Scala, 90 89 88 bytes

Saved 2 bytes thanks to Shaggy

n=>1 to n map{x=>val s=if(x<2)"╔╦╗"else if(x<n)"╠╬╣"else"╚╩╝";s(0)+(""+s(1))*(n-3)+s(2)}

Try it in Scastie!

I feel like there's a nicer way to repeat middle elements of a list n times, but this is all I can golf it for now.

n => //The input
1 to n //Make a range [1..n]
map{x=> //For each x in that range, make a line:
val s= //s is a string in the form "$left$inner$right"
 if(x<2)"╔╦╗" //For the top
 else if(x<n)"╠╬╣" //For the inner parts
 else"╚╩╝"; //For the bottom
s(0)+(""+s(1))*(n-3)+s(2)} //Repeat the inner part n-3 times

posted about 3 years ago

CC BY-SA 4.0

3y ago

user‭

1546 reputation 9 59 174 89

Python 3, 74 bytes

lambda n:"╔"+"╦"*(n-3)+"╗\n"+("╠"+"╬"*(n-3)+"╣\n")*(n-2)+"╚"+"╩"*(n-3)+"╝"

Try it online!

posted about 3 years ago

CC BY-SA 4.0

bastolski‭

181 reputation 0 6 18 2

C (gcc), 119 118 116 115 bytes

#define p(x,y,z) printf(i^1?i^c?#z:#y:#x)
i,j;f(c){for(i=1;i<=c;p(╗\n,╝\n,╣\n),i++)for(j=p(╔,╚,╠);j++<c;)p(╦,╩,╬);}

Try it online!

Function solution.

I started with a recursive solution but it didn't work out well... though I'm still convinced that this can be shaved down quite a bit with recursion.

EDIT:

Recursive version which currently landed at exactly 115 bytes too...

#define p(x,y,z) printf(i^1?i^c?#z:#x:#y);
c,j;f(i){for(c=c?c:i,j=p(╔,╚,╠)j++<c;)p(╦,╩,╬)p(╗\n,╝\n,╣\n)--i?f(i):0;}

posted about 3 years ago

CC BY-SA 4.0

3y ago

Lundin‭

1149 reputation 17 43 134 70

Ruby, 53 bytes

->n{n-=3;[?╔+?╦*n+?╗,*[?╠+?╬*n+?╣]*(n+1),?╚+?╩*n+?╝]}

Try it online!

posted about 3 years ago

CC BY-NC 4.0

snail_‭

681 reputation 4 28 73 18

Sclipting, (UTF-16) 80 bytes

갰減먩놔 먩놦①復먩놗겮꺕똀 먩놬①復①增疊먩놣겮꺕똀會먩놣겮꺕떠 먩놩⑴復먩놝

Explanation

Input n pushed on stack
갰減        Subtract 3
먩놔        "╔"
먩놦        "╦"
①復         String of above repeated n-3 times
먩놗겮꺕똀   "╗\n╠"

먩놬        "╬"
①復         String of above repeated n-3 times
①增疊       List of above repeated (n-3)+1 times
먩놣겮꺕똀   "╣\n╠"
會          Join the list with separater above

먩놣겮꺕떠   "╣\n╚"
먩놩        "╩"
⑴復        String of above repeated n-3 times (but finally remove n-3 from stack)
먩놝        "╝"

posted about 3 years ago

CC BY-SA 4.0

Moshi‭

1722 reputation 17 58 204 80

C (clang), 210 198 183 bytes

i,j,k,l;s(v,w,x,y,z){printf("%s",v);for(;w<x;w++){printf("%s",y);}puts(z);}main(){scanf("%i",&i);l=i-3;s("╔",j,l,"╦","╗");for(j=0;j<i-2;j++){s("╠",k,l,"╬","╣");}s("╚",j=0,l,"╩","╝");}

Try it online!

Still wondering if these characters could even count as a byte, probably two. ¯\_(ツ)_/¯

A special golfing trick I learned. If you don't want to repeat the loops, then make a function. For some reason, C (clang) is okay with such non-returning function that's not main(). Enough abuse for this.

posted about 3 years ago

CC BY-SA 4.0

3y ago

General Sebast1an‭

1211 reputation 53 97 129 328