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Challenges

Can you give me half?

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Challenge idea taken from: Eliseo D'Annunzio

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.

One example that fits the brief would be the following:

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

Scoring

This is a code challenge, where scoring is done in terms of number of unique characters used in the submission. Ties are broken by lowest bytecount.

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+5
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dc, 3 unique bytes, 138 bytes

This challenge is pretty easy if you aim for just 4 unique bytes. For example, you could push the stack depth repeatedly to get ascending integers, set the precision to something positive, divide and then print.

zzzzk/p


And surely it can't go any lower, right? After all, we have to increase the precision (k), print (p, n, f), do a floating-point operation (v, ^, /, .), and still get a number from somewhere...

Luckily though, there is actually a narrow path forward thanks to a and x. a pops a number off the stack, casts it to a long, and pushes its LSB back as a single-character string. Using x we can then execute that string as a command. This leaves us to choose a third command which should give us some helpful numbers that we can feed into a. The only options that can give us a range of numbers are z as well as the hex digits A through F. I haven't found a way to tame z, since it buries every intermediate result you get, so let's just see what useful characters we can get if we choose one of the hex digits.

Program Resulting character
AAa n
AAAAa f
AAAAAAa F
CCCa 4
DDDDa k
FFFFFFa i

At first, this might not look too promising, but if we choose F as our third command, we can use the i to change the input base to 15 and basically get a reroll on our commands. 🤞

Program Resulting character
FxFFFFFFaxFFFFFa /
FxFFFFFFaxFFFFFFFa ?

The ? doesn't help us, since we don't get any input, the / however is game-changing. By carefully placing the right constants on the stack before and after the base change we can later use / to efficiently calculate the ASCII value of any command. We use this to first set the precision to 15 and then sum, divide and print three pre-placed Fs (as in F k F F F + / n) to do the actual calculation.

Finally, here is an annotated version of the complete program:

F F F         # three 15s for calculating 0.5
FFFFFFFFFFFFF # "+" numerator
FFFFFF        # "+" denominator 1
F             # argument for k
FFFFFFFFFFF   # "k" numerator
FFFFF         # "k" denominator 1
F             # argument for i
FFFFFFax      # execute "i" to switch base
FFFFFax       # "k" division 1
F             # "k" denominator 2
FFFFFax       # "k" division 2
ax            # execute "k" to set precedence
FFFFFax       # "+" division 1
F             # "+" denominator 2
FFFFFax       # "+" division 2
ax            # execute "+" to add two 15s
FFFFFax       # divide 15 by 30
FFFFFFFFFFFF  # "n" numerator 1
FFF           # "n" denominator 1
FFFFFax       # "n" division 1
FFFF          # "n" denominator 2
FFFFFax       # "n" division 2
ax            # execute "n" to print the result


We can get rid of any whitespace by placing xs between constants since they act as no-ops on numbers.

FxFxFxFFFFFFFFFFFFFxFFFFFFxFxFFFFFFFFFFFxFFFFFxFxFFFFFFaxFFFFFaxFxFFFFFaxaxFFFFFaxFxFFFFFaxaxFFFFFaxFFFFFFFFFFFFxFFFxFFFFFaxFFFFxFFFFFaxax


Try it online!

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+4
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Embed ESCR, 6 unique characters, 19 bytes

[/ [*] [+ [*] [*]]]

Each [] pair of brackets encloses an in-line function. The first token in a function is the function name, and subsequent tokens, if any, the parameters. Functions can be nested.

The * function multiplies a series of numbers. The product is initialized to 1 before the first term. A side-effect is that just "*" without any parameter returns 1. The "+" function adds two 1s to make 2. The "/" function divides the first parameter by the second, so 1 divided by 2. The result is 1/2.

The spaces are enforced for readability, and to simplify the parsing rules. This NOT a golfing language.

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comment (1 comment)
+3
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Python 3, 7 distinct characters, 11 bytes

True/-~True


Try it online!

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+3
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Perl, 3 distinct, 12 bytes

s//yy//y/y//


Try it online!

Prepends yy to the start of \$_, and then transliterates them, dividing the count of each.

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+3
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C (gcc), 8 7 unique

cos(cos-cos)/(cos(cos-cos)-(-cos(cos-cos)))

Try it online!

-1 unique character thanks to @orthoplex‭

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+3
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Ruby, 7 unique, 12 bytes

$$.fdiv$$+$$ Ruby, 8 unique, 13 bytes: $$.fdiv $$+$$

Ruby, 9 unique, 22 bytes:

"i".size.fdiv"ii".size

Ruby, 9 unique, 24 bytes:

Math::E./Math::E+Math::E

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+3
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Clojure - 15 characters total, 5 unique

Inspired by the Embed ESCR solution above, I realized I could do something very similar in Clojure:

(/(*)(+(*)(*)))


This returns 1/2, which is a Ratio in Clojure.

(*) - * with no arguments returns 1.

(+...) - + returns the sum of its arguments

(/...) - / divides the first argument by the second, returning a Ratio.

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+3
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J, 4 3 distinct, 6 bytes

-:_-:_


Half of _ -: _ which is asks if infinity matches infinity. Thanks to @xash in the APL farm discord for saving the 3 unique bytes and orthoplex for catching my blunder.

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%#_ _ has unique 4 bytes, no? (2 comments)
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Javascript, 5 Distinct characters, 9 Bytes

~[]/~-~[]

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+2
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Scala, 15 bytes

'#'.toFloat/'F'


Try it in Scastie!

Characters in Scala kinda get casted to integers. This is usually annoying, but it's useful when golfing. # is 35 and F is 70, and the .toFloat prevents it from doing integer division and truncating to 0.

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+2
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Lua 5.4, 3 unique characters

#'#'/#'##'


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[...] scoring is done in terms of number of unique characters used in the submission.

1. Hashtag #
2. Single quote '
3. Slash /
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+2
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Julia 1.0, 16 bytes

(o=one(Int))/-~o


Try it online!

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+2
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Japt, 1 byte

½


Test it

Jelly, Vyxal, 1 byte

.


Try it online!

Yeah, golf langs still kind of kill this challenge.

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I think the second works in 05AB1E too (3 comments)
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Python 3, 5 unique bytes, 28 total

(()==())/((()==())+(()==()))


(, ), =, /, +

Uses (()==()) to get True with only 3 unique characters, as well as making the later grouping of (...+...) free.

Try it online!

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+2
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Raku (also Perl polyglot), 5 unique, 13 bytes

!()/(!()+!())


() or empty list coerces to its length which is 0 in math operations and !0 is True which coerces to 1

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+2
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Ruby, 10 distinct, 35 chars

t=/%/=~%/=%/
o=/%/=~%/==%/
t/o.to_f

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+2
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JS, 15 bytes

!![]/(-~[]-~[])

Basically typed random characters into js console and it worked.

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+1
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J, 4 unique, 4 char

-:=_


How it works:

-: results in half of what's on the right
= results in signum of what's on the right
_ is infinity

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+1
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MATL, 3 distinct, 3 total

lH/


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Computes 1/2. Reserved labels for constants can be found here

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+0
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Vyxal, 1 byte

.


Try it Online!

Very simple 1 byter

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