# Count up diagonally!

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 , 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

• Are leading spaces allowed, or does the first number have to be exactly on the left side of the screen? – Nathan.Eilisha Shiraini Jul 31 '17 at 9:51
• @NathanShiraini Leading newlines are allowed – Mr. Xcoder Jul 31 '17 at 9:52
• Related – Stephen Jul 31 '17 at 11:36
• @StepHen This one's a bit harder though, thanks for the reference. – Mr. Xcoder Jul 31 '17 at 11:40
• @Adnan You may have a leading newline or a consistent leading space on each line. – Mr. Xcoder Jul 31 '17 at 12:54

# 05AB1E, 87 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

• .O = pop a,b push connected_overlap(b) (deprecated) - Oh, I guess? – Magic Octopus Urn Jul 31 '17 at 14:49
• @MagicOctopusUrn Yeah, .O is extremely buggy and deprecated for over a year so I have no idea what works and what doesn't. I could swear that I needed Î, but that suddenly doesn't seem to be the case anymore (?). Thanks! :) – Adnan Jul 31 '17 at 14:55
• Btw, the Î was needed to reduce the maximum number of leading spaces to 1. – Adnan Jul 31 '17 at 14:57
• I... Wait... What, how...? – Magic Octopus Urn Jul 31 '17 at 15:31

# 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).

• Can you use a literal \x0b in your code to save a byte? – Dom Hastings Jul 31 '17 at 12:57
• @DomHastings maybe, I don't know how though – Rod Jul 31 '17 at 13:12
• I've just tested it and it appears to work. For getting the character into the file to test, I used Sublime Text and did a find and replace in regex mode for \\v and replaced with \x0B which shows up a VT character in its place for scoring you can either post a reversible hexdump (xxd or something) or just state that: "\v is a literal vertical tab", I think that would be fair. Hope that helps! – Dom Hastings Jul 31 '17 at 18:46

# 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!
• Nice, Charcoal wins this again! – Mr. Xcoder Jul 31 '17 at 9:46
• 8 bytes – ASCII-only Jul 31 '17 at 10:35
• EDIT: Charcoal got outgolfed... Wow – Mr. Xcoder Jul 31 '17 at 15:59
• @Mr.Xcoder at least I know how my answer works. :-D – Charlie Jul 31 '17 at 16:35

# R, 7069 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

• I see 70 bytes there. Using scan() twice it can be reduced to 67 bytes for(i in scan():scan()){cat(rep(' ',F),i,'\n',sep='');F=F+nchar(i)}. – djhurio Jul 31 '17 at 10:35
• Unfortunately you have to reset F, otherwise the function can be used only once in a new sessions. F=0;for(i in scan():scan()){cat(rep(' ',F),i,'\n',sep='');F=F+nchar(i)} (71 byte) – djhurio Jul 31 '17 at 10:43
• @djhurio Inside a function, that is not necessary, since F is only modified in its own namespace. Also, I count 69 bytes, using nchar. – JAD Jul 31 '17 at 10:44
• But replacing \n for an actual newline works too, and that doesn't cost two bytes apparently. – JAD Jul 31 '17 at 10:50
• Nice, I thought of abusing the automatic spacing of cat, but I couldn't think straight and figure it out for some reason. – JAD Aug 1 '17 at 6:33

# C#, 9089 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));

}
}
}

• +1, now you don't have 5k precisely ;D – Mr. Xcoder Jul 31 '17 at 11:22
• 1 byte at i<=e to e>i – LiefdeWen Jul 31 '17 at 11:41
• @LiefdeWen Thanks :) – TheLethalCoder Jul 31 '17 at 11:46
• I believe you can save more 4 bytes by removing the i and reusing the s instead – auhmaan Jul 31 '17 at 12:22
• @auhmaan Thanks don't know why I never think of using the input variable. – TheLethalCoder Jul 31 '17 at 12:25

# 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!

• Wow, surprising recursive solution and out-golfs most python answers, +1. – officialaimm Jul 31 '17 at 10:58
• Very good job Ruud, your solution is also OS and console-independent by not using the vertical tab character like Rod did. – Raphaël Côté Aug 2 '17 at 17:03

# 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

• Wow, an answer that actually contains the word Diagonal – Mr. Xcoder Jul 31 '17 at 13:44
• You need to add Spacings -> 0 if you want this to be character-exact. – Mr.Wizard Jul 31 '17 at 22:31
• The input is only non-negative, not guaranteed to be positive. – user202729 Aug 1 '17 at 4:51
• Grid[(DiagonalMatrix@Range[1+##]/. 0->""+1)-1,Spacings->0]& is the shortest way I could find to fix those problems – ngenisis Aug 2 '17 at 22:51

# Jelly, 9 bytes

rD⁶ṁ$;¥\Y  Try it online! # 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 • This isn't valid, is it? But +1 just for taking the title literally. – Zacharý Aug 1 '17 at 13:45 # C, 16613495 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 • Welcome to PPCG! I believe you can rename argc and argv to one letter variables. – Stephen Aug 1 '17 at 0:14 • I think you can move the int s=0 to the for loop, as in for(int s=0;a<=b;a++). – Zacharý Aug 1 '17 at 13:29 • Ahh your right thanks, I updated the post! – Asleepace Aug 1 '17 at 17:58 • Using int i=s;while(i--) instead of for(int i=0;i<s;i++) for the inner loop will save two bytes. – Falken Aug 2 '17 at 10:25 • Ahhh your right forgot about log10 on 0 and negatives, I've updated the solution thanks! – Asleepace Aug 2 '17 at 20:31 ## 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;}  • 1. Could you move the int m=0,i=l,j to the first for loop to save a byte? 2. Can you change r+=t;r+=10 to r+=t+10? 3. I beat someone, yay. – Zacharý Jul 31 '17 at 13:05 • @Zacharý I can do r+=t+=10 but not r+=t+10, it gave me an error – HatsuPointerKun Jul 31 '17 at 13:13 • But r+=t+=10 does work? Wouldn't that affect t.size()? – Zacharý Jul 31 '17 at 13:14 • @Zacharý Yes, it works, with only +, it says it can't find an overload with int as parameter, but with += it uses the overload with the char – HatsuPointerKun Jul 31 '17 at 13:21 • Oh, could you move the ++i to the std::to_string(i) as std::to_string(i++) to save one more byte? – Zacharý Jul 31 '17 at 13:34 # 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 • Could +-⍨ be --? – Zacharý Jul 31 '17 at 11:46 • @Zacharý Yes, of course. Thanks. – Adám Jul 31 '17 at 11:47 # 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. # Pyth, 14 13 bytes V}FQ =k+*dlkN  Try it online! # 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)  # 05AB1E, 8 bytes Ÿʒ¾ú=þv¼  Try it online! -2 thanks to Adnan. • Ahh, that is very clever. You can replace vy by ʒ and gF by v to save 2 bytes. – Adnan Jul 31 '17 at 13:18 • @Adnan I didn't expect the good old ʒ trick to still be used... – Erik the Outgolfer Jul 31 '17 at 13:21 # JavaScript (ES8), 6967 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> # 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. # 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! • You do not need the braces for exec. m,n=input();s=0;exec(n-m+1)*"print s*' '+m;s+=len(m);m+=1;" suffices. – Mr. Xcoder Jul 31 '17 at 10:03 • I think you can change print s*' ' to print' '*s to save one byte. – Zacharý Aug 1 '17 at 13:35 # JavaScript, 57 bytes f=(x,y,s='')=>y>=x?s+ +f(x+1,y,s.replace(/./g,' ')+x):s  • 55 bytes: y=>g=(x,s='')=>y<x?s:s+'\n'+g(x+1,s.replace(/./g,' ')+x) Call with currying with the integers reversed: f(103)(95). – Shaggy Jul 31 '17 at 10:09 • 54 bytes: x=>y=>g=(s='')=>y<x?s:s+'\n'+g(s.replace(/./g,' ')+x++) Call as f(x)(y)(). – Shaggy Jul 31 '17 at 12:16 # 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. # Python 2, 7877 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 • Welcome to PPCG! nice first answer. sum(len(j)) for can become sum(len(j)for, -1 bytes – Mr. Xcoder Jul 31 '17 at 10:01 • To make this answer valid, you must replace range(a,b) with range(a,b+1), because Python has semi inclusive ranges. – Mr. Xcoder Jul 31 '17 at 10:02 • Indeed, I missed that. What's more surprising is that I did add that +1 when I made my tests! No wonder I had 2 bytes missing when I typed it into TiO... – Nathan.Eilisha Shiraini Jul 31 '17 at 10:05 # 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! • This doesn't produce correct output. – Erik the Outgolfer Jul 31 '17 at 9:50 • it's tricky, output to a file and then use more on that file – Giacomo Garabello Jul 31 '17 at 9:51 • @GiacomoGarabello You must provide the full code in order for us to be able to run your program. If you do not provide a working test ground / do not provide instructions on how to run your program such that it produces correct output, please delete this answer. – Mr. Xcoder Jul 31 '17 at 9:56 • Linefeed may return to begin of line, it depends. This does work when it does not. – user202729 Jul 31 '17 at 10:04 • @Mr.Xcoder Edited – Giacomo Garabello Jul 31 '17 at 10:04 # V, 16 bytes ÀñÙywÒ$pça/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.

# 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


# 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($0).count}.reduce(0,+)),i)}}  Try it online! ## Perl, 19 bytes Note: \x0b is counted as one byte. Along with others, I thought using cursor movements would be the shortest route, this does mean it doesn't work on TIO: print"$_\x0b"for<>..<>


### Usage

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

• Nice, haven't seen Perl at all in a while. Could you add a testing link? Additionally, I was wondering what the 1.. does there, since you are given two integers. – Mr. Xcoder Jul 31 '17 at 12:57
• @Mr.Xcoder Yeah, 1.. was me not fully reading the spec, that's fixed now! As for testing it online, because the output contains the vertical tab, it doesn't render as expected. Trying to see if I can find a renderer that does support control chars... If not, that might be my new project! – Dom Hastings Jul 31 '17 at 13:07

# 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

• Dang, I had just come up with the same solution as an improvement to my own answer. – Justin Mariner Jul 31 '17 at 17:33

# D, 133127126125121 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!

# Java 8, 79 bytes

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


Try it online!

• You can save a byte by changing (a,b)-> to b->a->. (And you could save three more bytes by going to Java 10 and changing String to var.) Try it online. – Kevin Cruijssen Jun 6 '18 at 9:31