23
\$\begingroup\$

The Binary Sierpinski Triangle sequence is the sequence of numbers whose binary representations give the rows of the Binary Sierpinski Triangle, which is given by starting with a 1 in an infinite row of zeroes, then repeatedly replacing every pair of bits with the xor of those bits, like so:

f(0)=      1                    =1
f(1)=     1 1                   =3
f(2)=    1 0 1                  =5
f(3)=   1 1 1 1                 =15
f(4)=  1 0 0 0 1                =17

More digits are given at OEIS: https://oeis.org/A001317

Input: A non-negative integer n in any format you like. (Must work for all n up to 30.)

Output: The nth term (0-indexed) of the sequence as a decimal number.

This is so try give the shortest answer in bytes of which your language is capable. No answer will be accepted. Standard loopholes apply (e.g. no hard-coding the sequence), except that you may use a language created/modified after this challenge was posted. (Do avoid posting another solution in a language that has already been used unless your solution is shorter.)

Leaderboard

The Stack Snippet at the bottom of this post generates the catalogue from the answers a) as a list of shortest solution per language and b) as an overall leaderboard.

To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:

## Language Name, N bytes

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

## Ruby, <s>104</s> <s>101</s> 96 bytes

If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:

## Perl, 43 + 2 (-p flag) = 45 bytes

You can also make the language name a link which will then show up in the snippet:

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

/* Configuration */

var QUESTION_ID = 67497; // Obtain this from the url
// It will be like http://XYZ.stackexchange.com/questions/QUESTION_ID/... on any question page
var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";
var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk";
var OVERRIDE_USER = 47050; // This should be the user ID of the challenge author.

/* App */

var answers = [], answers_hash, answer_ids, answer_page = 1, more_answers = true, comment_page;

function answersUrl(index) {
  return "http://api.stackexchange.com/2.2/questions/" +  QUESTION_ID + "/answers?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + ANSWER_FILTER;
}

function commentUrl(index, answers) {
  return "http://api.stackexchange.com/2.2/answers/" + answers.join(';') + "/comments?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + COMMENT_FILTER;
}

function getAnswers() {
  jQuery.ajax({
    url: answersUrl(answer_page++),
    method: "get",
    dataType: "jsonp",
    crossDomain: true,
    success: function (data) {
      answers.push.apply(answers, data.items);
      answers_hash = [];
      answer_ids = [];
      data.items.forEach(function(a) {
        a.comments = [];
        var id = +a.share_link.match(/\d+/);
        answer_ids.push(id);
        answers_hash[id] = a;
      });
      if (!data.has_more) more_answers = false;
      comment_page = 1;
      getComments();
    }
  });
}

function getComments() {
  jQuery.ajax({
    url: commentUrl(comment_page++, answer_ids),
    method: "get",
    dataType: "jsonp",
    crossDomain: true,
    success: function (data) {
      data.items.forEach(function(c) {
        if (c.owner.user_id === OVERRIDE_USER)
          answers_hash[c.post_id].comments.push(c);
      });
      if (data.has_more) getComments();
      else if (more_answers) getAnswers();
      else process();
    }
  });  
}

getAnswers();

var SCORE_REG = /<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/;

var OVERRIDE_REG = /^Override\s*header:\s*/i;

function getAuthorName(a) {
  return a.owner.display_name;
}

function process() {
  var valid = [];
  
  answers.forEach(function(a) {
    var body = a.body;
    a.comments.forEach(function(c) {
      if(OVERRIDE_REG.test(c.body))
        body = '<h1>' + c.body.replace(OVERRIDE_REG, '') + '</h1>';
    });
    
    var match = body.match(SCORE_REG);
    if (match)
      valid.push({
        user: getAuthorName(a),
        size: +match[2],
        language: match[1],
        link: a.share_link,
      });
    else console.log(body);
  });
  
  valid.sort(function (a, b) {
    var aB = a.size,
        bB = b.size;
    return aB - bB
  });

  var languages = {};
  var place = 1;
  var lastSize = null;
  var lastPlace = 1;
  valid.forEach(function (a) {
    if (a.size != lastSize)
      lastPlace = place;
    lastSize = a.size;
    ++place;
    
    var answer = jQuery("#answer-template").html();
    answer = answer.replace("{{PLACE}}", lastPlace + ".")
                   .replace("{{NAME}}", a.user)
                   .replace("{{LANGUAGE}}", a.language)
                   .replace("{{SIZE}}", a.size)
                   .replace("{{LINK}}", a.link);
    answer = jQuery(answer);
    jQuery("#answers").append(answer);

    var lang = a.language;
    lang = jQuery('<a>'+lang+'</a>').text();
    
    languages[lang] = languages[lang] || {lang: a.language, lang_raw: lang, user: a.user, size: a.size, link: a.link};
  });

  var langs = [];
  for (var lang in languages)
    if (languages.hasOwnProperty(lang))
      langs.push(languages[lang]);

  langs.sort(function (a, b) {
    if (a.lang_raw.toLowerCase() > b.lang_raw.toLowerCase()) return 1;
    if (a.lang_raw.toLowerCase() < b.lang_raw.toLowerCase()) return -1;
    return 0;
  });

  for (var i = 0; i < langs.length; ++i)
  {
    var language = jQuery("#language-template").html();
    var lang = langs[i];
    language = language.replace("{{LANGUAGE}}", lang.lang)
                       .replace("{{NAME}}", lang.user)
                       .replace("{{SIZE}}", lang.size)
                       .replace("{{LINK}}", lang.link);
    language = jQuery(language);
    jQuery("#languages").append(language);
  }

}
body { text-align: left !important}

#answer-list {
  padding: 10px;
  width: 290px;
  float: left;
}

#language-list {
  padding: 10px;
  width: 290px;
  float: left;
}

table thead {
  font-weight: bold;
}

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="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b">
<div id="language-list">
  <h2>Shortest Solution 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>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr>
  </tbody>
</table>
<table style="display: none">
  <tbody id="language-template">
    <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr>
  </tbody>
</table>

\$\endgroup\$
  • 8
    \$\begingroup\$ I'm not a big fan of must not output a wrong answer for any n. This basically forces languages that do not use arbitrary precision integers by default to check if the input is small enough... \$\endgroup\$ – Dennis Dec 23 '15 at 15:46
  • \$\begingroup\$ Please clarify if understood the rules correctly (see comments here and here) and if rounded output (e.g., 1.288490189e10 for input 33) counts as wrong. \$\endgroup\$ – Dennis Dec 23 '15 at 19:25
  • \$\begingroup\$ "Must work for all n up to 30, and must not output a wrong answer for any n.". This is self-contradictory - surely "must not output a wrong answer" is the same as "Must work" ??? \$\endgroup\$ – Digital Trauma Dec 23 '15 at 19:49
  • 5
    \$\begingroup\$ Due to overwhelming popular opposition to the unreasonable and soul-crushing burden of input validation, this requirement has been removed. You may output whatever garbage you want for large n. Enjoy! \$\endgroup\$ – quintopia Dec 23 '15 at 20:13
  • 2
    \$\begingroup\$ Rather than saying that the output shouldn't be wrong, I would recommend just saying that submissions have to support input up to the largest n for which nothing overflows. \$\endgroup\$ – Alex A. Dec 23 '15 at 20:17

25 Answers 25

14
\$\begingroup\$

05AB1E, 5 4 bytes

I proudly present to you, 05AB1E. Although it is very short, it is probably very bad at long challenges.

Thanks to ETHproductions for shaving off 1 byte :)

$Fx^

Explanation:

$      # Pushes 1 and input
 F     # Pops x, creates a for-loop in range(0, x)
  x    # Pops x, pushes x and 2x
   ^   # Bitwise XOR on the last two elements
       # Implicit, ends the for-loop
       # Implicit, nothing has printed so the last element is printed automatically
\$\endgroup\$
  • \$\begingroup\$ You know, a good way to golf a byte off of many programs in a custom language is to make a trailing } be automatically inserted. Then this would be 4 bytes. :) \$\endgroup\$ – ETHproductions Dec 23 '15 at 15:54
  • 1
    \$\begingroup\$ @ETHproductions Wait a minute, that already has been implemented :). Thanks for shaving off 1 byte haha. \$\endgroup\$ – Adnan Dec 23 '15 at 15:58
  • 2
    \$\begingroup\$ There's a bug in this code. How do I know? It's beating Dennis. \$\endgroup\$ – Arcturus Dec 23 '15 at 20:43
  • 2
    \$\begingroup\$ @Ampora Not only is it beating Dennis, it's beating Dennis's custom golfing language. ;) \$\endgroup\$ – ETHproductions Dec 23 '15 at 20:54
  • \$\begingroup\$ @Adnan Wow. You're on to something. \$\endgroup\$ – RK. Dec 24 '15 at 14:55
12
\$\begingroup\$

Pari/GP, 27 bytes

n->lift(Mod(x+1,2)^n)%(x-2)

The function takes the n-th power of the polynomial x+1 in the ring F2[x], lifts it to Z[x], and then evaluates it at 2.

Try it online!

\$\endgroup\$
6
\$\begingroup\$

Jelly, 6 bytes

1Ḥ^$³¡

Try it online!

The binary version that works with this revision of the Jelly interpreter has the xxd dump

0000000: 31 a8 5e 24 8b 80  1.^$..

How it works

1Ḥ^$³¡    Input: n

1         Set the left argument to 1.
 Ḥ        Multiple the left argument by two.
  ^       Hook; XOR it with its initial value.
   $      Create a monadic chain from the last two insructions.
    ³¡    Call the chain n times, updating the left argument after each call.
\$\endgroup\$
5
\$\begingroup\$

Haskell, 44 bytes

import Data.Bits
f n=iterate((2*)>>=xor)1!!n

In the ((->) r) monad, (f >>= g) x equals g (f x) x.

\$\endgroup\$
  • \$\begingroup\$ I think you can anonymize the last line to (iterate((2*)>>=xor)1!!) \$\endgroup\$ – xnor Dec 23 '15 at 19:38
  • \$\begingroup\$ I tried that, but it doesn't work, for dreaded monomorphism restriction reasons. \$\endgroup\$ – Lynn Dec 24 '15 at 0:33
  • \$\begingroup\$ However, that might apply as a legal expression, since the monomorphism restriction doesn't apply on expressions, but declarations. And expressions are considered legal answers, if I'm not mistaken. \$\endgroup\$ – proud haskeller May 21 '16 at 20:33
4
\$\begingroup\$

Matlab, 45 bytes

Solution:

@(i)2.^[0:i]*diag(mod(fliplr(pascal(i+1)),2))

Test:

ans(10)
ans =
1285

Explanation: pascal constructs Pascal triangle, but it starts from 1, so input should be i+1. fliplr flips array from left to right. mod(_,2) takes remainder after division by 2. diag extracts main diagonal.Multiplication using 2.^[0:i] converts vector to decimal

I'm glad, @flawr that I found Matlab competitor here :)

\$\endgroup\$
  • \$\begingroup\$ Seems to work with Octave as well. \$\endgroup\$ – Dennis Dec 23 '15 at 20:14
4
\$\begingroup\$

JavaScript (ES6), 23 bytes

f=x=>x?(y=f(x-1))^y*2:1

Based on the first formula on the OEIS page. If you don't mind the code taking almost forever to finish for an input of 30, we can shave off a byte:

f=x=>x?f(--x)^f(x)*2:1

Here's a non-recursive version, using a for loop in an eval: (32 bytes)

x=>eval("for(a=1;x--;a^=a*2);a")
\$\endgroup\$
  • \$\begingroup\$ The rules, as currently written, invalidate this answer, since f(35) returns 15. Also, fork bomb alert: I had to force-close Chromium to stop f(30) (original revision) from running. :P \$\endgroup\$ – Dennis Dec 23 '15 at 15:59
  • 1
    \$\begingroup\$ @Dennis Wait, so if I can't output any wrong value, what am I supposed to do with inputs higher than 30? \$\endgroup\$ – ETHproductions Dec 23 '15 at 16:04
  • \$\begingroup\$ I'm not sure (and I hope the rule gets changed), but something like f=x=>x?(y=f(x-(x<31)))^y*2:1 would work. \$\endgroup\$ – Dennis Dec 23 '15 at 16:09
  • \$\begingroup\$ @Dennis Ah, recurse infinitely = no output. I'll fix this when I get back to my computer. I hope that rule gets changed as well. \$\endgroup\$ – ETHproductions Dec 23 '15 at 16:11
  • \$\begingroup\$ The rule has been removed from the question. \$\endgroup\$ – Dennis Dec 23 '15 at 20:13
3
\$\begingroup\$

Matlab, 77 70 bytes

This function calculates the n-th row of the Pascal triangle via repeated convolution with [1,1] (a.k.a. binomial expansion or repeated multiplication with a binomial), and calculates the number from that.

function r=c(n);k=1;for i=1:n;k=conv(k,[1,1]);end;r=2.^(0:n)*mod(k,2)'
\$\endgroup\$
3
\$\begingroup\$

Ruby, 26 bytes

anonymous function with iteration.

->n{a=1;n.times{a^=a*2};a}

this recursive function is one byte shorter, but as it needs to be named to be able to refer to itself, it ends up one byte longer.

f=->n{n<1?1:(s=f[n-1])^s*2}
\$\endgroup\$
3
\$\begingroup\$

Ruby, 25 bytes

->n{eval"n^=2*"*n+?1*n=1}

Shorter than the other answers on here so far. It constructs this string:

n^=2*n^=2*n^=2*n^=2*1

Then it sets n=1 (this actually happens while the string is being made) and evaluates the above string, returning the result.

\$\endgroup\$
  • \$\begingroup\$ Does that *n=1 actually save anything over m=1;"m^=2*"*n+?1 \$\endgroup\$ – Martin Ender Dec 25 '15 at 8:13
  • \$\begingroup\$ No, but doing it with just one variable is very showy :) \$\endgroup\$ – Lynn Dec 25 '15 at 8:14
3
\$\begingroup\$

Samau, 4 bytes

Now Samau has built-in functions for XOR multiplication and XOR power.

▌3$ⁿ

Hex dump (Samau uses CP737 encoding):

dd 33 24 fc

Explanation:

▌       read a number
 3      push 3
  $     swap
   ⁿ    take the XOR power
\$\endgroup\$
  • \$\begingroup\$ Could this be reduced to 3 bytes by swapping the first two commands and eliminating the swap? \$\endgroup\$ – quintopia Jan 22 '16 at 20:33
  • \$\begingroup\$ @quintopia No. Samau automatically pushes the input onto the stack as a string , and reads a number from the string. If we swap the first two commands, it would try to read a number from 3, which is not a string. \$\endgroup\$ – alephalpha Jan 23 '16 at 4:48
  • \$\begingroup\$ why doesn't samau try to eval the string when possible? \$\endgroup\$ – quintopia Jan 23 '16 at 5:06
2
\$\begingroup\$

CJam, 10 bytes

1ri{_2*^}*

Test it here.

Simple iterative solution using bitwise XOR.

\$\endgroup\$
2
\$\begingroup\$

Pure Bash (no external utilities), 37

Bash integers are 64-bit signed, so this works for inputs up to and including 62:

for((x=1;i++<$1;x^=x*2)){
:
}
echo $x
\$\endgroup\$
2
\$\begingroup\$

Python 2.7.6, 38 33 bytes

Thanks to Dennis for shaving off a few bytes!

x=1
exec'x^=x*2;'*input()
print x
\$\endgroup\$
  • 1
    \$\begingroup\$ Welcome to Programming Puzzles & Code Golf! exec'x^=x*2;'*input() saves a few bytes over the for approach. \$\endgroup\$ – Dennis Dec 23 '15 at 18:20
  • \$\begingroup\$ This beats my Python entry, which I'll just leave here for posterity: f=lambda n:f(n-1)^2*f(n-1)if n>0 else 1 \$\endgroup\$ – Jack Brounstein Dec 23 '15 at 22:33
2
\$\begingroup\$

Pyth, 7 bytes

uxyGGQ1

Try it online: Demonstration

Explanation:

u    Q1   apply the following statement input-times to G=1:
 xyGG        update G with (2*G xor G)
\$\endgroup\$
2
\$\begingroup\$

Perl 5, 23 bytes

$.^=2*$.for 1..<>;say$.

Try it online!

\$\endgroup\$
2
\$\begingroup\$

MIPS, 28 bytes

Input in $a0, output in $v0.

0x00400004  0x24020001          li      $v0, 1
0x00400008  0x10800005  loop:   beqz    $a0, exit
0x0040000c  0x00024021          move    $t0, $v0
0x00400010  0x00021040          sll     $v0, $v0, 1
0x00400014  0x00481026          xor     $v0, $v0, $t0
0x00400018  0x2084ffff          addi    $a0, $a0, -1
0x0040001c  0x08100002          j       loop
\$\endgroup\$
1
\$\begingroup\$

Mathematica, 40 24 bytes

Nest[BitXor[#,2#]&,1,#]&
\$\endgroup\$
1
\$\begingroup\$

k4, 26 bytes

{x{2/:~(=). 0b\:'1 2*x}/1}

0b\: converts a number to a boolean vector (i.e. bitstring), XOR is implemented as "not equal", 2/: converts a bitstring back to a number by treating it as a polynomial to evaluate, and x f/y with x an integer is f applied to y first, and then to its successive outputs x times.

Sample run:

  {x{2/:~(=). 0b\:'1 2*x}/1}'!5                                                                                                                                                                                    
1 3 5 15 17
\$\endgroup\$
1
\$\begingroup\$

Ruby, 31 26 bytes

EDIT: Changed to a different language entirely! All golfing suggestions welcome!

This program bitwise XORs the previous element of the sequence with twice itself, i.e. f(n) = f(n-1) ^ 2*f(n-1).

->n{v=1;n.times{v^=2*v};v}
\$\endgroup\$
1
\$\begingroup\$

MATL, 15 bytes

Similar to @flawr's answer:

i:1w"TToX+]2\XB

EDIT (May 20, 2016) Try it online! with X+ replaced by Y+ to conform to version 18.0.0 of the language.

Example

>> matl i:1w"TToX+]2\XB
> 5
51

Explanation

i              % input                                                     
:              % vector of values 1, 2, ... to previous input                           
1              % number literal                                            
w              % swap elements in stack                                    
"              % for                                                       
    TTo        % vector [1 1]
    X+         % convolution                                               
]              % end                                                       
2\             % modulo 2
XB             % convert from binary to decimal              
\$\endgroup\$
1
\$\begingroup\$

brainfuck, 87 bytes

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

Try it online!

Assumes infinite sized cells (otherwise it can’t get past 7, which is conveniently 255). The Pascal’s triangle mod 2 method is actually much longer because of the costly mod 2 operation, while XOR is much easier to implement.

\$\endgroup\$
0
\$\begingroup\$

APL, 31 bytes

{({2⊥⊃~1 2=.{(64⍴2)⊤⍺×⍵}⍵}⍣⍵)1}

This is almost certainly awful code, but I'm a complete APL newb. I expect anyone with any skill could get rid of all the D-functions and shorten it considerably. The logic is more or less the same as my k4 answer -- multiply by 1 or 2, convert to bits with , XOR using not equals, convert back to a number with , wrap the whole thing in a function and ask for a specific number of iterations using . I have no idea why the result coming out of the inner product is enclosed, but cleans that up.

\$\endgroup\$
  • \$\begingroup\$ You should be able to save a byte by changing ~1 2=. to 1 2≠. \$\endgroup\$ – Zacharý Jul 3 '17 at 13:12
  • \$\begingroup\$ And, which APL system is this on? If it's on Dyalog, you should be able to make it {({2⊥⊃1 2≠.((64⍴2)⊤×)⍵}⍣⍵)1} [28 bytes] \$\endgroup\$ – Zacharý Jul 3 '17 at 13:14
0
\$\begingroup\$

Seriously, 12 bytes

2,╣`2@%`Mεj¿

Hex Dump:

322cb960324025604dee6aa8

Try it online

Since Seriously does not include any means of doing a bitwise xor, this solution takes the challenge completely literally, directly computing the given row of the triangle. This method gives correct answers up to n=1029 (after which there is not enough memory to compute the given row of Pascal's triangle).

Explanation:

 ,                       get input
  ╣                 push the nth row of pascal's triangle
   `2@%`M           take each element of the row mod 2
         εj         join all the binary digits into a string
2          ¿        interpret it as a base 2 number
\$\endgroup\$
0
\$\begingroup\$

Pyt, 40 10 bytes

Đ0⇹Řć2%ǰ2Ĩ

Explanation:

Observing that the Binary Sierpinski Triangle is equivalent to Pascal's Triangle mod 2,

                      Implicit input
Đ                     Duplicate input
 0⇹Ř                  Push [0,1,2,...,input]
    ć2%               Calculate the input-th row of Pascal's Triangle mod 2
       ǰ              Join elements of the array into a string
        2Ĩ            Interpret as a binary number
                      Implicit print

Try it online!

\$\endgroup\$
0
\$\begingroup\$

Stax, 5 bytes

±s┤ε─

Run and debug online!

Port of the Jelly answer.

Uses ASCII representation to explain:

ODcH|^
O         Put 1 under top of stack
 D        Repeat for times specified by input
  cH|^    Xor the number with itself doubled
          Implicit output
\$\endgroup\$

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