# ASCII Art Octagons

Given an input integer n > 1, output an ASCII-art octagon with side lengths composed of n characters. See examples below:

n=2
##
#  #
#  #
##

n=3
###
#   #
#     #
#     #
#     #
#   #
###

n=4
####
#    #
#      #
#        #
#        #
#        #
#        #
#      #
#    #
####

n=5
#####
#     #
#       #
#         #
#           #
#           #
#           #
#           #
#           #
#         #
#       #
#     #
#####

and so on.


You can print it to STDOUT or return it as a function result.

Any amount of extraneous whitespace is acceptable, so long as the characters line up appropriately.

### Rules and I/O

• Input and output can be given by any convenient method.
• You can use any printable ASCII character instead of the # (except space), but the "background" character must be space (ASCII 32).
• Either a full program or a function are acceptable.
• Standard loopholes are forbidden.
• This is so all usual golfing rules apply, and the shortest code (in bytes) wins.
• Can we use different output characters, or does it need to be consistent? – Emigna Nov 6 '18 at 18:13
• @Emigna Different characters are fine. – AdmBorkBork Nov 6 '18 at 18:26
• – Charlie Nov 7 '18 at 11:06

# 05AB1E, 3 bytes

7ÝΛ


Try it online!

Explanation

      # implicit input as length
# implicit input as string to print
7Ý    # range [0...7] as directions
Λ   # canvas print


See this answer to understand the 05AB1E canvas.

• Surely this should be 5 bytes? Or do code golf challenges see bytes and characters as interchangeable – Doug Nov 7 '18 at 17:10
• @Doug: It is 3 bytes in 05ab1e's code page – Emigna Nov 7 '18 at 17:11
• Oh, cool! Thanks for the docs link! – Doug Nov 7 '18 at 17:58
• >:( damnit, adnan – ASCII-only Nov 11 '18 at 5:14

# JavaScript (ES6), 114106105104 103 bytes

n=>(g=x=>v=x*2>w?w-x:x,F=x=>~y?#
[~x?(h=g(x--))*g(y)>0&h+v!=n|n>h+v:(y--,x=w,2)]+F(x):'')(y=w=--n*3)


Try it online!

### How?

This builds the output character by character.

Given the input $$\n\$$, we compute:

$$n'=n-1\\w=3n'$$

For each character at $$\(x,y)\$$, we compute $$\(h,v)\$$:

$$h=w/2-\left|x-w/2\right|\\v=w/2-\left|y-w/2\right|$$

The cells belonging to the octagon satisfy one of the following conditions:

• ($$\h=0\$$ OR $$\v=0\$$) AND $$\h+v\ge n'\$$ (in red below)
• $$\h+v=n'\$$ (in orange below)

For example, with $$\n=4\$$ (and $$\n'=3\$$):

$$\begin{matrix}(0,0)&(1,0)&(2,0)&\color{red}{(3,0)}&\color{red}{(4,0)}&\color{red}{(4,0)}&\color{red}{(3,0)}&(2,0)&(1,0)&(0,0)\\ (0,1)&(1,1)&\color{orange}{(2,1)}&(3,1)&(4,1)&(4,1)&(3,1)&\color{orange}{(2,1)}&(1,1)&(0,1)\\ (0,2)&\color{orange}{(1,2)}&(2,2)&(3,2)&(4,2)&(4,2)&(3,2)&(2,2)&\color{orange}{(1,2)}&(0,2)\\ \color{red}{(0,3)}&(1,3)&(2,3)&(3,3)&(4,3)&(4,3)&(3,3)&(2,3)&(1,3)&\color{red}{(0,3)}\\ \color{red}{(0,4)}&(1,4)&(2,4)&(3,4)&(4,4)&(4,4)&(3,4)&(2,4)&(1,4)&\color{red}{(0,4)}\\ \color{red}{(0,4)}&(1,4)&(2,4)&(3,4)&(4,4)&(4,4)&(3,4)&(2,4)&(1,4)&\color{red}{(0,4)}\\ \color{red}{(0,3)}&(1,3)&(2,3)&(3,3)&(4,3)&(4,3)&(3,3)&(2,3)&(1,3)&\color{red}{(0,3)}\\ (0,2)&\color{orange}{(1,2)}&(2,2)&(3,2)&(4,2)&(4,2)&(3,2)&(2,2)&\color{orange}{(1,2)}&(0,2)\\ (0,1)&(1,1)&\color{orange}{(2,1)}&(3,1)&(4,1)&(4,1)&(3,1)&\color{orange}{(2,1)}&(1,1)&(0,1)\\ (0,0)&(1,0)&(2,0)&\color{red}{(3,0)}&\color{red}{(4,0)}&\color{red}{(4,0)}&\color{red}{(3,0)}&(2,0)&(1,0)&(0,0)\end{matrix}$$

• Wow, this is awesome! I think $h + v \geq n'$ can be simplified to $h+v>n'$, although I'm not sure if that helps the golfing logic at all. – Giuseppe Nov 6 '18 at 22:20
• @Giuseppe It could indeed be simplified that way if both conditions were tested. But in the code, the cases $hv=0$ and $hv\neq0$ are separated. However, I'm actually testing the opposite condition ($n'>h+v$), which already is 1 byte shorter. – Arnauld Nov 6 '18 at 22:29
• @Giuseppe Your comment prompted me to have a closer look at the formula and I finally saved a byte by writing it a bit differently. :) – Arnauld Nov 6 '18 at 22:41
• heh, well your comment about $hv=0$ prompted me to go look at my port of your logic and save another couple of bytes! – Giuseppe Nov 6 '18 at 22:44

# Charcoal, 5 bytes

ＧＨ*Ｎ#


My first answer with Charcoal!

Explanation:

ＧＨ*Ｎ#      //Full program
ＧＨ          //Draw a hollow polygon
*         //with 8 sides
Ｎ       //of side length from input
#      //using '#' character


Try it online!

• For those who prefer verbose Charcoal, that's PolygonHollow(:*, InputNumber(), "#");. – Neil Nov 6 '18 at 18:51

# Canvas, 1514 12 bytes

／⁸⇵╷+×＋：⤢ｎ╬┼


Try it here!

Explanation:

/             a diagonal of length n
⁸            the input,
⇵           ceiling divided by 2, (storing the remainder)
╷          minus one
#×        repeat "#" that many times
+       append that to the diagonal
:⤢n    overlap that with its transpose
╬┼  quad-palindromize with the overlap being the remainder stored earlier


# R, 122117 115 bytes

function(n){n=n-1
m=matrix(0,y<-3*n+1,y)
v=t(h<-(w=3*n/2)-abs(row(m)-1-w))
m[h*v&h+v-n|h+v<n]=' '
write(m,1,y,,"")}


Try it online!

Ports the logic from Arnauld's answer, specifically this revision in case there are further improvements. Another 2 bytes saved thanks to Arnauld's suggestion of inverting the logic!

• -2 bytes by doing it the other way around (I can't do h*v&h+v-n in JS because & is a bitwise operator; but it's a logical one in R, so that works). – Arnauld Nov 7 '18 at 15:05
• @Arnauld thanks! – Giuseppe Nov 7 '18 at 15:39

# Python 2, 96 bytes

a=b=n=input()
while a>2-n-n:a-=1;b-=a/~-n+1;s=(-~b*' '+'#').ljust(n);print s+s[-1]*(n-2)+s[::-1]


Try it online!

# Python 2, 81 bytes

a=d=n=input()-1
while a<=n:print' '*a+'#'+' #'[a==n]*(3*n-a+~a)+'#';d-=1;a-=d/n+1


Try it online!

Python 2, 75 bytes

a=d=n=input()-1
while a<=n:print' '*a+' '[a==n]*(3*n-a+~a);d-=1;a-=d/n+1


Try it online!

If mixing output characters is OK.

# Powershell, 91 bytes

param($n)($s=' '*--$n+'#'*$n+'#')
--$n..0+,0*$n+0..$n|%{' '*$_+"#$(' '*(3*$n-2*$_+2))#"}$s


# PowerShell, 107 97 bytes

param($n)($z=$n-1)..1+,0*$n+1..$z|%{" "*$_+"#"+($x=" "*($z-$_))+(" ","#")[!($_-$z)]*($n-2)+"$x#"}  Try it online! If there was a cheap way to reverse the first half, this answer would feel a lot better. It builds the left half, then the core (which is either x #'s or spaces), then mirrors the left's logic to make the right. Fun fact, you don't need to copy over trailing white-space. Unrolled and explained: param($n)
($z=$n-1)..1 + ,0*$n + 1..$z |%{  #Range that repeats 0 n times in the middle
" "*$_ + "#" +($x=" "*($z-$_)) +  #Left side
(" ","#")[!($_-$z)]*($n-2) + #Core that swaps when it's the first or last row "$x#"}                            #Right side which is left but backwards


# APL (Dyalog Unicode), 46 bytesSBCS

(' '@~5 6∊⍨1⊥⊢∘,)⌺3 3⊢<(⍉⌽⌊⊢)⍣2∘(∘.+⍨∘⍳¯2+3×⊢)


This solution was provided by Adám - thanks!

Try it online!

# APL (Dyalog Unicode), 61 bytesSBCS

(((⊃∘' #'¨1+5∘=+6∘=)⊢)1⊥⊢∘,)⌺3 3⊢<(((⊖⌊⊢)⌽⌊⊢)(∘.+⍨(⍳¯2+3×⊢)))


Try it online!

Thanks to Adám for his help!

The idea is to find the "diamond" that lies partly in the square and apply an edge-detect filter to "outline" the octagone.

• – Adám Nov 7 '18 at 13:51
• You can't actually use Classic here because of ⌺. Rather count 1 byte/char by referring to SBCS as per Meta. – Adám Nov 7 '18 at 13:52
• @Adám Thanks! I don't know how to edit the header, can you do it for me? – Galen Ivanov Nov 7 '18 at 14:01
• What do you mean by editing the header? – Adám Nov 7 '18 at 14:22
• Edit and copy from here. – Adám Nov 7 '18 at 14:40

# C (clang), -DP=printf(-DF=for(i+ 179 = 199 180 bytes

i;*m="%*s%*s\n";g(n){P"%*s",n,H;F;--i;)P H;P"\n");}f(n){g(n);F;--i;)P m,i,(H,3*n-i+~i,H;F-2;i--;)P"#%*s\n",3*n-3,H;F;--i;)P m,n-i,(H,n+i+i-1,H;g(n);}


Try it online!

Ungolfed:

f(n){
int i;
printf("%*d",n,0);
for(i=0;i<n-1;i++){
printf("0");
}
printf("\n");
for(i=1;i<n;i++){
printf("%*d%*d\n",n-i,0,n+i+i-1,0);
}
for(i=0;i<n-2;i++){
printf("0%*d\n",n+n+n-3,0);
}
for(i=n-1;i>0;i--){
printf("%*d%*d\n",n-i,0,n+i+i-1,0);
}
printf("%*d",n,0);
for(i=0;i<n-1;i++){
printf("0");
}
}


-19 bytes thanks to @ceilingcat

# Python 2, 130 bytes

def f(n):
a=[' '*~-n+n*'#']
b=[' '*(n-i-2)+'#'+' '*(n+2*i) +'#'for i in range(n-2)]
return a+b+['#%*s'%(3*n-3,'#')]*n+b[::-1]+a


Try it online!

On mobile, so not incredibly golfed.

• You can remove the space after (n+2*i). – Zacharý Nov 8 '18 at 13:26

## Batch, 260 bytes

@echo off
set s=
for /l %%i in (1,1,%1)do call set s= %%s%%
echo %s% %s: =#%
call:c %1,-1,3
for /l %%i in (1,1,%1)do echo   #%s:~2%%s%%s:~2%#
call:c 3,1,%1
echo %s% %s: =#%
exit/b
:c
for /l %%i in (%*)do call echo %%s:~,%%i%%#%%s:~%%i%%%s%%%s:~%%i%%#


Outputs two leading spaces on each line. Explanation: Batch has no string repetition operator, limited string slicing capability and requires separate statements to perform arithmetic. It was therefore golfiest to make up a string of the input length in spaces (Batch can at least translate these to #s for the top and bottom lines) and then slice from or to a specific position ranging from 3 to the length to generate the diagonals (this is what the last line of the script achieves).

# Ruby, 96 bytes

->n{[*(n-=2).step(z=n*3+2,2),*[z]*n,*z.step(n,-2)].map{|x|([?#]*2*('# '[x<=>n]*x)).center(z+2)}}


Try it online!

Not very golfed yet. Might golf if I find the time.

# Red, 171 bytes

func[n][c:(a: n - 1)* 2 + n
b: collect[loop c[keep pad/left copy"^/"c + 1]]s: 1x1 s/1: n
foreach i[1x0 1 0x1 -1x1 -1x0 -1 0x-1 1x-1][loop a[b/(s/2)/(s/1): #"#"s: s + i]]b]


Try it online!

## Explanation:

Red[]
f: func [ n ] [
a: n - 1                                         ; size - 1
c: a * 2 + n                                     ; total size of widht / height
b: collect [                                     ; create a block
loop c [                                     ; composed of size - 1 rows
keep pad/left copy "^/" c + 1            ; of empty lines of size c (and a newline)
]
]
s: a * 1x0 + 1                                   ; starting coordinate
foreach i [ 1x0 1 0x1 -1x1 -1x0 -1 0x-1 1x-1 ] [ ; for each offset for the 8 directions
loop a [                                     ; repeat n - 1 times
b/(s/2)/(s/1): #"#"                      ; set the array at current coordinate to "#"
s: s + i                                 ; next coordinate
]
]
b                                                ; return the block
]


# Perl 5, 201 197 188 187 186 bytes:

$a=<>;$b=3*$a-4;$c='$"x($e-$_)."#".$"x$f."#\n"';$e=($b-$a)/2+1;$d=$"x$e."#"x$a.$/;$f=$a;print$d,(map{(eval$c,$f+=2)[0]}1..$a-2),("#".$"x$b."#\n")x$a,(map{$f-=2;eval$c}reverse 1..$a-2),$d


Try it online!

Reads the size of the octagon from first line of STDIN.

• Welcome to PPCG! You can probably shave off a few bytes here and there by using tricks found in this post. – user45941 Nov 8 '18 at 5:14
• @Mego Yep. I was able to save 4 bytes by using $" instead of " ". – Nathan Mills Nov 8 '18 at 17:14 # Perl 5, 176 bytes $f=$a=<>;$b=3*$a-4;$c='$"x($e-$_)."#".$"x$f."#\n"';$e=$a-1;$d=$"x$e."#"x$a.$/;print$d,(map{(eval$c,$f+=2)[0]}1..$a-2),("#".$"x$b."#\n")x$a,(map{$f-=2;eval$c}reverse 1..$a-2),$d  Based on Nathan Mills' answer above (which I have insufficient rep to comment on!). $e can be simplified to $a-1 saving 6 bytes; $f can be chain assigned; saving two bytes; Not sure where the other two come from!

While $e can be replaced with $a-1 in the two places it occurs, the extra brackets needed means this only breaks even.

Ungolfed:

$f =$a = <>;
$b = 3 *$a - 4;
$c = '$"x($e-$_)."#".$"x$f."#\n"';
$e =$a - 1;
$d =$" x $e . "#" x$a . $/; print$d, ( map { ( eval $c,$f += 2 )[0] } 1 .. $a - 2 ), ( "#" .$" x $b . "#\n" ) x$a,
( map { $f -= 2; eval$c } reverse 1 .. $a - 2 ),$d


# Perl 6, 76 73 bytes

-3 bytes thanks to Jo King

{(' 'x$_-1~\*x$_,{S/.\S(.)+/* {' 'x$0}*/}...*)[^$_,$_ xx$_-2,[R,] ^$_;*]}  Try it online! Returns a list of lines. # C (gcc), 158153 150 bytes O,c,t,g;o(n){for(O=2*~-n,t=c=O+n;t--;puts(""))for(g=c;g--;)putchar(33-(!t|t>c-2?g<n-1|g>O:t<n-1|t>O?t+O-g&&t-O-g&&~c+g+t+n+n&&c-g-t+n-3+n:g&&g<c-1));}  Try it online! • @ceilingcat Thank you. – Jonathan Frech Nov 8 '18 at 21:00 • @ceilingcat Thank you. – Jonathan Frech Jan 21 '19 at 7:53 # Python 3, 224 bytes n=int(input()) z=" "*(n-1)+"#"*n+" "*(n-1) print(z) for i in range(n-2):print(" "*(n-i-2)+"#"+" "*(i*2+n)+"#") print((("#"+" "*(n*3-4)+"#\n")*n)[:-1]) for i in range(n-3,-1,-1):print(" "*(n-i-2)+"#"+" "*(i*2+n)+"#") print(z)  Try it online! # Perl 5, 170168 166 bytes $a=<>-1;$\="#\n";print$x=$_=$"x$a."#"x$a;if(s/^( *)  #*/$1 #$1 /){print}while (s/ #/#  /){print}$z=$_;for(1..$a){print$_=$z}while(s/# (\s{$a})/ #$1/){print}print$x


This works by the magic of regex. The "if" is only needed to deal with the pathological case of n=2, which otherwise outputs something like:

 ##
##
#  #
##


probably this can be coded away.

I think there may be a lot more to gain by creating a string up to the mid-point then reversing it. Of course we then need to insert/delete an extra space if n is odd (or use thin-space :p).

Ungolfed

$a = <> -1; # Subtracting one is very useful!$\ = "#\n";                          # Every line ends with a '#' let perl provide.
$x=$_ = " " x $a. "#" x$a;          # The horiz line (one short)
print;                               # print it plus the extra #
if(s/^( *)  #*/$1 #$1 /){print}    # create a hole and remove a leading space(if n=2 this fails)
while (s/ #/#  /){                   # make the hole bigger
print;                           # and print (with a trailing #)
}
$z=$_;                               # store $_ for later use for (1 ..$a) {                      # nice that we don't have to do 2..$a but not golf-nice$_ =$z; # restore$_ (we could use $z but since we have print; # to restore somewhere, doing it here saves us bytes) } while (s/# (\s{$a})/ #$1/){ # now move the # to the right and reduce the trailing spaces print; } print$x;                            # and finish...


I think this can probably be golfed a bit more, quite apart from significant changes like pushing onto $@ and printing that at the end. [Golfed spaces around .. and moved print before assigns in two cases, saving on semicolons.] • saved 20 bytes reordering some instructions TIO, why \s and not just a space in last regex – Nahuel Fouilleul Jan 22 '19 at 10:56 # J, 5945 41 bytes h=.|.,],~,:@{.#~_2+# ' #'{~h@|.@|:@h@=@i.  Try it online! Will add explanation tonight. # Perl 5, 98 bytes $_=2x$_.1x$_.$/;s/2//;s/.$/ #/,y/1/ /while$a.=$_,$b=$_.$b,s/2[#1]/# /;$_=$a.$_ x("@F"-2).\$b;y/2/ /
`

TIO