# Enthusiastically Russianify a String

Many of you may have interacted with people from Russia on the internet at some point, and a subset of you may have noticed the slightly odd method they have of expressing themselves.

e.g. удали игру нуб)))

where the ))) are added for emphasis on the previous statement, I have been working on a theory that the ratio of )'s to the rest of the string is directly proportional to the amount of implied emphasis, however I oftentimes find it difficult to compute the ratio on the fly, as I am also trying to cope with a slew of abuse, so I would like the shortest possible code to help me calculate what the resulting string should be, for a value of enthusiasm between 0 and 500%, given the original, unenthusiastic string, this will aid my research greatly as I will not have to type out bulky scripts every time I wish to test my hypothesis.

So, the challenge:

write a full program or function, which, provided two arguments, a string of unknown length, and a number, in either integer format (between 0 and 500) or in decimal format (between 0 and 5, with 2 points of accuracy) will

• return/display the original string, suffixed with a number of )'s
• the number will be the calculated as a ratio of the input number to the string length.
• so if the number 200, or 2.00 was provided, 200% of the string must be suffixed as )'s
• the number of brackets rounded to in decimal situations does not matter.
• script is required to support Printable ASCII characters.
• only has to support one input number format, of your choice.

Examples:

"codegolf" 125      = codegolf))))))))))
"codegolf" 75       = codegolf))))))
"noob team omg" 0.5 = noob team omg))))))
"hi!" 4.99          = hi!)))))))))))))))

Example code (PowerShell) (with decimal input):

Function Get-RussianString ([string]$InputStr,[decimal]$Ratio){
$StrLen =$InputStr.Length

# Jelly, 7 bytes

ȮL×Ċ”)x

Try it online!

Uses the decimal format.

### How?

ȮL×Ċ”)x - Main link: string, decimal
Ȯ       - print string
L      - length(string)
×     - multiply by the decimal
Ċ    - ceiling (since rounding method is flexible)
”)  - a ')' character
x - repeated that many times
- implicit print
• @ConnorLSW I just noticed that this will print the required string as a full program, but that the specification states "return" - is this OK? Jan 30 '17 at 15:34
• any standard accepted output format is fine Jan 30 '17 at 16:51
• no worries - this is my first challenge so there's a few of these things that I missed, i've updated it in the question to be more clear - thanks for asking. Jan 30 '17 at 17:04

## Common Lisp, 5952 50

Parentheses? I am in.

(lambda(s n)(format()"~a~v@{)~}"s(*(length s)n)0))

### Details

(lambda(s n)               ; two arguments (string and ratio)
(format ()               ; format as string
"~a~v@{)~}"      ; control string (see below)
s                ; first argument (string)
(* (length s) n) ; second argument (number of parens)
0))              ; one more element, the value does not matter

### Format control string

• ~a : pretty print argument (here the given string)
• ~v@{...~} : iteration block, limited to V iteration, where V is taken as an argument, namely the (* ...) expression. The iteration is supposed to iterate over a list, but when you add the @ modifier, the list is the remaining list of arguments to the format function. There must be at least one element in the iterated list (otherwise we exit, disregarding V). That is why there is an additional argument to format (0).

Since no element in the list is consumed by the format, the loop is infinite but fortunately, it is also bounded by V, a.k.a. the number of parentheses to be printed.

Edit: thanks to Michael Vehrs for pointing out that there is no need to round the numerical argument (the question allows to truncate/round however we want, so the default behavior works here).

• (())/10 not enough parentheses Jan 31 '17 at 13:40
• Who thought this language is a good idea? Jan 31 '17 at 16:14
• Scheme's format accepts a decimal argument to v. Maybe Common Lisp's does, too? Feb 1 '17 at 8:13
• @MichaelVehrs Indeed, thanks a lot. Feb 1 '17 at 8:19
• @coredump Actually, I should have said "Guile's format accepts ...", since standard Scheme format does not support ~r; and Guile's format follows the example of Common Lisp's. Feb 1 '17 at 10:14

# JavaScript ES6, 3831 30 bytes

s=>n=>s+')'.repeat(s.length*n)

f=s=>n=>s+')'.repeat(s.length*n)

console.log(f("hi!")(4.99))

• Nice, I think that's the shortest possible. You can save a byte through currying: s=>n=>s+')'.repeat(s.length*n) (it would then be called like f("hi!")(4.99)) Jan 30 '17 at 15:30

# Python 2, 29 bytes

lambda s,p:s+len(s)*p/100*')'

s in the string, p is the percentage (integer).

Try it online!

# 05AB1E, 9 8 bytes

g*ï')×¹ì

Try it online!

g*       # Length, multiplied by emphasis.
ï')×   # Covnerted to an integer, push that many parenthesis.
¹ì # Prepend original string.

Works for both integer and decimal, arguments order: f(String, Double)

• -1 in the new version of 05AB1E, where the ï is done implicitly for × with float argument. Sep 30 '19 at 11:17
• And here’s a 7-byter that works on both legacy and modern 05AB1E: sg*F')«. Sep 30 '19 at 20:19

# Python, 30 bytes

lambda s,r:s+')'*int(len(s)*r)

Uses the decimal input.

Try it online!

# Pyth, 8 bytes

*\)s*lpz

Online Test! Takes the excitement ratio first, then the string to be enthused about.

Explanation:

pz  print out the enthused string
l    ... and get its length
*...Q multiply that by the ratio
s      floor to get an integer, let's call this S
\)       single-character string ")"
* ")" S   multiply that integer by the string, which gives a string of )s of length S.
implicitly print that string of S )s.

# PowerShell, 33 bytes

$a,$b=$args;$a+')'*($b*$a.Length)

Try it online!

Supports decimal format.

# R, 6246 42 bytes

Anonymous function that takes string a and decimal n, prints output to stdout.

pryr::f(cat(a,rep(")",n*nchar(a)),sep=""))

# Pyth, 9 bytes

*s*lpzE")

Takes two lines of input: string and ratio (decimal).

Try it on pyth.herokuapp.com

### Explanation

A denotes a function's first argument, B its second argument.

*s*lpzE")
pz     # print the input string
lAA     # take the length of the printed string
E    # read the next line of input (the emphasis ratio)
*AAAB    # multiply the length by the ratio
sAAAAA    # floor the result
*AAAAAA")  # repeat ")" n times
# implicit print

# TI-Basic, 33 bytes

Takes decimal input.

Prompt Str1,A
")
For(I,0,9
Ans+Ans
End
Str1+sub(Ans,1,AI

# Perl 5, 29 bytes

This version of the compiler was written on 1/27/2017 at 11:12 pm, which might have been a few minutes after this question was posted. So here's a version which works on the oldest version of the compiler, written an hour earlier: iS$,N?S$;')'*(N*l(S$)) (22 bytes) # PostgreSQL, 102 bytes create function q(text,int)returns text as $$select rpad(1,(100+2)*length(1)/100,')')$$language sql ### Details Uses the integer input format. This simply right-pads the input string with parens out to the target length. create function q(text,int) returns text as $$select rpad(1, -- Pad the string input (100 + 2) * -- to 100 + int input % ... length(1) / 100, -- ...of the input string ')') -- with ) characters$$ language sql Called with select q('codegolf', 125), q('codegolf', 75); select q('noob team omg', 50), q('hi!', 499); # Bash + coreutils, 45 bytes echo$1seq -s\) $[${#1}*$2/100+1]|tr -cd \) Try it online! Integer input. • echo$1jot -b \) -s '' $[${#1}*$2/100] 40 bytes :) try it out Mar 24 '19 at 20:01 # Groovy, 27 bytes Straightforward solution {s,r->s+')'*(s.length()*r)} Test program : def f = {s,r->s+')'*(s.length()*r)} println f("hi!", 4.99) println f("noob team omg", 0.5) # Rebol, 39 bytes func[a b][append/dup a")"b * length? a] # Clojure, 40 bytes Quite boring solution : #(reduce str %(repeat(*(count %)%2)")")) Just reduces str function on a list of closing parentheses with a string as initial parameter. See it online : https://ideone.com/5jEgWS Not-so-boring solution (64 bytes) : #(.replace(str(nth(iterate list(symbol %))(*(count %)%2)))"(""") Converts input string to a symbol (to get rid of quotes) and repeatedly applies function list on it generating infinite sequence like this: (a (a) ((a)) (((a))) ... ). Takes nth element converts it to string and replaces all opening parentheses with nothing. See it online : https://ideone.com/C8JmaU • #(.replaceAll(str(nth(iterate list %)(*(count %)%2)))"[(\"]""") 1 byte less (yay). I wanted to do comp but can't get it below 70 bytes. Feb 2 '17 at 11:34 • You can change ")" to \) to save a byte. May 5 '17 at 12:05 # SimpleTemplate, 92 bytes Takes the string as the first parameter and the "ratio" as the second. The ratio is between 0 and 5, with 2 decimal places. {@echoargv.0}{@callstrlen intoL argv.0}{@set*Y argv.1,L}{@callstr_repeat intoO")",Y}{@echoO} As you can see, it is non-optimal. The 2 {echo} there could be reduced to 1. Due to a bug in the compiler, this code can't be reduced much further. Ungolfed: {@echo argv.0} {@call strlen into length argv.0} {@set* ratio argv.1, length} {@call str_repeat into parenthesis ")", ratio} {@echo parenthesis} If no bug existed, the code would look like this, 86 bytes: {@callstrlen intoL argv.0}{@set*Y argv.1,L}{@callstr_repeat intoO")",Y}{@echoargv.0,O} ## C# Interactive,77 67 bytes string r(string s,int p)=>s+new string(')',(int)(s.Length*p/100d)); C# interactive is sweet. • If you are using C# Interactive that needs to be in the header otherwise, in C#, you should include the using System; or fully qualify Math. Also, not sure if you can do it in interactive, but you could compile to a Func<string, Func<int, string>> to save bytes i.e. s=>p=>s+new... Jan 31 '17 at 15:57 • Also you probably don't need the call to Math.Round just casting to an int should call Floor and the OP said either Floor or Ceiling is fine Jan 31 '17 at 15:58 # Stax, 7 bytes é┴eó¡µf Run and debug it # International Phonetic Esoteric Language, 24 bytes A reimplementation of the old answer into the new specification. Input is as a positive integer first, then the string. ɪ{100}viħɖf0ɑ")"x1esøɒʕo Explanation: ɪ{100}viħɖf0ɑ")"x1esøɒʕo ɪ (push input int) {100} (push 100) v (divide int//100) iħ (push input str, push len) ɖ (rotate top 3 clockwise) f (multiply int//100 * len) 0 (push 0) ɑ (begin loop, from 0 to int//100 * len) ")" (push ")") x (concat) 1esø (increment index) ɒ (end loop) ʕ (list to str) o (print) # International Phonetic Esoteric Language, 18 bytes (WIP language) (OLD) I discovered some bugs in my interpreter because of this challenge. Takes input as string then integer. iɢ291tʃɪðθœ<)>ɲqɶo No TIO interpreter yet, but is runnable by cloning the repository above, and calling python3 main.py "code here". iɢ291tʃɪðθœ<)>ɲqɶo i ; push string input ɢ ; peek, push string length 291tʃ ; push (1 + 9)^2 ɪ ; push number input ðθ ; push [num / 100 * len] œ ; start loop: pop, run ceil(n) times <)> ; push string ")" ɲ ; swap top 2 elements of the stack q ; pop, pop, push concatenated strings ɶ ; end loop o ; output Sample cases: python3 main.py "iɢ291tʃɪðθœ<)>ɲqɶo" codegolf 125 codegolf)))))))))) python3 main.py "iɢ291tʃɪðθœ<)>ɲqɶo" iɢ291tʃɪðθœ<)>ɲqɶo 300 iɢ291tʃɪðθœ<)>ɲqɶo)))))))))))))))))))))))))))))))))))))))))))))))))))))) python3 main.py "iɢ291tʃɪðθœ<)>ɲqɶo" less than 1.00 50 less than 1.00))))))) ## SmileBASIC, 29 bytes INPUT S$,N?S$;")"*(LEN(S$)*N)
• since 3*4.99 = 14.97, only 14 or 15 would be acceptable as answers, the 29 bytes version should work fine though, sorry! Jan 30 '17 at 15:50

# Gol><> (Golfish), 17 bytes

i:a=?v
R*Il~/Hr)`

The top line reads characters (i) until it finds a newline (ASCII 10, a), then goes down (v).

Then we discard one character (the newline) with ~, push the length of the stack (l), read a float (I), multiply the two, and repeatedly (R) push the character ")" that many times. Finally, reverse the stack (r), output it and halt (H).

# PHP, 50 bytes

<?=str_pad($s=$argv[1],strlen($s)*++$argv[2],")");

takes string and decimal number as command line arguments; cuts padding. Run with -r;

breakdown

<?=                     // print ...
$s=$argv[1],            // string=argument 1
strlen($s)*++$argv[2],  // to string length*(1+argument 2)
")"                     // using ")" as padding string
);

# Ruby, 25 bytes

->(s,n){s+')'*(s.size*n)}

I'm using lambdas. The test program would be something like:

f=->(s,n){s+')'*(s.size*n)}
f.("codegolf", 1.5)        # => "codegolf))))))))))))"
f.("hi!", 4.99)            # => "hi!))))))))))))))"

# Clojure, 68 bytes

An anonymous function that accepts decimal input.

(fn [s n] (print (str s (reduce str (repeat (* n (count s)) ")")))))

Literally the first Lisp program I've ever written! I'm already having fun.

• Welcome to the world of Lisp! :P In Clojure, you can use the condensed form of anonymous functions #(...), and you can get rid of the print (since function returns should be acceptable). You can change reduce to apply for the str function, and you can change ")" to \), which does the same thing. So, the final code should be: #(str %(apply str(repeat(*(count %)%2)\))))). May 5 '17 at 11:58
• Also, the current state of your code doesn't work, (#(...) "codegolf" 125) must add 125 percent of the length of "codegolf" instead of 125 times the length of "codegolf". So, the fixed program would be: #(str %(apply str(repeat(*(count %)%2 1/100)\)))), which is 49 bytes. May 5 '17 at 12:01