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

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 2

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

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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 3

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 4

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 5

Par, 5 bytes

✶Σ⌐Σ²

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

Answer 6

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

---

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

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

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 9

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 10

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 11

CJam, 10 bytes

ri2bW%2b_*

Try it online

Answer 12

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 13

PHP, 45 bytes

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

Answer 14

Ruby, 35 bytes

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

Answer 15

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 16

05AB1E, 4 bytes

bRCn

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Explanation

     Implicit input
b    Convert to binary
 R   Reverse
  C  Convert from binary
   n Square
     Implicit output

Answer 17

MathGolf, 4 bytes

àxå²

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Explanation:

à     # Convert the (implicit) input-integer to a binary-string
 x    # Reverse this string
  å   # Convert it from a binary-string back to an integer
   ²  # Take the square of this integer
      # (after which the entire stack joined together is output implicitly as result)

Answer 18

Wolfram Language (Mathematica), 40 bytes

#+##&~Fold~Reverse@IntegerDigits[#,2]^2&

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Answer 19

Vyxal, 4 bytes

bṘB²

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bṘB²
bṘB  # Convert to binary, reverse, convert back to decimal
   ² # Square

Answer 20

Desmos, 124 72 bytes

f(n)=total(mod(floor(n/2^l),2)2^k/2^l)^2
k=floor(log_2(n+0^n))
l=[k...0]

Try it on Desmos!

-52 bytes thanks to Aiden Chow

Answer 21

Haskell, 38 bytes

(#0)
0#s=s^2
n#s=div n 2#(2*s+n`mod`2)

Attempt This Online!

Answer 22

Pyth - 9 bytes

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

^i_jQK2KK

Test Suite.

Answer 23

Pyth, 9 bytes

^i_.BQ2 2

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

Answer 24

𝔼𝕊𝕄𝕚𝕟, 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 25

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 26

C (gcc), 38 bytes

r;f(n){for(r=n%2;n/=2;r+=r+n%2);r*=r;}

Reverse all the bits, square the result. Pretty standard.

-1 byte thanks to ceilingcat!

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Answer 27

Perl 5 -p, 38 bytes

$_=(oct"0b".reverse sprintf"%b",$_)**2

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Answer 28

Splunk (v8.2.6) Search Processing Language, 182 170 bytes:

|inputcsv b|foreach *[eval x=<<FIELD>>]|eval p=mvrange(0,floor(log(x,2)+1))|mvexpand p|eval x=floor(x/pow(2,p))%2|stats list(x) as x|eval x=tonumber(mvjoin(x,""),2),x=x*x

Big data. Smallish code.

Am learning Splunk for work, and decided to try to have a bit of fun with it.

Input is expected in a file at $SPLUNK_HOME/var/run/splunk/csv/b.csv. The file must contain the number in base 10 followed by a newline. The output is the search result, which is analogous to an SQL SELECT. This is as close to our standard I/O rules as I could reasonably do in SPL, but I'm willing to work something out if this is problematic.

With whitespace:

| inputcsv b
| foreach * [eval x=<<FIELD>>] 
| eval p=mvrange(0,floor(log(x,2)+1))
| mvexpand p
| eval x=floor(x/pow(2,p))%2
| stats list(x) as x
| eval x=tonumber(mvjoin(x,""),2),x=x*x

Explanation (I plan to flesh this out more later):

| inputcsv b                            ``` Read in b ```
| foreach * [eval x=<<FIELD>>]          ``` Get the value into a field "x" (Splunk reads our "CSV" file in as an event with one field, named the value we want, with a value of the empty string.) ``` 
| eval p=mvrange(0,floor(log(x,2))+1)   ``` Create a multivalue field "p" which contains the numbers 0 to the number of bits required to represent our number, the order here also means we get the reverse for free ```
| mvexpand p                            ``` Expand our single event with a multivalue field "p" into multiple events, each with a certain value of p ```
| eval x=floor(x/pow(2,p))%2            ``` Calculate (x/(2**p))%2, and set that value back to x```
| stats list(x) as x                    ``` Collect the values of x back into a single multivalue event ```
| eval x=tonumber(mvjoin(x,""),2),x=x*x ``` Combine all the bits back into a string, and convert that string back into a numeric value, then square it```
```result is a single event with field x, containing the square of the bit reversed input.```

Edit: forgot how math works (ceil instead of floor + 1 for the bit length calculation), and managed to only check with inputs that worked with my wrong math. Also sort happened to be a leftover from when I was doing the problem more verbosely, we get the reverse sort for free. Nets to -12.

Answer 29

LOLCODE, 192 bytes

HAI 1.3
I HAS A s ITZ 0
I HAS A x
GIMMEH x
IM IN YR l
s R SUM OF PRODUKT OF s 2 MOD OF x 2
x R QUOSHUNT OF x 2
BOTH SAEM x 0,O RLY?
YA RLY
GTFO
OIC
IM OUTTA YR l
VISIBLE PRODUKT OF s s
KTHXBYE

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Answer 30

cQuents, 10 bytes

K\rJ$)))^2

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First time using cQuents.

Explanation

    $      # Take the index.
   J )     # Convert to (default) base 2.
 \r   )    # Reverse it.
K      )   # Convert from (default) base 2 to base 10.
        ^2 # And square it.