Reverse and square

Question

In this challenge you will compute numbers from a curious sequence.

Your input is a single decimal nonnegative integer. Reverse the bits in this integer and then square the number to get the required output.

When reversing the bits you must not use any leading zeroes in the input. For example:

26 (base 10) = 11010 (base 2) -> 01011 (base 2) = 11 -> 11*11 = 121

The first 25 inputs/outputs of this sequence:

0: 0
1: 1
2: 1
3: 9
4: 1
5: 25
6: 9
7: 49
8: 1
9: 81
10: 25
11: 169
12: 9
13: 121
14: 49
15: 225
16: 1
17: 289
18: 81
19: 625
20: 25
21: 441
22: 169
23: 841
24: 9

Your solution should work for arbitrarily sized integers. If your language does not have a convenient built-in method of using those, implement your answer as if it does. You are then excused if your answer breaks for large numbers. However, do not use tricks/bounds that only work for a limited domain (such as a lookup table).

---

Your score is the number of bytes of source code.

-50% bonus if you never convert the number to/from binary. This is not limited to builtins, if you loop over the number bit by bit (either by shifting or masking or any other method), it will also count as conversion. I don't know whether this is actually possible, but it gives an incentive to spot a pattern in the sequence.

Smallest score wins.

Answer 1

Par, 5 bytes

✶Σ⌐Σ²

That's read-binary-reverse-binary-square.

Answer 2

Mathematica, 42 21 bytes

Thanks to alephalpha for halving the score.

#~IntegerReverse~2^2&

The actual reason I did this in Mathematica was because I wanted to look at a plot... it sure looks funny:

Answer 3

Japt, 29 28 11 7 bytes

^{(You can save the program as a 7-byte IEC_8859-1-encoded file, then upload it to the interpreter.)}

Japt is shortened JavaScript made by ETHproductions.

¢w n2 ²

Try it online!

Explanation:

1. ¢ is shortcut to Us2, which compiles to U.s(2). U is input (implicit), .s(2) called by a number, invokes .toString(2) (converts to binary, parses as string).

2. w compiles to .w(), which reverses the string (.split('').reverse().join('')).

3. n2 works as parseInt(<number>,2), i.e. converts binary to decimal.

4. ² invokes Math.pow(<number>,2), i.e. squares the number.

Answer 4

Minkolang 0.14, 43 bytes

Thanks to Mego for inspiring this.

n1{d1`,2$3*&$z2zd2%-2l$Md1%-;z2%*z2:{+}2;N.

Test the code here and check all test cases here.

Explanation

This uses this recurrence relation:

a(0) = 0
a(1) = 1
a(2n) = a(n)
a(2n+1) = a(n) + 2^(floor(log_2(n))+1)

If n is the input, then a(n) is the resulting number after its binary sequence has been flipped. 0 and 1 are obvious. For a(2n) = a(n), consider that x0 (where x is any sequence of binary digits) flipped is 0x, which is the same as x. For a(2n+1), the reasoning is a bit more complicated. x1 flipped is 1x, which is equal to x + 2^k for some k. This k is one more than the number of digits in x, which is floor(log_2(n))+1. The full formula follows, except that it's modified a bit. This is what I actually code:

a(0) = 0
a(1) = 1
a(n) = a(n//2) + (n%2) * 2^(floor(log_2(n - n%2)))

As Mego and I worked out in chat, floor(n/2) = (n - n%2)/2. Thus, log_2(floor(n/2))+1 = log_2(n - n%2). Furthermore, multiplying by (n%2) collapses both the odd and even parts into one expression.

Finally, without any further ado, here's the code, explained.

n                                              Take number from input
 1{                                            Start recursion that takes only one element
   d1`,                                        1 if top of stack 0 or 1, 0 otherwise
       2$3*                                    26
           &                                   Jump if top of stack is not zero
            $z                                 Store top of stack in register (z)

               zd2%-                           n - n%2
                    2l$M                       log_2(n - n%2)
                        d1%-                   floor(log_2(n - n%2))
              2             ;                  2^floor(log_2(n - n%2))
                             z2%               n%2
                                *              Multiply
                                 z2:           n//2
                                    {          Recurse
                                     +         Add
                                      }        Return
                                       2;N.    Square it, output as number, and stop.

Answer 5

Jolf, 7 bytes

Just run it. The input on the page doesn't work.

^C_Bj22

Explanation

^C_Bj22
    j   numeric input
   B    convert to binary (str)
  _     reverse
 C   2  parse as binary integer to base 10
^     2 square
        implicit output

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I added the Q command, which makes this 6 bytes: QC_Bj2

Answer 6

Python, 32 bytes

lambda x:int(bin(x)[:1:-1],2)**2

Try it online.

The code is pretty straightforward: bin(6), for example, gives 0b110, the binary representation of 6. [:1:-1] reverses the string and removes 0b. int converts the string to an integer from binary, and **2 squares it.

Answer 7

TeaScript, 9 bytes 11

TeaScript is Javascript for golfing

®x÷v¤)**2

Will golf more once I get back to my computer

Try it online!

Test all

Answer 8

J, 10 9 bytes

2^~|.&.#:

This is a tacit, monadic verb. Try it online!

Thanks to @randomra for golfing off 1 byte!

How it works

2^~|.&.#:  Right argument: y

       #:  Convert y to binary.
   |.      Reverse the digits.
     &.    Dual; apply the inverse of #:, i.e., convert back to integer.
 ^~        Apply power (^) with reversed argument order (~)...
2          to 2 and the previous result.

Answer 9

Seriously, 8 7 bytes

2;,¡R¿ª

Challenges like these are perfect for Seriously :)

Try it online

Explanation:

2;,¡    get a string representing the (decimal) input in binary, with a 2 on the bottom of the stack
R      reverse the string
¿    convert binary string to decimal int (using that extra 2 from earlier)
ª      square it

Answer 10

JavaScript, 64 63 56 53 bytes

n=>parseInt([...n.toString(2)].reverse().join``,2)**2

I realize I'm extra long, but hey, I can do it :P

Demo

Answer 11

CJam, 10 bytes

ri2bW%2b_*

Try it online

Answer 12

Perl 6, 21 bytes

{:2(.base(2).flip)²}

Example usage:

say {:2(.base(2).flip)²}(26); # 121

say (0..24).map: {:2(.base(2).flip)²};
# (0 1 1 9 1 25 9 49 1 81 25 169 9 121 49 225 1 289 81 625 25 441 169 841 9)

my &code = {:2(.base(2).flip)²};
say code 3; # 9

say chars code 10¹⁰⁰; # 140

Answer 13

Shell, 25

dc -e2o?p|rev|dc -e2i?d*p

Input/output via STDIN/STDOUT:

$ echo 26|dc -e2o?p|rev|dc -e2i?d*p
121
$ 

Answer 14

PHP, 45 bytes

echo pow(bindec(strrev(decbin($argv[1]))),2);

Answer 15

TI-Basic (TI-84 Plus CE), 42 bytes

Prompt X
0→S
While X
2S→S
If X/2≠int(X/2
S+1→S
End
S2

Answer 16

Pyth - 9 bytes

Straightforward conversions. I actually assigned 2 to a var which is pretty weird.

^i_jQK2KK

Test Suite.

Answer 17

, 12 chars / 21 bytes

⦅`ᶀ`+ᴙ(ïß2)²

Try it here (Firefox only).

Noncompetitive answer, 9 chars / 18 bytes

⦅Յ+ᴙ(ïⓑ)²

Try it here (Firefox only).

Answer 18

Pyth, 9 bytes

^i_.BQ2 2

This is a very simple pyth based answer similar to the Python one

Answer 19

Ruby, 35 bytes

->(x){x.to_s(2).reverse.to_i(2)**2}