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Husk, 10 bytes L►LfΛo<0-Ṗ Try it online! Explanation L►LfΛo<0-Ṗ Ṗ power set of input f get elements which match the following: Λo - all pairwise difference...
Python 3, 49 bytes lambda m,n:[[*range(n*_,n*_+n)]for _ in range(m)] Try it online!
Python 2, 46 bytes lambda m,n:[range(n*_,n*_+n)for _ in range(m)] Try it online!
C (gcc), 59 bytes i;f(m,n){while(i<m*n){putchar(i%n?32:13);putchar(48+i++);}} Attempt This Online! m is the number of rows. n is the number of columns.
BQN (CBQN), 10 bytes Anonymous function that takes m on the left and n on the right. {𝕨‿𝕩⥊↕𝕨×𝕩} ↕𝕨×𝕩 # list of range [0,m*n) 𝕨‿𝕩⥊ # reshape list to m*n Try it here!
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...
