@DJMcMayhem Generic algorithm generation: given m, n, and k at runtime, generate a program that can beat 80% of the other competitors. Shortest program in bytes wins

I am really loving Dlang

I should make that a challenge ^^

@noɥʇʎԀʎzɐɹƆ nice

Something like that, anyway

@noɥʇʎԀʎzɐɹƆ But wait, wouldn't people's answers be randomly invalidated due to some other answer a year later that makes the other answer fail the requirement (beat 80% of the other competitors.)

@noɥʇʎԀʎzɐɹƆ But it'd be annoying, right?

My idea is to make the winning criteria KotH

CMC: Given a floating point number r representing a radius, consider the circle with radius r centered at the origin. Return a list of lattice points that lie within the circle but which have Von Neumann neighbors that do not. You must do this in O(N) time.

@EsolangingFruit What is a Von Neumann neighbor?

@MilkyWay90 Neighbors above and below and to the left and right.

@EsolangingFruit So basically points in the circumference of the circle?

Yes, the points will enclose the circle.

okay

Not very throughly golfed, but 141 bytes

n=>{for(int i=0;i<n*n;i++)if(i/n*(i/n)+i%n*i%n<n*n&&"".Any(x=>(i%n+x%6-4)*(i%n+x%6-4)+(i/n+x%5-2)*(i/n+x%5-2)>n*n))WriteLine((i/n,i%n));}

C#

Actually 137, n=>{for(int i=0;i<n*n;i++)if(i/n*(i/n)+i%n*i%n<n*n&&"".Any(x=>(i%n+x%6-4)*(i%n+x%6-4)+(i/n+x%5-2)*(i/n+x%5-2)>n*n))WriteLine((i/n,i%n));}

I mean n=>{for(int i=0;i<n*n;i++)if(i/n*(i/n)+i%n*i%n<n*n&&"".Any(x=>(i%n+x%6-4)*(i%n+x%6-4)+(i/n+x%5-2)*(i/n+x%5-2)>n*n))Print((i/n,i%n));}

^ That's obviously at least n^2. Invalid.

@EsolangingFruit so, midpoint circle algorithm?

@EsolangingFruit do u have test case?

... who was removed?

ah, well if they were in fact fraudulent votes then I guess I wasn't entitled to them anyway

@Neil You can't be sure if they vote on random posts to try to make their activity seems legit.