# Split, flip and recombine integers

## Background

It's well known in mathematics that integers can be put into a one-to-one correspondence with pairs of integers. There are many possible ways of doing this, and in this challenge, you'll implement one of them and its inverse operation.

Your input is a positive integer n > 0. It is known that there exist unique non-negative integers a, b ≥ 0 such that n == 2a * (2*b + 1). Your output is the "flipped version" of n, the positive integer 2b * (2*a + 1).

You can assume that the input and output fit into the standard unsigned integer datatype of your language.

## Rules and scoring

You can write either a full program or a function. The lowest byte count wins, and standard loopholes are disallowed.

## Test cases

These are given in the format in <-> out, since the function to be implemented is its own inverse: if you feed the output back to it, you should get the original input.

1 <-> 1
2 <-> 3
4 <-> 5
6 <-> 6
7 <-> 8
9 <-> 16
10 <-> 12
11 <-> 32
13 <-> 64
14 <-> 24
15 <-> 128
17 <-> 256
18 <-> 48
19 <-> 512
20 <-> 20
28 <-> 40
30 <-> 384
56 <-> 56
88 <-> 224
89 <-> 17592186044416


Here is a Stack Snippet to generate both a regular leaderboard and an overview of winners by language. 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 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 leaderboard snippet:

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


var QUESTION_ID=70299,OVERRIDE_USER=32014;function answersUrl(e){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"https://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="<h1>"+e.body.replace(OVERRIDE_REG,"")+"</h1>")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery("#answer-template").html();t=t.replace("{{PLACE}}",n+".").replace("{{NAME}}",e.user).replace("{{LANGUAGE}}",e.language).replace("{{SIZE}}",e.size).replace("{{LINK}}",e.link),t=jQuery(t),jQuery("#answers").append(t);var o=e.language;/<a/.test(o)&&(o=jQuery(o).text()),s[o]=s[o]||{lang:e.language,user:e.user,size:e.size,link:e.link}});var t=[];for(var o in s)s.hasOwnProperty(o)&&t.push(s[o]);t.sort(function(e,s){return e.lang>s.lang?1:e.lang<s.lang?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;
body{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;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="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <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><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><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>

• Funny, this problem was posted on anarchy golf last week Jan 27 '16 at 16:51
• @feersum Oh, I wasn't aware. What a coincidence. Jan 27 '16 at 16:53
• Obligatory xkcd.com/153 Jan 27 '16 at 22:56

# Jelly, 1716 15 bytes

BUi1µ2*³:2*×Ḥ’$ Try it online! ### How it works BUi1µ2*³:2*×Ḥ’$    Main link. Input: n

B                  Convert n to base 2.
U                 Reverse the array of binary digits.
i1               Get the first index (1-based) of 1.
This yields a + 1.
µ              Begin a new, monadic chain. Argument: a + 1
2*            Compute 2 ** (a+1).
³:          Divide n (input) by 2 ** (a+1).
: performs integer division, so this yields b.
2*        Compute 2 ** b.
$Combine the two preceding atoms. Ḥ Double; yield 2a + 2. ’ Decrement to yield 2a + 1. × Fork; multiply the results to the left and to the right.  • Wait, are you implementing operators in Jelly to suit the problem? In that case LOL Jan 28 '16 at 13:00 • I am not. The only commit to Jelly after this challenge was posted was an update of the documentation, and all the operator used in this answer have been implemented for at least a month. Feel free to verify Jan 28 '16 at 14:45 • No worries, I'm not familiar with the rules or anything, I just thought it was cool that golfing has come to people inventing their own languages! Jan 28 '16 at 14:57 • And there's a lot of them. Check out What programming languages have been created by PPCG users?. Jan 28 '16 at 15:01 # Pyth, 16 15 bytes *hyJ/PQ2^2.>QhJ  1 byte thanks to Dennis Test suite Explanation: *hyJ/PQ2^2.>QhJ Implicit: Q = eval(input()) PQ Take the prime factorization of Q. / 2 Count how many 2s appear. This is a. J Save it to J. y Double. h +1. .>QhJ Shift Q right by J + 1, giving b. ^2 Compute 2 ** b. * Multiply the above together, and print implicitly.  # MATL, 22 bytes Yft2=XK~)pq2/2w^Ks2*Q*  Try it online! ### Explanation Yf % factor t % duplicate 2= % compare to 2 (yields a logical array) XK % save a copy of that to variable K ~) % keep only values != 2 in the factors array p % multiply that factors q2/ % product - 1 / 2 2w^ % 2^x K % load variable K (the logical array) s % sum (yields the number of 2s) 2*Q % count * 2 + 1 * % multiply both values  • Very nice! You can use Q for 1+ (this has been introduced recently) and q for 1-. That also saves a space (which you could save with H anyway). See here Jan 27 '16 at 17:20 • @LuisMendo Thanks. Didn't know that feature. Jan 27 '16 at 17:30 • Well done beating Luis using MATL! Jan 27 '16 at 20:22 # Python 2, 39 bytes lambda n:2*len(bin(n&-n))-5<<n/2/(n&-n)  n & -n gives the largest power of 2 that divides n. It works because in two's-complement arithmetic, -n == ~n + 1. If n has k trailing zeros, taking its complement will cause it to have k trailing ones. Then adding 1 will change all the trailing ones to zeroes, and change the 2^k bit from 0 to 1. So -n ends with a 1 followed by k 0's (just like n), while having the opposite bit from n in all higher places. • can you shortly explain how does n&-n work ? i see what this trick do but not how :( Jan 27 '16 at 22:10 • n&-n returns the highest power of 2 that divdes n. – Neil Jan 27 '16 at 22:13 • @Erwan I explained about n & -n. Jan 27 '16 at 22:32 • I had gotten the corresponding program on Anarchy golf n=1;exec"c=n&-n;print n,':',2*len(bin(c))-5<<n/2/c;n+=1;"*100, but it's two chars behind the best solution. – xnor Jan 27 '16 at 22:33 • @xnor Hint: the 59-byte solution (well, mine at least) does not work for all values of n. Jan 27 '16 at 22:35 # MATL, 25 26 bytes :qt2w^w!2*Q*G=2#f2*q2bq^*  This uses current release (10.2.1) of the language/compiler. Try it online! ### Explanation Pretty straightforward, based on brute force. Tries all combinations of a and b, selects the appropriate one and does the required computation. :q % implicit input "n". Generate row vector [0,1,...,n-1], say "x" t2w^ % duplicate and compute 2^x element-wise w!2*Q % swap, transpose to column vector, compute 2*x+1 * % compute all combinations of products. Gives 2D array G=2#f % find indices where that array equals n 2*q2bq^* % apply operation to flipped values  • Hah! :-P Beaten in your own language... first time? Jan 27 '16 at 20:22 • @StewieGriffin Yep! A nice milestone :-) Jan 28 '16 at 0:48 # Julia, 41 bytes n->2^(n>>(a=get(factor(n),2,0)+1))*(2a-1)  This is an anonymous function that accepts an integer and returns an integer. To call it, assign it to a variable. We define a as 1 + the exponent of 2 in the prime factorization of n. Since factor returns a Dict, we can use get with a default value of 0 in case the prime factorization doesn't contain 2. We right bit shift n by a, and take 2 to this power. We multiply that by 2a-1 to get the result. # Perl 5, 40 bytes 38 bytes plus 2 for -p $i++,$_/=2until$_%2;$_=2*$i+1<<$_/2-.5  -p reads the STDIN into the variable $_.

$i++,$_/=2until$_%2 increments $i (which starts at 0) and halves $_ until $_ is nonzero mod 2. After that, $_ is the odd factor of the original number, and $i is the exponent of 2.

$_=2*$i+1<<$_/2-.5 — The right-hand side of the = is just the formula for the number sought: {1 more than twice the exponent of 2} times {2 to the power of {half the odd factor minus a half}}. But "times {2 to the power of…}" is golfed as "bit-shifted leftward by…". And that right-hand side is assigned to $_.

And -p prints \$_.

# C, 49 bytes

a;f(n){for(a=0;n%2-1;a++)n/=2;return 2*a+1<<n/2;}


## JavaScript ES6, 36 33 bytes

n=>63-2*Math.clz32(b=n&-n)<<n/b/2


My understanding is that Math.clz32 is going to be shorter than fiddling around with toString(2).length.

Edit: Saved 3 bytes thanks to @user81655.

• Nice. You could also save a few bytes by setting n&-n to a variable: n=>63-2*Math.clz32(x=n&-n)<<n/x/2 Jan 28 '16 at 0:08
• @user81655 Thanks; I only wish I could have used n&=-n, but I need n again...
– Neil
Jan 28 '16 at 8:54

# PARI/GP, 38 bytes

f(n)=k=valuation(n,2);(2*k+1)<<(n>>k\2)


Note that >> and \ have the same precedence and are computed left-to-right, so the last part can be n>>k\2 rather than (n>>k)\2. The ungolfed version would make k lexical with my:

f(n)=
{
my(k=valuation(n,2));
(2*k+1) << ((n>>k)\2);
}


# Pip, 24 23 bytes

Wa/2=Ya//:2i+:2++i*2**a


Try it online!

Loops, int-dividing a by 2 and incrementing i by 2, until the int division is different from the equivalent float division (thus indicating a had become odd). The way the loop is set up, a is now half of that odd number, rounded down: i.e., if the odd factor of the original input is 2k+1, a is now k. Meanwhile, if the largest power-of-two factor of the original input is 2^n, then i is now 2n. It remains only to increment i to 2n+1 and multiply that by 2^k.

### Original 24-byte version

2**(a/2//YaBA-a)*#TB:y*y


Try it online!

2**(a/2//YaBA-a)*#TB:y*y
a               First command-line argument
BA             Bitwise AND
-a           with negative a
This gives the largest power of 2 that divides a
Y                Yank that number into y
a/2//                 Divide a by 2 and int-divide by y
This has the same effect as dividing by y and then int-dividing
by 2: since a/y is an odd number 2k+1, a/y//2 is k
2**(           )          Take 2 to that power
Now we need to multiply by 2n+1, where y is 2^n
y*y  Square y, giving 2^(2n)
TB:     Convert to binary: 1 followed by (2n) 0s
#        Get the length: 2n+1
*         Multiply