Backhanded Compliments

Clash Royale CLAN TAG#URR8PPP

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This challenge is created in celebration of my first esoteric language, Backhand!

Backhand is a one dimensional language with a non-linear pointer flow. The pointer moves three steps at a time, only executing every third instruction.

The program 1 1 + O @ will add 1+1 and output 2 before terminating. The intermediate instructions are skipped, so 1<>1()+O@ is exactly the same program.

When the pointer is about to step off the end of the tape, it instead reverses direction and steps the other way, so 1.@1.O+. is the same program. Note that it only counts the end instruction once. This allows us to compress most linear programs down, such as 1O+1@

Your challenge here is to write a program or function that take a string, and output the instructions that would be executed if the program was interpreted like Backhand (you don't need to handle any actual Backhand instructions). You are to only output until the pointer lands on the last character of the string (at which point the execution would normally go backwards).

But wait, that's not all! When your program itself is interpreted in this fashion, the resulting code should output one of the below:

• (Mostly) works


• Turing complete


• 
  'Recreational' (quotes can be either ' or ", but not both)


• Perfectly okay


• Only a few bugs

For example, if your source code is code 2 backhand, then the program ce cankb od2ahd should output one of these phrases.

Test cases:

"1 1 + O @" -> "11+O@"
"1O+1@" -> "11+O@"
"HoreWll dlo!" -> "Hello World!"
"abcdefghijklmnopqrstuvwxyz" -> "adgjmpsvyxurolifcbehknqtwz"
"0123456789" -> "0369" (not "0369630369")
"@" -> "@"
"io" -> "io" (Cat program in Backhand)
"!?O" -> "!?O" (Outputs random bits forever in Backhand)
"---!---!" -> "-!-----!"

And a reference program written in, of course, Backhand (this might be a bit buggy).

Rules.

• 
  Standard Loopholes are forbidden


• The input of the first program will contain only printable ASCII and newlines (that is, bytes 0x20-0x7E as well as 0x0A)


• You can choose whether your second program is converted from your first by bytes or by UTF-8 characters.


• Second program:
  
  
  • Case does not matter, so your output could be pErFectLy OKay if you want.
  
  
  • Any amount of trailing/leading whitespace (newline, tabs, spaces) are also okay.
  
  
  • The second program should be the same language as the first, though not necessarily the same format (program/function)
  
  
  • I'm happy to include suggestions from the comments on extra phrases (as long as they're not too short)
  
  


• As this is code-golf, your aim is to get the shortest answer for your language!


• In two weeks, I'll award a 200 bounty to the shortest Backhand answer.

code-golf source-layout

share|improve this question

edited 8 mins ago

asked 8 hours ago

Jo King

16.6k24190

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sandbox (deleted)
  – Jo King
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Suggested testcase: "---!---!" (or any string where the last character appears more than once)
  – TFeld
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  When your program itself is interpreted in this fashion - interpreted by what?
  – ngm
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  So let's say I write an R program (because that's pretty much all I do around here.) My R program has to transform the Backhanded code into the sequence of Backhanded instructions. In addition, my R program when input into into itself has to become another R program that outputs on of those strings when run (in the case of R, interpreted by an R interpreter). Is this correct?
  – ngm
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ngm Yes. -----
  – user202729
  5 hours ago
  

  
  
  
  

 | 
show 1 more comment

up vote
4
down vote

favorite

1

This challenge is created in celebration of my first esoteric language, Backhand!

Backhand is a one dimensional language with a non-linear pointer flow. The pointer moves three steps at a time, only executing every third instruction.

The program 1 1 + O @ will add 1+1 and output 2 before terminating. The intermediate instructions are skipped, so 1<>1()+O@ is exactly the same program.

When the pointer is about to step off the end of the tape, it instead reverses direction and steps the other way, so 1.@1.O+. is the same program. Note that it only counts the end instruction once. This allows us to compress most linear programs down, such as 1O+1@

Your challenge here is to write a program or function that take a string, and output the instructions that would be executed if the program was interpreted like Backhand (you don't need to handle any actual Backhand instructions). You are to only output until the pointer lands on the last character of the string (at which point the execution would normally go backwards).

But wait, that's not all! When your program itself is interpreted in this fashion, the resulting code should output one of the below:

• (Mostly) works


• Turing complete


• 
  'Recreational' (quotes can be either ' or ", but not both)


• Perfectly okay


• Only a few bugs

For example, if your source code is code 2 backhand, then the program ce cankb od2ahd should output one of these phrases.

Test cases:

"1 1 + O @" -> "11+O@"
"1O+1@" -> "11+O@"
"HoreWll dlo!" -> "Hello World!"
"abcdefghijklmnopqrstuvwxyz" -> "adgjmpsvyxurolifcbehknqtwz"
"0123456789" -> "0369" (not "0369630369")
"@" -> "@"
"io" -> "io" (Cat program in Backhand)
"!?O" -> "!?O" (Outputs random bits forever in Backhand)
"---!---!" -> "-!-----!"

And a reference program written in, of course, Backhand (this might be a bit buggy).

Rules.

• 
  Standard Loopholes are forbidden


• The input of the first program will contain only printable ASCII and newlines (that is, bytes 0x20-0x7E as well as 0x0A)


• You can choose whether your second program is converted from your first by bytes or by UTF-8 characters.


• Second program:
  
  
  • Case does not matter, so your output could be pErFectLy OKay if you want.
  
  
  • Any amount of trailing/leading whitespace (newline, tabs, spaces) are also okay.
  
  
  • The second program should be the same language as the first, though not necessarily the same format (program/function)
  
  
  • I'm happy to include suggestions from the comments on extra phrases (as long as they're not too short)
  
  


• As this is code-golf, your aim is to get the shortest answer for your language!


• In two weeks, I'll award a 200 bounty to the shortest Backhand answer.

code-golf source-layout

share|improve this question

edited 8 mins ago

asked 8 hours ago

Jo King

16.6k24190

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sandbox (deleted)
  – Jo King
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Suggested testcase: "---!---!" (or any string where the last character appears more than once)
  – TFeld
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  When your program itself is interpreted in this fashion - interpreted by what?
  – ngm
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  So let's say I write an R program (because that's pretty much all I do around here.) My R program has to transform the Backhanded code into the sequence of Backhanded instructions. In addition, my R program when input into into itself has to become another R program that outputs on of those strings when run (in the case of R, interpreted by an R interpreter). Is this correct?
  – ngm
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ngm Yes. -----
  – user202729
  5 hours ago
  

  
  
  
  

 | 
show 1 more comment

up vote
4
down vote

favorite

1

up vote
4
down vote

favorite

1

1

This challenge is created in celebration of my first esoteric language, Backhand!

Backhand is a one dimensional language with a non-linear pointer flow. The pointer moves three steps at a time, only executing every third instruction.

The program 1 1 + O @ will add 1+1 and output 2 before terminating. The intermediate instructions are skipped, so 1<>1()+O@ is exactly the same program.

When the pointer is about to step off the end of the tape, it instead reverses direction and steps the other way, so 1.@1.O+. is the same program. Note that it only counts the end instruction once. This allows us to compress most linear programs down, such as 1O+1@

Your challenge here is to write a program or function that take a string, and output the instructions that would be executed if the program was interpreted like Backhand (you don't need to handle any actual Backhand instructions). You are to only output until the pointer lands on the last character of the string (at which point the execution would normally go backwards).

But wait, that's not all! When your program itself is interpreted in this fashion, the resulting code should output one of the below:

• (Mostly) works


• Turing complete


• 
  'Recreational' (quotes can be either ' or ", but not both)


• Perfectly okay


• Only a few bugs

For example, if your source code is code 2 backhand, then the program ce cankb od2ahd should output one of these phrases.

Test cases:

"1 1 + O @" -> "11+O@"
"1O+1@" -> "11+O@"
"HoreWll dlo!" -> "Hello World!"
"abcdefghijklmnopqrstuvwxyz" -> "adgjmpsvyxurolifcbehknqtwz"
"0123456789" -> "0369" (not "0369630369")
"@" -> "@"
"io" -> "io" (Cat program in Backhand)
"!?O" -> "!?O" (Outputs random bits forever in Backhand)
"---!---!" -> "-!-----!"

And a reference program written in, of course, Backhand (this might be a bit buggy).

Rules.

• 
  Standard Loopholes are forbidden


• The input of the first program will contain only printable ASCII and newlines (that is, bytes 0x20-0x7E as well as 0x0A)


• You can choose whether your second program is converted from your first by bytes or by UTF-8 characters.


• Second program:
  
  
  • Case does not matter, so your output could be pErFectLy OKay if you want.
  
  
  • Any amount of trailing/leading whitespace (newline, tabs, spaces) are also okay.
  
  
  • The second program should be the same language as the first, though not necessarily the same format (program/function)
  
  
  • I'm happy to include suggestions from the comments on extra phrases (as long as they're not too short)
  
  


• As this is code-golf, your aim is to get the shortest answer for your language!


• In two weeks, I'll award a 200 bounty to the shortest Backhand answer.

code-golf source-layout

share|improve this question

edited 8 mins ago

asked 8 hours ago

Jo King

16.6k24190

This challenge is created in celebration of my first esoteric language, Backhand!

Backhand is a one dimensional language with a non-linear pointer flow. The pointer moves three steps at a time, only executing every third instruction.

The program 1 1 + O @ will add 1+1 and output 2 before terminating. The intermediate instructions are skipped, so 1<>1()+O@ is exactly the same program.

When the pointer is about to step off the end of the tape, it instead reverses direction and steps the other way, so 1.@1.O+. is the same program. Note that it only counts the end instruction once. This allows us to compress most linear programs down, such as 1O+1@

Your challenge here is to write a program or function that take a string, and output the instructions that would be executed if the program was interpreted like Backhand (you don't need to handle any actual Backhand instructions). You are to only output until the pointer lands on the last character of the string (at which point the execution would normally go backwards).

But wait, that's not all! When your program itself is interpreted in this fashion, the resulting code should output one of the below:

• (Mostly) works


• Turing complete


• 
  'Recreational' (quotes can be either ' or ", but not both)


• Perfectly okay


• Only a few bugs

For example, if your source code is code 2 backhand, then the program ce cankb od2ahd should output one of these phrases.

Test cases:

"1 1 + O @" -> "11+O@"
"1O+1@" -> "11+O@"
"HoreWll dlo!" -> "Hello World!"
"abcdefghijklmnopqrstuvwxyz" -> "adgjmpsvyxurolifcbehknqtwz"
"0123456789" -> "0369" (not "0369630369")
"@" -> "@"
"io" -> "io" (Cat program in Backhand)
"!?O" -> "!?O" (Outputs random bits forever in Backhand)
"---!---!" -> "-!-----!"

And a reference program written in, of course, Backhand (this might be a bit buggy).

Rules.

• 
  Standard Loopholes are forbidden


• The input of the first program will contain only printable ASCII and newlines (that is, bytes 0x20-0x7E as well as 0x0A)


• You can choose whether your second program is converted from your first by bytes or by UTF-8 characters.


• Second program:
  
  
  • Case does not matter, so your output could be pErFectLy OKay if you want.
  
  
  • Any amount of trailing/leading whitespace (newline, tabs, spaces) are also okay.
  
  
  • The second program should be the same language as the first, though not necessarily the same format (program/function)
  
  
  • I'm happy to include suggestions from the comments on extra phrases (as long as they're not too short)
  
  


• As this is code-golf, your aim is to get the shortest answer for your language!


• In two weeks, I'll award a 200 bounty to the shortest Backhand answer.

code-golf source-layout

code-golf source-layout

share|improve this question

edited 8 mins ago

asked 8 hours ago

Jo King

16.6k24190

share|improve this question

edited 8 mins ago

asked 8 hours ago

Jo King

16.6k24190

share|improve this question

share|improve this question

edited 8 mins ago

edited 8 mins ago

edited 8 mins ago

asked 8 hours ago

Jo King

16.6k24190

asked 8 hours ago

Jo King

16.6k24190

asked 8 hours ago

Jo King

16.6k24190

16.6k24190

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sandbox (deleted)
  – Jo King
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Suggested testcase: "---!---!" (or any string where the last character appears more than once)
  – TFeld
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  When your program itself is interpreted in this fashion - interpreted by what?
  – ngm
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  So let's say I write an R program (because that's pretty much all I do around here.) My R program has to transform the Backhanded code into the sequence of Backhanded instructions. In addition, my R program when input into into itself has to become another R program that outputs on of those strings when run (in the case of R, interpreted by an R interpreter). Is this correct?
  – ngm
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ngm Yes. -----
  – user202729
  5 hours ago
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sandbox (deleted)
  – Jo King
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Suggested testcase: "---!---!" (or any string where the last character appears more than once)
  – TFeld
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  When your program itself is interpreted in this fashion - interpreted by what?
  – ngm
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  So let's say I write an R program (because that's pretty much all I do around here.) My R program has to transform the Backhanded code into the sequence of Backhanded instructions. In addition, my R program when input into into itself has to become another R program that outputs on of those strings when run (in the case of R, interpreted by an R interpreter). Is this correct?
  – ngm
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ngm Yes. -----
  – user202729
  5 hours ago
  

  
  
  
  

Sandbox (deleted)
– Jo King
7 hours ago

Sandbox (deleted)
– Jo King
7 hours ago

1

1

Suggested testcase: "---!---!" (or any string where the last character appears more than once)
– TFeld
6 hours ago

Suggested testcase: "---!---!" (or any string where the last character appears more than once)
– TFeld
6 hours ago

When your program itself is interpreted in this fashion - interpreted by what?
– ngm
6 hours ago

When your program itself is interpreted in this fashion - interpreted by what?
– ngm
6 hours ago

4

4

So let's say I write an R program (because that's pretty much all I do around here.) My R program has to transform the Backhanded code into the sequence of Backhanded instructions. In addition, my R program when input into into itself has to become another R program that outputs on of those strings when run (in the case of R, interpreted by an R interpreter). Is this correct?
– ngm
5 hours ago

So let's say I write an R program (because that's pretty much all I do around here.) My R program has to transform the Backhanded code into the sequence of Backhanded instructions. In addition, my R program when input into into itself has to become another R program that outputs on of those strings when run (in the case of R, interpreted by an R interpreter). Is this correct?
– ngm
5 hours ago

1

1

@ngm Yes. -----
– user202729
5 hours ago

@ngm Yes. -----
– user202729
5 hours ago

 | 
show 1 more comment

5 Answers
5

active

oldest

votes

up vote
2
down vote

Python 2, 163 bytes

# p r i n t ' P e r f e c t l y o k a y ' # 
s=list(enumerate(input()))
S=(s+s[-2:0:-1]+s)[::3]
print''.join(zip(*S[:S.index(s[-1])+1])[1])

Try it online!

share|improve this answer

answered 6 hours ago

TFeld

12k2833

add a comment | 

up vote
2
down vote

Python 2, 130 bytes

p='r'#i n t ' P e r f e c t l y o k a y ' # 
s=input();S=s+s[-2:0:-1]+s
print S[:len(~-len(s)%3and S or s):3]

Try it online!

share|improve this answer

answered 5 hours ago

user202729

13k12549

add a comment | 

up vote
2
down vote

R, 187 bytes

 # c a t ( ' P e r f e c t l y o k a y ' ) #
g=function(x,n=nchar(x),i=c(1:n,(n-1):1,2:n),j=seq(1,3*n-2,3),k=i[j][1:which(i[j]==n)[1]])cat(substring(x,k,k),sep='') 

Try it online!

The single space at the end is needed so that the n never gets printed when the program is applied to itself.

Explanation

Part 1:

Ungolfed:

 # c a t ( ' P e r f e c t l y o k a y ' ) #
g <- function(x) 
 n <- nchar(x) # number of characters in string
 i <- c(1:n, (n - 1):1, 2:n) # index: 1 to n, n-1 back to 1, 2 to n
 j <- seq(1, 3 * n - 2, 3) # every third element of i
 k <- i[j][1:which(i[j] == n)[1]] # the elements of i at indices j, up to the first appearance of n
 cat(substring(x, k, k), sep = "") # extract from x the characters at indices k, and paste them together

Part 2:

The function produces this when it acts on the entire program:

cat('Perfectly okay')#=ni(ncr)=1,-:2)=q,n,,i]:i(j=assi(k)e' 

share|improve this answer

edited 2 hours ago

answered 2 hours ago

ngm

2,44920

add a comment | 

up vote
1
down vote

05AB1E, 41 bytes

“€€Ã€€Ð€€µ€€Ý€€“/g i që¬?3U[ûDXè?XIg<Ö#X3+U

Can without a doubt be golfed..

Try it online. (PS: Switch the language from 05AB1E (legacy) to 05AB1E for test cases 0123456789 and @. The legacy version is much faster, but it shows incorrect results for number inputs with leading zeros or single-char inputs.)

The 'backhanded' program will become:

“Ãеݓ q?[XX<XU

Which will output perfectly okay in full lowercase.

Try it online.

Explanation base program:

“€€Ã€€Ð€€µ€€Ý€€“ # Push the string "the pointed who the ethernet been the"
 / # And remove it from the stack again
g # Push the length of the (implicit) input
 i # If the length is exactly 1:
 q # Stop the program (and output the input implicitly)
 ë # Else:
 ¬? # Take the input implicitly again, and output its head
 3U # Push and pop 3, and store it in variable `X`
 [ # Start an infinite loop
 û # Palindromize the string at the top of the stack
 # i.e. "1O+1@" becomes "1O+1@1+O1" the first iteration,
 # and "1O+1@1+O1O+1@1+O1" the next iteration, etc.
 D # Duplicate the palindromized string
 Xè # Index variable `X` in the string
 ? # Output the character
 X # Check if variable `X`
 Ö # is divisible by
 Ig< # The length of the input - 1
 # # And if it is: stop the infinite loop
 X3+ # Then increase `X` by 3
 U # And pop and store it as the new `X`

Explanation 'backhanded' program:

“Ãеݓ # Push the string "perfectly okay"
 q # Exit the program (and implicitly output the top of the stack)
?[XX<XU # No-ops

See this 05AB1E tip of mine (section How to use the dictionary?) to understand why “€€Ã€€Ð€€µ€€Ý€€“ is the pointed who the ethernet been the and “Ãеݓ is perfectly okay.

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Kevin Cruijssen

30.9k553168

add a comment | 

up vote
1
down vote

JavaScript (ES6), 130 bytes

Early attempt. Not very satisfying.

f =/*> " P e r f e c t l y o k a*/y=>""+/**/(g=p=>(c=y[p])?m++%3?g(p+d):y[p+1]?c+g(p+d):c:g(p-d-d,d=-d))(m=0,d=1)

Try it online!

When the code is processed by itself, the following characters are isolated:

f =/*> " P e r f e c t l y o k a*/y=>""+/**/…
^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^

which gives:

f=>"Perfectly okay"//…

share|improve this answer

edited 26 mins ago

answered 2 hours ago

Arnauld

65.6k583277

add a comment | 

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

Python 2, 163 bytes

# p r i n t ' P e r f e c t l y o k a y ' # 
s=list(enumerate(input()))
S=(s+s[-2:0:-1]+s)[::3]
print''.join(zip(*S[:S.index(s[-1])+1])[1])

Try it online!

share|improve this answer

answered 6 hours ago

TFeld

12k2833

add a comment | 

up vote
2
down vote

Python 2, 163 bytes

# p r i n t ' P e r f e c t l y o k a y ' # 
s=list(enumerate(input()))
S=(s+s[-2:0:-1]+s)[::3]
print''.join(zip(*S[:S.index(s[-1])+1])[1])

Try it online!

share|improve this answer

answered 6 hours ago

TFeld

12k2833

add a comment | 

up vote
2
down vote

up vote
2
down vote

Python 2, 163 bytes

# p r i n t ' P e r f e c t l y o k a y ' # 
s=list(enumerate(input()))
S=(s+s[-2:0:-1]+s)[::3]
print''.join(zip(*S[:S.index(s[-1])+1])[1])

Try it online!

share|improve this answer

answered 6 hours ago

TFeld

12k2833

Python 2, 163 bytes

# p r i n t ' P e r f e c t l y o k a y ' # 
s=list(enumerate(input()))
S=(s+s[-2:0:-1]+s)[::3]
print''.join(zip(*S[:S.index(s[-1])+1])[1])

Try it online!

share|improve this answer

answered 6 hours ago

TFeld

12k2833

share|improve this answer

share|improve this answer

answered 6 hours ago

TFeld

12k2833

answered 6 hours ago

TFeld

12k2833

answered 6 hours ago

TFeld

12k2833

12k2833

add a comment | 

add a comment | 

up vote
2
down vote

Python 2, 130 bytes

p='r'#i n t ' P e r f e c t l y o k a y ' # 
s=input();S=s+s[-2:0:-1]+s
print S[:len(~-len(s)%3and S or s):3]

Try it online!

share|improve this answer

answered 5 hours ago

user202729

13k12549

add a comment | 

up vote
2
down vote

Python 2, 130 bytes

p='r'#i n t ' P e r f e c t l y o k a y ' # 
s=input();S=s+s[-2:0:-1]+s
print S[:len(~-len(s)%3and S or s):3]

Try it online!

share|improve this answer

answered 5 hours ago

user202729

13k12549

add a comment | 

up vote
2
down vote

up vote
2
down vote

Python 2, 130 bytes

p='r'#i n t ' P e r f e c t l y o k a y ' # 
s=input();S=s+s[-2:0:-1]+s
print S[:len(~-len(s)%3and S or s):3]

Try it online!

share|improve this answer

answered 5 hours ago

user202729

13k12549

Python 2, 130 bytes

p='r'#i n t ' P e r f e c t l y o k a y ' # 
s=input();S=s+s[-2:0:-1]+s
print S[:len(~-len(s)%3and S or s):3]

Try it online!

share|improve this answer

answered 5 hours ago

user202729

13k12549

share|improve this answer

share|improve this answer

answered 5 hours ago

user202729

13k12549

answered 5 hours ago

user202729

13k12549

answered 5 hours ago

user202729

13k12549

13k12549

add a comment | 

add a comment | 

up vote
2
down vote

R, 187 bytes

 # c a t ( ' P e r f e c t l y o k a y ' ) #
g=function(x,n=nchar(x),i=c(1:n,(n-1):1,2:n),j=seq(1,3*n-2,3),k=i[j][1:which(i[j]==n)[1]])cat(substring(x,k,k),sep='') 

Try it online!

The single space at the end is needed so that the n never gets printed when the program is applied to itself.

Explanation

Part 1:

Ungolfed:

 # c a t ( ' P e r f e c t l y o k a y ' ) #
g <- function(x) 
 n <- nchar(x) # number of characters in string
 i <- c(1:n, (n - 1):1, 2:n) # index: 1 to n, n-1 back to 1, 2 to n
 j <- seq(1, 3 * n - 2, 3) # every third element of i
 k <- i[j][1:which(i[j] == n)[1]] # the elements of i at indices j, up to the first appearance of n
 cat(substring(x, k, k), sep = "") # extract from x the characters at indices k, and paste them together

Part 2:

The function produces this when it acts on the entire program:

cat('Perfectly okay')#=ni(ncr)=1,-:2)=q,n,,i]:i(j=assi(k)e' 

share|improve this answer

edited 2 hours ago

answered 2 hours ago

ngm

2,44920

add a comment | 

up vote
2
down vote

R, 187 bytes

 # c a t ( ' P e r f e c t l y o k a y ' ) #
g=function(x,n=nchar(x),i=c(1:n,(n-1):1,2:n),j=seq(1,3*n-2,3),k=i[j][1:which(i[j]==n)[1]])cat(substring(x,k,k),sep='') 

Try it online!

The single space at the end is needed so that the n never gets printed when the program is applied to itself.

Explanation

Part 1:

Ungolfed:

 # c a t ( ' P e r f e c t l y o k a y ' ) #
g <- function(x) 
 n <- nchar(x) # number of characters in string
 i <- c(1:n, (n - 1):1, 2:n) # index: 1 to n, n-1 back to 1, 2 to n
 j <- seq(1, 3 * n - 2, 3) # every third element of i
 k <- i[j][1:which(i[j] == n)[1]] # the elements of i at indices j, up to the first appearance of n
 cat(substring(x, k, k), sep = "") # extract from x the characters at indices k, and paste them together

Part 2:

The function produces this when it acts on the entire program:

cat('Perfectly okay')#=ni(ncr)=1,-:2)=q,n,,i]:i(j=assi(k)e' 

share|improve this answer

edited 2 hours ago

answered 2 hours ago

ngm

2,44920

add a comment | 

up vote
2
down vote

up vote
2
down vote

R, 187 bytes

 # c a t ( ' P e r f e c t l y o k a y ' ) #
g=function(x,n=nchar(x),i=c(1:n,(n-1):1,2:n),j=seq(1,3*n-2,3),k=i[j][1:which(i[j]==n)[1]])cat(substring(x,k,k),sep='') 

Try it online!

The single space at the end is needed so that the n never gets printed when the program is applied to itself.

Explanation

Part 1:

Ungolfed:

 # c a t ( ' P e r f e c t l y o k a y ' ) #
g <- function(x) 
 n <- nchar(x) # number of characters in string
 i <- c(1:n, (n - 1):1, 2:n) # index: 1 to n, n-1 back to 1, 2 to n
 j <- seq(1, 3 * n - 2, 3) # every third element of i
 k <- i[j][1:which(i[j] == n)[1]] # the elements of i at indices j, up to the first appearance of n
 cat(substring(x, k, k), sep = "") # extract from x the characters at indices k, and paste them together

Part 2:

The function produces this when it acts on the entire program:

cat('Perfectly okay')#=ni(ncr)=1,-:2)=q,n,,i]:i(j=assi(k)e' 

share|improve this answer

edited 2 hours ago

answered 2 hours ago

ngm

2,44920

R, 187 bytes

 # c a t ( ' P e r f e c t l y o k a y ' ) #
g=function(x,n=nchar(x),i=c(1:n,(n-1):1,2:n),j=seq(1,3*n-2,3),k=i[j][1:which(i[j]==n)[1]])cat(substring(x,k,k),sep='') 

Try it online!

The single space at the end is needed so that the n never gets printed when the program is applied to itself.

Explanation

Part 1:

Ungolfed:

 # c a t ( ' P e r f e c t l y o k a y ' ) #
g <- function(x) 
 n <- nchar(x) # number of characters in string
 i <- c(1:n, (n - 1):1, 2:n) # index: 1 to n, n-1 back to 1, 2 to n
 j <- seq(1, 3 * n - 2, 3) # every third element of i
 k <- i[j][1:which(i[j] == n)[1]] # the elements of i at indices j, up to the first appearance of n
 cat(substring(x, k, k), sep = "") # extract from x the characters at indices k, and paste them together

Part 2:

The function produces this when it acts on the entire program:

cat('Perfectly okay')#=ni(ncr)=1,-:2)=q,n,,i]:i(j=assi(k)e' 

share|improve this answer

edited 2 hours ago

answered 2 hours ago

ngm

2,44920

share|improve this answer

share|improve this answer

edited 2 hours ago

edited 2 hours ago

edited 2 hours ago

answered 2 hours ago

ngm

2,44920

answered 2 hours ago

ngm

2,44920

answered 2 hours ago

ngm

2,44920

2,44920

add a comment | 

add a comment | 

up vote
1
down vote

05AB1E, 41 bytes

“€€Ã€€Ð€€µ€€Ý€€“/g i që¬?3U[ûDXè?XIg<Ö#X3+U

Can without a doubt be golfed..

Try it online. (PS: Switch the language from 05AB1E (legacy) to 05AB1E for test cases 0123456789 and @. The legacy version is much faster, but it shows incorrect results for number inputs with leading zeros or single-char inputs.)

The 'backhanded' program will become:

“Ãеݓ q?[XX<XU

Which will output perfectly okay in full lowercase.

Try it online.

Explanation base program:

“€€Ã€€Ð€€µ€€Ý€€“ # Push the string "the pointed who the ethernet been the"
 / # And remove it from the stack again
g # Push the length of the (implicit) input
 i # If the length is exactly 1:
 q # Stop the program (and output the input implicitly)
 ë # Else:
 ¬? # Take the input implicitly again, and output its head
 3U # Push and pop 3, and store it in variable `X`
 [ # Start an infinite loop
 û # Palindromize the string at the top of the stack
 # i.e. "1O+1@" becomes "1O+1@1+O1" the first iteration,
 # and "1O+1@1+O1O+1@1+O1" the next iteration, etc.
 D # Duplicate the palindromized string
 Xè # Index variable `X` in the string
 ? # Output the character
 X # Check if variable `X`
 Ö # is divisible by
 Ig< # The length of the input - 1
 # # And if it is: stop the infinite loop
 X3+ # Then increase `X` by 3
 U # And pop and store it as the new `X`

Explanation 'backhanded' program:

“Ãеݓ # Push the string "perfectly okay"
 q # Exit the program (and implicitly output the top of the stack)
?[XX<XU # No-ops

See this 05AB1E tip of mine (section How to use the dictionary?) to understand why “€€Ã€€Ð€€µ€€Ý€€“ is the pointed who the ethernet been the and “Ãеݓ is perfectly okay.

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Kevin Cruijssen

30.9k553168

add a comment | 

up vote
1
down vote

05AB1E, 41 bytes

“€€Ã€€Ð€€µ€€Ý€€“/g i që¬?3U[ûDXè?XIg<Ö#X3+U

Can without a doubt be golfed..

Try it online. (PS: Switch the language from 05AB1E (legacy) to 05AB1E for test cases 0123456789 and @. The legacy version is much faster, but it shows incorrect results for number inputs with leading zeros or single-char inputs.)

The 'backhanded' program will become:

“Ãеݓ q?[XX<XU

Which will output perfectly okay in full lowercase.

Try it online.

Explanation base program:

“€€Ã€€Ð€€µ€€Ý€€“ # Push the string "the pointed who the ethernet been the"
 / # And remove it from the stack again
g # Push the length of the (implicit) input
 i # If the length is exactly 1:
 q # Stop the program (and output the input implicitly)
 ë # Else:
 ¬? # Take the input implicitly again, and output its head
 3U # Push and pop 3, and store it in variable `X`
 [ # Start an infinite loop
 û # Palindromize the string at the top of the stack
 # i.e. "1O+1@" becomes "1O+1@1+O1" the first iteration,
 # and "1O+1@1+O1O+1@1+O1" the next iteration, etc.
 D # Duplicate the palindromized string
 Xè # Index variable `X` in the string
 ? # Output the character
 X # Check if variable `X`
 Ö # is divisible by
 Ig< # The length of the input - 1
 # # And if it is: stop the infinite loop
 X3+ # Then increase `X` by 3
 U # And pop and store it as the new `X`

Explanation 'backhanded' program:

“Ãеݓ # Push the string "perfectly okay"
 q # Exit the program (and implicitly output the top of the stack)
?[XX<XU # No-ops

See this 05AB1E tip of mine (section How to use the dictionary?) to understand why “€€Ã€€Ð€€µ€€Ý€€“ is the pointed who the ethernet been the and “Ãеݓ is perfectly okay.

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Kevin Cruijssen

30.9k553168

add a comment | 

up vote
1
down vote

up vote
1
down vote

05AB1E, 41 bytes

“€€Ã€€Ð€€µ€€Ý€€“/g i që¬?3U[ûDXè?XIg<Ö#X3+U

Can without a doubt be golfed..

Try it online. (PS: Switch the language from 05AB1E (legacy) to 05AB1E for test cases 0123456789 and @. The legacy version is much faster, but it shows incorrect results for number inputs with leading zeros or single-char inputs.)

The 'backhanded' program will become:

“Ãеݓ q?[XX<XU

Which will output perfectly okay in full lowercase.

Try it online.

Explanation base program:

“€€Ã€€Ð€€µ€€Ý€€“ # Push the string "the pointed who the ethernet been the"
 / # And remove it from the stack again
g # Push the length of the (implicit) input
 i # If the length is exactly 1:
 q # Stop the program (and output the input implicitly)
 ë # Else:
 ¬? # Take the input implicitly again, and output its head
 3U # Push and pop 3, and store it in variable `X`
 [ # Start an infinite loop
 û # Palindromize the string at the top of the stack
 # i.e. "1O+1@" becomes "1O+1@1+O1" the first iteration,
 # and "1O+1@1+O1O+1@1+O1" the next iteration, etc.
 D # Duplicate the palindromized string
 Xè # Index variable `X` in the string
 ? # Output the character
 X # Check if variable `X`
 Ö # is divisible by
 Ig< # The length of the input - 1
 # # And if it is: stop the infinite loop
 X3+ # Then increase `X` by 3
 U # And pop and store it as the new `X`

Explanation 'backhanded' program:

“Ãеݓ # Push the string "perfectly okay"
 q # Exit the program (and implicitly output the top of the stack)
?[XX<XU # No-ops

See this 05AB1E tip of mine (section How to use the dictionary?) to understand why “€€Ã€€Ð€€µ€€Ý€€“ is the pointed who the ethernet been the and “Ãеݓ is perfectly okay.

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Kevin Cruijssen

30.9k553168

05AB1E, 41 bytes

“€€Ã€€Ð€€µ€€Ý€€“/g i që¬?3U[ûDXè?XIg<Ö#X3+U

Can without a doubt be golfed..

Try it online. (PS: Switch the language from 05AB1E (legacy) to 05AB1E for test cases 0123456789 and @. The legacy version is much faster, but it shows incorrect results for number inputs with leading zeros or single-char inputs.)

The 'backhanded' program will become:

“Ãеݓ q?[XX<XU

Which will output perfectly okay in full lowercase.

Try it online.

Explanation base program:

“€€Ã€€Ð€€µ€€Ý€€“ # Push the string "the pointed who the ethernet been the"
 / # And remove it from the stack again
g # Push the length of the (implicit) input
 i # If the length is exactly 1:
 q # Stop the program (and output the input implicitly)
 ë # Else:
 ¬? # Take the input implicitly again, and output its head
 3U # Push and pop 3, and store it in variable `X`
 [ # Start an infinite loop
 û # Palindromize the string at the top of the stack
 # i.e. "1O+1@" becomes "1O+1@1+O1" the first iteration,
 # and "1O+1@1+O1O+1@1+O1" the next iteration, etc.
 D # Duplicate the palindromized string
 Xè # Index variable `X` in the string
 ? # Output the character
 X # Check if variable `X`
 Ö # is divisible by
 Ig< # The length of the input - 1
 # # And if it is: stop the infinite loop
 X3+ # Then increase `X` by 3
 U # And pop and store it as the new `X`

Explanation 'backhanded' program:

“Ãеݓ # Push the string "perfectly okay"
 q # Exit the program (and implicitly output the top of the stack)
?[XX<XU # No-ops

See this 05AB1E tip of mine (section How to use the dictionary?) to understand why “€€Ã€€Ð€€µ€€Ý€€“ is the pointed who the ethernet been the and “Ãеݓ is perfectly okay.

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Kevin Cruijssen

30.9k553168

share|improve this answer

share|improve this answer

edited 4 hours ago

edited 4 hours ago

edited 4 hours ago

answered 4 hours ago

Kevin Cruijssen

30.9k553168

answered 4 hours ago

Kevin Cruijssen

30.9k553168

answered 4 hours ago

Kevin Cruijssen

30.9k553168

30.9k553168

add a comment | 

add a comment | 

up vote
1
down vote

JavaScript (ES6), 130 bytes

Early attempt. Not very satisfying.

f =/*> " P e r f e c t l y o k a*/y=>""+/**/(g=p=>(c=y[p])?m++%3?g(p+d):y[p+1]?c+g(p+d):c:g(p-d-d,d=-d))(m=0,d=1)

Try it online!

When the code is processed by itself, the following characters are isolated:

f =/*> " P e r f e c t l y o k a*/y=>""+/**/…
^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^

which gives:

f=>"Perfectly okay"//…

share|improve this answer

edited 26 mins ago

answered 2 hours ago

Arnauld

65.6k583277

add a comment | 

up vote
1
down vote

JavaScript (ES6), 130 bytes

Early attempt. Not very satisfying.

f =/*> " P e r f e c t l y o k a*/y=>""+/**/(g=p=>(c=y[p])?m++%3?g(p+d):y[p+1]?c+g(p+d):c:g(p-d-d,d=-d))(m=0,d=1)

Try it online!

When the code is processed by itself, the following characters are isolated:

f =/*> " P e r f e c t l y o k a*/y=>""+/**/…
^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^

which gives:

f=>"Perfectly okay"//…

share|improve this answer

edited 26 mins ago

answered 2 hours ago

Arnauld

65.6k583277

add a comment | 

up vote
1
down vote

up vote
1
down vote

JavaScript (ES6), 130 bytes

Early attempt. Not very satisfying.

f =/*> " P e r f e c t l y o k a*/y=>""+/**/(g=p=>(c=y[p])?m++%3?g(p+d):y[p+1]?c+g(p+d):c:g(p-d-d,d=-d))(m=0,d=1)

Try it online!

When the code is processed by itself, the following characters are isolated:

f =/*> " P e r f e c t l y o k a*/y=>""+/**/…
^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^

which gives:

f=>"Perfectly okay"//…

share|improve this answer

edited 26 mins ago

answered 2 hours ago

Arnauld

65.6k583277

JavaScript (ES6), 130 bytes

Early attempt. Not very satisfying.

f =/*> " P e r f e c t l y o k a*/y=>""+/**/(g=p=>(c=y[p])?m++%3?g(p+d):y[p+1]?c+g(p+d):c:g(p-d-d,d=-d))(m=0,d=1)

Try it online!

When the code is processed by itself, the following characters are isolated:

f =/*> " P e r f e c t l y o k a*/y=>""+/**/…
^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^

which gives:

f=>"Perfectly okay"//…

share|improve this answer

edited 26 mins ago

answered 2 hours ago

Arnauld

65.6k583277

share|improve this answer

share|improve this answer

edited 26 mins ago

edited 26 mins ago

edited 26 mins ago

answered 2 hours ago

Arnauld

65.6k583277

answered 2 hours ago

Arnauld

65.6k583277

answered 2 hours ago

Arnauld

65.6k583277

65.6k583277

add a comment | 

add a comment | 

 

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