Create a binary ruler

Question

Given a number n, generate the first n columns of this pattern:

                               #
               #               #
       #       #       #       #
   #   #   #   #   #   #   #   #
 # # # # # # # # # # # # # # # #
################################ ...

The height of the column at (1-indexed) n is the number of trailing 0 bits in n's binary representation, plus one. As a result, the bottom layer has every column filled in, the second layer every second column, the third layer every fourth column, etc.

Rules

• You may input and output through any standard method.
• You may assume the input is an integer between 1 and 999, inclusive.
• The output may contain any amount of whitespace, as long as the pattern is intact.
• The pattern must be 1-indexed, and in the same format as shown here.
• You may use any single non-whitespace character in place of #, but you may not change the space character.

Test cases

1
#

2
 #
##

3
 # 
###

4
   #
 # #
####

5
   # 
 # # 
#####

7
   #   
 # # # 
#######

32
                               #
               #               #
       #       #       #       #
   #   #   #   #   #   #   #   #
 # # # # # # # # # # # # # # # #
################################

A few larger test cases can be found here.

Scoring

This is code-golf, so the shortest code in bytes in each language wins.

Answer 1

Python 2, 54 bytes

i=n=input()
while i:i-=1;print((' '*~-2**i+'#')*n)[:n]

Try it online!

Prints with lots of leading whitespace. Each row i counting down from n repeats a pattern of 2**i-1 spaces followed by a #. This pattern is repeated up to the width of the ruler, which is the input n. This is done by multiplying the pattern string by n and taking the first n characters with [:n].

The pattern can be made by string formatting for an equal length alternative.

i=n=input()
while i:i-=1;print('%%%ds'%2**i%'#'*n)[:n]

A cute slicing method is longer.

n=input();s=~-2**n*' '+'#'
exec"s=s[1::2]*2;print s[:n];"*n

Answer 2

Python 3, 74 bytes

n=int(input())
a=1
while a<n:a*=2
while a:print(("%%%dd"%a%4*n)[:n]);a//=2

Try it online!

Answer 3

V, 17, 16 bytes

é#Àñä}Är {ñÎÀlD

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Hexdump:

00000000: e923 c0f1 e416 7dc4 7220 7bf1 cec0 6c44  .#....}.r {...lD

Thanks to @KritixiLithos for saving one byte!

This algorithm is horribly inefficient, but it should work in theory for any size input.

It works by generating the first n iterations of the following pattern:

#

 #
##

   #
 # #
####

       #
   #   #
 # # # #
########

               #
       #       #
   #   #   #   #
 # # # # # # # #
################

                               #
               #               #
       #       #       #       #
   #   #   #   #   #   #   #   #
 # # # # # # # # # # # # # # # #
################################

And then chopping off all but the first n columns. As such, this will produce a ton of leading whitespace, but the OP said:

The output may contain any amount of whitespace, as long as the pattern is intact

Explanation:

é#                      " Insert an '#'
  Àñ           ñ        " 'N' times:
    ä<C-v>}             "   Duplicate every line blockwise (duplicating horizontally instead of vertically)
           Ä            "   Duplicate the top line. This conveniently puts us on the first non-whitespace character (that is, '#')
            r           "   Replace this character with a space
              {         "   Move to the beginning of the buffer
                Î       " On every line:
                 Àl     "   Move 'N' characters to the right ('l' for right, makes sense, right?)
                   D    "   And delete everything after the cursor

Answer 4

JavaScript (ES6), 61 58 bytes

f=(n,c=n,s='')=>c?f(n,c>>1,s+s+' ')+`
`+(s+1).repeat(c):''

Saved 1 byte thanks to @ETHProductions, then 2 more bytes when I saw that any character could be used.

A recursive solution.

Test cases:

f=(n,c=n,s='')=>c?f(n,c>>1,s+s+' ')+`
`+(s+1).repeat(c):''

console.log(f(32));
console.log(f(1));  
console.log(f(5));  
console.log(f(7));  

Animation:

f=(n,c=n,s='')=>c?f(n,c>>1,s+s+' ')+`
`+(s+1).repeat(c):''

for(let i = 1 ; i < 64 ; i++) {
  setTimeout(_=>document.querySelector('pre').textContent=f(i), i*40);
}

<pre>

Answer 5

APL (Dyalog), 21 bytes

'# '[1+⊖0⍪∨⍀⊖2⊥⍣¯1⍳⎕]

Try it online!

'# '[…`] index the string with

 ⎕ get input

 ⍳ that many integers

 2⊥⍣¯1 convert to binary, using as many digits as needed (one number in each column)

 ⊖ flip upside down

 ∨⍀ vertical cumulative OR reduction

 0⍪ concatenate zeros on top

 ⊖ flip upside down (i.e. back up again)

 1+ add one (for 1-based indexing)

Answer 6

Jelly, 11 10 bytes

Rọ2‘4ẋz⁶ṚY

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1 byte saved after the OP added a relaxation that the character doesn't have to be #.

Answer 7

Octave, 45 bytes

@(n)[[sort(cummin(de2bi(g=0:n)'));g|1]+32 '']

Try it on Octave Online!

Instead of '#' prints '!'.

Answer 8

PHP, 139 bytes

for($f=array_fill(0,$l=1+log($a=$argn,2)," ");$n++<$a;)for($h=1+strspn(strrev(decbin($n)),$i=0);$i<$h;)$f[$l-++$i][$n]=3;echo join("
",$f);

Try it online!

Answer 9

Japt, 20 17 bytes

Saved 3 bytes thanks to @Shaggy and @ETHproductions

õ_¤q1 o Ä ço÷z w

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Explanation:

Input: 5

õ_¤q1 o Ä ço÷z w
õ           Ã       // Create a range [1...Input] [1,2,3,4,5]
 _                  // Map; At each item:
  ¤                 //   Convert to binary        ["1","10","11","100","101"]
   q1               //   Split on "1"             [["",""],["","0"],["","",""],["","00"],["","0",""]]
      o             //   Get the last item        ["","0","","00",""]
        Ä           //   Add 1                    [["1","01","1","001","1"]]
          ço        //   Fill with "o"            ["o","oo","o","ooo","o"]
             ·      // Join with new-lines        ["o\noo\no\nooo\no"]
              z     // Rotate 90 degrees          ["ooooo\n o o \n o   "]
                w   // Reverse                    ["   o \n o o \nooooo"]

Answer 10

C, 84 74 bytes

f(i,l,m){putchar(32+3*!(i&m));i<l?f(i+1,l,m):m?putchar(10),f(1,l,m>>1):1;}

Ungolfed:

void f(int counter, int length, int mask) {
    putchar((counter&mask) ? ' ' : '#');
    if(counter<length) {
        f(counter+1, length, mask);
    } else if(mask) {
        putchar('\n');
        f(1, length, mask>>1);
    }
}

Test with:

int main() {
    f(1, 32, 1023);
    putchar('\n');
    f(1, 1, 1023);
    putchar('\n');
    f(1, 999, 1023);
    putchar('\n');
}

Explanation

Once again, recursion takes less characters in C than iteration, so the two loops are expressed as the two recursional invocations.

Also, C is a great language for playing tricks with boolean expressions, allowing the decision of whether to put a blank or a # to be expressed by the expression 32+3*!(i&m). A space has the ASCII value of 32, the # is ASCII 35, so we get a blank if any of the bits in the mask is set in i.

Answer 11

Pyth, 15 Bytes

j_.tm*Nhx_.Bd1S

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explanation

j_.tm*Nhx_.Bd1S
    m         SQ   # map over the numbers from 0 to the implicit input (lambda variable: d)
          .Bd      # Convert d to a binary string: (12 -> 1100)
         _         # reverse: (1100 -> 0011)
        x    1     # get the location of the first 1 ( 2 )
     *Nh           # make one more than that " quotation marks (""")
 _.t               # transpose the list of quotation mark strings and reverse it
j                  # join on newline
    

Answer 12

Python 2, 47 bytes

f=lambda n:n and' '.join(f(n/2))+'\n'+n*'#'or''

Try it online!

Answer 13

JavaScript (ES8), 71 bytes

The padStart() function was introduced in ECMAScript 2017!

N=>eval(`for(s='',n=1;n<=N;n*=2)s='\\n'+'#'.padStart(n).repeat(N/n)+s`)

f=
N=>eval(`for(s='',n=1;n<=N;n*=2)s='\\n'+'#'.padStart(n).repeat(N/n)+s`)

console.log(f(+prompt()))

---

JavaScript (ES6), 77 bytes

N=>eval(`for(s='',n=1;n<=N;n*=2)s='\\n'+(' '.repeat(n-1)+'#').repeat(N/n)+s`)

f=
N=>eval(`for(s='',n=1;n<=N;n*=2)s='\\n'+(' '.repeat(n-1)+'#').repeat(N/n)+s`)

console.log(f(+prompt()))

Answer 14

Mathematica, 69 bytes

Rotate[Grid["#"~Table~#&/@(IntegerExponent[2*#,2]&/@Range[#])],Pi/2]&

Answer 15

(WESRRMICGSE): 237 bytes

IF(ROW()<=FLOOR(LOG(COUNTA(R1C:R[-1]C,R[1]C:R[1024]C)+1,2),1)+2,LEFT(REPT(REPT(" ",FLOOR(POWER(2,LOG(COUNTA(R1C:R[-1]C,R[1]C:R[1024]C)+1,2)-ROW()+2),1)-1) &"#",COUNTA(R1C:R[-1]C,R[1]C:R[1024]C)+1),COUNTA(R1C:R[-1]C,R[1]C:R[1024]C)+1),"")

Alright. 'splaining time.

First off, replace every COUNTA(R1C:R[-1]C,R[1]C:R[1024]C)+1 with simply [i], for input. the token counts the number of cells, not including itself, that contain a formula, and then adds one, to include itself. Since WESRRMICGSE drags a formula according to the input you give it, this token always results in the input.

we have:

IF(ROW()<=FLOOR(LOG([i],2),1)+3,LEFT(REPT(REPT(" ",FLOOR(POWER(2,LOG([i],2)-ROW()+2),1)-1) &"#",[i]),[i]),"")

This is much more readable. You're going to see the FLOOR(LOG([i],2),1) token a lot, which simply means take the nearest power of 2 which is less than the input ([i]) number. eg: 4->4, 5->4, 6->4, 7->4, 8->8 ...etc. I'll replace that with GS[[i]]

IF(ROW()<=GS[[i]]+3,LEFT(REPT(REPT(" ",,FLOOR(POWER(2,LOG([i],2)-ROW()+2),1),1)-1) &"#",[i]),[i]),"")

better. breaking down the if clause, we're testing if the row is less than or equal to GS[[i]]+3, because all rulers' height is equal to the GS[[i]]+1, this selects the rows which are equal to the height of the ruler. +1 for 1-indexing rows, and +1 again for WESRRMICGSE offset.

The FALSE result yields an empty cell (""), and a true result yields LEFT(REPT(REPT(" ",,FLOOR(POWER(2,LOG([i],2)-ROW()+2),1),1)-1) &"#",[i]),[i])

currently still editing, stay tuned

Answer 16

Haskell, 64 62 bytes

l m=([2..2^m]>>" ")++'#':l m
f n=unlines$take n.l<$>[n,n-1..0]

Try it online! Example usage: f 10.

Answer 17

k, 33 bytes

`0:|" #"{(1+!x){~y!x}/:(x>)(2*)\1}

This only seems to work in AW's interpreter.

The oK version (which you can try online) seems to have a bug, requiring a slight change to make it work:

`0:|" #"{(1+!x){~y!x}/:{x>y}[x](2*)\1}

Answer 18

C#, 174 bytes

This method has two parameters, an input for the ruler length, and an output which is the ruler as string.

Golfed:

void R(int n,out string s){var l=new int[++n];int i,x=n,y=0;for(s="";x-->1;)for(i=0;0==(l[x]=(x>>i++&1)*i);y=y<i?i:y);for(y++;y-->0;s+='\n')for(x=0;++x<n;s+=y<l[x]?'#':' ');}

Indented:

void R(int n,out string s){                       // Return the result in an out parameter.
    var l=new int[++n];                           // Use a 1-based array.
    int i,x=n,y=0;                                //
    for(s="";x-->1;)                              // For each number x on the ruler
        for(i=0;0==(l[x]=(x>>i++&1)*i);y=y<i?i:y) // ... find lowest set bit of x, counting the maximum value.
            ;                                     //
    for(y++;y-->0;s+='\n')                        // Count down each line.
        for(x=0;++x<n;s+=y<l[x]?'#':' ')          // Output # for numbers that are tall enough.
            ;                                     //
}

Try it online!

Answer 19

Charcoal, 27 23 bytes

↶Ｆ…·¹Ｎ«Ｊι⁰#Ｗ¬﹪鲫Ａ÷ι²ι#

Try it online! Link is to verbose version of code. Edit: Saved 4 bytes by switching to JumpTo.

Answer 20

J, 38 bytes

3 :'|.|:''#''#~,.(1+|.i.1:)@#:"0>:i.y'

Not great. Lmk if the byte count is off -- I'm on my phone.

Answer 21

Husk, 13 bytes

↔T' ↑moR'#→İr

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I saw a ruler sequence builtin(İr), so I just went ahead and found a challenge I could cheat on.

Answer 22

Java (OpenJDK 8), 91 bytes

n->{int i=10,j;String s="";for(;i-->0;s+="\n")for(j=0;j++<n;)s+=j>>i<<i<j?' ':35;return s;}

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Ungolfed:

n->{
    int i=10,j; // Since we are allowed extra whitespace, set columns always to 10
    String s = "";
    for(;i-->0;s+="\n")      // Every iteration add a newline, i=9..0
        for(j=0;j++<n;)      // j=1..n+1
            s+= j>>i<<i<j    // if j has less than i trailing 0s in binary form
                ?' '         // add a space else
                :35          // # (java handles ternary return types weirdly)
}

Answer 23

CJam, 34 bytes

ri{2bW%0+0#)}%_'#f*\:e>f{Se]}zW%N*

Meh.

Answer 24

C (gcc), 70 bytes

f(x,l,n){for(l=512;l/=2;)for(n=0;n++<=x;)putchar(n^x+1?n%l?32:35:10);}

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Answer 25

Perl 5, 60 + 1 (-n) = 61 bytes

$n=0|1+(log)/log 2;say(($"x(2**$n-1),X)x($_/2**$n))while$n--

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Answer 26

Japt -R, 11 bytes

õ_&n)¤çTÃz3

Try it

õ_&n)¤çTÃz3     :Implicit input of integer U
õ               :Range [1,U]
 _              :Map each Z
  &n            :  Bitwise AND with -Z
    )           :  Group that operation
     ¤          :  Result to binary string
      ç         :  Fill with
       T        :    0
        Ã       :End map
         z3     :Rotate 90° clockwise 3 times
                :Implicit output joined with newlines