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caird coinheringaahing
caird coinheringaahing
12:12 AM
I got 15 bytes but I think I can get rid of the no-ops
I'll try again tomorrow, gtg
 
MD XF
MD XF
Huh not bad
There's no way to remove the no-ops with the current method, it's one byte too long
Use i (INCREMENT) instead of 1 +, gets 10 bytes. Who should post it?
 
caird coinheringaahing
caird coinheringaahing
You can if you want. I just realised that I could do it in 10 before looking here.
Plus I'm sure the community will be glad to have you back, even for 1 week
 
MD XF
MD XF
Heh I don't think I am was that involved in the community. I could make it CW, since we collaborated
 
caird coinheringaahing
caird coinheringaahing
@MDXF yeah but then you don't get rep. You take it. I just gave you the idea
 
MD XF
MD XF
¯\_(ツ)_/¯ if you say so. Thanks :D
0
A: Sum of all integers from 1 to n

MD XFTriangular, 10 bytes $\:_%i/2*< Ungolfed: $ \ : _ % i / 2 * < Try it online! The code, without directionals, is read as $:i*2_%. $ reads an integer x, stack contains {x}. : duplicates it, stack contains {x,x}. i increments the top of stack, stack contains {x,x+1}. * multiplies the ...

 

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