Do two numbers contain unique powers of 2

Question

Break two numbers up into their powers of 2, if they share any, return a falsey value. Otherwise, return a truthy value.

If one input is 0, the answer will always be truthy. if one input is 255, the answer will be falsey if the other is not 0.

Example

Given 5 and 3

5 = 2⁰ + 2²
3 = 2⁰ + 2¹

Both share 2⁰, return falsey.

Given 8 and 17

8  = 2³
17 = 2⁰ + 2⁴

Neither share any powers, return truthy.

I/O

Input and output can be through any standard means

Input will always be two integers in the range 0 <= n <= 255

Output should be one truthy or falsey value

Test Cases

Truthy:
1, 2
3, 4
2, 4
8, 17
248, 7

Falsey:
1, 3
6, 4
8, 8

Standard Loopholes apply.

This is code-golf, so fewest bytes wins!

Answer 1

Actually, 2 bytes

&Y

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Explanation:

&Y
&   bitwise AND
 Y  boolean negation

Answer 2

x86 Machine Code, 5 bytes

20 C8 0F 95 C0

takes inputs in al and cl, and returns output in al, and is equivalent to:

and al, cl
setnz al

Answer 3

Jelly, 2 bytes

&¬

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Explanation:

&¬
&   bitwise AND
 ¬  logical negation

Answer 4

Ruby, 12 bytes

->a,b{a&b<1}

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Answer 5

Mathematica, 12 bytes

BitAnd@##<1&

Pure function taking two positive integers as argument and returning True or False.

Answer 6

Python, 18 bytes

lambda a,b:not a&b

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Answer 7

Bash, 10 bytes

!(($1&$2))

Input is via command-line arguments. Output is via exit code, so 0 is truthy and 1 if falsy.

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Answer 8

05AB1E, 2 bytes

&_

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Answer 9

JavaScript (ES6), 12 bytes

a=>b=>!(a&b)

Takes input in the currying format.

let f=

a=>b=>!(a&b)

console.log(f(1)(2));
console.log(f(3)(4));
console.log(f(2)(4));
console.log(f(8)(17));
console.log(f(248)(7));
console.log(f(1)(3));
console.log(f(6)(4));
console.log(f(8)(8));

Answer 10

CJam, 23 4 bytes

q~&!

It turns out my previous solution contained an implementation of what was essentially just bitwise AND (which is silly because it used & anyway).

Answer 11

CJam, 4 bytes

{&!}

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Answer 12

R, 28 bytes

f=function(a,b)!bitwAnd(a,b)

Answer 13

R, 23 bytes

if allowed - this version can get multiple inputs as a1,a2,a3..... b1,b2,b3.....

!bitwAnd(scan(),scan())

Answer 14

GolfScript, 3 bytes

~&!

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Answer 15

Pyke, 3 bytes

.&!

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.&  -  bitwise_and(input_1, input_2)
  ! - not ^

Answer 16

MATL, 3 bytes

Z&~

Try it at MATL Online.

Explanation

    % Implicitly grab both inputs
Z&  % Bit-wise AND
~   % Negate the result and implicitly display

Answer 17

Pari/GP, 18 bytes

x->y->!bitand(x,y)

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Answer 18

Javascript, 28

(a,b)=>!(256&(-~a|-~b)||a&b)

Answer 19

QBIC, 12 bytes

?(: and :)=0

This returns -1 for truthy inputs, and 0otherwise. "Bitwise" is something QBasic isn't very good at, and I haven't built any of this stuff into QBIC either.

Answer 20

D, 13 bytes

(x,y)=>!(x&y)

Here, (arg1,...)=>result is the D lambda syntax, and the ! and & are common notations for boolean negation and bitwise AND.

You can find the whole program coupled with the tests at tio.run.

Answer 21

PHP, 25 Bytes

print 1 for true and nothing for false

<?=!($argv[1]&+$argv[2]);

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Answer 22

Lua, 19 bytes

a,b=...print(a&b<1)

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Answer 23

C (gcc), 18 bytes

Uses exit-code 0 as truthy.

f(a,b){exit(a&b);}

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Answer 24

Pyth, 6 bytes

!.&hQe

Test it online!

Explanations

!.&hQe

   hQ     Get first input number
     e    Get second (implicit) input number
 .&       Bitwise and
!         Negate

Answer 25

Chip, 80 bytes

 AZ BZ CZ DZ EZ FZ GZ HZ
,[',[',[',[',[',[',[',['*Z~S
`--)--)--)--)--)--)--)~aef

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Takes input as two consecutive bytes. (In the TIO, the characters 'A' and '$' are used, corresponding to values 65 and 36.)

This solution has eight chunks, one per bit, plus an extra chunk at the end. Looking at these chunks:

A is bit 0x01 of the input. (The letters A though H correspond to a bit each.) Z holds on to a value for the next cycle while producing the previous value. Hence, A and Z are the current and previous values of bit 0x01.

[ is an AND-gate; it simply ANDs its two inputs together. So the full row of such gates is performing a bitwise AND of the two bytes.

Finally, ) is an OR-gate. All eight AND results are OR'd together, giving us a True is there is a match, and a False otherwise. We want the inverse of that, so use a NOT-gate (~) before passing it to the output.

We output via a, e, and f. The a, which corresponds to bit 0x01 of the output, receives the result of the above computation. The other two bits e and f are unconditionally on, together forming 0x30. Therefore, the output is either 0x30 or 0x31, which are the codes for ASCII characters '0' and '1'.

The extra stuff is housekeeping, it prevents output for the first cycle (there is no previous value at this point, so there is no reason to do a comparison).

---

Alternate solution, also 80 bytes

A~.B~.C~.D~.E~.F~.G~.H~.
Z~<Z~<Z~<Z~<Z~<Z~<Z~<Z~<*Z~S
 a[--[--[--[--[--[--[--'ef

This is very similar in style, but uses a different algorithm. Where above we AND each pair of bits, and use OR to reduce the result to a single value, here we OR the negation of each pair of bits, then use AND to reduce.

One thing to note: we use an implicit OR here, instead of a proper gate as above. We could have used implicit OR'ing above as well, but there is no byte-savings there.

Answer 26

Java 13 bytes

a->b->(a&b)<1

Answer 27

Wise, 11 bytes

&::^~![&&]|

Due to a bug this does not currently work on tio, however it does work on the desktop version.

Here is a slightly longer version that works for both:

&::^~![?~&:]|

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Explanation

Most of our code goes into a logical not, in fact only the first byte is required if you want to get the opposite answer.

&             #Bitwise and
 ::^~![?~&:]| #Logical not

Answer 28

Oracle SQL, 34 bytes

-- Begin test cases
WITH T (x, y, desired) AS (
  SELECT   1,  2, 1 FROM DUAL UNION ALL
  SELECT   3,  4, 1 FROM DUAL UNION ALL
  SELECT   2,  4, 1 FROM DUAL UNION ALL
  SELECT   8, 17, 1 FROM DUAL UNION ALL
  SELECT 248,  7, 1 FROM DUAL UNION ALL
  SELECT   1,  3, 0 FROM DUAL UNION ALL
  SELECT   6,  4, 0 FROM DUAL UNION ALL
  SELECT   8,  8, 0 FROM DUAL)
-- End test cases
SELECT T.*, -SIGN(BITAND(x, y)) + 1 result FROM T
/