The Digit Triangles

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 $nequiv 0pmod 4$, output it in the triangle shape ◣

If $nequiv 1pmod 4$, output it in the triangle shape ◤

If $nequiv 2pmod 4$, output it in the triangle shape ◥

If $nequiv 3pmod 4$, output it in the triangle shape ◢

Example:

Input: $n=13$

Because $13equiv 1pmod 4$, 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:

Input: $n=2$

All possible outputs:

12    21    12    21    11    22

 1     2     2     1     2     1

edited 6 hours ago

asked yesterday

Kevin Cruijssen

41.1k566212

$begingroup$
Can we use other values for different triangles, like 1 for ◤, etc?
$endgroup$
– Embodiment of Ignorance
yesterday

$begingroup$
@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).
$endgroup$
– Erik the Outgolfer
yesterday

1

$begingroup$
If n==13, can the topmost row be '33333333333333333' (or, equivalently, '31211101987654321')?
$endgroup$
– Chas Brown
yesterday

$begingroup$
@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.
$endgroup$
– Kevin Cruijssen
yesterday

$begingroup$
@ChasBrown Yes, both of those are fine. I only gave three possible examples for the $n=13$, but all six options (like the $n=5$ test case) are valid outputs.
$endgroup$
– Kevin Cruijssen
yesterday

add a comment |

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 $nequiv 0pmod 4$, output it in the triangle shape ◣

If $nequiv 1pmod 4$, output it in the triangle shape ◤

If $nequiv 2pmod 4$, output it in the triangle shape ◥

If $nequiv 3pmod 4$, output it in the triangle shape ◢

Example:

Input: $n=13$

Because $13equiv 1pmod 4$, 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:

Input: $n=2$

All possible outputs:

12    21    12    21    11    22

 1     2     2     1     2     1

edited 6 hours ago

asked yesterday

Kevin Cruijssen

41.1k566212

$begingroup$
Can we use other values for different triangles, like 1 for ◤, etc?
$endgroup$
– Embodiment of Ignorance
yesterday

$begingroup$
@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).
$endgroup$
– Erik the Outgolfer
yesterday

1

$begingroup$
If n==13, can the topmost row be '33333333333333333' (or, equivalently, '31211101987654321')?
$endgroup$
– Chas Brown
yesterday

$begingroup$
@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.
$endgroup$
– Kevin Cruijssen
yesterday

$begingroup$
@ChasBrown Yes, both of those are fine. I only gave three possible examples for the $n=13$, but all six options (like the $n=5$ test case) are valid outputs.
$endgroup$
– Kevin Cruijssen
yesterday

add a comment |

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 $nequiv 0pmod 4$, output it in the triangle shape ◣

If $nequiv 1pmod 4$, output it in the triangle shape ◤

If $nequiv 2pmod 4$, output it in the triangle shape ◥

If $nequiv 3pmod 4$, output it in the triangle shape ◢

Example:

Input: $n=13$

Because $13equiv 1pmod 4$, 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:

Input: $n=2$

All possible outputs:

12    21    12    21    11    22

 1     2     2     1     2     1

edited 6 hours ago

asked yesterday

Kevin Cruijssen

41.1k566212

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 $nequiv 0pmod 4$, output it in the triangle shape ◣

If $nequiv 1pmod 4$, output it in the triangle shape ◤

If $nequiv 2pmod 4$, output it in the triangle shape ◥

If $nequiv 3pmod 4$, output it in the triangle shape ◢

Example:

Input: $n=13$

Because $13equiv 1pmod 4$, 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:

Input: $n=2$

All possible outputs:

12    21    12    21    11    22

 1     2     2     1     2     1

code-golf ascii-art number integer

edited 6 hours ago

asked yesterday

Kevin Cruijssen

41.1k566212

edited 6 hours ago

asked yesterday

Kevin Cruijssen

41.1k566212

edited 6 hours ago

asked yesterday

Kevin Cruijssen

41.1k566212

asked yesterday

Kevin Cruijssen

41.1k566212

asked yesterday

Kevin Cruijssen

41.1k566212

$begingroup$
Can we use other values for different triangles, like 1 for ◤, etc?
$endgroup$
– Embodiment of Ignorance
yesterday

$begingroup$
@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).
$endgroup$
– Erik the Outgolfer
yesterday

1

$begingroup$
If n==13, can the topmost row be '33333333333333333' (or, equivalently, '31211101987654321')?
$endgroup$
– Chas Brown
yesterday

$begingroup$
@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.
$endgroup$
– Kevin Cruijssen
yesterday

$begingroup$
@ChasBrown Yes, both of those are fine. I only gave three possible examples for the $n=13$, but all six options (like the $n=5$ test case) are valid outputs.
$endgroup$
– Kevin Cruijssen
yesterday

add a comment |

$begingroup$
Can we use other values for different triangles, like 1 for ◤, etc?
$endgroup$
– Embodiment of Ignorance
yesterday

$begingroup$
@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).
$endgroup$
– Erik the Outgolfer
yesterday

1

$begingroup$
If n==13, can the topmost row be '33333333333333333' (or, equivalently, '31211101987654321')?
$endgroup$
– Chas Brown
yesterday

$begingroup$
@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.
$endgroup$
– Kevin Cruijssen
yesterday

$begingroup$
@ChasBrown Yes, both of those are fine. I only gave three possible examples for the $n=13$, but all six options (like the $n=5$ test case) are valid outputs.
$endgroup$
– Kevin Cruijssen
yesterday

Can we use other values for different triangles, like 1 for ◤, etc?

– Embodiment of Ignorance
yesterday

@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).

– Erik the Outgolfer
yesterday

If n==13, can the topmost row be '33333333333333333' (or, equivalently, '31211101987654321')?

– Chas Brown
yesterday

@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
yesterday

@ChasBrown Yes, both of those are fine. I only gave three possible examples for the $n=13$, but all six options (like the $n=5$ test case) are valid outputs.

– Kevin Cruijssen
yesterday

add a comment |

19 Answers
19

active

oldest

votes

Japt, 8 bytes

Returns an array of lines.

õ ¬å+ zU

Try it

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

edited yesterday

answered yesterday

Shaggy

19.1k21667

1

$begingroup$
Is the ú necessary? It seems the rotate does this implicitly?
$endgroup$
– Kevin Cruijssen
yesterday

$begingroup$
@KevinCruijssen, hmm ... so it does. I always forget that; rarely get to use z.
$endgroup$
– Shaggy
yesterday

1

$begingroup$
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.. ;)
$endgroup$
– Kevin Cruijssen
yesterday

add a comment |

JavaScript (ES6), 93 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)

Try it online!

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 $nbmod 4$:

 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    | ##... | ####. | .#### | ...##

          | ###.. | ###.. | ..### | ..###

          | ####. | ##... | ...## | .####

          | ##### | #.... | ....# | #####

edited yesterday

answered yesterday

Arnauld

79.5k796330

add a comment |

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)

Try it online!

answered 20 hours ago

xnor

92.7k18188447

add a comment |

Canvas, 8 bytes

ŗ］∑［］⁸［↷

Try it here!

answered yesterday

dzaima

15.8k22059

add a comment |

Perl 6, 94 bytes

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

Try it online!

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

answered 23 hours ago

Jo King

25.2k359128

add a comment |

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.

answered yesterday

Neil

82k745178

add a comment |

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));}

Try it online!

answered 22 hours ago

Embodiment of Ignorance

2,098125

add a comment |

Ruby, 95 82 79 bytes

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

Try it online!

3 bytes saved by G B.

edited 11 hours ago

answered yesterday

Kirill L.

5,6431525

add a comment |

APL+WIN, 45 bytes

Prompts for integer

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

Try it online! Courtesy of Dyalog Classic

answered yesterday

Graham

2,59678

add a comment |

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]]}

Try it online!

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.

edited yesterday

answered yesterday

Veskah

1,133214

add a comment |

Jelly, 12 bytes

D€Ẏ;z⁶U$⁸¡Y

Try it online!

answered yesterday

Erik the Outgolfer

32.7k429105

add a comment |

05AB1E (legacy), 14 12 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»

Try it online!

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

edited yesterday

answered yesterday

Emigna

47.1k433143

$begingroup$
You can save 2 bytes changing LJη€S to LSη, since S implicitly flattens.
$endgroup$
– Kevin Cruijssen
yesterday

$begingroup$
@KevinCruijssen: Oh yeah, thanks! I'd forgotten about that. I tried €S which didn't work out very well ;)
$endgroup$
– Emigna
yesterday

add a comment |

Stax, 10 bytes

Ç√çZ╟84Γó║

Run and debug it

answered yesterday

recursive

5,6091322

add a comment |

PowerShell, 105 101 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))

Try it online!

Less golfed:

param($n)

$x=1..$n-join''

$x|% toCharArray |% {

    ($s+="$_")

} | sort -Descending:($n%4-in1,2) |% PadLeft ($x.Length*($n%4-ge2))

edited 13 hours ago

answered 14 hours ago

mazzy

2,8851317

add a comment |

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])&

Try it online!

edited yesterday

answered yesterday

J42161217

13.4k21251

add a comment |

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

Try it online!

answered yesterday

Chas Brown

5,0641523

add a comment |

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]]

Try it online!

edited 16 hours ago

answered 17 hours ago

Galen Ivanov

7,18211034

add a comment |

perl 5, 117 bytes

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

TIO

edited 11 hours ago

answered 11 hours ago

Nahuel Fouilleul

2,875211

add a comment |

R, 175 172 bytes

function(n)write(`[<-`(m<-matrix(S<-el(strsplit(x<-Reduce(paste0,1:n,""),"")),y<-nchar(x),y),upper.tri(m)["if"(n%%4%in%2:3,1:y,y:1),"if"(n%%4%in%1:2,1:y,y:1)]," "),1,y,,"")

Try it online!

A horrible in-line mess!

-3 bytes by changing the rotation condition

Ungolfed and simplified a bit:

function(n){

Concat <- Reduce(paste0,1:n,init="")				# concatenate all the digits, with init = "" to handle the n=1 case

digits <- el(strsplit(Concat,""))				# split the string to get component digits

mat    <- matrix(digits,nchar(Concat),nchar(Concat))		# create a matrix with the digits running down the columns,

								# for example, n = 4 is [[1,1,1,1],[2,2,2,2],[3,3,3,3],[4,4,4,4]]

mat[upper.tri(mat)[(n%%4 condiditon),(n%%4 condition)]] <- " "	# conditioned on (n %% 4), replace the values in mat with spaces

write(mat,1,nchar(Concat),,"")					# print the transpose of mat

}

edited 7 hours ago

answered 7 hours ago

Giuseppe

17k31052

add a comment |

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19 Answers
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19 Answers
19

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oldest

votes

Japt, 8 bytes

Returns an array of lines.

õ ¬å+ zU

Try it

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

edited yesterday

answered yesterday

Shaggy

19.1k21667

1

$begingroup$
Is the ú necessary? It seems the rotate does this implicitly?
$endgroup$
– Kevin Cruijssen
yesterday

$begingroup$
@KevinCruijssen, hmm ... so it does. I always forget that; rarely get to use z.
$endgroup$
– Shaggy
yesterday

1

$begingroup$
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.. ;)
$endgroup$
– Kevin Cruijssen
yesterday

add a comment |

Japt, 8 bytes

Returns an array of lines.

õ ¬å+ zU

Try it

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

edited yesterday

answered yesterday

Shaggy

19.1k21667

1

$begingroup$
Is the ú necessary? It seems the rotate does this implicitly?
$endgroup$
– Kevin Cruijssen
yesterday

$begingroup$
@KevinCruijssen, hmm ... so it does. I always forget that; rarely get to use z.
$endgroup$
– Shaggy
yesterday

1

$begingroup$
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.. ;)
$endgroup$
– Kevin Cruijssen
yesterday

add a comment |

Japt, 8 bytes

Returns an array of lines.

õ ¬å+ zU

Try it

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

edited yesterday

answered yesterday

Shaggy

19.1k21667

Japt, 8 bytes

Returns an array of lines.

õ ¬å+ zU

Try it

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

edited yesterday

answered yesterday

Shaggy

19.1k21667

edited yesterday

answered yesterday

Shaggy

19.1k21667

answered yesterday

Shaggy

19.1k21667

answered yesterday

Shaggy

19.1k21667

1

$begingroup$
Is the ú necessary? It seems the rotate does this implicitly?
$endgroup$
– Kevin Cruijssen
yesterday

$begingroup$
@KevinCruijssen, hmm ... so it does. I always forget that; rarely get to use z.
$endgroup$
– Shaggy
yesterday

1

$begingroup$
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.. ;)
$endgroup$
– Kevin Cruijssen
yesterday

add a comment |

1

$begingroup$
Is the ú necessary? It seems the rotate does this implicitly?
$endgroup$
– Kevin Cruijssen
yesterday

$begingroup$
@KevinCruijssen, hmm ... so it does. I always forget that; rarely get to use z.
$endgroup$
– Shaggy
yesterday

1

$begingroup$
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.. ;)
$endgroup$
– Kevin Cruijssen
yesterday

Is the ú necessary? It seems the rotate does this implicitly?

– Kevin Cruijssen
yesterday

@KevinCruijssen, hmm ... so it does. I always forget that; rarely get to use z.

– Shaggy
yesterday

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.. ;)

– Kevin Cruijssen
yesterday

add a comment |

JavaScript (ES6), 93 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)

Try it online!

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 $nbmod 4$:

 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    | ##... | ####. | .#### | ...##

          | ###.. | ###.. | ..### | ..###

          | ####. | ##... | ...## | .####

          | ##### | #.... | ....# | #####

edited yesterday

answered yesterday

Arnauld

79.5k796330

add a comment |

JavaScript (ES6), 93 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)

Try it online!

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 $nbmod 4$:

 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    | ##... | ####. | .#### | ...##

          | ###.. | ###.. | ..### | ..###

          | ####. | ##... | ...## | .####

          | ##### | #.... | ....# | #####

edited yesterday

answered yesterday

Arnauld

79.5k796330

add a comment |

JavaScript (ES6), 93 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)

Try it online!

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 $nbmod 4$:

 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    | ##... | ####. | .#### | ...##

          | ###.. | ###.. | ..### | ..###

          | ####. | ##... | ...## | .####

          | ##### | #.... | ....# | #####

edited yesterday

answered yesterday

Arnauld

79.5k796330

JavaScript (ES6), 93 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)

Try it online!

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 $nbmod 4$:

 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    | ##... | ####. | .#### | ...##

          | ###.. | ###.. | ..### | ..###

          | ####. | ##... | ...## | .####

          | ##### | #.... | ....# | #####

edited yesterday

answered yesterday

Arnauld

79.5k796330

edited yesterday

answered yesterday

Arnauld

79.5k796330

answered yesterday

Arnauld

79.5k796330

answered yesterday

Arnauld

79.5k796330

add a comment |

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)

Try it online!

answered 20 hours ago

xnor

92.7k18188447

add a comment |

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)

Try it online!

answered 20 hours ago

xnor

92.7k18188447

add a comment |

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)

Try it online!

answered 20 hours ago

xnor

92.7k18188447

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)

Try it online!

answered 20 hours ago

xnor

92.7k18188447

answered 20 hours ago

xnor

92.7k18188447

answered 20 hours ago

xnor

92.7k18188447

answered 20 hours ago

xnor

92.7k18188447

add a comment |

Canvas, 8 bytes

ŗ］∑［］⁸［↷

Try it here!

answered yesterday

dzaima

15.8k22059

add a comment |

Canvas, 8 bytes

ŗ］∑［］⁸［↷

Try it here!

answered yesterday

dzaima

15.8k22059

add a comment |

Canvas, 8 bytes

ŗ］∑［］⁸［↷

Try it here!

answered yesterday

dzaima

15.8k22059

Canvas, 8 bytes

ŗ］∑［］⁸［↷

Try it here!

answered yesterday

dzaima

15.8k22059

answered yesterday

dzaima

15.8k22059

answered yesterday

dzaima

15.8k22059

answered yesterday

dzaima

15.8k22059

add a comment |

Perl 6, 94 bytes

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

Try it online!

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

answered 23 hours ago

Jo King

25.2k359128

add a comment |

Perl 6, 94 bytes

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

Try it online!

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

answered 23 hours ago

Jo King

25.2k359128

add a comment |

Perl 6, 94 bytes

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

Try it online!

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

answered 23 hours ago

Jo King

25.2k359128

Perl 6, 94 bytes

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

Try it online!

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

answered 23 hours ago

Jo King

25.2k359128

answered 23 hours ago

Jo King

25.2k359128

answered 23 hours ago

Jo King

25.2k359128

answered 23 hours ago

Jo King

25.2k359128

add a comment |

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.

answered yesterday

Neil

82k745178

add a comment |

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.

answered yesterday

Neil

82k745178

add a comment |

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.

answered yesterday

Neil

82k745178

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.

answered yesterday

Neil

82k745178

answered yesterday

Neil

82k745178

answered yesterday

Neil

82k745178

answered yesterday

Neil

82k745178

add a comment |

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));}

Try it online!

answered 22 hours ago

Embodiment of Ignorance

2,098125

add a comment |

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));}

Try it online!

answered 22 hours ago

Embodiment of Ignorance

2,098125

add a comment |

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));}

Try it online!

answered 22 hours ago

Embodiment of Ignorance

2,098125

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));}

Try it online!

answered 22 hours ago

Embodiment of Ignorance

2,098125

answered 22 hours ago

Embodiment of Ignorance

2,098125

answered 22 hours ago

Embodiment of Ignorance

2,098125

answered 22 hours ago

Embodiment of Ignorance

2,098125

add a comment |

Ruby, 95 82 79 bytes

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

Try it online!

3 bytes saved by G B.

edited 11 hours ago

answered yesterday

Kirill L.

5,6431525

add a comment |

Ruby, 95 82 79 bytes

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

Try it online!

3 bytes saved by G B.

edited 11 hours ago

answered yesterday

Kirill L.

5,6431525

add a comment |

Ruby, 95 82 79 bytes

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

Try it online!

3 bytes saved by G B.

edited 11 hours ago

answered yesterday

Kirill L.

5,6431525

Ruby, 95 82 79 bytes

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

Try it online!

3 bytes saved by G B.

edited 11 hours ago

answered yesterday

Kirill L.

5,6431525

edited 11 hours ago

answered yesterday

Kirill L.

5,6431525

answered yesterday

Kirill L.

5,6431525

answered yesterday

Kirill L.

5,6431525

add a comment |

APL+WIN, 45 bytes

Prompts for integer

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

Try it online! Courtesy of Dyalog Classic

answered yesterday

Graham

2,59678

add a comment |

APL+WIN, 45 bytes

Prompts for integer

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

Try it online! Courtesy of Dyalog Classic

answered yesterday

Graham

2,59678

add a comment |

APL+WIN, 45 bytes

Prompts for integer

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

Try it online! Courtesy of Dyalog Classic

answered yesterday

Graham

2,59678

APL+WIN, 45 bytes

Prompts for integer

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

Try it online! Courtesy of Dyalog Classic

answered yesterday

Graham

2,59678

answered yesterday

Graham

2,59678

answered yesterday

Graham

2,59678

answered yesterday

Graham

2,59678

add a comment |

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]]}

Try it online!

edited yesterday

answered yesterday

Veskah

1,133214

add a comment |

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]]}

Try it online!

edited yesterday

answered yesterday

Veskah

1,133214

add a comment |

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]]}

Try it online!

edited yesterday

answered yesterday

Veskah

1,133214

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]]}

Try it online!

edited yesterday

answered yesterday

Veskah

1,133214

edited yesterday

answered yesterday

Veskah

1,133214

answered yesterday

Veskah

1,133214

answered yesterday

Veskah

1,133214

add a comment |

Jelly, 12 bytes

D€Ẏ;z⁶U$⁸¡Y

Try it online!

answered yesterday

Erik the Outgolfer

32.7k429105

add a comment |

Jelly, 12 bytes

D€Ẏ;z⁶U$⁸¡Y

Try it online!

answered yesterday

Erik the Outgolfer

32.7k429105

add a comment |

Jelly, 12 bytes

D€Ẏ;z⁶U$⁸¡Y

Try it online!

answered yesterday

Erik the Outgolfer

32.7k429105

Jelly, 12 bytes

D€Ẏ;z⁶U$⁸¡Y

Try it online!

answered yesterday

Erik the Outgolfer

32.7k429105

answered yesterday

Erik the Outgolfer

32.7k429105

answered yesterday

Erik the Outgolfer

32.7k429105

answered yesterday

Erik the Outgolfer

32.7k429105

add a comment |

05AB1E (legacy), 14 12 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»

Try it online!

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

edited yesterday

answered yesterday

Emigna

47.1k433143

$begingroup$
You can save 2 bytes changing LJη€S to LSη, since S implicitly flattens.
$endgroup$
– Kevin Cruijssen
yesterday

$begingroup$
@KevinCruijssen: Oh yeah, thanks! I'd forgotten about that. I tried €S which didn't work out very well ;)
$endgroup$
– Emigna
yesterday

add a comment |

05AB1E (legacy), 14 12 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»

Try it online!

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

edited yesterday

answered yesterday

Emigna

47.1k433143

$begingroup$
You can save 2 bytes changing LJη€S to LSη, since S implicitly flattens.
$endgroup$
– Kevin Cruijssen
yesterday

$begingroup$
@KevinCruijssen: Oh yeah, thanks! I'd forgotten about that. I tried €S which didn't work out very well ;)
$endgroup$
– Emigna
yesterday

add a comment |

05AB1E (legacy), 14 12 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»

Try it online!

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

edited yesterday

answered yesterday

Emigna

47.1k433143

05AB1E (legacy), 14 12 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»

Try it online!

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

edited yesterday

answered yesterday

Emigna

47.1k433143

edited yesterday

answered yesterday

Emigna

47.1k433143

answered yesterday

Emigna

47.1k433143

answered yesterday

Emigna

47.1k433143

$begingroup$
You can save 2 bytes changing LJη€S to LSη, since S implicitly flattens.
$endgroup$
– Kevin Cruijssen
yesterday

$begingroup$
@KevinCruijssen: Oh yeah, thanks! I'd forgotten about that. I tried €S which didn't work out very well ;)
$endgroup$
– Emigna
yesterday

add a comment |

$begingroup$
You can save 2 bytes changing LJη€S to LSη, since S implicitly flattens.
$endgroup$
– Kevin Cruijssen
yesterday

$begingroup$
@KevinCruijssen: Oh yeah, thanks! I'd forgotten about that. I tried €S which didn't work out very well ;)
$endgroup$
– Emigna
yesterday

You can save 2 bytes changing LJη€S to LSη, since S implicitly flattens.

– Kevin Cruijssen
yesterday

@KevinCruijssen: Oh yeah, thanks! I'd forgotten about that. I tried €S which didn't work out very well ;)

– Emigna
yesterday

add a comment |

Stax, 10 bytes

Ç√çZ╟84Γó║

Run and debug it

answered yesterday

recursive

5,6091322

add a comment |

Stax, 10 bytes

Ç√çZ╟84Γó║

Run and debug it

answered yesterday

recursive

5,6091322

add a comment |

Stax, 10 bytes

Ç√çZ╟84Γó║

Run and debug it

answered yesterday

recursive

5,6091322

Stax, 10 bytes

Ç√çZ╟84Γó║

Run and debug it

answered yesterday

recursive

5,6091322

answered yesterday

recursive

5,6091322

answered yesterday

recursive

5,6091322

answered yesterday

recursive

5,6091322

add a comment |

PowerShell, 105 101 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))

Try it online!

Less golfed:

param($n)

$x=1..$n-join''

$x|% toCharArray |% {

    ($s+="$_")

} | sort -Descending:($n%4-in1,2) |% PadLeft ($x.Length*($n%4-ge2))

edited 13 hours ago

answered 14 hours ago

mazzy

2,8851317

add a comment |

PowerShell, 105 101 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))

Try it online!

Less golfed:

param($n)

$x=1..$n-join''

$x|% toCharArray |% {

    ($s+="$_")

} | sort -Descending:($n%4-in1,2) |% PadLeft ($x.Length*($n%4-ge2))

edited 13 hours ago

answered 14 hours ago

mazzy

2,8851317

add a comment |

PowerShell, 105 101 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))

Try it online!

Less golfed:

param($n)

$x=1..$n-join''

$x|% toCharArray |% {

    ($s+="$_")

} | sort -Descending:($n%4-in1,2) |% PadLeft ($x.Length*($n%4-ge2))

edited 13 hours ago

answered 14 hours ago

mazzy

2,8851317

PowerShell, 105 101 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))

Try it online!

Less golfed:

param($n)

$x=1..$n-join''

$x|% toCharArray |% {

    ($s+="$_")

} | sort -Descending:($n%4-in1,2) |% PadLeft ($x.Length*($n%4-ge2))

edited 13 hours ago

answered 14 hours ago

mazzy

2,8851317

edited 13 hours ago

answered 14 hours ago

mazzy

2,8851317

answered 14 hours ago

mazzy

2,8851317

answered 14 hours ago

mazzy

2,8851317

add a comment |

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])&

Try it online!

edited yesterday

answered yesterday

J42161217

13.4k21251

add a comment |

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])&

Try it online!

edited yesterday

answered yesterday

J42161217

13.4k21251

add a comment |

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])&

Try it online!

edited yesterday

answered yesterday

J42161217

13.4k21251

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])&

Try it online!

edited yesterday

answered yesterday

J42161217

13.4k21251

edited yesterday

answered yesterday

J42161217

13.4k21251

answered yesterday

J42161217

13.4k21251

answered yesterday

J42161217

13.4k21251

add a comment |

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

Try it online!

answered yesterday

Chas Brown

5,0641523

add a comment |

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

Try it online!

answered yesterday

Chas Brown

5,0641523

add a comment |

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

Try it online!

answered yesterday

Chas Brown

5,0641523

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

Try it online!

answered yesterday

Chas Brown

5,0641523

answered yesterday

Chas Brown

5,0641523

answered yesterday

Chas Brown

5,0641523

answered yesterday

Chas Brown

5,0641523

add a comment |

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]]

Try it online!

edited 16 hours ago

answered 17 hours ago

Galen Ivanov

7,18211034

add a comment |

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]]

Try it online!

edited 16 hours ago

answered 17 hours ago

Galen Ivanov

7,18211034

add a comment |

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]]

Try it online!

edited 16 hours ago

answered 17 hours ago

Galen Ivanov

7,18211034

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]]

Try it online!

edited 16 hours ago

answered 17 hours ago

Galen Ivanov

7,18211034

edited 16 hours ago

answered 17 hours ago

Galen Ivanov

7,18211034

answered 17 hours ago

Galen Ivanov

7,18211034

answered 17 hours ago

Galen Ivanov

7,18211034

add a comment |

perl 5, 117 bytes

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

TIO

edited 11 hours ago

answered 11 hours ago

Nahuel Fouilleul

2,875211

add a comment |

perl 5, 117 bytes

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

TIO

edited 11 hours ago

answered 11 hours ago

Nahuel Fouilleul

2,875211

add a comment |

perl 5, 117 bytes

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

TIO

edited 11 hours ago

answered 11 hours ago

Nahuel Fouilleul

2,875211

perl 5, 117 bytes

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

TIO

edited 11 hours ago

answered 11 hours ago

Nahuel Fouilleul

2,875211

edited 11 hours ago

answered 11 hours ago

Nahuel Fouilleul

2,875211

answered 11 hours ago

Nahuel Fouilleul

2,875211

answered 11 hours ago

Nahuel Fouilleul

2,875211

add a comment |

R, 175 172 bytes

function(n)write(`[<-`(m<-matrix(S<-el(strsplit(x<-Reduce(paste0,1:n,""),"")),y<-nchar(x),y),upper.tri(m)["if"(n%%4%in%2:3,1:y,y:1),"if"(n%%4%in%1:2,1:y,y:1)]," "),1,y,,"")

Try it online!

A horrible in-line mess!

-3 bytes by changing the rotation condition

Ungolfed and simplified a bit:

function(n){

Concat <- Reduce(paste0,1:n,init="")				# concatenate all the digits, with init = "" to handle the n=1 case

digits <- el(strsplit(Concat,""))				# split the string to get component digits

mat    <- matrix(digits,nchar(Concat),nchar(Concat))		# create a matrix with the digits running down the columns,

								# for example, n = 4 is [[1,1,1,1],[2,2,2,2],[3,3,3,3],[4,4,4,4]]

mat[upper.tri(mat)[(n%%4 condiditon),(n%%4 condition)]] <- " "	# conditioned on (n %% 4), replace the values in mat with spaces

write(mat,1,nchar(Concat),,"")					# print the transpose of mat

}

edited 7 hours ago

answered 7 hours ago

Giuseppe

17k31052

add a comment |

R, 175 172 bytes

function(n)write(`[<-`(m<-matrix(S<-el(strsplit(x<-Reduce(paste0,1:n,""),"")),y<-nchar(x),y),upper.tri(m)["if"(n%%4%in%2:3,1:y,y:1),"if"(n%%4%in%1:2,1:y,y:1)]," "),1,y,,"")

Try it online!

A horrible in-line mess!

-3 bytes by changing the rotation condition

Ungolfed and simplified a bit:

function(n){

Concat <- Reduce(paste0,1:n,init="")				# concatenate all the digits, with init = "" to handle the n=1 case

digits <- el(strsplit(Concat,""))				# split the string to get component digits

mat    <- matrix(digits,nchar(Concat),nchar(Concat))		# create a matrix with the digits running down the columns,

								# for example, n = 4 is [[1,1,1,1],[2,2,2,2],[3,3,3,3],[4,4,4,4]]

mat[upper.tri(mat)[(n%%4 condiditon),(n%%4 condition)]] <- " "	# conditioned on (n %% 4), replace the values in mat with spaces

write(mat,1,nchar(Concat),,"")					# print the transpose of mat

}

edited 7 hours ago

answered 7 hours ago

Giuseppe

17k31052

add a comment |

R, 175 172 bytes

function(n)write(`[<-`(m<-matrix(S<-el(strsplit(x<-Reduce(paste0,1:n,""),"")),y<-nchar(x),y),upper.tri(m)["if"(n%%4%in%2:3,1:y,y:1),"if"(n%%4%in%1:2,1:y,y:1)]," "),1,y,,"")

Try it online!

A horrible in-line mess!

-3 bytes by changing the rotation condition

Ungolfed and simplified a bit:

function(n){

Concat <- Reduce(paste0,1:n,init="")				# concatenate all the digits, with init = "" to handle the n=1 case

digits <- el(strsplit(Concat,""))				# split the string to get component digits

mat    <- matrix(digits,nchar(Concat),nchar(Concat))		# create a matrix with the digits running down the columns,

								# for example, n = 4 is [[1,1,1,1],[2,2,2,2],[3,3,3,3],[4,4,4,4]]

mat[upper.tri(mat)[(n%%4 condiditon),(n%%4 condition)]] <- " "	# conditioned on (n %% 4), replace the values in mat with spaces

write(mat,1,nchar(Concat),,"")					# print the transpose of mat

}

edited 7 hours ago

answered 7 hours ago

Giuseppe

17k31052

R, 175 172 bytes

function(n)write(`[<-`(m<-matrix(S<-el(strsplit(x<-Reduce(paste0,1:n,""),"")),y<-nchar(x),y),upper.tri(m)["if"(n%%4%in%2:3,1:y,y:1),"if"(n%%4%in%1:2,1:y,y:1)]," "),1,y,,"")

Try it online!

A horrible in-line mess!

-3 bytes by changing the rotation condition

Ungolfed and simplified a bit:

function(n){

Concat <- Reduce(paste0,1:n,init="")				# concatenate all the digits, with init = "" to handle the n=1 case

digits <- el(strsplit(Concat,""))				# split the string to get component digits

mat    <- matrix(digits,nchar(Concat),nchar(Concat))		# create a matrix with the digits running down the columns,

								# for example, n = 4 is [[1,1,1,1],[2,2,2,2],[3,3,3,3],[4,4,4,4]]

mat[upper.tri(mat)[(n%%4 condiditon),(n%%4 condition)]] <- " "	# conditioned on (n %% 4), replace the values in mat with spaces

write(mat,1,nchar(Concat),,"")					# print the transpose of mat

}

edited 7 hours ago

answered 7 hours ago

Giuseppe

17k31052

edited 7 hours ago

answered 7 hours ago

Giuseppe

17k31052

answered 7 hours ago

Giuseppe

17k31052

answered 7 hours ago

Giuseppe

17k31052

add a comment |

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.

…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.

…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.

…Avoid asking for help, clarification or responding to other answers (use comments instead).

draft saved

draft discarded

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.

…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.

…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.

…Avoid asking for help, clarification or responding to other answers (use comments instead).

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draft discarded

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This page is only for reference, If you need detailed information, please check here

The Digit Triangles

Challenge:

Challenge rules:

General rules:

Test cases:

Challenge:

Challenge rules:

General rules:

Test cases:

Challenge:

Challenge rules:

General rules:

Test cases:

Challenge:

Challenge rules:

General rules:

Test cases:

19 Answers 19

Japt, 8 bytes

JavaScript (ES6), 93 89 bytes

Commented

Shape summary

Python 2, 94 bytes

Canvas, 8 bytes

Perl 6, 94 bytes

Charcoal, 17 bytes

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

Ruby, 95 82 79 bytes

APL+WIN, 45 bytes

PowerShell, 108 bytes

Jelly, 12 bytes

05AB1E (legacy), 14 12 10 bytes

Stax, 10 bytes

PowerShell, 105 101 95 bytes

Wolfram Language (Mathematica), 137 bytes

Python 2, 116 bytes

Red, 155 bytes

perl 5, 117 bytes

R, 175 172 bytes

Your Answer

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19 Answers 19

19 Answers 19

Japt, 8 bytes

Japt, 8 bytes

Japt, 8 bytes

Japt, 8 bytes

JavaScript (ES6), 93 89 bytes

Commented

Shape summary

JavaScript (ES6), 93 89 bytes

Commented

Shape summary

JavaScript (ES6), 93 89 bytes

Commented

Shape summary

JavaScript (ES6), 93 89 bytes

Commented

Shape summary

Python 2, 94 bytes

Python 2, 94 bytes

Python 2, 94 bytes

Python 2, 94 bytes

Canvas, 8 bytes

Canvas, 8 bytes

Canvas, 8 bytes

Canvas, 8 bytes

Perl 6, 94 bytes

Perl 6, 94 bytes

Perl 6, 94 bytes

Perl 6, 94 bytes

Charcoal, 17 bytes

Charcoal, 17 bytes

Charcoal, 17 bytes

Charcoal, 17 bytes

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

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

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

19 Answers
19

19 Answers
19

19 Answers
19