkyWay90
yesterday
$begingroup$
@Jonah Oh, missed that
$endgroup$
– MilkyWay90
yesterday
$begingroup$
in case (b), if you do the append, does the instruction pointer move right one or two characters?
$endgroup$
– Sparr
yesterday
$begingroup$
in case (b), if you do the append, does the instruction pointer move right one or two characters?
$endgroup$
– Sparr
yesterday
|
show 11 more comments
8 Answers
8
active
oldest
votes
$begingroup$
Haskell, 77 71 bytes
f@(d:e)#(p:q)=f:[[e,e],[f,f++[q!!0]]]!!p!!d#q
_#_=[]
a!b=tail$a#cycle b
Try it online!
answered yesterday
niminimi
32.6k32489
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 82 bytes
m=>n=>{for(int i=0;m!="";Print(m=n[i++]<49?m.Substring(1):m[0]>48?m+n[i]:m))n+=n;}
Try it online!
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
$endgroup$
$begingroup$
what are the significance of the 48 and 49, out of curiosity?
$endgroup$
– Jonah
yesterday
1
$begingroup$
@Jonah 48 is the ASCII value of0, and 49 is the ASCII value of1
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
shouldn't you use 0 and 1 instead here :P
$endgroup$
– ASCII-only
yesterday
$begingroup$
@ASCII-only I'm using a string, not an array.
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@EmbodimentofIgnorance why not use aListandSkip, or something like that
$endgroup$
– ASCII-only
yesterday
|
show 1 more comment
$begingroup$
J, 65 bytes
(([:(][echo)(}.@[)`([,{.@[#1{],])@.({.@]));1|.])&>/^:(0<0#@{>)^:5
Try it online!
I may golf this further later. Note the 5 at the end would be infinity _ in the actual program, but I've left it there to make running the non-halting examples easier.
answered yesterday
JonahJonah
2,6611017
$endgroup$
add a comment |
$begingroup$
Python 2, 96 82 bytes
def g(d,p):
while d:
c=p[0];p=p[1:]+[c];d=[d[1:],d+[p[0]]*d[0]][c]
if d:yield d
Try it online!
Stealing a bit from Emodiment of Ignorance's answer...
A generator which uses lists of 1's and 0's for input / output.
answered yesterday
Chas BrownChas Brown
5,2091523
$endgroup$
add a comment |
$begingroup$
Jelly, 40 bytes
;€Ø2œịxØ1œị$Ʋ$Ḋ€2,1œị$?1¦ṙ€1$2¦µ⁺1ịGṄƲ¿Ḣ
Try it online!
I’ve assumed trailing newlines are ok. I’ve also gone with a list of two lists of zeros and ones as input, and output to stdout.
answered yesterday
Nick KennedyNick Kennedy
1,36649
$endgroup$
add a comment |
$begingroup$
Python 3, 74 bytes
def f(d,p):
while d:c,*p=p+p[:1];d=(d[1:],d+p[:1]*d[0])[c];d and print(d)
Try it online!
Arguments: d: data, p: program.
answered yesterday
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
add a comment |
$begingroup$
Python 1, 78 bytes
a,b=input()
while a:
b=b[1:]+b[:1];a=[a[1:],a+b[:1]*a[0]][b[0]]
if a:print a
Try it online!
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
$endgroup$
$begingroup$
Nice! A niggle: for data '100', program '0', this will print the empty string once: but rule c says "If the data is not empty now, output the data."
$endgroup$
– Chas Brown
yesterday
$begingroup$
@ChasBrown Small typo, I'm waiting for clarification from the OP if trailing newlines are ok
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@ChasBrown The OP says multiple trailing newlines are allowed, see here
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
But after switching to arrays of 1's and 0s, now you're printing an empty array[]instead of a newline on e.g., data[1,0,0], program[0].
$endgroup$
– Chas Brown
yesterday
$begingroup$
python 1? python 2 doesn't work?
$endgroup$
– ASCII-only
yesterday
add a comment |
$begingroup$
05AB1E, 24 21 bytes
[¹Nèi¬i¹N>è«}ë¦}DõQ#=
Takes the program as first input and data as second input.input.
Try it online.
Explanation:
[ # Start an infinite loop:
¹Nè # Get the N'th digit of the first (program) input
# (NOTES: N is the index of the infinite loop;
# indexing in 05AB1E automatically wraps around)
i # If this digit is 1:
¬ # Push the head of the current data (without popping it)
# (will take the second (data) input implicitly if it's the first iteration)
i } # If this head is 1:
¹N>è # Get the (N+1)'th digit of the first (program) input
« # And append it to the current data
ë } # Else (the digit is a 0 instead):
¦ # Remove the first digit from the current data
# (will take the second input (data) implicitly if it's the first iteration)
DõQ # If the current data is an empty string:
# # Stop the infinite loop
= # Print the current data with trailing newline (without popping it)
answered 16 hours ago
Kevin CruijssenKevin Cruijssen
42.4k570217
$endgroup$
add a comment |
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8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Haskell, 77 71 bytes
f@(d:e)#(p:q)=f:[[e,e],[f,f++[q!!0]]]!!p!!d#q
_#_=[]
a!b=tail$a#cycle b
Try it online!
answered yesterday
niminimi
32.6k32489
$endgroup$
add a comment |
$begingroup$
Haskell, 77 71 bytes
f@(d:e)#(p:q)=f:[[e,e],[f,f++[q!!0]]]!!p!!d#q
_#_=[]
a!b=tail$a#cycle b
Try it online!
answered yesterday
niminimi
32.6k32489
$endgroup$
add a comment |
$begingroup$
Haskell, 77 71 bytes
f@(d:e)#(p:q)=f:[[e,e],[f,f++[q!!0]]]!!p!!d#q
_#_=[]
a!b=tail$a#cycle b
Try it online!
answered yesterday
niminimi
32.6k32489
$endgroup$
Haskell, 77 71 bytes
f@(d:e)#(p:q)=f:[[e,e],[f,f++[q!!0]]]!!p!!d#q
_#_=[]
a!b=tail$a#cycle b
Try it online!
answered yesterday
niminimi
32.6k32489
edited yesterday
answered yesterday
niminimi
32.6k32489
answered yesterday
niminimi
32.6k32489
answered yesterday
niminimi
32.6k32489
32.6k32489
add a comment |
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 82 bytes
m=>n=>{for(int i=0;m!="";Print(m=n[i++]<49?m.Substring(1):m[0]>48?m+n[i]:m))n+=n;}
Try it online!
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
$endgroup$
$begingroup$
what are the significance of the 48 and 49, out of curiosity?
$endgroup$
– Jonah
yesterday
1
$begingroup$
@Jonah 48 is the ASCII value of0, and 49 is the ASCII value of1
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
shouldn't you use 0 and 1 instead here :P
$endgroup$
– ASCII-only
yesterday
$begingroup$
@ASCII-only I'm using a string, not an array.
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@EmbodimentofIgnorance why not use aListandSkip, or something like that
$endgroup$
– ASCII-only
yesterday
|
show 1 more comment
$begingroup$
C# (Visual C# Interactive Compiler), 82 bytes
m=>n=>{for(int i=0;m!="";Print(m=n[i++]<49?m.Substring(1):m[0]>48?m+n[i]:m))n+=n;}
Try it online!
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
$endgroup$
$begingroup$
what are the significance of the 48 and 49, out of curiosity?
$endgroup$
– Jonah
yesterday
1
$begingroup$
@Jonah 48 is the ASCII value of0, and 49 is the ASCII value of1
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
shouldn't you use 0 and 1 instead here :P
$endgroup$
– ASCII-only
yesterday
$begingroup$
@ASCII-only I'm using a string, not an array.
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@EmbodimentofIgnorance why not use aListandSkip, or something like that
$endgroup$
– ASCII-only
yesterday
|
show 1 more comment
$begingroup$
C# (Visual C# Interactive Compiler), 82 bytes
m=>n=>{for(int i=0;m!="";Print(m=n[i++]<49?m.Substring(1):m[0]>48?m+n[i]:m))n+=n;}
Try it online!
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
$endgroup$
C# (Visual C# Interactive Compiler), 82 bytes
m=>n=>{for(int i=0;m!="";Print(m=n[i++]<49?m.Substring(1):m[0]>48?m+n[i]:m))n+=n;}
Try it online!
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
edited yesterday
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
2,866127
$begingroup$
what are the significance of the 48 and 49, out of curiosity?
$endgroup$
– Jonah
yesterday
1
$begingroup$
@Jonah 48 is the ASCII value of0, and 49 is the ASCII value of1
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
shouldn't you use 0 and 1 instead here :P
$endgroup$
– ASCII-only
yesterday
$begingroup$
@ASCII-only I'm using a string, not an array.
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@EmbodimentofIgnorance why not use aListandSkip, or something like that
$endgroup$
– ASCII-only
yesterday
|
show 1 more comment
$begingroup$
what are the significance of the 48 and 49, out of curiosity?
$endgroup$
– Jonah
yesterday
1
$begingroup$
@Jonah 48 is the ASCII value of0, and 49 is the ASCII value of1
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
shouldn't you use 0 and 1 instead here :P
$endgroup$
– ASCII-only
yesterday
$begingroup$
@ASCII-only I'm using a string, not an array.
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@EmbodimentofIgnorance why not use aListandSkip, or something like that
$endgroup$
– ASCII-only
yesterday
$begingroup$
what are the significance of the 48 and 49, out of curiosity?
$endgroup$
– Jonah
yesterday
$begingroup$
what are the significance of the 48 and 49, out of curiosity?
$endgroup$
– Jonah
yesterday
1
1
$begingroup$
@Jonah 48 is the ASCII value of
0, and 49 is the ASCII value of 1$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@Jonah 48 is the ASCII value of
0, and 49 is the ASCII value of 1$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
shouldn't you use 0 and 1 instead here :P
$endgroup$
– ASCII-only
yesterday
$begingroup$
shouldn't you use 0 and 1 instead here :P
$endgroup$
– ASCII-only
yesterday
$begingroup$
@ASCII-only I'm using a string, not an array.
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@ASCII-only I'm using a string, not an array.
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@EmbodimentofIgnorance why not use a
List and Skip, or something like that$endgroup$
– ASCII-only
yesterday
$begingroup$
@EmbodimentofIgnorance why not use a
List and Skip, or something like that$endgroup$
– ASCII-only
yesterday
|
show 1 more comment
$begingroup$
J, 65 bytes
(([:(][echo)(}.@[)`([,{.@[#1{],])@.({.@]));1|.])&>/^:(0<0#@{>)^:5
Try it online!
I may golf this further later. Note the 5 at the end would be infinity _ in the actual program, but I've left it there to make running the non-halting examples easier.
answered yesterday
JonahJonah
2,6611017
$endgroup$
add a comment |
$begingroup$
J, 65 bytes
(([:(][echo)(}.@[)`([,{.@[#1{],])@.({.@]));1|.])&>/^:(0<0#@{>)^:5
Try it online!
I may golf this further later. Note the 5 at the end would be infinity _ in the actual program, but I've left it there to make running the non-halting examples easier.
answered yesterday
JonahJonah
2,6611017
$endgroup$
add a comment |
$begingroup$
J, 65 bytes
(([:(][echo)(}.@[)`([,{.@[#1{],])@.({.@]));1|.])&>/^:(0<0#@{>)^:5
Try it online!
I may golf this further later. Note the 5 at the end would be infinity _ in the actual program, but I've left it there to make running the non-halting examples easier.
answered yesterday
JonahJonah
2,6611017
$endgroup$
J, 65 bytes
(([:(][echo)(}.@[)`([,{.@[#1{],])@.({.@]));1|.])&>/^:(0<0#@{>)^:5
Try it online!
I may golf this further later. Note the 5 at the end would be infinity _ in the actual program, but I've left it there to make running the non-halting examples easier.
answered yesterday
JonahJonah
2,6611017
edited yesterday
answered yesterday
JonahJonah
2,6611017
answered yesterday
JonahJonah
2,6611017
answered yesterday
JonahJonah
2,6611017
2,6611017
add a comment |
add a comment |
$begingroup$
Python 2, 96 82 bytes
def g(d,p):
while d:
c=p[0];p=p[1:]+[c];d=[d[1:],d+[p[0]]*d[0]][c]
if d:yield d
Try it online!
Stealing a bit from Emodiment of Ignorance's answer...
A generator which uses lists of 1's and 0's for input / output.
answered yesterday
Chas BrownChas Brown
5,2091523
$endgroup$
add a comment |
$begingroup$
Python 2, 96 82 bytes
def g(d,p):
while d:
c=p[0];p=p[1:]+[c];d=[d[1:],d+[p[0]]*d[0]][c]
if d:yield d
Try it online!
Stealing a bit from Emodiment of Ignorance's answer...
A generator which uses lists of 1's and 0's for input / output.
answered yesterday
Chas BrownChas Brown
5,2091523
$endgroup$
add a comment |
$begingroup$
Python 2, 96 82 bytes
def g(d,p):
while d:
c=p[0];p=p[1:]+[c];d=[d[1:],d+[p[0]]*d[0]][c]
if d:yield d
Try it online!
Stealing a bit from Emodiment of Ignorance's answer...
A generator which uses lists of 1's and 0's for input / output.
answered yesterday
Chas BrownChas Brown
5,2091523
$endgroup$
Python 2, 96 82 bytes
def g(d,p):
while d:
c=p[0];p=p[1:]+[c];d=[d[1:],d+[p[0]]*d[0]][c]
if d:yield d
Try it online!
Stealing a bit from Emodiment of Ignorance's answer...
A generator which uses lists of 1's and 0's for input / output.
answered yesterday
Chas BrownChas Brown
5,2091523
edited yesterday
answered yesterday
Chas BrownChas Brown
5,2091523
answered yesterday
Chas BrownChas Brown
5,2091523
answered yesterday
Chas BrownChas Brown
5,2091523
5,2091523
add a comment |
add a comment |
$begingroup$
Jelly, 40 bytes
;€Ø2œịxØ1œị$Ʋ$Ḋ€2,1œị$?1¦ṙ€1$2¦µ⁺1ịGṄƲ¿Ḣ
Try it online!
I’ve assumed trailing newlines are ok. I’ve also gone with a list of two lists of zeros and ones as input, and output to stdout.
answered yesterday
Nick KennedyNick Kennedy
1,36649
$endgroup$
add a comment |
$begingroup$
Jelly, 40 bytes
;€Ø2œịxØ1œị$Ʋ$Ḋ€2,1œị$?1¦ṙ€1$2¦µ⁺1ịGṄƲ¿Ḣ
Try it online!
I’ve assumed trailing newlines are ok. I’ve also gone with a list of two lists of zeros and ones as input, and output to stdout.
answered yesterday
Nick KennedyNick Kennedy
1,36649
$endgroup$
add a comment |
$begingroup$
Jelly, 40 bytes
;€Ø2œịxØ1œị$Ʋ$Ḋ€2,1œị$?1¦ṙ€1$2¦µ⁺1ịGṄƲ¿Ḣ
Try it online!
I’ve assumed trailing newlines are ok. I’ve also gone with a list of two lists of zeros and ones as input, and output to stdout.
answered yesterday
Nick KennedyNick Kennedy
1,36649
$endgroup$
Jelly, 40 bytes
;€Ø2œịxØ1œị$Ʋ$Ḋ€2,1œị$?1¦ṙ€1$2¦µ⁺1ịGṄƲ¿Ḣ
Try it online!
I’ve assumed trailing newlines are ok. I’ve also gone with a list of two lists of zeros and ones as input, and output to stdout.
answered yesterday
Nick KennedyNick Kennedy
1,36649
answered yesterday
Nick KennedyNick Kennedy
1,36649
answered yesterday
Nick KennedyNick Kennedy
1,36649
answered yesterday
Nick KennedyNick Kennedy
1,36649
1,36649
add a comment |
add a comment |
$begingroup$
Python 3, 74 bytes
def f(d,p):
while d:c,*p=p+p[:1];d=(d[1:],d+p[:1]*d[0])[c];d and print(d)
Try it online!
Arguments: d: data, p: program.
answered yesterday
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
add a comment |
$begingroup$
Python 3, 74 bytes
def f(d,p):
while d:c,*p=p+p[:1];d=(d[1:],d+p[:1]*d[0])[c];d and print(d)
Try it online!
Arguments: d: data, p: program.
answered yesterday
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
add a comment |
$begingroup$
Python 3, 74 bytes
def f(d,p):
while d:c,*p=p+p[:1];d=(d[1:],d+p[:1]*d[0])[c];d and print(d)
Try it online!
Arguments: d: data, p: program.
answered yesterday
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
Python 3, 74 bytes
def f(d,p):
while d:c,*p=p+p[:1];d=(d[1:],d+p[:1]*d[0])[c];d and print(d)
Try it online!
Arguments: d: data, p: program.
answered yesterday
Erik the OutgolferErik the Outgolfer
33k429106
answered yesterday
Erik the OutgolferErik the Outgolfer
33k429106
answered yesterday
Erik the OutgolferErik the Outgolfer
33k429106
answered yesterday
Erik the OutgolferErik the Outgolfer
33k429106
33k429106
add a comment |
add a comment |
$begingroup$
Python 1, 78 bytes
a,b=input()
while a:
b=b[1:]+b[:1];a=[a[1:],a+b[:1]*a[0]][b[0]]
if a:print a
Try it online!
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
$endgroup$
$begingroup$
Nice! A niggle: for data '100', program '0', this will print the empty string once: but rule c says "If the data is not empty now, output the data."
$endgroup$
– Chas Brown
yesterday
$begingroup$
@ChasBrown Small typo, I'm waiting for clarification from the OP if trailing newlines are ok
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@ChasBrown The OP says multiple trailing newlines are allowed, see here
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
But after switching to arrays of 1's and 0s, now you're printing an empty array[]instead of a newline on e.g., data[1,0,0], program[0].
$endgroup$
– Chas Brown
yesterday
$begingroup$
python 1? python 2 doesn't work?
$endgroup$
– ASCII-only
yesterday
add a comment |
$begingroup$
Python 1, 78 bytes
a,b=input()
while a:
b=b[1:]+b[:1];a=[a[1:],a+b[:1]*a[0]][b[0]]
if a:print a
Try it online!
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
$endgroup$
$begingroup$
Nice! A niggle: for data '100', program '0', this will print the empty string once: but rule c says "If the data is not empty now, output the data."
$endgroup$
– Chas Brown
yesterday
$begingroup$
@ChasBrown Small typo, I'm waiting for clarification from the OP if trailing newlines are ok
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@ChasBrown The OP says multiple trailing newlines are allowed, see here
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
But after switching to arrays of 1's and 0s, now you're printing an empty array[]instead of a newline on e.g., data[1,0,0], program[0].
$endgroup$
– Chas Brown
yesterday
$begingroup$
python 1? python 2 doesn't work?
$endgroup$
– ASCII-only
yesterday
add a comment |
$begingroup$
Python 1, 78 bytes
a,b=input()
while a:
b=b[1:]+b[:1];a=[a[1:],a+b[:1]*a[0]][b[0]]
if a:print a
Try it online!
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
$endgroup$
Python 1, 78 bytes
a,b=input()
while a:
b=b[1:]+b[:1];a=[a[1:],a+b[:1]*a[0]][b[0]]
if a:print a
Try it online!
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
edited yesterday
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,866127
2,866127
$begingroup$
Nice! A niggle: for data '100', program '0', this will print the empty string once: but rule c says "If the data is not empty now, output the data."
$endgroup$
– Chas Brown
yesterday
$begingroup$
@ChasBrown Small typo, I'm waiting for clarification from the OP if trailing newlines are ok
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@ChasBrown The OP says multiple trailing newlines are allowed, see here
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
But after switching to arrays of 1's and 0s, now you're printing an empty array[]instead of a newline on e.g., data[1,0,0], program[0].
$endgroup$
– Chas Brown
yesterday
$begingroup$
python 1? python 2 doesn't work?
$endgroup$
– ASCII-only
yesterday
add a comment |
$begingroup$
Nice! A niggle: for data '100', program '0', this will print the empty string once: but rule c says "If the data is not empty now, output the data."
$endgroup$
– Chas Brown
yesterday
$begingroup$
@ChasBrown Small typo, I'm waiting for clarification from the OP if trailing newlines are ok
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@ChasBrown The OP says multiple trailing newlines are allowed, see here
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
But after switching to arrays of 1's and 0s, now you're printing an empty array[]instead of a newline on e.g., data[1,0,0], program[0].
$endgroup$
– Chas Brown
yesterday
$begingroup$
python 1? python 2 doesn't work?
$endgroup$
– ASCII-only
yesterday
$begingroup$
Nice! A niggle: for data '100', program '0', this will print the empty string once: but rule c says "If the data is not empty now, output the data."
$endgroup$
– Chas Brown
yesterday
$begingroup$
Nice! A niggle: for data '100', program '0', this will print the empty string once: but rule c says "If the data is not empty now, output the data."
$endgroup$
– Chas Brown
yesterday
$begingroup$
@ChasBrown Small typo, I'm waiting for clarification from the OP if trailing newlines are ok
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@ChasBrown Small typo, I'm waiting for clarification from the OP if trailing newlines are ok
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@ChasBrown The OP says multiple trailing newlines are allowed, see here
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
@ChasBrown The OP says multiple trailing newlines are allowed, see here
$endgroup$
– Embodiment of Ignorance
yesterday
$begingroup$
But after switching to arrays of 1's and 0s, now you're printing an empty array
[] instead of a newline on e.g., data [1,0,0] , program [0].$endgroup$
– Chas Brown
yesterday
$begingroup$
But after switching to arrays of 1's and 0s, now you're printing an empty array
[] instead of a newline on e.g., data [1,0,0] , program [0].$endgroup$
– Chas Brown
yesterday
$begingroup$
python 1? python 2 doesn't work?
$endgroup$
– ASCII-only
yesterday
$begingroup$
python 1? python 2 doesn't work?
$endgroup$
– ASCII-only
yesterday
add a comment |
$begingroup$
05AB1E, 24 21 bytes
[¹Nèi¬i¹N>è«}ë¦}DõQ#=
Takes the program as first input and data as second input.input.
Try it online.
Explanation:
[ # Start an infinite loop:
¹Nè # Get the N'th digit of the first (program) input
# (NOTES: N is the index of the infinite loop;
# indexing in 05AB1E automatically wraps around)
i # If this digit is 1:
¬ # Push the head of the current data (without popping it)
# (will take the second (data) input implicitly if it's the first iteration)
i } # If this head is 1:
¹N>è # Get the (N+1)'th digit of the first (program) input
« # And append it to the current data
ë } # Else (the digit is a 0 instead):
¦ # Remove the first digit from the current data
# (will take the second input (data) implicitly if it's the first iteration)
DõQ # If the current data is an empty string:
# # Stop the infinite loop
= # Print the current data with trailing newline (without popping it)
answered 16 hours ago
Kevin CruijssenKevin Cruijssen
42.4k570217
$endgroup$
add a comment |
$begingroup$
05AB1E, 24 21 bytes
[¹Nèi¬i¹N>è«}ë¦}DõQ#=
Takes the program as first input and data as second input.input.
Try it online.
Explanation:
[ # Start an infinite loop:
¹Nè # Get the N'th digit of the first (program) input
# (NOTES: N is the index of the infinite loop;
# indexing in 05AB1E automatically wraps around)
i # If this digit is 1:
¬ # Push the head of the current data (without popping it)
# (will take the second (data) input implicitly if it's the first iteration)
i } # If this head is 1:
¹N>è # Get the (N+1)'th digit of the first (program) input
« # And append it to the current data
ë } # Else (the digit is a 0 instead):
¦ # Remove the first digit from the current data
# (will take the second input (data) implicitly if it's the first iteration)
DõQ # If the current data is an empty string:
# # Stop the infinite loop
= # Print the current data with trailing newline (without popping it)
answered 16 hours ago
Kevin CruijssenKevin Cruijssen
42.4k570217
$endgroup$
add a comment |
$begingroup$
05AB1E, 24 21 bytes
[¹Nèi¬i¹N>è«}ë¦}DõQ#=
Takes the program as first input and data as second input.input.
Try it online.
Explanation:
[ # Start an infinite loop:
¹Nè # Get the N'th digit of the first (program) input
# (NOTES: N is the index of the infinite loop;
# indexing in 05AB1E automatically wraps around)
i # If this digit is 1:
¬ # Push the head of the current data (without popping it)
# (will take the second (data) input implicitly if it's the first iteration)
i } # If this head is 1:
¹N>è # Get the (N+1)'th digit of the first (program) input
« # And append it to the current data
ë } # Else (the digit is a 0 instead):
¦ # Remove the first digit from the current data
# (will take the second input (data) implicitly if it's the first iteration)
DõQ # If the current data is an empty string:
# # Stop the infinite loop
= # Print the current data with trailing newline (without popping it)
answered 16 hours ago
Kevin CruijssenKevin Cruijssen
42.4k570217
$endgroup$
05AB1E, 24 21 bytes
[¹Nèi¬i¹N>è«}ë¦}DõQ#=
Takes the program as first input and data as second input.input.
Try it online.
Explanation:
[ # Start an infinite loop:
¹Nè # Get the N'th digit of the first (program) input
# (NOTES: N is the index of the infinite loop;
# indexing in 05AB1E automatically wraps around)
i # If this digit is 1:
¬ # Push the head of the current data (without popping it)
# (will take the second (data) input implicitly if it's the first iteration)
i } # If this head is 1:
¹N>è # Get the (N+1)'th digit of the first (program) input
« # And append it to the current data
ë } # Else (the digit is a 0 instead):
¦ # Remove the first digit from the current data
# (will take the second input (data) implicitly if it's the first iteration)
DõQ # If the current data is an empty string:
# # Stop the infinite loop
= # Print the current data with trailing newline (without popping it)
answered 16 hours ago
Kevin CruijssenKevin Cruijssen
42.4k570217
edited 16 hours ago
answered 16 hours ago
Kevin CruijssenKevin Cruijssen
42.4k570217
answered 16 hours ago
Kevin CruijssenKevin Cruijssen
42.4k570217
answered 16 hours ago
Kevin CruijssenKevin Cruijssen
42.4k570217
42.4k570217
add a comment |
add a comment |
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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$begingroup$
This seems a duplicate of /questions/40300/simulate-a-cyclic-tag-system
$endgroup$
– Sanchises
yesterday
$begingroup$
@Sanchises Seems like a borderline duplicate to that, but you have to get the result at a certain generation and that is for any cyclic tag system.
$endgroup$
– MilkyWay90
yesterday
$begingroup$
in the first test case,
100goes to10on cmd0, whose definition is "delete the left-most bit in the data." wouldn't the leftmost bit of100be1?$endgroup$
– Jonah
yesterday
$begingroup$
@Jonah Oh, missed that
$endgroup$
– MilkyWay90
yesterday
$begingroup$
in case (b), if you do the append, does the instruction pointer move right one or two characters?
$endgroup$
– Sparr
yesterday