Code Golf: Running Water

Question

The challenge

The shortest code by character count to identify and mark water depressions in the ASCII representation of a land from input.

Input will be an ASCII representation of a landscape, having hills, valleys and flat lands. The program should simulate what the landscape would look like if if was flooded - filling all valleys with water (character x).

The landscape will always start and stop with the character _ and will be at least 2 characters long, making the shortest input __.

A hill is defined as a raise, and should not be filled with water:

  __
_/  \_

A valley is defined as a depression and will be filled with water until a flatland is encountered:

_    _
 \__/

Input can be assumed clean and will be composed only from the characters space (), newline (\n), underscore (_), and forward and backward slashes (/ and \). Input can be seen as a continuous line, and any input that contains ambiguous line input such as _/_ or

_   _
 \_/
 / \

Is considered invalid.

Regarding underwater caves, water level should be maintained if cave level goes above water level.

Test cases

Input:
    __/\__
          \__              
             \       ___       ___________
             /      /   \_     \_
             \_____/      \__  _/
                             \/
Output:

    __/\__
          \__              
             \       ___       ___________
             /xxxxxx/   \xxxxxx\_
             \xxxxx/      \xxxxx/
                             \/

---

Input:
                                         __       ___
                                        /  \_____/
                                       / _______
                         ________     /  \     /
                   _____/        \   /__  \    \_
    ____          /               \    /__/   __/
        \_       /                 \     ____/
          \______\                 /____/

Output:
                                         __       ___
                                        /  \xxxxx/
                                       / _______
                         ________     /  \     /
                   _____/        \xxx/__  \xxxx\_
    ____          /               \xxxx/__/xxxxx/
        \xxxxxxxx/                 \xxxxxxxxx/
          \xxxxxx\                 /xxxx/

---

Input:
                                                      __     _
    _       ____                    ____        _____/  \   /
     \     /    \        __________/    \    __/  ___   /___\
      \___/      \       \               \   \___/  /_
                 /________\               \___________\

Output:
                                                      __     _
    _       ____                    ____        _____/  \xxx/
     \xxxxx/    \xxxxxxxxxxxxxxxxxx/    \xxxxxx/  ___   /xxx\
      \xxx/      \xxxxxxx\               \xxx\___/xx/_
                 /xxxxxxxx\               \xxxxxxxxxxx\

Code count includes input/output (i.e full program).

Answer 1

C - 741 621 600 characters (but handles the new cases properly)

$ gcc water.c && ./a.out < test6.txt 
                                     __       ___    
                                    /  \xxxxx/       
                                   / _______         
                     ________     /  \     /         
               _____/        \xxx/__  \xxxx\_        
____          /               \xxxx/__/xxxxx/        
    \xxxxxxxx/                 \xxxxxxxxx/           
      \xxxxxx\                 /xxxx/                

#include<stdio.h>
char d[99][99],*p,*e,*z,*s=d,c,S=' ',D='-',O='.',U='_';n,w,x,N=99,i;
g(y){for(i=0;!i;p+=N,e+=N){i=*p==D;for(z=p;z!=e;z+=y){if(*z!=O&&*z!=
D)break;*z=*z==O?S:U;}}}f(char*n,int r){if(*n==O||*n==D){*n=r>0?'x':
S;int k;for(k=0;k<9;k++)f(n+k/3*N-N+k%3-1,r+k/3-1);}}main(){for(p=s;
gets(p);p+=N,n++){x=strlen(p)-1;w=x>w?x:w;}for(p=s,e=d[N];p<s+N;p++)
{for(i=1,z=p;z<e;z+=N)c=*z,c==0?*z=c=S:0,i?c==S?*z=O:c==U?*z=D:0:0,(
c=='/'&&z[1]!=U)||(c=='\\'&&z[-1]!=D)||c==U?i=1-i:0;}p=s;e=s+w;g(1);
p=s+w;e=s;g(-1);for(p=s;p<s+w;p++){for(z=p;*z==S;z+=N);f(z,1);}for(i
=0;i<n;i++)printf("%.*s\n",w+1,d[i]);}

Answer 2

Python, 702 805 794 778 758 754 710 651

Handles DigitalRoss's test cases, as well as large test cases such as http://pastie.org/708764.

Example run

$ python runningwater.py < test4.txt                   
                                           ____________________________
                                          /           
                                   _      \        __
                                  / \xxxxx/       /  \
                  ___       _____/  /xxx/        /    \
____________     /   \xxxxx/   ____/xxx/ __     /xxxxxx\ 
            \xxx/    /xxxxx\__ \xxxxxx/ /xx\___/xxxxxxx/
                 ___/xxxxxxxxx\____    /xxxxxxxxxxxxxx/
                /xxxxx/      \xxxxx\__/x/    \xxxxxxx/
               /xxxxx/        \xxxxxxxx/      \xxxxx/
               \xxxxx\    _________            \xxx/
                  \xxx\  /xxxxxxxxx\           /xx/
                     \x\ \x\   /\ \x\         /xx/
    __________        \x\ \x\_/x/ /x/        /xx/
   /xxxxxxxxxx\        \x\ \xxx/ /x/        /xx/
  /xxxxxxxxxxxx\        \x\ \x/ /x/        /xx/
  \xxxxxxxxxxxxx\        \x\   /x/        /xx/
         \xxxxxxx\        \x\_/x/        /xx/
     ____/xxx/ \xx\        \xxx/        /xx/
     \xxxxxx/   \xx\___________________/xx/
      \xx/       \xxxxxxxxxxxxxxxxxxxxxxx/

Code

import sys
q=sys.stdin.readlines()
e=enumerate
s=type
k=int
o=[]
t=[0]*max(map(len,q))
n=1
L=[]
l={}
for p,d in e(q):
 w=a=0;o+=[[]]
 for i,c in e(d):
  T=t[i];C=[[c,T]];D=d[i+1:];b=0;o[-1]+=C;L+=C
  if c in'_ ':
   if('/'in D or '\\'in D)*(T%2-1)*w*p:
    for j in range(max(i-1,0),min(i+2,len(o[p-1]))):R=o[p-1][j][0];b=R*(k==s(R))or b
    for x in L:x[0]=b*(x[0]==a)or x[0]
    a=C[0][0]=b or a or n
  elif c in'\\/':w=1;a=0;n+=1
  D=d[i-1]+c;t[i-1]+=(D=='/_');t[i]+=(c in'_/\\')+(D=='_\\')
for i,a in e(o):
 for c,r in a:
  if(r==0)*(s(c)==k):l[c]=1
 for j,(c,r)in e(a):
  if(c in l)-1:a[j]=q[i][j],0
 print''.join((k==s(x))*'x'or x for x,r in a),

Answer 3

Perl, 534 545 550 566 569 567 578 594 596

sub i{$a=1;$a^=substr(x.$l[$_],$_[0],3)=~/^(.[_y]|.\/[^_]|[^_]\\)/for 0..$r-1;
$a}sub f{$c=$e-$s;$_=$l[$r];$f=s/(.{$s})(.{0,$c})/$1<$2>/;(/[ _x]>/&i$e-1and$f=
/>[ _xy]*[\\\/]/,$e=$+[0]-2)or/[ _]*>/,$e=$-[0]-1;(/<[ _x]/&i$s and$f&=
/[\\\/][ _xy]*</,$s=$-[0])or/<[ _]*/,$s=$+[0]-1;$f&$s<$e&&substr($l[$r],$s,$e-$s
)=~s!([\\/][ _xy]*)([\\/][ _]*)!($t=$1)=~y/ _/xy/,$t.$2!eg,$r--&&&f}$q=@l=<>;
while($q--){i$-[0]+1and substr($l[$r--],$-[1],length$1)=~y/_y/x/,$s=$-[0],$e=
$+[0],$q&&f while$l[$r=$q]=~m~\\/|[\\/]([_y]+)[\\/]~g}y/y/x/,print for@l

This handles all the test cases that I've seen. Newlines are optional and are only there for formatting.

Call it as e.g. perl water.pl test.txt.

Here's another funny edge case (for my algorithm anyway) not in any of the previous examples:

__      _
  \__  /
    /_/

The verbose version I'd put up earlier still fails on that.

Answer 4

Ruby, 794 759 769 752 715 692 663 655 626  616

Additional test cases:   http://pastie.org/708281   and   http://pastie.org/708288   and   http://pastie.org/708310

Compressed except for indent:

def g i,j,&f
  t=[-1,0,1]
  t.each{|r|next if@w[i][j,1]=='_'&&r>0
    t.each{|c|a=i+r
      b=j+c
      if a>=0&&b>=0&&a<@r&&b<@c
        @t[a]||=[]
        if r!=0&&c!=0
          k=@w[a][j,1]
          l=@w[i][b,1]
          next if/[\/\\]/=~k+l&&((k!=l)||((r<=>0)==(c<=>0)?k!='\\': k!='/'))
        end
        e=@w[a][b,1]
        z,@t[a][b]=@t[a][b],1
        return 1if !z&&(e==' '||r>=0&&e=='_')&&yield(a,b,f)
      end}}
  nil
end
w=$stdin.readlines
@c=w.map{|e|e.size}.max-1
@w=w=w.map{|e|e.chomp.ljust@c}
z=w.map{|e|e.dup}
@r=w.size
@r.times{|r|@m=r
  @c.times{|c|e=w[r][c,1]
    z[r][c]='x'if(e==' '||e=='_')&&(@t=[]
      !g(r,c){|u,v,f|u>=@m and v==0||v==@c-1||g(u,v,&f)})&&(@t=[]
      g(r,c){|u,v,f|u==0||g(u,v,&f)})}}
puts z