Count up diagonally!

Question

We have lots of horizontal axis for numbers, but I honestly think they're kind of boring. Your task today is to build me a portion of a diagonal axis between two distinct non-negative integers given as input.

How to build a diagonal axis?

• Let's take an example, with the input 0, 5. Our axis should look like this:

  0
   1
    2
     3
      4
       5
  

• However, our axis should look nice for numbers that have more digits too! If the input is, for instance 0, 14, the new axis should be:

  0
   1
    2
     3
      4
       5
        6
         7
          8
           9
            10
              11
                12
                  13
                    14
  

• The idea is that the first digit of next number on the axis must always be placed exactly after the last digit of the previous number. To understand the idea even better, here is another example with 997, 1004:

  997
     998
        999
           1000
               1001
                   1002
                       1003
                           1004
  

---

Rules

• You may assume that input is in ascending or descending order (you may choose between 5,3 and 3,5).

• You may also assume that the difference between the two integers is lower than 100.

• You may have a leading newline or a consistent leading space (on each line). Trailing spaces / newlines are fine as well.

• Default Loopholes are forbidden.

• You can take input and provide output by any standard mean.

• This is code-golf, so the shortest code in bytes in every language wins!

---

Other Test Cases

• 1, 10:

  1
   2
    3
     4
      5
       6
        7
         8
          9
           10
  

• 95, 103:

  95
    96
      97
        98
          99
            100
               101
                  102
                     103
  

• 999999, 1000009:

  999999
        1000000
               1000001
                      1000002
                             1000003
                                    1000004
                                           1000005
                                                  1000006
                                                         1000007
                                                                1000008
                                                                       1000009
  

Answer 1

05AB1E, 8 7 6 bytes

Thanks to Magic Octopus Urn for saving a byte!

It somehow works, but honestly I have no idea why.

Code

Ÿvy.O=

Uses the 05AB1E encoding. Try it online!

Explanation

Ÿ          # Create the range [a, .., b] from the input array
 vy        # For each element
   .O      #   Push the connected overlapped version of that string using the
                 previous version of that string. The previous version initially
                 is the input repeated again. Somehow, when the input array is
                 repeated again, this command sees it as 1 character, which gives
                 the leading space before each line outputted. After the first
                 iteration, it reuses on what is left on the stack from the
                 previous iteration and basically attaches (or overlaps) itself 
                 onto the previous string, whereas the previous string is replaced 
                 by spaces and merged into the initial string. The previous string
                 is then discarded. We do not have to worry about numbers overlapping 
                 other numbers, since the incremented version of a number never
                 overlaps entirely on the previous number. An example of 123 and 456:

                 123
                    456

                 Which leaves us "   456" on the stack.
     =     #   Print with a newline without popping

Answer 2

Python 2, 43 bytes

lambda a,b:'\v'.join(map(str,range(a,b+1)))

Makes use of vertical tab to make the ladder effect. The way thet \v is rendered is console dependent, so it may not work everywhere (like TIO).

Answer 3

Charcoal, 9 8 bytes

Ｆ…·ＮＮ⁺¶ι

Try it online!

Link is to the verbose version of the code. Input in ascending order.

• 1 byte saved thanks to ASCII-only!

Answer 4

R, 70 69 61 bytes

function(a,b)for(i in a:b){cat(rep('',F),i,'
');F=F+nchar(i)}

Function that takes the start and end variable as arguments. Loops over the sequence, and prints each element, prepended with enough spaces. F starts as FALSE=0, and during each iteration, the amount of characters for that value is added to it. F decides the amount of spaces printed.

Try it online!

-8 bytes thanks to @Giuseppe

Answer 5

Python 2, 58 54 bytes

def f(a,b,s=''):print s;b<a or f(a+1,b,' '*len(s)+`a`)

Try it online!

Answer 6

C#, 90 89 85 bytes

s=>e=>{var r="";for(int g=0;e>s;g+=(s+++"").Length)r+="".PadLeft(g)+s+"\n";return r;}

Saved 1 byte thanks to @LiefdeWen.
Saved 4 bytes thanks to @auhmaan.

Try it online!

Full/Formatted version:

namespace System
{
    class P
    {
        static void Main()
        {
            Func<int, Func<int, string>> f = s => e =>
            {
                var r = "";
                for (int g = 0; e > s; g += (s++ + "").Length)
                    r += "".PadLeft(g) + s + "\n";

                return r;
            };

            Console.WriteLine(f(0)(5));
            Console.WriteLine(f(0)(14));
            Console.WriteLine(f(997)(1004));
            Console.WriteLine(f(1)(10));
            Console.WriteLine(f(95)(103));
            Console.WriteLine(f(999999)(1000009));

            Console.ReadLine();
        }
    }
}

Answer 7

Mathematica, 59, bytes

Grid[(DiagonalMatrix@Range[1+##]/. 0->""+1)-1,Spacings->0]&

input

[10,15]

-3 bytes @JungHwanMin
problem with 0 fixed (see comments for details)
thanx to @ngenisis

Answer 8

Jelly, 9 bytes

rD⁶ṁ$;¥\Y

Try it online!

Answer 9

C, 166 134 95 82 Bytes

New Answer

Just as a function not as a whole program.

f(a,b){int s=0,i;while(a<=b){i=s;while(i--)printf(" ");s+=printf("%i\n",a++)-1;}}

Thanks to Falken for helping knock off 13 Bytes (and fix a glitch)!

Thanks to Steph Hen for helping knock off 12 Bytes!

Thanks to Zacharý for help knock off 1 Byte!

Old Answers

Got rid of the int before main and changed const char*v[] to char**v and got rid of return 0;

main(int c,char**v){int s=0;for(int a=atoi(v[1]);a<=atoi(v[2]);a++){for(int i=0;i<s;i++)printf(" ");printf("%i\n",a);s+=log10(a)+1;}}


int main(int c,const char*v[]){int s=0;for(int a=atoi(v[1]);a<=atoi(v[2]);a++){for(int i=0;i<s;i++)printf(" ");printf("%i\n",a);s+=log10(a)+1;}return 0;}

This is my first time golfing and I wanted to try something in C. Not sure if I formatted this correctly, but I had fun making it!

int main(int c, const char * v[]) {
    int s = 0;
    for(int a=atoi(v[1]); a<=atoi(v[2]); a++) {
        for(int i=0; i<s; i++) printf(" ");
        printf("%i\n",a);
        s += log10(a)+1;
    }
    return 0;
}

Explanation

int s = 0; // Number of spaces for each line

for(int a=atoi(argv[1]); a<=atoi(argv[2]); a++) { // Loop thru numbers

for(int i=0; i<s; i++) printf(" "); // Add leading spaces

printf("%i\n",a); // Print number

s += log10(a)+1; // Update leading spaces

Usage

Answer 10

Mathematica, 48 bytes

Rotate[""<>Table[ToString@i<>" ",{i,##}],-Pi/4]&

since there are so many answers, I thought this one should be included

input

[0,10]

output

Answer 11

Retina, 81 78 bytes

.+
$*
+`\b(1+)¶11\1
$1¶1$&
1+
$.& $.&
 (.+)
$.1$* 
+1`( *)(.+?)( +)¶
$1$2¶$1$3

Try it online! Takes input as a newline-separated list of two integers. Edit: Saved 3 bytes by stealing the range-expansion code from my answer to Do we share the prime cluster? Explanation:

.+
$*

Convert both inputs to unary.

+`\b(1+)¶11\1
$1¶1$&

While the last two elements (a, b) of the list differ by more than 1, replace them with (a, a+1, b). This expands the list from a tuple into a range.

1+
$.& $.&

Convert back to decimal in duplicate.

 (.+)
$.1$* 

Convert the duplicate copy to spaces.

+1`( *)(.+?)( +)¶
$1$2¶$1$3

Cumulatively sum the spaces from each line to the next.

Answer 12

APL (Dyalog), 25 24 bytes

-1 thanks to Zacharý.

Assumes ⎕IO←0 for zero based counting. Takes the lower bound as left argument and the upper bound as right argument.

{↑⍵↑⍨¨-+\≢¨⍵}(⍕¨⊣+∘⍳1--)

Try it online!

(…) apply the following tacit function between the arguments:

 - subtract the upper lower from the upper bound

 1- subtract that from one (i.e. 1 + ∆)

 ⊣+∘⍳ left lower bound plus the integers 0 through that

 ⍕¨ format (stringify) each

{…} apply the following anonymous on that (represented by ⍵):

 ≢¨ length of each (number)

 +\ cumulative sum

 - negate

 ⍵↑⍨¨ for each stringified number, take that many characters from the end (pads with spaces)

 ↑ mix list of strings into character matrix

Answer 13

C++, 167 165 bytes

-2 bytes thanks to Zacharý

#include<string>
#define S std::string
S d(int l,int h){S r;for(int m=0,i=l,j;i<=h;){for(j=0;j<m;++j)r+=32;S t=std::to_string(i++);r+=t;r+=10;m+=t.size();}return r;}

Answer 14

Python 2, 65 63 62 61 bytes

-2 bytes Thanks to @Mr. Xcoder: exec doesn't need braces

-1 bye thanks to @Zacharý: print s*' ' as print' '*s

def f(m,n,s=0):exec(n-m+1)*"print' '*s+`m`;s+=len(`m`);m+=1;"

Try it online!

Answer 15

Pyth, 14 13 bytes

V}FQ
=k+*dlkN

Try it online!

Answer 16

LOGO, 53 bytes

[for[i ? ?2][repeat ycor[type "\ ]pr :i fd count :i]]

There is no "Try it online!" link because all online LOGO interpreter does not support template-list.

That is a template-list (equivalent of lambda function in other languages).

Usage:

apply [for[i ? ?2][repeat ycor[type "\ ]pr :i fd count :i]] [997 1004]

(apply calls the function)

will print

997
   998
      999
         1000
             1001
                 1002
                     1003
                         1004

Note:

This uses turtle's ycor (Y-coordinate) to store the number of spaces needed to type, therefore:

• The turtle need to be set to home in its default position and heading (upwards) before each invocation.
• window should be executed if ycor gets too large that the turtle moves off the screen. Description of window command: if the turtle is asked to move past the boundary of the graphics window, it will move off screen., unlike the default setting wrap, which if the turtle is asked to move past the boundary of the FMSLogo screen window, it will "wrap around" and reappear at the opposite edge of the window.

Explanation:

for[i ? ?2]        Loop variable i in range [?, ?2], which is 2 input values
repeat ycor        That number of times
type "\            space character need to be escaped to be typed out.
pr :i              print the value of :i with a newline
fd count :i        increase turtle's y-coordinate by the length of the word :i. (Numbers in LOGO are stored as words)

Answer 17

Japt, 12 bytes

òV
£¯Y ¬ç +X

Takes input in either order and always returns the numbers in ascending order, as an array of lines.

Try it online! with the -R flag to join the array with newlines.

Explanation

Implicit input of U and V.

òV
£

Create inclusive range [U, V] and map each value to...

¯Y ¬ç

The values before the current (¯Y), joined to a string (¬) and filled with spaces (ç).

+X

Plus the current number. Resulting array is implicitly output.

Answer 18

JavaScript (ES8), 69 67 62 bytes

Takes input as integers, in ascending order, using currying syntax. Returns an array of strings.

x=>y=>[...Array(++y-x)].map(_=>s="".padEnd(s.length)+x++,s="")

---

Try it

o.innerText=(f=

x=>y=>[...Array(++y-x)].map(_=>s="".padEnd(s.length)+x++,s="")

)(i.value=93)(j.value=105).join`\n`
oninput=_=>o.innerText=f(Math.min(i.value,j.value))(Math.max(i.value,j.value)).join`\n`

label,input{font-family:sans-serif}input{margin:0 5px 0 0;width:100px;}

<label for=i>x: </label><input id=i type=number><label for=j>y: </label><input id=j type=number><pre id=o>

Answer 19

05AB1E, 8 bytes

Ÿʒ¾ú=þv¼

Try it online!

-2 thanks to Adnan.

Answer 20

C (gcc), 41 38 bytes

-3 bytes Thanks to ASCII-only

t(x,v){while(x<=v)printf("%d\v",x++);}

Works on RedHat6, accessed via PuTTY

Try it online!

Answer 21

Python 2, 78 77 79 bytes

def f(a,b):
 for i in range(a,b+1):print sum(len(`j`)for j in range(i))*' '+`i`

Try it online!

f(A, B) will print the portion of the axis between A and B inclusive.

First time I answer a challenge!

Uses and abuses Python 2's backticks to count the number of spaces it has to add before the number.

-1 byte thanks to Mr.Xcoder

+2 because I forgot a +1

Answer 22

JavaScript, 57 bytes

f=(x,y,s='')=>y>=x?s+`
`+f(x+1,y,s.replace(/./g,' ')+x):s

Answer 23

Python 2, 60 59 bytes

-1 byte thanks to Mr.Xcoder for defining my s=0 as an optional variable in my function.

def f(l,u,s=0):
 while l<=u:print' '*s+`l`;s+=len(`l`);l+=1

Try it online!

I think it is possible to transfer this into a lambda version, but I do not know how. I also think that there is some sort of mapping between the spaces and the length of the current number, but this I also did not figure out yet. So I think there still is room for improvement.

What i did was creating a range from the lowerbound lto the upper bound u printing each line with a space multiplied with a number s. I am increasing the multiplier with the length of the current number.

Answer 24

MATL, 11 bytes

vii&:"t~@Vh

Try it online!

Explanation

This works by generating a string for each number and concatenating it with a logically-negated copy of the previous string. Thus char 0 is prepended 0 as many times as the length of the previous string. Char 0 is displayed as a space, and each string is displayed on a different line

v       % Concatenate stack (which is empty): pushes []
ii      % Input two numbers
&:      % Range between the two numbers
"       % For each
  t     %   Duplicate
  ~     %   Logical negation. This gives a vector of zeros
  @     %   Push current number
  V     %   Convert to string
  h     %   Concatenate with the vector of zeros, which gets automatically 
        %   converted into chars.
        % End (implicit). Display stack (implicit), each string on a diferent
        % line, char 0 shown as space

Answer 25

D, 133 127 126 125 121 119 bytes

import std.conv,std.stdio;void f(T)(T a,T b,T s=0){for(T j;j++<s;)' '.write;a.writeln;if(a-b)f(a+1,b,s+a.text.length);}

Jelly and APL were taken.

Try it online!

If you're fine with console-dependent results (goes off the same principle as Giacomos's C answer) here's one for 72 71 bytes:

import std.stdio;void f(T)(T a,T b){while(a<=b){a++.write;'\v'.write;}}

How? (Only D specific tricks)

• f(T)(T a,T b,T s=0) D's template system can infer types
• for(T j;j++<s;) Integers default to 0.
• ' '.write;a.writeln D lets you call fun(arg) like arg.fun (one of the few golfy things D has)
• a.text.length Same as above, and D also allows you to call a method with no parameters as if it was a property (text is conversion to string)
• One thing that might be relevant (I didn't use this though) newlines can be in strings!

Answer 26

V, 16 bytes

ÀñÙywÒ $pça/jd

Try it online!

This would be way easier if I could take start end - start but I think that's changing the challenge a bit too much.

This takes the start number as input in the buffer and the end number as an argument. It actually creates the ladder from start to start + end and then deletes everything after the end number.

Answer 27

Perl, 19 bytes

Note: \x0b is counted as one byte.

Along with others, I thought using cursor movements would be the shortest route, this does mean it doesn't work on TIO:

print"$_\x0b"for<>..<>

Usage

perl -e 'print"$_\x0b"for<>..<>' <<< '5
10'
5
 6
  7
   8
    9
     10

Answer 28

Swift 4, 115 bytes

I think nobody would have posted a Swift solution anyway...

func f(l:Int,b:Int){for i in l...b{print(String(repeating:" ",count:(l..<i).map{String($0).count}.reduce(0,+)),i)}}

Try it online!

Answer 29

Japt, 10 9 bytes

òV åÈç +Y

Test it online! Returns an array of lines; -R flag included to join on newlines for easier viewing.

Explanation

 òV åÈ   ç +Y
UòV åXY{Xç +Y}   Ungolfed
                 Implicit: U, V = inputs, P = empty string
UòV              Create the range [U, U+1, ..., V-1, V].
    åXY{     }   Cumulative reduce: Map each previous result X and current item Y to:
        Xç         Fill X with spaces.
           +Y      Append Y.
                 Implicit: output result of last expression

---

Old version, 10 bytes:

òV £P=ç +X

Test it online!

 òV £  P= ç +X
UòV mX{P=Pç +X}  Ungolfed
                 Implicit: U, V = inputs, P = empty string
UòV              Create the range [U, U+1, ..., V-1, V].
    mX{       }  Map each item X to:
         Pç        Fill P with spaces.
            +X     Append X.
       P=          Re-set P to the result.
                   Implicitly return the same.
                 Implicit: output result of last expression

Answer 30

Java 8, 79 bytes

(a,b)->{for(String s="";a<=b;System.out.printf("%"+s.length()+"d\n",a++))s+=a;}

Try it online!

Answer 31

Haskell, 39 bytes

a%b=scanl((.show).(++).(' '<$))""[a..b]

Try it online!

Answer 32

QBIC, 25 bytes

[:,:|A=space$(_lA|)+!a$?A

Explanation:

[:     FOR a = <input 1 from cmd line>
,:|    TO <input 2 from cmd line>
A=         SET A$ to 
 space$      a number of spaces equal to 
 (_lA|)      the current length of A$ (starts as "" = 0)
+!a$         and add a cast-to string of the current loop iterator
?A     PRINT A$
           Loop auto-closed by QBIC

Answer 33

Ruby, 18 bytes

->a,b{[*a..b]*?\v}

Just for reference, but does not work on TIO:

Try it online!

Answer 34

PowerShell, 49 bytes

$a,$b=$args;$a..$b|%{$l=($z=" "*$l+$_).length;$z}

Try it online!

Takes input as two command-line arguments, $a and $b, forms a range .. out of them, then loops |%{} over that range. Each iteration, we construct a string $z of a certain number of spaces " "*$l concatenated with our current number $_. We then pull out the .length of that, re-store it into $l for next time, and leave $z on the pipeline. Output is implicit.

Answer 35

Bash + coreutils, 10

seq -s^K $@

Here ^K is a literal vertical tab and control character. The xxd dump of this script is as follows - use xxd -r to regenerate the actual script:

00000000: 7365 7120 2d73 0b20 2440                 seq -s. $@

Input range is given as two integers at the command-line.

Not sure if much of an explanation is needed here - seq simply does the counting (from a to b, passed in the $@ parameter), with -s specifying the separator as a vertical tab. This causes the next number to be printed after the previous one, down one line, as required.

Answer 36

Haskell, 46 bytes

a%b=[([a..x-1]>>=show>>" ")++show x|x<-[a..b]]

Try it online! Usage: 0%20. Returns a list of lines. Use putStr . unlines $ 0%20 for pretty-printing.

Answer 37

dc, 51 bytes

sj0skp[32Plkll1+dsl<S]sS[0sldZlk+sklSx1+pdlj>M]dsMx

Try it online!

This pretty much has to be suboptimal, my brain is clearly still in Monday mode. Pretty much just macro S printing spaces, and macro M doing the rest of the work. Register l restarts at zero spaces before running S, and compares against register k, which we always increment by the number of digits of top-of-stack.

Answer 38

Perl 5, 43 42 31 + 1 (-a) 0 = 43 31 bytes

map{say$l=$l=~y// /cr.$_}<>..<>

Try it online!

Answer 39

WendyScript, 50 bytes

<<f=>(x,y){<<s=""#i:x->y+1{s+i#i!=0{s+=" "i\=10}}}

f(95, 103)

Try it online!

Only issue is once you get past 100000, numbers are outputted in scientific notation (which I should probably change to always display in full form). The online syntax highlighter also dislikes "" but it is parsed correctly.

Answer 40

Tcl, 80 bytes

proc P a\ b {while \$a<=$b {puts [format %[incr i [string len $a]]d $a];incr a}}

Try it online!

Answer 41

SmileBASIC, 43 bytes

INPUT A,B
FOR I=A TO B?" "*C;I;
C=CSRX?NEXT

Explanation

?" "*C;I;    'Print C (initially 0) spaces followed by I, with no newline.
C=CSRX       'Set C to the horizontal cursor position.
?            'Print a newline.

Answer 42

SNOBOL4 (CSNOBOL4), 85 bytes

	M =INPUT
	N =INPUT
O	OUTPUT =S M
	S =S DUPL(' ',SIZE(M))
	M =LT(M,N) M + 1 :S(O)
END

Try it online!

Answer 43

PHP 7.1, 66 bytes

for([,$i,$z]=$argv;$i<=$z;)printf("%".($e+=strlen($i))."d
",$i++);

Run with -nr or try them online.

Answer 44

Python 3, 85 bytes

f=lambda a,b,n,z:'\n'+' '*(a+n-z)+str(a)+f(a+1,b,n+len(str(a))-1,z) if a!=b+1 else ''

Try it online!

Answer 45

K3, 29 bytes

{`0:{(1_0+\#:'$:'x)$'x}x+!y-x}

example:

  {`0:{(1_0+\#:'$:'x)$'x}x+!y-x}[1;15]
1
 2
  3
   4
    5
     6
      7
       8
        9
         10
           11
             12
               13
                 14

Answer 46

JavaScript (ES6) + CSS, 72 + 12 = 84 bytes

a=>b=>{for(i=0;a<=b;)document.write(`<x>${"<br>".repeat(i++)+a++}</x>`)}

x{float:left

Takes input in currying syntax with the smaller number first. Outputs the result to the page in HTML.

Test Snippet

let f=

a=>b=>{for(i=0;a<=b;)document.write(`<x>${"<br>".repeat(i++)+a++}</x>`)}

f(0)(8)
f(5)(14)
f(998)(1003)

x{float:left

Answer 47

Excel VBA, 41 Bytes

Anonymous VBE immediate window function that takes input from range A1:B1 and outputs to the VBE immediate window

For i=[A1]To[B1]:?spc(j)i:j=j+len(i):Next

Answer 48

Sinclair ZX81/Timex TS1000/1500 BASIC ~73 tokenized BASIC bytes

 1 LET X=SGN PI
 2 LET Y=CODE "="
 3 LET A=NOT PI
 4 FOR I=X TO Y
 5 PRINT TAB A;I
 6 LET A=A+LEN STR$ I
 7 NEXT I

This newer solution has fewer bytes but slightly slower. SGN PI is 1, CODE "=" is 20 and NOT PI is 0; the PRINT TAB command does what you might expect, except a TAB in ZX81 land is 1 space; A is increased by the length of the I variable once converted to a string.

Old answer (~98 tokenized bytes)

 1 LET X=1
 2 LET Y=20
 3 LET A$=""
 4 FOR I=X TO Y
 5 PRINT A$;I
 6 FOR T=1 TO LEN STR$ I
 7 LET A$=A$+" "
 8 NEXT T
 9 NEXT I

I'm using the most expensive methods here in terms of bytes (i.e., SGN PI is 1, and counts as two bytes in ZX81 land, whereas 1 counts as four bytes), so I could cut the byte count by increasing the typing and making the listing less readable, and slower.

The 'spacing' works by converting the I number to a string and taking the string literal length, using that in a FOR/NEXT loop as the destination [line 6]. X is the start number, and Y is the end number.

There are a few limitations to consider, such as only having a 32 columns display by default, and only allowing 22 rows without some POKEing. Also, the screen doesn't scroll when it is full unless you tell it to, with the command SCROLL.

Answer 49

><>, 56+3 = 59 bytes

1</+1$,a\.0e:/o" "\
:1>$:a)?/~}r+\1-:?/~r:n&:&:&=?;&1+ao

Try It Online

+3 bytes for -v. Not sure whether I actually need to add this for a function, but better to err on the side of caution

How It Works

1<................\     Initialises the stack with the space counter on top of the inputted numbers
................../~...

...................                  Print the first number
....................r:n&:&:&=?;&1+ao If the first number is equal to the end, exit program. Otherwise increment the first number and print a newline

../+1$,a\... Count the number of digits in the number
:1>$:a)?/... By incrementing a counter and dividing the number by 10 until it's below 10

........\.0e:/... Add the amount of digits to the counter
.......?/~}r+\...

............./o" "\     Print the counter amount of spaces and loop around again
.............\1-:?/~...

Answer 50

NotQuiteThere, 5 bytes

yr-11

Try it online!

Uses the vertical tab trick in Rod's python answer, so doesn't really work on TIO. Input is taken in descending order (11, 7)