Logic Gates Manually

Question

Make a program that simulates the basic logic gates.

Input: An all-caps word followed by 2 1 digit binary numbers, separated by spaces, such as OR 1 0. The gates OR, AND, NOR, NAND, XOR, and XNOR are needed.

Output: What the output of the entered logic gate would be given the two numbers: either 1 or 0.

Examples:
AND 1 0 becomes 0
XOR 0 1 becomes 1
OR 1 1 becomes 1
NAND 1 1 becomes 0

This is codegolf, so the shortest code wins.

Answer 1

Jelly, 13 10 bytes

OSị“ʋN’BŻ¤

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Port of Peter Taylor's answer.

Answer 2

Python 2, 38 bytes

lambda s:sum(map(ord,s))*3%61%37%9%7%2

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A good ol' modulo chain applied to the sum of ASCII values of the input string, making for a solution that's just overfitting. The total ASCII value is distinct for each possible input, except that those with 0 1 and 1 0 give the same result, which is works out because all the logic gates used are symmetric.

The *3 separates out otherwise-adjacent values for inputs that different only in the bits, since these make it hard for the mod chain to split up. The length and size of numbers in the mod chain creates roughly the right amount of entropy to fit 18 binary outputs.

A shorter solution is surely possible using hash(s) or id(s), but I avoided these because they are system-dependent.

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Python 2, 50 bytes

lambda s:'_AX0NRD'.find((s*9)[35])>>s.count('0')&1

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A slightly more principled solution. Each logic gate gives a different result for each count of zeroes in the input, encodable as a three-bit number from 1 to 6. Each possible logic gate is mapped to the corresponding number by taking (s*9)[35], which are all distinct. For OR, this winds up reading one of the bits so the character could be 0 or 1, but it turns out to work to check if it's 0, and a 1 will correctly give a 1 result anyway.

Answer 3

JavaScript (ES6), 39 bytes

s=>341139>>parseInt(btoa(s),34)%86%23&1

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How?

We can't parse spaces with parseInt(), no matter which base we're working with. So we inject a base-64 representation of the input string instead. This may generate = padding characters (which can't be parsed with parseInt() either), but these ones are guaranteed to be located at the end of the string and can be safely ignored.

We parse as base \$34\$ and apply a modulo \$86\$, followed by a modulo \$23\$, which gives the following results. This includes inaccuracies due to loss of precision. The final result is in \$[0..19]\$, with the highest truthy index being at \$18\$, leading to a 19-bit lookup bitmask.

 input      | to base-64     | parsed as base-34 | mod 86 | mod 23 | output
------------+----------------+-------------------+--------+--------+--------
 "AND 0 0"  | "QU5EIDAgMA==" |  1632500708709782 |   26   |    3   |    0
 "AND 0 1"  | "QU5EIDAgMQ==" |  1632500708709798 |   42   |   19   |    0
 "AND 1 0"  | "QU5EIDEgMA==" |  1632500708866998 |   34   |   11   |    0
 "AND 1 1"  | "QU5EIDEgMQ==" |  1632500708867014 |   50   |    4   |    1
 "OR 0 0"   | "T1IgMCAw"     |     1525562056532 |   52   |    6   |    0
 "OR 0 1"   | "T1IgMCAx"     |     1525562056533 |   53   |    7   |    1
 "OR 1 0"   | "T1IgMSAw"     |     1525562075028 |   58   |   12   |    1
 "OR 1 1"   | "T1IgMSAx"     |     1525562075029 |   59   |   13   |    1
 "XOR 0 0"  | "WE9SIDAgMA==" |  1968461683492630 |   48   |    2   |    0
 "XOR 0 1"  | "WE9SIDAgMQ==" |  1968461683492646 |   64   |   18   |    1
 "XOR 1 0"  | "WE9SIDEgMA==" |  1968461683649846 |   56   |   10   |    1
 "XOR 1 1"  | "WE9SIDEgMQ==" |  1968461683649862 |   72   |    3   |    0
 "NAND 0 0" | "TkFORCAwIDA=" | 61109384461626344 |   62   |   16   |    1
 "NAND 0 1" | "TkFORCAwIDE=" | 61109384461626350 |   70   |    1   |    1
 "NAND 1 0" | "TkFORCAxIDA=" | 61109384461665650 |   64   |   18   |    1
 "NAND 1 1" | "TkFORCAxIDE=" | 61109384461665656 |   72   |    3   |    0
 "NOR 0 0"  | "Tk9SIDAgMA==" |  1797025468622614 |   76   |    7   |    1
 "NOR 0 1"  | "Tk9SIDAgMQ==" |  1797025468622630 |    6   |    6   |    0
 "NOR 1 0"  | "Tk9SIDEgMA==" |  1797025468779830 |   84   |   15   |    0
 "NOR 1 1"  | "Tk9SIDEgMQ==" |  1797025468779846 |   14   |   14   |    0
 "XNOR 0 0" | "WE5PUiAwIDA=" | 66920415258533864 |    0   |    0   |    1
 "XNOR 0 1" | "WE5PUiAwIDE=" | 66920415258533870 |    8   |    8   |    0
 "XNOR 1 0" | "WE5PUiAxIDA=" | 66920415258573170 |    2   |    2   |    0
 "XNOR 1 1" | "WE5PUiAxIDE=" | 66920415258573176 |   10   |   10   |    1

Answer 4

CJam (13 bytes)

q1bH%86825Yb=

Assumes that input is without a trailing newline.

Online test suite

This is just a simple hash which maps the 24 possible inputs into 17 distinct but consistent values and then looks them up in a compressed table.

Python 2 (36 bytes)

lambda s:76165>>sum(map(ord,s))%17&1

This is just a port of the CJam answer above. Test suite using xnor's testing framework.

Answer 5

05AB1E, 13 12 10 8 bytes

ÇO₁*Ƶï%É

Port of @mazzy's alternative calculation mentioned in the comment on his Powershell answer (*256%339%2 instead of *108%143%2).

Try it online or verify all test cases.

Explanation:

Ç            # Convert each character in the (implicit) input to a unicode value
 O           # Sum them together
  ₁*         # Multiply it by 256
    Ƶï%      # Then take modulo-339
        É    # And finally check if it's odd (short for %2), and output implicitly

See this 05AB1E tip of mine (section How to compress large integers?) to understand why Ƶï is 339.

Answer 6

Perl 6, 20 bytes

*.ords.sum*108%143%2

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A port of maddy's approach. Alternatively, *256%339%2 also works.

Perl 6, 24 bytes

*.ords.sum*3%61%37%9%7%2

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A port of xnor's answer. I'll have a go at finding a shorter one, but I think that's probably the best there is.

Answer 7

Charcoal, 32 bytes

§01÷⌕⪪”＆⌈4Ｙ⍘ＬH⦄vü|⦃³Ｕ}×▷” Ｓ∨⁺ＮＮ⁴

Try it online! Link is to verbose version of code. Explanation: The compressed string expands to a list of the supported operations so that the index of the given operation is then shifted right according to the inputs and the bit thus extracted becomes the result.

XOR     001
AND     010
OR      011
NOR     100
NAND    101
XNOR    110
inputs  011
        010

74-byte version works for all 16 binary operations, which I have arbitrarily named as follows: ZERO AND LESS SECOND GREATER FIRST XOR OR NOR XNOR NFIRST NGREATER NSECOND NLESS NAND NZERO.

§10÷÷⌕⪪”＆↖VρＳ´↥cj/v⊗J[Rf↓⪫？9ＫO↘Ｙ⦄;↙W´Ｃ>η=⁴⌕✳AＫＸIＢ｜⊖\`⊖:Ｂ�Ｊ/≧vF@$h⧴” Ｓ∨Ｎ²∨Ｎ⁴

Try it online! Link is to verbose version of code.

Answer 8

J, 21 bytes

2|7|9|37|61|3*1#.3&u:

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Port of xnor's Python 2 solution.

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J, 30 bytes

XNOR=:=/
NAND=:*:/
NOR=:+:/
".

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Some bit of fun with eval ". and standard library (which already includes correct AND, OR, XOR).

---

J, 41 bytes

({7 6 9 8 14 1 b./)~1 i.~' XXNNA'E.~_2&}.

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More J-style approach.

How it works

A very general J trick is hidden here. Often, the desired function has the structure "Do F on one input, do H on the other, and then do G on both results." Then it should operate like (F x) G H y. In tacit form, it is equivalent to (G~F)~H:

x ((G~F)~H) y
x (G~F)~ H y
(H y) (G~F) x
(H y) G~ F x
(F x) G H y

If G is an asymmetric primitive, just swap the left and right arguments of the target function, and we can save a byte.

Now on to the answer above:

({7 6 9 8 14 1 b./)~1 i.~' XXNNA'E.~_2&}.

1 i.~' XXNNA'E.~_2&}.  Processing right argument (X): the operation's name
                _2&}.  Drop two chars from the end
1 i.~' XXNNA'E.~       Find the first match's index as substring
                       Resulting mapping is [OR, XOR, XNOR, NOR, NAND, AND]

7 6 9 8 14 1 b./  Processing left argument (Y): all logic operations on the bits
7 6 9 8 14 1 b.   Given two bits as left and right args, compute the six logic functions
               /  Reduce by above

X{Y  Operation on both: Take the value at the index

Answer 9

Mathematica, 55 bytes

Symbol[ToCamelCase@#][#2=="1",#3=="1"]&@@StringSplit@#&

Pure function. Takes a string as input and returns True or False as output. Since Or, And, Nor, Nand, Xor, and Xnor are all built-ins, we use ToCamelCase to change the operator to Pascal case, convert it to the equivalent symbol, and apply it to the two arguments.

Answer 10

Powershell, 36 34 bytes

Inspired by xnor, but the sequence *108%143%2 is shorter then original *3%61%37%9%7%2

$args|% t*y|%{$n+=108*$_};$n%143%2

Test script:

$f = {

 $args|% t*y|%{$n+=108*$_};$n%143%2
#$args|% t*y|%{$n+=3*$_};$n%61%37%9%7%2   # sequence by xnor

}

@(
    ,("AND 0 0", 0)
    ,("AND 0 1", 0)
    ,("AND 1 0", 0)
    ,("AND 1 1", 1)
    ,("XOR 0 0", 0)
    ,("XOR 0 1", 1)
    ,("XOR 1 0", 1)
    ,("XOR 1 1", 0)
    ,("OR 0 0", 0)
    ,("OR 0 1", 1)
    ,("OR 1 0", 1)
    ,("OR 1 1", 1)
    ,("NAND 0 0", 1)
    ,("NAND 0 1", 1)
    ,("NAND 1 0", 1)
    ,("NAND 1 1", 0)
    ,("NOR 0 0", 1)
    ,("NOR 0 1", 0)
    ,("NOR 1 0", 0)
    ,("NOR 1 1", 0)
    ,("XNOR 0 0", 1)
    ,("XNOR 0 1", 0)
    ,("XNOR 1 0", 0)
    ,("XNOR 1 1", 1)

) | % {
    $s,$e = $_
    $r = &$f $s
    "$($r-eq$e): $s=$r"
}

Output:

True: AND 0 0=0
True: AND 0 1=0
True: AND 1 0=0
True: AND 1 1=1
True: XOR 0 0=0
True: XOR 0 1=1
True: XOR 1 0=1
True: XOR 1 1=0
True: OR 0 0=0
True: OR 0 1=1
True: OR 1 0=1
True: OR 1 1=1
True: NAND 0 0=1
True: NAND 0 1=1
True: NAND 1 0=1
True: NAND 1 1=0
True: NOR 0 0=1
True: NOR 0 1=0
True: NOR 1 0=0
True: NOR 1 1=0
True: XNOR 0 0=1
True: XNOR 0 1=0
True: XNOR 1 0=0
True: XNOR 1 1=1

Answer 11

JavaScript (Node.js), 106 94 bytes

x=>([a,c,d]=x.split` `,g=[c&d,c^d,c|d]["OR".search(a.slice(1+(b=/N[^D]/.test(a))))+1],b?1-g:g)

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Link to the code and all 24 cases.

+9 for forgotten to map the XNOR case.

Answer 12

Java 10, 30 28 bytes

s->s.chars().sum()*108%143%2

Port of @mazzy's Powershell answer.

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Answer 13

JavaScript (Node.js), 45 bytes

Just a port of xnor's superb Python 2 answer posted upon consent, please give that answer upvotes instead of this.

x=>Buffer(x).reduce((a,b)=>a+b)*3%61%37%9%7%2

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Answer 14

Attache, 55 bytes

{Eval!$"${Sum!Id''Downcase!SplitAt!_}${N=>__2}"}@@Split

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A rather brutish solution. Converts the input to the relevant Attache command and evaluates it. (Attache has built-ins for each of the 6 logic gates.)

Answer 15

Ruby, 20 bytes

->s{76277[s.sum%17]}

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How it works:

Basically the same as Peter Taylor's answer, but Ruby makes it easier. The magic number is different but the idea was the same.