Can you give me half? (No digits) [closed]

Question

The task is to provide code that evaluates to 0.5 numerically, i.e. the output must be recognized by your chosen language as a numeric value (Number, float, double, etc), not as a string. The catch, the characters 0 through to 9 cannot be used.

PLEASE NOTE: This IS NOT a golfing challenge, this is a popularity contest, so creative answers are encouraged. The more obscure and convoluted answers are also encouraged.

One example that fits the brief would be the following:

((++[[]][[~~""]]<<++[[]][[~~""]]))**((--[[]][[~~""]])) which works out to 0.5 in JavaScript.

A bounty of 150 will be awarded as an added incentive for the most voted creative answer. Good luck!

Any questions, feel free to ask.

Answer 1

Desmos, 23 bytes

\frac{\{\}}{\{\}+\{\}}

(newline is required)

Looks crazy, but it really isn't once you figure out that \{\} is just 1.

The actual crazy part is that it doesn't render properly on Desmos, though it still outputs 0.5.

Try It On Desmos!

Answer 2

RAMDISP, 19 bytes

[P[[[[]]][/[[]]];]]

RAMDISP is my own programming language, but i made it before this challenge, so i think it's allowed. [i also had to fix a bug i left in it, which i only found after trying to do this challenge]

explanation:

[P - pipes through the following array
  [
    [[[]]] - the value 2.
    [/ - divide
      [[]] - the value 1.
    ] - by the incoming number.
    ; - print the result
  ]
]

Answer 3

Python 3, 25 bytes

print(len("l")/len("pr"))

Try it online!

Short, sweet, and with only 11 distinct bytes too.

Answer 4

Charcoal, 57 bytes

Ｉ⁻Ｌα⊘Ｌ







https://www.youtube.com/watch?v=AyOqGRjVtls

Try it online!

Answer 5

05AB1E, 18 bytes

тT*;тT*·<ŸтT*/ΩΔ;t

Try it online!

Choose a random float in \$[0.5, 2)\$, and iterate \$x\mapsto \sqrt{x \over 2}\$ until convergence.

In theory the iteration converges to \$0.5\$ for all positive real numbers, but there might be floating point issues leading to results close to \$0.5\$.

---

05AB1E, 7 bytes

XΔNÌoz-

Try it online!

$$ 1-\sum^\infty_{k=2}{1 \over 2^k} = 1-{1 \over 2} = {1 \over 2} $$

X        # start at 1
 Δ       # until the result doesn't change:
  NÌ     #   iteration index N + 2
    o    #   2 ** (N+2)
     z   #   1 / 2**(N+2)
      -  #   subtract this from the current value

Answer 6

<>^v

≈:) :) :) :) :( :( :( )≈)?/;

≈    Duplicate top of stack (by default stack is [0])
then the :) and :( do nothing, because : executes only the next characted if the top two elements of the stack are not equal, so it skips over the parentheses.

) Increment top of stack -> stack is now [0, 1]
≈ Duplicate top of stack -> stack is now [0, 1, 1]
) Increment top of stack -> stack is now [0, 1, 2]
? Reverse stack -> stack is now [0, 2, 1]
/ Set top of stack to top of stack / second element of stack -> stack is now [0, 1, 0.5]
; Print top of stack

run online

Answer 7

Barrel, 2 bytes

½

Evaluates to, and prints, 0.5. Not necessarily terribly imaginative, but probably one of the best solutions to this problem in Barrel.

Answer 8

R, 57 bytes

function(`+`=function(x)as.double(as.roman(x)))+'I'/+'II'

Try it online!

As odd as it may sound, R has a partial support for Roman numerals.

So, in this code we:

• abuse the plus sign by turning it into a function accepting Roman numerals strings and returning the numeric value
• we aesthetically compute +I / +II (we actually calculate 1 / 2 because the abused plus sign act as a roman-to-double parser)
• then we return the result

Answer 9

Red, 82 bytes

print divide length? next system/locale/days to-float length? system/locale/months

Try it online!

Divides the number of days in a week minus one to the number of months. The current Red built doesn't need to-float.

Answer 10

R, 82 bytes

function()round(mean(runif(sqrt(exp(exp(pi)))) - rnorm(sqrt(exp(exp(pi))))),pi/pi)

Try it online!

Generates the mean of (\$\sqrt{e ^ {e ^ \pi}}\$ random uniform numbers in the range (0, 1) minus \$\sqrt{e ^ {e ^ \pi}}\$ random numbers from a normal distribution with mean 0 and standard deviation 1) and rounds it to 1 decimal place.

Answer 11

ZX Spectrum Basic, 8 bytes

PRINT SGN PI/INT EXP SGN PI

This is actually a well-known compression technique for numeric constants. Plain PRINT 0.5 would require 10 bytes (1 byte for the PRINT keyword, 3 bytes for the ASCII string 0.5, 6 bytes for the float constant)

Answer 12

Vyxal

\/C⌐\\C/-

Explanation:

\/C          . gets the codepoint of / 
   ⌐         . 1 - x
    \\C      . gets the codepoint of \
       /     . division (to get -0.5)
        -    . subtract from implicit 0

Try it Online!

Answer 13

C# (Visual C# Interactive Compiler), 30 bytes

Console.Write('$'/(float)'H');

Try it online!

Division of Ascii Codes: '$' = 36, 'H'(alf) = 72

Answer 14

Python 3, 203 bytes

((True+True)*(True+True+True)**(True+True+True+True)*(True+True+True+True+True)/(True+True+True+True)-(True+True)**(True+True+True)*(True+True+True+True+True)**(True+True))/((True+True)*(True+True)+True)

Heh.

Answer 15

Python 3

len('check')/len('check-mate')
# Output: 0.5

Answer 16

Ruby, 27 bytes

->{($.**=$.).to_f/($.<<$.)}

Try it online!

Using the variable $. which is initially set to zero, and the interesting property 0⁰==1.

Answer 17

Python 3, 14 bytes

d=True;d/(d+d)

True is basically just 1 in Python, so this is basically just

1/(1+1)

which obviously evaluates to 0.5.

Answer 18

JavaScript (Node.js), 61 bytes

~[] / ~~(Math[`SQRT${-~-~[]}`] * Math[`SQRT${-~-~[]}`]) * ~[]

Try it online!

The ~~ just before the Math.SQRT2 multiplication is needed because Math.SQRT2 * Math.SQRT2 = 2.0000000000000004.

Answer 19

Python 3

Not very creative but ehh

ord('!')/ord('B')

edit: as pointed out by EasyasPi (thanks), doesn't work for python 2

Answer 20

Dyalog APL, 5 bytes

÷⌊*≡⍬

Not very creative, but golfy! Try it online.

÷       Inverse              ->   1/2
⌊       Floor                ->   2
*       Exponent [1]         ->   2.71828
≡       Depth                ->   1
⍬       Empty list           ->   []

[1]: Default base is e, so e^1

Answer 21

JAVA

As a simple java project. I haven't looked at the others, so it is entirely possible (and likely) that this is the umpteenth answer to contain this solution. This code, placed inside src/Run.java of a fresh project, will return o.5.

The actual algorithm is only 2 (sorry, two XD) lines; The rest is just getting it to print. I know you asked for whimsical, but I was impressed with the simplicity of this, so I used it, for better or worse....

This algorithm takes advantage of the fact that any number, when divided by (itself plus itself) will give one half.

public class Run {

    public static void main(String[] args) {
        double x = Math.random();
        double y = x/(x+x);
        System.out.println(y); 
       
    }

}

Btw, Im not sure how to do the giant text in md, so I just used a title hash. Ill be back to fix once I figure it out.

EDIT: Although I guess using java for this two line solution is a bit whimsical! LOL

Answer 22

JavaScript

t=true;
console.log((t) / ((t) ^ (t + t) ^ (t)) ** (t));

JavaScript works well, here is a simple numerically deficient example getting you the half or it.