Is it double speak?

Question

In an earlier challenge I asked code golfers to produce strings which copy each character in a string. For example:

TThhiiss  iiss  ddoouubbllee  ssppeeaakk!!

This challenge is simply to detect if some text meets the definition of a double speak string.

• There is an even number of characters.
• When split into pairs, every pair consists of two of the same character.

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The challenge

• It's code golf, do it in few bytes.
• Use any language you choose.
• Please include a link to an online interpreter.
• The code will accept some text.
  • For simplicity, the input will only consist of printable ASCII characters
• It will return an indication of whether or not the input is double speak. It could be:
  • A boolean
  • Strings ('true', 'false', 'yes', 'no' etc)
  • Integers 0 or 1

---

Test Cases:

• aba - false
• abba - false
• aabb - true
• aaabb - false
• tthhiiss - true
• ttthhhiiisss - false

Answer 1

Python 3, 24 bytes

lambda s:s[::2]==s[1::2]

Try it online!

Answer 2

brainfuck, 20 bytes

Saved 1 byte thanks to Jo King.

+>,[>,[-<->]<[<],]<.

Try it online!

Readable output!

Takes input two characters at a time, and moves away from the 1 on the tape if any pair doesn't match. EOF is treated as 0 and thus handled automatically.

Output is a null byte if the string is not double speak, and 0x01 if it is. The readable version outputs these as characters at a cost of 14 bytes.

Answer 3

MATL, 4 bytes

Heda

Input is a string, enclosed with single qoutes. Output is 0 for double speak, 1 otherwise.

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Explanation

Consider input 'TThhiiss iiss ddoouubbllee ssppeeaakk!!' as an example.

H    % Push 2
     % STACK: 2
     % Implicit input (triggered because the next function requires two inputs): string 
     % STACK: 'TThhiiss  iiss  ddoouubbllee  ssppeeaakk!!', 2
e    % Reshape as a 2-column matrix of chars, in column-major order. Pads with char(0)
     % if needed. Note that char(0) cannot be present in the input
     % STACK: ['This is double speak!';
               'This is double speak!']
d    % Difference of each column
     % STACK: [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
a    % Any: gives 0 if and only if all elements are 0
     % STACK: 0
     % Implicit display

Answer 4

05AB1E, 6 5 2 bytes

ιË

Input as a list of characters.

-3 bytes by porting @Shaggy's Japt answer, so make sure to upvote him!

Try it online or verify a few more test cases.

Explanation:

ι   # Uninterleave the (implicit) input-list of characters
    #  i.e. ["t","t","t","t","e","e","s","s","t","t","!","!","!"]
    #   → [["t","t","e","s","t","!","!"],["t","t","e","s","t","!"]]
 Ë  # Check if both inner lists are equal
    #  → 0 (falsey)
    # (after which the result is output implicitly)

Answer 5

Japt, 4 bytes

ó r¶

Try it

ó r¶     :Implicit input of string
ó        :Uniterleave
  r      :Reduce by
   ¶     :  Testing equality

Alternative

ó
¥o

Try it

Answer 6

Retina, 9 bytes

(.)\1

^$

Try it online.

Explanation:

Remove all pair of the same characters:

(.)\1

Check if there are no characters left:

^$

Answer 7

Jelly, 3 bytes

ŒœE

Try it online!

Answer 8

Stax, 5 bytes

■◄┼$Δ

Run and debug it

Procedure:

• Calculate run-lengths.
• Get GCD of array.
• Is even?

Answer 9

PHP, 58 56 bytes

function f($s){return!$s?:$s[0]==$s[1]&f(substr($s,2));}

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As a recursive function.

PHP, 61 56 52 bytes

while(''<$l=$argn[$i++])$r|=$l!=$argn[$i++];echo!$r;

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Or standalone program. Input string via STDIN, output is truthy (1) if it is double speak, and falsey (0) if it is not double speak.

-4 bytes thx to @Night2!

Answer 10

x86 machine code, 9 7 bytes

D1 E9       SHR  CX, 1          ; divide length in half 
AD          LODSW               ; load next two chars into AH/AL 
3A E0       CMP  AH, AL         ; compare AH and AL 
E1 FB       LOOPE -5            ; if equal, continue loop

Input string in SI, input string length in CX. Output ZF if is double speak.

Or 14 bytes as a complete PC DOS executable:

B4 01       MOV  AH, 01H        ; DOS read char from STDIN (with echo) 
CD 21       INT  21H            ; read first char into AL
92          XCHG DX, AX         ; put first char into DL
B4 08       MOV  AH, 08H        ; DOS read char from STDIN (no echo) 
CD 21       INT  21H            ; read second char into AL
3A C2       CMP  AL, DL         ; compare first and second char 
74 F3       JE   -13            ; if the same, continue loop 
C3          RET                 ; otherwise exit to DOS 

Input is via STDIN, either pipe or interactive. Will echo the "de-doubled" input until a non-doubled character is detected, at which point will exit (maybe bending I/O rules a little bit, but this is just a bonus answer).

Build and test ISDBL2.COM using xxd -r:

00000000: b401 cd21 92b4 08cd 213a c274 f3c3       ...!....!:.t..

Original 24 bytes complete PC DOS executable:

D1 EE       SHR  SI, 1          ; SI to DOS PSP (080H) 
AD          LODSW               ; load string length into AL 
D0 E8       SHR  AL, 1          ; divide length in half 
8A C8       MOV  CL, AL         ; put string length into BL 
        CLOOP: 
AD          LODSW               ; load next two chars into AH/AL 
3A E0       CMP  AH, AL         ; compare AH and AL 
E1 FB       LOOPE CLOOP         ; if equal, continue loop
        DONE: 
B8 0E59     MOV  AX, 0E59H      ; BIOS tty function in AH, 'Y' in AL 
74 02       JZ   DISP           ; if ZF, result was valid double 
B0 4E       MOV  AL, 'N'        ; if not, change output char to N 
        DISP: 
B4 0E       MOV  AH, 0EH 
CD 10       INT  10H 
C3          RET                 ; return to DOS

Input from command line, output to screen 'Y' if double, 'N' if not.

Build and test ISDBL.COM using xxd -r:

00000000: d1ee add0 e88a c8ad 3ae0 e1fb b859 0e74  ........:....Y.t
00000010: 02b0 4eb4 0ecd 10c3                      ..N.....

Credits:

• -2 bytes thx to @ErikF!

Answer 11

Lua, 67 66 63 59 33 32 bytes

-25 bytes thanks to Giuseppe
-1 byte thanks to val

print(#(...):gsub("(.)%1","")<1)

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Removes every doubled character, then checks if the result is empty.

Answer 12

JavaScript, 28 bytes

s=>s.every((x,y)=>x==s[y|1])

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23 bytes using wastl's regex

s=>/^((.)\2)*$/.test(s)

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Answer 13

Perl 5, 15 bytes

$_=/^((.)\2)*$/

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Outputs 1 for double-speak, nothing for non-double-speak.

Answer 14

MathGolf, 2 bytes

½=

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Basically the same as the 05AB1E answer, ½ splits the string into even and odd characters, then check for equality. Passes for the empty string.

Answer 15

Haskell, 28 23 bytes

f(x:y:z)|x==y=f z
f[]=1

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Very straightforward. Double speak is only empty or a repeated character prepended to double speak.

Less straightforward now. Outputs via presence or absence of an error, per meta consensus; no error means double speak. Pattern matching fails when the first two characters differ or when there are an odd number of characters. Thanks to Laikoni for these savings!

Answer 16

Julia 1.0, 25 bytes

s->s[1:2:end]==s[2:2:end]

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Answer 17

V (vim), 7 bytes

ӈ±
ø^$

Try it online! or Verify test cases

Hexdump:

00000000: d388 b10a d85e 24                        .....^$

Just two regexes. Explanation:

Ó   " Remove all occurrences...
 ˆ  "   Any character
  ± "   Followed by itself
    "   This regex is actually just the compressed form of (.)\1
ø   " Count the number of matches
 ^$ "   An empty line

Answer 18

J, 13 11 10 bytes

-:2#_2{.\]

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-2 bytes thanks to Adám

-1 byte thanks to miles

TLDR explanation: Is the input the same as every other character of the input doubled?

Answer 19

PowerShell, 64 59 bytes

filter f($n){$a,$b,$r=$n;$a-eq$b-and$(if($r){f $r}else{1})}

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Recursive function, no regex. Takes input as a char-array (see TIO link). Peels off the first two elements into $a and $b, stores the remaining into $r. If we still have elements remaining, recurse along with $a -eq $b. Otherwise just check whether $a -eq $b. Output is implicit.

-5 bytes thanks to mazzy

Answer 20

Brachylog, 5 bytes

ġ₂z₂=

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Succeeds or fails.

ġ₂       The at-most-length-2 chunks of the input,
  z₂     which have equal length, zipped together,
    =    are equal.

Answer 21

PowerShell, 39 38 bytes

!$($args|?{+$p*($p="$_"[$p-eq$_])};$p)

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where $p contains a previous char.

No recursion, no regex :). Takes input as a char-array via a splatting string (see TIO link).

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PowerShell, 48 bytes

for(;$b-eq$a-and$args){$a,$b,$args=$args}$b-eq$a

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No recursion, no regex and no pipe :D. It also takes input as a char-array via a splatting string. It uses $b-eq$a instead $a-eq$b for a case when a last char has #0 code.

Answer 22

Shakespeare Programming Language, 204 156 bytes

-48 bytes thanks to Jo King (mostly by changing the output method)

A.Ajax,.Puck,.Act I:.Scene I:.[Exeunt][Enter Ajax and Puck]Ajax:Open mind.Puck:Open
mind.Is I worse zero?If soSpeak thy.Is you as big as I?If soLet usAct I.

Try it online!

Exits with error if the input is double speak, and with warning if it is not double speak (which is allowed by default).

Answer 23

Keg, 19 17 characters

?{!1<|=[|0.(_)]}1

Explanation:

?             # read input

{             # while
    !1<       # stack length greater than 1?
|             # end of while condition and beginning of while block
    =         # compare the 2 top values in the stack
    [         # if (the condition is the top of stack)
    |         # end of then block and beginning of else block
        0.    # output 0
        (_)   # clear stack (discard top of stack in for loop stack length times)
    ]         # end if
}             # end while

1             # stack is already empty, push a truthy value

              # implicitly output the stack content if there was no explicit output

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Answer 24

Brain-Flak, 26, 22 bytes

({<({}[{}])>{()<>}{}})

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Outputs 1 for false and 0 for true.

Readable version:

({
    <({}[{}])>
    {
        ()
        <>
    }
    {}
})

I originally had this:

{
    ({}[{}])

    {
        <>([()])<>{{}}
    }{}
}
<>({}())

Which is 10 bytes longer.

Answer 25

JavaScript, 26 23 bytes

s=>/^((.)\2)+$/.test(s)

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Recursive Solution, 30 bytes

Thanks to Arnauld for a fix at the cost of 0 bytes.

f=([x,y,...s])=>x?x==y&f(s):!y

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Answer 26

R, 53 34 bytes

-19 bytes thanks to Giuseppe

function(a)gsub("(.)\\1","",a)==""

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Answer 27

QuadR, 11 bytes

''≡⍵
(.)\1

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''≡⍵ the result is an empty string when

(.)\1 a character followed by itself

 is replaced by nothing

Answer 28

C# (Visual C# Interactive Compiler), 40 bytes

s=>s.Where((c,i)=>c!=$"{s}"[i|1]).Any()

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Each character is tested against it's neighbor for inequality. If any inequalities exist, the result is false.

Since input is limited to printable ASCII characters, a control character is appended to input which is compared to the last character of an odd length string.

Outputs are reversed.

-3 bytes inspired by @Oliver

Answer 29

Red, 36 bytes

func[s][parse s[any[copy t skip t]]]

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Longer alternative:

Red, 40 bytes

func[s][(extract s 2)= extract next s 2]

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Answer 30

Prolog (SWI), 60 45 bytes

thanks to Unrelated String

+[].
+[A,A|T]:- +T.
-X:-string_chars(X,Y),+Y.

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Converting it from a string to a list of atoms kind of ruined the score, but well..

Answer 31

Brain-Flak, 36 10 42 30 bytes

(<>)<>{({}[{}]<>){{}{}}{}<>}<>

Outputs 0 if it is double speak, and nothing if it isn't

-12 bytes thanks to Nitrodon

How it works

(<>) Pushes 0 to the second stack
<>   Swaps back to the first stack
{    Begins a loop that will run until the stack is empty
({}[{}]<>)    Pops the top two items off of the first stack and pushes their difference to the second stack
{{}{}}{}      If the difference is zero, it gets popped and nothing happens. 
If there difference is one, the zero that was put on at the beginning gets popped, so nothing will get outputted at the end 
<>   Swaps back to the first stack
}    End loop
<>   Swap to the second stack for output

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Answer 32

Zsh, 36 bytes

My Zsh answer to the previous challenge can be found here.

Exits truthy (0) if NOT double speak, and falsy (1) if double speak. (As allowed in a comment.)

for a b (${(s::)1})r+=${a#$b}
[ $r ]

---

for a b (${(s::)1})r+=${a#$b}
         ${(s::)1}             # split $1 characterwise
for a b (         )            # take pairs of characters from ${(s::)1}, assign to $a and $b
                      ${a   }  # first character
                      ${ #$b}  # remove second character as prefix
                   r+=         # append to $r as string
[ $r ]                         # exit truthy if $r is non-empty

Try it online!

Answer 33

Mathematica, 31 bytes

AllTrue[Length/@Split@#,EvenQ]&

This works by converting the input (an array of characters, according to @attinat) into a list of characters, splitting this list into sublists of contiguous identical elements, and checking that each sublist has an even number of elements.

Answer 34

Scala, 44 bytes

_.grouped(2).forall(t=>t.size>1&&t(0)==t(1))

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Answer 35

Cubix, 18 bytes

;U;;A-!/@/n\nW.\O?

Try it online! Outputs 1 if True and nothing if False.

I suppose that it is fitting that this is twice the length to my answer to the previous question

    ; U
    ; ;
A - ! / @ / n \
n W . \ O ? . .
    . .
    . .

Watch it run

• A Put all input onto the stack
• - subtract the TOS and start of the test loop
• ! test result for 0
• @ if not 0, halt (FALSE for double speak)
• /;U;; if 0, pop subtraction result and top two items from the stack, u-turn sends it in the right direction.
• !\n/? an ignored "if 0", reflect, negate, reflect and test for EOI (now 1)
• O\/@ if positive (1) output and reflect a couple of times onto the halt (TRUE for double speak)
• nW if negative, reverse the negation and shift lane onto the start of the test loop.

Answer 36

Gaia, 3 bytes

2Zy

Almost as trivial as the solution to the other challenge. Errors with an input size < 2, which was never confirmed to be allowed, so might be invalid.

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2Z   Unzip input by 2
  y  All equal?

Unzipping by 2 basically undoes the result of interleaving 2 strings. So if two equal strings are interleaved, then it is doublespeak.

Answer 37

K (oK), 12 bytes

Solution:

&/1==/'0N 2#

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Explanation:

&/1==/'0N 2# / the solution
       0N 2# / reshape into Nx2 grid
    =/'      / equals (=) over (/) each (')
  1=         / equal to 1 (ie a match)
&/           / take the minimum

Extra:

• (~).+0N 2# works(ish) for 10 bytes, but not if the only difference is a character on the end of one of the strings :(

Answer 38

PowerShell, 27 24 bytes

^{-3 bytes thanks to mazzy}

!($args-creplace"(.)\1")

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Uses the regex method going around. If the string is emptied out, it will return true, otherwise false.

Answer 39

SNOBOL4 (CSNOBOL4), 63 bytes

	I =INPUT
R	I LEN(1) . X
	I X X =	:S(R)
	OUTPUT =IDENT(I) 1
END

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Matches the first character LEN(1) and saves it . to X. If I matches X concatenated with itself, it replaces that substring and repeats. If the remaining string is empty, IDENT(I,<implicit empty string>), then 1 is output, else nothing is output.

Answer 40

Charcoal, 10 bytes

⬤⪪Ｓ²⁼ι⮌…ι²

Try it online! Link is to verbose version of code. Outputs Charcoal's default boolean format, which is - for true and nothing for false. Explanation:

  Ｓ         Input string
 ⪪ ²        Split into substrings of length up to 2
⬤           All substrings
        ι   Current substring
       … ²  Extended to length 2
      ⮌     Reversed
    ⁼       Equals
     ι      Current substring
            Implicitly print

Answer 41

Whitespace, 73 bytes

[N
S S N
_Create_Label_LOOP][S S S N
_Push_0][S N
S _Duplicate_0][T   N
T   S _Read_input_as_character][T   T   T   _Retrieve][S N
S _Duplicate_input][S S S T S T S N
_Push_10][T S S T   _Subtract][N
T   S S N
_If_0_Jump_to_Label_EXIT][S S S N
_Push_0][S N
S _Duplicate_0][T   N
T   S _Read_input_as_character][T   T   T   _Retrieve][T    S S T   _Subtract][N
T   S N
_If_0_Jump_to_Label_LOOP][S S S N
_Push_0][T  N
S T _Print_as_integer][N
S S S N
_Create_Label_EXIT]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Since Whitespace inputs one character at a time, the input should contain a trailing newline so it knows when to stop reading characters and the input is done.

Outputs 0 for falsey, or nothing for truthy.

Try it online (with raw spaces, tabs and new-lines only).

Explanation in pseudo-code:

Start LOOP:
  Character c = STDIN as character
  If(c == '\n'):
    Stop program
  Character d = STDIN as character
  If(c == d):
    Go to next iteration of LOOP
  Print 0

Answer 42

APL+WIN, 20 15 bytes

5 bytes saved thanks to Adam.

Prompts for input of string:

n≡(2×2|⍳⍴n)/n←⎕

Try it online! Courtesy of Dyalog Classic

Answer 43

Stax, 6 5 bytes

╩3╦x╨

Run and debug it at staxlang.xyz!

Outputs integers. Zero for double speak and nonzero (one now!) otherwise, as allowed here. Replace 1I below with |e if you want the opposite behavior, but I think this version looks nicer.

Unpacked (6 bytes) and explanation

:G|g1I
:G        Array of run lengths
  |g      GCD
    1I    Is odd? Could use 2% instead

The GCD of run lengths will be even exactly when all run lengths are.

Answer 44

PHP, 37 40 bytes

+3 bytes to fix '0' issue

<?=''==preg_replace('/(.)\1/','',$argn);

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Similar to other RegEx answers.

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PHP, 55 bytes

for(;($l=$argn[$i++])==$argn[$i++]&&$l.$l;);echo$l=='';

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Loops to the end of the string as long as every even character (0th, 2nd, 4th, etc ...) is equal to the character after it, else stops the loop. Finally, checks if it has arrived at the end of the string, which means the string is a double speak.

---

Outputs 1 for double speak and nothing for not double speak.

Answer 45

dzaima/APL, 14 11 bytes

⊢≡⊢⌿⍨2 0⍴⍨≢

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-3 thanks to Adám!

Answer 46

Kotlin, 56 52 bytes

{it.chunked(2).filter{it.toHashSet().size>1}.size<1}

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Thank you Khuldraeseth na'Barya for saving 4 bytes

Answer 47

Pip, 31 bytes

a:qFb,#a{b%2?x(ab+1)Qa@b?xi:1}i

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slightly different approach with fold operator, 31 bytes

a:qFb,#a{I!b%2i:$Q[a@b(ab+1)]}i

outputs 0 if double speak, 1 otherwise

Answer 48

INTERCAL, 192 bytes

PLEASE,1<-#2DOCOMEFROM(2)DOWRITEIN,1DO.5<-#1$',1SUB#1'~#256PLEASE(1)NEXTDOREADOUT#1DOGIVEUP(1)DO(1002)NEXTDO.5<-#1$',1SUB#2'~,1SUB#2DO(3)NEXTDOREADOUT#0PLEASEGIVEUP(3)DO(1002)NEXT(2)DOFORGET#2

Try it online!

Output is done with INTERCAL's native "butchered Roman numerals", so the true output of 1 prints as \nI\n, and the false output of 0 prints as _\n\n.

I don't feel like writing out a full explanation at the moment, but the ungolfed code is here, I lifted the control flow from something I wrote earlier, and the gist of what it does is read two characters at a time from the input through the usually unhelpful "Turing Tape" I/O until either the first resulting number is 256 (in which case the entire input has been validated and has even length, so it is double-speak), or the second resulting number is not 0 (in which case the second character is different from the first or does not exist, and the input is not double-speak).

I think I might be able to restructure this to cut down on redundancy, but I'm not quite feeling up to that at the moment either.

Answer 49

C (gcc), 50 bytes

Returns 0 if double-speak (or empty string), truthy otherwise.

f(s,i)char*s;{for(i=*s;i&(i=*s++)&&i==*s++;);s=i;}

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If the input string is assumed to always be at least 2 characters (gives incorrect result for single-character strings):

C (gcc), 38 bytes

f(char*s){while(*s&&*s++==*s++);s=*s;}

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Answer 50

Java (JDK), 64 bytes

s->{int i=s.length;for(;i>1;)i=s[--i]==s[--i]?i:i|1;return i<1;}

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Explanations

s->{
 int i=s.length;
 for(;i>1;)                 // while the length is 2 or greater.
  i=s[--i]==s[--i]?i:i|1;   // check the 2 previous values. If they're identical, don't do anything. Else, make i odd so that it fails at the return.
 return i<1;                // i will be 0 only when the size is even and all characters were doubled.
}

Answer 51

Perl 6, 14 bytes

{!S:g/(.)$0//}

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Replaces all pairs of identical characters with nothing and then boolean NOTs the result to return true if is an empty string (i.e. all characters were next to an identical one), or false otherwise.

Answer 52

PHP, 51 50 38 bytes

echo preg_match("/^((.)\2)+$/",$argn);

Checks whether input matches pairs of chars where the chars in each pair are the same (using back ref).

Run like this:

echo 'dd00uubbllee' | php -nR 'echo preg_match("/^((.)\\2)+$/",$argn);';echo

• -1 byte thanks to @manatwork
• -12 bytes by using preg_match with back references.

Answer 53

Malbolge (unlimited memory access variant), Around 4 megabytes

You asked the golfers, but for the second time forgot about the bowlers.

This is too big to include in the answer for obvious reason so here is gist link.

You might want to use the fast interpreter to test this program, as it's hellishly slow (hellishly, get it?). I'm going to include the TIO.run link after Dennis (hopefully) takes on my issue on TIO tracker.

// Edit: Nope, no TIO link as the answer size limit is 65536 bytes, and no abusing url shorteners because they just refuse to shorten it

Answer 54

Pushy, 13 bytes

LL2%}2/:=};P#

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L              \ Push len(input)
 L2%           \ Take (len(input) + 1) % 2. This will be 1 only if the input had even length.
    }          \ Move this to the bottom of the stack for later.
     2/:  ;    \ Length of original input, divided by 2, times do:
        =      \    Pop top two characters, and check equality. (Pushes 0 or 1)
         }     \    Move this to the bottom of the stack for later.
           P   \ Push the stack's product. If the input had odd length, or any two characters
               \ did not match, then the stack will contain a 0, so the product will be 0.
               \ Else, it will be 1.
            #  \ Print the product.  

Answer 55

naz, 50 bytes

2a2x1v1x1f1r3x1v2e2x2v1r3x2v1e0m1o0x1x2f0m1a1o0x1f

Works for any input string, provided it's passed as a file terminated with the control character STX (U+0002).

Explanation (with 0x commands removed)

2a2x1v                       # Set variable 1 equal to 2
1x1f1r3x1v2e                 # Function 1
                             # Read a byte and jump to function 2 if it equals variable 1
            2x2v             # Otherwise, store it in variable 2
                1r3x2v1e     # Read another byte
                             # Jump back to the start of function 1 if it equals variable 2
                        0m1o # Otherwise, output 0
1x2f0m1a1o                   # Function 2
                             # Output 1
1f                           # Call function 1

Answer 56

Burlesque, 8 bytes

J2en)J==

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                           [AABBCC]         [ABC]
J   #Duplicate             [AABBCC,AABBCC]  [ABC,ABC]
2en #Every other character [AABBCC,ABC]     [ABC,AC]
)J  #Duplicated            [AABBCC,AABBCC]  [ABC,AACC]
==  #Is the same           [1]              [0]

Burlesque, 11 bytes

=[{L[2dv}al

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=[   # Group consecutive like values
{
 L[  # Length of (group)
 2dv # Divisible by 2
}al  # All

Answer 57

Befunge-98 (PyFunge), 11 bytes

~#@~# -_#q1

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Returns 1 if double speak, 0 if not.

10 byte solution which returns 0 for double speak and non-0 otherwise:

~#q~#1-_#q

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Answer 58

Python 3, 75 bytes

lambda s:all[([s[2*i]==s[2*i+1]for i in range(int(len(s)/2))]),0][len(s)%2]

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Answer 59

Kotlin, 59 57 bytes

{s->(0..s.length-1 step 2).fold(""){t,c->t+s[c]+s[c]}==s}

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-1 byte thanks to Khuldraeseth na'Barya

-1 byte changing it -> s

---

Slightly lazy attempt - basically just took my answer from the doublespeak question, applied it to every second char of the input and see if the result is the input.

Answer 60

Wolfram Language (Mathematica), 27 26 24 bytes

2==GCD@@Length/@Split@#&

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        Length/@Split@#&    (*for the lengths of runs of characters*)
2==GCD@@                    (*check that they are all even*)

Answer 61

Ruby -n, 21 14+1 = 15 bytes

p~/^((.)\2)*$/

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Answer 62

Lua, 90 87 86 82 bytes

a,e=1,os.exit;(...):gsub('.',load's=...a=a~=1 and(a==s and 1 or e(1))or s')e(a==1)

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Take input as argument, use exit code as output.

Explanation

This is based on abusing few Lua features:

• os.exit could accept either number, allowing short code for exit or boolean, making last check possible (true meaning success, false meaning failure).
• gsub can be also used to iterate over string in callback-style, not only for doing replacement.
• load is shorter than function(...) end.
• ... is often shorter than arg[1].

Whole idea hidden behind those tricks:

1. For each character: either remember if nothing is already it or perform check: if current is different from remembered, fail. Otherwise, reset remembered value and continue.
2. When done, make sure that there's nothing pending in the buffer (e(a==1)). This is required to fail on strings with odd length.

Answer 63

Gema, 17 characters

?$1=
?=0@end
\Z=1

Outputs 1 on double speak and 0 on not.

Sample run:

bash-5.0$ echo -n 'TThhiiss  iiss  ddoouubbllee  ssppeeaakk!!' | gema '?$1=;?=0@end;\Z=1'
1

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Gema, 12 characters

?$1=
?=@fail

Terminates with exit code 0 on double speak and 2 on not.

Sample run:

bash-5.0$ echo -n 'TThhiiss  iiss  ddoouubbllee  ssppeeaakk!!' | gema '?$1=;?=@fail'
bash-5.0$ echo $?
0

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Answer 64

C (tcc), 59 55 50 bytes

f(char *s){while(*s==*++s)s++;return!*--s&&s!="";}

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Thanks to Jo King for the 4 bytes and Unrelated String for saving 5 bytes

Can it be even shorter?

Answer 65

Kotlin, 4649 bytes

-3 bytes thanks to @cubic lettuce

{it.chunked(2).all{it.length>1&&it[0]==it[1]}}

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{it.length%2==0&&it.chunked(2).all{it[0]==it[1]}}

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Answer 66

T-SQL 2012, 76 bytes

Looping using the following logic.

Finding the first match on the first character in the rest of the string.

If the match is on the second position, the first 2 characters are removed.

if the match is on a later position, character 0 to 1 is removed, resulting in null string.

If there are no matches, characters 3-4 are removed. Repeating this will eventually result in a null string.

The final string will be an empty string or a null string.

Empty string is "double speak"(1). Null string is not "double speak"(0).

DECLARE @ varchar(max)='ccbbddeeff'

WHILE @>''SET
@=stuff(@,-3/~charindex(left(@,1),@,2),2,'')PRINT
iif(@=0,1,0)

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Answer 67

Rust, 73 bytes

|s:&str|!s.chars().enumerate().any(|(i,c)|s.chars().nth(i|1).unwrap()!=c)

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Answer 68

C++ (gcc), 48 47 45 43 bytes

-2 bytes thanks to @ceilingcat
-2 bytes thanks to @JoKing

recursive function assuming zero terminated char table

int f(char*s){return!*s||*s==s[1]&&f(s+2);}

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Answer 69

EDIT:

This solution is totally broken as demonstrated by the aabbbb test case. Thanks to @manatwork for pointing this out. Do not pass Go! Do not collect 200 dollars!

Bash and Gnu utils, 46 45 bytes

fold -1|uniq -c|grep -qv "2 "&&echo 0||echo 1

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-1 byte by skipping the first space in grep expression

Answer 70

JavaScript (Node.js), 26 bytes

s=>!s.replace(/(.)\1/g,'')

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Answer 71

Rust, 61 58 39 bytes

|s:&[u8]|s.chunks(2).all(|n|n[0]==n[1])

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This is a closure that takes the input as a byte slice (b prefix on a string), e.g.

println!("{}", f(b"aabb"));

Answer 72

C++ (gcc), 156 145 bytes

#include <iostream>
int f(){
    char a[2]; while(std::cin.get(a, 3)) if(a[0] != a[1]) return 0; return 1;
}
int main()
{
    std::cout << f();
}

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Answer 73

Oracle SQL, 47 bytes

select*from t where regexp_like(x,'^((.)\2)+$')

It works with an assumption that input data is stored in a table t(x), e.g.

with t(x) as (select 'hheelllloo' from dual)

Returns either original string or no rows.

Answer 74

Swift, 109 bytes

func a(b:String){let c=Array(b),l=c.count;var d=0,i=0;while i<l-1{if c[i]==c[i+1]{d+=1};i+=2;};print(d*2==l)}

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Answer 75

C#, 190 bytes

public class P{public static void Main(string[]a){bool d=true;if(a[0].Length%(double)2>0)d=false;else for(int i=0;i<a[0].Length;i+=2){d=a[0][i]==a[0][i+1]?d:false;}System.Console.Write(d);}}

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Answer 76

Clojure, 48 bytes

(fn[x](every? #(apply = %)(partition 2 2[\0]x)))

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Answer 77

GolfScript, 7 bytes

2/zip~=

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Explanation

2/      # Split input into parts of 2,            e.g. tt,ee,ss,tt
  zip   # Read the list horizontally,         yielding test, test
     ~  # Put the two strings onto the stack, yielding test,test
      = # Check whether they are equal,       yielding 1

# False case:
#  tteesst
#->tt,ee,ss,t
#->test,tes
#->0
```