Search
C (gcc), 43 38 bytes f(i){return i>1?f(i%2?3*i+1:i/2)+1:0;} Attempt This Online! Credits to @Moshi for shortening the code.
Lundin just suggested in a comment under another question that I tag the challenge Looping Counter as kolmogorov-complexity. Now I'm not sure if it actually qualifies for that tag, for the followi...
Wolfram Language (Mathematica), 38 bytes Tr[1^ResourceFunction["Collatz"]@#]-1& Don't Try it online! Doesn't work on TIO due to using the Collatz builtin which needs to be downloaded from ...
C (gcc), 48 43 bytes s;r(n,m){s+=rand()%m+1;return--n?r(n,m):s;} Try it online! Previous 48 bytes version using loop: i,s;r(n,m){for(;i<n;i++)s+=rand()%m+1;return s;}
Python 3, 59 bytes lambda n,m:sum(choices(range(m),k=n))+n from random import* Try it online!
Zsh, 61 bytes n=2;s=s=%q\;printf\ n=\$n\$n\\\;\$s\ \$s;printf n=$n$n\;$s $s Attempt This Online! A trivial modification of this quine: s=s=%q\;printf\ \$s\ \$s;printf $s $s. n goes from 2 to 22 ...
J, 35 char <:#-:`(>:@:*&3)@.(2&|)^:(1&<)^:(<_) How it works: NB. <: subtract one from number result on right NB. # count number of items from list result on righ...
HQ9+, 2 bytes Two quines. QQ
J, 13 char '*',^:(<_)'*' How it works: What's in parenthesis indicates to the verb ^: that the verb to the left of ^: has to be performed to the object on the right of the ^: an infinite nu...
J, 7 char *./~:q: How it works: 'q:' produces prime factors of number on right '~:' replaces the first instance of unique numbers with a 1, the rest 0 '*./' tests for all ones Sample runs: ...
J, 29 char ('!',~'Hello, ',1!:1<1)1!:2<4 Sample run ('!',~'Hello, ',1!:1<1)1!:2<4 torres Hello, torres!
J, 46 char 'ABCDE'{~<:0 128 192 224 240 I.>:".{.}.;.1'.', Sample runs 'ABCDE'{~<:0 128 192 224 240 I.>:".{.}.;.1'.', '127.255.255.255' A 'ABCDE'{~<:0 128 192 224 240 I...
Python 3, 38 36 33 bytes lambda x:{n:x.count(n)for n in x} -2 bytes thanks to @Razetime Another -3 bytes thanks to @orthoplex Attempt it online!
J, 8 bytes ~.,:#/.~ Tacit function, this is the de facto method for this problem in J. Attempt it online!
J, 24 char ([: +/ [: (* +/\ )"1 ~. ="0 1 ]) Sample Runs ([:+/[:(*+/\)"1~.="0 1]) 1 1 2 2 2 1 1 1 3 3 1 2 1 2 3 3 4 5 1 2 ([:+/[:(*+/\)"1~.="0 1]) 3 7 5 4 9 2 3 2 6 6 1 1 1 1 1 1 2 ...
J, 9 bytes 1#.]=&><\ This is the 7 byte APL solve but makes use of #. in place of +/"1. I came up with 1#.]=[\ first, but bubbler pointed out it breaks when non zeros are present. At...
Python 3.8 (pre-release), 69 bytes def f(x,y={},z=[]): for i in x:y[i]=y.get(i,0)+1;z+=[y[i]] return z Try it online! Bonus: theoretical answer if python allowed named assignment with su...
Python 3, 74 bytes def f(a): d={x:0 for x in a};r=[] for x in a:d[x]+=1;r+=[d[x]] return r Try it online!
Vyxal, 4 bytes KƛtO Try it Online! Explained KƛtO Kƛ # For each prefix of the input tO # How many times does the tail occur in the prefix?
Myby, 12 5 bytes primf=primfd primf : prime factors = : equals primfd : unique prime factors Evaluated as a monadic fork in J (f y) g (h y). The test cases (retest...
Factor, 95 bytes USING: kernel ranges sequences ; IN: r : r ( n q -- n ) [ [1..b] [ ] ] dip map-reduce ; inline
Factor, 122 bytes USING: kernel sequences sequences.windowed ; IN: c : c ( s -- s ) dup length [ dup last [ = ] curry count ] rolling-map ;
Factor, 147 bytes USING: kernel math math.text.utils sequences ; IN: n : n ( n -- s ) 0 swap [ [ 1 + dup dup 1 digit-groups sum mod 0 > ] loop dup ] replicate nip ;
