Embedded Triangles!

Question

Your task: given an integer n, output an embedded triangle to the nth depth.

Standard methods of input and output, input in any convenient format, output should be as a string. Examples are 0-indexed, you can use 1-indexed if you want. Standard loopholes apply

The left and right sides of each figure is 2^n, when 0 indexed. The top is twice that.

Examples:

n=0
--
\/

n=1
----
\/ /
 \/

n=2
--------
\/ /   /
 \/   /
  \  /
   \/

This is code-golf, so shortest code wins!

_{This is NOT a duplicate of the Embedded Hexagons! challenge. I don't think any answers from that question could be trivially copied over to this one.}

Answer 1

Retina, 44 bytes

^
\/
+`( *)/1
$1/$1 $1/
*`.
-
;+(*G`
\\..
 \

Try it online!

I only want to print successful replacements, but I don't really know how to do that.

As I'm no longer allowed to add another answer, here's a Batch answer for 192 bytes:

@echo off
set d=-
set l=\
set r=/
for /l %%i in (1,1,%1)do @call set r=%%r%%%%d:-= %%/&call set d=%%d%%-%%d%%
echo -%d%
:l
echo %l%%r%
set l= %l%
set r=%r:~2%
if not "%r%"=="" goto l

And here's a JavaScript (ES7) port of @lanlock4's excellent Mathematica answer for 124 bytes:

f=
n=>[...Array(2**n+1)].map((_,i)=>[...Array(2<<n)].map((_,j)=>i?i+~j?i>j?` `:i+j-(i+j&-i-j)?` `:`/`:`\\`:`-`).join``).join`
`

<input type=number min=0 oninput=o.textContent=f(this.value)><pre id=o>

And here's a Charcoal answer for 16 bytes:

ＦＥＮＸ²ι«ιι↙↙ι↑↖ι→

Try it online! Link is to verbose version of code. Explanation:

  Ｎ                 Input number
 Ｅ                  Map over implicit range
    ²               Literal 2
     ι              Current index
   Ｘ                Exponentiate
Ｆ     «             Loop over resulting powers of 2
       ι            Print right
        ι           Print right again
         ↙          Move down and left
          ↙ι        Print down and left
            ↑       Move up
             ↖ι     Print up and left
               →    Move right

Answer 2

Charcoal, 27 bytes

ＮαＷα«Ｘ²ι↙ＡＸ²⁻ι¹β↙β↑↖β→Ａ⁻α¹α

Try it online!

Just modified my submission from the hexagon challenge a bit to make a triangle.

Answer 3

Mathematica, 105 bytes

""<>Table[Which[i<1,"-",i==j,"\\",IntegerQ@Log2[i+j-1]&&j>i,"/",1>0," "],{i,0,2^#},{j,2*2^#}]~Riffle~"\n"&

where \n is replaced with a newline.

Explanation: Makes an array of characters (indexed from i = 0 to 2^n vertically, j = 1 to 2^(n+1) horizontally) according to these rules:

• If i<1 (so i = 0), put a "-" (for the top row)
• If i==j, put a "\" (escaped, for the left-hand side)
• If Log2[i+j-1] is an integer (so i+j-1 is a power of 2) and j>i, put a "/" (for the other edges)
• And put " " everywhere else.

Then ""<>...~Riffle~"\n"& turns this array into a string.

Answer 4

JavaScript (JavaScript Shell), 136 132 bytes

t=2**readline();p=print;p(v='-'.repeat(t*2));for(y=0;y<t;y++)p(' '.repeat(y)+'\\'+v.replace(/-/g,(_,i)=>i&-~i?' ':'/').slice(y*2+1))

Try it:

readline = () => prompt();
print = (...args) => output += (args.join(' ') + '\n');
output = '';

t=2**readline();p=print;p(v='-'.repeat(t*2));for(y=0;y<t;y++)p(' '.repeat(y)+'\\'+v.replace(/-/g,(_,i)=>i&-~i?' ':'/').slice(y*2+1))

document.write(`<pre>${output}</pre>`);

Answer 5

C, 200 bytes

#define P printf(
i,j,k,l;f(n){for(i=0,k=pow(2,n);i++<k;)P"--");for(i=0;i<k;++i){l=P"\n%*c",i+1,92);for(j=0;j<n;++j)if(2*(int)pow(2,j)+1-i>l)l+=P"%*c",2*(int)pow(2,j)+1-i-l,47);P"%*c",2*k-i-l+1,47);}}

Try it online!

Answer 6

JavaScript (ES6), 108 95 93 92 bytes

let f =

n=>(g=x=>y>1<<n?'':x<2<<n?' /\\-'[y?~x+y?+!(k=x+y,x<y|k&-k-k):2:3]+g(x+1):`
`+g(!++y))(y=0)

console.log(f(0))
console.log(f(1))
console.log(f(2))
console.log(f(3))

Answer 7

05AB1E, 45 bytes

I made it! (It's a fun challenge, I don't get why they VTC'd it)

<oLð×ε¨'\«}IL<oð×ε¨'/«}JDgU.sR2ô€нøJ»X>'-×s‚»

Try it online!

Answer 8

Canvas, 22 bytes

╵ｒＨ２；＾／Ｌ ×；＋ｎ｝ｌ＼ｎＬ-×；∔

Try it here!

Explanation

'rH2;^/L ×;+n}l\nL-×;∔
'                      increment input integer n
 r                     range 0..n
  H          }         push an empty string and start a loop
   2;^                 get 2^i
      /                antidiagonal of that length
       L               get width(don't pop)
         ×;_           add that many spaces to the left
            n          overlap with prev iteration
              l        get height(don't pop)
               \       create a diagonal of that length 
                n      and overlap with the created sides
                 L     get width(don't pop)
                  -×;∔ add that many minuses to the top
       

Answer 9

PHP, 137 Bytes

for(;$i<1+2**$a=$argn;++$i)for($n=0;$n<$m=2**($a+1);$s[]=2**++$n)echo$n%$m?"":"\n",$i?$i-1!=$n?$i<=$n&in_array($i+$n,$s)?"/":" ":"\\":"-";

Online Version

Expanded

for(;$i<1+2**$a=$argn;++$i)
for($n=0;$n<$m=2**($a+1);$s[]=2**++$n) # fill array for sum compare after loop
echo$n%$m?"":"\n", # print line feed
$i  # false for first line
 ?$i-1!=$n # row number - 1 equal column print \
  ?$i<=$n&in_array($i+$n,$s) # if row number lte column and sum of both in array print /
   ?"/"
   :" " # else print space
  :"\\"
 :"-"; # fill first line

A recursive way for 199 Bytes

function f($a){global$r;for(;$i<1+2**$a;$i++)for($n=0;$n<$m=2**($a+1);$n++)$r[$i][$n]=$i?$i-1!=$n?$i<=$n&&$i+$n==2**($a+1)?"/":" ":"\\":"-";$a?f($a-1):0;}f($argn);echo join("\n",array_map("join",$r));