Code Golf: Lasers

Answer 1

Perl, 177 Characters

The first linebreak can be removed; the other two are mandatory.

$/=%d=split//,' >/^\v';$_=<>;$s='#';{
y/v<^/>v</?do{my$o;$o.=" 
"while s/.$/$o.=$&,""/meg;y'/\\'\/'for$o,$s;$_=$o}:/>x/?die"true
":/>#/?die"false
":s/>(.)/$s$d{$1}/?$s=$1:1;redo}

Explanation:

$/ = %d = (' ' => '>', '/' => '^', '\\' => 'v');

If a right-moving beam runs into an {empty space, up-angled mirror, down-angled mirror} it becomes a {right-moving beam, up-moving beam, down-moving beam}. Initialize $/ along the way -- fortunately "6" is not a valid input char.

$_ = <>;

Read the board into $_.

$s="#";

$s is the symbol of whatever the beam is sitting on top of now. Since the laser emitter is to be treated like a wall, set this to be a wall to begin with.

if (tr/v<^/>v</) {
  my $o;
  $o .= "\n" while s/.$/$o .= $&, ""/meg;
  tr,/\\,\\/, for $o, $s;
  $_ = $o;
}

If the laser beam is pointing any way except right, rotate its symbol, and then rotate the whole board in place (also rotating the symbols for the mirrors). It's a 90 degree left rotation, accomplished effectively by reversing the rows while transposing rows and columns, in a slightly fiendish s///e with side effects. In the golfed code, the tr is written in the form y''' which allows me to skip backslashing one backslash.

die "true\n" if />x/; die "false\n" if />#/;

Terminate with the right message if we hit the target or a wall.

$s = $1 if s/>(.)/$s$d{$1}/;

If there's an empty space in front of the laser, move forward. If there's a mirror in front of the laser, move forward and rotate the beam. In either case, put the "saved symbol" back into the old beam location, and put the thing we just overwrote into the saved symbol.

redo;

Repeat until termination. {...;redo} is two characters less than for(;;){...} and three less than while(1){...}.

Answer 2

C# 3.0

259 chars

bool S(char[]m){var w=Array.FindIndex(m,x=>x<11)+1;var s=Array.FindIndex(m,x=>x>50&x!=92&x<119);var t=m[s];var d=t<61?-1:t<63?1:t<95?-w:w;var u=0;while(0<1){s+=d;u=m[s];if(u>119)return 0<1;if(u==47|u==92)d+=d>0?-w-1:w+1;else if(u!=32)return 0>1;d=u>47?-d:d;}}

Slightly more readable:

bool Simulate(char[] m)
{
    var w = Array.FindIndex(m, x => x < 11) + 1;
    var s = Array.FindIndex(m, x => x > 50 & x != 92 & x < 119);
    var t = m[s];
    var d = t < 61 ? -1 : t < 63 ? 1 : t < 95 ? -w : w;
    var u = 0;
    while (0 < 1)
    {
        s += d;
        u = m[s];
        if (u > 119)
            return 0 < 1;
        if (u == 47 | u == 92)
            d += d > 0 ? -w - 1 : w + 1;
        else if (u != 32)
            return 0 > 1;
        d = u > 47 ? -d : d;
    }
}

The main waste of chars seems to be in finding the width of the map and the position of the laser source. Any ideas how to shorten this?

Answer 3

Python - 152

Reads input from a file called "L"

A=open("L").read()
W=A.find('\n')+1
D=P=-1
while P<0:D+=1;P=A.find(">^<v"[D])
while D<4:P+=[1,-W,-1,W][D];D=[D,D^3,D^1,4,5][' \/x'.find(A[P])]
print D<5

To read from stdin replace the first line with this

import os;A=os.read(0,1e9)

If you need lowercase true/false change the last line to

print`D<5`.lower()

Answer 4

C++: 388 characters

#include<iostream>
#include<string>
#include<deque>
#include<cstring>
#define w v[y][x]
using namespace std;size_t y,x,*z[]={&y,&x};int main(){string p="^v<>",s;deque<string>v;
while(getline(cin,s))v.push_back(s);while(x=v[++y].find_first_of(p),!(x+1));int 
i=p.find(w),d=i%2*2-1,r=i/2;do while(*z[r]+=d,w=='/'?d=-d,0:w==' ');while(r=!r,
!strchr("#x<^v>",w));cout<<(w=='x'?"true":"false");}

(318 without headers)

---

How it works:

First, all lines are read in, then, the laser is found. The following will evaluate to 0 as long as no laser arrow was found yet, and the same time assign to x the horizontal position.

x=v[++y].find_first_of(p),!(x+1)

Then we look what direction we found and store it in i. Even values of i are top/left ("decreasing") and odd values are bottom/right ("increasing"). According to that notion, d ("direction") and r ("orientation") are set. We index pointer array z with orientation and add the direction to the integer we get. The direction changes only if we hit a slash, while it remains the same when we hit a back-slash. Of course, when we hit a mirror, then we always change orientation (r = !r).

Answer 5

PostScript, 359 bytes

First attempt, lots of room for improvement...

/a[{(%stdin)(r)file 99 string readline not{exit}if}loop]def a{{[(^)(>)(<)(v)]{2
copy search{stop}if pop pop}forall}forall}stopped/r count 7 sub def pop
length/c exch def[(>)0(^)1(<)2(v)3>>exch get/d exch def{/r r[0 -1 0 1]d get
add def/c c[1 0 -1 0]d get add def[32 0 47 1 92 3>>a r get c get .knownget
not{exit}if/d exch d xor def}loop a r get c get 120 eq =

Answer 6

F#, 36 lines, very readable

Ok, just to get an answer out there:

let ReadInput() =
    let mutable line = System.Console.ReadLine()
    let X = line.Length 
    let mutable lines = []
    while line <> null do
        lines <- Seq.to_list line :: lines
        line <- System.Console.ReadLine()
    lines <- List.rev lines
    X, lines.Length, lines

let X,Y,a = ReadInput()
let mutable p = 0,0,'v'
for y in 0..Y-1 do
    for x in 0..X-1 do 
        printf "%c" a.[y].[x]
        match a.[y].[x] with 
        |'v'|'^'|'<'|'>' -> p <- x,y,a.[y].[x]
        |_ -> ()
    printfn ""

let NEXT = dict [ '>', (1,0,'^','v')
                  'v', (0,1,'<','>')
                  '<', (-1,0,'v','^')
                  '^', (0,-1,'>','<') ]
let next(x,y,d) =
    let dx, dy, s, b = NEXT.[d]
    x+dx,y+dy,(match a.[y+dy].[x+dx] with
               | '/' -> s
               | '\\'-> b
               | '#'|'v'|'^'|'>'|'<' -> printfn "false"; exit 0
               | 'x' -> printfn "true"; exit 0
               | ' ' -> d)

while true do
    p <- next p    

Samples:

##########
#   / \  #
#        #
#   \   x#
# >   /  #
##########
true

##########
#   v x  #
# /      #
#       /#
#   \    #
##########
false

#############
#     #     #
# >   #     #
#     #     #
#     #   x #
#     #     #
#############
false

##########
#/\/\/\  #
#\\//\\\ #
#//\/\/\\#
#\/\/\/x^#
##########
true

##########
#   / \  #
#        #
#/    \ x#
#\>   /  #
##########
false

##########
#  /    \#
# / \    #
#/    \ x#
#\^/\ /  #
##########
false