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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
Ruby 3.1.0, 274 259 249 bytes -10 bytes thanks to Paradox in the Ruby discord BEGIN{END{alias module class begin rescue retry ensure end if self elsif defined? for$.in true do super while f...
brainfuck, 31 bytes ++++++++++[[>]>-[<+>---]<[.<]>] Try it online!
Haskell, 35 bytes main=mapM putStrLn$iterate('*':)"*" Attempt This Online! 55 -> 39, with orthoplex's idea. 39 -> 35 from orthoplex.
Lua 5.4.4, 99 bytes _=0s:gsub('.',function(c)_=({d=_-1,i=_+1;s=_*_})[c]or print(_)or _ _=({[256]=0,[-1]=0})[_]or _ end) Try it online! Thanks to @orthoplex for more shortening.
MATL, 16 bytes
1`1yh1
Given a number
Simple challenge: Read a byte, swap bit 7 with 0, 6 with 1, 5 with 2 and 4 with 3 and then output the byte. For example the byte 0b1001'1100 would turn into 0b0011'1001. Rules: Shortest code wi...
Sclipting, (UTF-16) 26 bytes 갠긂갠밂乘감뒄뀢감雙갿및剩 Uses a formula I found here unsigned char b; // reverse this (8-bit) byte b = (b * 0x0202020202ULL & 0x010884422010ULL) % 1023;
I was surprised by your post, as I don't recall any discussions about changing that, and then I wondered how to change that. I looked around in the site-level and category-level settings both here...
On most Codidact communities the posts list has a "Random" button among the sorting buttons: The only exceptions are the Codidact Proposals community, and Code Golf: This suggests that it is ...
Python (3.6 and up), 32 bytes (unsigned integer I/O) lambda x:int(f'{x:08b}'[::-1],2) Input and output are int objects. This also reverses the bits in integers larger than 255, implicitly infer...
Ruby, 53 51 bytes ->i{i.chars.sort_by{"tibdfghklpqyj".index(_1)||-1}} Try it online! Works in Ruby 2.7 and Ruby 3. Explanation ->i{...} is a short way to define a 1-argument lambda...
ARM Thumb machine code, 6 bytes 0: fa90 f0a0 rbit r0, r0 // reverse bits in 32-bit register 0 4: 4770 bx lr // return if bytes have to be octets (8-bit): ARM Thumb machine code...
Input is a number, you have to decide if it is part of the mandelbrot set or not, after at least 16 iterations.
This is done by applying this formula:
Python 3, 68 bytes def g(n): while True:print(n);d=[*map(int,str(n))];n+=min(d)+max(d) Outputs the n-based sequence indefinitely, starting with n.
Dyalog APL, 14 bytes
{6-+/⍵*⍨6÷⍨⍳5}
Not bruteforce! An exact implementation of the formula
Dyalog APL, 9 bytes +/⊢,2 2≡⊢ Takes the input as a pair +/ sum reduce ⊢ the input , concatenated with 2 2≡⊢ whether the input is equal to the list 2 2 (1 if true, 0 if false)
SageMath, 68 66 64 Byte. 62 if you don't count the m= g=gcd;m=lambda a,b,c:min(g(a,b),g(a,c),g(b,c))<2or g(g(b,c),a)>1 Returns False for borromean coprimes and True for all other natural ...
You roll
jq, 38 bytes
while(1;.+("\(.)"|explode|max+min-96))
Try it online!
prints the infinite sequence starting with
Sclipting, (UTF-16) 2 bytes 反 Yay for builtins~
Background Inspired by this challenge that is also a mathematical English translator. Challenge Write a program that translates a mathematical expression using English with the following specifi...
APL (Dyalog Unicode), 1 byte / Try it online! / is called Replicate.
