Talk interpreter

Question

"Talk" is a baroquified accumulator-based language that is created in response to Dennis's quote on talk.tryitonline.net.

Waiting for someone to create an esolang called talk. 

. The "Talk" language has 4 commands:

• 00 If the accumulator is 0, set the accumulator to 0.
• 01 If the accumulator is 0, set the accumulator to 1.
• 10 If the accumulator is 1, set the accumulator to 0.
• 11 If the accumulator is 1, set the accumulator to 1.

Input:

• The input can be taken via any acceptable input method by our standard I/O rules.

• There are two inputs, the initial accumulator value and the program. You can merge these two inputs into one input or split your input into valid commands (e.g. taking them as a list; e.g. [00, 01, 00]) if you like.

Output:

• On the end of a command execution, the accumulator is outputted implicitly.

Rules:

• The input can be a single string or character list.
• As this is code-golf, the shortest answer, in bytes, wins.
• We take digits or strings/characters.

  Test cases:

0 0001111101 -> 1
0 000100 -> 1
0 11001000 -> 0

Leaderboards

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# [><>](http://esolangs.org/wiki/Fish), 121 bytes

var QUESTION_ID=190819;
var OVERRIDE_USER=8478;
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body{text-align:left!important}#answer-list{padding:10px;float:left}#language-list{padding:10px;float:left}table thead{font-weight:700}table td{padding:5px}

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="https://cdn.sstatic.net/Sites/codegolf/primary.css?v=f52df912b654"> <div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td><a href="{{LINK}}">{{SIZE}}</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td><a href="{{LINK}}">{{SIZE}}</a></td></tr></tbody> </table> 

• xnor: So there's two inputs, the sequence of commands and the initial accumulator value?
• Jo King: Some test cases that don't change the accumulator, start with 1 in the accumulator or have no instructions would be good
• Luis Mendo: Can Talk really be considered a programming language?
• Luis Mendo: @A_ That comment was probably meant as a joke. Back in the day, a language called Lang would have a URL lang.tryitonline.net (now it's tio.run/#lang). So a language called Talk would cause confusion with the URL for the then-recently created chatroom, which is talk.tryitonline.net
• GammaFunction: In the future, please refrain from changing I/O after a number of answers have been posted. I came back today and having mapped inputs allowed makes this an entirely different challenge from what I answered.

Answer 1

Jelly, 3 bytes

y@/

Input is a single list: the accumulator, followed by the pairs.

Try it online!

How it works

The y atom performs transliteration; [a,b]yc replaces a with b, so it returns b if a=c and c if a≠c.

y@/ folds/reduces the input by y with swapped arguments, performing one transliteration per pair.

• user85052: This is the only Jelly answer that I have seen so far that only uses ASCII characters.
• Dennis: There were a few. Look ma, no Unicode!

Answer 2

Python 3, 43 bytes

lambda s:re.sub("00|11","",s)[-1]
import re

Try it online!

The function takes a single string as input, where the first character is the initial state and the rest of the string represents the commands. This solution can be easily ported to other languages that have better support for regular expressions.

The difficult part is to prove the solution yields the correct outcome. To see this, we need a deep analysis of the commands. Firstly, we can see the commands have the following properties:

• Property (1): commands 00 and 11 retain the accumulator state.
• Property (2): commands 01 and 10 make the accumulator state the same as the second bit regardless of its original state.

Therefore, the final accumulator state is:

• Case 1: If no 01 or 10 command exists, the final state is the same as the initial state.
• Case 2: Otherwise, the last bit of the last 10 or 01 command.

Next we will show the solution yields the correct outcome in both cases. We will prove the statement for the final state 0 and the final state of 1 can be proved analogously. If the final state is 0 the input is in either of the following forms:

• ^0{2k+1}11(11|00)*

  For Case 1, the input string s must start with 2k+1 0s, followed by 11 and 00 commands. Eliminating 00s and 11s yields a single 0, which is the final state.

• .+10{2k+1}11(11|00)*

  For Case 2, the input string ends with a 10 command, followed by zero or more 00 and 11 s. This pattern is equivalent to a 1 followed by 2k+1 0s, and then zero or more 11s and 00s. Eliminating 00s and 11s leaves behind the last one of the 2k+1 0s at the end of the string, which represents the final state.

Based on all the above, after eliminating 00s and 11s simultaneously in one single pass (01001 is a counter-example if 00 is eliminated in one pass and then 11 in another pass) from the input s, the last character is the final state. Hence the correctness of the solution is proved.

• GammaFunction: Welcome to PPCG! Excellent answer, and a nice formal proof to go with it!
• Joel: Thanks. I feel that people may doubt such a simple solution yields the correct outcome at the first glance. So there is a need to provide a proof for that.

Answer 3

Perl 6, 17 bytes

{m/.)>[(.)0ドル]*$/}

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Takes advantage of "You can merge these two inputs into one input if you like" by taking input as the accumulator value concatenated with the commands e.g. 1,[00,11] is 10011. If this isn't okay, than it's only 5 extra bytes to take it as f(accumulator, commands). Returns a match object that can be coerced to a string.

Explanation:

{ } # Anonymous code block
 m/ / # Find the first match from the input
 .)> # Capture a number
 [ ]* # Followed by any number of
 (.)0ドル # Pairs of identical characters
 $ # Ending the string

Basically this works because the 00 and 11 commands do literally nothing, while the 01 and 10 commands just set the accumulator to the second digit of the command. If there are no commands, then it takes the initial value of the accumulator instead.

Answer 4

Zsh, 33 bytes

The character list is passed as arguments, the initial value of the accumulator is passed as stdin.

read a
for x y;a=$[x^a?a:y]
<<<$a

Try it online!

---

39 bytes: If the commands must be a single string

Input is accumulator commands as arguments.

for x y (${(s::)2})1=$[x^1ドル?1ドル:y]
<<<1ドル

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---

For fun, here's a 50 byte recursive one-liner (TIO):

<<<${${2+`f $[1ドル^${2[1]}?1ドル:${2[2]}] ${2:2}`}:-1ドル}

Answer 5

Python 3, 52 bytes

f=lambda a,s:s and f([s[1],a][s[0]==s[1]],s[2:])or a

Try it online!

Fixed inconsistent return type thanks to Chas Brown

Takes input as two strings; the accumulator and the code.

• TwilightSparkle: Oh no, that was fast.
• Chas Brown: Nice but it has this potential problem - f(1,'11')==f(1,'01') is False; it sometimes returns an int and sometimes a str. So maybe specify it takes acc input as a string?
• negative seven: @ChasBrown Good call, that's much simpler than what I was thinking about.
• Jitse: Nice, but since your function is recursive, it cannot be anonymous. It should be 52 bytes.

Answer 6

Brachylog, (削除) 11 (削除ここまで) 9 bytes

tġ2≠stt|h

Try it online!

Since it's been long enough that I've been able to forget the notion of printing the accumulator after each command, I've formulated a significantly less naïve solution with some inspiration from Jo King's Perl answer.

 | The output is
 tt the last element of the last element of
t the last element of the input
 ġ2 split into length-2 slices
 ≠s with equal pairs removed.
 | If there is no such element, the input
 h 's first element is the output.

Old solution:

Brachylog, (削除) 18 (削除ここまで) 16 bytes

ġ2tc{th~h?tt|h}l

Try it online!

-2 bytes from changing the input format.

Answer 7

JavaScript (ES6), 27 bytes

Takes input as (a)(code), where code is a is list of 2-bit integers.

a=>c=>c.map(x=>a^=x==a+1)|a

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---

JavaScript (ES6), (削除) 47 (削除ここまで) 40 bytes

Takes input as (a)(code), where code is a string.

a=>c=>c.replace(/../g,x=>a^=x%4==a+1)&&a

Try it online!

How?

All possible cases are summarized below. The only two cases where we need to toggle the accumulator are \$(a=0,x=01_2)\$ and \$(a=1,x=10_2)\$.

 a | x (bin) | int(x) % 4 | a + 1 | equal?
----+---------+------------+-------+--------
 0 | "00" | 0 % 4 = 0 | 1 | N
 1 | "00" | 0 % 4 = 0 | 2 | N
 0 | "01" | 1 % 4 = 1 | 1 | Y
 1 | "01" | 1 % 4 = 1 | 2 | N
 0 | "10" | 10 % 4 = 2 | 1 | N
 1 | "10" | 10 % 4 = 2 | 2 | Y
 0 | "11" | 11 % 4 = 3 | 1 | N
 1 | "11" | 11 % 4 = 3 | 2 | N

Answer 8

sed -E, (削除) 26 (削除ここまで) 19 bytes

A whopping -7 bytes from @Cowsquack by realizing removing all pairs works as well.

s/(.)1円//g
s/.*\B//

Takes input concatenated together on stdin. Inspired by Jo King's Perl answer. (削除) Strip trailing pairs (削除ここまで) Remove all pairs, then get last digit.

(削除) Try it online! (削除ここまで) Try it online!

• user41805: The last line can be simply s/.*\B//, but anyways changing the approach slightly gives an even shorter 19 bytes Try it online!
• GammaFunction: Huh, I didn't think that s/(.)1円//g would work, since it could remove the end of one pair and the start of the next, but it still works out. Excellent!
• Joel: @GammaFunction s/(.)1円//g is equivalent to s/00|11//g as shown in my solution.

Answer 9

Retina 0.8.2, (削除) 18 (削除ここまで) 11 bytes

(.)1円
!`.$

Try it online! Link includes test cases. Takes input concatenated. Saved 6 bytes thanks to @CowsQuack for pointing out that removing all doubled characters and then taking the last remaining character works, although in fact the port of @JoKing's original answer could have been golfed by 3 bytes even without that trick.

• user41805: 12 bytes Try it online!
• Neil: @Cowsquack D'oh, once you think about separating the stages, that's a 2-byte saving already, then another byte saved because you can use !`.$, and then another 4 bytes because you don't have to limit to trailing pairs...

Answer 10

Haskell, 29 bytes

Defines an unnamed function on the first line with type (Foldable t, Eq b) => b -> t [b] -> b. For the purposes of this code golf, we can instantiate it as Char -> [String] -> Char where the first argument is the accumulator and the second is a list of string with each string being a single command.

foldl(#)
a#[x,y]|a==x=y|1>0=a

Try it online!

• cole: Same bytecount if you define it using prefix notation. I can't believe I wrote almost the exact same answer at the same time, even including the type signature explanation...

Answer 11

Python 3, 38 bytes

lambda l:[y for*x,y in l if[y]!=x][-1]

Try it online!

Based on Joel's solution. Takes input as a list of the initial accumulator value (length-one string) followed by the commands (length-two strings). Finds the last command with two unequal values, and outputs its second character.

To make this fall through to the initial accumulator value when there are no such commands, we make it so that the single-char initial value string passes the test. We do so by checking if a singleton list with the last character is unequal to a list of all preceding characters, which is passed by any length-one string or length-two string with two different characters.

Answer 12

Perl 5 -p, (削除) 37 (削除ここまで) 33 bytes

$\=<>;s/(.)(.)/$\=2ドルif$\==1ドル/ge}{

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Input is two lines: first line is the command sequence, second is the accumulator.

Answer 13

Python 2, 56 bytes

f=lambda a,c:f([a,1,0,a][int(c[:2],2)],c[2:])if c else a

Try it online!

Answer 14

Jelly, (削除) 8 (削除ここまで) 6 bytes

EÐḟṪṪo

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-2 bytes thanks to Nick Kennedy informing me of a rules change. (His proposed golf, EÐḟFȯṪ, seems somewhat more clever but has the same length as my previous solution minus s2.) The input format now takes the commands as a list of two-character strings, but the testing footer translates from the old format for convenience.

Translated from my newer Brachylog solution.

Old version:

Jelly, 13 bytes

ḢẎ=9a8o
s2ç@ƒ

Try it online!

I'm not 100% sure this is correct, but it succeeds on all three test cases. Takes the commands as the left argument and the initial accumulator as the right argument.

• Nick Kennedy: The input is allowed to be split into a list, so you could have EÐḟFȯṪ with the input as e.g. [[0,0],[0,1],[1,1],[1,1],[0,1]].
• Unrelated String: Wow, the changes to the spec really were quite large...

Answer 15

APL (Dyalog Unicode), 16 bytes

(+⌷13⍴0 1,9/⊢)/⌽

A tacit function which takes initial accumulator value and the program as a single integer list.

Maps the relevant sums of instruction and accumulator to an array.

Table: (a→accumulator, i→instruction)

a i a+i result
00 0 0 0
01 0 1 1
10 1 11 0
11 1 12 1

All other cases return the same value, so they are assigned to the preexisting value of a.

Try it online!

Explanation

(+⌷13⍴0 1,9/⊢)/⌽ accumulator→a
 /⌽ reduce the reversed array using th following function: (reducing happens from the right)
 9/⊢) replicate a 9 times
 13⍴0 1, concatenate with (0,1) and expand to 13 elements → (0 1 a a a a a a a a a 0 1)
(+⌷ sum first two elements and find element at that index in the array

Solution which accepts a string

{{(13⍴0 1,9/⍺)[⍺+⍵]}/⌽(⍎ ̈((⍴⍵)⍴1 0)⊂('0',⍵))}

It was pretty ambiguous how the input was supposed to be taken, so I decided I'd leave this one in as well.

This was the original dfn, which was golfed down using Adám and Bubbler's advice.

Answer 16

Regex (JS flavor), 32 bytes

^(0(..)*01|1((..)*01)?)(00|11)*$

Try it online!

Matches if output is 1, doesn't if output is 0. Global and multiline flags enabled in link to verify multiple test cases at once, but not necessary for solution. Did I format this submission correctly?

Answer 17

Python, 111 bytes

def f(a,b):
 c=a
 for i in range(0,len(b)-1,2):
 c=(not b[i])*(c or b[i] or b[i+1]) or c*b[i]*b[i+1]
 return c

Ungolfed. EDIT: AHHH Someone beat me to it!

Answer 18

Haskell, 36 bytes

f(x:y:s)=f s.last.(:[y|x/=y])
f _=id

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Takes input as f(string)(char) where the character is the accumulator and the string is the list of commands.

Answer 19

05AB1E, 3 bytes

ø`:

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Zip, dump on the stack, replace.

Answer 20

Stax, 3 bytes

F|t

Run and debug it

For each instruction, perform character translation.

Answer 21

Keg, -ir, 16 bytes

"(!;1⁄2|':"=['_"|_

Try it online!

Explained:

1. Takes the implicit input and right shifts the accumulators value to the bottom

2. Repeat the following (length of stack - 1 divided by 2) times

2.1. Shift the accumulator back to the top

2.2. Compare for equality with the first part of the command

2.2.1. If true, replace the accumulator, otherwise pop the replacement

Input is taken as the initial acc value concatenated with the source. E.g.

010011000

• First char is acc value
• Rest is program

Answer 22

CJam, 8 bytes

{{)er}/}

Try it online!

Takes input on the stack in the form of "0" ["00" "01" "11" "11" "01"]. Does string replacement using every command.

Answer 23

Headass, 22 bytes

U[{N-()]PNE:U(])U[:U;}

Try It Online!

U[{N-()]PNE:U(])U[:U;} full program
U[ save initial state of accumulator to storage register
 { : } while
 N-() input remains
 U(]) ; if input == accumulator state
 U[ set accumulator to next input
 : else
 U discard next input
 ]P print final state of accumulator
 NE go to code block 1
 block does not exist, exits with error

There is room for golfing maybe if I can avoid storing anything in the comparison register (i.e. having it remain 0 for the entirety of the program and never using (), but it might not be possible.

Answer 24

Regex (any flavor), 15 bytes

(^|0)1(00|11)*$

Try it on regex101!

It was pointed out before that the problem can be simplified:

• 00 and 11 are no-ops. 01 unconditionally sets the value to 1 while 10 sets to 0.
• The output is 1 if and only if the input is 1 and the entire code is no-op or the last non-no-op command is 01.

The regex precisely does the job: (00|11)*$ after discarding no-ops at the end, the last thing found is (^|0)1 the input 1 or a 01 command.

1(00|11)*$ does not work because the 1 may be part of another no-op command 11.

Answer 25

Bash, (削除) 58 (削除ここまで) 40 bytes

Add one byte for a full program: change f to 0ドル.

((1ドル=2ドル-a?a:3,1ドル))&&f 1ドル ${@:4}||echo 1ドル

(削除) 58 bytes (削除ここまで) Try it online!

The ternary will return false when 1ドル is set to 0, but the ,1 at the end ensures the whole ((expression)) will return true, except a syntax error.

When all the arguments are consumed, a syntax error happens and the recursion ends.

---

Answer 26

Charcoal, 16 bytes

F⪪η2F=θ§ι0≔§ι1θθ

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

F⪪η2

Split the instructions into pairs of digits and loop over them.

F=θ§ι0

If the accumulator is equal to the first digit...

≔§ι1θ

... then assign the second digit to it.

θ

Print the accumulator at the end of the loop.

Answer 27

MATL, (削除) 13 (削除ここまで) 12 bytes

!dh2Ol4$Ys0)

Takes the input as a 2-column matrix where each row is a command, and a number

Try it online! Or verify all test cases

Answer 28

Jelly, 7 bytes

fØ1⁄2Ḃ9;Ṫ

A dyadic Link accepting the program as a list of integers on the left and the initial accumulator on the right which yields an integer.

Try it online! Or see a test-suite

• user85052: I am removing input-mapping because @GammaFunction recommended me to do this.
• Jonathan Allan: @A_ ah OK cant delete on mobile so will have to address this later
• Jonathan Allan: @A_ fixed up to work with 0,1,2,3 the unmapped version of the instructions as digits is that OK?
• user85052: Yes, that's OK.

Answer 29

PHP, 38 bytes

<?=strtr($argn,['00'=>'',11=>''])[-1];

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Basically port of Jo King's idea.

Answer 30

Runic Enchantments, 28 bytes

/~@/i~/i<
/=?/~iR:l}i{l1-=?!

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Takes input as a series of space separated bytes (Runic does not understand lists). The first byte is the initial state and every other byte is the program. No validation is performed (i.e. it assumes only valid programs are given as input and it doesn't care what value is used to represent 0 and 1).