Code Golf: Lasers

Answer 1

Ruby - 146 Chars

A=$<.read
W=A.index('
')+1
until
q=A.index(">^<v"[d=d ?d+1:0])
end
while d<4
d=[d,d^3,d^1,4,5][(' \/x'.index(A[q+=[1,-W,-1,W][d]])or 4)]
end
p 5>d

Answer 2

Ruby, 176 characters

x=!0;y=0;e="^v<>#x";b=readlines;b.map{|l|(x||=l=~/[v^<>]/)||y+=1};c=e.index(b[y][x])
loop{c<2&&y+=c*2-1;c>1&&x+=2*c-5;e.index(n=b[y][x])&&(p n==?x;exit);c^='  \/'.index(n)||0}

I used a simple state machine (like most posters), nothing fancy. I just kept whittling it down using every trick I could think of. The bitwise XOR used to change direction (stored as an integer in the variable c) was a big improvement over the conditionals I had in earlier versions.

I have a suspicion that the code that increments x and y could be made shorter. Here is the section of the code that does the incrementing:

c<2&&y+=c*2-1;c>1&&x+=(c-2)*2-1

Edit: I was able to shorten the above slightly:

c<2&&y+=c*2-1;c>1&&x+=2*c-5

The current direction of the laser c is stored as follows:

0 => up
1 => down
2 => left
3 => right

The code relies on this fact to increment x and y by the correct amount (0, 1, or -1). I tried rearranging which numbers map to each direction, looking for an arrangement that would let me do some bitwise manipulation to increment the values, because I have a nagging feeling that it would be shorter than the arithmetic version.

Answer 3

C + ASCII, 197 characters:

G[999],*p=G,w,z,t,*b;main(){for(;(*p++=t=getchar()^32)>=0;w=w|t-42?w:p-G)z=t^86?t^126?t^28?t^30?z:55:68:56:75,b=z?b:p;for(;t=z^55?z^68?z^56?z^75?0:w:-w:-1:1;z^=*b)b+=t;puts(*b^88?"false":"true");}

This C solution assumes an ASCII character set, allowing us to use the XOR mirror trick. It's also incredibly fragile - all the input lines must be the same length, for example.

It breaks under the 200 character mark - but dang it, still haven't beaten those Perl solutions!

Answer 4

C89 (209 characters)

#define M(a,b)*p==*#a?m=b,*p=1,q=p:
*q,G[999],*p=G;w;main(m){for(;(*++p=getchar())>0;)M(<,-1)M
(>,1)M(^,-w)M(v,w)!w&*p<11?w=p-G:0;for(;q+=m,m=*q&4?(*q&1?
-1:1)*(m/w?m/w:m*w):*q&9?!puts(*q&1?"false":"true"):m;);}

Explanation

This monstrosity will probably be difficult to follow if you don't understand C. Just a forewarning.

#define M(a,b)*p==*#a?m=b,*p=1,q=p:

This little macro checks if the current character (*p) is equal to whatever a is in character form (*#a). If they are equal, set the movement vector to b (m=b), mark this character as a wall (*p=1), and set the starting point to the current location (q=p). This macro includes the "else" portion.

*q,G[999],*p=G;
w;

Declare some variables. * q is the light's current location. * G is the game board as a 1D array. * p is the current read location when populating G. * w is the board's width.

main(m){

Obvious main. m is a variable storing the movement vector. (It's a parameter to main as an optimization.)

    for(;(*++p=getchar())>0;)

Loop through all characters, populating G using p. Skip G[0] as an optimization (no need to waste a character writing p again in the third part of the for).

        M(<,-1)
        M(>,1)
        M(^,-w)
        M(v,w)

Use the aforementioned macro to define the lazer, if possible. -1 and 1 correspond to left and right, respectively, and -w and w up and down.

        !w&*p<11
            ?w=p-G
            :0;

If the current character is an end-of-line marker (ASCII 10), set the width if it hasn't already been set. The skipped G[0] allows us to write w=p-G instead of w=p-G+1. Also, this finishes off the ?: chain from the M's.

    for(;
        q+=m,

Move the light by the movement vector.

        m=
        *q&4
            ?(*q&1?-1:1)*(
                m/w?m/w:m*w
            )

Reflect the movement vector.

            :*q&9
                ?!puts(*q&1?"false":"true")
                :m
        ;

If this is a wall or x, quit with the appropriate message (m=0 terminates the loop). Otherwise, do nothing (noop; m=m)

    );
}

Answer 5

353 chars in Ruby:

314 277 chars now!

OK, 256 chars in Ruby and now I'm done. Nice round number to stop at. :)

247 chars. I can't stop.

223 203 201 chars in Ruby

d=x=y=-1;b=readlines.each{|l|d<0&&(d="^>v<".index l[x]if x=l.index(/[>^v<]/)
y+=1)};loop{c=b[y+=[-1,0,1,0][d]][x+=[0,1,0,-1][d]]
c==47?d=[1,0,3,2][d]:c==92?d=3-d:c==35?(p !1;exit):c<?x?0:(p !!1;exit)}

With whitespace:

d = x = y = -1
b = readlines.each { |l|
  d < 0 && (d = "^>v<".index l[x] if x = l.index(/[>^v<]/); y += 1)
}

loop {
  c = b[y += [-1, 0, 1, 0][d]][x += [0, 1, 0, -1][d]]

  c == 47 ? d = [1, 0, 3, 2][d] :
  c == 92 ? d = 3 - d :
  c == 35 ? (p !1; exit) :
  c < ?x ? 0 : (p !!1; exit)
}

Slightly refactored:

board = readlines

direction = x = y = -1
board.each do |line|
  if direction < 0
    x = line.index(/[>^v<]/)
    if x
      direction = "^>v<".index line[x]
    end
    y += 1
  end
end

loop do
  x += [0, 1, 0, -1][direction]
  y += [-1, 0, 1, 0][direction]

  ch = board[y][x].chr
  case ch
  when "/"
    direction = [1, 0, 3, 2][direction]
  when "\\"
    direction = 3 - direction
  when "x"
    puts "true"
    exit
  when "#"
    puts "false"
    exit
  end
end

Answer 6

F#, 255 chars (and still rather readable!):

Ok, after a night's rest, I improved this a lot:

let a=System.Console.In.ReadToEnd()
let w,c=a.IndexOf"\n"+1,a.IndexOfAny[|'^';'<';'>';'v'|]
let rec n(c,d)=
 let e,s=[|-w,2;-1,3;1,0;w,1|].[d]
 n(c+e,match a.[c+e]with|'/'->s|'\\'->3-s|' '->d|c->printfn"%A"(c='x');exit 0)
n(c,"^<>v".IndexOf a.[c])

Let's talk through it line by line.

First, slurp all the input into a big one-dimensional array (2D arrays can be bad for code golf; just use a 1D array and add/subtract the width of one line to the index to move up/down a line).

Next we compute 'w', the width of an input line, and 'c', the starting position, by indexing into our array.

Now let's define the 'next' function 'n', which takes a current position 'c' and a direction 'd' which is 0,1,2,3 for up,left,right,down.

The index-epsilon 'e' and the what-new-direction-if-we-hit-a-slash 's' are computed by a table. For example, if the current direction 'd' is 0 (up), then the first element of the table says "-w,2" which means we decrement the index by w, and if we hit a slash the new direction is 2 (right).

Now we recurse into the next function 'n' with (1) the next index ("c+e" - current plus epsilon), and (2) the new direction, which we compute by looking ahead to see what's in the array in that next cell. If the lookahead char is a slash, the new direction is 's'. If it's a backslash, the new direction is 3-s (our choice of encoding 0123 makes this work). If it's a space, we just keep going in the same direction 'd'. And if it's any other character 'c', then the game ends, printing 'true' if the char was 'x' and false otherwise.

To kick things off, we call the recursive function 'n' with the initial position 'c' and the starting direction (which does the initial encoding of direction into 0123).

I think I can probably still shave a few more characters off it, but I am pretty pleased with it like this (and 255 is a nice number).