Building Demolition

Question

Challenge

Write a full program/function to Demolish a given building at the rate 1 floor per second.

INPUT

Input is a building through STDIN(or whatever is called in your language) or through an ARGUMENT of a function. t=0 s

   |
  |#|
  {#}
 |# #|
|# # #|
|# # #|
|# # #|
TTTTTTT  

input consists of :

• | marks start and end of a floor.
• {,} marks the floor with explosives.
• # is a window which is for decoration .
• is space, which is everywhere inside the floors where # is not there.
• T marks ground(cannot be demolished).
• * marks the detonated floor.

Input rules :

• building starts from top with one | and ends on ground(with no(T)=no (char used in ground floor)).
• a window # is at every odd place inside every floor.
• T marks the end of your input.
• only one floor consists of explosives.
• every floor is made of odd no. of chars.
• as you ascend down the floors, the floors can either be of equal size of that of previous floor or can be 2 chars bigger.
• Input can be taken as an array of chars or strings.

Output :

t=0.5 s

   |
  |#|
  ***
 |# #|
|# # #|
|# # #|
|# # #|
TTTTTTT  

t=1 s

   |
  |#|
 |# #|
|# # #|
|# # #|
|# # #|
TTTTTTT  

t=1.5 s

   |
  ***
 |# #|
|# # #|
|# # #|
|# # #|
TTTTTTT  

t=2 s

   |
 |# #|
|# # #|
|# # #|
|# # #|
TTTTTTT  

t=2.5 s

   *
 |# #|
|# # #|
|# # #|
|# # #|
TTTTTTT  

t=3 s

 |# #|
|# # #|
|# # #|
|# # #|
TTTTTTT  

t=3.5 s

 *****
|# # #|
|# # #|
|# # #|
TTTTTTT  

t=4 s

|# # #|
|# # #|
|# # #|
TTTTTTT  

t=4.5 s

*******
|# # #|
|# # #|
TTTTTTT  

t=6.5 s

*******
TTTTTTT  

t=7 s

TTTTTTT  

Output rules :

• every floor ends with one newline character.
• ground may/may not have a trailing newline.
• output the building(or whats left of it) every 0.5 s.
• Output is like animation which is modified every 0.5 s and is displayed.
  (Also feel free to post GIFs with your answer)

This is a Code-Golf so the code in shortest bytes wins!
last date of submission is exactly 5 days from now

(But you know what ? Winning is not everything, feel free to try this challenge in your favorite language, even after a winner has been declared :] ).

EDIT : I changed the way you take input (my bad actually).

Answer 1

Vim, 41 38 bytes

qw:sl 500m␍q/{␍qqjk^v$r*@wdd:-␍@w@qq@q

Here, ^ is used for a literal caret; ␍ is used for CTRL-M.

Explanation

qw:sl 500m␍q sleeps half a second, while recording the half second sleep as macro w. /{␍ moves to the floor with explosives. qq begins recording macro q, which will recursively call itself.

jk moves down and up; this generates an error if you're on the last line (ground); the error terminates the recursive macro. ^v$r* replaces everything from the first non-whitespace character to the end of the line with *'s. @w waits half a second, then dd deletes the current floor. :-␍ moves up a floor without terminating the macro if you're on the top line. @w then waits another half second, and @q calls the q macro (initially empty).

q@q stops recording the macro q, then calls it, triggering the recursion.

Animation

Answer 2

JavaScript (ES6), 208 198 bytes

f=
(t,s=f=>setTimeout(f,500),v=t.value.split(/(\S.*\n)/),i=v.findIndex(s=>/}/.test(s)),c=_=>t.value=v.join``,d=_=>c(v.splice(--i,2),v[3]&&s(e,i?--i:++i)),e=_=>c(v[i]=v[i].replace(/./g,'*'),s(d)))=>s(e)

<textarea id=t rows=9>
   |
  |#|
  {#}
 |# #|
|# # #|
|# # #|
|# # #|
TTTTTTT
</textarea><input type=button value=Go! onclick=f(t)>

Answer 3

Java 7, 589 477 476 bytes

import java.util.*;void x(List<String>b,int x)throws Exception{Thread.sleep(500);int i=0,l=b.size(),z=x;String w;for(;i<l;i++){System.out.println(w=b.get(i));if(w.contains("{"))x=i;}System.out.println();w=b.get(x);i=w.contains("*")?1:0;if(i>0)b.remove(x);else b.set(x,z<0?r(w,'{','}'):r(w,'|','|'));if(l>1)x(b,i>0&x>0?x-1:x);}String r(String s,char y,char z){int a,b;return s.substring(0,a=s.indexOf(y))+s.substring(a,b=s.lastIndexOf(z)+1).replaceAll(".","*")+s.substring(b);}

Ok, it's a mess/long, but this challenge has so many annoying things for Java.. Printing multi-line; Thread.sleep(500) which requires a throws Exception; replacing a substring between two delimiters with an equal amount of *; etc.
All this causes the program to be pretty big.. It can defintely be golfed some more, maybe even halved with a different approach, but at least there is an answer now. ;)

Ungolfed:

void x(List<String>b, int x) throws Exception{
  Thread.sleep(500);
  int i = 0,
      l = b.size(),
      z = x;
  String w;
  for(;i<l; i++){
    System.out.println(w=b.get(i));
    if(w.contains("{")){
      x = i;
    }
  }
  System.out.println();
  w = b.get(x);
  i = s.contains("*")
       ? 1
       : 0;
  if(i>0){
    b.remove(x);
  }
  else{
    b.set(x, z < 0
              ? r(s, '{', '}')
              : r(s, '|', '|'));
  }
  if(l>1){
    x(b, i > 0 & x > 0
          ? x-1
          : x);
  }
}

String r(String s, chary, char z){
    int a, b;
    return s.substring(0, a=s.indexOf(y)) + s.substring(a, b=s.lastIndexOf(z) + 1).replaceAll(".", "*") + s.substring(b);
}

Test code:

import java.util.*;

class M{
  void x(List<String>b,int x)throws Exception{Thread.sleep(500);int i=0,l=b.size(),z=x;String w;for(;i<l;i++){System.out.println(w=b.get(i));if(w.contains("{"))x=i;}System.out.println();w=b.get(x);i=w.contains("*")?1:0;if(i>0)b.remove(x);else b.set(x,z<0?r(w,'{','}'):r(w,'|','|'));if(l>1)x(b,i>0&x>0?x-1:x);}String r(String s,char y,char z){int a,b;return s.substring(0,a=s.indexOf(y))+s.substring(a,b=s.lastIndexOf(z)+1).replaceAll(".","*")+s.substring(b);}

  public static void main(String[] a){
    try{
        List<String> l = new ArrayList(){{
            add("   |   ");
            add("  |#|  ");
            add("  |#|  ");
            add(" {# #} ");
            add("|# # #|");
            add("|# # #|");
            add("|# # #|");
            add("TTTTTTT");
        }};
        new M().c(l, -1);
    }
    catch(Exception e){}
  }
}

Try it here. (On ideone it outputs at once and ignores the sleep..)

Answer 4

C, 314 287 281 271 bytes

s,e,t,i;f(char**b){char*p;do{system("CLS");i=0;do{while(!(p=b[i]))i++;if(!*b&&e==1)e=i;while(*p<33)putchar(*p++);if(!t)e=*p=='{'?i:e,s+=2;do{putchar(e==i&&t%2&&*p>16?'*':*p);}while(*p&&*p++>16);}while(*b[i++]-84);if(t++%2)b[e]=0,*b&&e>0?e--:e++;Sleep(500);}while(t<s-1);}

-10 after changing != to - and avoiding char literals when it is possible, as well as isspace (many thanks to H Walters). But ungolfed code remains unchanged.

s,e,t,i;f(char**b){char*p;do{system("CLS");i=0;do{while(!(p=b[i]))i++;if(!*b&&e==1)e=i;while(isspace(*p))putchar(*p++);if(!t)e=*p=='{'?i:e,s+=2;do{putchar(e==i&&t%2&&*p>0xF?'*':*p);}while(*p&&*p++>0xF);}while(*b[i++]!='T');if(t++%2)b[e]=0,*b&&e>0?e--:e++;Sleep(500);}while(t<s-1);}

-6 bytes after applying coma and deletion of {} after two if.

s,e,t,i;f(char**b){char*p;do{system("CLS");i=0;do{while(!(p=b[i]))i++;if(!*b&&e==1)e=i;while(isspace(*p))putchar(*p++);if(!t){s+=2;e=*p=='{'?i:e;}do{putchar(e==i&&t%2&&*p>0xF?'*':*p);}while(*p&&*p++>0xF);}while(*b[i++]!='T');if(t++%2){b[e]=0;e=*b&&e>0?e-1:e+1;}Sleep(500);}while(t<s-1);}

-26 bytes after little optimization, removal unnecessary parentheses, as well as change local variables to global (with auto 0 initialization) and b[0] by *b.

f(char**b){int s=0,e=0,t=0;char*p;do{system("CLS");int i=0;do{if(!t){s+=2;if(strchr(b[i],'}'))e=i;printf(b[i]);}else{while(!(p=b[i]))i++;if(!b[0]&&e==1)e=i;do{putchar((e==i&&t%2&&!isspace(*p))?'*':*p);}while(*p&&*p++!='\n');}}while(b[i++][0]!='T');if(t%2){b[e]=0;e=(b[0]&&e)?e-1:e+1;}t++;Sleep(500);}while(--s>1);}

Test code with ungolfed f:

#include <stdio.h>
#include <windows.h> // to use Sleep and system

s, e, t, i;
f(char**b)
{
    char*p;
    do{
        system("CLS");
        i = 0;
        do
        {
            while (!(p=b[i]))i++; // skip demolished floors
            if (!*b && e==1) e = i;
            while (isspace(*p)) putchar(*p++); // print spaces 
            if (!t){ s += 2; e = *p == '{' ? i : e; } // find a bomb and count steps at the first iteration
            do{
                putchar(e == i && t%2 &&*p>0xF ? '*' : *p); // print floor or * for next floor at odd step
            } while (*p && *p++ >0xF); // >0xF is instead of !='\n'
        } while (*b[i++] != 'T'); // until the ground
        if (t++ % 2)
        {
            b[e] = 0; // delete the demolished floor
            e = *b&&e>0 ? e-1 : e+1; // and determine next floor to be demolished
        }
        Sleep(500);
    } while (t<s-1);
}

int main(void)
{
    char * arr[] = { "   |\n",
                     "  |#|\n",
                     "  {#}\n",
                     " |# #|\n",
                     "|# # #|\n",
                     "|# # #|\n",
                     "|# # #|\n",
                     "TTTTTTT" };
    f(arr);
}

Answer 5

Haskell, 245 221 bytes

import System.Posix.Unistd
r=reverse
c#l|(i,j)<-span(<'!')l=i++(c<$j)
f[b]=[[b]]
f s|(h@(a:b),d:c)<-break(elem '{')s=(h++'*'#d:c):(h++c):f(init h++'{'#last h:c)|1<2=r<$>(f$r s)
mapM((usleep 500000>>).mapM putStrLn).init.f

Usage example:

mapM((usleep 500000>>).mapM putStrLn).init.f $ ["  |"," {#}"," |#|","|# #|","|# #|","TTTTT"]

How it works:

c#l|(i,j)<-span(<'!')l=i++(c<$j)    -- helper function that takes a char c and
                                    -- a string l and keeps the leading spaces
                                    -- of l and replaces the rest with c

                                    -- f takes a building (list of strings) and
                                    -- makes the sequence of collapsing buildings
                                    -- as a list of list of strings
f[b]=[[b]]                          -- base case. Stop when there's only a single
                                    -- floor left
f s                                 -- if the building has at least two floors
   |(h@(a:b),d:c)<-break(elem '{')s --   if there is at least one floor above the
                                    --   explosion
        =(h++'*'#d:c) :             --     return the current building with the
                                    --     floor to explode replaced by *
        (h++c) :                    --     followed by the current building with
                                    --     the exploded floor removed 
        f(init h++'{'#last h:c)     --     followed by a recursive call
                                    --     with the next to explode floor marked
                                    --     with '{'
   |1<2=r<$>(f$r s)                 --   if all top floors have exploded, reverse
                                    --   the left over floors, let them explode
                                    --   and reverse the floors of each building
                                    --   again.

                      f             -- let the building collapse
                 init               -- f comes with an additional building with
                                    -- a single floor of * only -> drop it
mapM(     )                         -- for every building
     (usleep 500000>>)              --   delay 0.5 sec
             mapM putStrLn          --   and print the floors

Note: there's also threadDelay from GHC.Conc instead of usleep from System.Posix.Unistd which is a little bit shorter, but it only works with the GHC compiler, so it wouldn't be a generic Haskell answer.

Answer 6

Perl, 153 bytes

for(@a=<>;$a[$i]!~/{/;$i++){}for(;$a[0]!~/^T/;`sleep .5`){if($P=!$P){$a[$i]=~s/(\S.*\S|\S)/"*" x length($1)/e}else{splice(@a,$i,1);if($i){$i--}}print@a}

This uses GNU sleep command to wait for 500 milliseconds.

Ungolfed version

for(@a=<>;$a[$i]!~/{/;$i++){}
for(;$a[0]!~/^T/;`sleep .5`){
    if($P=!$P){
       $a[$i]=~s/(\S.*\S|\S)/"*" x length($1)/e
    } else { 
       splice(@a,$i,1);
       if($i){$i--}
    }
    print @a
 }

Answer 7

PHP, 286 282 274 234 229 bytes

<?for($a=$argv,array_shift($a);!strstr($a[+$i++],"{"););while($a[0][0]!=T){$x=&$a[$i-=$i>0];$x=str_pad(substr($x,0,strspn($x," ")),strlen($x),"*");eval($p='echo join("\n",$a),"\n\n";usleep(5e5);');array_splice($a,$i,1);eval($p);}

takes input as list of strings from command line arguments (no newline character!)

save to file, run with php <filename> ' |' ' |#|' ' {#}' ' |# #|' '|# # #|' '|# # #|' '|# # #|' 'TTTTTTT'

breakdown

<?
for($a=$argv,array_shift($a);   // import input
    !strstr($a[+$i++],"{"););   // find explosives
while($a[0][0]!=T)              // loop while ground not reached:
{
    $x=&$a[$i-=$i>0];               // move up if possible, reference floor
    $x=str_pad(
        substr($x,0,strspn($x," ")  // keep leading spaces
    ),strlen($x),"*");              // replace rest with asterisks
                                    // print and wait
    eval($p='echo join("\n",$a),"\n\n";usleep(5e5);');
    array_splice($a,$i,1);          // remove current floor
    eval($p);                       // print and wait
}