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There's a cool magic trick that works using the power of binary. The effect of the trick is as follows:

  1. An audience member chooses some natural number in the range of 1 to x where x is chosen by the magician.

  2. The magician hands the audience member some special cards. Each card contains some numbers from 1 to x.

  3. The audience member selects the cards which contain their number.

  4. Almost instantly, the magician can determine the original number selected.


Specification:

The numbers used for the cards are determined based on binary place value. Each card is first labeled with a power of 2. The first card becomes 1, the second becomes 2, the third becomes 4, and so on.

From now on, I will refer to card n as the card labeled with n.

To determine whether a number k is on card n, determine whether k in binary has at 1 at place value n. Consider the numbers k=13 and n=4.

K in binary is 1101. The second digit (n=4) is 1, so k=13, n=4 is a valid combination.

Goal:

Given two natural numbers 0 < n < 128 and 0 < k < 128, determine whether n appears on card k. Any reasonable input and output is allowed. Standard loopholes are banned.

This is code-golf, so the fewest bytes wins.

Test cases

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  • 4
    \$\begingroup\$ Better add the test cases directly here. \$\endgroup\$ – Mr. Xcoder Aug 15 '17 at 14:19
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    \$\begingroup\$ Also, what is determine whether k in binary has at 1 at place value n supposed to mean? Did you mean that the character in the binary representation of k at index n is 1? \$\endgroup\$ – Mr. Xcoder Aug 15 '17 at 14:22
  • \$\begingroup\$ @Mr.Xcoder Probably "at" should be "a" but not sure. \$\endgroup\$ – Erik the Outgolfer Aug 15 '17 at 14:23
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    \$\begingroup\$ Also, what did you mean by The second digit (n=4). Did you refer to the fourth digit? I am not sure I understand. \$\endgroup\$ – Mr. Xcoder Aug 15 '17 at 14:25
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    \$\begingroup\$ Does the truthy/falsy value have to be consistent? \$\endgroup\$ – Erik the Outgolfer Aug 15 '17 at 14:34
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Pyth, 5 bytes

._.&E

Try it here.

Notice the cute face ._.


How?

._.&EQ  -  Q means input and is implicit.

  .&    - Bitwise AND between:
    E     - The second input and
     Q    - The first input.
._      - Sign. 0 if it equals 0, 1 otherwise.

If inconsistent values are allowed:

Pyth, 3 bytes

No cute face this time ._.

.&E

Try it here.

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  • 2
    \$\begingroup\$ +2 for cute face, -1 for removing it \$\endgroup\$ – IsThisJavascript Aug 15 '17 at 15:25
1
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Jelly, 1 byte

&

Try it online!

Bitwise AND Builtin

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  • \$\begingroup\$ After I have seen your comments, I am not sure enitrely you understood. binary place value is 0 for 1, 1 for 2, 2 for 4, 3 for 8 (2^binary_place_value = n). Is that what you do in your answer? \$\endgroup\$ – Mr. Xcoder Aug 15 '17 at 14:30
  • \$\begingroup\$ @Mr.Xcoder Now I am not entirely side I understand your comment :P I'm fairly sure this is correct since it checks if the 4s place is truthy, not if the 4th digit from the right is truthy. \$\endgroup\$ – HyperNeutrino Aug 15 '17 at 14:31
  • \$\begingroup\$ @Mr.Xcoder The 4s place is the place where it adds 4 if it's 1, so 2^2. 2^n = binary_place_value \$\endgroup\$ – Erik the Outgolfer Aug 15 '17 at 14:32
  • \$\begingroup\$ If this is the intended input, then certainly simply & will do since the range is [1,x] and "Any reasonable input and output is allowed." implies truthy/falsey is fine. \$\endgroup\$ – Jonathan Allan Aug 15 '17 at 14:46
  • \$\begingroup\$ @JonathanAllan Cool, thanks. \$\endgroup\$ – HyperNeutrino Aug 15 '17 at 14:47
0
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05AB1E, 2 bytes

Try it online!

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0
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Proton, 3 bytes

(&)

Try it online!

(Waiting for pull from TIO before it works on TIO but you can verify it by cloning the repository)

Functional operator of the Bitwise AND operator

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