Generate a one-path maze

Question

Given an odd integer N (5 <= N <= 51), generate a maze with side length N that follows the following requirements:

The maze must be composed of the characters | - and +. It must use the | character as a vertical wall, the - character as a horizontal wall, and if that wall turns, the character + should be used.

The width of the path must be one character.

The maze must turn at least four times.

The maze must have outer walls, that break at two points: the start and the end.

The maze must consist of one non-breaking path, from the start to the end.

For example, the following is a valid maze: (N = 5)

+---+
|   |
| | |
  |  
--+--

And for N = 7:

--+---+
  |   |
| | | |
| | | |
| | | |
|   |  
+---+--

Remember, this is code-golf, so the code with the fewest number of bytes wins.

It's not a maze, it's a labyrinth english.stackexchange.com/a/144103/199361 — edc65, Commented Oct 30, 2016 at 20:54
I'm not clear on "The maze must consist of one non-breaking path, from the start to the end." Does this mean that there is only one-path and it's from start to end, or that the number of paths from start to end is 1? Can there be other paths with dead ends? Separate loops? — xnor, Commented Oct 30, 2016 at 22:39

Jonathan Allan · Accepted Answer · 2016-10-31 14:55:18Z

Jelly, 36 35 34 33 32 bytes

2*×5H_2Bị⁾ |
_3”-ẋ”+;;Çsẋ2U3¦ṁµY

TryItOnline!

Builds a nighter™ the other way around to the examples like:

+---- |
|     |
| ----+
|     |
+---- |
|     |
| ----+

How?

2*×5H_2Bị⁾ | - Link 1, pipes & spaces: n      e.g. 7
2*           - 2 raised to the nth power      e.g. 128
  ×5         - multiply by 5                  e.g. 640
    H        - halve                          e.g. 320
     _2      - subtract 2                     e.g. 318
       B     - convert to binary              e.g. [1,0,0,1,1,1,1,1,0]
        ị    - index into (1-based)
         ⁾ | - char list " |"                 e.g. " ||     |"

_3”-ẋ”+;;Çsẋ2U3¦ṁµY - Main link: n            e.g. 7
_3                  - n - 3                   e.g. 4
  ”-                - char list "-"
    ẋ               - repeat                  e.g. "----"
     ”+             - char list "+"
       ;            - concatenate             e.g. "+----"
         Ç          - call last link (1) as a monad
        ;           - concatenate             e.g. "+---- ||     |"" 
          s         - split into chunks of n  e.g. ["+---- |","|     |"]
           ẋ2       - repeat 2 times          e.g. ["+---- |",
                                                    "|     |",
                                              +-->  "+---- |",
                                              |     "|     |"]
              3¦    - apply to index 3:       |
             U      -    upend                +---  "| ----+"
                ṁ   - mould like n (i.e. repeat to fill)
                 µ  - monadic chain separation
                  Y - join with line feeds

(each byte saved involved fairly non-trivial changes, see the edit history if you're interested, although I just noticed that Link 1 is the same byte count as the more conventional repeat and join: _2⁶ẋ“ ||“|”j)

edc65 · Accepted Answer · 2016-10-31 17:39:09Z

5

JavaScript (ES6), 86 92 116

~~Almost a kolmogorv-complexity challenge ...~~ With a little lateral thinking (inspired by @Neil's answer), it can be very shorter. Just turn 90°

n=>`|${s=' '[r='repeat'](n-2)}|
| ${h='-'[r](n-3)}+
|${s}|
+${h} |
`[r](n).slice(n*~n)

Test

f=
n=>`|${s=' '[r='repeat'](n-2)}|
| ${h='-'[r](n-3)}+
|${s}|
+${h} |
`[r](n).slice(n*~n)

function update() {
  var i=+I.value
  O.textContent=i&1? f(i): 'even'
}

update()

<input id=I value=7 min=5 max=49 step=2 type=number oninput="update()"><pre id=O><pre>

edited Oct 31, 2016 at 17:39

answered Oct 30, 2016 at 21:59

edc65

32.2k3 gold badges36 silver badges90 bronze badges

\$\begingroup\$ @Neil wow every day I learn something new. Thanks \$\endgroup\$
– edc65
Commented Oct 30, 2016 at 22:03
\$\begingroup\$ Amazing! I count 86 bytes, btw \$\endgroup\$
– ETHproductions
Commented Oct 31, 2016 at 16:40
\$\begingroup\$ @ETHproductions right. Thanks \$\endgroup\$
– edc65
Commented Oct 31, 2016 at 17:39

Add a comment |

Neil · Accepted Answer · 2016-10-31 00:39:55Z

4

Batch, 216 bytes

@echo off
set s=
for /l %%i in (4,1,%1)do call set s=-%%s%%
set b= %s:-= %
set r=! %s%+
set s=+%s% !
call:e
for /l %%i in (5,2,%1)do call:l
:l
echo ^|%b%^|
set t=%s%
set s=%r%
set r=%t%
:e
echo %s:!=^|%

|s are awkward in Batch so I ~~reflected~~ rotated the examples instead.

edited Oct 31, 2016 at 0:39

answered Oct 30, 2016 at 22:00

Neil

177k12 gold badges74 silver badges281 bronze badges

\$\begingroup\$ Do you mean rotate 90°? I tried it and saved a lot in JS. Thanks again \$\endgroup\$
– edc65
Commented Oct 30, 2016 at 22:28
\$\begingroup\$ @edc65 I was originally going for a reflection, but you're right in that I ended up with a rotation instead. \$\endgroup\$
– Neil
Commented Oct 31, 2016 at 0:39

Add a comment |

Community · Accepted Answer · 2017-04-13 12:39:10Z

3

PHP, 99 Bytes

up to down gates

for($s=str_pad("\n| ",$c=$argv[1],"-")."+\n";$i++<$c;)echo$i%2?$s=strrev($s):str_pad("|",$c-1)."|";

PHP, 157 Bytes

left right gates

<?=($p=str_pad)($n="\n",$c=1+$argv[1],"+---").$p($n,$c-1,"|   ")." #|"[$c%4].str_repeat($p($n,$c,"| "),$c-5).$p($n,$c-1,"  | ")."|# "[$c%4].$p($n,$c,"--+-");

@Titus Thank you for minialize the bytes

edited Apr 13, 2017 at 12:39

CommunityBot

1

answered Oct 30, 2016 at 22:36

Jörg Hülsermann

13.3k3 gold badges15 silver badges36 bronze badges

1

\$\begingroup\$ save 3 bytes by assigning `$n="\n" \$\endgroup\$
– Titus
Commented Oct 30, 2016 at 22:55
1

\$\begingroup\$ $p("",$c*($c-4),...) instead of str_repeat(...,$c-4) (-2) \$\endgroup\$
– Titus
Commented Oct 30, 2016 at 22:59
1

\$\begingroup\$ ... and 3 more bytes with $p($n,$c-1," | ") instead of $p("\n ",$c-1," | ") \$\endgroup\$
– Titus
Commented Oct 30, 2016 at 23:03
1

\$\begingroup\$ unnecessary parens at ($c)%4 (-2), remove $g from the code (-1) \$\endgroup\$
– Titus
Commented Oct 30, 2016 at 23:11
1

\$\begingroup\$ up-to-down-version; unnecessary 3rd parameter for str_pad (-4), $c=$argv[1]-1 instead of ($c=$argv[1])-1, <=$c instead of <$c and $c instead of $c-1 (-3) \$\endgroup\$
– Titus
Commented Oct 30, 2016 at 23:21

| Show 7 more comments

Arnauld · Accepted Answer · 2016-10-31 15:33:10Z

3

JavaScript (ES6), 87 bytes

A recursive function. Outputs a few trailing spaces.

f=(n,w=n,s=' -'[n&1].repeat(w-3),c='|+'[n&1])=>n?`
`+(n&2?c+s+' |':'| '+s+c)+f(n-1,w):s

Test

f=(n,w=n,s=' -'[n&1].repeat(w-3),c='|+'[n&1])=>n?`
`+(n&2?c+s+' |':'| '+s+c)+f(n-1,w):s

console.log(f(9));

answered Oct 31, 2016 at 15:33

Arnauld

198k20 gold badges179 silver badges650 bronze badges

\$\begingroup\$ Oh wow, I just tried recursion and ended up 9 bytes longer. Well done :-) \$\endgroup\$
– ETHproductions
Commented Oct 31, 2016 at 15:37

Add a comment |

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

2

Ruby 72 or 69 bytes

Lambda function. As shown, returns a newline-separated string. Delete the *$/ to return an array of strings.

->n{(1..n).map{|i|"|%s |+%s || %s|| %s+"[i%4*5,5]%(" -"[i%2]*(n-3))}*$/}

Draws a maze rotated 90 degrees from the examples. For each line, a format string is selected (for example +%s | for the 1st line (there is no zeroth line) and the %s is substituted with an appopriate number of - or spaces using the % operator (equivalent to sprintf, but shorter.)

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 31, 2016 at 0:05

Level River St

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

Java 7, 228 bytes

String c(int n){String a="+x |",b="|y|\n",x,y,r=x=y="";int i=0;for(;i++<n-2;){x+=i>1?"-":"";y+=" ";}a=a.replace("x",x);b=b.replace("y",y);for(i=0;i<n;i++)r+=i%4<1?a+"\n":(i-2)%4==0?new StringBuffer(a).reverse()+"\n":b;return r;}

Used a similar vertical output as @JonathanAllan's Jelly answer.

Ungolfed & test code:

Try it here.

class M{
  static String c(int n){
    String a = "+x |",
           b = "|y|\n",
           x = "",
           y = "",
           r = "";
    int i = 0;
    for (; i++ < n-2;){
      x += i > 1
            ? "-"
            : "";
      y += " ";
    }
    a = a.replace("x", x);
    b = b.replace("y", y);
    for(i = 0; i < n; i++){
      r += i % 4 < 1
            ? a+"\n"
            : (i-2) % 4 == 0
               ? new StringBuffer(a).reverse()+"\n"
               : b;
    }
    return r;
  }

  public static void main(String[] a){
    System.out.println(c(7));
    System.out.println();
    System.out.println(c(25));
  }
}

Output:

+---- |
|     |
| ----+
|     |
+---- |
|     |
| ----+


+---------------------- |
|                       |
| ----------------------+
|                       |
+---------------------- |
|                       |
| ----------------------+
|                       |
+---------------------- |
|                       |
| ----------------------+
|                       |
+---------------------- |
|                       |
| ----------------------+
|                       |
+---------------------- |
|                       |
| ----------------------+
|                       |
+---------------------- |
|                       |
| ----------------------+
|                       |
+---------------------- |

excuse, but Given an odd integer N (5 <= N <= 51), generate a maze with side length N. You seem to have a different top and bottom side length for yours... — Destructible Lemon, Commented Nov 9, 2016 at 23:38
@DestructibleWatermelon I read past that. My code still complied to all rules/requirements though. Ah well, I've roll-backed to my original answer which used the same width as height. — Kevin Cruijssen, Commented Nov 10, 2016 at 15:55

Jonathan Allan · Accepted Answer · 2016-10-30 23:29:11Z

1

Python 2, 89 bytes

def f(n):x='+'+'-'*(n-3)+' |';y='|'+' '*(n-2)+'|';print'\n'.join(([x,y,x[::-1],y]*n)[:n])

repl.it

Builds an internal wall, x, like '+---- |' and an internal corridor, y, like '| |'
Then builds a list of [x,y,x[::-1],y] (x[::-1] is a reverse of x)
Then repeats that list n times (as a single list), with *n, and truncates it to the first n entries, with (...)[:n], joins the list with line feeds, with '\n'.join(...), and prints the result.

answered Oct 30, 2016 at 23:29

Jonathan Allan

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rnso · Accepted Answer · 2016-10-31 04:30:28Z

Racket 187 bytes

Using display pattern by @JonathanAllan

(let*((u "+---- |")(v "|     |")(sr(λ(s)(list->string(reverse(string->list s)))))(g #t)(d displayln)
(p(λ()(d(if g u(sr u)))(set! g(if g #f #t)))))(for((i(ceiling(/ n 2))))(p)(d v))(p))

Ungolfed:

(define (f n)
  (let* ((sr (lambda(s)
               (list->string
                (reverse
                 (string->list s)))))
         (u "+---- |")
         (v "|     |")
         (g #t)
         (d displayln)
         (p (lambda()
              (d (if g u (sr u)))
              (set! g (if g #f #t)))))
    (for ((i (ceiling (/ n 2))))
      (p)
      (d v))
    (p)))

Testing:

(f 10)

Output:

+---- |
|     |
| ----+
|     |
+---- |
|     |
| ----+
|     |
+---- |
|     |
| ----+

Community · Accepted Answer · 2017-03-16 16:03:59Z

GNU sed 140 bytes

Includes +1 for -r

s/1{5}//
h
s/^/+---+\n|   |\n| | |\n  |  \n--+--/
/1/!b
:w
s/^..(..)[^$]*/\1&/gm
s/11//
/1/bw
G
:h
s/\n[^\n]*\n([^\n]*\n)/&\1/
s/1//
/1/bh

Try it Online!

Takes input in unary (see this consensus).

Basically it inserts the size 5 maze, then appends the 2nd and 3rd character of each line to the beginning as many times as needed. Then duplicates the 3rd line (alternating | and ) as many times as needed.

The only interesting thing that I used is the m option on line 6 which allows ^ and $ to match respectively (in addition to the normal behavior) the empty string after a newline, and the empty string before a newline.

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

T-SQL, 123/79 bytes

Golfed:

DECLARE @i INT=7

,@ INT=0z:PRINT
STUFF(CHOOSE(@%4+1,'+- |','|  |','| -+','|  |'),3,0,REPLICATE(IIF(@%2=0,'-',' '),@i))SET
@+=1IF @<@i GOTO z

Ungolfed:

DECLARE @i INT = 9

DECLARE @i INT=7

,@ INT=0
z:
  PRINT 
    STUFF(CHOOSE(@%4+1,'+- |','|  |','| -+','|  |'),3,0,
      REPLICATE(IIF(@%2=0,'-',' '),@i))
  SET @+=1
IF @<@i GOTO z

If you cheat and only make a narrow labyrinth the script can be golfed down to 79 bytes:

Golfed:

DECLARE @i INT = 9

,@ INT=0z:PRINT CHOOSE(@%4+1,'+- |','|  |','| -+','|  |')SET @+=1IF @<@i GOTO z

Ungolfed:

DECLARE @i INT = 9

,@ INT=0
z:
  PRINT CHOOSE(@%4+1,'+- |','|  |','| -+','|  |')
  SET @+=1
IF @<@i GOTO z

Fiddle for the long answer

Community · Accepted Answer · 2017-04-13 12:39:00Z

0

JavaScript (ES6), 96 bytes

f=(n,w=n-3)=>(h="-".repeat(w),n&2?`+${h} |`:`| ${h}+`)+(n>1?`
| ${" ".repeat(w)}|
`+f(n-2,w):"")

I was hoping recursion would turn out to be the shortest route, and apparently it is...

edited Apr 13, 2017 at 12:39

CommunityBot

1

answered Oct 31, 2016 at 15:36

ETHproductions

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Stack Exchange Network

Generate a one-path maze

12 Answers 12

Jelly, 36 35 34 33 32 bytes

How?

JavaScript (ES6), 86 92 116

Batch, 216 bytes

PHP, 99 Bytes

PHP, 157 Bytes

JavaScript (ES6), 87 bytes

Test

Ruby 72 or 69 bytes

Java 7, 228 bytes

Python 2, 89 bytes

Racket 187 bytes

GNU sed 140 bytes

T-SQL, 123/79 bytes

JavaScript (ES6), 96 bytes

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Generate a one-path maze

12 Answers 12

Jelly, 36 35 34 33 32 bytes

How?

JavaScript (ES6), 86 92 116

Batch, 216 bytes

PHP, 99 Bytes

PHP, 157 Bytes

JavaScript (ES6), 87 bytes

Test

Ruby 72 or 69 bytes

Java 7, 228 bytes

Python 2, 89 bytes

Racket 187 bytes

GNU sed 140 bytes

T-SQL, 123/79 bytes

JavaScript (ES6), 96 bytes

Your Answer

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