# Challenge

Given a binary number as input through any means, "simplify" the number using a full program or a function.

# Input

[binary]
• binary is a number in binary that is over 0.

# Output

Take the input, convert it to base 10 without using a builtin, then if that number contains only 1s and 0s, convert it into a base 10 number as if it were another binary number. Repeat the process until the number cannot be read in binary and output that number.

# Other information

• If the input is 1, simply output 1. Your program should not go on infinitely simplifying 1.

• This is code golf, so shortest answer in bytes by Tuesday (November 17th) wins.

• If anything is confusing, leave a comment specifying what I need to clear up and I will edit it accordingly.

• Builtins for base conversion are not allowed.

# Examples

Input | Output

1 | 1
1010 | 2
1011 | 3
1100100 | 4
1100101 | 5
1111110011 | 3
• Could use a couple test cases. – isaacg Nov 11 '15 at 2:10
• Is the input an ASCII string, or actually 1's and 0's? – Tom Carpenter Nov 11 '15 at 2:15
• @TomCarpenter 1s and 0s. – The_Basset_Hound Nov 11 '15 at 2:20
• @isaacg Added ways to get 1-5 as output. – The_Basset_Hound Nov 11 '15 at 2:21
• Are functions which convert a string to a given base allowed? – isaacg Nov 11 '15 at 2:40

# Pyth, 20 16 bytes

u?-GTGu+yNsTG0z

4 bytes thanks to Jakube

Half of the code (u+yNsTG0) is simply the base conversion code.

Test Suite

u?-GTGu+yNsTG0z
z = input() (The string of 1s and 0s)
T = 10
u              z    Apply until the value stops changing, starting with z
G is the current value, a string of 0s and 1s.
?-GT               If G - T, e.g., G with the digits 1 and 0 removed is not empty,
G              Return G, to end the iteration.
u     G0     Else, reduce over G with initial value 0.
yN         Double the running total
+  sT       and add the next digit, cast to an int.
             Convert to string.

The input 1 is handled by the fact that u notices the value has stopped changing.

• Congratulations, you outgolfed Dennis! For the moment... – Conor O'Brien Nov 11 '15 at 2:51
• @CᴏɴᴏʀO'Bʀɪᴇɴ The secret is Pyth. – isaacg Nov 11 '15 at 2:52

# CJam, 24 23 bytes

q{:~{1$++}*s__,(As*-!}g Try it online in the CJam interpreter. ### How it works q Read all input. { }g Do: :~ Evaluate each character. Maps '0' -> 0 and '1' -> 1. { }* Fold; for each integer but the first: 1$                      Copy the second-topmost integer.
++                    Add all three integers on the stack.
s__             Cast to string and push two copies.
,(           Calculate string length and subtract 1.
As         Push the string "10".
*        Repeat the string length-1 times.
-       Remove its elements from the string representation
of the integer.
!      Apply logical NOT.
If !' pushed 1, repeat the loop.
• Do you have to repeat the "10" string length-1 times, or could you skip the decrement? – DLosc Nov 11 '15 at 3:34
• Subtracting 1 from the length turns "10" into "" if the integer has a single digit. This makes sure the code doesn't get into an infinite loop. – Dennis Nov 11 '15 at 3:35
• Fascinating, captain. }:^| – DLosc Nov 11 '15 at 3:37

## Pip, 28 27 bytes

Ta=1|aRMta:$+(^a)*2**RV,#aa Takes input as a command-line argument. We want to loop until a=1 or a contains some character(s) besides 0's and 1's. This latter condition is tested by RM'ing all characters in t = 10 from a. If there's anything left, the condition is truthy. Inside the loop, the conversion works as follows: a:$+(^a)*2**RV,#a

,#a  range(len(a))
RV     reversed
2**       2 to the power of each element
(^a)*          multiplied item-wise with each digit in split(a)
$+ Sum a: and assign back to a Putting a at the end auto-prints it. A recursive solution in 28 bytes: a<2|aRMt?a(f$+(^a)*2**RV,#a)

## Python 2, 52

f=lambda n:n>1<'2'>max(n)and f(n%10+2*f(n/10))or n

It's easier to think of this as two recursive functions:

g=lambda n:n and n%10+2*g(n/10)
f=lambda n:n>1<'2'>max(n)and f(g(n))or n

The function g converts a decimal value to binary, and the function f applies g repeatedly is long as its argument is made of digits 0 and 1 ('2'>max(n)) and is not 1. The golfed code collapses them into a single function by inserting the definition of g(n) for f(n), replacing the recursive call to g with f. The base case of n=0 of g is automatically handled by the check n>1.

• Nice :) The only thing is that the usual problem applies - the pesky L from repr... – Sp3000 Nov 11 '15 at 7:47

# Prolog, 220 212 bytes

:-use_module(library(clpfd)).
x(B,N):-reverse(B,C),foldl(y,C,0-0,_-N).
y(B,J-M,I-N):-B in 0..1,N#=M+B*2^J,I#=J+1.
b(N,I):-N>47,N<50,I is(N-48).
p(N):-N>1,number_codes(N,L),maplist(b,L,Y),x(Y,B),p(B);write(N).

Explanation
p is the main function and performs the following steps (with help from b,x,y):

• checks if current number is bigger than 1
• converts integer to list of ascii representations of digits
• checks that all numbers are 0 or 1
• converts ascii list to binary integer list
• converts binary integer list to decimal number
• recurses
• prints when a predicate fails.

Edit: Saved 8 bytes by unifying the p-clauses with OR.

# Mathematica 107 106

With a byte saved by DLosc.

j@d_:=(p=0;v=IntegerDigits@d;
Which[d<2,1,Complement[v,{0,1}]=={},j@Fold[#+#2 2^p++&,0,Reverse@v],1<2,d])

Break the input into its digits. If the input is 1, output 1.

If the input is a number consisting of 0's and 1's, convert that to decimal and run it through again.

Otherwise, return the input.

j[1]

1

j[11010001]

209

j[1111110001]

1009

j[1111110011]

3

The first step yields 1011 which in turn yields 3.

Here we test starting with 1011.

j[1011]

3

# Javascript, 132, 123 bytes

Well, it's not the best answer, but..

FYI, if an invalid input is given, it displays the same to the user.

• You could save 19 bytes by using forinstead of while and setting values directly in the statement (this also reduces some {}), discarding some ;, using ES6 function description, increment i inline. It'll look like this: c=x=>{for(r=0;x&&!/[2-9]/.test(x);x=r)for(i=0;x>0;r+=x%10*Math.pow(2,i++),x=parseInt(x/10));alert(x)};c(prompt()). – insertusernamehere Nov 11 '15 at 14:00
• 114: function c(x){while(x^0&&!/[2-9]/.test(x)){for(i=r=0;x;i++)r+=x%10*Math.pow(2,i),x=0|x/10;x=r}alert(x)}c(prompt()) – Mama Fun Roll Nov 13 '15 at 6:03
• @insertusernamehere, thanks for the suggestion, but I didn't understand the c=x=> at the start, didn't work on Chrome or Firefox console. :( @ןnɟuɐɯɹɐןoɯ, could not wrap my head around the XOR condition and the x=0|x/10‌ instead of parseInt, I've incorporated the rest of the changes. Thanks.. – LearningDeveloper Nov 16 '15 at 5:36
• @GauthamPJ I'm sorry, somehow the code got broken while copying and contained unprintable characters. Here's the correct version: c=x=>{for(r=0;x!=0&&!/[2-9]/.test(x);x=r)for(i=r=0;x;)r+=x%10*Math.pow(2,i++),x=parseInt(x/10);alert(x)};c(prompt()). It definitely runs in Firefox 42, try this fiddle. Note that this more golfed version and also your original code don't work for 1 and will run into an endless loop. c=x=> is like function c(x){} see "Arrow functions". – insertusernamehere Nov 16 '15 at 9:44

# JavaScript ES6, 52

As a function. The function argument must be either a string of binary digits or a number whose decimal representation contains only 1 and 0.

Test running the snippet below in an EcmaScript 6 compliant browser - implementing arrow functions, template strings and spread operator (I use Firefox)

f=s=>s<2|[...s+''].some(c=>(n+=+c+n,c>1),n=0)?s:f(n)

// To test
console.log=(...x)=>O.innerHTML+=x+'\n';

// Basic test cases
;[[1,1],[1010,2],[1011,3],[1100100,4],[1100101,5],[1111110011,3]]
.forEach(t=>console.log(t[0]+' -> '+f(t[0])+' expected '+t[1]))

function longtest() {
var o=[],i;
for (i=1;i<1e6;i++)
b=i.toString(2),v=f(b),v!=i?o.push(b+' '+v):0;
O.innerHTML=o.join\n
}
Click to run the long test <button onclick="longtest()">go</button>
<pre id=O></pre>

• Really liking n+=+c+n for the binary conversion. So elegant... – nderscore Nov 11 '15 at 15:14

# Mathematica, 625955 48 bytes

Saved 7 bytes thanks to Martin Büttner.

#//.a_/;Max[b=IntegerDigits@a]<2:>Fold[#+##&,b]&

# Javascript (ES7) 87807877 74 bytes

Snippet demo for supporting browsers (currently only Firefox nightly supports the exponential operator)

f=x=>[...x].reverse(i=y=j=0).map(z=>(j|=z,y+=z*2**i++))&&j<2&y>1?f(y+[]):x
<input type="text" id="x" value="1111110011"><button onclick="o.innerHTML=f(x.value)">Run</button><div id="o"></div>

f=x=>
[...x].reverse(i=y=j=0) // reverse string as array, initialize vars
.map(z=>( // iterate over the all chatacters
j|=z, // keep track of whether a digit higher than 1 is encountered
y+=z*2**i++ // build decimal result from binary
))&&
j<2&y>1? // if we encountered only 1's and 0's and result > 1
f(y+[]) // then call recursively and cast to a string
:x // else return x

# Javascript (ES6) 81 bytes

Snippet demo for supporting browsers

f=x=>[...x].reverse(i=y=j=0).map(z=>y+=z*Math.pow(2,i++,j|=z))&&j<2&y>1?f(y+[]):x
<input type="text" id="x" value="1111110011"><button onclick="o.innerHTML=f(x.value)">Run</button><div id="o"></div>

# PHP, 210 204 bytes

It's my first time posting here, so hope you guys will like it ! Even if it's obviously not the best way to write it, I'm still glad to show it off here !

## The Code

<?function j($a){$c=0;if($a==1){return 1;}else{if(preg_match("#^[01]+$#",$a)){$b=strlen($a);$a=str_split($a);foreach($a as$d){$c+=($d==0?0:2**($b-1));$b--;}return j($c);}else{return$a;}}}echo j($_GET[0]);

I've made a recursive function "j" that will first check if the input is equal to 1. If so, the function returns 1 as expected, else it'll split the number in an array to calculate the decimal value, but only if the number is a binary one. If it's not, it'll return the number as is.

## Ungolfed code

<?
function j($a) {$c = 0;
if ($a == 1) { return 1; } else { if (preg_match("#^[01]+$#", $a) {$b = strlen($a);$a = str_split($a); foreach ($a as $d) {$c += ($d == 0 ? 0 : 2 ** ($b - 1));
$b--; } return j($c);
}
else {
return $a; } } } echo j($_GET[0]);

I've used a "foreach" statement instead of my initial "for" one, allowing me a gain of 6 bytes but I'm pretty sure there's a lot more to do.

# PHP, 114 112 bytes

also works for 0. Run with -r.

for($n=$argv[1];count_chars($s="$n",3)<2&$s>1;)for($i=$n=0;""<$c=$s[$i++];)$n+=$n+$c;echo$s;

count_chars($s,3) returns a string containing all characters from the string (like array_unique does for arrays). For binary numbers, this will be 0, 1 or 01. For other numbers, this will contain a digit larger than 1, so <2will return true only for binary numbers. &$s>1 is needed for the special case 1.

f=y=>y>1&/^[01]+$/.test(y)?f(''+[...y].reverse().reduce(((p,v,i)=>p+v*Math.pow(2,i)),0)):y ## Demo Only works in ES6-compliant browsers such as Firefox and Microsoft Edge f=y=>y>1&/^[01]+$/.test(y)?f(''+[...y].reverse().reduce(((p,v,i)=>p+v*Math.pow(2,i)),0)):y

// Snippet stuff
$(form).submit((e) => { document.getElementById(y).textContent = f(document.getElementById(x).value); e.preventDefault() }) <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <form> <label>Input: <input pattern=^[01]+$ required id=x>
</label>
<button type=submit>Go</button>
<p>Output:
<output id=y></output>
</p>
</form>

• If you use eval, you might be able to pull off an implicit return. – Mama Fun Roll Nov 11 '15 at 5:08
• 5 bytes shorter with eval and anonymous functions – Downgoat Nov 11 '15 at 5:15
• @ןnɟuɐɯɹɐןoɯ For some reason the eval'd function does not work with 1. because it doesn't enter the loop I assume – rink.attendant.6 Nov 11 '15 at 5:20
• @nderscore Thanks, but recursion was 4 bytes shorter :-) – rink.attendant.6 Nov 11 '15 at 5:46

# Scala, 128 bytes

def b(s:String):String=if(s.matches("[10]{2,}"))b(""+s.reverse.zipWithIndex.collect{case('1',i)=>Math.pow(2,i)}.sum.toInt)else s

## Matlab (115)

@(a)num2str(sum((fliplr(a)-48).*arrayfun(@(x)2^x,0:nnz(a)-1)));a=ans(input('','s'));while(find(a<50))a=ans(a);end,a

• The anonymous function is number type conversion (bin2dec)