The (Easy) Road to Code

Question

Input:

• An integer \$n\$, which is guaranteed to be \$\ge3\$.
• An integer \$d\$, which is one of \$[-1,0,1]\$.

Output:

A road of size \$n\$, which will be in a north-west direction if \$d=-1\$; a north direction if \$d=0\$; or a north-east direction if \$d=1\$. The road will always be three spaces wide (or five in general if we'd include the outer borders). In addition, there will be a road separator line on the bottom section, and after that alternating while going upwards.

Some examples:

Input: \$n=7, d=1\$
Output:

      / / /
     /   /
    / / /
   /   /
  / / /
 /   /
/ / /

Input: \$n=4, d=-1\$ Output:

\   \
 \ \ \
  \   \
   \ \ \

Challenge rules:

• Any amount of leading/trailing spaces and/or newlines are acceptable, as long as it prints the required road somewhere on the screen.
• Instead of the options \$[-1,0,1]\$ you are also allowed to use the options \$[0,1,2]\$ or \$[1,2,3]\$ instead. Also, you can choose which of the three options map to which of the three directions. (Make sure to mention which options you've used if it differs from the \$[-1,0,1]\$ for [north-west, north, north-east] respectively that is used in this challenge description!)
• Any reasonable output format is acceptable. The most common is of course to print it to STDOUT, but returning it as a string or 2D character list is fine as well.

General rules:

• This is code-golf, so shortest answer in bytes wins.
  Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
• Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
• Default Loopholes are forbidden.
• If possible, please add a link with a test for your code (i.e. TIO).
• Also, adding an explanation for your answer is highly recommended.

Test cases:

The two examples above, and:

Input: \$n=10, d=0\$
Output:

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

Answer 1

Python 2, 79 78 73 72 bytes

n,d=input()
c='|\/'[d]
i=n
while i:print' '*(n-i*d)+c,i%2*c or' ',c;i-=1

Try it online!

Takes [1,0,-1] for [north-west, north, north-east]

_{-1 byte, thanks to Neil}

Answer 2

Python 2, 66 bytes

n,d=input()
s=' '+'|\/'[d]
for c in(s*n)[n:]:print' '*n+s,c+s;n+=d

Try it online!

Uses d=-1 for NE, d=0 for N, and d=1 for NW. Takes advantage of leading spaces being allowed. The rule that the bottom segment of the road have a separator made it tricky to get the parity right; it's achieved by the slicing (s*n)[n:] taking the second half of the 2n alternations between space and a the road character.

Answer 3

1. Python 3.5, 122 120 bytes

Script takes two params: n, d.

d: 0, 1, 2 -> \ | /

tio.run

import sys;(n,d)=[*map(int,sys.argv[1:3])];c="\\|/"[d]
for i in range(n):j=n+~i;print(" "*(i,0,j)[d],c,c*(j%2<1)or" ",c)

output:

$ ./script.py 6 2
      /   /
     / / /
    /   /
   / / /
  /   /
 / / /
$ ./script.py 6 1
 |   |
 | | |
 |   |
 | | |
 |   |
 | | |
$ ./script.py 6 0
 \   \
  \ \ \
   \   \
    \ \ \
     \   \
      \ \ \

Explanation

# parse input params
(n,d)=[*map(int,sys.argv[1:3])]

# select char for "road"
c="\\|/"[d]

# loop n-times
for i in range(n):

    # complement (how many lines to end)
    j=n+~i

    # print
    #   space i or 0 or j times
    #   road, center of road if j is even else space, road
    print(" "*(i,0,j)[d], c, c*(j%2<1) or " ", c)

edit: -2 bytes thanks to Kevin Cruijssen

Answer 4

PowerShell, 88 82 80 74 71 bytes

^{-8 bytes thanks to Mazzy
-6 bytes thanks to AdmBorkBork and Mazzy
-3 bytes thanks to AdmBorkBork}

param($n,$d)$n..1|%{' '*($_,($n-$_))[$d]+($y='/\|'[$d])," $y"[$_%2],$y}

Try it online!

Uses [0,1,2] for NW, NE, N. Uses d as a list index twice to first get the spacing method (Drops off the list when using 2 which returns 0) and then which character to use when drawing the lines. Appends a list to the string of spaces (which innately has spaces between the members when appended as such) which builds the road. Also swaps between an open lane or a dash based on modulo math.

Answer 5

C (gcc), 100 90 bytes

f(a,b,c,d){for(c=a,d="\\|/"[b+1];c--;)printf("%*c %c %c\n",b>0?c+1:b?a-c:0,d,c%2?32:d,d);}

Try it online!

---

Well written version

Answer 6

Charcoal, 33 29 23 bytes

↷ＮＮη⊘⊕ηＵＥ¹¦¹Ｆ⟦²±²⟧«Ｊι⁰η

Try it online! Link is to verbose version of code. Now that all of the underlying Charcoal bugs seem to be fixed I can advertise the optimal code. Explanation:

↶Ｎ

Rotate the cursor direction anticlockwise according to the first input, so that 1 becomes north east, 2 north and 3 north west.

Ｎη

Input the length of the road.

⊘⊕ηＵＥ¹¦¹

Print half the length of the road and then stretch it giving the road separator.

Ｆ⟦²±²⟧«Ｊι⁰η

Print the sides of the road.

@KevinCruijssen subsequently submitted a harder version of this question which was since deleted but users with enough rep can see it here: The (Hard) Road to Code The stretching trick I used in this answer isn't applicable to that question, so instead I wrote the following 47 45-byte program:

Ｆ³«Ｊ×ι⊘⊕θ⁰≔…⟦¹ ⟧⊕﹪ι²ιＦη«↶§κ⁰Ｆ⊖§κ¹§ιⅉＰ§ιⅉ↑↷§κ⁰

Try it online! Link is to verbose version of code. Explanation:

Ｆ³«

Loop over the sides and separator.

Ｊ×ι⊘⊕θ⁰

Jump to the start of the side.

≔…⟦¹ ⟧⊕﹪ι²ι

Make an array containing a 1 and a space, but remove the space again if we're drawing the sides rather than the separator.

Ｆη«

Loop over each road section.

↶§κ⁰

Rotate the cursor direction accordingly.

Ｆ⊖§κ¹

Loop over one less than the length of the road section...

§ιⅉ

... and print alternating elements of the array. The alternation is achieved by cyclically indexing into the array with the Y-coordinate of the cursor.

Ｐ§ιⅉ

Print the last row of this road section, but without moving the cursor...

↑

... so that the cursor can be moved up ready for the next section instead.

↷§κ⁰

Rotate the cursor direction back ready for the next section.

Answer 7

Kotlin, 96 92 bytes

{n,d->val c="\\|/"[d];(0..n-1).map{println(" ".repeat(n-it*(d-1))+ "$c ${" $c"[it%2]} $c")}}

Accepts [0, 1, 2] instead of [-1, 0, 1]

It works similarly to solutions in other languages, but unfortunately Kotlin doesn’t really shine in this one.

val c=“\\|/“; retrieves the char to use in road building by taking advantage of the fact that Kotlin treats Strings like an Array of chars (as it should, looking at you Java)

Try it online!

Answer 8

TSQL code, 171 117 bytes

I realized this could be written much shorter as code.

DECLARE @n INT=7,@d INT=0

,@ INT=0a:PRINT
space(@n-@*@d)+stuff(replicate(substring('\|/',@d+2,1)+' ',3),3,@%2,space(@%2))SET
@+=1IF @n>@ GOTO a

TSQL query, 137 bytes

USE master
DECLARE @n INT=6,@ INT=-1

SELECT space(@n-y*@)+z+iif(y%2=1,z,'  ')+z 
FROM(SELECT top(@n)row_number()over(order
by @)y,' '+substring('\|/',@+2,1)z FROM spt_values)x

USE master is not necessary, if your database is master already. Some users however have a different default database.

Script when trying it out is slightly different. I had to replace space ascii-32 with ascii-160, the spaces were not shown.

Try it out

Made some tweaks and realized that I could replace

order by 1/0

with

order by @

Answer 9

Canvas, 23 22 bytes

B}⁴H‟ｍ↶⁷？／|∙/╋］＼|∙\╋］｝

Try it here!

Uses the direction inputs 0, 1 and 2.

Explanation:

...‟    push "--¶¶- ¶¶--"
    m   mold the horizontal length to the 1st input - a horizontal road
     ↶  rotate counter-clockwise - so the separator is on the bottom
        also replaces the dashes with bars

⁷?     ]     ]}  switch over the 2nd input:
                 default case (input 2):
  /                pad with spaces to a "/" diagonal
   |∙/╋            replace "|" with "/"
                 case 0:
        \          pad with spaces to a "\" diagonal
         |∙\╋      replace "|" with ""

case 1 is empty, but because of a bug, the trailing "}" is still required

Answer 10

JavaScript (ES8),  90 87  85 bytes

Takes input as (direction)(length). Expects \$0\$ for North-West, \$1\$ for North or \$2\$ for North-East.

d=>g=(n,w=n)=>n?(c='/|\\'[d]).padStart([n--,,w-n][d])+` ${n&1?' ':c} ${c}
`+g(n,w):''

Try it online!

---

JavaScript (ES6), 90 bytes

This one draws the output character by character with a little more maths instead of the .padStart() method.

Takes input as (direction)(length). Expects \$1\$ for North-West, \$2\$ for North or \$3\$ for North-East.

d=>n=>(g=x=>y?` /|\\
`[x+5?17+y%2*4>>--x+5-[,n+1-y,n,y][d]&1&&d:(x=n,y--,4)]+g(x):'')(y=n)

Try it online!

Answer 11

Jelly,  31  30 bytes

⁶ẋẋƤḂ};€ị“\/|”ẋ3KƊ}Ṛ⁹¡⁶-2¦€ÐeṚ

A dyadic Link accepting the length on the left and the negated-direction* on the right which yields a 2d array of characters.
* [north-west, north, north-east]=[1, 0, -1]

Try it online! (footer calls the Link, joins with newline characters and prints the result)

How?

⁶ẋẋƤḂ};€ị“\/|”ẋ3KƊ}Ṛ⁹¡⁶-2¦€ÐeṚ - Link: integer, L; integer, D
⁶                              - space character
 ẋ                             - repeat L times
     }                         - using the right argument (D):
    Ḃ                          -   mod 2 (0 if D = 0, 1 otherwise ...A would work too)
   Ƥ                           - for each prefix (of the list of spaces):
  ẋ                            -   repeat (D mod 2 times)
                  }            - using the right argument (D):
                 Ɗ             -   last three links as a monad:
        ị“\/|”                 -     index into "\/|" (1-indexed & modular)
              ẋ3               -     repeat three times - e.g. "\\\"
                K              -     join with spaces          "\ \ \"
      ;€                       - concatenate €ach (altered prefix with that list)
                     ¡         - repeat action...
                    ⁹          - ...number of times: right argument (-1 repeats 0 times)
                   Ṛ           - ...action: reverse (i.e. reverse if north-west)
                           Ðe  - for even indexes:
                         ¦€    -   sparse application for €ach...
                       -2      -   ...to indices: [-2] (the middle of the road)
                      ⁶        -   ...action: a space character
                             Ṛ - reverse

Answer 12

Python 2, 127 bytes

n,d=input()
c='|/\\'[d]
for i in range(n):t=0if d==0else(i if d<0else n-i);print(' '*t+'{} {} {}'.format(c,' 'if i%2==0else c,c))

Try it online!

First two lines taken from @TFeld.

Answer 13

Python 2, 93 bytes

n,d=input();p=''
while n:g='|/\\'[d];n-=1;print['',' '*n,p][d]+g+' %s '%(g,' ')[n%2]+g;p+=' '

Try it online!

Answer 14

J, 64 bytes

' \|/'{~(_2&|*[:i.[*1 _1{~3=])|."_1((5+[){."1(5,~[)|.@$]*#:@689)

Try it online!

• 1 North-West
• 2 North
• 3 North-East

Will try to golf it later

Answer 15

C# (Visual C# Interactive Compiler), 129 107 103 93 bytes

a=>b=>{for(char i=a,W="/|\\"[b+1];i>0;)Write("{0,"+(a-i*b+5)+"} {1} {0}\n",W,--i%2<1?W:' ');}

-1 for north-east, 1 for north-west, 0 for north

Try it online!

Answer 16

Vyxal, 91 bytes

£:0=[¥(n₂[`| | |`|`|   |`],)|1=[¥(ðn*n₂[`\ \ \\`|`\   \\`]+,)|¥(ð¥‹n-*n₂[`/ / /`|`/   /`]+,

Try it Online!

A little better.

Answer 17

Red, 157 bytes

func[s d][p: pick"\|/"d set[a b]reduce pick[[4 1]0[5 + s -1]]d 
repeat n s[print pad/left form reduce pick[[p" "p][p p p]]odd?(n + pick[1 0]odd? s)a: a + b]]

Try it online!

• 1 North-West
• 2 North
• 3 North-East

Answer 18

Japt, 53 bytes

W="|/\\"gV;Æ"{W} {YÉ %2?W:' } {W}"ùVÄ?V?Y+5:5:U-Y+4Ãw

Try it Online!

Answer 19

Swift 4.2, 112 108 bytes

^{-4 bytes thanks to Sriotchilism O'Zaic}

let c=["\\","|","/"][d];(0..<n).map{print((0..<n-$0*(d-1)).map{_ in" "}.joined()+c+" \($0%2==0 ?c:" ") "+c)}

[0,1,2] instead of [-1,0,1].

Try it online!

Explanation

let c=["\\","|","/"][d]

Determines the street sign to use. (I bet this can be shortened somehow)

(0..<n).map{

Iterates over the length of the street.

(0..<n-$0*(d-1)).map{_ in" "}.joined()

Adds the spaces in front of the street.

\($0%2==0 ?c:" ")"

Adds the middle sign to every other line.

My very first one ☝️, I'm happy about any optimization advices. Learned mostly from that thread: Tips for golfing in Swift.

Answer 20

Perl 6, 66 bytes

->\n,\d{{(' 'x--$ *d%-+^n~(' ',<\ | />[d+1])[1,(n+$++)%2,1])xx n}}

Try it online!

Returns a list of lines.

Explanation

->\n,\d{                                                         }  # Block taking n, d
        {                                                       }   # Reset $ variables
         (                                                 )xx n    # Repeat n times
          ' 'x  # Space times
              --$ *d  # (-1,-2,-3,...) if d=1
                      # (0,0,0,...) if d=0
                      # (1,2,3,...) if d=-1
                    %-+^n  # modulo -~n = n+1
                         ~  # Concat
                          (' ',<\ | />[d+1])  # Space or line
                                            [1,(n+$++)%2,1]  # Alternate middle char

Answer 21

Ruby, 90 bytes

n,a=*$*.map{|i|-i.to_i}
c='|\/'[a]
n.upto(-1){|i|puts' '*(i*a-n)+c+' '+[' ',c][i%2]+' '+c}

Try it online!

UPDATE: Thank you, Kevin, for pointing out that my original submission was missing spaces between the road edges and markers (i.e., 3-width roads instead of 5-width.) That said, there might be a shorter fix for that than c+' '+...+' '+c.

Directional information: -1, 0, 1 maps to { north-west, north, north-east }

Arguments: this reads command-line arguments such as 4 -1 (a four row road, slanted to the north-west).

Additional note: This was tested locally with Ruby v2.6.1, and it appears that Try It Online uses Ruby v2.5.3. I have no reason to think it would not work with all other Ruby versions.

Answer 22

Java (JDK), 116 bytes

d->n->{for(int c="\\|/".charAt(d),i=n;i-->0;)System.out.printf("%"+(d<1?n-i:d>1?i+1:1)+"c%2c %c%n",c,i%2<1?c:32,c);}

Try it online!

Takes d as one of 0, 1 or 2.

Answer 23

Japt -R, 40 bytes

Æ=Vg"|\\/";²i1S+iX%2?S:U)iSp[TZÊ-YY]gVÃw

Try it online!

Explanation:

Æ=Vg"|\\/";²i1S+iX%2?S:U)iSp[TZÊ-YY]gVÃw
                                            // U = road size
                                            // V = direction
Æ                                     Ã     // Loop X through [0...U]
 =        ;                                 //   Set U to:
    "|\\/"                                  //       "|\/"
  Vg                                        //            [V]
           ²                                //  Double U
            i1          )                   //  Insert at index 1:
              S+                            //    " " +
                 X%2?S:U                    //         X%2?" ":U
                i                           //                  + " "
                         i                  //  Insert at index 0:
                          Sp                //    " " repeated this many times:
                            [TZÊ-YY]        //      [0,U.length - X,X]
                                    gV      //                        [V]
                                        w   // Reverse output
      
  

Answer 24

Stax, 23 bytes

ù╒←∩♂2Ωⁿ¡├∩22♀Ç○=÷â╧δÖ↑

Run and debug it

Input is two integers separated by a space.
The first is d. -1 specifies north-east. 0 for north. 1 for north-west. The second parameter is n. These two values will get implicitly parsed from input and placed on the input stack with n on top. Additionally d will be accessible from the stax's x register since it's the first implicitly parsed value.

For example, input "1 7"

            Main stack      Explanation
            ----------      -----------
;           [7]             peek from input stack
m                           map over integers [1..n] using the rest of the program
                            output each produced value implicitly
  ;         [1 7]           peek from input stack
  +         [8]             add
  .*"       [8 [42 34]]     2 character string literal with codepoints 42 and 34
  @         [42]            get the element at the specified modular index (8)
  0         [42 0]          literal zero
  "|\/"x@   [42 0 92]       get codepoint from string literal at input d (from x register)
  \         [42 [0 92]]     make pair
  :B        ['\ \ \ ']      encode in custom base; it's binary with custom codepoints
  _x*       ['\ \ \ ' 1]    (loop variable) * (input d)
  ;^        ['\ \ \ ' 1 8]  peek from input stack and increment
  %         ['\ \ \ ' 1]    modulus
  6+        ['\ \ \ ' 7]    add 6
  )         [' \ \ \ ']     left-pad to length

Run this one

Answer 25

perl -M5.010 -Mfeature=signatures, 97 bytes

sub f($n,$d){$x=qw[| / \\][$d];say" "x($d<1?$_*-$d:$n-$_),$x,$",($n-$_)%2?$":$x,$",$x for 1..$n;}