Negate The String

Question

Disclaimer: This isn't my challenge but ThisGuy said I was OK to post.

Occasionally I want to make a word into it's opposite, like happiness goes to unhappiness. Unfortunately when this happens, my brain will sometimes go blank. Then one day, after yet another this happening, I thought to my self "This is what programs are for!"

As the English language has many exceptions, I have created a list which contains the prefix for the starting letter

q or h -> dis- (honest -> dishonest) l -> il- (legal -> illegal) m or p -> im- (mature -> immature) r -> ir- (responsible -> irresponsible) everything else -> un- (worthy -> unworthy)

Task

Given an input as a string, make the string into its negative and output the result. You can assume that all the inputs given will fit the above rules. Submissions may be programs or functions, not snippets.

Input

A single string, taken either as a parameter or from STDIN

Output

The negated form of that string, conforming to the above rules

How to Win

This is a code-golf so shortest code wins

Can we assume we'll never get a word that starts with a q without a u? — Business Cat
– Business Cat, Commented Apr 5, 2017 at 20:06
Off the top of my head, qadi, qat, the aforementioned qi, qirsh, and qwerty. (I play a lot of Scrabble) — AdmBorkBork
– AdmBorkBork, Commented Apr 5, 2017 at 20:23
@wsbltc Well, there's a fair few, but they're pretty much all borrowed words from other languages so questionable whether they really count as English. So can we assume that a q is always followed by a u in the string or not? — Business Cat
– Business Cat, Commented Apr 5, 2017 at 20:24

ovs · Accepted Answer · 2017-04-06 21:40:07Z

Python, 55 bytes

lambda n:'ddiiiiuiimmlrnss'[5-'rlmphq'.find(n[0])::7]+n

Try it online!

We need to handle 7 different starting letters:
g -> dis, h -> dis, p -> im, m -> im, l -> il, r -> ir and everything else -> un

We can store all these negations in a single string and extract the right one through slicing:

 d i s d i s i m i m i l i r u n 'ddiiiiuiimmlrnss'[i::7]

Now we need to calculate the start index i. 'rlmphq'.find returns 0 for 'r', 5 for q and -1 for everything not contained in the string. To get the needed value from 0 to 6 we still need to subtract the return value from 5, resulting in this code:

'ddiiiiuiimmlrnss'[5-'rlmphq'.find(n[0])::7]

Can someone explain how the hell is this works? I understand what's going on with the slice notation, but what are the magic strings ddiiiiuiimmlrnss and rlmphq and the number 5 for, why is the slice skip 7? — Keatinge
– Keatinge, Commented Apr 6, 2017 at 19:51

Riley · Accepted Answer · 2017-04-05 21:29:48Z

GNU sed, 50 bytes

Includes +1 for -r

s/^q|^h/dis&/ s/^l|^r|^m/i&&/ s/^p/imp/ t s/^/un/

Nothing fancy. Uses & in the replacement to combine a few substitutions, and t to skip the last one if one of the first substitutions happens.

Try it online!

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

Jelly, 30 bytes

1ị“qmlrrhp”iị“3bµWI⁼ṡ÷ʠ$»œs5¤;

Try it online!

How?

1ị“qmlrrhp”iị“3bµWI⁼ṡ÷ʠ$»œs5¤; - Main link: string 1ị - 1 index into the string (first character of the string) i - 1-based index of 1st occurrence of that in (else zero): “qmlrrhp” - char list "qmlrrhp" ị - index into (1-based and modulo): ¤ - nilad followed by link(s) as a nilad: “3bµWI⁼ṡ÷ʠ$» - compressed string "dis"+"imili"+"run"="disimilirun" œs5 - split "equally" into 5: ["dis","im","il","ir","un"] ; - concatenate with the string

Note that the repeated r in “qmlrrhp” is in the 5th index, which, if referenced, would result in prepending un, so it could equally well be anything other than h or p.

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

TI-Basic, 104 bytes

Prompt Str0 sub(Str0,1,1→Str2 Str0 If Str2="Q" or Str2="H "DIS"+Ans If Str2="L "IL"+Ans If Str2="M" or Str2="P "IM"+Ans If Str2="R "IR"+Ans If Ans=Str0 "UN"+Ans Ans

Requires all capital letters.

Explanation:

Prompt Str0 # 4 bytes, input user string to Str0 sub(Str0,1,1→Str2 # 12 bytes, store first character in Str2 Str0 # 3 bytes, store Str0 in Ans If Str2="Q" or Str2="H # 14 bytes, if the first letter was Q or H "DIS"+Ans # 8 bytes, store DIS+Ans in Ans If Str2="L # 7 bytes, If the first letter was L "IL"+Ans # 7 bytes, store IL+Ans in Ans If Str2="Q" or Str2="H # 14 bytes, if the first letter was Q or H "IM"+Ans # 7 bytes, store DIS+Ans in Ans If Str2="R # 7 bytes, if the first letter was R "IR"+Ans # 7 bytes, store IR+Ans in Ans If Ans=Str0 # 6 bytes, if Ans has not been changed (first letter was none of the above) "UN"+Ans # 7 bytes, store UN+Ans in Ans Ans # 1 byte, implicitly return Ans

\$\begingroup\$ 81 bytes \$\endgroup\$

MarcMush
– MarcMush

2021-06-06 22:08:58 +00:00
Commented Jun 6, 2021 at 22:08 — MarcMush
– MarcMush, Commented Jun 6, 2021 at 22:08

sporkl · Accepted Answer · 2017-04-07 20:16:34Z

///, 59 56 bytes

/^/\/\/#//#qu/disqu^h/dish^l/ill^m/imm^p/ipp^r/irr^/un/#

Try it online!

Input goes after the very last #.

How it works:

I made an optimization that reduced the size to 56 bytes, but since that complicates things I will explain the original version, then explain the golf.

/#qu/disqu//#h/dish//#l/ill//#m/imm//#p/ipp//#r/irr//#/un/# |everything after the ` |` is not code. /#qu/disqu/ |replace `qu` with `disqu` /#h/dish/ |replace `h` with `dish`. /#l/ill/ |replace `l` with `ill`. /#m/imm/ |replace `m` with `imm`. /#p/ipp/ |replace `p` with `ipp`. /#r/irr/ |replace `r` with `irr`. /#/un/ |replace everything else with `un`. # |safety control

Intuition: The challenge is simple enough, just add the negative depending on the beginning of the word. However, in ///, you cannot just concatenate if [...], you can only replace something following a specific pattern. So in this program, the positive word beginnings are replaced with the negative word beginnings. The # was added to make sure that once a new beginning was added, no more new beginnings would be added. The # also made it possible to do 'everything else: un'.

The golf incorporates a new substitution at he beginning: /^/\/\/#/. This replaces all ^ with //#, which was a common pattern in the original version.

Neil · Accepted Answer · 2017-04-05 20:42:39Z

Batch, 114 bytes

@set/pw= @set p=un @for %%p in (dis.q dis.h il.l im.m im.p ir.r)do @if .%w:~,1%==%%~xp set p=%%~np @echo %p%%w%

Checks the first character of the word against the list of custom prefixes and if so changes the prefix from the default of un. Special-casing qu is possible at a cost of 21 bytes.

Okx · Accepted Answer · 2017-04-06 13:24:36Z

Javascript, 72 71 66 61 60 59 bytes

~~w=>('dis....il.im...im.dis.ir'.split('.')[w.charCodeAt(0)-104]||'un')+w~~

~~w=>'un.dis.dis.il.im.im.ir'.split('.')['qhlmpr'.indexOf(w[0])+1]+w~~

~~Yeah, it's still longer than an existing solution. :)~~

w=>['un','dis','im','il','ir']['qmlrhp'.search(w[0])%4+1]+w

In case this needs any explanation, I'm taking advantage of the q/h and m/p pairs by combining their index in the search string with a mod 4, then using that as the lookup into the prefix array.

Great answer. Save 1 byte using search instead of indexOf. And some more, I think, using & instead of % — edc65
– edc65, Commented Apr 6, 2017 at 10:53
Thank you! I didn't know about search. Can't see how to make the & trick work - would be perfect if my array was only 4 elements. — Steve Bennett
– Steve Bennett, Commented Apr 6, 2017 at 11:07

forrert · Accepted Answer · 2017-04-06 15:40:07Z

3

JavaScript (71 64 61 bytes)

w=>({q:a='dis',h:a,l:'il',m:b='im',p:b,r:'ir'}[w[0]]||'un')+w

Edits:

Saved 7 bytes thanks to @ErtySeidohl (charAt(0) -> [0])
Saved 3 bytes thanks to @edc65 (assigning shared prefixes to variables)

var f = w=>({q:a='dis',h:a,l:'il',m:b='im',p:b,r:'ir'}[w[0]]||'un')+w; function onChange() { var word = event.target.value; var output = f(word); document.getElementById('output').innerHTML = output; }

Input Word: <input type='text' oninput='onChange()'/><br/> Output Word: <span id="output">

edited Apr 6, 2017 at 15:40

answered Apr 5, 2017 at 23:22

forrert

1315 bronze badges

1

\$\begingroup\$ If you don't care about backward compatibility with IE7, couldn't you use w[0] instead of w.charAt(0)? \$\endgroup\$

Erty Seidohl
– Erty Seidohl

2017-04-06 02:21:18 +00:00
Commented Apr 6, 2017 at 2:21
\$\begingroup\$ @ErtySeidohl Thanks! Just learned something new ;-) \$\endgroup\$

forrert
– forrert

2017-04-06 04:41:27 +00:00
Commented Apr 6, 2017 at 4:41
\$\begingroup\$ I'm new here, but is it legitimate to provide an answer starting with just w=>...? The actual function definition would include let f=w=>... ? (Probably covered in an FAQ somewhere...) \$\endgroup\$

Steve Bennett
– Steve Bennett

2017-04-06 07:10:53 +00:00
Commented Apr 6, 2017 at 7:10
\$\begingroup\$ @SteveBennett yes it's legitimate. Naming the function is not relevant \$\endgroup\$

edc65
– edc65

2017-04-06 10:32:08 +00:00
Commented Apr 6, 2017 at 10:32
1

\$\begingroup\$ w=>({q:a='dis',h:a,l:'il',m:b='im',p:b,r:'ir'}[w[0]]||'un')+w 3 bytes less \$\endgroup\$

edc65
– edc65

2017-04-06 10:52:47 +00:00
Commented Apr 6, 2017 at 10:52

Add a comment |

nimi · Accepted Answer · 2017-04-05 21:01:40Z

Haskell, 71 bytes

f l=maybe"un"id(lookup(l!!0)$zip"qhlmpr"$words"dis dis il im im ir")++l

Usage example: f "legal"-> "illegal". Try it online!

Build a lookup table of prefix/replacement pairs for looking up the first char of the input string with a default value of "un" if not found.

user61980 · Accepted Answer · 2017-04-05 21:13:24Z

Mathematica, 107 bytes

StringReplace[StartOfString~~x:#:>#2<>x&@@@{{"q"|"h","dis"},{"l","il"},{"m"|"p","im"},{"r","ir"},{_,"un"}}]

Explanation:

StartOfString~~x:#:>#2<>x& is a pure function where the first argument is a string pattern to match at the beginning of the string and the second argument is a string to prepend to the match. It returns a delayed rule suitable for use within StringReplace. This is then applied to each of the pairs {{"q"|"h","dis"},{"l","il"},{"m"|"p","im"},{"r","ir"},{_,"un"}} resulting in the list of rules

{ StartOfString~~x:"q"|"h":>"dis"<>x, StartOfString~~x:"l":>"il"<>x, StartOfString~~x:"m"|"p":>"im"<>x, StartOfString~~x:"r":>"ir"<>x, StartOfString~~x_:>"un"<>x }

Finally this list is passed into StringReplace which gives an operator on strings.

\$\begingroup\$ Does Mathmatica have a builtin for everything? \$\endgroup\$

user67719
– user67719

2017-04-05 21:14:18 +00:00
Commented Apr 5, 2017 at 21:14 — user67719
– user67719, Commented Apr 5, 2017 at 21:14

Okx · Accepted Answer · 2017-04-06 08:04:31Z

Retina, 54 bytes

^[^hqlmpr] un$+ ^[hq] dis$+ ^l il$+ ^[mp] im$+ ^r ir$+

Explanation:

 {implicit replace stage} ^[^hqlmpr] Append un to words starting with none of: hqlmpr un$+ ^[hq] Append dis to words starting with h or q dis$+ ^l Append il to words starting with l il$+ ^[mp] Append il to words starting with m or p im$+ ^r Append ir to words starting with r ir$+

First time I've used Retina. It's a pretty neat language.

Try it online!

\$\begingroup\$ Very nice first try at a language! +1 \$\endgroup\$

Arjun
– Arjun

2017-04-06 08:22:23 +00:00
Commented Apr 6, 2017 at 8:22 — Arjun
– Arjun, Commented Apr 6, 2017 at 8:22
\$\begingroup\$ And with that you have 2500 rep! Congratulations! \$\endgroup\$

Arjun
– Arjun

2017-04-06 08:23:02 +00:00
Commented Apr 6, 2017 at 8:23 — Arjun
– Arjun, Commented Apr 6, 2017 at 8:23
\$\begingroup\$ And your user index is 26600! \$\endgroup\$

Arjun
– Arjun

2017-04-06 08:23:38 +00:00
Commented Apr 6, 2017 at 8:23 — Arjun
– Arjun, Commented Apr 6, 2017 at 8:23
\$\begingroup\$ So much perfect base-10 ! \$\endgroup\$

Arjun
– Arjun

2017-04-06 08:24:13 +00:00
Commented Apr 6, 2017 at 8:24 — Arjun
– Arjun, Commented Apr 6, 2017 at 8:24

Steadybox · Accepted Answer · 2017-04-05 20:13:56Z

C, 109 107 bytes

f(char*s){printf("%s%s",*s-109&&*s-112?*s-108?*s-114?*s-104&&*s-113|s[1]-117?"un":"dis":"ir":"il":"im",s);}

Try it online!

Jörg Hülsermann · Accepted Answer · 2017-04-05 21:38:49Z

PHP, 101 Bytes

echo preg_match("#^(qu|[hlmpr])#",$argn,$t)?[qu=>dis,h=>dis,l=>il,m=>im,p=>im,r=>ir][$t[1]]:un,$argn;

Online Version

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

Excel 78 bytes

=TRIM(MID(" disdisil im im ir un",IFERROR(FIND(LEFT(A1),"qhlmpr"),7)*3,3))&A1

I found some close contenders using different methods that scored 81 bytes:

=IFERROR(CHOOSE(FIND(LEFT(A1),"qhlmpr"),"dis","dis","il","im","im","ir"),"un")&A1

And 84 bytes:

=IFERROR(TRIM(MID("im disi"&LEFT(A1),MOD(FIND(LEFT(F1),"qlmhrp"),3)*3+1,3)),"un")&A1

idrougge · Accepted Answer · 2017-04-06 08:19:10Z

REXX, 78 bytes

arg a s.=un s.q=dis s.h=s.q s.l=il s.m=im s.p=im s.r=ir p=left(a,1) say s.p||a

Saves a few bytes by replying in UPPERCASE, e.g. potent -> IMPOTENT.

Denis Ibaev · Accepted Answer · 2017-04-06 08:33:45Z

Perl, 49 + 1 (`-p` flag) = 50 bytes

s|^[hq]|dis$&|||s|^[lmr]|i$&$&|||s|p|imp|||s||un|

Using:

perl -pe 's|^[hq]|dis$&|||s|^[lmr]|i$&$&|||s|p|imp|||s||un|' <<< responsible

Try it online.

NikoNyrh · Accepted Answer · 2017-04-06 08:48:48Z

Clojure, 65 bytes

#(str(get{\q"dis"\h"dis"\l"il"\m"im"\p"im"\r"ir"}(first %)"un")%)

Well this is boring... but I couldn't make it any shorter. At least there is very little whitespace.

Redouane Red · Accepted Answer · 2017-04-06 10:45:38Z

OCaml, 85

(fun s->(match s.[0]with|'q'|'h'->"dis"|'l'->"il"|'m'|'p'->"im"|'r'->"ir"|_->"un")^s)

Anonymous function, uses pattern matching on its first char.

emanresu A · Accepted Answer · 2021-06-06 05:27:37Z

0

Vyxal, 38 bytes

«÷c4w«?hḟ:u=[`un`|4%«⟑Ẇf:«2ẇ`Ġ⟨`p$i]?+

Try it Online!

Messy.

answered Jun 6, 2021 at 5:27

emanresu A

46.2k5 gold badges112 silver badges257 bronze badges

Add a comment |

mazzy · Accepted Answer · 2021-06-06 08:30:19Z

PowerShell, 66 62 bytes

%4+1 inspired by @Steve Bennett

param($s)(echo un dis im il ir)[('qmlrhp'|% I*f $s[0])%4+1]+$s

Try it online!

Stack Exchange Network

Negate The String

Task

Input

Output

How to Win

20 Answers 20

Python, 55 bytes

GNU sed, 50 bytes

Jelly, 30 bytes

How?

TI-Basic, 104 bytes

///, 59 56 bytes

Batch, 114 bytes

Javascript, 72 71 66 61 60 59 bytes

JavaScript (71 64 61 bytes)

Haskell, 71 bytes

Mathematica, 107 bytes

Explanation:

Retina, 54 bytes

C, 109 107 bytes

PHP, 101 Bytes

Excel 78 bytes

REXX, 78 bytes

Perl, 49 + 1 (`-p` flag) = 50 bytes

Clojure, 65 bytes

OCaml, 85

Vyxal, 38 bytes

PowerShell, 66 62 bytes

Linked

Hot Network Questions

Negate The String

Task

Input

Output

How to Win

20 Answers 20

Python, 55 bytes

GNU sed, 50 bytes

Jelly, 30 bytes

How?

TI-Basic, 104 bytes

///, 59 56 bytes

Batch, 114 bytes

Javascript, 72 71 66 61 60 59 bytes

JavaScript (71 64 61 bytes)

Haskell, 71 bytes

Mathematica, 107 bytes

Explanation:

Retina, 54 bytes

C, 109 107 bytes

PHP, 101 Bytes

Excel 78 bytes

REXX, 78 bytes

Perl, 49 + 1 (-p flag) = 50 bytes

Clojure, 65 bytes

OCaml, 85

Vyxal, 38 bytes

PowerShell, 66 62 bytes

Linked

Related

Hot Network Questions

Perl, 49 + 1 (`-p` flag) = 50 bytes