The Digit Triangles

Question

Challenge:

Input: A positive integer \$n\$

Output:

Create a list in the range \$[1,n]\$, and join it together to a string (i.e. \$n=13\$ would be the string 12345678910111213).

Now we output a triangle using the prefixes or suffixes of this string, in one of the following four orientations based on the input integer:

• If \$n\equiv 0\pmod 4\$, output it in the triangle shape ◣
• If \$n\equiv 1\pmod 4\$, output it in the triangle shape ◤
• If \$n\equiv 2\pmod 4\$, output it in the triangle shape ◥
• If \$n\equiv 3\pmod 4\$, output it in the triangle shape ◢

Example:

Input: \$n=13\$

Because \13ドル\equiv 1\pmod 4\$, the shape will be ◤. Here three possible valid outputs:

12345678910111213 11111111111111111 12345678910111213
1234567891011121 2222222222222222 2345678910111213
123456789101112 333333333333333 345678910111213
12345678910111 44444444444444 45678910111213
1234567891011 5555555555555 5678910111213
123456789101 666666666666 678910111213
12345678910 77777777777 78910111213
1234567891 8888888888 8910111213
123456789 999999999 910111213
12345678 11111111 10111213
1234567 0000000 0111213
123456 111111 111213
12345 11111 11213
1234 1111 1213
123 222 213
12 11 13
1 3 3

Challenge rules:

• As you can see at the three valid outputs above, only the correct shape and using all the digits in the correct order is important. Apart from that you're free to use prefixes or suffixes; reverses/reflects; diagonal printing; etc. etc. Any of the six possible outputs for each shape is allowed (see the test case below to see all valid outputs based on the shape). This allows languages with rotation builtins to use it, but those without can also use an alternative approach of using the prefixes in the correct size from top-to-bottom, or using the prefixes for two of the shapes but suffixes for the other two shapes. Choosing the most appropriate output options for your language is part of the golfing process. :)
• Input is guaranteed to be a positive integer. For \$n=1\$ we simply output 1.
• Any amount of leading/trailing newlines/spaces are allowed, as long as it prints the correct triangle (without vertical nor horizontal delimiters!) somewhere on the screen.

General rules:

• This is code-golf, so shortest answer in bytes wins.
  Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
• Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
• Default Loopholes are forbidden.
• If possible, please add a link with a test for your code (i.e. TIO).
• Also, adding an explanation for your answer is highly recommended.

Test cases:

Input: \$n=5\$
All possible valid outputs:

12345 54321 12345 54321 11111 55555
1234 5432 2345 4321 2222 4444
123 543 345 321 333 333
12 54 45 21 44 22
1 5 5 1 5 1

Input: \$n=6\$
All possible outputs:

123456 654321 123456 654321 111111 666666
 12345 65432 23456 54321 22222 55555
 1234 6543 3456 4321 3333 4444
 123 654 456 321 444 333
 12 65 56 21 55 22
 1 6 6 1 6 1

Input: \$n=7\$
All possible outputs:

 1 1 7 7 7 1
 12 21 67 76 66 22
 123 321 567 765 555 333
 1234 4321 4567 7654 4444 4444
 12345 54321 34567 76543 33333 55555
 123456 654321 234567 765432 222222 666666
1234567 7654321 1234567 7654321 1111111 7777777

Input: \$n=8\$
All possible outputs:

1 1 8 8 8 1
12 21 78 87 77 22
123 321 678 876 666 333
1234 4321 5678 8765 5555 4444
12345 54321 45678 87654 44444 55555
123456 654321 345678 876543 333333 666666
1234567 7654321 2345678 8765432 2222222 7777777
12345678 87654321 12345678 87654321 11111111 88888888

Input: \$n=1\$
Only possible output:

1

Input: \$n=2\$
All possible outputs:

12 21 12 21 11 22
 1 2 2 1 2 1

• Gymhgy: Can we use other values for different triangles, like 1 for ◤, etc?
• Erik the Outgolfer: @EmbodimentofIgnorance Unfortunate example, since that's what the spec says. I think you wanted to ask if we can change the order of the four arrangements as long as we keep it consistent (I think that would be a no).
• Chas Brown: If n==13, can the topmost row be '33333333333333333' (or, equivalently, '31211101987654321')?
• Kevin Cruijssen: @EmbodimentofIgnorance Sorry, but I'd say no in this case. The shapes and their corresponding mod 4 are strict pairs for this challenge. So you may not switch the four shapes for the four mod 4 cases. But good question nonetheless.
• Kevin Cruijssen: @ChasBrown Yes, both of those are fine. I only gave three possible examples for the \$n=13\,ドル but all six options (like the \$n=5\$ test case) are valid outputs.

Answer 1

JavaScript (ES6), (削除) 93 (削除ここまで) 89 bytes

Returns a matrix of characters.

n=>[...(g=n=>n?g(n-1)+n:'')(n)].map((d,y,a)=>a.map(_=>y-(n&2)*y--<0?' ':d)).sort(_=>-n%2)

Try it online!

Alternate pattern (same size):

n=>[...(g=n=>n?g(n-1)+n:'')(n)].map((_,y,a)=>a.map(d=>y-(n&2)*y--<0?' ':d)).sort(_=>-n%2)

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Commented

n => // n = input
 [... // split the result of ...
 ( g = n => // ... a call to the recursive function g, taking n
 n ? // if n is not equal to 0:
 g(n - 1) // append the result of a recursive call with n - 1
 + n // append n
 : // else:
 '' // stop recursion and return an empty string
 )(n) // initial call to g
 ].map((d, y, a) => // for each digit d at position y in this array a[]:
 a.map(_ => // for each character in a[]:
 y - // we test either y < 0 if (n AND 2) is not set
 (n & 2) // or -y < 0 (i.e. y > 0) if (n AND 2) is set
 * y-- < 0 // and we decrement y afterwards
 ? // if the above condition is met:
 ' ' // append a space
 : // else:
 d // append d
 ) // end of inner map()
 ) // end of outer map()
 .sort(_ => -n % 2) // reverse the rows if n is odd

Shape summary

Below is a summary of the base shape (generated by the nested map loops) and the final shape (after the sort) for each \$n\bmod 4\$:

 n mod 4 | 0 | 1 | 2 | 3
----------+-------+-------+-------+-------
 n & 2 | 0 | 0 | 2 | 2
----------+-------+-------+-------+-------
 test | y < 0 | y < 0 | y > 0 | y > 0
----------+-------+-------+-------+-------
 base | #.... | #.... | ##### | #####
 shape | ##... | ##... | .#### | .####
 | ###.. | ###.. | ..### | ..###
 | ####. | ####. | ...## | ...##
 | ##### | ##### | ....# | ....#
----------+-------+-------+-------+-------
 n % 2 | 0 | 1 | 0 | 1
----------+-------+-------+-------+-------
 reverse? | no | yes | no | yes
----------+-------+-------+-------+-------
 final | #.... | ##### | ##### | ....#
 shape | ##... | ####. | .#### | ...##
 | ###.. | ###.. | ..### | ..###
 | ####. | ##... | ...## | .####
 | ##### | #.... | ....# | #####

• Chau Giang: thank you for your detail explaining.

Answer 2

Python 2, 94 bytes

n=0;s=''
exec"n+=1;s+=`n`;"*input()
K=k=len(s)
while k:k-=1;print s[k^n/-2%-2:].rjust(n%4/2*K)

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Answer 3

Japt, 8 bytes

Returns an array of lines.

õ ¬å+ zU

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Saved 2 bytes thanks to Kevin.

õ ¬å+ zU :Implicit input of integer U
õ :Range [1,U]
 ¬ :Join to a string
 å+ :Cumulatively reduce by concatenation
 zU :Rotate clockwise by 90 degrees U times

• Kevin Cruijssen: Is the ú necessary? It seems the rotate does this implicitly?
• Shaggy: @KevinCruijssen, hmm ... so it does. I always forget that; rarely get to use z.
• Kevin Cruijssen: Well, I don't know Japt at all. Was just curious what the output looked like without the padding for fun, and saw it worked exactly the same.. ;)

Answer 4

Canvas, 8 bytes

ŗ]∑[]8[↷

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Answer 5

Perl 6, 94 bytes

{[o](|(&reverse xx$_/2+.5),|(*>>.flip xx$_/2+1))([\~](my@a=[~](1..$_).comb)>>.fmt("%{+@a}s"))}

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Anonymous code block that takes a number and returns a list of lines.

Answer 6

Charcoal, 17 bytes

Nθ≔⭆θ⊕ιηGLLηη⟲⊗θ‖

Try it online! Link is to verbose version of code. Explanation:

Nθ

Input n.

≔⭆θ⊕ιη

Create a string by concatenating the numbers 1 to n.

GLLηη

Fill a triangle of that length with the string.

⟲⊗θ

Rotate the triangle anticlockwise by n*90 degrees.

‖

Reflect everything, thus ending up with a triangle that is rotated clockwise by n*90 degrees.

Answer 7

C# (Visual C# Interactive Compiler), 128 bytes

n=>{var s="";int i=0,j,p;for(;i<n;s+=++i);for(p=j=s.Length;;)Write("{0,"+(n%4>1?p:0)+"}\n",new string(s[--j],-~n%4>1?j+1:p-j));}

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• ceilingcat: Suggest n%4/2*p instead of n%4>1?p:0

Answer 8

Ruby, (削除) 95 (削除ここまで) (削除) 82 (削除ここまで) 79 bytes

->n{r=(0...z=/$/=~s=[*1..n]*'').map{|i|" "*n[1]*i+s[0,z-i]};-n&2>0?r:r.reverse}

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3 bytes saved by G B.

Answer 9

R, (削除) 152 (削除ここまで) (削除) 139 (削除ここまで) (削除) 137 (削除ここまで) 134 bytes

function(n,N=nchar(s<-Reduce(paste0,1:n,'')),A=!n%%4%%3)for(i in 1:N)cat(rep('',(M=c(N-i+1,i))[1+!A]*(n%%4>1)),substr(s,1,M[1+A]),'
')

Unrolled code :

function(n){
 s = Reduce(paste0,1:n,'') # concat the digits from 1 to n into a string s
 N = nchar(s) # store the length of s
 A = !n%%4%%3 # A = TRUE if n MOD 4 == 1 or 2 
 for(i in 1:N){ # for 1 to N (length of s)
 M = c(N-i+1,i) # store N-i+1 and i into a vector
 nSpaces = M[1+!A]*(n%%4>1) # if n MOD 4 == 1 or 2 then nSpaces = i else nSpaces = N-i+1, 
 # but if n MOD 4 == 0 or 1, then force nSpaces = 0
 nDigits = M[1+A] # if n MOD 4 == 1 or 2 then nDigits = N-i+1 else nDigits = i
 prfx = rep('',) # create a character vector repeating '' nSpaces times
 sufx = substr(s,1,M[1+A]) # substring s by taking the first nDigits characters
 cat(pr,su,'\n') # print prfx and sufx using space as separator for the values 
 # contatenation (cat function default) and append a newline
 }

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• Giuseppe: ...it hasn't been my day for golfing, clearly.
• digEmAll: @Giuseppe: ahah been there...and then you usually outgolfed me :P

Answer 10

APL+WIN, 45 bytes

Prompts for integer

m←⊃(⍳⍴s)↑ ̈⊂s←(⍕⍳n←⎕)~' '⋄⍎(' ⊖⍉⌽'[1+4|n]),'m'

Try it online! Courtesy of Dyalog Classic

Answer 11

PowerShell, 108 bytes

param($n)0..($y=($x=-join(1..$n)).length-1)|%{' '*(0,0,$_,($z=$y-$_))[$n%4]+-join$x[0..($_,$z,$z,$_)[$n%4]]}

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A bit rough around the edges but works. Joins the digits 1 to n into a string then iterates from 0 to the length of that string-1. Each time, it uses list indexing to swap to the correct spacing method and number range used to slice our new string.

Answer 12

Jelly, 12 bytes

D€Ẏ;\z6U8ドル¡Y

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Answer 13

05AB1E (legacy), (削除) 14 (削除ここまで) (削除) 12 (削除ここまで) 10 bytes

Using the legacy verion as the rewrite is extremely slow on this for some reason.

Saved 2 bytes thanks to Kevin Cruijssen

LSηsFRζ}J»

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Explanation

L # push range [1 ... input]
 S # split to list of individual digits
 η # calculate prefixes
 sF } # input times do:
 R # reverse list
 ζ # and transpose it
 J # join to list of strings
 » # and join on newlines

• Kevin Cruijssen: You can save 2 bytes changing LJη€S to LSη, since S implicitly flattens.
• Emigna: @KevinCruijssen: Oh yeah, thanks! I'd forgotten about that. I tried €S which didn't work out very well ;)

Answer 14

Stax, 10 bytes

Ç√çZ╟84Γó║

Run and debug it

Answer 15

PowerShell, (削除) 105 (削除ここまで) (削除) 101 (削除ここまで) 95 bytes

-4 bytes thanks Arnauld for the trick with Sort.

param($n)($x=1..$n-join'')|% t*y|%{($s+="$_")}|sort -d:($n%4-in1,2)|% *ft($x.Length*($n%4-ge2))

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Less golfed:

param($n)
$x=1..$n-join''
$x|% toCharArray |% {
 ($s+="$_")
} | sort -Descending:($n%4-in1,2) |% PadLeft ($x.Length*($n%4-ge2))

Answer 16

R, (削除) 175 (削除ここまで) (削除) 172 (削除ここまで) 154 bytes

function(n)write(c(" ",0:9)[1+(x=utf8ToInt(Reduce(paste0,1:n,""))-47)*!upper.tri(diag(y<-sum(x|1)))["if"(n%%4>1,1:y,y:1),"if"(!n%%4%%3,y:1,1:y)]],1,y,,"")

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A horrible in-line mess!

-3 bytes by changing the rotation condition

-17 bytes thanks to digEmAll's suggestion, and another byte after golfing that further

• digEmAll: I like this upper.triangle approach and can be shortened to 155 bytes... maybe even more, I'm sure I'm missing something obvious...
• Giuseppe: @digEmAll ah, much improved, but still long :-(

Answer 17

Wolfram Language (Mathematica), 137 bytes

(t=Table[Table[" ",If[(m=#~Mod~4)>1,Tr[1^s]-k,0]]<>ToString/@s[[;;k]],{k,Length[s=Join@@IntegerDigits/@Range@#]}];If[0<m<3,Reverse@t,t])&

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Answer 18

Python 2, 116 bytes

n=input()
s=''.join(map(str,range(1,n+1)));L=len(s)
p=-~n/2%2;i=~-L*p+1
exec'print s[:i].rjust(n/2%2*L);i+=1-2*p;'*L

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Answer 19

Red, 155 bytes

func[n][b: copy""repeat i n[append b i]repeat i l:
length? b[t:[l - i + 1]set[k m]pick[i t[l t][l i]]n % 4 + 1
print pad/left copy/part b do do m do do k]]

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Answer 20

perl 5, 117 bytes

$p=$_++&2?'/ ':'$/';$s='(.*\d.\n)';$r=$_--&2?$s.'\K$':"^(?=$s)";$_=(join"",1..$_).$/;1while s,$r,'1ドル=~s/\d'."$p/r",ee

TIO

Answer 21

PHP, (削除) 116 (削除ここまで) (削除) 111 (削除ここまで) 109 bytes

for($m=$l=strlen($s=join(range(1,$n=$argn)));$l--;)printf('%'.($n&2?$m:-$l).'.'.($n-1&2?$m-$l:$l+1)."s
",$s);

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Run with php -nF input from STDIN.

$ echo 6|php -nF t.php

123456
 12345
 1234
 123
 12
 1

Answer 22

Java (JDK), (削除) 247 (削除ここまで) (削除) 209 (削除ここまで) (削除) 188 (削除ここまで) (削除) 186 (削除ここまで) (削除) 160 (削除ここまで) 148 bytes

i->{String s="";int l=0,w;for(;l<i;s+=++l);for(w=s.length(),l=0;l<w;)System.out.printf("%"+(1>i%4/2?1:w)+"s\n",s.substring(0,1>~-i%4/2?w-l++:++l));}

Try it online!

-38 bytes thanks to @KevinCruijssen
-21 bytes by letting printf handle the padding.
-2 bytes by doing substring before replace, allowing us to increment l in one location rather than two.
-26 bytes - with printf doing the padding, the string full of spaces was no longer necessary, and the digit strings could be generated in a shorter way apparently.
-12 bytes by not messing about with single digits instead of printing substrings of the perfectly servicable digit string we already have.

Ungolfed

input->{
 // Lambda expression with target type of IntConsumer
 String allDigits = "";
 int lineLength, line = 0;
 // Collect a list of all digits in order.
 for (;line < input; allDigits += ++line) {}
 // Save the max length of a line, and create a string of that many spaces.
 for (lineLength=allDigits.length(), line=0; line < lineLength;) {
 System.out.printf( "%" // construct a format string to handle the padding
 + ( 1 > input%4/2
 ? 1 // No padding
 : w // Yes padding
 )
 + "s\n"
 , allDigits.substring( 0
 , 1 > (i-1)%4/2
 ? w - l++
 : ++l
 ) // A string containing only the digit we want.
 );
 }
}

• Kevin Cruijssen: Nice answer. There are a bunch of things to golf more, though: The spaces after the for( can be removed. new String(new char[w=s.length()]).replace('0円',' ') can be " ".repeat(w=s.length()) using Java 11+. You can remove the parenthesis around the ternary-checks. 1>(i-1)%4/2 can be 1>~-i%4/2. w-1-l++ can be w+~l++. And you don't have to count the trailing semicolon in the byte-count. Which all combined becomes 209 bytes.