Play Zip, Zap, Zop

Question

There's a little improv warm up game where you arrange yourselves in a circle and send zips, zaps, and zops around by pointing to a person and saying the next word in the sequence, then they do the same until all of you are warmed up or whatever.

Your task is to create a program that gives the next word in sequence given an input word. (Zip --> Zap --> Zop --> Zip) Since there's a lot of different ways to say these three words and flairs that can be added to them, your program should imitate case and letter duplication and carry suffixes.

To elaborate, your input will be one or more Zs, then one or more Is, As, or Os (all the same letter), then one or more Ps, (all letters up to this point may be in mixed case) followed by some arbitrary suffix (which may be empty). You should leave the runs of Zs and Ps, as well as the suffix exactly as received, but then change the Is to As, As to Os, or Os to Is, preserving case at each step.

Example Test Cases

zip         ==> zap
zAp         ==> zOp
ZOP         ==> ZIP
ZiiP        ==> ZaaP
ZZaapp      ==> ZZoopp
zzzzOoOPppP ==> zzzzIiIPppP
Zipperoni   ==> Zapperoni
ZAPsky      ==> ZOPsky
ZoPtOn      ==> ZiPtOn
zipzip      ==> zapzip
zapzopzip   ==> zopzopzip
zoopzaap    ==> ziipzaap

Rules and Notes

You may use any convenient character encoding for input and output, provided that it supports all ASCII letters and that it was created prior to this challenge.
You may assume the input word is some variant of Zip, Zap, or Zop. All other inputs result in undefined behavior.
- Valid inputs will full-match the regex Z+(I+|A+|O+)P+.* (in mixed case)

Happy Golfing!

@Joshua This is invalid according to the description (see "all the same letter"). — Arnauld, Feb 27 at 16:23
@Arnauld: And the test case for zoopzaap disagrees with the description. — Joshua, Feb 27 at 16:25
@Joshua Why? This only applies to the vowels between the leading z's and the first p. The suffix may contain anything. — Arnauld, Feb 27 at 16:27

Arnauld · Accepted Answer · 2019-02-27 15:18:03Z

JavaScript (Node.js), 69 63 57 54 bytes

s=>Buffer(s).map(c=>s|c%4<1?s=c:c+c*90%320%34%24-8)+''

Try it online!

How?

We process the input string \$s\$ character by character.

We reuse \$s\$ as a flag: as soon as a numeric value is stored in it, we know that we must not update anything else.

To identify "p" (112) and "P" (80), we use the fact that their ASCII codes are multiples of \$4\$ and the ASCII codes of the other letters at the beginning of the string ("z", "Z" and vowels) are not.

To turn a vowel with ASCII code \$c\$ into its counterpart \$n\$ while leaving z and Z unchanged, we use the following function:

$$n=c+((((90\times c) \bmod 320)\bmod 34)\bmod 24)-8$$

 letter | ASCII code |  * 90 | % 320 | % 34 | % 24 | - 8 | new letter
--------+------------+-------+-------+------+------+-----+-----------------------
   'i'  |     105    |  9450 |  170  |   0  |   0  |  -8 | 105 -  8 =  97 -> 'a'
   'a'  |      97    |  8730 |   90  |  22  |  22  |  14 |  97 + 14 = 111 -> 'o'
   'o'  |     111    |  9990 |   70  |   2  |   2  |  -6 | 111 -  6 = 105 -> 'i'
   'z'  |     122    | 10980 |  100  |  32  |   8  |   0 | 122 +  0 = 122 -> 'z'
   'I'  |      73    |  6570 |  170  |   0  |   0  |  -8 |  73 -  8 =  65 -> 'A'
   'A'  |      65    |  5850 |   90  |  22  |  22  |  14 |  65 + 14 =  79 -> 'O'
   'O'  |      79    |  7110 |   70  |   2  |   2  |  -6 |  79 -  6 =  73 -> 'I'
   'Z'  |      90    |  8100 |  100  |  32  |   8  |   0 |  90 +  0 =  90 -> 'Z'

Commented

s =>                  // s = input string
  Buffer(s)           // convert it to a Buffer of ASCII codes
  .map(c =>           // for each ASCII code c in s:
    s |               //   if s is numeric
    c % 4 < 1 ?       //   or c is either 'p' or 'P':
      s = c           //     turn s into a numeric value and yield c
    :                 //   else:
      c +             //     update c
        c * 90 % 320  //     by applying the transformation function
        % 34 % 24     //     (see above)
        - 8           //
  ) + ''              // end of map(); coerce the Buffer back to a string

@ThomasHirsch It was found with a brute-force search function. The first step (multiplication + 1st modulo) makes sure that the parameters give identical result for both lowercase and uppercase. The 2nd step (the next 2 modulos and the subtraction) checks whether the correct delta values can be obtained from there. — Arnauld, Feb 27 at 23:19

Arnauld · Accepted Answer · 2019-02-28 12:56:15Z

6

C (gcc), 81 ... 61 48 46 bytes

Saved 2 bytes thanks to @Grimy

Port of my JS answer. Outputs by modifying the input string.

f(char*s){for(;*++s%4;*s+=*s*90%320%34%24-8);}

Try it online!

Commented

f(char * s) {       // f = function taking the input string s
  for(;             //   for each character *s in s:
    *++s % 4;       //     advance the pointer; exit if *s is either 'p' or 'P' (it's safe 
                    //     to skip the 1st character, as it's guaranteed to be 'z' or 'Z')
    *s +=           //     update the current character:
      *s * 90 % 320 //       apply a transformation formula that turns
      % 34 % 24     //       a vowel into the next vowel in the sequence
      - 8           //       while leaving 'z' and 'Z' unchanged
  );                //   end of for()
}                   // end of function

edited Feb 28 at 12:56

answered Feb 27 at 0:38

Arnauld

105k8 gold badges115 silver badges402 bronze badges

\$\begingroup\$ -2 bytes \$\endgroup\$ – Grimmy Feb 28 at 12:50
\$\begingroup\$ @Grimy Nice catch, thanks! (I did try *++s%4 at some point but overlooked the resulting optimization...) \$\endgroup\$ – Arnauld Feb 28 at 13:02
1

\$\begingroup\$ Further -3 bytes. This one should also be applicable to your JS answer. \$\endgroup\$ – Grimmy Feb 28 at 14:37
\$\begingroup\$ @Grimy It's different enough from mine, so you may want to post this as a separate answer. \$\endgroup\$ – Arnauld Feb 28 at 14:48

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recursive · Accepted Answer · 2019-02-26 22:23:41Z

5

Stax, 19 bytes

Ç╛√êΣ%,╖FP╚`=Lh←⌡·ƒ

Run and debug it

answered Feb 26 at 22:23

recursive

8,56615 silver badges32 bronze badges

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Neil · Accepted Answer · 2019-02-27 01:17:36Z

Retina 0.8.2, 21 bytes

iT`Io`A\OIia\oi`^.+?p

Try it online! Transliterates letters up to and including the first p, although the z and p aren't in the transliteration section so aren't affected. The first O is quoted because it normally expands to 13567 and the second o is quoted because it too is magic; in the first part of the transliteration it expands to the other string. The resulting transliteration is therefore from IAOIiaoi to AOIiaoi then removing the duplicate source letters results in IAOiao to AOIaoi.

Grimmy · Accepted Answer · 2019-02-28 15:11:12Z

4

C, 43 bytes

f(char*s){for(;*++s%4;*s^=*s%16*36%98%22);}

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Based on Arnauld's answer. I did a brute-force search to find the shortest formula that turns a => o, o => i, i => a, z => z.

edited Feb 28 at 15:11

answered Feb 28 at 14:53

Grimmy

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ASCII-only · Accepted Answer · 2019-02-26 23:52:30Z

3

Japt, 22 20 bytes

-2 bytes thanks to Oliver

r"%v+"_d`i¬iao¯`pu}"

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edited Feb 26 at 23:52

answered Feb 26 at 23:05

ASCII-only

4,68714 silver badges38 bronze badges

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CT Hall · Accepted Answer · 2019-02-27 15:21:39Z

3

R, 110 76 bytes

-36 bytes thanks to Krill

This function takes an input of one string.

function(a)sub(s<-sub('z+(.+?)p.*','\\1',a,T),chartr('aioAIO','oaiOAI',s),a)

Try it online!

edited Feb 27 at 15:21

answered Feb 27 at 2:25

CT Hall

59111 bronze badges

1

\$\begingroup\$ Collecting the string from pieces in R tends to be very long... But you save a lot of bytes by first extracting a substring s that we'll be translating, and then replacing it with a translated copy: 76 bytes \$\endgroup\$ – Kirill L. Feb 27 at 8:46
\$\begingroup\$ @KirillL., Ah that's the clever trick I was trying to find. \$\endgroup\$ – CT Hall Feb 27 at 15:18

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Jo King · Accepted Answer · 2019-02-27 11:16:14Z

2

Perl 6, 41 33 bytes

{S:i{.+?p}=$/~~tr/iaoIAO/aoiAOI/}

Try it online!

Simple case-insensitive substitution to shift the vowel section.

edited Feb 27 at 11:16

answered Feb 26 at 22:59

Jo King

36.1k4 gold badges79 silver badges151 bronze badges

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TFeld · Accepted Answer · 2019-02-28 06:23:57Z

2

Python 3.8 (pre-release), 81 78 bytes

lambda s,p='iaoIAO':''.join(('aoiAOI'+c)[(p:=p*(ord(c)%4)).find(c)]for c in s)

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Python 2, 98 bytes

lambda s:''.join(('aoiAOI'+c)[(-~-('p'in s[:i].lower())*'iaoIAO').find(c)]for i,c in enumerate(s))

Try it online!

edited Feb 28 at 6:23

answered Feb 27 at 13:47

TFeld

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Olivier Grégoire · Accepted Answer · 2019-03-02 16:39:54Z

2

Java (JDK), 52 bytes

s->{for(int i=0;s[++i]%4>0;)s[i]^=s[i]%16*36%98%22;}

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Port of Grimy's C answer. Make sure to upvote his answer!

edited Mar 2 at 16:39

answered Feb 27 at 14:56

Olivier Grégoire

10.4k2 gold badges19 silver badges45 bronze badges

\$\begingroup\$ -3 bytes by using a port of Grimy's C answer instead: s->{for(int i=0;s[++i]%4>0;)s[i]^=s[i]%16*36%98%22;} 52 bytes. \$\endgroup\$ – Kevin Cruijssen Mar 1 at 15:44

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Nahuel Fouilleul · Accepted Answer · 2019-02-27 10:56:07Z

1

Perl 5, 31 bytes

s/.+?p/$&=~y,iaoIAO,aoiAOI,r/ei

TIO

edited Feb 27 at 10:56

answered Feb 27 at 10:49

Nahuel Fouilleul

5,3771 gold badge5 silver badges15 bronze badges

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Giuseppe · Accepted Answer · 2019-02-27 15:38:57Z

1

SNOBOL4 (CSNOBOL4), 183 bytes

	INPUT (BREAK('Pp') SPAN('Pp')) . P REM . S
A	P 'a' ='o'	:S(A)
Z	P 'A' ='O'	:S(Z)F(P)
I	P 'i' ='a'	:S(I)
K	P 'I' ='A'	:S(K)F(P)
O	P 'o' ='i'	:S(O)
L	P 'O' ='I'	:S(L)
P	OUTPUT =P S
END

Try it online!

answered Feb 27 at 15:38

Giuseppe

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Chas Brown · Accepted Answer · 2019-02-28 06:01:34Z

1

Python 2, 84 bytes

f=lambda s,r='':s[0]in'pP'and r+s or f(s[1:],r+('aoiAOI'+s[0])['iaoIAO'.find(s[0])])

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edited Feb 28 at 6:01

answered Feb 28 at 5:55

Chas Brown

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Embodiment of Ignorance · Accepted Answer · 2019-03-02 17:03:59Z

1

C# (Visual C# Interactive Compiler), 60 bytes

n=>{for(int i=0;n[i]%4>0;)n[i]^=(char)(n[i++]%16*36%98%22);}

Based off of Grimy's C answer.

Try it online!

edited Mar 2 at 17:03

answered Feb 27 at 3:03

Embodiment of Ignorance

6,6084 silver badges33 bronze badges

1

\$\begingroup\$ This unfortunately doesn't work as it replaces the vowels in the suffix as well. \$\endgroup\$ – Emigna Feb 27 at 7:04
\$\begingroup\$ As @Emigna states above, this replaces all aoi-vowels, instead of just the ones before the first p/P. One thing to golf however: ("iao".IndexOf((char)(c|32))+1)%4 can be -~"iao".IndexOf((char)(c|32))%4 \$\endgroup\$ – Kevin Cruijssen Feb 27 at 12:04

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der bender · Accepted Answer · 2019-03-25 16:08:11Z

C/C++ (VC++ compiler) 192bytes

this is a rather naive try but anyway

void f(char*I){int c[]={-8,14,6},B=1,v[]={105,97,111},j=0;for(*I;*I>0&B;I++){if(*I==80|*I==112){B=0;break;}if(*I==90|*I==122){}else{for(j;j<3;j++){if(*I==v[j]|*I==v[j]-32){*I+=c[j];break;}}}}}

some slightly more readable Version is this

#include "stdafx.h"

void f(char * theString)
{
    signed int change[] = {'a'-'i','o'-'a','o'-'i'}; // add this to the vowel to get the next one
    char theVowels[] = {'i','a','o'};
    int breaker = 1;
    printf("Input %s\n",theString);
    for (int i = 0;(theString[i] != '\0') && breaker; i++)
    {
        switch (theString[i])
        {
            case 'Z': /*fall through*/
            case 'z': break;
            case 'P': /*fall through*/
            case 'p': breaker = 0;
                      break; 
            default: 
            {
                for (int j = 0; j < 3; j++)
                {
                    if ((theString[i] == theVowels[j]) || (theString[i]==(theVowels[j]-'a'+'A')))
                    {
                        theString[i] += change[j];
                        break;
                    }
                }
            }
            break;
        }

    }
    printf("Output %s\n",theString);
}
int main()
{
    char theString[]= "zzzzIIIIp0815-4711"; // a test string
    f(theString);
    return 0;
}

\$\begingroup\$ tio: tio.run/… \$\endgroup\$ – der bender Mar 25 at 16:13 — der bender, Mar 25 at 16:13
\$\begingroup\$ tio golfed : tio.run/… \$\endgroup\$ – der bender Mar 25 at 16:16 — der bender, Mar 25 at 16:16

Kirill L. · Accepted Answer · 2019-02-27 07:36:02Z

0

Ruby, 40 bytes

$_[$&]=$&.tr"IAOiao","AOIaoi"if/z+.+?p/i

Try it online!

answered Feb 27 at 7:36

Kirill L.

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Kevin Cruijssen · Accepted Answer · 2019-02-27 12:01:02Z

05AB1E (legacy), 22 bytes

l'pkIg‚£ć…iaoDÀ‚Du+`‡ì

Try it online or verify all test cases.

Can definitely be golfed a bit more..

Uses the legacy version of 05AB1E instead of the Elixir rewrite, because + merges fields in same-length lists, whereas the new version would need a pair-zip-join ‚øJ instead.

Explanation:

l                       # Lowercase the (implicit) input-string
                        #  i.e. "ZipPeroni" → "zipperoni"
 'pk                   '# Get the index of the first "p"
                        #  i.e. "zipperoni" → 2
    Ig‚                 # Pair it with the length of the entire input-string
                        #  i.e. 2 and "zipperoni" → [2,9]
       £                # Split the (implicit) input-string into parts of that size
                        #  i.e. "ZipPeroni" and [2,9] → ["Zi","pPeroni"]
        ć               # Extract head: push the head and remainder separately to the stack
                        #  i.e. ["Zi","pPeroni"] → ["pPeroni"] and "Zi"
         …iao           # Push string "iao"
             DÀ         # Duplicate, and rotate it once towards the left: "aoi"
               ‚        # Pair them up: ["iao","aoi"]
                Du      # Duplicate and transform it to uppercase: ["IAO","AOI"]
                  +     # Python-style merge them together: ["iaoIAO","aoiAOI"]
                   `    # Push both strings to the stack
                    ‡   # Transliterate; replacing all characters at the same indices
                        #  i.e. "Zi", "iaoIAO" and "aoiAOI" → "Za"
                     ì  # Prepend it to the remainder (and output implicitly)
                        #  i.e. ["pPeroni"] and "Za" → ["ZapPeroni"]

Titus · Accepted Answer · 2019-03-01 08:27:22Z

0

PHP, 30 bytes

too simple for a TiO:

<?=strtr($argn,oiaOIA,iaoIAO);

Run as pipe with -nF. For PHP 7.2, put the string literals in quotes.

answered Mar 1 at 8:27

Titus

13.8k1 gold badge14 silver badges41 bronze badges

\$\begingroup\$ This transliterates all oia vowels instead of only the ones before the first p/P, doesn't it? \$\endgroup\$ – Kevin Cruijssen Mar 1 at 15:39

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Play Zip, Zap, Zop

Example Test Cases

Rules and Notes

18 Answers 18

JavaScript (Node.js), 69 63 57 54 bytes

How?

Commented

C (gcc), 81 ... 61 48 46 bytes

Commented

Stax, 19 bytes

Retina 0.8.2, 21 bytes

C, 43 bytes

Japt, 22 20 bytes

R, 110 76 bytes

Perl 6, 41 33 bytes

Python 3.8 (pre-release), 81 78 bytes

Python 2, 98 bytes

Java (JDK), 52 bytes

Perl 5, 31 bytes

SNOBOL4 (CSNOBOL4), 183 bytes

Python 2, 84 bytes

C# (Visual C# Interactive Compiler), 60 bytes

C/C++ (VC++ compiler) 192bytes

Ruby, 40 bytes

05AB1E (legacy), 22 bytes

PHP, 30 bytes

Your Answer

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Play Zip, Zap, Zop

Example Test Cases

Rules and Notes

18 Answers 18

JavaScript (Node.js), 69 63 57 54 bytes

How?

Commented

C (gcc), 81 ... 61 48 46 bytes

Commented

Stax, 19 bytes

Retina 0.8.2, 21 bytes

C, 43 bytes

Japt, 22 20 bytes

R, 110 76 bytes

Perl 6, 41 33 bytes

Python 3.8 (pre-release), 81 78 bytes

Python 2, 98 bytes

Java (JDK), 52 bytes

Perl 5, 31 bytes

SNOBOL4 (CSNOBOL4), 183 bytes

Python 2, 84 bytes

C# (Visual C# Interactive Compiler), 60 bytes

C/C++ (VC++ compiler) 192bytes

Ruby, 40 bytes

05AB1E (legacy), 22 bytes

PHP, 30 bytes

Your Answer

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