Draw some mountain peaks

Question

Write a program or function that draws a mountain range, where each subsequent larger mountain peak is "behind" the ones in front, and alternates which side is visible.

This is a mountain range of size 1

/\

This is a mountain range of size 2

 /\
/\ \

This is a mountain range of size 3

  /\
 / /\
/ /\ \

This is a mountain range of size 4

   /\
  /\ \
 / /\ \
/ /\ \ \

This is a mountain range of size 5

    /\
   / /\
  / /\ \
 / / /\ \
/ / /\ \ \

And so on.

Input

A single positive integer in any convenient format, n > 0.

Output

An ASCII-art representation of the mountain range, following the above rules. Leading/trailing newlines or other whitespace are optional, provided that the peaks line up appropriately.

Rules

• Either a full program or a function are acceptable. If a function, you can return the output rather than printing it.
• Standard loopholes are forbidden.
• This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.

Answer 1

Charcoal, 16 bytes

ＮλＦλ«Ｐ↘⁻λι←↙¹‖Ｔ→

Try it online!

How?

Ｎλ inputs the size of the largest mountain into λ. Ｆλ« runs a loop over values of ι from 0 through λ-1. (The closing » is implied at the end of the program.)

Inside the loop, Ｐ↘⁻λι calculates λ-ι and draws, without moving the cursor afterward, a line of that length going southeast. Based on its direction, this line will consist of \ characters. ← moves one step to the west, and ↙¹ draws a line of length 1 going southwest (made of /). Finally, ‖Ｔ→ horizontally reflects the drawing, transforming characters as appropriate: \ becomes / and / becomes \.

Adding the dump instruction Ｄ at the beginning of the loop (try it) allows us to see the progression:

    /\
   /  
  /   
 /    
/     
    /\    
   /\ \   
  /    \  
 /      \ 
/        \
    /\    
   / /\   
  / /\ \  
 / /    \ 
/ /      \
    /\    
   /\ \   
  / /\ \  
 / /\ \ \ 
/ /    \ \
    /\    
   / /\   
  / /\ \  
 / / /\ \ 
/ / /\ \ \

Answer 2

JavaScript (ES6), 75 bytes

for(n=prompt(s="/\\");n--;s=n%2?s+' \\':'/ '+s)console.log(" ".repeat(n)+s)

The full program is currently slightly shorter than the recursive function:

f=n=>n?" ".repeat(--n)+`/\\
`+f(n).replace(/\S.+/g,x=>n%2?x+" \\":"/ "+x):""

Answer 3

Python 2, 67 bytes

n=input()
s='/\\'
while n:n-=1;print' '*n+s;s=['/ '+s,s+' \\'][n%2]

Prints line by line, accumulating the string s by alternately adding a slash to the left or right based on the current parity of n. Prefixes with n spaces.

An alternative way to update was the same length:

s=n%2*'/ '+s+~n%2*' \\'
s=['/ '+s,s+' \\'][n%2]

A recursive method was longer (70 bytes).

f=lambda n,s='/\\':n*'_'and' '*~-n+s+'\n'+f(n-1,[s+' \\','/ '+s][n%2])

Answer 4

Haskell, 77 bytes

0%_=""
n%s=(' '<$[2..n])++s++'\n':(n-1)%(cycle[s++" \\","/ "++s]!!n)
(%"/\\")

Usage:

putStrLn $ f 5
    /\
   / /\
  / /\ \
 / / /\ \
/ / /\ \ \

Prints line by line, accumulating the string s by alternately adding a slash to the left or right based on the current parity of n. Prefixes with n-1 spaces.

Answer 5

Batch, 202 bytes

@echo off
set/af=%1^&1
set m=/\
set s=
for /l %%i in (2,1,%1)do call set s= %%s%%
for /l %%i in (2,1,%1)do call:l
:l
echo %s%%m%
set s=%s:~1%
set/af^^=1
if %f%==1 (set m=%m% \)else set m=/ %m%

Takes input as a command-line parameter. Falls through to execute the last loop.

Answer 6

Ruby, 61 bytes

A pretty straightforward port of ETHproductions' JavaScript answer.

->n{s="/\\"
(puts" "*n+s
s=n%2>0?s+" \\":"/ "+s)until 0>n-=1}

See it on repl.it: https://repl.it/EPU5/1

Ungolfed

->n{
  s = "/\\"
  ( puts " "*n+s
    s = n%2 > 0 ? s+" \\" : "/ "+s
  ) until 0 > n -= 1
}

Answer 7

Haskell, 117 107 105 97 90 bytes

b!1=["/\\"]
b!n|m<-(1-b)!(n-1)=map(' ':)m++[[("/ "++),(++" \\")]!!b$last m]
(unlines.(1!))

Try it on Ideone. Edit: Saved 8 bytes with an idea from Neil.

Ungolfed version:

p b 1 = ["/\\"]
p b n = let m = p (1-b) (n-1)
            k = last m
            r = map (' ':) m
        in if b == 1
           then r ++ [k ++ " \\"]
           else r ++ ["/ " ++ k]
f n = unlines(p 1 n)

Recursive approach. The shape for n is generated by adding a space in front of each line of the n-1 shape and taking the last line of n-1 and add "/ " before if n is odd or " \" after if n is even ... or so I thought before noticing that this last step is reversed for all recursive steps when the final n is odd. Therefore a flag b is passed which alternates each recursive call and determines if the next mountain part is added left or right.

Answer 8

Java 7,130 bytes

String f(int n,String s){String l="";for(int i=1;i++<n;l+=" ");return n>1?n%2<1?l+s+"\n"+f(--n,s+" \\"):l+s+"\n"+f(--n,"/ "+s):s;}

Ungolfed

class Mountain {
 public static void main(String[] args) {
    System.out.println(f( 5 , "/\\" ) );
  }
 static String f(int n,String s){
    String l = "";
    for (int i = 1; i++ < n; l += " ") ;
      return n > 1? n % 2 < 1?l + s + "\n" + f(--n , s + " \\")
                           :l + s + "\n" + f(--n , "/ " + s)
                            :s;
    }

}

Answer 9

Canvas, 7 bytes

Ｈ＼║ｎ↔｝↕

Try it here! I guess Canvas is now the ideal language for this challenge.

Answer 10

C++ 138(function)

Function :-

#define c(X) cout<<X;
#define L(X,Y) for(X=0;X<Y;X++)  

void M(int h){int l=1,r=1,j,H=h,i;L(i,h){for(j=H;j>0;j--)c(" ")L(j,l)c(" /")L(j, r)c("\\ ")c("\ n")(h%2)?(i%2)?r++:l++:(i%2)?l++:r++;H--;}  

Full program :-

#include<conio.h>
#include<iostream>

using namespace std;

#define c(X) cout<<X;
#define L(X,Y) for(X=0;X<Y;X++)

void M(int h)
{
    int l=1,r=1,j,H=h,i;
    
    L(i, h)
    {
        for (j = H;j > 0;j--)
            c(" ")
        L(j, l)
            c(" /")
        L(j, r)
            c("\\ ")
        c("\n")
        
        (h % 2) ? (i % 2) ? r++ : l++ :(i % 2) ? l++ : r++;
        H--;
    }
}

int main()
{
    int h;
    cin >> h;
    M(h);
    _getch();
    return 0;
}  

NOTE : the function _getch() may have different prototype names across different compilers.