Output the language name times the number of characters [closed]

Question

Closed. This question needs details or clarity. It is not currently accepting answers.

Want to improve this question? Add details and clarify the problem by editing this post.

Closed 6 years ago.

Improve this question

Challenge:

Your task is to use any language to output that language's name times the number of character's in its name.

Use the common name or acronym of the language.
If there are multiple versions of a language, you need not include the version number.
The capitalization matters.

Examples:

PHP -> PHPPHPPHP
Java -> JavaJavaJavaJava
C++ -> C++C++C++
Python 3 -> PythonPythonPythonPythonPythonPython
JavaScript -> JavaScriptJavaScriptJavaScriptJavaScriptJavaScriptJavaScriptJavaScriptJavaScriptJavaScriptJavaScript

Tasks that depend on a language's name are a form of disadvantaging some languages by arbitrary criteria. — Martin Ender, Commented Dec 17, 2017 at 15:08
I think this challenge would have been a lot more interesting if the repetition count was supposed to be our bytecount... — Maya, Commented Dec 17, 2017 at 16:53
And because it hasn't been said yet: Welcome to PPCG! I hope you stick around despite your probably confusing first impression of people telling you that this is not a good challenge while one answer after the other arrives. — Laikoni, Commented Dec 17, 2017 at 21:06
IMO this is unclear. Use the common name suggests that, for example, JS could be used for JavaScript. But what's to stop me saying that my language "Language" is more commonly known as "L"? — FlipTack, Commented Dec 24, 2017 at 19:25

Community · Accepted Answer · 2020-06-17 09:04:33Z

2

sed, 10 bytes

isedsedsed

I hope two trailing newlines are OK :P

Try it online!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 17, 2017 at 16:48

Maya

5,7131 gold badge26 silver badges39 bronze badges

Add a comment |

Erik the Outgolfer · Accepted Answer · 2017-12-17 17:49:33Z

2

itflabtijtslwi, 31 bytes

/I/itflabtijtslwi//J/IIIIIII/JJ

Try it online!

answered Dec 17, 2017 at 17:49

Erik the Outgolfer

40.6k5 gold badges45 silver badges122 bronze badges

Add a comment |

Arnauld · Accepted Answer · 2017-12-17 19:31:02Z

2

Bubblegum, 14 13 bytes

Saved 1 byte thanks to @ovs

00000000: 732a 4d4a ca49 4d2f cda5 1603 00         s*MJ.IM/.....

Try it online!

edited Dec 17, 2017 at 19:31

answered Dec 17, 2017 at 18:26

Arnauld

198k20 gold badges179 silver badges650 bronze badges

\$\begingroup\$ 13 bytes, generated with zopfli. \$\endgroup\$
– ovs
Commented Dec 17, 2017 at 19:24
\$\begingroup\$ @ovs Thanks! I was pretty sure I was doing it wrong. \$\endgroup\$
– Arnauld
Commented Dec 17, 2017 at 19:34

Add a comment |

Conor O'Brien · Accepted Answer · 2017-12-17 21:06:06Z

Hexadecimal Stacking Pseudo-Assembly Language, 636 bytes

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

Try it online!

Formatted:
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Explanation

; initialize 77 with 0x002D (45)
20 002D
40 77 (FF)

; label AAAA
00 AAAA

; print out the text
20006540000020006740000020006140000020007540000020006740000020006e40000020006140000020004c400000200020400000
20007940000020006c40000020006240000020006d40000020006540000020007340000020007340000020004140000020002d400000
20006f40000020006440000020007540000020006540000020007340000020005040000020002040000020006740000020006e400000
20006940000020006b40000020006340000020006140000020007440000020005340000020002040000020006c400000200061400000
20006d400000200069400000200063400000200065400000200064400000200061400000200078400000200065400000200048400000
140000

; decrement
    ; move counter to stack 99
        42 77 (FF)
        40 99 (FF)
    ; push 1 to 77
        20 0001
        40 77 (FF)
    ; move counter to stack 77
        42 99 (FF)
        40 77 (FF)
    ; subtract
        22 77 (FF)

; duplicate
42 77 (FF)
40 77 (FF)

; terminate if counter is 0
03 77 (FF)
04 (0001)

; redo
01 AAAA

KSmarts · Accepted Answer · 2017-12-17 21:31:18Z

2

Hodor, 38 bytes

hodor.hod("HodorHodorHodorHodorHodor")

It's actually shorter than the pure Javascript equivalent!

Try it online!

answered Dec 17, 2017 at 21:31

KSmarts

2,00011 silver badges11 bronze badges

Add a comment |

Shaun Bebbers · Accepted Answer · 2017-12-18 11:34:45Z

2

PHP (9 bytes)

PHPPHPPHP

Try it online

answered Dec 18, 2017 at 11:34

Shaun Bebbers

2,28216 silver badges22 bronze badges

2

\$\begingroup\$ Hah. can't believe str_repeat(3,'PHP') is the long alternative. \$\endgroup\$
– Magic Octopus Urn
Commented Dec 19, 2017 at 14:55
1

\$\begingroup\$ Excuse my brevity with this answer; it's hardly a difficult challenge and by the questions posed above, it's apparently a perfectly legal one as well. \$\endgroup\$
– Shaun Bebbers
Commented Dec 19, 2017 at 17:19

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

2

TI-BASIC, 16 bytes

"TI-BASIC
Disp Ans,Ans
Ans

Alternatively (19 bytes):

For(I,1,3
Disp "TI-BASIC
End

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 18, 2017 at 13:18

Timtech

12.6k2 gold badges45 silver badges61 bronze badges

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

Standard ML (MLton), 55 54 46 bytes

val s="Standard ML";s^s^s;print(it^it^it^s^s);

Try it online!

Output:

Standard MLStandard MLStandard MLStandard MLStandard MLStandard MLStandard MLStandard MLStandard MLStandard MLStandard ML

Previous version: (55 bytes)

"Standard ML";print(concat(map(fn _=>it)(explode it)));

Try it online! (No worries, the code is harmless, eventhough it contains explode it.)

Explanation:

"Standard ML"; binds the string to the implicit value identifier it. We then explode it to convert the string into a list of characters and map a function fn _=>it over the list which replaces each character with the string. concat combines the list of strings to a single string and print outputs the resulting string.

MD XF · Accepted Answer · 2017-12-18 22:37:27Z

2

Bash, 14 bytes

echo bash{,,,}

Try it online!

ksh, 12 bytes

echo ksh{,,}

Try it online!

sh, 10 bytes

echo sh{,}

Try it online!

Although, for 9 bytes, I can just do echo shsh.

edited Dec 18, 2017 at 22:37

answered Dec 18, 2017 at 22:28

MD XF

13.9k4 gold badges66 silver badges104 bronze badges

Add a comment |

MD XF · Accepted Answer · 2017-12-19 20:34:50Z

2

Underload, 28 bytes

(Underload)::::::::********S

Try it online!

answered Dec 19, 2017 at 20:34

MD XF

13.9k4 gold badges66 silver badges104 bronze badges

\$\begingroup\$ (Underload)::**::**S for 20 bytes. \$\endgroup\$
– Esolanging Fruit
Commented Oct 22, 2019 at 15:44

Add a comment |

MD XF · Accepted Answer · 2017-12-20 20:19:12Z

brainfuck, 125 119 bytes

+++++++++[>+++++++++++[>+++++++++[>+>+<<-]>>+<<<-]>>-.>++++.<-.++++++++.>----.<---.>+++++++.<---.++++++++.>[-]<[-]<<<-]

First BF, I think it's still golfable but it may be a bit BF to golf BF ;)

Explanation:

+++++++++                        (_009  000  000  000  000)
[
 >+++++++++++                    ( 009 _011  000  000  000)
 [
  >+++++++++                     ( 009  011 _009  000  000)
  [
   >+>+<<-                       ( 009  011 _008  001  001)
  ]                              ( 009  011 _000  009  009)
  >>+<<<-                        ( 009 _010  000  009  010)
 ]                               ( 009 _000  000  099  110)
 >>-.                            ( 009  000  000 _098  110) Output: b
 >++++.                          ( 009  000  000  098  114) Output: r
 <-.                             ( 009  000  000 _097  114) Output: a
 ++++++++.                       ( 009  000  000 _105  114) Output: i
 >----.                          ( 009  000  000  105 _110) Output: n
 <---.                           ( 009  000  000 _102  110) Output: f
 >+++++++.                       ( 009  000  000  102 _117) Output: u
 <---.                           ( 009  000  000 _099  117) Output: c
 ++++++++.                       ( 009  000  000 _107  117) Output: k
 >[-]<[-]<<<-                    (_008  000  000  000  000)
]                                (_000  000  000  000  000)

-6 for removing redundant clear commands

Try it online!

Shaun Bebbers · Accepted Answer · 2017-12-21 10:40:02Z

2

Sinclair ZX81/Timex TS1000/1500 BASIC, ~33 tokenized BASIC bytes for the listing

 LET A$="BASIC"
 1 FOR I=SGN PI TO LEN A$
 2 PRINT A$;
 3 NEXT I

Enter the language name into the var stack using direct mode to save one BASIC line, and save bytes in the listing itself. I assume the A$ variable is five bytes for the string and two bytes for the pointer, but I'm not 100% sure how the this works on the ZX81 - I will find out and update the byte count.

Sinclair ZX81/Timex TS1000/1500 BASIC (non-dynamic solution) ~21 tokenized BASIC bytes (symbolic listing)

 LET A$="BASIC"
 1 PRINT A$;A$;A$;A$;A$

Using the same technique as above, entering the A$ variable in direct mode and then simply printing the string five times. To run either solution, don't use the command RUN or the var stack is initialised. Use:

 GOTO 1

edited Dec 21, 2017 at 10:40

answered Dec 18, 2017 at 10:57

Shaun Bebbers

2,28216 silver badges22 bronze badges

1

\$\begingroup\$ Save 5 bytes by using SGN PI instead of 1. And by my count the second one is only 36 bytes. \$\endgroup\$
– Neil
Commented Dec 18, 2017 at 11:23
\$\begingroup\$ Yes, I will optimise futher; also one could use INPUT A$ in line one, and will therefore work with any language inputted into the A$ string \$\endgroup\$
– Shaun Bebbers
Commented Dec 18, 2017 at 11:25
\$\begingroup\$ @Neil - how are you counting 36 bytes in the second listing? \$\endgroup\$
– Shaun Bebbers
Commented Dec 19, 2017 at 14:24
\$\begingroup\$ The first line is 4 bytes for the line number and length, 1 for the token, 10 normal bytes, 1 for the newline. The second is similar, except it has 14 normal bytes. I make that 36 bytes in total. \$\endgroup\$
– Neil
Commented Dec 19, 2017 at 16:43
\$\begingroup\$ It's a good job I put a ~ in front of my byte count isn't it. Somehow I mis-counted. Twice. But I'd rather work out proper byte counts on 8 bit machines, i.e., how much actual memories do my listings take? \$\endgroup\$
– Shaun Bebbers
Commented Dec 19, 2017 at 17:21

| Show 1 more comment

Community · Accepted Answer · 2020-06-17 09:04:33Z

2

Befunge-93, 27 Bytes

Edit: Much cleaner solution ~~of the same length~~ suggested by James Holderness

0"39-egnufeB">:#,_$1+:9`#@_

Try It Online

Old 30 byte version:

?
_g7-00p"39-egnufeB">:#,_g`#@

Try It Online

It feels weird to use the ? command for something that isn’t random, but that’s the only direction changing instruction that had an ASCII value low enough

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 18, 2017 at 18:07

Jo King♦

47.5k5 gold badges125 silver badges186 bronze badges

1

\$\begingroup\$ Surely it's easier to just keep the counter on the stack: Try it online! \$\endgroup\$
– James Holderness
Commented Dec 20, 2017 at 15:25
\$\begingroup\$ @JamesHolderness Not if you're printing the "-93" as well. Then it's the same length (but still a much cleaner solution) \$\endgroup\$
– Jo King ♦
Commented Dec 24, 2017 at 15:34

Add a comment |

Mr. Xcoder · Accepted Answer · 2017-12-25 06:20:20Z

2

;#*, 420 bytes

;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;#

Try it online! (uses ;#+ instead)

edited Dec 25, 2017 at 6:20

answered Dec 17, 2017 at 15:01

Mr. Xcoder

42.5k9 gold badges81 silver badges214 bronze badges

\$\begingroup\$ You can golf this approach. \$\endgroup\$
– Conor O'Brien
Commented Dec 17, 2017 at 19:03
\$\begingroup\$ @ConorO'Brien Well I intended this to be a ;# answer but since that is not considered a language (it cannot perform a primality test), I used ;#+. Will look at the docs and see what I can do \$\endgroup\$
– Mr. Xcoder
Commented Dec 17, 2017 at 19:10
\$\begingroup\$ This is a kolgomorov complexity challenge, so primality testing isn't required. This can be a ;# answer if you still want to \$\endgroup\$
– caird coinheringaahin g ♦
Commented Dec 25, 2017 at 1:20
\$\begingroup\$ @cairdcoinheringaahing Changed to ;#* \$\endgroup\$
– Mr. Xcoder
Commented Dec 25, 2017 at 6:23

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

2

Java 11, 19 bytes

v->"Java".repeat(4)

Try it online.

Java 8, 21 bytes

v->"JavaJavaJavaJava"

Well...

Try it here.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 18, 2017 at 8:31

Kevin Cruijssen

131k13 gold badges144 silver badges384 bronze badges

2

\$\begingroup\$ JAVA! JAVA! JAVA! JAVA! \$\endgroup\$
– Engineer Toast
Commented Dec 19, 2017 at 19:22

Add a comment |

Daniel · Accepted Answer · 2017-12-17 14:58:25Z

1

Pyth, 7 bytes

*4"Pyth

My first Pyth answer!

Thanks to Mr. Xcoder for a neat trick.

Try it online!

edited Dec 17, 2017 at 14:58

answered Dec 17, 2017 at 14:55

Daniel

6,7191 gold badge21 silver badges58 bronze badges

\$\begingroup\$ *4"Pyth for 7 bytes \$\endgroup\$
– Mr. Xcoder
Commented Dec 17, 2017 at 14:57
\$\begingroup\$ @Mr.Xcoder, neat! \$\endgroup\$
– Daniel
Commented Dec 17, 2017 at 14:58

Add a comment |

caird coinheringaahin g · Accepted Answer · 2017-12-17 15:02:21Z

1

Jelly, 7 bytes

“¢³ƒ»ẋ5

Try it online!

edited Dec 17, 2017 at 15:02

answered Dec 17, 2017 at 14:52

caird coinheringaahin g♦

50.3k11 gold badges128 silver badges354 bronze badges

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

1

APL, 10 7 bytes

∊(⍴c)⍴⊂c←⎕

APL => APLAPLAPL

previous answer above prompted for screen input but if no input is required then we have the trivial for 7 bytes

9⍴'APL'

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 17, 2017 at 15:27

Graham

6,8461 gold badge15 silver badges11 bronze badges

Add a comment |

Giuseppe · Accepted Answer · 2017-12-17 16:16:39Z

1

SNOBOL4 (CSNOBOL4), 30 bytes

 OUTPUT =DUPL('SNOBOL4',7)
END

Try it online!

answered Dec 17, 2017 at 16:16

Giuseppe

28.8k3 gold badges31 silver badges105 bronze badges

Add a comment |

caird coinheringaahin g · Accepted Answer · 2017-12-17 16:19:33Z

1

Deorst, 11 bytes

'Deorst'l6*

Try it online!

answered Dec 17, 2017 at 16:19

caird coinheringaahin g♦

50.3k11 gold badges128 silver badges354 bronze badges

Add a comment |

Steadybox · Accepted Answer · 2017-12-17 16:44:18Z

1

PHP, 59 bytes

Not the best way to do it (and there's a better php solution), but here's my go at it:

<?php $s="php";$l=strlen($s);for($i=0;$i<$l;$i++){echo $s;}

Try it online!

edited Dec 17, 2017 at 16:44

Steadybox

16.5k5 gold badges41 silver badges84 bronze badges

answered Dec 17, 2017 at 16:25

RAZERZ

1112 bronze badges

Add a comment |

Maya · Accepted Answer · 2017-12-17 16:52:25Z

1

rk, 22 bytes

rk:start
print: "rkrk"

Try it online!

answered Dec 17, 2017 at 16:52

Maya

5,7131 gold badge26 silver badges39 bronze badges

\$\begingroup\$ Wait someone else knows about this language :o \$\endgroup\$
– MD XF
Commented Dec 17, 2017 at 23:51

Add a comment |

Luis Mendo · Accepted Answer · 2017-12-17 16:55:32Z

1

Octave, 26 bytes

disp(repmat('Octave',1,6))

Try it online!

answered Dec 17, 2017 at 16:55

Luis Mendo

105k9 gold badges135 silver badges372 bronze badges

Add a comment |

Maya · Accepted Answer · 2017-12-17 17:00:46Z

1

MY-BASIC, 33 bytes

for x=1 to 8
print"MY-BASIC"
next

Try it online!

answered Dec 17, 2017 at 17:00

Maya

5,7131 gold badge26 silver badges39 bronze badges

Add a comment |

Erik the Outgolfer · Accepted Answer · 2017-12-17 17:28:35Z

1

05AB1E, 10 bytes

"05AB1E"6×

Try it online!

answered Dec 17, 2017 at 17:28

Erik the Outgolfer

40.6k5 gold badges45 silver badges122 bronze badges

Add a comment |

ovs · Accepted Answer · 2017-12-17 18:07:10Z

1

Befunge-98 (FBBI), 22 bytes

7v
"<_@#:-1,k6"Befunge

Try it online!

with -98 included:

av
"<_@#:-1,k9"Befunge-98

Try it online!

answered Dec 17, 2017 at 18:07

ovs

60.6k3 gold badges47 silver badges161 bronze badges

\$\begingroup\$ -2 bytes if you increment instead of decrementing (idea from james holderness' comment on my Befunge-93 answer) \$\endgroup\$
– Jo King ♦
Commented Dec 24, 2017 at 15:44

Add a comment |

ovs · Accepted Answer · 2017-12-17 18:12:18Z

1

pl, 4 bytes

plpl

Try it online!

answered Dec 17, 2017 at 18:12

ovs

60.6k3 gold badges47 silver badges161 bronze badges

Add a comment |

ovs · Accepted Answer · 2017-12-17 18:16:57Z

1

Cy, 12 bytes

"CyCy" print

Try it online!

answered Dec 17, 2017 at 18:16

ovs

60.6k3 gold badges47 silver badges161 bronze badges

Add a comment |

sergiol · Accepted Answer · 2017-12-17 19:24:44Z

1

Tcl, 14 bytes

puts TclTclTcl

Try it online!

answered Dec 17, 2017 at 19:24

sergiol

3,4561 gold badge15 silver badges31 bronze badges

Add a comment |

ბიმო · Accepted Answer · 2017-12-17 20:16:03Z

1

Husk, 8 6 bytes

‼D¨Ż%k

-2 bytes thanks to @Zgarb!

Try it online or alternatively DD¨Ż%k would work too..

Explanation

‼D¨Ż%k
  ¨Ż%k  -- compressed string: "Husk"
‼D      -- duplicate twice: "HuskHuskHuskHusk"

edited Dec 17, 2017 at 20:16

answered Dec 17, 2017 at 15:44

ბიმო

16.9k2 gold badges41 silver badges104 bronze badges

2

\$\begingroup\$ ‼D works too. \$\endgroup\$
– Zgarb
Commented Dec 17, 2017 at 20:08

Add a comment |

Output the language name times the number of characters [closed]

Challenge:

97 Answers 97

sed, 10 bytes

itflabtijtslwi, 31 bytes

Bubblegum, 14 13 bytes

Hexadecimal Stacking Pseudo-Assembly Language, 636 bytes

Explanation

Hodor, 38 bytes

PHP (9 bytes)

TI-BASIC, 16 bytes

Standard ML (MLton), 55 54 46 bytes

Output:

Previous version: (55 bytes)

Explanation:

Bash, 14 bytes

ksh, 12 bytes

sh, 10 bytes

Underload, 28 bytes

brainfuck, 125 119 bytes

Sinclair ZX81/Timex TS1000/1500 BASIC, ~33 tokenized BASIC bytes for the listing

Sinclair ZX81/Timex TS1000/1500 BASIC (non-dynamic solution) ~21 tokenized BASIC bytes (symbolic listing)

Befunge-93, 27 Bytes

;#*, 420 bytes

Java 11, 19 bytes

Java 8, 21 bytes

Pyth, 7 bytes

Jelly, 7 bytes

APL, 10 7 bytes

SNOBOL4 (CSNOBOL4), 30 bytes

Deorst, 11 bytes

PHP, 59 bytes

rk, 22 bytes

Octave, 26 bytes

MY-BASIC, 33 bytes

05AB1E, 10 bytes

Befunge-98 (FBBI), 22 bytes

pl, 4 bytes

Cy, 12 bytes

Tcl, 14 bytes

Husk, 8 6 bytes

Explanation

Not the answer you're looking for? Browse other questions tagged code-golfstringkolmogorov-complexity or ask your own question.

Linked

Related

Hot Network Questions

Not the answer you're looking for? Browse other questions tagged
code-golf
string
kolmogorov-complexity
or ask your own question.