Unzip some numbers

Question

Your task is to create a program or function that generates a zipper with a length of 10 lines. Each line of the zipper is represented by two dashes --:

--
--
--
--
--
--
--
--
--
--

The program/function will take a percentage (divisible by 10) as input, and the output will be the zipper "unzipped" (dashes separated) from the top by the percentage, revealing the 1-indexed, lowest level repeated 2 times, with all prior levels repeated 4, 6, 8, ...etc. times, while keeping the bottom of the zipper centered.

Examples

>>10%
-11-
 --
 --
 --
 --
 --
 --
 --
 --
 --

>>50%
-1111111111-
 -22222222-
  -333333-
   -4444-
    -55-
     --
     --
     --
     --
     --

>>100%
-11111111111111111111-
 -222222222222222222-
  -3333333333333333-
   -44444444444444-
    -555555555555-
     -6666666666-
      -77777777-
       -888888-
        -9999-
         -10-

The input (percentage) can be formatted however you like (50%, .5, 50, 5 [zero implied], etc), and it will always be in the range of 0 to 100 and divisible by 10. The spacing in the examples must be preserved in your output.

Answer 1

Python 2 - 184 151 146 Bytes

def r(n):
  n/=10
  for i in range(1,11):
    if n<0:print"%s--"%p
    elif i>9:print" %s-10-"%p
    else:p=" "*~-i;print"%s-%s-"%(p,`i`*2*n);n-=1

The last number kinda messed with me a little. I might be able to remove the second if statement if I look at it later.

EDIT: Thx to mbomb007 for removing 3 bytes. Thanks to charredgrass for the formatting tips to help remove lots and lots of bytes! :-D Thanks to TheBikingViking for helping with yet another two bytes!

Answer 2

Python 2, 74 bytes

n=input()
x=0
exec"print' '*x+'-'+`x+1`*(n-x<<1-x/9)+'-';x=min(x+1,n);"*10

Saved two bytes by exec-ifying a loop, thanks to Dennis.

EDIT: I took a slightly different approach and saved two more bytes.

Answer 3

PowerShell v2+, 130 120 116 110 bytes

param($n)$i=0;10..1|%{" "*$i+(("-"+("$($i+1)"*([math]::Max($n-10+$_,0))*2)),"-10")[$n-$_-eq9]+"-";$i+=$i-ne$n}

_{Edit 1 - Golfed 10 bytes by eliminating $x variable and slightly redoing how the string is formulated.
Edit 2 - Golfed another 4 bytes by redoing how input happens and by redoing how $i is calculated each loop.
Edit 3 - Saved 6 bytes by OP allowing input as 0..10, so no need to divide by 10.}

Surprisingly difficult!

Takes input as 1,5, etc., stored in $n. Sets helper $i variable (one of the very rare times that a variable needs to be initialized to 0 in PowerShell), and then starts a loop from 10 to 1.

Each iteration, we set start our string with a number of spaces equal to $i, followed by a pseudo-ternary (... , ...)[]. Inside the pseudo-ternary, we select a string of either - with a number of digits (the higher of $n-10+$_ or 0, multiplied by 2), or the string -10 -- the selection is based on whether we're at the 10th iteration and our input was 100. We concatenate that with a final -. That resultant string is placed onto the pipeline.

Finally, we increment $i, and this was really tricky. We wound up using a binary-cast-to-int trick to only increment $i up until it reaches $n, and then keeping it at the same value thereafter. This ensures we've reached the "end" of the zipper indentation at the appropriate level.

Once the loop has finished, the resultant strings are all accumulated on the pipeline and output is implicit.

Examples

PS C:\Tools\Scripts\golfing> .\unzip-some-numbers.ps1 70
-11111111111111-
 -222222222222-
  -3333333333-
   -44444444-
    -555555-
     -6666-
      -77-
       --
       --
       --

PS C:\Tools\Scripts\golfing> .\unzip-some-numbers.ps1 100
-11111111111111111111-
 -222222222222222222-
  -3333333333333333-
   -44444444444444-
    -555555555555-
     -6666666666-
      -77777777-
       -888888-
        -9999-
         -10-

Answer 4

Pyth, 37 34 bytes

=h/QTjm+*;thS,dQj<*`d20eS,0y-Qd"--"ST
=h/QTjm+*;thS,dQj*`d/y-Qdl`d"--"ST

Test suite.

Answer 5

Python, 95 84 bytes

I wasn't aware lambdas were legal, thanks @Dr Green Eggs and Iron Man

lambda p:'\n'.join(' '*min(p,l)+'-'+(2-(l==9))*(p-l)*str(l+1)+'-'for l in range(10))

Answer 6

MATL, 48 41 35 bytes

0:9"@GhX<Z"45h@QVG@-E@8>-OhX>Y"h45h

Try it online!

Answer 7

Ruby, 74 bytes

Uses the implied-zero format specified in the question, so 40% is f[4] if the anonymous function is assigned to f. If a full percentage is needed, +6 bytes for n/=10;

Try it online!

->n{10.times{|i|puts' '*[i,n].min+"-#{i>8?10:(i+1).to_s*2*(n-i)if i<n}-"}}

Answer 8

Python 3, 98, 90, 87, 85 bytes.

Takes the number already divided by 10. I can probably remove some of the parens, but this is pretty closed to fully golfed.

lambda n:'\n'.join(' '*min(i,n)+'-%s-'%(str(i+1)*(n>i>8or(n-i)*2))for i in range(10))

Answer 9

Haskell, 83 bytes

f n=unlines[[[1..y]>>" ",take(2*(n-y))$cycle$show$y+1]>>=(++"-")|y<-min n<$>[0..9]]

Answer 10

Javascript es7, 105 bytes

A=>[...Array(10)].map((a,i)=>' '.repeat((d=A<i)?A:i)+('-'+(i+1+'').repeat(d?0:i<9?2*(A-i):1)+'-')).join`
`

---

call with

f=A=>[...Array(10)].map((a,i)=>' '.repeat((d=A<i)?A:i)+('-'+(i+1+'').repeat(d?0:i<9?2*(A-i):1)+'-')).join`
`


f(10)
f(5)
f(0)

Answer 11

Python 2.7, 113 108 77 bytes

n=input()
for x in range(0,10):
v=str(x+1)*(n-x)2
if xn>89:v='10'
print' '*min(x,n)+'-'+v+'-'

First time playing golf. Gonna keep going, try to get it <100.
Assumes input is 1-10.

Edit: Used some tricks from @LeakyNun 's answer (thanks), then crunched it a bit further and got... basically the same answer :/ Didn't know about the `int` string conversion, and 2 - boolean let me get rid of the if statement, which let me get rid of the whole v variable. Very cool.

My version:

n=input()
for x in range(10):print' '*min(x,n)+'-'+`x+1`*(n-x)*(2-(x>8))+'-'

Answer 12

Python 2, 81 bytes

n=input()/10
for i in range(10):print(' '*min(i,n)+'-'+2*(n-i)/-~(i>8)*`i+1`+'-')

Ideone it!

Answer 13

Python 2.7, 110 99 95 91 bytes:

G=input();i=1;exec"print' '*[~-i,G][i>G]+'-%s-'%[`i`*(2*-~G-i*2),'10'][(i>9)*G>9];i+=1;"*10

A full program that takes input by an integer in the inclusive range [1,10], where 10 means 100% and 1 means 10%. Can probably be golfed down a bit more.

Try It Online! (Ideone)

Answer 14

PHP 5.3, 92 91 bytes

<?for(;$i<10;$i++)echo'
'.str_pad('-'.str_repeat($i+1,$n>9&&$i>8?:($n-$i)*2).'-',22,' ',2);

• for PHP 5.3 with register_globals=1 and short_open_tags=1 (and error_reporting=0)
  call in cli with php-cgi -f <filename> n=<number>
• number from 0 to 10

• to call in web browser with <scriptpath>?n=<number>: prepend <pre>
• for 4.0.1 < PHP < 5.3: replace ?: with ?1: (+1)
• for PHP>=5.4: replace the first $n with ($n=$_GET[n]) (+11)

ungolfed as a function (any PHP>=4.0.1)

function unzip($n) // $n from 0 to 10
{
    for($i=0;$i<10;$i++)                // $i = line number -1
        $s.='
'.str_pad(                              // pad the result of the following on both sides
            '-'.                            // prepend '-'
            str_repeat($i+1,                // print line number
                ($n>9&&$i>8 ? 1             // in tenth line, if $n is 10: once
                : ($n-$i)*2)                    // else  2*($n-$i) times
            ).'-'                           // append '-'
            , 22,' ', STR_PAD_BOTH);    // pad to 22 width with SPC on both sides
    return $s;
}

test suite

echo'<table border=1><tr>';
for($i=0;$i<11;$i++)echo'<th>',$i*10,'%</th>';
echo'</tr><tr>';
for($i=0;$i<11;$i++)echo'<td><pre>', unzip($i), '</pre></td>';
echo '</table>';

^{Now this is new to me: PHP beats JavaScript.}
This approach is golfed to the min, I think.

Answer 15

Julia, 73 bytes

x->join([" "^min(x,a)*"-$("$(a+1)"^(x>a?a>8?1:2(x-a):0))-"for a=0:9],'
')

First Julia answer! Tips are appreciated.

Try it online!

Answer 16

Perl, 122 bytes

$k=<>;for(my $i=0;$i++<10;){$s="-"."$i"x($k/10-$i+1);$p=length $s;$l=$l>$p?$l:$p;printf "%${l}s%s\n",$s,scalar reverse $s}

Answer 17

Common Lisp (Lispworks), 314 bytes

(defun f()(let((s(string(read))))(let((n(/(parse-integer(subseq s 0(1-(length s))))10)))(if(> n 0)(progn(dotimes(r n)(progn(dotimes(c r)#1=(format t" "))(format t"-")(if(=(1+ r)10)(format t"10")(dotimes(j(* 2(- n r)))(format t"~S"(1+ r))))(format t"-~%")))(dotimes(d(- 10 n))(dotimes(c n)#1#)(format t"--~%")))))))

ungolded:

    (defun f ()
      (let ((s (string (read))))
        (let ((n (/ (parse-integer (subseq s 0 (1- (length s)))) 10)))
          (if (> n 0)
              (progn
                (dotimes (r n)
                  (progn
                    (dotimes (c r)
                      (format t " "))
                    (format t "-")
                    (if (= (1+ r) 10)
                        (format t "10")
                      (dotimes (j (* 2 (- n r)))
                        (format t "~S" (1+ r))))
                    (format t "-~%")))
                (dotimes (d (- 10 n))
                  (dotimes (c n)
                    (format t " "))
                  (format t "--~%")))))))

Usage:

    CL-USER 2515 > (f)
    10%
    -11-
     --
     --
     --
     --
     --
     --
     --
     --
     --
    NIL

    CL-USER 2516 > (f)
    50%
    -1111111111-
     -22222222-
      -333333-
       -4444-
        -55-
         --
         --
         --
         --
         --
    NIL

    CL-USER 2517 > (f)
    100%
    -11111111111111111111-
     -222222222222222222-
      -3333333333333333-
       -44444444444444-
        -555555555555-
         -6666666666-
          -77777777-
           -888888-
            -9999-
             -10-
    NIL

Answer 18

APL, 46 bytes

{↑(⍳10){(''↑⍨1-⍺⌊a+1),1⌽'--',⍵⍴⍕⍺}¨10↑2×⌽⍳a←⍵}

The argument should be given as the percentage divided by 10 (that is: a simple integer, in the range [0,10]).