Even-Odd chunks

Question

(Inspired by the Keg utility of this challenge)

Given a non-empty input string, e.g. s c 1= e(a"E"), split the input into even-odd chunks.

Example (Feel free to suggest more)

I can only think of this test case, fee free to suggest more.

This input string, when mapped to its code points, yields the list [115, 32, 99, 32, 49, 61, 32, 101, 40, 97, 34, 69, 34, 41]. When applied modulo-2 for every item, this returns [1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1].

In this list let's find the longest possible chunk that is consistent with even and odd code points:

[1, 0, 1, 0, 1], [1, 0, 1, 0, 1, 0, 1, 0, 1]

For the first chunk, this yields [1, 0, 1, 0, 1] because this is the longest chunk that follows the pattern

Odd Even Odd Even Odd Even ...

or

Even Odd Even Odd Even Odd ...

. Adding another codepoint into [1, 0, 1, 0, 1, 1] breaks the pattern, therefore it is the longest possible even-odd chunk that starts from the beginning of the string.

Using this method, we should split the input into chunks so that this rule applies. Therefore the input becomes (the ; here is simply a separator; this can be any separator that is not an empty string, Including the string itself):

s c 1;= e(a"E")

However, returning a list of strings is also permitted.

Rules

• This is code-golf so the shortest solution wins. Let it be known that flags don't count towards being in the pattern. They also don't count towards byte count in this challenge.

• The input will only be in ASCII, and the mapping will always be in ASCII (as far as I can tell most golflangs use a superset of ASCII).

Answering some of the comments

• You may output strings as lists of codepoints.

• "Any separator" includes the input string itself.

• You may insert other characters like the MATL answer, such as alphanumeric characters.

• You may not use integers as input instead of ASCII. Doing that will trivialize the challenge.

Answer 1

05AB1E, 12 6 4 5 bytes

Ç.¬+È

Input as a string, output as a 2D list of ASCII codepoint integers.

-6 bytes thanks to @Grimmy.
+1 byte now that the allowed I/O rules are finally settled..

Try it online or try it online with output as a list of strings instead.

Explanation:

Ç      # Transform the (implicit) input-string to a list of codepoint integers
 .¬    # Split this list of integers at:
   +   #  Sum the two codepoint at both sides of the potential split
    È  #  And check whether it is odd
       # (after which the resulting 2D integer list is output implicitly)

Answer 2

Japt -R, 9 bytes

óÈcv ¦Ycv

Try it

óÈcv ¦Ycv     :Implicit input of string
ó             :Partition
 È            :Between characters X & Y where
  c           :Charcode of X
   v          :Parity
     ¦        :Is not equal to
      Ycv     :Parity of charcode of Y
              :Implicit output, joined by newlines

Answer 3

Python 2, 72 71 bytes

lambda s:''.join(x+(ord(x)-~ord(y))%2*';'for x,y in zip(s,s[1:]))+s[-1]

Try it online!

1 byte thx to Jonathan Allan

Answer 4

Charcoal, 19 bytes

⭆θ⁺׶∧κ¬﹪⁺℅ι℅§θ⊖κ²ι

Try it online! Link is to verbose version of code. Explanation:

 θ                  Input string
⭆                   Map over characters and join
    ¶               Literal newline
   ×                Repeated by
      κ             Current index
     ∧              Logical And
       ¬            Logical Not
           ι        Current character
          ℅         Ordinal
         ⁺          Plus
              θ     Input string
             §      Indexed by
                κ   Current index
               ⊖    Decremented
            ℅       Ordinal
        ﹪           Modulo
                 ²  Literal 2
  ⁺                 Concatenated with
                  ι Current character
                    Implicitly print

Answer 5

Jelly 6 bytes

(7 if we must return a list of strings - add Ọ.)

ḂI¬Żœṗ

A monadic Link accepting a list of integers which yields a list of lists of integers.

Try it online!

How?

ḂI¬Żœṗj⁷ - Main Link: list of characters (A)
Ḃ        - least significant bit (vectorises)
 I       - incremental differences
  ¬      - logical NOT (vectorises)
   Ż     - prepend a zero
    œṗ   - at truthy indices (of left) partition (right=A)

---

Original challenge 7 bytes

żOḂMḊƲƝ

A full program using 2 as the separator.

Try it online!

How?

żOḂMḊƲƝ - Main Link: list of characters, S    e.g. ['4','5','7','4','0']
                                                   (from program argument '45740')
      Ɲ - for neighbours of S:                     ['4','5'] ['5','7'] ['7','4'] ['4','0']
     Ʋ  -   last four links as a monad:
 O      -     to ordinal (vectorises)              [52,53]   [53,55]   [55,52]   [52,48]
  Ḃ     -     least significant bit (vectorises)   [0,1]     [1,1]     [1,0]     [0,0]
   M    -     maximal indices                      [2]       [1,2]     [1]       [1,2]
    Ḋ   -     dequeue                              []        [2]       []        [2]
        - }                                        [[],[2],[],[2]]
ż       - (S) zip with (that)                      [['4',[]],['5',[2]],['7',[]],['4',[2]],['0']]
        - implicit, smashing print                 4527420

Answer 6

C (clang), 49 bytes

f(char*s){for(;*s;)putchar(*s)+*++s&1||puts("");}

Try it online!

Answer 7

MATL, 10 bytes

"I@2\XI=f@

Outputs each character on a different line. The separator is 1.

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Explanation

"      % Input: string (implicit). For each
  I    %   Push contents of clipboard I, initially 3
  @    %   Push current character
  2\   %   Modulo 2 of (code point) of that character. Gives 0 or 1
  XI   %   Copy result into clipboard I
  =    %   Equal? This compares the current and previous contents of
       %   clipboard I. Gives true or false
  f    %   Find. This outputs indices of true entries. Gives 1 or []
  @    %   Push current character
       % End (implicit)
       % Display stack (implicit), bottom to top. Each entry is displayed
       % on a different line. [] is not shown and doesn't start a new line

Answer 8

J, 18 14 bytes

<;.1~1,2=/\2|]

Try it online!

-4 bytes thanks to Bubbler

• 2|] remainders mod 2
• 2=/\ consecutive pairs of those, are the equal?
• 1, start it off with a group
• <;.1~ cut into groups using the first element as a delimiter, ie, starting a new group whenever we consecutive items are equal, odd odd or even even

Answer 9

C (clang), 90 ... 67 66 bytes

b;f(char*s){for(b=*s&1;*s;putchar(*s++))b^~*s&1?b=!b:putchar(59);}

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Saved 9 10 11 bytes thanks to @ceilingcat!!!

Answer 10

JavaScript (Node.js), 94 97 95 bytes

+1 byte for allowing more than 1 cut. Had to completely rewrite my code but had a lot of help from tsh's answer

s=>(r=[...s]).map(c=>c.charCodeAt()%2).flatMap((c,i,l)=>c-l[i-1]&1?b.push(r[i])&&[]:[b=[r[i]]])

Try it online!

Answer 11

JavaScript (Node.js), 53 bytes

s=>s.flatMap((c,i)=>c-s[i-1]&1?l.push(c)&&[]:[l=[c]])

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