CMC: print the sequence. only print n when if the first n can perfectly divided by first number of n then n is divided by first num of n and repeat until the length of n is 1.
Print random characters indefinitely
Task
Continuously print a random character (a-z, A-Z, 0-9) not separated by a newline (\n).
Expected output
b7gFDRtgFc67h90h8H76f5dD55f7GJ6GRT86hG7TH6T7302f2f4 ...
Note: output should be randomised.
Requirements
Output must be continuous (i.e. never endin...
@Fmbalbuena there's no reason to specifically call out forbidden loopholes - unless you have some to suggest, that is. you don't have to specify forbidden loopholes it's just a good idea if there's any major ones
why are you so desperate for my answer dude? chill
i'm pretty sure that sequence can be reduced to something simpler to check against. since it could be broken down to some form of floor(x/10) in some cases, or smth like that
it also seems familiar as a problem statement (not quite this one but there's some sort of sequence derived from first digits that i can't remember right now)
i mean there's probably some easier way to determine if a number meets that criteria recursively than just doing that recursively
also, what does 'no, not because different' mean on the sandbox comment? it's not the same challenge but i believe that sandboxed post is just a more specific form of the one I linked
and that there's not really any new optimizations that could be applied to the smaller one
actually nah this goes as a comment
@Fmbalbuena this is also an annoying thing to test without a set of test cases. part of why i'm beginning to think there's a better way to do this is how rare numbers that meet those criteria seem to be
Background
Slowsort is an in-place, stable sorting algorithm that has worse-than-polynomial time complexity. The pseudocode for Slowsort looks like this:
procedure slowsort(A[], i, j) // Sort array range A[i ... j] in-place.
if i ≥ j then
return
m := floor( (i+j)/2 )
...
CMC print the sequence. only print n when if the first n can perfectly divided by first number of n plus 1 then n is divided by first num of n and repeat until the length of n is 1.
only diff is n can perfectly divided by first number of n plus 1
Implement xorshift128+
I don't have time to finish writing this draft at the moment.
Might also consider something like xoroshiro if it's more interesting to implement, but since xorshift128+ is so common it seems like a clearer choice.
(Possible idea: Given initial seed, determine number of iter...
@emanresuA I was somewhat busy today, but I assure you we'll have >25 candidates by tomorrow. I guess we'll have to choose which one to use or discard after that.
@PyGamer0 if you want to go the Vyxal route, where the codepage is fairly meaningful (not just random letters), then create the language first and make the codepage fit around it
...went to look up a good canonical example of pro-drop and til that the pro stands for pronoun obvious in retrospect but all along i thought it was like pro-drop versus anti-drop
I wish all of you a reflective advent season. If anyone like to play some code golf, here your challenge: Write the query that generate a similar christmas tree in less signes than I did. But also any other generated Christmas ASCII art in SQL is welcome.
WITH recursive cnst as (
SELECT 32 as...
Ok, 11260 bytes in my language "99 bottles of beer on the wall"
The way the language works: If the program is the text of "99 bottles of beer on the wall", print the text of "99 bottles of beer on the wall" Otherwise, interpret it as golfscript
@Fmbalbuena if anybody wants to solve this one in piet (interpreting beer-shaped as 'code when viewed as image looks like beer') i'll buy you a beer when legal/appropriate