Write a domino effect

Question

Using the fewest Unicode characters, write a function that accepts three parameters:

• Total number of dominoes
• nth affected domino
• Topple direction of the affected domino (0 or L for left, 1 or R for right)

Once a domino is toppled, it must also topple the remaining dominoes in the same direction.

You should output the dominoes with | representing a standing domino and \ and / representing a domino toppled to the left and right respectively.

Examples

10, 5, 1 should return ||||//////
6, 3, 0 should return \\\|||

Answer 1

Ruby, 38 (46) characters

e=->n,k,r{k-=r;'\|'[r]*k+'|/'[r]*n-=k}

This function takes the direction as an integer (1 for right, 0 for left). A function that takes a string is 8 characters longer:

d=->n,k,r{n-=k;r<?r??\\*k+?|*n :?|*~-k+?/*-~n}

Usage examples:

puts e[10, 5, 1] # or d[10, 5, 'r']
||||//////
puts e[10, 5, 0] # or d[10, 5, 'l']
\\\\\|||||

Answer 2

J - 32 26 char

J can't handle more than two arguments without using a list, and it can't handle non-homogenous lists without boxing. So having the input as a list of three integers is ideal. The parameter order is the reverse of the standard one: 0 for left or 1 for right, then position, then total number of dominoes. The reason for this is because J will end up going through them right-to-left.

{`(('|/\'{~-@>:,:<:)1+i.)/

Here's what's going on. F`G/ applied to a list x,y,z will evaluate x F (y G z). y G z constructs both possible ways the dominoes could have toppled, and then F uses x to select which of the two to use.

Below is a back-and-forth with the J REPL that explains how the function is built together: indented lines are input to the REPL, and responses are flush with the left margin. Recall that J evaluates strictly right to left unless there are parens:

   1 ] 3 (]) 10            NB. ] ignores the left argument and returns the right
10
   1 ] 3 (] 1+i.) 10       NB. hook: x (F G) y  is  x F (G y)
1 2 3 4 5 6 7 8 9 10
   1 ] 3 (>: 1+i.) 10      NB. "greater than or equal to" bitmask
1 1 1 0 0 0 0 0 0 0
   1 ] 3 (-@>: 1+i.) 10    NB. negate
_1 _1 _1 0 0 0 0 0 0 0
   1 ] 3 (<: 1+i.) 10      NB. "less than or equal to"
0 0 1 1 1 1 1 1 1 1
   1 ] 3 ((-@>:,:<:)1+i.) 10          NB. laminate together
_1 _1 _1 0 0 0 0 0 0 0
 0  0  1 1 1 1 1 1 1 1
   1 ] 3 (('|/\'{~-@>:,:<:)1+i.) 10   NB. turn into characters
\\\|||||||
||////////
   1 { 3 (('|/\'{~-@>:,:<:)1+i.) 10   NB. select left or right version
||////////
   {`(('|/\'{~-@>:,:<:)1+i.)/ 1 3 10  NB. refactor
||////////
   {`(('|/\'{~-@>:,:<:)1+i.)/ 0 3 10
\\\|||||||

At the expense of a few characters, we can make the order the standard order: just append @|. to the end of the function:

   |. 10 3 1
1 3 10
   {`(('|/\'{~-@>:,:<:)1+i.)/@|. 10 3 1
||////////

Adapting this to work with a string argument for direction would be much more costly, however.

Answer 3

Haskell, 70

f R i l=(i-1)#'|'++(l-i+1)#'/'
f L i l=i#'\\'++(l-i)#'|'
(#)=replicate

assuming there is a type Direction, which has constructors R and L.

Answer 4

PowerShell, 66

filter d($n,$k,$d){"$('\|'[$d])"*($k-$d)+"$('|/'[$d])"*($n-$k+$d)}

Probably the same idea every one else had.

• Takes either 0 or 1 as the direction parameter (for left and right, respectively)

Answer 5

Golfscript (44 53)

My first ever Golfscript program. Took me way longer than it should have and can probably be done in a smarter, more concise way (I'm sure someone will prove that :) ):

:d;:j;:^,{:x j<d&'\\'{x^j)->d!&'/''|'if}if}%

A sample input is 10 5 0.

Ungolfed:

:d;:j;:^      # save input in variables and discard from stack, except total length ^
,             # create an array of numbers of length ^
{             # start block for map call
  :x          # save current element (= index) in variable
  j<          # check whether we are left of the first knocked over domino
  d           # check whether the direction is to the left
  &           # AND both results
  '\\'        # if true, push a backslash (escaped)
  {           # if false, start a new block
    x^j)->    # check whether we are on the right of the knocked over domino
    d!        # check whether the direction is to the right
    &         # AND both results
    '/'       # if true, push a slash
    '|'       # if false, push a non-knocked over domino
    if
  }
  if
}%            # close block and call map

Answer 6

JS (ES6) - 79 74 72 65 62

thanks to @nderscore!

The 3rd param is a boolean (0: left / 1: right)

d=(a,b,c)=>"\\|"[a-=--b,c].repeat(c?b:a)+"|/"[c].repeat(c?a:b)

// Test
d(10,3,1); // => "||////////"
d(10,3,0); // => "\\\\\\\\||"

Answer 7

Python - 45 52

This requires 1 for right and 0 for left.

x=lambda n,k,d:'\\|'[d]*(k-d)+"|/"[d]*(n-k+d)

Here's a version that takes r and l correctly, at 58:

def x(n,k,d):d=d=='r';return'\\|'[d]*(k-d)+"|/"[d]*(n-k+d)

Some usage examples...

>>> print(x(10,3,0))
\\\|||||||
>>> print(x(10,3,1))
||////////
>>> print(x(10,5,1))
||||//////
>>> print(x(10,5,0))
\\\\\|||||
>>> print(x(10,3,0))
\\\|||||||

Answer 8

GolfScript, 28 23 characters

'\\'@*2$'|/'*$-1%1>+@/=

Arguments on top of stack, try online:

> 10 5 1
||||//////

> 10 5 0
\\\\\|||||

Answer 9

Haskell, 42 bytes

(n%k)b=["\\|/"!!(b-div(k-b-c)n)|c<-[1..n]]

Try it online!

Takes input like (%) n k b for n dominos, k'th domino toppled, direction b.

Finds the character at each position c ranging from 1 to n by using an arithmetic expression to compute the character index 0, 1, or 2.

Test cases taken from here.

---

Haskell, 44 bytes

(n%k)b=take n$drop(n+n*b+b-k)$"\\|/"<*[1..n]

Try it online!

An interesting strategy that turned out a bit longer. Generates the string "\\|/"<*[1..n] with n consecutive copies of each symbol, then takes a slice of n contiguous characters with start position determined arithmetically.

Answer 10

Python2/3 - 54

That last added on rule was quite nice (the 0/1 instead of 'l'/'r'). Made mine actually smaller than the existing python solution. 0 is left, 1 is right

def f(a,b,c):d,e='\|/'[c:2+c];h=b-c;return d*h+e*(a-h)

# Usage:
print(f(10,5,1)) # => ||||//////
print(f(10,5,0)) # => \\\\\|||||

Answer 11

Haskell, 57 bytes

4 bytes saved thanks to this tip

f 0 b _=[]
f a b c=last("|/":["\\|"|b>c])!!c:f(a-1)(b-1)c

Try it online!

Haskell, 69 61 60 58 bytes

(0!b)_=[]
(a!b)c|b==c=a!b$c+1|1>0="\\|/"!!c:((a-1)!(b-1))c

Try it online!

Not a very complex answer but it beats both of the existing Haskell answers.

Answer 12

R, 75 68 61 57 bytes

An anonymous function. I'll post a fuller explanation if there is interest.

function(t,n,d)cat(c("\\","|","/")[(1:t>n-d)+1+d],sep="")

Try it online!

Answer 13

Haskell, 51 bytes

f a b c=("\\|"!!c<$[1..b-c])++("|/"!!c<$[b-c..a-1])

a = number of dominoes, b = 1-based index of the touched one, c = direction (0 is left and 1 is right).

Try it online!

Answer 14

Python 2.7, 68 65 61 59 58 chars

Use d=1 for left and d=0 for right

f=lambda a,p,d:['|'*(p-1)+'/'*(a-p+1),'\\'*p+'|'*(a-p)][d]

Note: Thanks to @TheRare for further golfing it.

Answer 15

Perl, 67 65 Characters

sub l{($t,$p,$d)=@_;$p-=$d;($d?'|':'\\')x$p.($d?'/':'|')x($t-$p)}

Assign the first three params (total, position, direction as an integer [0 left, 1 right]). Extras go into the ether. Subtract 1 from the position if we're headed right so the domino in position X is flipped, too.

Answer 16

JavaScript (ES6) 61 63

Edit It was buggy - shame on me.

Not so different from @xem, but found it myself and it's shorter. Parameter d is 0/1 for left/right

F=(a,p,d,u='|'.repeat(--p),v='\\/'[d].repeat(a-p))=>d?u+v:v+u

Test In Firefox console

for(i=1;i<11;i+=3) console.log('L'+i+' '+F(10,i,0) + ' R'+i+' '+ F(10,i,1))

Output

L1 \\\\\\\\\\ R1 //////////
L4 \\\\\\\||| R4 |||///////
L7 \\\\|||||| R7 ||||||////
L10 \||||||||| R10 |||||||||/

Answer 17

Javascript, 46 characters

Seems like cheating to do 0=l and 1=r but there is is. Shrunk it with a little recursion.

f=(a,p,d)=>a?'\\|/'[(p-d<1)+d]+f(a-1,p-1,d):''

edit: missed an obvious character

Answer 18

J, 23 21 19 bytes

'|/\'{~{:-(>i.)~`-/

Try it online!

Input is a list of integers in the standard order.

Answer 19

05AB1E, 19 bytes

αα©„\|³è×¹®-„|/³è×J

I still have the feeling it's a bit long, but it works.. And better than the initial 23 bytes solution I had with if-else construction, which I quickly dropped..

Input order is the same as in the challenge: total length, index, 1/0 for left/right respectively.

Try it online or verify both test cases.

Explanation:

α                     # Take the absolute difference of the first two (implicit) inputs
                      #  i.e. 10 and 5 → 5
                      #  i.e. 6 and 3 → 3
 α                    # Then take the absolute difference with the third (implicit) input
                      #  i.e. 5 and 1 → 4
                      #  i.e. 3 and 0 → 3
  ©                   # Store this number in the register (without popping)
   „\|                # Push "\|"
      ³è              # Use the third input to index into this string
                      #  i.e. 1 → "|"
                      #  i.e. 0 → "\"
        ×             # Repeat the character the value amount of times
                      #  i.e. 4 and "|" → "||||"
                      #  i.e. 3 and "\" → "\\\"
         ¹®-          # Then take the first input, and subtract the value from the register
                      #  i.e. 10 and 4 → 6
                      #  i.e. 6 and 3 → 3
            „|/       # Push "|/"
               ³è     # Index the third input also in it
                      #  i.e. 1 → "/"
                      #  i.e. 0 → "|"
                 ×    # Repeat the character the length-value amount of times
                      #  i.e. 6 and "/" → "//////"
                      #  i.e. 3 and "|" → "|||"
                  J   # Join the strings together (and output implicitly)
                      #  i.e. "||||" and "//////" → "||||//////"
                      #  i.e. "///" and "|||" → "///|||"

Answer 20

PHP, 89 Characters

function o($a,$p,$d){for($i=0;$i<$a;$i++)echo$d==0?($i+1>$p)?'|':'\\':($i+1<$p?'|':'/');}

Just because I love PHP.

EDIT: The following code does the same.

function dominoes ($number, $position, $direction) {
    for ($i=0; $i<$number; $i++){
        if ($direction==0) {
            if (($i+1) > $position) {
                echo '|';
            } else {
                echo '\\';
            }
        } else {
            if (($i+1) < $position) {
                echo '|';
            } else {
                echo '/';
            }
        }
    }
}

Answer 21

PHP - 64

function f($a,$b,$c){for($w='\|/';++$i<=$a;)echo$w[$c+($i>$b)];}

A simple loop, and echo-ing the character.

Generates a Notice: Undefined variable: i, here's another version silenting the error (65 characters) :

function f($a,$b,$c){for($w='\|/';@++$i<=$a;)echo$w[$c+($i>$b)];}

And a version withtout any error (69 characters) :

function f($a,$b,$c){for($w='\|/',$i=0;++$i<=$a;)echo$w[$c+($i>$b)];}

---

Other functions in PHP :

sprintf / printf padding

function f($a,$b,$c){printf("%'{${0*${0}=$c?'|':'\\'}}{$a}s",sprintf("%'{${0*${0}=$c?'/':'|'}}{${0*${0}=$a-$b+$c}}s",''));}

padding via str_pad / str_repeat functions

function f($a,$b,$c){$f='str_repeat';echo$f($c?'|':'\\',$b-$c).$f($c?'/':'|',$a-$b+$c);}
function f($a,$b,$c){echo str_pad(str_repeat($c?'|':'\\',$b-$c),$a,$c?'/':'|');}

using both printf and str_repeat functions

function f($a,$b,$c){printf("%'{${0*${0}=$c?'|':'\\'}}{$a}s",str_repeat($c?'/':'|',$a-$b+$c));}
function f($a,$b,$c){$w='\|/';printf("%'$w[$c]{$a}s",str_repeat($w[$c+1],$a-$b+$c));}

Answer 22

Scala 75 characters

def f(l:Int,p:Int,t:Char)=if(t=='l')"\\"*p++"|"*(l-p) else "|"*(l-p):+"/"*p

Answer 23

CJam - 20

q~
:X-_"\|"X=*o-"|/"X=*

The main code is on the second line, the first line is just for getting the parameters from the standard input (otherwise you need to put the parameters in the code).

Try it at http://cjam.aditsu.net/

Examples:

12 4 1
|||/////////

8 5 0
\\\\\|||

Explanation:

:X stores the last parameter (0/1 direction) in variable X
- subtracts X from the knock-over position, obtaining the length of the first sequence of characters (let's call it L)
_ makes a copy of L
"\|"X= gets the character to use first: \ for X=0 and | for X=1
* repeats that character L times
o prints out the string, removing it from the stack
- subtracts L from the number of dominoes, obtaining the length of the second sequence of characters (let's call it R)
"|/"X= gets the character to use next: | for X=0 and / for X=1
* repeats that character R times

Answer 24

Common Lisp

This won't win in a code golf, but it highlights Common Lisp's justification format directive:

(lambda (n p d &aux (x "\\|/"))
   (format t "~v,,,v<~v,,,v<~>~>" n (aref x d) (+ d (- n p)) (aref x (1+ d))))

The arithmetic isn't bad: n is the total number of dominoes; p is the position of the first toppled domino; d is either 0 or 1, representing left and right (as allowed in the comments), and is used as an index into x; x is a string of \, |, and /. The format string uses two (nested) justification directives, each of which allows for a padding character. Thus:

(dotimes (d 2)
  (dotimes (i 10)
    ((lambda (n p d &aux (x "\\|/"))
       (format t "~v,,,v<~v,,,v<~>~>" n (aref x d) (+ d (- n p)) (aref x (1+ d))))
     10 (1+ i) d)
    (terpri)))

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

Answer 25

PHP, 55 bytes

for([,$c,$n,$d]=$argv;$c--;)echo"\|/"[$d+(++$i+$d>$n)];

• requires PHP 7.1 or later
• replace [...]=$argv with list(...)=$argv for PHP 5.5 or later
• for older PHP, insert $s="\|/", before that and replace the other "\|/" with $s.

Run with php -nr '<code>' <total> <N> <numeric direction> or try it online.

Answer 26

C - 62 64 ('l':left, 'r':right)

f(a,b,c){for(b=a-b;a--;)putchar(c&2?a<b?47:124:a>b-2?92:124);}

C - 58 (0:left,1:right)

f(a,b,c){for(b=a-b;a--;)putchar(124-(a<b-!c?c*77:!c*32));}

recursive approach, but still, same length

f(a,b,c){a&&f(a-1,b,c)+putchar(124-(a<=b+!c?!c*32:c*77));}

awfully long (75) but most complex one (repeat function integrated into it)

f(a,b,c){c>1?a&&f(a-1,putchar(b),2):f(b+c,c?92:124,2)+f(a-b-c,c?124:47,2);}

Answer 27

PowerShell (107 95)

filter d($a,$b,$c){-join(1..$a|%{if($_-ge$b-and$c){'/'}elseif($_-le$b-and!$c){'\'}else{'|'}})}

If we're allowed to treat the third parameter as an integer (0=l,1=r), then I can shave a little off here.

Test input:

d 10 5 r
||||//////
d 10 5 l
\\\\\|||||

Update:

d 10 5 1
||||//////
d 10 5 0
\\\\\|||||

If anyone by chance wants an explanation I will gladly add one. And as always I'm open to suggestions.

Original:

function d($a,$b,$c){-join(1..$a|%{if($_-ge$b-and$c-eq'r'){'/'}elseif($_-le$b-and$c-eq'l'){'\'}else{'|'}})}

Answer 28

C++ 181

#define C(x) cin>>x;
#define P(x) cout<<x;
int n,k,i;char p;
int main(){C(n)C(k)C(p)
for(;i<n;i++){if(p=='r'&&i>=k-1)P('/')else if(p=='l'&&i<=k-1)P('\\')else P('|')}
return 0;}

Answer 29

VBScript 101 95 74

Edit: Looks like a lot of answers accept bool directions. For char directions, replace -d with +(d="r") for a score increase of 6.

function f(a,p,d):for x=1to a:f=f+mid("/|\",3-d+(x>p),1):next:end function

Answer 30

C# 92('\':left, '/':right)

String d(int c, int p, char d){ String r="";for(int i=0;i<c;i++)r+=(i>=c-p)?d:'|';return r;}

Answer 31

AWK, 76 61

{ORS="";for(i=1;i<$2;i++)print"|";for(i=$2;i<=$1;i++)print($3=="r"?"/":"\\")}

Usage:

echo 10 5 r | awk '{ORS="";for(i=1;i<$2;i++)print"|";for(i=$2;i<=$1;i++)print($3=="r"?"/":"\\")}'

Although I'm wondering if I can nest ternary operators to make this shorter (although I'm playing around in bash right now and it doesn't seem like you can).

edit -- shorter version:

{ORS="";for(i=0;i<$1;i++)print i<$2?"|":($3=="r"?"/":"\\");}

Answer 32

PHP - 105,97, 96

 function a($r,$l,$f){$a=str_repeat('|',$l-$f);$b=str_repeat($r?'/':'\\',$f);echo$r?$a.$b:$b.$a;}

Example results:

a(true,10,4);  -> \\\\||||||
a(false,10,5); -> |||||/////
a(false,10,2); -> ||||||||//

Answer 33

Javascript, 81 85 characters

function e(a,b,c){l='repeat';d='|'[l](--a-b++);return c>'q'?d+"/"[l](b):"\\"[l](b)+d}

First time trying codegolf, was fun thanks :)

Answer 34

JavaScript - 85 chars

function d(a,b,c){for(r=c?"\\":"/",p="",b=a-b;a--;)p+=c?a<b?"|":r:a>b?"|":r;return p}

1 = Left, 0 = Right

d(10,3,1)
\\\|||||||
d(10,3,0)
||////////
d(10,7,1)
\\\\\\\|||
d(10,7,0)
||||||////

Answer 35

vb.net (~75c)

Dim f=Function(l,p,d)(If(d="l",StrDup(p,"\"),"")& StrDup(l-p-If(d="l",1,0),"|")).PadRight(l,"/")

Answer 36

Clojure, 81 chars

(defn g[a,p,d](apply str(map #(nth "\\|/"(+(if(>= % (- p d)) 1 0) d))(range a))))

Answer 37

JavaScript (ES5) - 86 chars

function f(a,p,d){r='';while(a-->p+1)r+='|';while(a-->=0)r=d>'q'?r+'/':'\\'+r;return r}

Answer 38

Autoit 245 Bytes

Local $1 = "",$dir = "l",$amout = 100,$pos = 33
for $i=1 to $pos
    if $dir = "r" then
    $1 = $1 & "|"
Else
    $1 = $1 & "\"

EndIf
next
for $i=1 to $amout-$pos
    if $dir = "r" then
    $1 = $1 & "/"
Else
    $1 = $1 & "|"
EndIf
next
ConsoleWrite($1 & @cr)

Answer 39

Python 71 59

def k(n,p,d):c,r='\/'[d],n-p;return[c*p+'|'*r,'|'*p+c*r][d]

Answer 40

Haskell (72)

Using guards and the characters 'r' and 'l' as directions:

r=replicate;k a o d|d=='r'=r(a-o)'|'++r o '/'|d=='l'=r o '\\'++r(a-o)'|'

Answer 41

Perl : 66 characters

My first code golf submission! Thanks for the fun challenge.

sub k{($n,$p,$r)=@_;$n-=$p;$d=$r?"|"x--$p.'/'x++$n:'\\'x$p.'|'x$n}

Explanation with whitespace:

sub k{
  ($n,$p,$r) = @_;      # get function arguments: <# dominos> <position> <0=left, 1=right>
  $n -= $p;             # total number of dominos - position
  $d = $r ? "|"x--$p . '/'x++$n : '\\'x$p . '|'x$n
                        # if direction = 1, make pos - 1 '|'s and (total - position  + 1) '/'s
                        # if direction = 0, make pos '\'s and (total - position) '|'s
                        # return is the last action of the function, it returns the string in $d.
                        # Semicolon at end of line can be left off, as it is the last statement of the function
}

Example usage and output:

my $result = k(10,5,1); print "$result\n";
||||//////
my $result = k(10,5,0); print "$result\n";
\\\\\|||||

Answer 42

Python, 69

My first code golf submission, with Python in 69 characters:

def d(a,b,c):print "|"*(b-1)+"/"*((a+1)-b) if c else "\\"*b+"|"*(a-b)

Usage: d(10,5,1)