50
\$\begingroup\$

With the big crash of the universal economy also the demand for custom made planets plunged. The Magratheans had to look after more steady revenues also from a broader class of customers. Therefore, they invented the have-your-own chain of mountain (or short havoc-o-mountains) for people with smaller budget who could not afford a complete planet.

The mountains are build according to the customer's plan (a.k.a. strings of digits and dots) and delivered using ascii-art (consisting of , /, \, ^ and v).

Task

Write a complete program which takes input (single string) either from STDIN or as argument and outputs to STDOUT. This puzzle is a code-golf so please show some attempt at golfing.

Input

A string of dots and digits providing the basis for the mountain chain. Each string is exactly as long as necessary to support the mountains and each peak is given by a digit instead of a dot, indicating the height of the peak.

Output

An ascii version of the mountain chain.

  • Each digit in the input represents exactly one peak (^) at exactly the height indicated by the digit (i.e. 9 is the highest height).
  • There must not be additional peaks in the output (i.e. at places where there is a dot in the input).
  • Mountains are of triangular shape, i.e. slopes are created using / and \ characters.
  • Passes where two mountains overlap are shaped using the character v.
  • No superfluous newlines nor blank lines.
  • Padding lines with trailing spaces is optional.

You may assume that the input provided is valid, i.e. there always exists a solution according to the rules (e.g. an input of 13.. would not result in a valid configuration and may be ignored). Moreover, on each side there are exactly as many dots such that the mountains must no be cropped.

Examples

The first line shows the input, all other lines constitute the desired output. (Actually the mountains look much better in my console than here.)

1
^

11
^^

1.2.
  ^
^/ \

.2.3..
   ^
 ^/ \
/    \

.2..3..
    ^
 ^ / \
/ v   \

...4...3...3..
   ^
  / \  ^   ^ 
 /   \/ \ / \
/        v   \
\$\endgroup\$
  • 1
    \$\begingroup\$ What a combination of poetry and art! I love it. \$\endgroup\$ – devnull Feb 26 '14 at 11:02
  • \$\begingroup\$ Is printing extra newlines okay? In other words, for an input of 1, is \n\n\n\n\n\n\n\n^ allowed? \$\endgroup\$ – durron597 Feb 26 '14 at 11:51
  • \$\begingroup\$ @durron597 The output should have no superfluous newlines, have a look at the examples. \$\endgroup\$ – Howard Feb 26 '14 at 12:15
  • \$\begingroup\$ What about trailing space characters? Is it OK if all lines are the same length as the original string, padded out with spaces? \$\endgroup\$ – Paul Prestidge Feb 27 '14 at 1:37
  • \$\begingroup\$ @Chron Yes, that's OK. \$\endgroup\$ – Howard Feb 27 '14 at 7:40

13 Answers 13

11
\$\begingroup\$

Javascript: 272 268 233 232 201 192 189 188 178 180 characters

Thanks to @Sam for reducing it from 268 to 233 characters, and for @manatwork for another 1 char. @VadimR for pointing out a bug.

p=prompt(r=t='');s=' ';for(d=10;d--;r=s+q+s,t+=q.trim()?q+'\n':'')for(q='',i=0;i<p.length;)q+=' \\/v^'[p[i++]==d?4:(/\^|\\/.test(r[i-1])+2*/\^|\//.test(r[i+1]))*(r[i]==s)];alert(t)

Properly idented and somewhat ungolfed version with comments:

// The output initialization is just a golfing trick suggested by @manatwork.
input = prompt(state = output = '');
space = ' ';

// Repeat for each line, from the top (the highest peak, highest digit) to the floor (digit 1). Start at 10 to avoid a bug.
for (digit = 10; digit--;

      // Update the state of our automaton, at the end of the iteration.
      // Add a space after and before to simplify the future pattern recognization.
      state = space + line + space,

      // Add the line to the output if it is not an empty line, at the end of the iteration.
      output += line.trim() ? q + '\n' : '')
{ // This curly brace was added for readability, it is not in the golfed source.

  // Analyze each character in the current state to produce a new state, like a cellular automaton.
  for (line = '', i = 0; i < input.length;)
  { // This curly brace was added for readability, it is not in the golfed source.
    line +=

        // If the input is the current digit number, evaluate to 4 and put a peak in this character.
        // Otherwise evaluate this expression with those rules:
        // 1 means that the hill is higher only at right in the previous iteration, we do climb it to the right in this one.
        // 2 means that the hill is higher only at left in the previous iteration, we do climb it to the left in this one.
        // 3 means that the hill is higher at both sides in the previous iteration, we are in a v-shaped valley.
        // 0 means nothing to do here. If the middle is not a space, it will be multiplied by 0 and become 0.
        ' \\/v^'[input[i++] == digit ? 4 : (/\^|\\/.test(state[i - 1]) + 2 * /\^|\//.test(state[i + 1])) * (r[i] == space)];
    } // This curly brace was added for readability, it is not in the golfed source.
} // This curly brace was added for readability, it is not in the golfed source.

// Give the final output.
alert(output);

As you may note from the code, this works as a cellular automaton, where each cells checks for a number in the input, looks to itself and to its two neighbours to decide what the next iteration will be. At each moment a cell may be a ^, /, \, v or . The input provided in the test cases produces the expected output.

Note that using the alert box sucks, since it normally does not have a monospaced font. You may copy & paste the text from the alert box to somewhere else for a better appreciation of the output, or you may replace the last line alert by console.log, but since this is code-golf, alert is shorter.

Further, it does not validates anything in the input. It simply considers unrecognized characters as spaces the same way it does to . (in fact . is an unrecognized character too).

\$\endgroup\$
  • \$\begingroup\$ There is an old golfing trick to reduce 1 character: initialize variables with empty string as prompt()'s parameter. \$\endgroup\$ – manatwork Feb 27 '14 at 17:14
  • \$\begingroup\$ @manatwork Thank you. Done. \$\endgroup\$ – Victor Stafusa Feb 27 '14 at 17:18
  • \$\begingroup\$ Excuse me, maybe I'm missing something, but I'm getting consistent results in both FF and Chromium. I launch a browser, run JS code from revision#14, and get error message. Then I run code from revision#1 - it runs OK. Again I run 14's code - and no error message, it runs OK. So revision#14's code can not be run by itself? \$\endgroup\$ – user2846289 Mar 2 '14 at 16:25
  • 1
    \$\begingroup\$ @VadimR Thanks, fixed. That was a side-effect for testing it with a poluted environment. Needed to prefix the code with delete r; delete s; delete q; delete p; delete t; delete i; delete d; to ensure that it was not poluted. \$\endgroup\$ – Victor Stafusa Mar 2 '14 at 18:28
  • \$\begingroup\$ q.trim()?q+'\n':'' could be q.trim()&&q+'\n', saving two. Also, i<p.length could just be p[i]. \$\endgroup\$ – Nicholas Pipitone Oct 26 '18 at 18:32
6
\$\begingroup\$

Ruby, 208 201 189

Very fun challenge! Here's an alternative Ruby solution.

gets.size.times{|x|0.upto(h=$_[x].to_i-1){|d|r=$*[h-d]||=' '*~/$/
[x+d,x-d].map{|o|r[o]=r[o]>?!??v:o<x ??/:?\\if r[o]<?w}
d<1?r[x]=?^:r[x-d+1,w=2*d-1]=?w*w}}
puts$*.reverse.*($/).tr(?w,' ')

As a bonus, here's a Ruby implementation of Victor's very clever "cellular automaton" algorithm, at 162 characters:

s=gets
9.downto(1){|h|$0=(-1..s.size).map{|x|$_=$0[x,3]
s[x]=="#{h}"??^:~/  [\^\/]/??/:~/[\^\\]  /??\\:~/[\^\\] [\^\/]/??v:' '}*''
$*<<$0[1..-2]if$0=~/\S/}
puts$*

Example output:

....5.....6..6.....
          ^  ^
    ^    / \/ \
   / \  /      \
  /   \/        \
 /               \
/                 \
\$\endgroup\$
  • 1
    \$\begingroup\$ I think you may use $/ for newline. \$\endgroup\$ – Howard Feb 27 '14 at 7:43
4
\$\begingroup\$

C# - 588 characters - not as good as Ray's 321 though!

class P{static void Main(string[] a){char[,] w=new char[a[0].Length+1,10];int x=0;foreach(char c in a[0]){if(c!='.'){int h=int.Parse(c+"");if(w[x,h]=='\0')w[x,h]='^';int s=1;for(int l=h-1;l>0;l--){for(int m=x-s;m<=x+s;m++){if(w[m,l]!='\0'){if(w[m,l]=='^')w[m,l]='/';if(w[m,l]=='\\')w[m,l]='v';}else{if(m==x-s)w[m,l]='/';else if(m==x+s)w[m,l]='\\';else w[m,l]='\0';}bool t=false;for(int f=9;f>0;f--){if(t)w[m,f]='\0';if(w[m,f]!='\0')t=true;}}s++;}}x++;}for(int k=9;k>0;k--){string u="";for(int j=0;j<w.GetLength(0);j++){u+=w[j,k];}if(u.Replace("\0","")!="")System.Console.WriteLine(u);}}}

Example output:

F:\>mountains ".2..3..4..."
       ^
    ^ / \
 ^ / v   \
/ v       \

Or a longer more complex one...

F:\>mountains ".2..3..6.....5...3......1..3..4....2."
       ^
      / \    ^
     /   \  / \               ^
    /     \/   \ ^         ^ / \
 ^ /            v \       / v   \  ^
/ v                \    ^/       \/ \

Brilliant puzzle... not as easy as it seems... loved it!

\$\endgroup\$
  • 2
    \$\begingroup\$ "Complex one" is ill-formed, there's no peak for "3". \$\endgroup\$ – user2846289 Mar 1 '14 at 17:20
  • \$\begingroup\$ All the 3s are there. If you're talking about the first one, it's part of the slope. \$\endgroup\$ – Hein Wessels Oct 26 '18 at 9:30
4
\$\begingroup\$

APL, 65 bytes

⍉⌽↑⌽¨h↑¨'^/v\'[1+(~×a)×2+×2+/2-/0,0,⍨h←¯1+⊃⌈/a-↓|∘.-⍨⍳⍴a←11|⎕d⍳⍞]

this symbol returns raw (not evaluated) input as a character array.

Solving interactively, in an APL session:

      s←'...4...3...3..' ⍝ let's use s instead of ⍞
      ⎕d ⍝ the digits
0123456789
      ⎕d⍳s ⍝ the indices of s in ⎕d or 11-s if not found
11 11 11 5 11 11 11 4 11 11 11 4 11 11
      11|⎕d⍳s ⍝ modulo 11, so '.' is 0 instead of 11
0 0 0 5 0 0 0 4 0 0 0 4 0 0
      a←11|⎕d⍳s ⍝ remember it, we'll need it later
      ⍴a ⍝ length of a
14
      ⍳⍴a
1 2 3 4 5 6 7 8 9 10 11 12 13 14
      ⍝ ∘.-    subtraction table
      ⍝ ∘.-⍨A  same as: A ∘.- A
      ⍝ |      absolute value
      |∘.-⍨⍳⍴a
 0  1  2  3 4 5 6 7 8 9 10 11 12 13
 1  0  1  2 3 4 5 6 7 8  9 10 11 12
 2  1  0  1 2 3 4 5 6 7  8  9 10 11
 ...
13 12 11 10 9 8 7 6 5 4  3  2  1  0
      ⍝ ↓      split the above matrix into rows
      ⍝ a-     elements of "a" minus corresponding rows
      ⍝ ⊃⌈/    max them together
      ⊃⌈/a-↓|∘.-⍨⍳⍴a
2 3 4 5 4 3 3 4 3 2 3 4 3 2
      ⍝ This describes the desired landscape,
      ⍝ except that it's a little too high.
      ⍝ Add -1 to correct it:
      ¯1+⊃⌈/a-↓|∘.-⍨⍳⍴a
1 2 3 4 3 2 2 3 2 1 2 3 2 1
      ⍝ Perfect!  Call it "h":
      h←¯1+⊃⌈/a-↓|∘.-⍨⍳⍴a
      0,⍨h ⍝ append a 0 (same as h,0)
1 2 3 4 3 2 2 3 2 1 2 3 2 1 0
      0,0,⍨h ⍝ also prepend a 0
0 1 2 3 4 3 2 2 3 2 1 2 3 2 1 0
      2-/0,0,⍨h ⍝ differences of pairs of consecutive elements
¯1 ¯1 ¯1 ¯1 1 1 0 ¯1 1 1 ¯1 ¯1 1 1 1
      ⍝ this gives us slopes between elements
      2+/2-/0,0,⍨h ⍝ sum pairs: left slope + right slope
¯2 ¯2 ¯2 0 2 1 ¯1 0 2 0 ¯2 0 2 2
      ×2+/2-/0,0,⍨h ⍝ signum of that
¯1 ¯1 ¯1 0 1 1 ¯1 0 1 0 ¯1 0 1 1
      2+×2+/2-/0,0,⍨h ⍝ add 2 to make them suitable for indexing
1 1 1 2 3 3 1 2 3 2 1 2 3 3
      ⍝ Almost ready.  If at this point we replace
      ⍝ 1:/ 2:v 3:\, only the peaks will require fixing.
      ~×a ⍝ not signum of a
1 1 1 0 1 1 1 0 1 1 1 0 1 1
      (~×a)×2+×2+/2-/0,0,⍨h ⍝ replace peaks with 0-s
1 1 1 0 3 3 1 0 3 2 1 0 3 3
      ⍝ Now replace 0:^ 1:/ 2:v 3:\
      ⍝ We can do this by indexing a string with the vector above
      ⍝ (and adding 1 because of stupid 1-based indexing)
      '^/v\'[1+(~×a)×2+×2+/2-/0,0,⍨h]
///^\\/^\v/^\\
      ⍝ Looks like our mountain, only needs to be raised according to h
      r←'^/v\'[1+(~×a)×2+×2+/2-/0,0,⍨h] ⍝ name it for convenience
      h¨↑r ⍝ extend r[i] with spaces to make it h[i] long
 /  /   /    ^     \    \   /   ^    \   v  /   ^    \   \
      ↑⌽¨h¨↑r ⍝ reverse each and mix into a single matrix
/
 /
  /
   ^
  \
 \
 /
  ^
 \
v
 /
  ^
 \
\
      ⍉⌽↑⌽¨h¨↑r ⍝ reverse and transpose to the correct orientation
   ^
  / \  ^   ^
 /   \/ \ / \
/        v   \
\$\endgroup\$
3
\$\begingroup\$

Ruby, 390 characters

Whew, this one was tricky.

I ended up having to append to a new string for each character, using a variable s that meant "skip next character" which was needed for processing ^ and \.

This output exactly the given sample output for all of the test cases.

m=[gets.chomp]
a=m[0].scan(/\d/).max.to_i
m[0].gsub!(/./){|n|n==?. ? ' ':a-n.to_i}
s=nil
until a==0
o=''
m[-1].chars{|c|o+=case c
when ?0;?^
when ' ';t=s;s=nil;t ? '':' '
when /\d/;(c.to_i-1).to_s
when ?^;s=1;o.slice! -1;"/ \\"
when ?/;t=s;s=nil;t ? "#{o.slice! -1;' '}":o.slice!(-1)=='\\' ? 'v ':"/ "
when ?\\;s=1;' \\'
when ?v;' '
end}
m.push o
a-=1
end
puts (m[1..-1]*"\n").gsub /\d/,' '

Chart of what the variables mean:

m | The mountain array.
a | The highest height of a mountain. Used for counting when to stop.
s | Whether or not to skip the next character. 1 for yes, nil for no.
o | Temp string that will be appended to mountain.
t | Temp variable to hold the old value of s.

I'm sure I could golf it down much more, but I have to go now. Shall be improved later!

\$\endgroup\$
  • \$\begingroup\$ I'm struggling with the input .2.2. and can't see why it doesn't work. \$\endgroup\$ – Howard Feb 27 '14 at 7:46
2
\$\begingroup\$

Java, 377 407

Edit: @Victor pointed out that this needed to be a complete program, so I added a few dozen characters to make it compilable and runnable. Just pass the "purchase order" as the first param when executing the program, like so: java M ..3.4..6..4.3..

I think this is similar in spirit to other answers, basically just traverses the "mountain order" repeatedly for every possible height, and builds the mountains from the tops down. That way I only have to deal with four conditions if not building a peak -- either an up slope '/', down slope '\, joint 'v', or empty ' '. I can discover that simple by looking at the three spaces centered "above" my current position in my top-down build.

Note that like other submissions, I treat anything other than a number as equivalent to '.' in the input, for brevity.

Golfed version:

class M{public static void main(String[]m){char[]n=m[0].toCharArray();int e=n.length,h=9,x=-1,p;char[][]o=new char[11][e];char l,r,u;boolean a,b,c;for(;h>=0;h--){for(p=0;p<e;p++){if(n[p]-49==h){o[h][p]=94;if(x==-1)x=h;}else{l=(p>0)?o[h+1][p-1]:0;r=(p<e-1)?o[h+1][p+1]:0;u=o[h+1][p];a=l>91&&l<99;b=r==94||r==47;c=u<33;o[h][p]=(char)((a&&b)?'v':(c&&b)?47:(c&&a)?92:32);}}if(x>=h)System.out.println(o[h]);}}}

Human readable form (and without some of the equivalent transmogrifications to achieve golf form):

class Magrathea2 {
    public static void main(String[] mountain) {
        String out = "";
        char[][] output = new char[11][mountain[0].length()];
        int height = 9; int maxheight = -1;
        int position = 0;
        char left,right,up;
        char[] mount = mountain[0].toCharArray();
        for (; height >= 0; height--) {
            for (position=0; position < mount.length; position++) {
                if (mount[position]-49 == height) {
                    output[height][position] = '^';
                    if (maxheight==-1) {
                        maxheight=height;
                    }
                } else { // deal with non-numbers as '.'
                    left=(position>0)?output[height+1][position-1]:0;
                    right=(position<mount.length-1)?output[height+1][position+1]:0;
                    up=output[height+1][position];
                    if ((left=='^'||left=='\\')&&(right=='^'||right=='/')) {
                        output[height][position]='v';
                    } else if ((up==' '||up==0)&&(right=='/'||right=='^')) {
                        output[height][position]='/';
                    } else if ((up==' '||up==0)&&(left=='\\'||left=='^')) {
                        output[height][position]='\\';
                    } else {
                        output[height][position]=' ';
                    }
                }
            }
            if (maxheight >= height) {
                out+=new String(output[height]);
                if (height > 0) {
                    out+="\n";
                }
            }
        }
        System.out.println(out);
    }
}

Enjoy.

Example output:

$ java M ..3..4...6...5....1
         ^
        / \  ^
     ^ /   \/ \
  ^ / v        \
 / v            \
/                \^
\$\endgroup\$
  • \$\begingroup\$ The question mentions Write a complete program, so please, add the missing class X{public static void main(String[]z){. \$\endgroup\$ – Victor Stafusa Mar 1 '14 at 21:25
  • \$\begingroup\$ Right on. I got misdirected by the next section of that sentence -- "or as argument" and missed the complete program part. I'll update it shortly. \$\endgroup\$ – ProgrammerDan Mar 2 '14 at 1:36
2
\$\begingroup\$

Perl 6, 264 224 216 206 200 194 124 bytes

$_=get;my$a=10;((s:g/$a/^/;s:g/\s\.\s/ v /;s:g'\.\s'/ ';s:g/\s\./ \\/;$!=say TR/.1..9/ /;tr'^\\/v' ')if .match(--$a)|$!)xx 9

Thanks to @JoKing for showing a s/// solution. This is golfed a bit further after fixing the tr/// bug in Perl 6.

My original solution with subst:

my$t=get;for 9...1 {if $t.match($_)|$! {$t=$t.subst($_,'^',:g).subst(' . ',' v ',:g).subst('. ','/ ',:g).subst(' .',' \\',:g);$!=say $t.subst(/<[\.\d]>/,' ',:g);$t.=subst(/<[^\\/v]>/,' ',:g)};}

Ungolfed:

my $t=slurp;
my $s;
for 9...1 {
    if $t.match($_)||$s {                    # match number or latched
        $t=$t.subst($_,'^',:g)               # peaks
        .subst(' . ',' v ',:g)               # troughs
        .subst('. ','/ ',:g)                 # up slope
        .subst(' .',' \\',:g);               # down slope
        $s=say $t.subst(/<[\.\d]>/,' ',:g);  # clean, display, latch
        $t=$t.subst(/<[^\\/v]>/,' ',:g)      # wipe for next line
    }
}

Output:

...4...3...33..4..4....2.3.22.33.5..22...333.222.3..
                                 ^                  
   ^           ^  ^             / \                 
  / \  ^   ^^ / \/ \     ^    ^^   \     ^^^     ^  
 /   \/ \ /  v      \  ^/ \^^/      ^^  /   \^^^/ \ 
/        v           \/               \/           \
\$\endgroup\$
  • 1
    \$\begingroup\$ I don't think Perl strictly needs a main function, the entry point can just be the first thing outside a function. \$\endgroup\$ – Nissa Oct 25 '18 at 23:38
  • \$\begingroup\$ I used main for parameter handling. Now using stdin. Thanks. \$\endgroup\$ – donaldh Oct 26 '18 at 9:24
  • \$\begingroup\$ A procedural solution. I'm sure someone can do better with regexes and hyperops. \$\endgroup\$ – donaldh Oct 26 '18 at 9:43
  • 1
    \$\begingroup\$ 131 bytes using s/// and tr///. I think that last one can use tr instead of s but I can't quite figure out to translate backslashes. Maybe the first one too \$\endgroup\$ – Jo King Oct 26 '18 at 11:56
  • \$\begingroup\$ Nice work @JoKing – I got into a topic mess when I tried to use s/// and TR///. I see that avoiding blocks is the answer. \$\endgroup\$ – donaldh Oct 26 '18 at 14:38
1
\$\begingroup\$

Perl, 254 218 212

$s=<>;sub f{9-$i-$_[0]?$":pop}for$i(0..8){$h=1;$_=$s;s!(\.*)(\d?)!$D=($w=length$1)+$h-($2||1);join'',(map{($x=$_-int$D/2)<0?f--$h,'\\':$x?f++$h,'/':$D%2?f--$h,v:f$h,'/'}0..$w-1),$2?f$h=$2,'^':''!ge;print if/\S/}
$s=<>;
sub f{9-$i-$_[0]?$":pop}
for$i(0..8){
    $h=1;
    $_=$s;
    s!(\.*)(\d?)!
        $D=($w=length$1)+$h-($2||1);
        join'',(map{
            ($x=$_-int$D/2)<0
                ?f--$h,'\\'
                :$x
                    ?f++$h,'/'
                    :$D%2
                        ?f--$h,v
                        :f$h,'/'
        }0..$w-1),$2
            ?f$h=$2,'^'
            :''
    !ge;
    print if/\S/
}

Edit: actually it's a bug-fix to work with ProgrammerDan's ..3..4...6...5....1 example, but, in the process, some bytes were off. And online test: https://ideone.com/P4XpMU

\$\endgroup\$
1
\$\begingroup\$

C# - 321 319

using System.Linq;class P{static void Main(string[]p){int h=p[0].Max()-48,i=h,j,n=p[0].Length;char[]A=new char[n+2],B=A;for(;i-->0;){for(j=0;j++<n;){var r=(A[j+1]==47|A[j+1]==94);B[j]=(char)(p[0][j-1]==i+49?94:i+1<h?A[j]==0?(A[j-1]>90&A[j-1]<95)?r?118:92:r?47:0:0:0);}A=(char[])B.Clone();System.Console.WriteLine(B);}}}

Ungolfed and commented:

using System.Linq;

class P
{
    static void Main(string[] p)
    {
        int h = p[0].Max() - 48,    // Getting the height. Codes for 0 to 9 are 48 to 57, so subtract 48 and hope no one will input anything but dots and numbers.
            i = h,
            j,                      // Declaring some iterators here, saves a few chars in loops.
            n = p[0].Length;
        char[] A = new char[n+2],   // Creating an array of char with 2 extra members so as not to check for "index out of bounds" exceptions
               B = A;               // B is referencing the same array as A at this point. A is previous row, B is the next one.
        for (;i-->0;)               // Looping from top to the bottom of the mountain
        {
            for (j = 0; j++ < n;)   // Looping from left to right.
            {
                var r = (A[j + 1] == 47 | A[j + 1] == 94);  // This bool is used twice, so it saves a few characters to make it a variable

                // Here's the logic
                B[j] = (char)(p[0][j - 1] == i + 49 ? 94    // If at this position in the string we have a number, output "^"
                                           : i + 1 < h ?    // And if not, check if we're on the top of the mountain
                                             A[j] == 0 ?    // If we're not at the top, check if the symbol above is a space (0, actually)
                                            (A[j - 1] > 90 & A[j - 1] < 95) ?   // If there's nothing above, we check to see what's to the left ( ^ or \ )
                                             r ?            // And then what's to the right ( ^ or / )
                                             118            // If there are appropriate symbols in both locations, print "v"
                                           : 92             // If there's only a symbol to the left, print "\"
                                           : r              // Otherwise check if there's a symbol to the right, but not to the left
                                           ? 47             // And if there is, print "/"
                                           : 0 : 0 : 0);    // Print nothing if there aren't any symbols above, to the left and to the right,
                                                            // or there's a "^" right above, or we're at the top of the mountain
            }
            A=(char[])B.Clone();    // Clone arrays to iterate over the next line
            System.Console.WriteLine(B);
        }
    }
}

Example:

C:\>program .2..3..4...
        ^
     ^ / \
  ^ / v   \
 / v       \

I think it outputs an extra space before each line, though.

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1
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CJam, 128 117 112 106 104 bytes

CJam is a bit younger than this challenge so this answer does not compete. This was a very nice challenge though! From the little I know about J and APL, I think a submission in those would be impressively short.

WlW++"."Waer{_{~U(e>:U}%\W%}2*;W%]z{$W=}%_$W=S*\:L,2-,\f{\_)L=(~"^/ ^^/ \v ^ \\"S/2/@L>3<_$0=f-{=}/t}zN*

Here is a test case, which I think contains all possible possible combinations of slopes, peaks and troughs:

...4...3...33..4..4....2.3.22.33.5..22...333.222.3..

which yields

                                 ^                  
   ^           ^  ^             / \                 
  / \  ^   ^^ / \/ \     ^    ^/   \     ^^^     ^  
 /   \/ \ /  v      \  ^/ \^^/      \^  /   \^^^/ \ 
/        v           \/               \/           \

Test it here.

I'll add an explanation for the code later.

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1
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Python, 297 234 218

-63 bytes thanks to Jo King
-16 bytes with r=s.replace instead of lambda

s=input()
r=s.replace
q=0
j=''.join
for i in range(9):
 if`9-i`in s or q:q=s=r(`9-i`,'^');s=r(' . ',' v ');s=r('. ','/ ');s=r(' .',' \\');print j([x,' '][x in'0123456789.']for x in s);s=j([x,' '][x in'/\^v']for x in s)

Takes input from STDIN. Ungolfed, simplified:

s=input() # Take input
r=lambda y,z: s.replace(y,z) # Function for quick s.replace(a, b)
j=lambda x: ''.join(x)
q=0 # Acts like boolean
for i in range(9): # Count to 9
 if `9-i`in s or q: # When digit has been found or found previously (no newlines at start)
  q=s=r(`9-i`,'^') # Digit to ^, set q to non-zero value for always executing from now on
  s=r(' . ',' v ') # ' . ' to ' v '
  s=r('. ','/ ') # '. ' to '/ '
  s=r(' .',' k') # ' .' to 'k'. K is a placeholder, since \\ takes two chars and `[...]`[2::5] fails
  print j([x,' '][x in'0123456789.']for x in s) # Print without '0123456789.'
  s=j([x,' '][x in'/\^v']for x in s) # Wipe (delete '/^\v`)
\$\endgroup\$
  • \$\begingroup\$ 234 bytes \$\endgroup\$ – Jo King Oct 26 '18 at 12:43
  • 1
    \$\begingroup\$ Yeah, I tried the s.replace method myself, but it doesn't work. You're just performing replacements on the original string since strings are immutable \$\endgroup\$ – Jo King Oct 27 '18 at 0:30
0
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Powershell, 148 145 bytes

It's a nice challenge!

param($s)9..1|?{($p+=$s-match$_)}|%{"$_,^; \. , v ;\. ,/ ; \., \;\^|\\|/|v, "-split';'|%{$x=$s-replace'\.|\d',' '
$s=$s-replace($_-split',')}
$x}

Less golfed test script:

$f = {

param($s)
9..1|?{($p+=$s-match$_)}|%{      # loop digits form 9 downto 1, execute to the end as soon as a suitable digit met
    $s=$s-replace$_,'^'          # replace current digit with '^'
    $s=$s-replace' \. ',' v '    # replace ' . '  with ' v '
    $s=$s-replace'\. ','/ '      # replace '. ' with '/ '
    $s=$s-replace' \.',' \'      # replace ' .' with ' \'
       $s-replace'\.|\d',' '     # replace all dots and digits with ' ' and push to output. Don't store this replacement
    $s=$s-replace'\^|\\|/|v',' ' # prepeare to the next step: replace ^ \ / and v to space
}

    # Example:
    #     $s="...4...3...3.."
    # 4 : $s="...^...3...3.." output: "   ^          "
    # 4 : $s="... ...3...3.."
    # 3 : $s="../ \..^...^.." output: "  / \  ^   ^  "
    # 3 : $s="..   .. ... .."
    # 2 : $s="./   \/ \./ \." output: " /   \/ \ / \ "
    # 2 : $s=".        .   ."
    # 1 : $s="/        v   \" output: "/        v   \"
    # 1 : $s="              "

}

@(
    ,("1",
      "^")

    ,("11",
      "^^")

    ,("1.2.",
    "  ^ ",
    "^/ \")

    ,(".2.3..",
      "   ^  ",
      " ^/ \ ",
      "/    \")

    ,(".2..3..",
      "    ^  ",
      " ^ / \ ",
      "/ v   \")

    ,("...4...3...3..",
      "   ^          ",
      "  / \  ^   ^  ",
      " /   \/ \ / \ ",
      "/        v   \")

    ,("...4...3...33..4..4....2.3.22.3..5...22...333.222.3..",
      "                                 ^                   ",
      "   ^           ^  ^             / \                  ",
      "  / \  ^   ^^ / \/ \     ^    ^/   \      ^^^     ^  ",
      " /   \/ \ /  v      \  ^/ \^^/      \^^  /   \^^^/ \ ",
      "/        v           \/                \/           \")

    ,(".2..3..6.....5...3......1..3..4....2.",
      "       ^                             ",
      "      / \    ^                       ",
      "     /   \  / \               ^      ",
      "    ^     \/   \ ^         ^ / \     ",
      " ^ /            v \       / v   \  ^ ",
      "/ v                \    ^/       \/ \")
) | % {
    $s,$expected = $_
    $result = &$f $s
    "$result"-eq"$expected"
    $s
    $result
}

Output:

True
1
^
True
11
^^
True
1.2.
  ^
^/ \
True
.2.3..
   ^
 ^/ \
/    \
True
.2..3..
    ^
 ^ / \
/ v   \
True
...4...3...3..
   ^
  / \  ^   ^
 /   \/ \ / \
/        v   \
True
...4...3...33..4..4....2.3.22.3..5...22...333.222.3..
                                 ^
   ^           ^  ^             / \
  / \  ^   ^^ / \/ \     ^    ^/   \      ^^^     ^
 /   \/ \ /  v      \  ^/ \^^/      \^^  /   \^^^/ \
/        v           \/                \/           \
True
.2..3..6.....5...3......1..3..4....2.
       ^
      / \    ^
     /   \  / \               ^
    ^     \/   \ ^         ^ / \
 ^ /            v \       / v   \  ^
/ v                \    ^/       \/ \
\$\endgroup\$
0
\$\begingroup\$

Pip -l, 100 bytes

Y#aZGMXaFi,#aIh:+a@i{(yi--h):4j:0Wh-j&++(yi-++jh-j)(yi+jh-j):2}RV Z(J*y)R`.(?=.*[^0])`0R,6;^" /\v^^"

(The language is newer than the question, but probably isn't going to beat the APL submission anyway. Although I hope it will get much shorter.)

Takes input via command-line argument. Try it online!

\$\endgroup\$

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