25
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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 \\\|||

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  • \$\begingroup\$ Should the third parameter be a string or will a bool/int do like 0:left , 1:right? \$\endgroup\$ – user80551 Jul 9 '14 at 18:11
  • \$\begingroup\$ Your example suggests that if there are 10 dominoes, and 5 are knocked right, we should display six of the ten dominoes knocked over. \$\endgroup\$ – algorithmshark Jul 9 '14 at 18:21
  • 1
    \$\begingroup\$ @algorithmshark I think we should show the result if the fifth domino is knocked right. \$\endgroup\$ – user80551 Jul 9 '14 at 18:22
  • \$\begingroup\$ @rybo111 Can you allow the third parameter to be an int as that can make comparison operations shorter. Simply if(third_parameter) instead of if(third_paramter=='l') \$\endgroup\$ – user80551 Jul 9 '14 at 18:42
  • \$\begingroup\$ Can we choose the order of the parameters? \$\endgroup\$ – Justin Jul 9 '14 at 18:52

42 Answers 42

14
\$\begingroup\$

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']
\\\\\|||||
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  • \$\begingroup\$ why are there only 5 dominoes knocked left in the second example? \$\endgroup\$ – Clyde Lobo Jul 10 '14 at 16:06
  • 1
    \$\begingroup\$ @ClydeLobo Because you start at position 5 and knock the domino to the left, which in turn knocks over the 4 dominoes to its left, for a total of 5. In the first example, starting at position 5 knocks over 6 dominoes: The one at position 5 plus the 5 to its right. \$\endgroup\$ – Ventero Jul 10 '14 at 16:10
8
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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.

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8
\$\begingroup\$

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.

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  • \$\begingroup\$ I know it's been a while since you wrote this answer, but the way it is structured is very cool. I really like how you made use of gerunds and / and also the way you build two outputs and select the desired one. I guess I feel like this lacks the recognition it deserves. \$\endgroup\$ – cole Jan 4 '18 at 3:04
  • \$\begingroup\$ What @cole said, I was awed. \$\endgroup\$ – FrownyFrog Jan 4 '18 at 7:13
7
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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)
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6
\$\begingroup\$

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
\$\endgroup\$
  • 1
    \$\begingroup\$ Proof done ;-) although I am not happy with my solution yet. \$\endgroup\$ – Howard Jul 10 '14 at 6:06
  • 1
    \$\begingroup\$ Some tips: you may choose d to be 0/1 instead of 'l'/'r' which gives you some shorter code. Otherwise, if you store d'l'= in a variable oyu may use it instead of the second comparison with d. In the term x i j you can save both whitespaces if you use a non-alphanumeric variable name instead of i. \$\endgroup\$ – Howard Jul 10 '14 at 6:12
  • \$\begingroup\$ @Howard Thanks for the tips! I chose 'l'/'r' because at the time I didn't see yet that we are free to use integers. The non-alphanumeric trick is slick, thanks! Maybe I'll update the answer later. \$\endgroup\$ – Ingo Bürk Jul 10 '14 at 6:14
4
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GolfScript, 28 23 characters

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

Arguments on top of stack, try online:

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

> 10 5 0
\\\\\|||||
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  • \$\begingroup\$ Amazing. Love to learn from all these golfscript solutions :) \$\endgroup\$ – Ingo Bürk Jul 10 '14 at 6:15
4
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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))
\\\|||||||
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4
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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); // => "\\\\\\\\||"
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  • 1
    \$\begingroup\$ this entry could be a reference card for ECMAScript 6 :D \$\endgroup\$ – bebe Jul 9 '14 at 19:00
  • \$\begingroup\$ @bebe haha, and it's not even its final form. ES6 can be very dirty. \$\endgroup\$ – xem Jul 9 '14 at 19:08
  • 1
    \$\begingroup\$ 65: d=(a,b,c)=>"\\"[r="repeat"](!c&&a-b+1)+"|"[r](--b)+"/"[r](c&&a-b) \$\endgroup\$ – nderscore Jul 10 '14 at 3:47
  • 1
    \$\begingroup\$ great! I also found this crazy thing but it's longer (67): d=(a,b,c,d=a-b+1)=>"\\|"[c].repeat(c?b-1:d)+"|/"[c].repeat(c?d:b-1) \$\endgroup\$ – xem Jul 10 '14 at 7:04
  • \$\begingroup\$ I don't think repeat aliasing is worth. [r='repeat'][r] 15 chars. .repeat.repeat 14 chars \$\endgroup\$ – edc65 Jul 10 '14 at 15:33
3
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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)) # => \\\\\|||||
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3
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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.

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2
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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.

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  • 1
    \$\begingroup\$ Why not d and'\\'...or'/'...? \$\endgroup\$ – seequ Jul 9 '14 at 19:21
  • \$\begingroup\$ You could also do ('\\'...,'/'...)[d] \$\endgroup\$ – seequ Jul 10 '14 at 7:59
  • \$\begingroup\$ @TheRare I would need two of those lists. \$\endgroup\$ – user80551 Jul 10 '14 at 8:55
  • \$\begingroup\$ I don't think so. f=lambda a,p,d:('|'*(p-1)+'/'*(a-p+1),'\\'*p+'|'*(a-p))[d] \$\endgroup\$ – seequ Jul 10 '14 at 9:11
  • \$\begingroup\$ @TheRare Also, I don't think your code works when falling left. Could you give a test case to prove? \$\endgroup\$ – user80551 Jul 10 '14 at 9:16
2
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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

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2
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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 |||||||||/
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  • 1
    \$\begingroup\$ Should it be --p? \$\endgroup\$ – nderscore Jul 10 '14 at 16:48
  • \$\begingroup\$ @nderscore yes it should, got parameters wrong, silly me. \$\endgroup\$ – edc65 Jul 11 '14 at 9:36
2
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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.

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  • 1
    \$\begingroup\$ replace $p--if$d with $p-=$d to lose two characters :) \$\endgroup\$ – chinese perl goth Aug 4 '14 at 14:35
2
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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.

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2
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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!

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2
\$\begingroup\$

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!

\$\endgroup\$
  • \$\begingroup\$ Defining an infix operator also works for more than two inputs: (a#b)c= .... \$\endgroup\$ – Laikoni Oct 13 '18 at 8:28
1
\$\begingroup\$

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));}
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1
\$\begingroup\$

Scala 75 characters

def f(l:Int,p:Int,t:Char)=if(t=='l')"\\"*p++"|"*(l-p) else "|"*(l-p):+"/"*p
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1
\$\begingroup\$

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

\$\endgroup\$
1
\$\begingroup\$

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)))

\|||||||||
\\||||||||
\\\|||||||
\\\\||||||
\\\\\|||||
\\\\\\||||
\\\\\\\|||
\\\\\\\\||
\\\\\\\\\|
\\\\\\\\\\
//////////
|/////////
||////////
|||///////
||||//////
|||||/////
||||||////
|||||||///
||||||||//
|||||||||/
\$\endgroup\$
1
\$\begingroup\$

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) < $position) {
                echo '|';
            } else {
                echo '/';
            }
        }
    }
}
\$\endgroup\$
  • \$\begingroup\$ Got a more detailed version? \$\endgroup\$ – Martijn Jul 10 '14 at 14:59
  • 1
    \$\begingroup\$ @Martijn , I edited my post to include one. \$\endgroup\$ – TribalChief Jul 11 '14 at 3:17
  • \$\begingroup\$ Now I can see what it does. Nothing too fancy, but +1 :) \$\endgroup\$ – Martijn Jul 11 '14 at 7:04
  • \$\begingroup\$ Thanks! @NPlay 's solution(s) look fancy though! \$\endgroup\$ – TribalChief Jul 11 '14 at 7:08
  • \$\begingroup\$ a couple of golfing tips: 1) Unnecessary parentheses at ($i+1>$p). 2) Rewriting your ternary expression to $d?($i+1<$p?'|':'/'):$i+1>$p?'|':'\\' saves another 3 bytes. Or just remove ==0 and invert directions. 3) With $i++<$a you can remove $i++ from the post condition and use $i instead of $i+1 (-6 bytes). 4) $i=0 is not necessary; but you´d have to suppress notices (option --n) if you remove it (-4 bytes). \$\endgroup\$ – Titus Oct 12 '18 at 15:57
1
\$\begingroup\$

J, 23 21 19 bytes

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

Try it online!

Input is a list of integers in the standard order.

\$\endgroup\$
1
\$\begingroup\$

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 "|||" → "///|||"
\$\endgroup\$
0
\$\begingroup\$

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;}
\$\endgroup\$
  • 1
    \$\begingroup\$ You don't actually need to explicitly return 0 from main. \$\endgroup\$ – zennehoy Jul 10 '14 at 16:27
  • \$\begingroup\$ It doesn't compile for me because cin and cout are not in the global namespace - what compiler are you using? Also, C(n)>>k>>p would be shorted than C(n)C(k)C(p) wouldn't it? And if the define for P() could stringify the argument wouldn't that save characters for all the quotes? And when you compare p to 'l' and 'r': 0 and 1 would be shorter - specifically >0 instead of =='r' and <1 instead of =='l' (assuming you are ok using numbers instead of r/l - if not <'r' is still shorter than =='l' and >'l' is still shorter than =='r') \$\endgroup\$ – Jerry Jeremiah Aug 4 '14 at 0:48
  • \$\begingroup\$ @JerryJeremiah for cin and cout need "using namespace std". \$\endgroup\$ – bacchusbeale Aug 6 '14 at 1:37
  • \$\begingroup\$ I know but since it is only used twice it is shorter to qualify the functions. Neither way works on my compiler without the include. \$\endgroup\$ – Jerry Jeremiah Aug 6 '14 at 9:21
0
\$\begingroup\$

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); -> ||||||||//
\$\endgroup\$
0
\$\begingroup\$

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 :)

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  • \$\begingroup\$ Might as well modify the function to be an ES6 function, as string repeat is ES6 (doesn;t work in Chrome). \$\endgroup\$ – Matt Jul 11 '14 at 2:03
0
\$\begingroup\$

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)
||||||////
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0
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Clojure, 81 chars

(defn g[a,p,d](apply str(map #(nth "\\|/"(+(if(>= % (- p d)) 1 0) d))(range a))))
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0
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vb.net (~75c)

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