Two lasers between two mirrors

Question

What if we have a corridor comprised of two parallel mirrors?

|          |
|          |
|          |
|          |
|          |
|          |
|          |
|          |
|          |
|          |

Now, we shine a laser down it...

|  \       |
|   \      |
|    \     |
|     \    |
|      \   |
|       \  |
|        \ |
|         \|
|         /|
|        / |

Oh, look. It bounced, towards the end, there.

What if we draw two lasers BUT going in the opposite direction?

|  \    /  |
|   \  /   |
|    \/    |
|    /\    |
|   /  \   |
|  /    \  |
| /      \ |
|/        \|
|\        /|
| \      / |

Hmm, they didn't seem to meet, there. That's convenient. What happens if both lasers take up the same space?

|  \     / |
|   \   /  |
|    \ /   |
|     X    |
|    / \   |
|   /   \  |
|  /     \ |
| /       \|
|/        /|
|\       / |

I guess that was pretty obvious, huh?

---

Drawing these diagrams by hand is pretty laborious (trust me on this). Perhaps some code could do it for us?

• Write some code to output two parallel mirrors, with two bouncing, intersecting lasers.
• Input (all integers):
  • The width of the corridor
  • The length of the corridor
  • Starting position of the right-going laser (zero-indexed, must be less than width)
  • Starting position of the left-going laser (zero-indexed, must be less than width)
• Process
  • If a laser is right going, it will be drawn one space to the right on the following line.
  • If a laser is left going, it will be drawn one space to the left on the following line.
  • If a laser can not take it's sideways step, it will change it's direction, but not it's position.
  • If both laser are at the same index, print an upper-case X at that index.
• Output
  • A string with multiple lines
  • Each line starts and ends with a pipe character (|)
  • Right-going laser are denoted by a back slash (\)
  • Left-going laser are denoted by a forward slash (/)
  • The intersection of two lasers is denoted by an upper-case X.
• Any language
• I'd like to see TIO links
• Attempt to fix it in the smallest number of bytes

---

Test cases

width: 6 length: 10 right-going: 1 left-going: 4

| \  / |
|  \/  |
|  /\  |
| /  \ |
|/    \|
|\    /|
| \  / |
|  \/  |
|  /\  |
| /  \ |

width: 6 length: 10 right-going: 0 left-going: 1

|\/    |
|/\    |
|\ \   |
| \ \  |
|  \ \ |
|   \ \|
|    \/|
|    /\|
|   / /|
|  / / |

width: 4 length: 10 right-going: 2 left-going: 0

|/ \ |
|\  \|
| \ /|
|  X |
| / \|
|/  /|
|\ / |
| X  |
|/ \ |
|\  \|

width: 20 length: 5 right-going: 5 left-going: 15

|     \         /    |
|      \       /     |
|       \     /      |
|        \   /       |
|         \ /        |

width: 5 length: 6 right-going: 2 left-going: 2

|  X  |
| / \ |
|/   \|
|\   /|
| \ / |
|  X  |

width: 1 length: 2 right-going: 0 left-going: 0

|X|
|X|

Answer 1

Stax, 40 bytes

àù@○⌡┼PY¼îαφu^·A☺°É⌠■╟¡Åt^◘v(µ╩Ñ♣t{╓○xß╦

Run and debug it

Try it online!

Pretty sure this can be further golfed.

Input is given in the form of width [right-going left-going] length(per comment by @EngineerToast).

ASCII equivalent:

xHXdmzx);hi+x%Y92&;Hi-x%cy=41*47+&2ME:R\{|+m'||S

Answer 2

JavaScript (ES6), 149 bytes

Takes input in currying syntax (w)(h)([a,b]).

w=>h=>g=(p,d=[1,-1],s=Array(w).fill` `)=>h--?`|${p=p.map((x,i)=>~(k=d[i],s[x]='/X\\'[x-p[i^1]?k+1:1],x+=k)&&x<w?x:x+(d[i]=-k)),s.join``}|
`+g(p,d):''

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Commented

w => h =>                  // w = width, h = height
  g = (                    // g = recursive function taking:
    p,                     //   p[] = array holding the point coordinates
    d = [1, -1],           //   d[] = directions
    s = Array(w).fill` `   //   s = array of w spaces (we can't use a string because it's
  ) =>                     //       immutable in JS)
    h-- ?                  // if we haven't reached the last row yet:
      `|${                 //   append the left pipe
      p = p.map((x, i) =>  //   for each x at position i in p[]:
        ~(k = d[i],        //     k = direction for this point
          s[x] = '/X\\'[   //     insert either '/', 'X' or '\' at position x in s
            x - p[i ^ 1] ? //     if p[0] != p[1]:
              k + 1        //       use the direction
            :              //     else:
              1            //       force 'X'
          ], x += k        //     add k to x
        ) &&               //     if the result is not equal to -1
        x < w ?            //     and is less than w:
          x                //       use the current value of x
        :                  //     else:
          x + (d[i] = -k)  //       change the direction and restore the initial value of x
      ),                   //   end of map()
      s.join``}|\n` +      //   join and append s; append the right bar and a linefeed
      g(p, d)              //   followed by the result of a recursive call
    :                      // else:
      ''                   //   stop recursion

Answer 3

Python 2, 119 bytes

w,l,a,b=input()
exec"print'|%s|'%''.join(' \/X'[sum(i==k%(2*w)for k in[a,~b]+[~a,b]*2)]for i in range(w));a+=1;b-=1;"*l

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Answer 4

Python 2, 187 181 179 177 174 172 171 bytes

def f(w,l,a,b,A=1,B=-1):
 while l:l-=1;print'|%s|'%''.join(' \X/'[[0,A,B,2][(i==a)+2*(i==b)]]for i in range(w));a,A=[a,a+A,-A,A][-1<a+A<w::2];b,B=[b,b+B,-B,B][-1<b+B<w::2]

Try it online!

---

Recursive:

Python 2, 172 bytes

def f(w,l,a,b,A=1,B=-1):
 if not-1<a<w:A=-A;a+=A
 if not-1<b<w:B=-B;b+=B
 if l:print'|%s|'%''.join(' \X/'[[0,A,B,2][(i==a)+2*(i==b)]]for i in range(w));f(w,l-1,a+A,b+B,A,B)

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---

Recursive, alternative print:

Python 2, 172 bytes

def f(w,l,a,b,A=1,B=-1):
 if not-1<a<w:A=-A;a+=A
 if not-1<b<w:B=-B;b+=B
 if l:L=[' ']*w;L[a]=' \/'[A];L[b]=[' \/'[B],'X'][a==b];print'|%s|'%''.join(L);f(w,l-1,a+A,b+B,A,B)

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Answer 5

Python 2, 168 bytes

w,n,r,l=input()
R=L=1
exec"""
d=[~r*R,-~l*L].count
print'|%s|'%''.join(' /\X'[2*d(~x)|d(x+1)]for x in range(w))
if-1<r+R<w:r+=R
else:R=-R
if-1<l-L<w:l-=L
else:L=-L"""*n

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Answer 6

C (clang), 240 236 208 bytes

#define g(a,b) b?a++,a==x&&(a=x-1,b=0):a--,a==-1&&(a=0,b=1)
i,m,n,o,p,t[]={47,92};f(x,y,r,l){for(m=1,n=0;y--;puts("|"),g(r,m),g(l,n))for(printf("|"),i=0;i<x;o=i==r,p=i++==l,putchar(o*p?88:o?t[m]:p?t[n]:32));}

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f() takes parameters as follows:

x = width,
y = length,
r = Initially right-going line starting position
l = Initially left-going-line starting position

-4 Bytes. credits Kevin Cruijssen. Thanks

Answer 7

Canvas, 66 40 bytes

｛²Ｘø⁶╵［⁷／²２％［↔｝∔｝；ｘ╷？⁷∔⁷＋╷｝［ｊ｝｝ｎ⁶［|ＰＪｐ|ｐ

Try it here!

Answer 8

Charcoal, 56 50 bytes

↷ＰＩθＭ⊕η→ＩθＦ²«Ｊ⊕⎇ιεζ⁰ＦＩθ«✳§⟦↘↙⟧ι∨⁼ＫＫψX¿⁼ＫＫ|«¿ι→←≦¬ι

Try it online! Link is to verbose version of code. Edit: Saved 6 bytes by reducing reliance on pivoting. Explanation:

↷ＰＩθＭ⊕η→Ｉθ

Print the sides.

Ｆ²«

Loop over the two lasers.

Ｊ⊕⎇ιεζ⁰

Move to the start of the laser.

ＦＩθ«

Loop over the height.

✳§⟦↘↙⟧ι∨⁼ＫＫψX

Draw a \ or / in the appropriate direction, unless the square is not empty, in which case draw an X.

¿⁼ＫＫ|«

Have we hit a side?

¿ι→←≦¬ι

If so then take one step sideways and invert the direction of travel.

Answer 9

Java (JDK 10), 186 bytes

(w,h,r,l)->{var x="";for(int i=0,j,R=1,L=-1;i++<h;l+=L,l+=l<0|l>=w?L=-L:0,r+=R,r+=r<0|r>=w?R=-R:0,x+="|\n")for(j=0,x+="|";j<w;j++)x+="/X\\ ".charAt(j==r?j==l?1:R+1:j==l?L+1:3);return x;}

Try it online!

Answer 10

PHP, 177 169 166 bytes

[,$w,$h,$a,$b]=$argv;for($e=-$d=1;$h--;$s[$a+=$d]^L?:$a+=$d=-$d,$s[$b+=$e]^L?:$b+=$e=-$e){$s=str_pad("",$w)."|";$s[$b]="X\/"[$e];$s[$a]="X\/"[$a-$b?$d:0];echo"|$s
";}

requires PHP 7.1 for negative string indexes, PHP 5.5 or later for indexing string literals.
for PHP <7.1, remove ^L, replace "X\/" with "/X\\", :0 with +1:1, [$e] with [$e+1], remove ."|" and insert | before the newline. (+3 bytes)
for PHP < 5.5, replace "/X\\" with $p and insert $p="/X\\"; at the beginning. (+2 bytes)

takes input from command line arguments. Run with -nr or try them online.

Answer 11

Python 3, 162 bytes

from numpy import*
def f(w,h,u,v):
 v=w+w-v-1;T=eye(w);M=vstack([T,2*T[::-1]]*2*h)
 for r in M[u:u+h,:]+M[v:v+h,:]:print('|%s|'%''.join(' \/X'[int(i)]for i in r))

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Answer 12

Ruby, 117 bytes

->w,h,a{a[1]-=w;(1..h).map{s=' '*w;a.map!{|x|d=x<0?-1:1;s[x]='X\\/'[s[x][/ /]?d:0];x+=d;x==w ?-1:x<-w ?0:x};?|+s+?|}}

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Anonymous lambda taking input as width w, height h and an array of starting points a.

Answer 13

PowerShell, 243 233 222 205 bytes

param($w,$h,$a,$b)$l,$r,$s=1,-1,' \/'
1..$h|%{$p,$p[$b],$p[$a]=[char[]](' '*$w),$s[$r],($s[$l],"x")[!($a-$b)]
if($a+$l-in($z=0..($w-1))){$a+=$l}else{$l*=-1}if($b+$r-in$z){$b+=$r}else{$r*=-1}"|$(-join$p)|"}

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Oooof. those logic blocks are big and dirty and mostly duplicated. The next step would be rewriting them so they don't need the else statement.

Answer 14

Python 2, 165 164 bytes

w,h,x,y=input()
a,b,s=1,-1,' \/'
exec"""l=[' ']*w
l[x],l[y]=s[a],s[b]if x-y else'X'
if-1<x+a<w:x+=a
else:a=-a
if-1<y+b<w:y+=b
else:b=-b
print'|%s|'%''.join(l)
"""*h

Saved a byte thanks to Jonathan Frech.
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Answer 15

K (ngn/k), 58 bytes

{" \\/X|"4,'(+/3!1 2*(x#'e+2*|e:=2*x)(2*x)!z+(!y;-!y)),'4}

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anonymous function that accepts three arguments: x the width, y the length, z a pair of starting positions for the lasers

Answer 16

C (gcc), 169 bytes

A,B,c;f(w,l,a,b){for(A=1,B=-1;l--;a+=A,a<0|a==w?A=-A,a+=A:0,b+=B,b<0|b==w?B=-B,b+=B:0,puts("|"))for(c=-1;c<w;c++)putchar(c<0?'|':a^c?b^c?32:B>0?92:47:b^c?A>0?92:47:88);}

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Answer 17

Kotlin, 322 311 302 bytes

Changed how I put laser direction in string for 11 bytes. Moved assignment out of when for 9 bytes.

{w:Int,h:Int,r:Int,l:Int->{var a=""
var f=r
var d=1>0
var s=l
var t=!d
for(o in 1..h){a+="|"
for(c in 0..w-1)a+=when{c==f&&c==s->"X"
c==f&&d||c==s&&t->"\\"
c==f||c==s->"/"
else->" "}
a+="|\n"
if(d){if(++f==w){--f
d=!d}}else if(--f<0){f=0
d=!d}
if(t){if(++s==w){--s
t=!t}}else if(--s<0){s=0
t=!t}}
a}()}

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