50
\$\begingroup\$

Task

Given a String as input, your task is to output 42 only if the input String happens to be exactly the following :

abbcccddddeeeeeffffffggggggghhhhhhhhiiiiiiiiijjjjjjjjjjkkkkkkkkkkkllllllllllllmmmmmmmmmmmmmnnnnnnnnnnnnnnoooooooooooooooppppppppppppppppqqqqqqqqqqqqqqqqqrrrrrrrrrrrrrrrrrrsssssssssssssssssssttttttttttttttttttttuuuuuuuuuuuuuuuuuuuuuvvvvvvvvvvvvvvvvvvvvvvwwwwwwwwwwwwwwwwwwwwwwwxxxxxxxxxxxxxxxxxxxxxxxxyyyyyyyyyyyyyyyyyyyyyyyyyzzzzzzzzzzzzzzzzzzzzzzzzzz

It may output any other value, produce an error or not output at all, if the input does not equal the aforementioned String.


Winning Criterion

This is , so the shortest code in bytes wins!

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10
  • 1
    \$\begingroup\$ Many of the solutions provided here are wrong because they print 42 when the string is longer than the desired string and the prefix matches with the desired string. \$\endgroup\$
    – fR0DDY
    Mar 10, 2011 at 10:30
  • \$\begingroup\$ @froddy: What if the only characters? following the string (is|are) a line break? My usual input mechanism doesn't care whether the input is terminated by a line break or not but yield the same in both cases, for example. \$\endgroup\$
    – Joey
    Mar 10, 2011 at 13:22
  • \$\begingroup\$ @fR0DDY : There was no clear definition on how the rest of the input should be handled, so there's no 'wrong' here. \$\endgroup\$
    – PatrickvL
    Mar 10, 2011 at 15:46
  • 4
    \$\begingroup\$ @PatrickvL It does mention 'only' if the input is the given string. So abbcccddddeeeee...zzabc does not satisfy that i suppose and i can see some programs giving yes on that input. \$\endgroup\$
    – fR0DDY
    Mar 10, 2011 at 15:56
  • 2
    \$\begingroup\$ @fR0DDY : Let me put it another way : There's no specification on how input is delimited, so that's open to interpretation. There's also no mention of character encoding (I guess most of us assume the default of their environment - ANSI, UTF8 and UTF16LE will be the most popular ones). Also no mention how the input is presented - is it entered via the standard input, via a command-line parameter? So you see - having all this freedom gives way to some interpretation that you would mark as 'incorrect', while others would judge it 'compliant'. NOFI, but this is daily practise for some of us. \$\endgroup\$
    – PatrickvL
    Mar 10, 2011 at 16:10

80 Answers 80

20
\$\begingroup\$

Golfscript, 20

26,{.97+\{.}*}%=42`*

with new line, 21 chars (by Nabb)

26,{).[96+]*}%n+=42`*

Actually Nabb beat mine, here is original solution for with new line, 22 chars

26,{.97+\{.}*}%n+=42`*

This is simply generating source string and just comparing it against string from stdin.

\$\endgroup\$
3
  • 3
    \$\begingroup\$ 26,{).[96+]*}%n+=42`* for 21 (inc newline). \$\endgroup\$
    – Nabb
    Mar 11, 2011 at 2:09
  • \$\begingroup\$ Heh, it doesn't work without the n+ because the array isn't flat. You'll have to either keep your original 20 chars or add a ~ to flatten the inner terms of the array. \$\endgroup\$
    – Nabb
    Mar 12, 2011 at 4:46
  • \$\begingroup\$ @Nabb, heheh, I didn't realize newline made it different. \$\endgroup\$
    – YOU
    Mar 12, 2011 at 5:44
13
\$\begingroup\$

Ruby 1.9, 46 42 39 characters

p (?a..?z).map{|a|a*$.+=1}*""==gets&&42

Assumes the input isn't terminated with a newline.

\$\endgroup\$
2
  • \$\begingroup\$ What about the newline from gets? \$\endgroup\$
    – steenslag
    Mar 10, 2011 at 12:44
  • 2
    \$\begingroup\$ @steenslag: The specs don't say anything the input being terminated by a newline, so this solution assumes there is none. \$\endgroup\$
    – Ventero
    Mar 10, 2011 at 12:48
11
\$\begingroup\$

C program - 78 89

Edit: Do not print 42 when there are extra characters.

Assumes input does not have a trailing newline.

main(i,j){for(i=1;i<27;i++)for(j=i;j--;getchar()==96+i?0:exit(1));puts("42"+!!gets(&i));}

If the prefix does not match, the program exits. If the prefix matches but there is 1-3 or so extra characters, prints 2. Otherwise, produces undefined behavior.

This can be made one character shorter by changing exit(1) to fork(). Oh, and on an unrelated note, remember to save any open documents in case, for whatever reason, your system happens to lock up.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ This will print 42 if the string is longer than the desired string and the prefix matches with the desired string. \$\endgroup\$
    – fR0DDY
    Mar 10, 2011 at 8:37
8
\$\begingroup\$

PHP (60)

Assuming the input is provided in the commandline:

for(;$i<702;)$s.=chr(96.5+sqrt($i+=2));echo$s!=$argv[1]?:42;

Explanation: you can view the string as a triangle structure.

j     i   val
0     0   a
1   1-2   bb
2   3-5   ccc
3   6-9   dddd
4 10-14   eeeee
5 15-20   ffffff
      ...

Line j starts at index i = j*(j+1)/2 (that's the triangular number formula). Solving the quadratic equation results in index i being on line j = int((sqrt(8*i+1)-1)/2) and therefore containing character 97 + int((sqrt(8*i+1)-1)/2). The 0-350 index range allows us to simplify that to 96.5 + sqrt(2*(i+1)), but that no longer holds true for larger values.

Edit: Switched to commandline input as suggested in the comments.
Edit: Uses conditional operator to save a character

\$\endgroup\$
8
  • \$\begingroup\$ +1 , that works ;) Umm could you please elaborate as how $s.=chr(96.5+sqrt($i+=2)); works ? \$\endgroup\$
    – Clyde Lobo
    Mar 10, 2011 at 12:24
  • \$\begingroup\$ Edited. I hope it makes sense :-) \$\endgroup\$ Mar 10, 2011 at 12:45
  • \$\begingroup\$ for(;$i<702;)$s.=chr(96.5+sqrt($i+=2));echo($s==$argv[1])*42; Only 61 chars, assuming stdin input \$\endgroup\$
    – Viper_Sb
    Mar 10, 2011 at 16:11
  • \$\begingroup\$ @Viper_Sb: thanks for the hint; I wasn't sure about the rules so I mimicked Clyde's solution. I'll go with your suggestion. \$\endgroup\$ Mar 10, 2011 at 17:37
  • 1
    \$\begingroup\$ @powtac the language syntax doesn’t require a space here \$\endgroup\$ Dec 31, 2011 at 4:49
7
\$\begingroup\$

Perl, 35 43

map$s.=$_ x++$a,a..z;say 42if<>~~$s

Needs Perl 5.10 or later (run with -E), no newline in input.

I liked my side-effects regex better, but the shorter code has spoken. Here it is as a souvenir. Also intended for Perl 5.10 or later, but only for the advanced/experimental regex features, so only a p command-line option is needed.

$a=a;$_=/^(??{$b++;$a++."{$b}"}){26}$/&&42
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5
  • \$\begingroup\$ Excellent. I tried hard to beat this w/recursive regex but couldn't get below 43c. :-(( \$\endgroup\$ Mar 10, 2011 at 11:04
  • \$\begingroup\$ Why is it necessary to say $a++."{$b}" instead of just $a++.$b? \$\endgroup\$
    – Timwi
    Mar 14, 2011 at 0:11
  • \$\begingroup\$ @Timwi because I do need those braces to appear in the resulting string. I don't want to match on literal "d4", I want "dddd", expressed as regex "d{4}" \$\endgroup\$
    – J B
    Mar 14, 2011 at 6:55
  • \$\begingroup\$ Of course. Thanks for explaining! \$\endgroup\$
    – Timwi
    Mar 18, 2011 at 15:07
  • \$\begingroup\$ Came up with a 33 byte solution for this! Try it online! \$\endgroup\$ Dec 10, 2017 at 19:52
7
\$\begingroup\$

05AB1E, 7 bytes

AƶJQi42

Try it online!

Explanation

A       push lowercase alphabet
 ƶ      lift every letter, push it and multiply it by its index
  J     join the list
   Qi   does it equal the input?
     42 push 42 and output implicitly

Just going through some challenges to learn 05AB1E (and golfing in general). This challenge was marked as active yesterday and I found a short solution, so why not share? :)

\$\endgroup\$
1
  • 3
    \$\begingroup\$ Welcome to PPCG! \$\endgroup\$
    – Steadybox
    Nov 10, 2017 at 14:57
6
\$\begingroup\$

Haskell program - 71 67 64 57

Assumes no trailing newline, and does not output one either.

f x|x==[c|c<-['a'..'z'],_<-['a'..c]]="42"
main=interact f

Usage:

$ echo -n 'abbcccddddeeeeeffffffggggggghhhhhhhhiiiiiiiiijjjjjjjjjjkkkkkkkkkkkllllllllllllmmmmmmmmmmmmmnnnnnnnnnnnnnnoooooooooooooooppppppppppppppppqqqqqqqqqqqqqqqqqrrrrrrrrrrrrrrrrrrsssssssssssssssssssttttttttttttttttttttuuuuuuuuuuuuuuuuuuuuuvvvvvvvvvvvvvvvvvvvvvvwwwwwwwwwwwwwwwwwwwwwwwxxxxxxxxxxxxxxxxxxxxxxxxyyyyyyyyyyyyyyyyyyyyyyyyyzzzzzzzzzzzzzzzzzzzzzzzzzz' | { ./42; echo; }
42
$ echo -n 'something else' | { ./42; echo; }
42: 42.hs:1:0-54: Non-exhaustive patterns in function f


$
\$\endgroup\$
6
  • 1
    \$\begingroup\$ fwiw this code is also still very readable. \$\endgroup\$
    – Dan Burton
    Mar 11, 2011 at 5:13
  • 1
    \$\begingroup\$ Since zipWith stops when it reaches the end of the shorter list, you can replace ['a'..'z'] with ['a'..] and save 3 characters. \$\endgroup\$
    – hammar
    Jun 9, 2011 at 19:29
  • \$\begingroup\$ @hammar: Thanks. I saved 1 more char by using >>= (concatMap) in lieu of concat and zipWith. \$\endgroup\$
    – Joey Adams
    Jun 10, 2011 at 0:16
  • \$\begingroup\$ @hammar: Saved even more by using do notation instead of >>= and lambda :-) \$\endgroup\$
    – Joey Adams
    Jun 10, 2011 at 0:22
  • 2
    \$\begingroup\$ @Joey: Or even better: [c|c<-['a'..'z'],_<-['a'..c]] \$\endgroup\$
    – hammar
    Jun 10, 2011 at 0:44
4
\$\begingroup\$

J, 29

f=:42#~((>:#a.{~97+])i.26)-:]

example:

f 'oasijfiojasef'

f 23841235

f 'abbccc...'
42
\$\endgroup\$
1
  • \$\begingroup\$ I like J. And I find it ugly and horrible. And I like it. \$\endgroup\$
    – seequ
    Nov 3, 2014 at 16:40
4
\$\begingroup\$

D: 94 Characters

void f(S)(S s){S t;foreach(i;0..26)t~=array(repeat(cast(char)(i+'a'),i+1));s==t&&writeln(42);}

More Legibly:

void f(S)(S s)
{
    S t;

    foreach(i; 0 .. 26)
        t ~= array(repeat(cast(char)(i + 'a'), i + 1));

    s == t && writeln(42);
}
\$\endgroup\$
3
\$\begingroup\$

Delphi, 164 132

This one builds a string and just compares it against the first command-line argument. It's shorter and less tricky than my other submission :

var s:string;c,i:int8;begin repeat s:=s+Char(c+97);i:=i-1;c:=c+Ord(i<0);if i<0then i:=c;until c=26;Write(42*Ord(s=ParamStr(1)));end.

(Note, that this version assumes that the c and i variables start out initialized at 0, as is the case in my version of Delphi (2010).)

Like my other submission, this one needs less characters if the string-building doesn't take place in a function, like I did before :

Delphi, 181

program a;{$APPTYPE CONSOLE}function s(c,i:byte):string;begin if(i>0)then Exit(Char(c)+s(c,i-1));if(c<122)then Exit(s(c+1,c-95));end;begin if(s(97,1)=ParamStr(1))then Write(42);end.

Note that the output doesn't need a newline, so WriteLn() became Write().

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3
\$\begingroup\$

PHP - 45 characters

I'm surprise nobody posted any answer that used hashing. It's a very size effecient way of testing for exact string.

echo md5($argv[1],1)!='¯è a@ÛÚƒ:ïT�p'?:42;

The data is kind of hard to copy/paste since there is a null-byte in the middle of the code. Here's an hex-dump of the code for testing purposes.

65 63 68 6f 20 6d 64 35 28 24 61 72 67 76 5b 31 5d 2c 31 29 21 3d 27 af e8 a0 61 40 db da 7f 11 0f 83 3a ef 54 00 70 27 3f 3a 34 32 3b

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1
  • 2
    \$\begingroup\$ Clever! Though since there are technically other input values yielding this same hash, it doesn't quite satisfy the requirement of outputting 42 only if the input is in the specified format. \$\endgroup\$
    – mellamokb
    Jun 10, 2011 at 3:50
3
\$\begingroup\$

Scala 79

 if((for(i <- 1 to 26;j<-1 to i)yield(96+i).toChar).mkString==args(0))print(42)
\$\endgroup\$
3
\$\begingroup\$

Pyth, 14

*42qzsm*dhxGdG

Just constructs the necessary string, then compares with the input and multiplies by 42.

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3
  • \$\begingroup\$ 'compares with the input and multiplies by 42.' would have never thought of that on my own. You just help me shave 1 character off my solution. Thanks. \$\endgroup\$
    – AndoDaan
    Nov 4, 2014 at 21:43
  • \$\begingroup\$ Argh, I came too late. Can you explain how it works? The function m cause me trouble… \$\endgroup\$
    – Jim
    May 31, 2017 at 16:36
  • 1
    \$\begingroup\$ @Jim We start with the map function, m, which applies a function to each element of its input. The input is G, the alphabet. xGd finds the position of d, the character in the alphabet, in G, the alphabet. h increases that by one, and *d replicates the character as that many times. Outside the m function, s combines the resulting list of strings into a single string, then qz checks whether the result is equal to the input. Booleans are represented as 0 if False and 1 if true, so *42 resultings in a value of 42 if True and 0 if False. \$\endgroup\$
    – isaacg
    May 31, 2017 at 22:11
3
\$\begingroup\$

Brachylog (2), 15 bytes, language postdates challenge

⊇Ạ∧?o?ọtᵐ~⟦₁+₁₆

Try it online!

And now for an answer which works on a completely different principle to most seen here. This is a function submission (the question doesn't specify what sort of submission is desired, but functions are permitted by default).

Explanation

This answer works by defining a sort of string: those which a) contain all lowercase letters of the alphabet, b) are in sorted order, and c) for which taking the number of occurrences of each character in the string produces a sequence of consecutive integers starting from 1. (It should be clear that there are many such strings, but the one we want to special-case is the shortest.) Then if the string fulfils those criteria, we add 16 to the number of distinct characters in the string; this will produce 42 if the string is the one the question asks us to special-case, and at least 43 in all other cases. (If the string fails any of the criteria to belong to the category, the function will end in failure, which is kind-of like throwing an exception.)

Here's how to interpret the source code:

⊇Ạ∧?o?ọtᵐ~⟦₁+₁₆
⊇Ạ               {the input} contains all lowercase letters
  ∧              and
   ?o?           the input sorts to itself
                 {and; implied when two conditions overlap}
     ?ọ          the {character, number of occurrences} pairs for the input
       tᵐ        when the second element of each is taken
         ~       create an output that could have been produced by
          ⟦₁     outputting the list of integers from 1 to some input inclusive;
            +₁₆  add 16 to that input {and output it}
\$\endgroup\$
3
\$\begingroup\$

R, 60 58

if(readline()==paste0(rep(letters,1:26),collapse=""))cat(42)

if(scan(,"")==paste(rep(letters,1:26),collapse=""))cat(42)

Thanks for the suggestion by @giusppe

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1
  • \$\begingroup\$ I think just paste is fine here, and you can use scan(,"") instead of readline(). \$\endgroup\$
    – Giuseppe
    Dec 11, 2017 at 17:10
2
\$\begingroup\$

Python (84)

Assumes a trailing newline at the end of the input.

import sys
if''.join(c*chr(c+96)for c in range(27))+'\n'==sys.stdin.read():print 42
\$\endgroup\$
2
  • 1
    \$\begingroup\$ Why not use raw_input instead sys.stdin.read? \$\endgroup\$
    – Champo
    Mar 10, 2011 at 11:33
  • 1
    \$\begingroup\$ @Juan: raw_input only reads a single line; I wasn't sure whether "the input" would always be a single line or whether there could be disqualifying input on later lines. \$\endgroup\$ Mar 10, 2011 at 15:42
2
\$\begingroup\$

Python - 62 chars

print("".join(x*chr(x+96) for x in range(27))==raw_input())*42
\$\endgroup\$
1
  • 2
    \$\begingroup\$ Could be shorter in Python 3: print("".join(x*chr(x+96)for x in range(27))==input())*42. \$\endgroup\$
    – mbomb007
    Mar 3, 2015 at 14:46
2
\$\begingroup\$

Perl, 49 46 characters

to be used in a program, not on the command line

$..=chr($+96)x$ for 1..26;$.eq(pop)&&print '42'

join('',map$_ x++$x,'a'..'z')eq pop&&print'42'

Regards

rbo

Edit: Idea ripped from Ventero

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2
\$\begingroup\$

PHP 92 88 87 chars

function _($a){for($i=97;$i<123;$i++)for($j=$i-96;$j;$j--)$b.=chr($i);echo($b==$a)*42;}

EDIT

Replaced $j<0 with $j and return $b==$a?42:0; with echo $b==$a?42:0;

Replaced echo $b==$a?42:0; with echo($b==$a)*42;

\$\endgroup\$
2
  • 2
    \$\begingroup\$ Could be 80 chars if it weren't for all the freaking dollar signs. \$\endgroup\$
    – Joey Adams
    Mar 10, 2011 at 6:41
  • \$\begingroup\$ Also see my entry for a single-loop solution. \$\endgroup\$ Mar 10, 2011 at 12:17
2
\$\begingroup\$

ECLiPSe Prolog - 173

c(_,[],_):-!. c(A,M,N):-length(L,N),checklist('='(A),L),append(F,L,M),P is N-1,B is A-1,c(B,F,P). ?- read_string(end_of_file,351,S),string_list(S,L),c(122,L,26),writeln(42).
\$\endgroup\$
2
\$\begingroup\$

JavaScript (91 93 94 98 102 116)

Usage: a('string'), returns 42 if valid according to spec, or 0.

function a(z){for(i=r='';i++<26;)for(j=i;j--;)r+=String.fromCharCode(i+96);return(z==r)*42}

http://jsfiddle.net/g25M3/6/

Edit: Removed var and eliminated two spaces in for (.

Edit 2: Changed j>0 to j, and

  1. return (z==r)?42:0; to
  2. return z==r?42:0

Edit 3: Initialize i with i='', change

  1. (z==r)?42:0 to
  2. (z==r)*42

Edit 4: Change

  1. for(;i<27;i++) to
  2. while(i++<26)

Edit 5: Change

  1. i=r='';while(i++<26) to
  2. for(i=r='';i++<26;) and
  3. for(j=i;j;j--) to
  4. for(j=i;j--;)
\$\endgroup\$
12
  • \$\begingroup\$ Well you can reduce another 12 chars if you dont use var to declare variables ;) \$\endgroup\$
    – Clyde Lobo
    Mar 10, 2011 at 6:03
  • \$\begingroup\$ @Clyde: Thanks! Also found I could remove space between for (. \$\endgroup\$
    – mellamokb
    Mar 10, 2011 at 6:16
  • \$\begingroup\$ Yeah , was about to suggest the same ;) Oh and I coded a solution in PHP using the same logic as yours \$\endgroup\$
    – Clyde Lobo
    Mar 10, 2011 at 6:18
  • \$\begingroup\$ You can save another 2 chars by replacing return(z==r)?42:0; with return z==r?42:0 \$\endgroup\$
    – Clyde Lobo
    Mar 10, 2011 at 6:22
  • \$\begingroup\$ j>0 could be just j I think. \$\endgroup\$
    – YOU
    Mar 10, 2011 at 6:25
2
\$\begingroup\$

JavaScript 1.8, 99 chars

function c(x)(x.replace(/([a-z])\1*/g,function(m)!(a-m.length)*m.charCodeAt(0)-96-a++,a=1)==0)*a+15

I dare you to make sense of it :)

\$\endgroup\$
2
\$\begingroup\$

PHP - 59

Assumes at least 1 input is provided over cli

echo md5($argv[1])!='afe8a06140dbda7f110f833aef540070'?:42;

It more or less works, except that md5 is can technically have duplications with the hashing algo.

\$\endgroup\$
2
\$\begingroup\$

PowerShell v2+, 47 bytes

42*(-join(1..26|%{,[char]($_+96)*$_})-ceq$args)

Constructs a range 1..26, feeds that through a loop with |%{...}. Each iteration we use the comma operator to construct an array literal of the current [char] multiplied by the current loop number. We then -join that all together to construct the string abbcccdddd... and then use a case-sensitive -ceq comparison against our input $args, which will result in either $TRUE or $FALSE. In PowerShell, Boolean values can be implicitly cast as 1 or 0, respectively, which is what happens here with the 42*. Will print out 42 iff the input is abbccc...zzzzzzzzzzzzzzzzzzzzzzzzzz and will output 0 otherwise.

\$\endgroup\$
1
  • \$\begingroup\$ you could save a byte :) Try it online! \$\endgroup\$
    – mazzy
    Apr 4, 2019 at 8:45
2
\$\begingroup\$

K, 26 Bytes

{(::;42)x~,/(1+!26)#'.Q.a}
{(::;42)x~,/(1+!26)#'.Q.a}"hello"
{(::;42)x~,/(1+!26)#'.Q.a}"abbcccddddeeeeeffffffggggggghhhhhhhhiiiiiiiiijjjjjjjjjjkkkkkkkkkkkllllllllllllmmmmmmmmmmmmmnnnnnnnnnnnnnnoooooooooooooooppppppppppppppppqqqqqqqqqqqqqqqqqrrrrrrrrrrrrrrrrrrsssssssssssssssssssttttttttttttttttttttuuuuuuuuuuuuuuuuuuuuuvvvvvvvvvvvvvvvvvvvvvvwwwwwwwwwwwwwwwwwwwwwwwxxxxxxxxxxxxxxxxxxxxxxxxyyyyyyyyyyyyyyyyyyyyyyyyyzzzzzzzzzzzzzzzzzzzzzzzzzz"
42

Thanks

\$\endgroup\$
2
  • \$\begingroup\$ {$[x~.Q.a@&1+!26;42;]} for 22 bytes. \$\endgroup\$
    – mkst
    Dec 11, 2017 at 12:20
  • 1
    \$\begingroup\$ 42*(&!27)~-96+"j"$ for 18 bytes (porting my oK answer). \$\endgroup\$
    – mkst
    Jan 15, 2018 at 22:33
2
\$\begingroup\$

VBA 91

There weren't any VBA answers but this works:

Function f(t)
    For i = 1 To 26
        o = o & String(i, Chr(i + 96))
    Next
    f = -42 * (t = o)
End Function
\$\endgroup\$
3
  • \$\begingroup\$ Is it really impossible to remove any whitespace from this? \$\endgroup\$ May 31, 2017 at 16:51
  • 1
    \$\begingroup\$ Condensed Version, 61 bytes - o="":For i=1To 26:o=o &String(i,Chr(i+96)):Next:?-42*([A1]=o) \$\endgroup\$ Sep 9, 2017 at 19:20
  • \$\begingroup\$ @Challenger5 No, it's not impossible. Taylor's Scott comment is exactly that. It runs in the Immediate Window and takes the value from cell A1 as input. \$\endgroup\$ Nov 10, 2017 at 21:21
2
\$\begingroup\$

APL (Dyalog), 18 17 bytes

42/⍨⍞≡819⌶⎕A/⍨⍳26

Try it online!

Four obvious bytes can be saved if we are allowed to use uppercase.

42 42

/⍨ if (lit. replicated by)

 character input

 is identical to

819⌶ the lowercased

⎕AAlphabet

/⍨ replicated by

 one through

26 26

\$\endgroup\$
2
\$\begingroup\$

05AB1E, 7 bytes

AƶJQ42*

Try it online!


Wellp, 05AB1E keeps evolving.

\$\endgroup\$
1
\$\begingroup\$

Clojure - 61 chars

(fn[a](if(=(mapcat #(repeat%(char(+% 96)))(range 1 27))a)42))

Exploits the following facts:

  • Clojure can interpret any string automatically as a sequence of chars
  • I can use the range of numbers from 1 to 26 to both create the characters and repeat them the correct number or times to generate the "correct" input
\$\endgroup\$
1
  • \$\begingroup\$ would save 6 chars if Clojure allowed nested #()s..... important feature request for Clojure 1.4 I think! \$\endgroup\$
    – mikera
    Mar 10, 2011 at 13:58
1
\$\begingroup\$

Javascript 144

Probably can be significantly improved, recursion has always been a head far for me.

Compressed

function r(a,b,d,c){c++;if(b<d)a+=r(a,b+1,d,c);for(i=0;i<c;i++)a=String.fromCharCode(b)+a;return a}function q(a){if(q==r(a,97,122,0))return 42};

Less Compressed

function r(s, c, x, w){        
    w++;
    if(c < x)
        s += r(s, c + 1, x, w);
    for(i = 0; i < w; i++)
        s = String.fromCharCode(c) + s;              
    return s;
}
function q(z){
    if(q==r(z,97, 122, 0))
        return 42;            
}

alert(q("rgrg"));
\$\endgroup\$

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