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|

Suggested edge case: width: 1, length: whatever, right: 0, left: 0 — Arnauld, Apr 9, 2018 at 14:58

Community · Accepted Answer · 2020-06-17 09:04:33Z

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):''

Try it online!

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

Weijun Zhou · Accepted Answer · 2018-04-11 07:11:25Z

12

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

edited Apr 11, 2018 at 7:11

answered Apr 9, 2018 at 14:55

Weijun Zhou

3,62711 silver badges39 bronze badges

1

\$\begingroup\$ Might want to note the input format as width [right-going left-going] length \$\endgroup\$
– Engineer Toast
Apr 10, 2018 at 17:56

Add a comment |

xnor · Accepted Answer · 2018-04-10 01:14:12Z

11

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

Try it online!

edited Apr 10, 2018 at 1:14

answered Apr 9, 2018 at 22:30

xnor

141k26 gold badges271 silver badges621 bronze badges

\$\begingroup\$ Can you not golf \\/ to \/? Even though the backslash is interpreted twice, it will still not escape the slash. \$\endgroup\$
– Jonathan Frech
Apr 10, 2018 at 0:10
\$\begingroup\$ @JonathanFrech You're right, I thought being in a string-in-string would fail, but it indeed doesn't escape either time. \$\endgroup\$
– xnor
Apr 10, 2018 at 1:15
\$\begingroup\$ Oh jeez, my solution got painfully close to this idea -- working modulo 2w makes a lot of sense in retrospect. Very clever! \$\endgroup\$
– Lynn
Apr 10, 2018 at 15:02

Add a comment |

Lynn · Accepted Answer · 2018-04-09 14:38:58Z

9

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

Try it online!

answered Apr 9, 2018 at 14:38

Lynn

67.1k11 gold badges129 silver badges276 bronze badges

\$\begingroup\$ What a weird and unfortunate error, just as if it didn't take precious bytes to define d! \$\endgroup\$
– Erik the Outgolfer
Apr 10, 2018 at 23:55

Add a comment |

TFeld · Accepted Answer · 2018-04-09 14:52:03Z

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)

Try it online!

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)

Try it online!

I am once again amazed by the speed of first answers on code golf challenges. Nice one! :) — AJFaraday, Apr 9, 2018 at 13:45

GPS · Accepted Answer · 2018-04-10 12:55:38Z

8

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));}

Try it online!

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

edited Apr 10, 2018 at 12:55

answered Apr 10, 2018 at 7:23

GPS

3611 silver badge6 bronze badges

1

\$\begingroup\$ You can golf 3 bytes by changing the while to a for to remove the {} and one of the semi-colons. And 1 more byte by changing c&&d to c&d. Try it online 236 bytes. \$\endgroup\$
– Kevin Cruijssen
Apr 10, 2018 at 7:48
1

\$\begingroup\$ You're not supposed to solve a generalisation of the challenge, but the challenge as specified. Regarding extra inputs, I dug around meta a bit and found this: codegolf.meta.stackexchange.com/a/12696/79343 I'm thinking it must be written up somewhere else, too, but i can't find it atm. It is however the norm. \$\endgroup\$
– O.O.Balance
Apr 10, 2018 at 9:08
1

\$\begingroup\$ In your g macro, you can golf 2 bytes by changing a==-1 to a<0. \$\endgroup\$
– JohnWells
Jul 8, 2018 at 21:55
1

\$\begingroup\$ Actually got more in the macro, a++,a and a--,a can golf 2 bytes each to ++a and --a \$\endgroup\$
– JohnWells
Jul 8, 2018 at 22:35
1

\$\begingroup\$ 187 bytes \$\endgroup\$
– ceilingcat
Oct 12, 2019 at 3:05

| Show 4 more comments

dzaima · Accepted Answer · 2018-04-10 17:01:49Z

7

Canvas, 66 40 bytes

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

Try it here!

edited Apr 10, 2018 at 17:01

answered Apr 9, 2018 at 17:10

dzaima

20.2k2 gold badges39 silver badges75 bronze badges

Add a comment |

Neil · Accepted Answer · 2018-04-10 12:59:25Z

5

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.

edited Apr 10, 2018 at 12:59

answered Apr 10, 2018 at 12:50

Neil

158k12 gold badges69 silver badges268 bronze badges

\$\begingroup\$ This goes out of bounds when the input is "10 2 4 2" \$\endgroup\$
– Martijn Vissers
Apr 12, 2018 at 14:56
1

\$\begingroup\$ @MartijnVissers Well yes, if your width is 2, then your positions can only be 0 or 1... \$\endgroup\$
– Neil
Apr 12, 2018 at 15:54

Add a comment |

Olivier Grégoire · Accepted Answer · 2018-04-11 07:21:20Z

5

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!

edited Apr 11, 2018 at 7:21

answered Apr 10, 2018 at 16:10

Olivier Grégoire

14.1k3 gold badges30 silver badges56 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

3

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.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Apr 10, 2018 at 11:58

Titus

14.6k1 gold badge22 silver badges41 bronze badges

\$\begingroup\$ It's unfortunate that onlinephpfunctions.com does not save the correct PHP version in the sharing link... \$\endgroup\$
– Arnauld
Apr 10, 2018 at 12:03

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RootTwo · Accepted Answer · 2018-04-11 03:34:47Z

3

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))

Try it online!

answered Apr 11, 2018 at 3:34

RootTwo

2,0068 silver badges7 bronze badges

\$\begingroup\$ I like the formatting in your test suite, reliably showing the inputs against the output... Nice one ;) \$\endgroup\$
– AJFaraday
Apr 11, 2018 at 8:55

Add a comment |

Kirill L. · Accepted Answer · 2018-04-11 12:55:07Z

3

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+?|}}

Try it online!

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

answered Apr 11, 2018 at 12:55

Kirill L.

13.3k2 gold badges14 silver badges48 bronze badges

\$\begingroup\$ You kind of made my day by making it an expandable array, not just 2 starting points. \$\endgroup\$
– AJFaraday
Apr 11, 2018 at 12:57

Add a comment |

Veskah · Accepted Answer · 2019-03-07 14:00:03Z

2

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)|"}

Try it online!

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.

edited Mar 7, 2019 at 14:00

answered Apr 10, 2018 at 3:21

Veskah

4,07212 silver badges29 bronze badges

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Jonathan Frech · Accepted Answer · 2018-04-10 11:57:57Z

1

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.
Try it online!

edited Apr 10, 2018 at 11:57

Jonathan Frech

7,3211 gold badge20 silver badges43 bronze badges

answered Apr 9, 2018 at 21:58

user48543

1

\$\begingroup\$ \\/ is equivalent to \/. \$\endgroup\$
– Jonathan Frech
Apr 10, 2018 at 0:07

Add a comment |

ngn · Accepted Answer · 2018-04-10 13:00:15Z

1

K (ngn/k), 58 bytes

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

Try it online!

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

answered Apr 10, 2018 at 13:00

ngn

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Add a comment |

gastropner · Accepted Answer · 2018-07-08 10:24:44Z

1

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);}

Try it online!

answered Jul 8, 2018 at 10:24

gastropner

4,2831 gold badge12 silver badges20 bronze badges

\$\begingroup\$ 163 bytes \$\endgroup\$
– ceilingcat
Nov 27, 2018 at 1:35

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JohnWells · Accepted Answer · 2018-07-09 16:58:39Z

1

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}()}

Try it online!

edited Jul 9, 2018 at 16:58

answered Jul 9, 2018 at 0:20

JohnWells

6813 silver badges7 bronze badges

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Stack Exchange Network

Two lasers between two mirrors

Test cases

17 Answers 17

JavaScript (ES6), 149 bytes

Commented

Stax, 40 bytes

Python 2, 119 bytes

Python 2, 168 bytes

Python 2, 187 181 179 177 174 172 171 bytes

Python 2, 172 bytes

Python 2, 172 bytes

C (clang), 240 236 208 bytes

Canvas, 66 40 bytes

Charcoal, 56 50 bytes

Java (JDK 10), 186 bytes

PHP, 177 169 166 bytes

Python 3, 162 bytes

Ruby, 117 bytes

PowerShell, 243 233 222 205 bytes

Python 2, 165 164 bytes

K (ngn/k), 58 bytes

C (gcc), 169 bytes

Kotlin, 322 311 302 bytes

Your Answer

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code-golf
ascii-art
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Hot Network Questions

Two lasers between two mirrors

Test cases

17 Answers 17

JavaScript (ES6), 149 bytes

Commented

Stax, 40 bytes

Python 2, 119 bytes

Python 2, 168 bytes

Python 2, 187 181 179 177 174 172 171 bytes

Python 2, 172 bytes

Python 2, 172 bytes

C (clang), 240 236 208 bytes

Canvas, 66 40 bytes

Charcoal, 56 50 bytes

Java (JDK 10), 186 bytes

PHP, 177 169 166 bytes

Python 3, 162 bytes

Ruby, 117 bytes

PowerShell, 243 233 222 205 bytes

Python 2, 165 164 bytes

K (ngn/k), 58 bytes

C (gcc), 169 bytes

Kotlin, 322 311 302 bytes

Your Answer

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