Print a physical multiplication table












24












$begingroup$


Rectangles have this nice property - an $n times m$ rectangle consists of exactly $n times m$ characters!



A.. more interesting property is that the rectangles can be aligned nicely in a multiplication table - for example, a $3 times 3$ table:



# ## ###

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

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


Your challenge is to, given a number $n$ ($ n > 1$), output a formatted $n times n$ multiplication table.



Rules




  • You can take the input one above or below $n$

  • Default I/O rules apply

  • You can choose any non-whitespace character to represent the blocks; every other character (though newlines are special) is considered whitespace. The chosen character can be different for different inputs, but must be the same throughout the input

  • The result can have unneeded characters, as long as the table aligns up and there are no occurrences of the chosen character that aren't part of the required output

  • The separators must be 1 character wide/tall, and the rectangles must be packed (i.e. no separators between their characters)

  • The empty lines can be empty, padding isn't required

  • The result can be a string, matrix, vector of lines, array of character arrays, or anything 2Dish

  • You may alternatively output a matrix/vector-of-vectors/anything 2Dish of numbers, but the background & foreground must be 2 distinct numbers (which can vary input to input, but not throughout an output) and no other numbers can be present. Extra surrounding characters are allowed with this format too (though they must match the background number)

  • This is code-golf, shortest answer in bytes, per-language, wins!


Examples



For the input 2, a valid ascii-art output, with the character , is:



        ∙ ∙∙

Result: ∙ ∙∙.
∙ ∙∙


yes the period is there just to confuse you

Another valid answer as a number matrix, with 2 being the background number and 9 the foreground:



[[9,2,9,9,2,2],
[2,2,2,2,2,2],
[9,2,9,9,2,2],
[9,2,9,9,2,2]]


An invalid output example would be



#  # #


# # #

# # #


as the rectangles have separators in between them.



Example outputs for $4 times 4$:



# ## ### ####

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

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

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


1 0 1 1 0 1 1 1 0 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1









share|improve this question











$endgroup$












  • $begingroup$
    Can we have extra row/columns of background characters in front, rather than at the end of the table?
    $endgroup$
    – Kirill L.
    12 hours ago










  • $begingroup$
    @KirillL. sure, as long as the rows line up
    $endgroup$
    – dzaima
    12 hours ago
















24












$begingroup$


Rectangles have this nice property - an $n times m$ rectangle consists of exactly $n times m$ characters!



A.. more interesting property is that the rectangles can be aligned nicely in a multiplication table - for example, a $3 times 3$ table:



# ## ###

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

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


Your challenge is to, given a number $n$ ($ n > 1$), output a formatted $n times n$ multiplication table.



Rules




  • You can take the input one above or below $n$

  • Default I/O rules apply

  • You can choose any non-whitespace character to represent the blocks; every other character (though newlines are special) is considered whitespace. The chosen character can be different for different inputs, but must be the same throughout the input

  • The result can have unneeded characters, as long as the table aligns up and there are no occurrences of the chosen character that aren't part of the required output

  • The separators must be 1 character wide/tall, and the rectangles must be packed (i.e. no separators between their characters)

  • The empty lines can be empty, padding isn't required

  • The result can be a string, matrix, vector of lines, array of character arrays, or anything 2Dish

  • You may alternatively output a matrix/vector-of-vectors/anything 2Dish of numbers, but the background & foreground must be 2 distinct numbers (which can vary input to input, but not throughout an output) and no other numbers can be present. Extra surrounding characters are allowed with this format too (though they must match the background number)

  • This is code-golf, shortest answer in bytes, per-language, wins!


Examples



For the input 2, a valid ascii-art output, with the character , is:



        ∙ ∙∙

Result: ∙ ∙∙.
∙ ∙∙


yes the period is there just to confuse you

Another valid answer as a number matrix, with 2 being the background number and 9 the foreground:



[[9,2,9,9,2,2],
[2,2,2,2,2,2],
[9,2,9,9,2,2],
[9,2,9,9,2,2]]


An invalid output example would be



#  # #


# # #

# # #


as the rectangles have separators in between them.



Example outputs for $4 times 4$:



# ## ### ####

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

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

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


1 0 1 1 0 1 1 1 0 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1









share|improve this question











$endgroup$












  • $begingroup$
    Can we have extra row/columns of background characters in front, rather than at the end of the table?
    $endgroup$
    – Kirill L.
    12 hours ago










  • $begingroup$
    @KirillL. sure, as long as the rows line up
    $endgroup$
    – dzaima
    12 hours ago














24












24








24


2



$begingroup$


Rectangles have this nice property - an $n times m$ rectangle consists of exactly $n times m$ characters!



A.. more interesting property is that the rectangles can be aligned nicely in a multiplication table - for example, a $3 times 3$ table:



# ## ###

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

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


Your challenge is to, given a number $n$ ($ n > 1$), output a formatted $n times n$ multiplication table.



Rules




  • You can take the input one above or below $n$

  • Default I/O rules apply

  • You can choose any non-whitespace character to represent the blocks; every other character (though newlines are special) is considered whitespace. The chosen character can be different for different inputs, but must be the same throughout the input

  • The result can have unneeded characters, as long as the table aligns up and there are no occurrences of the chosen character that aren't part of the required output

  • The separators must be 1 character wide/tall, and the rectangles must be packed (i.e. no separators between their characters)

  • The empty lines can be empty, padding isn't required

  • The result can be a string, matrix, vector of lines, array of character arrays, or anything 2Dish

  • You may alternatively output a matrix/vector-of-vectors/anything 2Dish of numbers, but the background & foreground must be 2 distinct numbers (which can vary input to input, but not throughout an output) and no other numbers can be present. Extra surrounding characters are allowed with this format too (though they must match the background number)

  • This is code-golf, shortest answer in bytes, per-language, wins!


Examples



For the input 2, a valid ascii-art output, with the character , is:



        ∙ ∙∙

Result: ∙ ∙∙.
∙ ∙∙


yes the period is there just to confuse you

Another valid answer as a number matrix, with 2 being the background number and 9 the foreground:



[[9,2,9,9,2,2],
[2,2,2,2,2,2],
[9,2,9,9,2,2],
[9,2,9,9,2,2]]


An invalid output example would be



#  # #


# # #

# # #


as the rectangles have separators in between them.



Example outputs for $4 times 4$:



# ## ### ####

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

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

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


1 0 1 1 0 1 1 1 0 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1









share|improve this question











$endgroup$




Rectangles have this nice property - an $n times m$ rectangle consists of exactly $n times m$ characters!



A.. more interesting property is that the rectangles can be aligned nicely in a multiplication table - for example, a $3 times 3$ table:



# ## ###

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

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


Your challenge is to, given a number $n$ ($ n > 1$), output a formatted $n times n$ multiplication table.



Rules




  • You can take the input one above or below $n$

  • Default I/O rules apply

  • You can choose any non-whitespace character to represent the blocks; every other character (though newlines are special) is considered whitespace. The chosen character can be different for different inputs, but must be the same throughout the input

  • The result can have unneeded characters, as long as the table aligns up and there are no occurrences of the chosen character that aren't part of the required output

  • The separators must be 1 character wide/tall, and the rectangles must be packed (i.e. no separators between their characters)

  • The empty lines can be empty, padding isn't required

  • The result can be a string, matrix, vector of lines, array of character arrays, or anything 2Dish

  • You may alternatively output a matrix/vector-of-vectors/anything 2Dish of numbers, but the background & foreground must be 2 distinct numbers (which can vary input to input, but not throughout an output) and no other numbers can be present. Extra surrounding characters are allowed with this format too (though they must match the background number)

  • This is code-golf, shortest answer in bytes, per-language, wins!


Examples



For the input 2, a valid ascii-art output, with the character , is:



        ∙ ∙∙

Result: ∙ ∙∙.
∙ ∙∙


yes the period is there just to confuse you

Another valid answer as a number matrix, with 2 being the background number and 9 the foreground:



[[9,2,9,9,2,2],
[2,2,2,2,2,2],
[9,2,9,9,2,2],
[9,2,9,9,2,2]]


An invalid output example would be



#  # #


# # #

# # #


as the rectangles have separators in between them.



Example outputs for $4 times 4$:



# ## ### ####

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

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

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


1 0 1 1 0 1 1 1 0 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1
1 0 1 1 0 1 1 1 0 1 1 1 1






code-golf ascii-art array-manipulation






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 11 hours ago







dzaima

















asked 14 hours ago









dzaimadzaima

15.5k21857




15.5k21857












  • $begingroup$
    Can we have extra row/columns of background characters in front, rather than at the end of the table?
    $endgroup$
    – Kirill L.
    12 hours ago










  • $begingroup$
    @KirillL. sure, as long as the rows line up
    $endgroup$
    – dzaima
    12 hours ago


















  • $begingroup$
    Can we have extra row/columns of background characters in front, rather than at the end of the table?
    $endgroup$
    – Kirill L.
    12 hours ago










  • $begingroup$
    @KirillL. sure, as long as the rows line up
    $endgroup$
    – dzaima
    12 hours ago
















$begingroup$
Can we have extra row/columns of background characters in front, rather than at the end of the table?
$endgroup$
– Kirill L.
12 hours ago




$begingroup$
Can we have extra row/columns of background characters in front, rather than at the end of the table?
$endgroup$
– Kirill L.
12 hours ago












$begingroup$
@KirillL. sure, as long as the rows line up
$endgroup$
– dzaima
12 hours ago




$begingroup$
@KirillL. sure, as long as the rows line up
$endgroup$
– dzaima
12 hours ago










15 Answers
15






active

oldest

votes


















5












$begingroup$


R, 56 54 43 36 30 bytes





x=!!sequence(2:scan())-1;x%o%x


Try it online!



Takes input one above $n$ (so returns 3x3 matrix for $n = 4$).
Returns a matrix of $1s$ (foreground), and $0s$ (background) with an extra row/column of zeroes in front.



Thanks to digEmAll for -7 bytes.






share|improve this answer











$endgroup$













  • $begingroup$
    36 bytes using sequence (if I'm not missing something)
    $endgroup$
    – digEmAll
    12 hours ago










  • $begingroup$
    Thanks, BTW it can probably even be 30, if the extra blank row can be in front, rather than at the end.
    $endgroup$
    – Kirill L.
    12 hours ago












  • $begingroup$
    Oh, can they? I missed that !
    $endgroup$
    – digEmAll
    11 hours ago



















5












$begingroup$

MATL, 14 10 bytes



:"@:Fv]g&*


This answer uses 1 for the blocks and 0 for the background



Try it at MATL Online



Explanation



     % Implicitly grab the input as an integer, N
% STACK: { 3 }
: % Create an array from 1...N
% STACK: { [1, 2, 3] }
" % For each element M in this array
@: % Create an array from 1...M
% STACK (for 1st iteration): { [1] }
% STACK (for 2nd iteration): { [1; 0], [1, 2] }
% STACK (for 3rd iteration): { [1; 0; 1; 2; 0], [1, 2, 3] }
F % Push a zero to the stack
% STACK (for 1st iteration): { [1], 0 }
% STACK (for 2nd iteration): { [1; 0], [1, 2], 0 }
% STACK (for 3rd iteration): { [1; 0; 1; 2; 0], [1, 2, 3], 0 }
v % Vertically concatenate everything on the stack
% STACK (for 1st iteration): { [1; 0] }
% STACK (for 2nd iteration): { [1; 0; 1; 2; 0] }
% STACK (for 3rd iteration): { [1; 0; 1; 2; 0; 1; 2; 3; 0] }
]
g % Convert everything to be boolean (turns all non-zeros to 1)
% STACK: { [1; 0; 1; 1; 0; 1; 1; 1; 0] }
&* % Perform element-wise multiplication to expand this array out into the 2D grid
% Implicitly display the result





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$endgroup$





















    4












    $begingroup$


    Jelly, 8 bytes



    1ẋⱮj0ȧþ`


    Try it online!






    share|improve this answer









    $endgroup$





















      4












      $begingroup$


      Haskell, 49 bytes





      f n=map=<<flip(map.max)$[0^i|x<-[1..n],i<-[0..x]]


      Try it online!



      Generates a pattern like [1,0,1,0,0,1,0,0,0,...], then makes a 2D by taking the min of pairs. The pointfree weirdness saves 2 bytes over the more readable:





      f n|l<-do x<-[1..n];0:(1<$[1..x])=[[a*b|a<-l]|b<-l]


      Try it online!






      share|improve this answer











      $endgroup$









      • 1




        $begingroup$
        This can be shortened with my old triangular number trick: 43 bytes
        $endgroup$
        – Ørjan Johansen
        5 hours ago



















      4












      $begingroup$

      JavaScript (ES6),  73 72  69 bytes



      Returns a string made of 1's, spaces and line feeds.





      n=>(g=s=>n-->0?g(s+`${p+=1} `):s[~n]?(+s[~n]?s:'')+`
      `+g(s):'')(p='')


      Try it online!





      JavaScript (ES7),  87 83  82 bytes



      Saved 3 bytes thanks to @dzaima



      Returns a binary matrix, which is built cell by cell.





      n=>[...Array(n*(n+3)/2)].map((_,y,a)=>a.map(h=(z,x)=>(17+8*x)**.5%1&&(z||h(1,y))))


      Try it online!



      How?



      The width $w$ of the matrix is given by:



      $$w=T_n+n-1={n+1choose 2}+n-1=frac{n(n + 3)}{2}-1$$



      (NB: As allowed by the challenge rules, we output a matrix of width $w+1$ instead.)



      Similarly, the cell located at $(X,Y)$ is empty if the following quadratic admits an integer root for either $k=X$ or $k=Y$:



      $$frac{x(x+3)}{2}-1-k=0\
      x^2+3x-2-2k=0
      $$



      whose determinant is:



      $$Delta=9-4(-2-2k)=17+8k$$






      share|improve this answer











      $endgroup$













      • $begingroup$
        Not sure if it can save bytes, but the solution of that quadratic would be $frac{-3pmsqrt{17+8k}}2$, so there would be an integer root if $sqrtDelta$ is odd, that is, $Delta$ is an odd perfect square.
        $endgroup$
        – Erik the Outgolfer
        9 hours ago



















      3












      $begingroup$


      APL (Dyalog Unicode), 12 10 12 bytesSBCS





      ∘.×⍨∊,,⎕⍴1


      Try it online!



      Edit: -2 bytes from ngn. +2 bytes because the previous answers were invalid (with idea thanks to ngn and dzaima).



      Explanation



      ∘.×⍨∊,,⎕⍴1

      ⎕ Take input.
      ⍴1 An array of 1s of length our input. Example: 1 1 1
      0, Prepend a 0. Example: 0 1 1 1
      , Take all the prefixes and concatenate them. Example: 0 0 1 0 1 1 0 1 1 1
      ∊ Flatten the list. Example: 0 0 1 0 1 1 0 1 1 1
      ∘.×⍨ Turn the above list into a multiplication table of 0s and 1s
      by multiplying the list with itself.


      The output should look like:



      0 0 0 0 0 0 0 0 0 0
      0 0 0 0 0 0 0 0 0 0
      0 0 1 0 1 1 0 1 1 1
      0 0 0 0 0 0 0 0 0 0
      0 0 1 0 1 1 0 1 1 1
      0 0 1 0 1 1 0 1 1 1
      0 0 0 0 0 0 0 0 0 0
      0 0 1 0 1 1 0 1 1 1
      0 0 1 0 1 1 0 1 1 1
      0 0 1 0 1 1 0 1 1 1





      share|improve this answer











      $endgroup$









      • 1




        $begingroup$
        you could take input as to avoid the { }
        $endgroup$
        – ngn
        10 hours ago



















      3












      $begingroup$


      Wolfram Language (Mathematica), 41 bytes



      a#&/@(a=Flatten@Array[{0,1~Table~#}&,#])&


      Try it online!






      share|improve this answer











      $endgroup$





















        2












        $begingroup$


        Python 2, 67 bytes





        s='';n=input()
        while n:s='#'*n+' '+s;n-=1
        for c in s:print(c>' ')*s


        Try it online!



        Prints blank lines for empty lines, which the challenge allows.



        Same length (with input one above n):





        r=range(input())
        for n in r:print(' '.join(i*'#'for i in r)+'n')*n


        Try it online!






        share|improve this answer









        $endgroup$





















          2












          $begingroup$


          J, 17 bytes



          [:*/~2=/[:I.2+i.


          Try it online!






          share|improve this answer









          $endgroup$













          • $begingroup$
            did not know about this I. trick!
            $endgroup$
            – Jonah
            2 hours ago










          • $begingroup$
            @Jonah I was waiting for an opportunity to use it :)
            $endgroup$
            – Galen Ivanov
            26 mins ago



















          1












          $begingroup$

          Haskell, 69 68 bytes



          (a#b)0=
          (a#b)n=(a#b)(n-1)++b:(a<$[1..n])
          f n=((1#0)n#(0<$(1#0)n))n


          Returns a matrix of numbers.



          Try it online!



          Variants of f with the same byte count:



          f n=((#)<*>(0<$)$(1#0)n)n
          f n|l<-(1#0)n=(l#(0<$l))n





          share|improve this answer











          $endgroup$













          • $begingroup$
            Do the 0 row and column help?
            $endgroup$
            – dfeuer
            7 hours ago






          • 1




            $begingroup$
            @dfeuer: yes, they save a byte. See the first version of my answer.
            $endgroup$
            – nimi
            6 hours ago



















          1












          $begingroup$

          APL+WIN, 29 bytes



          m/⍉(m←¯1↓∊(⍳n),¨¯1)/(n,n←⎕)⍴1


          Explanation:



          (n,n←⎕)⍴1 prompt for integer n and create a nxn matrix of 1s

          (m←¯1↓∊(⍳n) replicate the columns by 1,2,.....n and insert 0s between each replication

          m/⍉ repeat replication and 0 insertion for the rows from above


          Example:



          ⎕:
          3
          1 0 1 1 0 1 1 1
          0 0 0 0 0 0 0 0
          1 0 1 1 0 1 1 1
          1 0 1 1 0 1 1 1
          0 0 0 0 0 0 0 0
          1 0 1 1 0 1 1 1
          1 0 1 1 0 1 1 1
          1 0 1 1 0 1 1 1





          share|improve this answer











          $endgroup$





















            1












            $begingroup$


            Jelly, 7 bytes



            ‘RÄṬ|þ`


            Try it online!



            Outputs a digit matrix, using $0$ for the rectangles and $1$ for the padding between them. The TIO link contains a footer which formats a digit matrix in a human-readable way by lining up the rows and columns.



            Explanation



            ‘RÄṬ|þ`
            R Take a range from 1 to
            ‘ {the input} plus 1
            Ä Cumulative sum; produces the first {input}+1 triangular numbers
            Ṭ Produce an array with 1s at those indexes, 0s at other indexes
            þ Create a table of {the array}
            ` with itself
            | using bitwise OR


            The number at cell $(x,y)$ of the resulting table will be $1$ if either $x$ or $y$ is a triangular number, or $0$ otherwise (because bitwise OR works like logical OR on 0 and 1). (We use R, range from 1, because Jelly uses 1-based indexing, so we don't have to worry about column 0 being incorrectly full of 0s; we have to add 1 to the input because the array produced by stops at the largest element given in the input, so we need to draw a line on the right-hand side and the bottom.) The gaps between triangular numbers are the consecutive integers, so the rectangular blocks formed by the gaps between the lines end up as the sizes requested by the question; and the use of an OR operation (in this case, bitwise) allows the lines to cross each other correctly.






            share|improve this answer











            $endgroup$













            • $begingroup$
              Why is this a community wiki?! If you want to waive rep you could just give it to Erik the Outgolfer
              $endgroup$
              – Jonathan Allan
              6 hours ago












            • $begingroup$
              I CW all my answers (unless I think they might get a bounty, in which case I use a temporary account for them). Aiming for a high reputation normally means aiming to make the site worse (I once repcapped every day for a week to prove that it was possible; it wasn't particularly difficult, and yet it involved a lot of shallow questions/answers that didn't really contribute to the site much). Additionally, gaining reputation is mostly a negative on the account because it makes the site nag you into doing moderation work; and gaining privileges increases the risk of accidentally gaining badges.
              $endgroup$
              – ais523
              6 hours ago










            • $begingroup$
              Also, I mostly disagree with the concept of ownership of posts on SE (although mostly with questions rather than answers, but you can't CW a question without moderator help). A CW marker very clearly says "if something is wrong here, feel free to edit it"; so I'd be applying a CW marker to everything even if it didn't waive reputation. SE is meant to be part wiki, after all, but people don't always use it as that.
              $endgroup$
              – ais523
              6 hours ago










            • $begingroup$
              RE "aiming for a high reputation" if you don't want that then just post when you think it is meaningful and avoid what you term "shallow" q&a. Either you feel this answer does add something to the site, in which case post it without CW, or you think it is "shallow", in which case just don't post it. RE "I mostly disagree with the concept of ownership of posts on SE" - well you're simply on the wrong site then.
              $endgroup$
              – Jonathan Allan
              5 hours ago












            • $begingroup$
              I think the answer adds something to the site, but if it's not CW, I feel compelled to post in a way that would gain me reputation rather than posting what I think would be interesting; back when I was a 20k user, I ended up really hating the site and almost got turned away from code golf in general because of it, so I deleted my account as a result. When I returned, I used to delete my account with every answer I posted (creating a new one for the next answer), but someone else pointed out that CWing every answer would have a similar effect, so nowadays I do that instead.
              $endgroup$
              – ais523
              5 hours ago



















            0












            $begingroup$


            Charcoal, 18 bytes



            ≔⪫EN×#⊕ι θEθ⭆θ⌊⟦ιλ


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



               N                Input number
            E Map over implicit range
            ι Current value
            ⊕ Incremented
            × Repetitions of
            # Literal `#`
            ⪫ Join with spaces
            ≔ θ Assign to variable
            θ Retrieve variable
            E Map over characters
            θ Retrieve variable
            ⭆ Replace characters with
            ⌊ Minimum of
            ⟦ List of
            ι Row character
            λ Column character
            Implicitly print each row on its own line





            share|improve this answer









            $endgroup$





















              0












              $begingroup$


              Perl 6, 35 bytes





              {[~](("
              "~[~] *xx$_)xx$_)>>.say}


              Try it online!



              Anonymous code block that takes a number and prints the multiplication table with *s and a leading newline.






              share|improve this answer









              $endgroup$





















                0












                $begingroup$


                Mouse-2002, 79 bytes



                Abusing Mouse's macros to repeat functionality.



                ?1+n:#P,n,j,k,b#P,j,y,k,#P,n,i,' ,#P,i,x,35;;;;$P0 2%:(1%.2%.-^4%3%!'2%.1+2%:)@


                Try it online!






                share|improve this answer









                $endgroup$













                  Your Answer





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                  15 Answers
                  15






                  active

                  oldest

                  votes








                  15 Answers
                  15






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  5












                  $begingroup$


                  R, 56 54 43 36 30 bytes





                  x=!!sequence(2:scan())-1;x%o%x


                  Try it online!



                  Takes input one above $n$ (so returns 3x3 matrix for $n = 4$).
                  Returns a matrix of $1s$ (foreground), and $0s$ (background) with an extra row/column of zeroes in front.



                  Thanks to digEmAll for -7 bytes.






                  share|improve this answer











                  $endgroup$













                  • $begingroup$
                    36 bytes using sequence (if I'm not missing something)
                    $endgroup$
                    – digEmAll
                    12 hours ago










                  • $begingroup$
                    Thanks, BTW it can probably even be 30, if the extra blank row can be in front, rather than at the end.
                    $endgroup$
                    – Kirill L.
                    12 hours ago












                  • $begingroup$
                    Oh, can they? I missed that !
                    $endgroup$
                    – digEmAll
                    11 hours ago
















                  5












                  $begingroup$


                  R, 56 54 43 36 30 bytes





                  x=!!sequence(2:scan())-1;x%o%x


                  Try it online!



                  Takes input one above $n$ (so returns 3x3 matrix for $n = 4$).
                  Returns a matrix of $1s$ (foreground), and $0s$ (background) with an extra row/column of zeroes in front.



                  Thanks to digEmAll for -7 bytes.






                  share|improve this answer











                  $endgroup$













                  • $begingroup$
                    36 bytes using sequence (if I'm not missing something)
                    $endgroup$
                    – digEmAll
                    12 hours ago










                  • $begingroup$
                    Thanks, BTW it can probably even be 30, if the extra blank row can be in front, rather than at the end.
                    $endgroup$
                    – Kirill L.
                    12 hours ago












                  • $begingroup$
                    Oh, can they? I missed that !
                    $endgroup$
                    – digEmAll
                    11 hours ago














                  5












                  5








                  5





                  $begingroup$


                  R, 56 54 43 36 30 bytes





                  x=!!sequence(2:scan())-1;x%o%x


                  Try it online!



                  Takes input one above $n$ (so returns 3x3 matrix for $n = 4$).
                  Returns a matrix of $1s$ (foreground), and $0s$ (background) with an extra row/column of zeroes in front.



                  Thanks to digEmAll for -7 bytes.






                  share|improve this answer











                  $endgroup$




                  R, 56 54 43 36 30 bytes





                  x=!!sequence(2:scan())-1;x%o%x


                  Try it online!



                  Takes input one above $n$ (so returns 3x3 matrix for $n = 4$).
                  Returns a matrix of $1s$ (foreground), and $0s$ (background) with an extra row/column of zeroes in front.



                  Thanks to digEmAll for -7 bytes.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 12 hours ago

























                  answered 12 hours ago









                  Kirill L.Kirill L.

                  5,4731525




                  5,4731525












                  • $begingroup$
                    36 bytes using sequence (if I'm not missing something)
                    $endgroup$
                    – digEmAll
                    12 hours ago










                  • $begingroup$
                    Thanks, BTW it can probably even be 30, if the extra blank row can be in front, rather than at the end.
                    $endgroup$
                    – Kirill L.
                    12 hours ago












                  • $begingroup$
                    Oh, can they? I missed that !
                    $endgroup$
                    – digEmAll
                    11 hours ago


















                  • $begingroup$
                    36 bytes using sequence (if I'm not missing something)
                    $endgroup$
                    – digEmAll
                    12 hours ago










                  • $begingroup$
                    Thanks, BTW it can probably even be 30, if the extra blank row can be in front, rather than at the end.
                    $endgroup$
                    – Kirill L.
                    12 hours ago












                  • $begingroup$
                    Oh, can they? I missed that !
                    $endgroup$
                    – digEmAll
                    11 hours ago
















                  $begingroup$
                  36 bytes using sequence (if I'm not missing something)
                  $endgroup$
                  – digEmAll
                  12 hours ago




                  $begingroup$
                  36 bytes using sequence (if I'm not missing something)
                  $endgroup$
                  – digEmAll
                  12 hours ago












                  $begingroup$
                  Thanks, BTW it can probably even be 30, if the extra blank row can be in front, rather than at the end.
                  $endgroup$
                  – Kirill L.
                  12 hours ago






                  $begingroup$
                  Thanks, BTW it can probably even be 30, if the extra blank row can be in front, rather than at the end.
                  $endgroup$
                  – Kirill L.
                  12 hours ago














                  $begingroup$
                  Oh, can they? I missed that !
                  $endgroup$
                  – digEmAll
                  11 hours ago




                  $begingroup$
                  Oh, can they? I missed that !
                  $endgroup$
                  – digEmAll
                  11 hours ago











                  5












                  $begingroup$

                  MATL, 14 10 bytes



                  :"@:Fv]g&*


                  This answer uses 1 for the blocks and 0 for the background



                  Try it at MATL Online



                  Explanation



                       % Implicitly grab the input as an integer, N
                  % STACK: { 3 }
                  : % Create an array from 1...N
                  % STACK: { [1, 2, 3] }
                  " % For each element M in this array
                  @: % Create an array from 1...M
                  % STACK (for 1st iteration): { [1] }
                  % STACK (for 2nd iteration): { [1; 0], [1, 2] }
                  % STACK (for 3rd iteration): { [1; 0; 1; 2; 0], [1, 2, 3] }
                  F % Push a zero to the stack
                  % STACK (for 1st iteration): { [1], 0 }
                  % STACK (for 2nd iteration): { [1; 0], [1, 2], 0 }
                  % STACK (for 3rd iteration): { [1; 0; 1; 2; 0], [1, 2, 3], 0 }
                  v % Vertically concatenate everything on the stack
                  % STACK (for 1st iteration): { [1; 0] }
                  % STACK (for 2nd iteration): { [1; 0; 1; 2; 0] }
                  % STACK (for 3rd iteration): { [1; 0; 1; 2; 0; 1; 2; 3; 0] }
                  ]
                  g % Convert everything to be boolean (turns all non-zeros to 1)
                  % STACK: { [1; 0; 1; 1; 0; 1; 1; 1; 0] }
                  &* % Perform element-wise multiplication to expand this array out into the 2D grid
                  % Implicitly display the result





                  share|improve this answer











                  $endgroup$


















                    5












                    $begingroup$

                    MATL, 14 10 bytes



                    :"@:Fv]g&*


                    This answer uses 1 for the blocks and 0 for the background



                    Try it at MATL Online



                    Explanation



                         % Implicitly grab the input as an integer, N
                    % STACK: { 3 }
                    : % Create an array from 1...N
                    % STACK: { [1, 2, 3] }
                    " % For each element M in this array
                    @: % Create an array from 1...M
                    % STACK (for 1st iteration): { [1] }
                    % STACK (for 2nd iteration): { [1; 0], [1, 2] }
                    % STACK (for 3rd iteration): { [1; 0; 1; 2; 0], [1, 2, 3] }
                    F % Push a zero to the stack
                    % STACK (for 1st iteration): { [1], 0 }
                    % STACK (for 2nd iteration): { [1; 0], [1, 2], 0 }
                    % STACK (for 3rd iteration): { [1; 0; 1; 2; 0], [1, 2, 3], 0 }
                    v % Vertically concatenate everything on the stack
                    % STACK (for 1st iteration): { [1; 0] }
                    % STACK (for 2nd iteration): { [1; 0; 1; 2; 0] }
                    % STACK (for 3rd iteration): { [1; 0; 1; 2; 0; 1; 2; 3; 0] }
                    ]
                    g % Convert everything to be boolean (turns all non-zeros to 1)
                    % STACK: { [1; 0; 1; 1; 0; 1; 1; 1; 0] }
                    &* % Perform element-wise multiplication to expand this array out into the 2D grid
                    % Implicitly display the result





                    share|improve this answer











                    $endgroup$
















                      5












                      5








                      5





                      $begingroup$

                      MATL, 14 10 bytes



                      :"@:Fv]g&*


                      This answer uses 1 for the blocks and 0 for the background



                      Try it at MATL Online



                      Explanation



                           % Implicitly grab the input as an integer, N
                      % STACK: { 3 }
                      : % Create an array from 1...N
                      % STACK: { [1, 2, 3] }
                      " % For each element M in this array
                      @: % Create an array from 1...M
                      % STACK (for 1st iteration): { [1] }
                      % STACK (for 2nd iteration): { [1; 0], [1, 2] }
                      % STACK (for 3rd iteration): { [1; 0; 1; 2; 0], [1, 2, 3] }
                      F % Push a zero to the stack
                      % STACK (for 1st iteration): { [1], 0 }
                      % STACK (for 2nd iteration): { [1; 0], [1, 2], 0 }
                      % STACK (for 3rd iteration): { [1; 0; 1; 2; 0], [1, 2, 3], 0 }
                      v % Vertically concatenate everything on the stack
                      % STACK (for 1st iteration): { [1; 0] }
                      % STACK (for 2nd iteration): { [1; 0; 1; 2; 0] }
                      % STACK (for 3rd iteration): { [1; 0; 1; 2; 0; 1; 2; 3; 0] }
                      ]
                      g % Convert everything to be boolean (turns all non-zeros to 1)
                      % STACK: { [1; 0; 1; 1; 0; 1; 1; 1; 0] }
                      &* % Perform element-wise multiplication to expand this array out into the 2D grid
                      % Implicitly display the result





                      share|improve this answer











                      $endgroup$



                      MATL, 14 10 bytes



                      :"@:Fv]g&*


                      This answer uses 1 for the blocks and 0 for the background



                      Try it at MATL Online



                      Explanation



                           % Implicitly grab the input as an integer, N
                      % STACK: { 3 }
                      : % Create an array from 1...N
                      % STACK: { [1, 2, 3] }
                      " % For each element M in this array
                      @: % Create an array from 1...M
                      % STACK (for 1st iteration): { [1] }
                      % STACK (for 2nd iteration): { [1; 0], [1, 2] }
                      % STACK (for 3rd iteration): { [1; 0; 1; 2; 0], [1, 2, 3] }
                      F % Push a zero to the stack
                      % STACK (for 1st iteration): { [1], 0 }
                      % STACK (for 2nd iteration): { [1; 0], [1, 2], 0 }
                      % STACK (for 3rd iteration): { [1; 0; 1; 2; 0], [1, 2, 3], 0 }
                      v % Vertically concatenate everything on the stack
                      % STACK (for 1st iteration): { [1; 0] }
                      % STACK (for 2nd iteration): { [1; 0; 1; 2; 0] }
                      % STACK (for 3rd iteration): { [1; 0; 1; 2; 0; 1; 2; 3; 0] }
                      ]
                      g % Convert everything to be boolean (turns all non-zeros to 1)
                      % STACK: { [1; 0; 1; 1; 0; 1; 1; 1; 0] }
                      &* % Perform element-wise multiplication to expand this array out into the 2D grid
                      % Implicitly display the result






                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited 11 hours ago

























                      answered 13 hours ago









                      SueverSuever

                      9,7721446




                      9,7721446























                          4












                          $begingroup$


                          Jelly, 8 bytes



                          1ẋⱮj0ȧþ`


                          Try it online!






                          share|improve this answer









                          $endgroup$


















                            4












                            $begingroup$


                            Jelly, 8 bytes



                            1ẋⱮj0ȧþ`


                            Try it online!






                            share|improve this answer









                            $endgroup$
















                              4












                              4








                              4





                              $begingroup$


                              Jelly, 8 bytes



                              1ẋⱮj0ȧþ`


                              Try it online!






                              share|improve this answer









                              $endgroup$




                              Jelly, 8 bytes



                              1ẋⱮj0ȧþ`


                              Try it online!







                              share|improve this answer












                              share|improve this answer



                              share|improve this answer










                              answered 14 hours ago









                              Erik the OutgolferErik the Outgolfer

                              32.5k429105




                              32.5k429105























                                  4












                                  $begingroup$


                                  Haskell, 49 bytes





                                  f n=map=<<flip(map.max)$[0^i|x<-[1..n],i<-[0..x]]


                                  Try it online!



                                  Generates a pattern like [1,0,1,0,0,1,0,0,0,...], then makes a 2D by taking the min of pairs. The pointfree weirdness saves 2 bytes over the more readable:





                                  f n|l<-do x<-[1..n];0:(1<$[1..x])=[[a*b|a<-l]|b<-l]


                                  Try it online!






                                  share|improve this answer











                                  $endgroup$









                                  • 1




                                    $begingroup$
                                    This can be shortened with my old triangular number trick: 43 bytes
                                    $endgroup$
                                    – Ørjan Johansen
                                    5 hours ago
















                                  4












                                  $begingroup$


                                  Haskell, 49 bytes





                                  f n=map=<<flip(map.max)$[0^i|x<-[1..n],i<-[0..x]]


                                  Try it online!



                                  Generates a pattern like [1,0,1,0,0,1,0,0,0,...], then makes a 2D by taking the min of pairs. The pointfree weirdness saves 2 bytes over the more readable:





                                  f n|l<-do x<-[1..n];0:(1<$[1..x])=[[a*b|a<-l]|b<-l]


                                  Try it online!






                                  share|improve this answer











                                  $endgroup$









                                  • 1




                                    $begingroup$
                                    This can be shortened with my old triangular number trick: 43 bytes
                                    $endgroup$
                                    – Ørjan Johansen
                                    5 hours ago














                                  4












                                  4








                                  4





                                  $begingroup$


                                  Haskell, 49 bytes





                                  f n=map=<<flip(map.max)$[0^i|x<-[1..n],i<-[0..x]]


                                  Try it online!



                                  Generates a pattern like [1,0,1,0,0,1,0,0,0,...], then makes a 2D by taking the min of pairs. The pointfree weirdness saves 2 bytes over the more readable:





                                  f n|l<-do x<-[1..n];0:(1<$[1..x])=[[a*b|a<-l]|b<-l]


                                  Try it online!






                                  share|improve this answer











                                  $endgroup$




                                  Haskell, 49 bytes





                                  f n=map=<<flip(map.max)$[0^i|x<-[1..n],i<-[0..x]]


                                  Try it online!



                                  Generates a pattern like [1,0,1,0,0,1,0,0,0,...], then makes a 2D by taking the min of pairs. The pointfree weirdness saves 2 bytes over the more readable:





                                  f n|l<-do x<-[1..n];0:(1<$[1..x])=[[a*b|a<-l]|b<-l]


                                  Try it online!







                                  share|improve this answer














                                  share|improve this answer



                                  share|improve this answer








                                  edited 11 hours ago

























                                  answered 11 hours ago









                                  xnorxnor

                                  92.3k18188446




                                  92.3k18188446








                                  • 1




                                    $begingroup$
                                    This can be shortened with my old triangular number trick: 43 bytes
                                    $endgroup$
                                    – Ørjan Johansen
                                    5 hours ago














                                  • 1




                                    $begingroup$
                                    This can be shortened with my old triangular number trick: 43 bytes
                                    $endgroup$
                                    – Ørjan Johansen
                                    5 hours ago








                                  1




                                  1




                                  $begingroup$
                                  This can be shortened with my old triangular number trick: 43 bytes
                                  $endgroup$
                                  – Ørjan Johansen
                                  5 hours ago




                                  $begingroup$
                                  This can be shortened with my old triangular number trick: 43 bytes
                                  $endgroup$
                                  – Ørjan Johansen
                                  5 hours ago











                                  4












                                  $begingroup$

                                  JavaScript (ES6),  73 72  69 bytes



                                  Returns a string made of 1's, spaces and line feeds.





                                  n=>(g=s=>n-->0?g(s+`${p+=1} `):s[~n]?(+s[~n]?s:'')+`
                                  `+g(s):'')(p='')


                                  Try it online!





                                  JavaScript (ES7),  87 83  82 bytes



                                  Saved 3 bytes thanks to @dzaima



                                  Returns a binary matrix, which is built cell by cell.





                                  n=>[...Array(n*(n+3)/2)].map((_,y,a)=>a.map(h=(z,x)=>(17+8*x)**.5%1&&(z||h(1,y))))


                                  Try it online!



                                  How?



                                  The width $w$ of the matrix is given by:



                                  $$w=T_n+n-1={n+1choose 2}+n-1=frac{n(n + 3)}{2}-1$$



                                  (NB: As allowed by the challenge rules, we output a matrix of width $w+1$ instead.)



                                  Similarly, the cell located at $(X,Y)$ is empty if the following quadratic admits an integer root for either $k=X$ or $k=Y$:



                                  $$frac{x(x+3)}{2}-1-k=0\
                                  x^2+3x-2-2k=0
                                  $$



                                  whose determinant is:



                                  $$Delta=9-4(-2-2k)=17+8k$$






                                  share|improve this answer











                                  $endgroup$













                                  • $begingroup$
                                    Not sure if it can save bytes, but the solution of that quadratic would be $frac{-3pmsqrt{17+8k}}2$, so there would be an integer root if $sqrtDelta$ is odd, that is, $Delta$ is an odd perfect square.
                                    $endgroup$
                                    – Erik the Outgolfer
                                    9 hours ago
















                                  4












                                  $begingroup$

                                  JavaScript (ES6),  73 72  69 bytes



                                  Returns a string made of 1's, spaces and line feeds.





                                  n=>(g=s=>n-->0?g(s+`${p+=1} `):s[~n]?(+s[~n]?s:'')+`
                                  `+g(s):'')(p='')


                                  Try it online!





                                  JavaScript (ES7),  87 83  82 bytes



                                  Saved 3 bytes thanks to @dzaima



                                  Returns a binary matrix, which is built cell by cell.





                                  n=>[...Array(n*(n+3)/2)].map((_,y,a)=>a.map(h=(z,x)=>(17+8*x)**.5%1&&(z||h(1,y))))


                                  Try it online!



                                  How?



                                  The width $w$ of the matrix is given by:



                                  $$w=T_n+n-1={n+1choose 2}+n-1=frac{n(n + 3)}{2}-1$$



                                  (NB: As allowed by the challenge rules, we output a matrix of width $w+1$ instead.)



                                  Similarly, the cell located at $(X,Y)$ is empty if the following quadratic admits an integer root for either $k=X$ or $k=Y$:



                                  $$frac{x(x+3)}{2}-1-k=0\
                                  x^2+3x-2-2k=0
                                  $$



                                  whose determinant is:



                                  $$Delta=9-4(-2-2k)=17+8k$$






                                  share|improve this answer











                                  $endgroup$













                                  • $begingroup$
                                    Not sure if it can save bytes, but the solution of that quadratic would be $frac{-3pmsqrt{17+8k}}2$, so there would be an integer root if $sqrtDelta$ is odd, that is, $Delta$ is an odd perfect square.
                                    $endgroup$
                                    – Erik the Outgolfer
                                    9 hours ago














                                  4












                                  4








                                  4





                                  $begingroup$

                                  JavaScript (ES6),  73 72  69 bytes



                                  Returns a string made of 1's, spaces and line feeds.





                                  n=>(g=s=>n-->0?g(s+`${p+=1} `):s[~n]?(+s[~n]?s:'')+`
                                  `+g(s):'')(p='')


                                  Try it online!





                                  JavaScript (ES7),  87 83  82 bytes



                                  Saved 3 bytes thanks to @dzaima



                                  Returns a binary matrix, which is built cell by cell.





                                  n=>[...Array(n*(n+3)/2)].map((_,y,a)=>a.map(h=(z,x)=>(17+8*x)**.5%1&&(z||h(1,y))))


                                  Try it online!



                                  How?



                                  The width $w$ of the matrix is given by:



                                  $$w=T_n+n-1={n+1choose 2}+n-1=frac{n(n + 3)}{2}-1$$



                                  (NB: As allowed by the challenge rules, we output a matrix of width $w+1$ instead.)



                                  Similarly, the cell located at $(X,Y)$ is empty if the following quadratic admits an integer root for either $k=X$ or $k=Y$:



                                  $$frac{x(x+3)}{2}-1-k=0\
                                  x^2+3x-2-2k=0
                                  $$



                                  whose determinant is:



                                  $$Delta=9-4(-2-2k)=17+8k$$






                                  share|improve this answer











                                  $endgroup$



                                  JavaScript (ES6),  73 72  69 bytes



                                  Returns a string made of 1's, spaces and line feeds.





                                  n=>(g=s=>n-->0?g(s+`${p+=1} `):s[~n]?(+s[~n]?s:'')+`
                                  `+g(s):'')(p='')


                                  Try it online!





                                  JavaScript (ES7),  87 83  82 bytes



                                  Saved 3 bytes thanks to @dzaima



                                  Returns a binary matrix, which is built cell by cell.





                                  n=>[...Array(n*(n+3)/2)].map((_,y,a)=>a.map(h=(z,x)=>(17+8*x)**.5%1&&(z||h(1,y))))


                                  Try it online!



                                  How?



                                  The width $w$ of the matrix is given by:



                                  $$w=T_n+n-1={n+1choose 2}+n-1=frac{n(n + 3)}{2}-1$$



                                  (NB: As allowed by the challenge rules, we output a matrix of width $w+1$ instead.)



                                  Similarly, the cell located at $(X,Y)$ is empty if the following quadratic admits an integer root for either $k=X$ or $k=Y$:



                                  $$frac{x(x+3)}{2}-1-k=0\
                                  x^2+3x-2-2k=0
                                  $$



                                  whose determinant is:



                                  $$Delta=9-4(-2-2k)=17+8k$$







                                  share|improve this answer














                                  share|improve this answer



                                  share|improve this answer








                                  edited 10 hours ago

























                                  answered 13 hours ago









                                  ArnauldArnauld

                                  78.8k795327




                                  78.8k795327












                                  • $begingroup$
                                    Not sure if it can save bytes, but the solution of that quadratic would be $frac{-3pmsqrt{17+8k}}2$, so there would be an integer root if $sqrtDelta$ is odd, that is, $Delta$ is an odd perfect square.
                                    $endgroup$
                                    – Erik the Outgolfer
                                    9 hours ago


















                                  • $begingroup$
                                    Not sure if it can save bytes, but the solution of that quadratic would be $frac{-3pmsqrt{17+8k}}2$, so there would be an integer root if $sqrtDelta$ is odd, that is, $Delta$ is an odd perfect square.
                                    $endgroup$
                                    – Erik the Outgolfer
                                    9 hours ago
















                                  $begingroup$
                                  Not sure if it can save bytes, but the solution of that quadratic would be $frac{-3pmsqrt{17+8k}}2$, so there would be an integer root if $sqrtDelta$ is odd, that is, $Delta$ is an odd perfect square.
                                  $endgroup$
                                  – Erik the Outgolfer
                                  9 hours ago




                                  $begingroup$
                                  Not sure if it can save bytes, but the solution of that quadratic would be $frac{-3pmsqrt{17+8k}}2$, so there would be an integer root if $sqrtDelta$ is odd, that is, $Delta$ is an odd perfect square.
                                  $endgroup$
                                  – Erik the Outgolfer
                                  9 hours ago











                                  3












                                  $begingroup$


                                  APL (Dyalog Unicode), 12 10 12 bytesSBCS





                                  ∘.×⍨∊,,⎕⍴1


                                  Try it online!



                                  Edit: -2 bytes from ngn. +2 bytes because the previous answers were invalid (with idea thanks to ngn and dzaima).



                                  Explanation



                                  ∘.×⍨∊,,⎕⍴1

                                  ⎕ Take input.
                                  ⍴1 An array of 1s of length our input. Example: 1 1 1
                                  0, Prepend a 0. Example: 0 1 1 1
                                  , Take all the prefixes and concatenate them. Example: 0 0 1 0 1 1 0 1 1 1
                                  ∊ Flatten the list. Example: 0 0 1 0 1 1 0 1 1 1
                                  ∘.×⍨ Turn the above list into a multiplication table of 0s and 1s
                                  by multiplying the list with itself.


                                  The output should look like:



                                  0 0 0 0 0 0 0 0 0 0
                                  0 0 0 0 0 0 0 0 0 0
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 0 0 0 0 0 0 0 0
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 0 0 0 0 0 0 0 0
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 1 0 1 1 0 1 1 1





                                  share|improve this answer











                                  $endgroup$









                                  • 1




                                    $begingroup$
                                    you could take input as to avoid the { }
                                    $endgroup$
                                    – ngn
                                    10 hours ago
















                                  3












                                  $begingroup$


                                  APL (Dyalog Unicode), 12 10 12 bytesSBCS





                                  ∘.×⍨∊,,⎕⍴1


                                  Try it online!



                                  Edit: -2 bytes from ngn. +2 bytes because the previous answers were invalid (with idea thanks to ngn and dzaima).



                                  Explanation



                                  ∘.×⍨∊,,⎕⍴1

                                  ⎕ Take input.
                                  ⍴1 An array of 1s of length our input. Example: 1 1 1
                                  0, Prepend a 0. Example: 0 1 1 1
                                  , Take all the prefixes and concatenate them. Example: 0 0 1 0 1 1 0 1 1 1
                                  ∊ Flatten the list. Example: 0 0 1 0 1 1 0 1 1 1
                                  ∘.×⍨ Turn the above list into a multiplication table of 0s and 1s
                                  by multiplying the list with itself.


                                  The output should look like:



                                  0 0 0 0 0 0 0 0 0 0
                                  0 0 0 0 0 0 0 0 0 0
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 0 0 0 0 0 0 0 0
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 0 0 0 0 0 0 0 0
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 1 0 1 1 0 1 1 1





                                  share|improve this answer











                                  $endgroup$









                                  • 1




                                    $begingroup$
                                    you could take input as to avoid the { }
                                    $endgroup$
                                    – ngn
                                    10 hours ago














                                  3












                                  3








                                  3





                                  $begingroup$


                                  APL (Dyalog Unicode), 12 10 12 bytesSBCS





                                  ∘.×⍨∊,,⎕⍴1


                                  Try it online!



                                  Edit: -2 bytes from ngn. +2 bytes because the previous answers were invalid (with idea thanks to ngn and dzaima).



                                  Explanation



                                  ∘.×⍨∊,,⎕⍴1

                                  ⎕ Take input.
                                  ⍴1 An array of 1s of length our input. Example: 1 1 1
                                  0, Prepend a 0. Example: 0 1 1 1
                                  , Take all the prefixes and concatenate them. Example: 0 0 1 0 1 1 0 1 1 1
                                  ∊ Flatten the list. Example: 0 0 1 0 1 1 0 1 1 1
                                  ∘.×⍨ Turn the above list into a multiplication table of 0s and 1s
                                  by multiplying the list with itself.


                                  The output should look like:



                                  0 0 0 0 0 0 0 0 0 0
                                  0 0 0 0 0 0 0 0 0 0
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 0 0 0 0 0 0 0 0
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 0 0 0 0 0 0 0 0
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 1 0 1 1 0 1 1 1





                                  share|improve this answer











                                  $endgroup$




                                  APL (Dyalog Unicode), 12 10 12 bytesSBCS





                                  ∘.×⍨∊,,⎕⍴1


                                  Try it online!



                                  Edit: -2 bytes from ngn. +2 bytes because the previous answers were invalid (with idea thanks to ngn and dzaima).



                                  Explanation



                                  ∘.×⍨∊,,⎕⍴1

                                  ⎕ Take input.
                                  ⍴1 An array of 1s of length our input. Example: 1 1 1
                                  0, Prepend a 0. Example: 0 1 1 1
                                  , Take all the prefixes and concatenate them. Example: 0 0 1 0 1 1 0 1 1 1
                                  ∊ Flatten the list. Example: 0 0 1 0 1 1 0 1 1 1
                                  ∘.×⍨ Turn the above list into a multiplication table of 0s and 1s
                                  by multiplying the list with itself.


                                  The output should look like:



                                  0 0 0 0 0 0 0 0 0 0
                                  0 0 0 0 0 0 0 0 0 0
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 0 0 0 0 0 0 0 0
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 0 0 0 0 0 0 0 0
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 1 0 1 1 0 1 1 1
                                  0 0 1 0 1 1 0 1 1 1






                                  share|improve this answer














                                  share|improve this answer



                                  share|improve this answer








                                  edited 9 hours ago

























                                  answered 10 hours ago









                                  Sherlock9Sherlock9

                                  8,06411860




                                  8,06411860








                                  • 1




                                    $begingroup$
                                    you could take input as to avoid the { }
                                    $endgroup$
                                    – ngn
                                    10 hours ago














                                  • 1




                                    $begingroup$
                                    you could take input as to avoid the { }
                                    $endgroup$
                                    – ngn
                                    10 hours ago








                                  1




                                  1




                                  $begingroup$
                                  you could take input as to avoid the { }
                                  $endgroup$
                                  – ngn
                                  10 hours ago




                                  $begingroup$
                                  you could take input as to avoid the { }
                                  $endgroup$
                                  – ngn
                                  10 hours ago











                                  3












                                  $begingroup$


                                  Wolfram Language (Mathematica), 41 bytes



                                  a#&/@(a=Flatten@Array[{0,1~Table~#}&,#])&


                                  Try it online!






                                  share|improve this answer











                                  $endgroup$


















                                    3












                                    $begingroup$


                                    Wolfram Language (Mathematica), 41 bytes



                                    a#&/@(a=Flatten@Array[{0,1~Table~#}&,#])&


                                    Try it online!






                                    share|improve this answer











                                    $endgroup$
















                                      3












                                      3








                                      3





                                      $begingroup$


                                      Wolfram Language (Mathematica), 41 bytes



                                      a#&/@(a=Flatten@Array[{0,1~Table~#}&,#])&


                                      Try it online!






                                      share|improve this answer











                                      $endgroup$




                                      Wolfram Language (Mathematica), 41 bytes



                                      a#&/@(a=Flatten@Array[{0,1~Table~#}&,#])&


                                      Try it online!







                                      share|improve this answer














                                      share|improve this answer



                                      share|improve this answer








                                      edited 7 hours ago

























                                      answered 9 hours ago









                                      shrapshrap

                                      913




                                      913























                                          2












                                          $begingroup$


                                          Python 2, 67 bytes





                                          s='';n=input()
                                          while n:s='#'*n+' '+s;n-=1
                                          for c in s:print(c>' ')*s


                                          Try it online!



                                          Prints blank lines for empty lines, which the challenge allows.



                                          Same length (with input one above n):





                                          r=range(input())
                                          for n in r:print(' '.join(i*'#'for i in r)+'n')*n


                                          Try it online!






                                          share|improve this answer









                                          $endgroup$


















                                            2












                                            $begingroup$


                                            Python 2, 67 bytes





                                            s='';n=input()
                                            while n:s='#'*n+' '+s;n-=1
                                            for c in s:print(c>' ')*s


                                            Try it online!



                                            Prints blank lines for empty lines, which the challenge allows.



                                            Same length (with input one above n):





                                            r=range(input())
                                            for n in r:print(' '.join(i*'#'for i in r)+'n')*n


                                            Try it online!






                                            share|improve this answer









                                            $endgroup$
















                                              2












                                              2








                                              2





                                              $begingroup$


                                              Python 2, 67 bytes





                                              s='';n=input()
                                              while n:s='#'*n+' '+s;n-=1
                                              for c in s:print(c>' ')*s


                                              Try it online!



                                              Prints blank lines for empty lines, which the challenge allows.



                                              Same length (with input one above n):





                                              r=range(input())
                                              for n in r:print(' '.join(i*'#'for i in r)+'n')*n


                                              Try it online!






                                              share|improve this answer









                                              $endgroup$




                                              Python 2, 67 bytes





                                              s='';n=input()
                                              while n:s='#'*n+' '+s;n-=1
                                              for c in s:print(c>' ')*s


                                              Try it online!



                                              Prints blank lines for empty lines, which the challenge allows.



                                              Same length (with input one above n):





                                              r=range(input())
                                              for n in r:print(' '.join(i*'#'for i in r)+'n')*n


                                              Try it online!







                                              share|improve this answer












                                              share|improve this answer



                                              share|improve this answer










                                              answered 9 hours ago









                                              xnorxnor

                                              92.3k18188446




                                              92.3k18188446























                                                  2












                                                  $begingroup$


                                                  J, 17 bytes



                                                  [:*/~2=/[:I.2+i.


                                                  Try it online!






                                                  share|improve this answer









                                                  $endgroup$













                                                  • $begingroup$
                                                    did not know about this I. trick!
                                                    $endgroup$
                                                    – Jonah
                                                    2 hours ago










                                                  • $begingroup$
                                                    @Jonah I was waiting for an opportunity to use it :)
                                                    $endgroup$
                                                    – Galen Ivanov
                                                    26 mins ago
















                                                  2












                                                  $begingroup$


                                                  J, 17 bytes



                                                  [:*/~2=/[:I.2+i.


                                                  Try it online!






                                                  share|improve this answer









                                                  $endgroup$













                                                  • $begingroup$
                                                    did not know about this I. trick!
                                                    $endgroup$
                                                    – Jonah
                                                    2 hours ago










                                                  • $begingroup$
                                                    @Jonah I was waiting for an opportunity to use it :)
                                                    $endgroup$
                                                    – Galen Ivanov
                                                    26 mins ago














                                                  2












                                                  2








                                                  2





                                                  $begingroup$


                                                  J, 17 bytes



                                                  [:*/~2=/[:I.2+i.


                                                  Try it online!






                                                  share|improve this answer









                                                  $endgroup$




                                                  J, 17 bytes



                                                  [:*/~2=/[:I.2+i.


                                                  Try it online!







                                                  share|improve this answer












                                                  share|improve this answer



                                                  share|improve this answer










                                                  answered 9 hours ago









                                                  Galen IvanovGalen Ivanov

                                                  7,16211034




                                                  7,16211034












                                                  • $begingroup$
                                                    did not know about this I. trick!
                                                    $endgroup$
                                                    – Jonah
                                                    2 hours ago










                                                  • $begingroup$
                                                    @Jonah I was waiting for an opportunity to use it :)
                                                    $endgroup$
                                                    – Galen Ivanov
                                                    26 mins ago


















                                                  • $begingroup$
                                                    did not know about this I. trick!
                                                    $endgroup$
                                                    – Jonah
                                                    2 hours ago










                                                  • $begingroup$
                                                    @Jonah I was waiting for an opportunity to use it :)
                                                    $endgroup$
                                                    – Galen Ivanov
                                                    26 mins ago
















                                                  $begingroup$
                                                  did not know about this I. trick!
                                                  $endgroup$
                                                  – Jonah
                                                  2 hours ago




                                                  $begingroup$
                                                  did not know about this I. trick!
                                                  $endgroup$
                                                  – Jonah
                                                  2 hours ago












                                                  $begingroup$
                                                  @Jonah I was waiting for an opportunity to use it :)
                                                  $endgroup$
                                                  – Galen Ivanov
                                                  26 mins ago




                                                  $begingroup$
                                                  @Jonah I was waiting for an opportunity to use it :)
                                                  $endgroup$
                                                  – Galen Ivanov
                                                  26 mins ago











                                                  1












                                                  $begingroup$

                                                  Haskell, 69 68 bytes



                                                  (a#b)0=
                                                  (a#b)n=(a#b)(n-1)++b:(a<$[1..n])
                                                  f n=((1#0)n#(0<$(1#0)n))n


                                                  Returns a matrix of numbers.



                                                  Try it online!



                                                  Variants of f with the same byte count:



                                                  f n=((#)<*>(0<$)$(1#0)n)n
                                                  f n|l<-(1#0)n=(l#(0<$l))n





                                                  share|improve this answer











                                                  $endgroup$













                                                  • $begingroup$
                                                    Do the 0 row and column help?
                                                    $endgroup$
                                                    – dfeuer
                                                    7 hours ago






                                                  • 1




                                                    $begingroup$
                                                    @dfeuer: yes, they save a byte. See the first version of my answer.
                                                    $endgroup$
                                                    – nimi
                                                    6 hours ago
















                                                  1












                                                  $begingroup$

                                                  Haskell, 69 68 bytes



                                                  (a#b)0=
                                                  (a#b)n=(a#b)(n-1)++b:(a<$[1..n])
                                                  f n=((1#0)n#(0<$(1#0)n))n


                                                  Returns a matrix of numbers.



                                                  Try it online!



                                                  Variants of f with the same byte count:



                                                  f n=((#)<*>(0<$)$(1#0)n)n
                                                  f n|l<-(1#0)n=(l#(0<$l))n





                                                  share|improve this answer











                                                  $endgroup$













                                                  • $begingroup$
                                                    Do the 0 row and column help?
                                                    $endgroup$
                                                    – dfeuer
                                                    7 hours ago






                                                  • 1




                                                    $begingroup$
                                                    @dfeuer: yes, they save a byte. See the first version of my answer.
                                                    $endgroup$
                                                    – nimi
                                                    6 hours ago














                                                  1












                                                  1








                                                  1





                                                  $begingroup$

                                                  Haskell, 69 68 bytes



                                                  (a#b)0=
                                                  (a#b)n=(a#b)(n-1)++b:(a<$[1..n])
                                                  f n=((1#0)n#(0<$(1#0)n))n


                                                  Returns a matrix of numbers.



                                                  Try it online!



                                                  Variants of f with the same byte count:



                                                  f n=((#)<*>(0<$)$(1#0)n)n
                                                  f n|l<-(1#0)n=(l#(0<$l))n





                                                  share|improve this answer











                                                  $endgroup$



                                                  Haskell, 69 68 bytes



                                                  (a#b)0=
                                                  (a#b)n=(a#b)(n-1)++b:(a<$[1..n])
                                                  f n=((1#0)n#(0<$(1#0)n))n


                                                  Returns a matrix of numbers.



                                                  Try it online!



                                                  Variants of f with the same byte count:



                                                  f n=((#)<*>(0<$)$(1#0)n)n
                                                  f n|l<-(1#0)n=(l#(0<$l))n






                                                  share|improve this answer














                                                  share|improve this answer



                                                  share|improve this answer








                                                  edited 12 hours ago

























                                                  answered 13 hours ago









                                                  niminimi

                                                  32.4k32389




                                                  32.4k32389












                                                  • $begingroup$
                                                    Do the 0 row and column help?
                                                    $endgroup$
                                                    – dfeuer
                                                    7 hours ago






                                                  • 1




                                                    $begingroup$
                                                    @dfeuer: yes, they save a byte. See the first version of my answer.
                                                    $endgroup$
                                                    – nimi
                                                    6 hours ago


















                                                  • $begingroup$
                                                    Do the 0 row and column help?
                                                    $endgroup$
                                                    – dfeuer
                                                    7 hours ago






                                                  • 1




                                                    $begingroup$
                                                    @dfeuer: yes, they save a byte. See the first version of my answer.
                                                    $endgroup$
                                                    – nimi
                                                    6 hours ago
















                                                  $begingroup$
                                                  Do the 0 row and column help?
                                                  $endgroup$
                                                  – dfeuer
                                                  7 hours ago




                                                  $begingroup$
                                                  Do the 0 row and column help?
                                                  $endgroup$
                                                  – dfeuer
                                                  7 hours ago




                                                  1




                                                  1




                                                  $begingroup$
                                                  @dfeuer: yes, they save a byte. See the first version of my answer.
                                                  $endgroup$
                                                  – nimi
                                                  6 hours ago




                                                  $begingroup$
                                                  @dfeuer: yes, they save a byte. See the first version of my answer.
                                                  $endgroup$
                                                  – nimi
                                                  6 hours ago











                                                  1












                                                  $begingroup$

                                                  APL+WIN, 29 bytes



                                                  m/⍉(m←¯1↓∊(⍳n),¨¯1)/(n,n←⎕)⍴1


                                                  Explanation:



                                                  (n,n←⎕)⍴1 prompt for integer n and create a nxn matrix of 1s

                                                  (m←¯1↓∊(⍳n) replicate the columns by 1,2,.....n and insert 0s between each replication

                                                  m/⍉ repeat replication and 0 insertion for the rows from above


                                                  Example:



                                                  ⎕:
                                                  3
                                                  1 0 1 1 0 1 1 1
                                                  0 0 0 0 0 0 0 0
                                                  1 0 1 1 0 1 1 1
                                                  1 0 1 1 0 1 1 1
                                                  0 0 0 0 0 0 0 0
                                                  1 0 1 1 0 1 1 1
                                                  1 0 1 1 0 1 1 1
                                                  1 0 1 1 0 1 1 1





                                                  share|improve this answer











                                                  $endgroup$


















                                                    1












                                                    $begingroup$

                                                    APL+WIN, 29 bytes



                                                    m/⍉(m←¯1↓∊(⍳n),¨¯1)/(n,n←⎕)⍴1


                                                    Explanation:



                                                    (n,n←⎕)⍴1 prompt for integer n and create a nxn matrix of 1s

                                                    (m←¯1↓∊(⍳n) replicate the columns by 1,2,.....n and insert 0s between each replication

                                                    m/⍉ repeat replication and 0 insertion for the rows from above


                                                    Example:



                                                    ⎕:
                                                    3
                                                    1 0 1 1 0 1 1 1
                                                    0 0 0 0 0 0 0 0
                                                    1 0 1 1 0 1 1 1
                                                    1 0 1 1 0 1 1 1
                                                    0 0 0 0 0 0 0 0
                                                    1 0 1 1 0 1 1 1
                                                    1 0 1 1 0 1 1 1
                                                    1 0 1 1 0 1 1 1





                                                    share|improve this answer











                                                    $endgroup$
















                                                      1












                                                      1








                                                      1





                                                      $begingroup$

                                                      APL+WIN, 29 bytes



                                                      m/⍉(m←¯1↓∊(⍳n),¨¯1)/(n,n←⎕)⍴1


                                                      Explanation:



                                                      (n,n←⎕)⍴1 prompt for integer n and create a nxn matrix of 1s

                                                      (m←¯1↓∊(⍳n) replicate the columns by 1,2,.....n and insert 0s between each replication

                                                      m/⍉ repeat replication and 0 insertion for the rows from above


                                                      Example:



                                                      ⎕:
                                                      3
                                                      1 0 1 1 0 1 1 1
                                                      0 0 0 0 0 0 0 0
                                                      1 0 1 1 0 1 1 1
                                                      1 0 1 1 0 1 1 1
                                                      0 0 0 0 0 0 0 0
                                                      1 0 1 1 0 1 1 1
                                                      1 0 1 1 0 1 1 1
                                                      1 0 1 1 0 1 1 1





                                                      share|improve this answer











                                                      $endgroup$



                                                      APL+WIN, 29 bytes



                                                      m/⍉(m←¯1↓∊(⍳n),¨¯1)/(n,n←⎕)⍴1


                                                      Explanation:



                                                      (n,n←⎕)⍴1 prompt for integer n and create a nxn matrix of 1s

                                                      (m←¯1↓∊(⍳n) replicate the columns by 1,2,.....n and insert 0s between each replication

                                                      m/⍉ repeat replication and 0 insertion for the rows from above


                                                      Example:



                                                      ⎕:
                                                      3
                                                      1 0 1 1 0 1 1 1
                                                      0 0 0 0 0 0 0 0
                                                      1 0 1 1 0 1 1 1
                                                      1 0 1 1 0 1 1 1
                                                      0 0 0 0 0 0 0 0
                                                      1 0 1 1 0 1 1 1
                                                      1 0 1 1 0 1 1 1
                                                      1 0 1 1 0 1 1 1






                                                      share|improve this answer














                                                      share|improve this answer



                                                      share|improve this answer








                                                      edited 12 hours ago

























                                                      answered 13 hours ago









                                                      GrahamGraham

                                                      2,52678




                                                      2,52678























                                                          1












                                                          $begingroup$


                                                          Jelly, 7 bytes



                                                          ‘RÄṬ|þ`


                                                          Try it online!



                                                          Outputs a digit matrix, using $0$ for the rectangles and $1$ for the padding between them. The TIO link contains a footer which formats a digit matrix in a human-readable way by lining up the rows and columns.



                                                          Explanation



                                                          ‘RÄṬ|þ`
                                                          R Take a range from 1 to
                                                          ‘ {the input} plus 1
                                                          Ä Cumulative sum; produces the first {input}+1 triangular numbers
                                                          Ṭ Produce an array with 1s at those indexes, 0s at other indexes
                                                          þ Create a table of {the array}
                                                          ` with itself
                                                          | using bitwise OR


                                                          The number at cell $(x,y)$ of the resulting table will be $1$ if either $x$ or $y$ is a triangular number, or $0$ otherwise (because bitwise OR works like logical OR on 0 and 1). (We use R, range from 1, because Jelly uses 1-based indexing, so we don't have to worry about column 0 being incorrectly full of 0s; we have to add 1 to the input because the array produced by stops at the largest element given in the input, so we need to draw a line on the right-hand side and the bottom.) The gaps between triangular numbers are the consecutive integers, so the rectangular blocks formed by the gaps between the lines end up as the sizes requested by the question; and the use of an OR operation (in this case, bitwise) allows the lines to cross each other correctly.






                                                          share|improve this answer











                                                          $endgroup$













                                                          • $begingroup$
                                                            Why is this a community wiki?! If you want to waive rep you could just give it to Erik the Outgolfer
                                                            $endgroup$
                                                            – Jonathan Allan
                                                            6 hours ago












                                                          • $begingroup$
                                                            I CW all my answers (unless I think they might get a bounty, in which case I use a temporary account for them). Aiming for a high reputation normally means aiming to make the site worse (I once repcapped every day for a week to prove that it was possible; it wasn't particularly difficult, and yet it involved a lot of shallow questions/answers that didn't really contribute to the site much). Additionally, gaining reputation is mostly a negative on the account because it makes the site nag you into doing moderation work; and gaining privileges increases the risk of accidentally gaining badges.
                                                            $endgroup$
                                                            – ais523
                                                            6 hours ago










                                                          • $begingroup$
                                                            Also, I mostly disagree with the concept of ownership of posts on SE (although mostly with questions rather than answers, but you can't CW a question without moderator help). A CW marker very clearly says "if something is wrong here, feel free to edit it"; so I'd be applying a CW marker to everything even if it didn't waive reputation. SE is meant to be part wiki, after all, but people don't always use it as that.
                                                            $endgroup$
                                                            – ais523
                                                            6 hours ago










                                                          • $begingroup$
                                                            RE "aiming for a high reputation" if you don't want that then just post when you think it is meaningful and avoid what you term "shallow" q&a. Either you feel this answer does add something to the site, in which case post it without CW, or you think it is "shallow", in which case just don't post it. RE "I mostly disagree with the concept of ownership of posts on SE" - well you're simply on the wrong site then.
                                                            $endgroup$
                                                            – Jonathan Allan
                                                            5 hours ago












                                                          • $begingroup$
                                                            I think the answer adds something to the site, but if it's not CW, I feel compelled to post in a way that would gain me reputation rather than posting what I think would be interesting; back when I was a 20k user, I ended up really hating the site and almost got turned away from code golf in general because of it, so I deleted my account as a result. When I returned, I used to delete my account with every answer I posted (creating a new one for the next answer), but someone else pointed out that CWing every answer would have a similar effect, so nowadays I do that instead.
                                                            $endgroup$
                                                            – ais523
                                                            5 hours ago
















                                                          1












                                                          $begingroup$


                                                          Jelly, 7 bytes



                                                          ‘RÄṬ|þ`


                                                          Try it online!



                                                          Outputs a digit matrix, using $0$ for the rectangles and $1$ for the padding between them. The TIO link contains a footer which formats a digit matrix in a human-readable way by lining up the rows and columns.



                                                          Explanation



                                                          ‘RÄṬ|þ`
                                                          R Take a range from 1 to
                                                          ‘ {the input} plus 1
                                                          Ä Cumulative sum; produces the first {input}+1 triangular numbers
                                                          Ṭ Produce an array with 1s at those indexes, 0s at other indexes
                                                          þ Create a table of {the array}
                                                          ` with itself
                                                          | using bitwise OR


                                                          The number at cell $(x,y)$ of the resulting table will be $1$ if either $x$ or $y$ is a triangular number, or $0$ otherwise (because bitwise OR works like logical OR on 0 and 1). (We use R, range from 1, because Jelly uses 1-based indexing, so we don't have to worry about column 0 being incorrectly full of 0s; we have to add 1 to the input because the array produced by stops at the largest element given in the input, so we need to draw a line on the right-hand side and the bottom.) The gaps between triangular numbers are the consecutive integers, so the rectangular blocks formed by the gaps between the lines end up as the sizes requested by the question; and the use of an OR operation (in this case, bitwise) allows the lines to cross each other correctly.






                                                          share|improve this answer











                                                          $endgroup$













                                                          • $begingroup$
                                                            Why is this a community wiki?! If you want to waive rep you could just give it to Erik the Outgolfer
                                                            $endgroup$
                                                            – Jonathan Allan
                                                            6 hours ago












                                                          • $begingroup$
                                                            I CW all my answers (unless I think they might get a bounty, in which case I use a temporary account for them). Aiming for a high reputation normally means aiming to make the site worse (I once repcapped every day for a week to prove that it was possible; it wasn't particularly difficult, and yet it involved a lot of shallow questions/answers that didn't really contribute to the site much). Additionally, gaining reputation is mostly a negative on the account because it makes the site nag you into doing moderation work; and gaining privileges increases the risk of accidentally gaining badges.
                                                            $endgroup$
                                                            – ais523
                                                            6 hours ago










                                                          • $begingroup$
                                                            Also, I mostly disagree with the concept of ownership of posts on SE (although mostly with questions rather than answers, but you can't CW a question without moderator help). A CW marker very clearly says "if something is wrong here, feel free to edit it"; so I'd be applying a CW marker to everything even if it didn't waive reputation. SE is meant to be part wiki, after all, but people don't always use it as that.
                                                            $endgroup$
                                                            – ais523
                                                            6 hours ago










                                                          • $begingroup$
                                                            RE "aiming for a high reputation" if you don't want that then just post when you think it is meaningful and avoid what you term "shallow" q&a. Either you feel this answer does add something to the site, in which case post it without CW, or you think it is "shallow", in which case just don't post it. RE "I mostly disagree with the concept of ownership of posts on SE" - well you're simply on the wrong site then.
                                                            $endgroup$
                                                            – Jonathan Allan
                                                            5 hours ago












                                                          • $begingroup$
                                                            I think the answer adds something to the site, but if it's not CW, I feel compelled to post in a way that would gain me reputation rather than posting what I think would be interesting; back when I was a 20k user, I ended up really hating the site and almost got turned away from code golf in general because of it, so I deleted my account as a result. When I returned, I used to delete my account with every answer I posted (creating a new one for the next answer), but someone else pointed out that CWing every answer would have a similar effect, so nowadays I do that instead.
                                                            $endgroup$
                                                            – ais523
                                                            5 hours ago














                                                          1












                                                          1








                                                          1





                                                          $begingroup$


                                                          Jelly, 7 bytes



                                                          ‘RÄṬ|þ`


                                                          Try it online!



                                                          Outputs a digit matrix, using $0$ for the rectangles and $1$ for the padding between them. The TIO link contains a footer which formats a digit matrix in a human-readable way by lining up the rows and columns.



                                                          Explanation



                                                          ‘RÄṬ|þ`
                                                          R Take a range from 1 to
                                                          ‘ {the input} plus 1
                                                          Ä Cumulative sum; produces the first {input}+1 triangular numbers
                                                          Ṭ Produce an array with 1s at those indexes, 0s at other indexes
                                                          þ Create a table of {the array}
                                                          ` with itself
                                                          | using bitwise OR


                                                          The number at cell $(x,y)$ of the resulting table will be $1$ if either $x$ or $y$ is a triangular number, or $0$ otherwise (because bitwise OR works like logical OR on 0 and 1). (We use R, range from 1, because Jelly uses 1-based indexing, so we don't have to worry about column 0 being incorrectly full of 0s; we have to add 1 to the input because the array produced by stops at the largest element given in the input, so we need to draw a line on the right-hand side and the bottom.) The gaps between triangular numbers are the consecutive integers, so the rectangular blocks formed by the gaps between the lines end up as the sizes requested by the question; and the use of an OR operation (in this case, bitwise) allows the lines to cross each other correctly.






                                                          share|improve this answer











                                                          $endgroup$




                                                          Jelly, 7 bytes



                                                          ‘RÄṬ|þ`


                                                          Try it online!



                                                          Outputs a digit matrix, using $0$ for the rectangles and $1$ for the padding between them. The TIO link contains a footer which formats a digit matrix in a human-readable way by lining up the rows and columns.



                                                          Explanation



                                                          ‘RÄṬ|þ`
                                                          R Take a range from 1 to
                                                          ‘ {the input} plus 1
                                                          Ä Cumulative sum; produces the first {input}+1 triangular numbers
                                                          Ṭ Produce an array with 1s at those indexes, 0s at other indexes
                                                          þ Create a table of {the array}
                                                          ` with itself
                                                          | using bitwise OR


                                                          The number at cell $(x,y)$ of the resulting table will be $1$ if either $x$ or $y$ is a triangular number, or $0$ otherwise (because bitwise OR works like logical OR on 0 and 1). (We use R, range from 1, because Jelly uses 1-based indexing, so we don't have to worry about column 0 being incorrectly full of 0s; we have to add 1 to the input because the array produced by stops at the largest element given in the input, so we need to draw a line on the right-hand side and the bottom.) The gaps between triangular numbers are the consecutive integers, so the rectangular blocks formed by the gaps between the lines end up as the sizes requested by the question; and the use of an OR operation (in this case, bitwise) allows the lines to cross each other correctly.







                                                          share|improve this answer














                                                          share|improve this answer



                                                          share|improve this answer








                                                          answered 7 hours ago


























                                                          community wiki





                                                          ais523













                                                          • $begingroup$
                                                            Why is this a community wiki?! If you want to waive rep you could just give it to Erik the Outgolfer
                                                            $endgroup$
                                                            – Jonathan Allan
                                                            6 hours ago












                                                          • $begingroup$
                                                            I CW all my answers (unless I think they might get a bounty, in which case I use a temporary account for them). Aiming for a high reputation normally means aiming to make the site worse (I once repcapped every day for a week to prove that it was possible; it wasn't particularly difficult, and yet it involved a lot of shallow questions/answers that didn't really contribute to the site much). Additionally, gaining reputation is mostly a negative on the account because it makes the site nag you into doing moderation work; and gaining privileges increases the risk of accidentally gaining badges.
                                                            $endgroup$
                                                            – ais523
                                                            6 hours ago










                                                          • $begingroup$
                                                            Also, I mostly disagree with the concept of ownership of posts on SE (although mostly with questions rather than answers, but you can't CW a question without moderator help). A CW marker very clearly says "if something is wrong here, feel free to edit it"; so I'd be applying a CW marker to everything even if it didn't waive reputation. SE is meant to be part wiki, after all, but people don't always use it as that.
                                                            $endgroup$
                                                            – ais523
                                                            6 hours ago










                                                          • $begingroup$
                                                            RE "aiming for a high reputation" if you don't want that then just post when you think it is meaningful and avoid what you term "shallow" q&a. Either you feel this answer does add something to the site, in which case post it without CW, or you think it is "shallow", in which case just don't post it. RE "I mostly disagree with the concept of ownership of posts on SE" - well you're simply on the wrong site then.
                                                            $endgroup$
                                                            – Jonathan Allan
                                                            5 hours ago












                                                          • $begingroup$
                                                            I think the answer adds something to the site, but if it's not CW, I feel compelled to post in a way that would gain me reputation rather than posting what I think would be interesting; back when I was a 20k user, I ended up really hating the site and almost got turned away from code golf in general because of it, so I deleted my account as a result. When I returned, I used to delete my account with every answer I posted (creating a new one for the next answer), but someone else pointed out that CWing every answer would have a similar effect, so nowadays I do that instead.
                                                            $endgroup$
                                                            – ais523
                                                            5 hours ago


















                                                          • $begingroup$
                                                            Why is this a community wiki?! If you want to waive rep you could just give it to Erik the Outgolfer
                                                            $endgroup$
                                                            – Jonathan Allan
                                                            6 hours ago












                                                          • $begingroup$
                                                            I CW all my answers (unless I think they might get a bounty, in which case I use a temporary account for them). Aiming for a high reputation normally means aiming to make the site worse (I once repcapped every day for a week to prove that it was possible; it wasn't particularly difficult, and yet it involved a lot of shallow questions/answers that didn't really contribute to the site much). Additionally, gaining reputation is mostly a negative on the account because it makes the site nag you into doing moderation work; and gaining privileges increases the risk of accidentally gaining badges.
                                                            $endgroup$
                                                            – ais523
                                                            6 hours ago










                                                          • $begingroup$
                                                            Also, I mostly disagree with the concept of ownership of posts on SE (although mostly with questions rather than answers, but you can't CW a question without moderator help). A CW marker very clearly says "if something is wrong here, feel free to edit it"; so I'd be applying a CW marker to everything even if it didn't waive reputation. SE is meant to be part wiki, after all, but people don't always use it as that.
                                                            $endgroup$
                                                            – ais523
                                                            6 hours ago










                                                          • $begingroup$
                                                            RE "aiming for a high reputation" if you don't want that then just post when you think it is meaningful and avoid what you term "shallow" q&a. Either you feel this answer does add something to the site, in which case post it without CW, or you think it is "shallow", in which case just don't post it. RE "I mostly disagree with the concept of ownership of posts on SE" - well you're simply on the wrong site then.
                                                            $endgroup$
                                                            – Jonathan Allan
                                                            5 hours ago












                                                          • $begingroup$
                                                            I think the answer adds something to the site, but if it's not CW, I feel compelled to post in a way that would gain me reputation rather than posting what I think would be interesting; back when I was a 20k user, I ended up really hating the site and almost got turned away from code golf in general because of it, so I deleted my account as a result. When I returned, I used to delete my account with every answer I posted (creating a new one for the next answer), but someone else pointed out that CWing every answer would have a similar effect, so nowadays I do that instead.
                                                            $endgroup$
                                                            – ais523
                                                            5 hours ago
















                                                          $begingroup$
                                                          Why is this a community wiki?! If you want to waive rep you could just give it to Erik the Outgolfer
                                                          $endgroup$
                                                          – Jonathan Allan
                                                          6 hours ago






                                                          $begingroup$
                                                          Why is this a community wiki?! If you want to waive rep you could just give it to Erik the Outgolfer
                                                          $endgroup$
                                                          – Jonathan Allan
                                                          6 hours ago














                                                          $begingroup$
                                                          I CW all my answers (unless I think they might get a bounty, in which case I use a temporary account for them). Aiming for a high reputation normally means aiming to make the site worse (I once repcapped every day for a week to prove that it was possible; it wasn't particularly difficult, and yet it involved a lot of shallow questions/answers that didn't really contribute to the site much). Additionally, gaining reputation is mostly a negative on the account because it makes the site nag you into doing moderation work; and gaining privileges increases the risk of accidentally gaining badges.
                                                          $endgroup$
                                                          – ais523
                                                          6 hours ago




                                                          $begingroup$
                                                          I CW all my answers (unless I think they might get a bounty, in which case I use a temporary account for them). Aiming for a high reputation normally means aiming to make the site worse (I once repcapped every day for a week to prove that it was possible; it wasn't particularly difficult, and yet it involved a lot of shallow questions/answers that didn't really contribute to the site much). Additionally, gaining reputation is mostly a negative on the account because it makes the site nag you into doing moderation work; and gaining privileges increases the risk of accidentally gaining badges.
                                                          $endgroup$
                                                          – ais523
                                                          6 hours ago












                                                          $begingroup$
                                                          Also, I mostly disagree with the concept of ownership of posts on SE (although mostly with questions rather than answers, but you can't CW a question without moderator help). A CW marker very clearly says "if something is wrong here, feel free to edit it"; so I'd be applying a CW marker to everything even if it didn't waive reputation. SE is meant to be part wiki, after all, but people don't always use it as that.
                                                          $endgroup$
                                                          – ais523
                                                          6 hours ago




                                                          $begingroup$
                                                          Also, I mostly disagree with the concept of ownership of posts on SE (although mostly with questions rather than answers, but you can't CW a question without moderator help). A CW marker very clearly says "if something is wrong here, feel free to edit it"; so I'd be applying a CW marker to everything even if it didn't waive reputation. SE is meant to be part wiki, after all, but people don't always use it as that.
                                                          $endgroup$
                                                          – ais523
                                                          6 hours ago












                                                          $begingroup$
                                                          RE "aiming for a high reputation" if you don't want that then just post when you think it is meaningful and avoid what you term "shallow" q&a. Either you feel this answer does add something to the site, in which case post it without CW, or you think it is "shallow", in which case just don't post it. RE "I mostly disagree with the concept of ownership of posts on SE" - well you're simply on the wrong site then.
                                                          $endgroup$
                                                          – Jonathan Allan
                                                          5 hours ago






                                                          $begingroup$
                                                          RE "aiming for a high reputation" if you don't want that then just post when you think it is meaningful and avoid what you term "shallow" q&a. Either you feel this answer does add something to the site, in which case post it without CW, or you think it is "shallow", in which case just don't post it. RE "I mostly disagree with the concept of ownership of posts on SE" - well you're simply on the wrong site then.
                                                          $endgroup$
                                                          – Jonathan Allan
                                                          5 hours ago














                                                          $begingroup$
                                                          I think the answer adds something to the site, but if it's not CW, I feel compelled to post in a way that would gain me reputation rather than posting what I think would be interesting; back when I was a 20k user, I ended up really hating the site and almost got turned away from code golf in general because of it, so I deleted my account as a result. When I returned, I used to delete my account with every answer I posted (creating a new one for the next answer), but someone else pointed out that CWing every answer would have a similar effect, so nowadays I do that instead.
                                                          $endgroup$
                                                          – ais523
                                                          5 hours ago




                                                          $begingroup$
                                                          I think the answer adds something to the site, but if it's not CW, I feel compelled to post in a way that would gain me reputation rather than posting what I think would be interesting; back when I was a 20k user, I ended up really hating the site and almost got turned away from code golf in general because of it, so I deleted my account as a result. When I returned, I used to delete my account with every answer I posted (creating a new one for the next answer), but someone else pointed out that CWing every answer would have a similar effect, so nowadays I do that instead.
                                                          $endgroup$
                                                          – ais523
                                                          5 hours ago











                                                          0












                                                          $begingroup$


                                                          Charcoal, 18 bytes



                                                          ≔⪫EN×#⊕ι θEθ⭆θ⌊⟦ιλ


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



                                                             N                Input number
                                                          E Map over implicit range
                                                          ι Current value
                                                          ⊕ Incremented
                                                          × Repetitions of
                                                          # Literal `#`
                                                          ⪫ Join with spaces
                                                          ≔ θ Assign to variable
                                                          θ Retrieve variable
                                                          E Map over characters
                                                          θ Retrieve variable
                                                          ⭆ Replace characters with
                                                          ⌊ Minimum of
                                                          ⟦ List of
                                                          ι Row character
                                                          λ Column character
                                                          Implicitly print each row on its own line





                                                          share|improve this answer









                                                          $endgroup$


















                                                            0












                                                            $begingroup$


                                                            Charcoal, 18 bytes



                                                            ≔⪫EN×#⊕ι θEθ⭆θ⌊⟦ιλ


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



                                                               N                Input number
                                                            E Map over implicit range
                                                            ι Current value
                                                            ⊕ Incremented
                                                            × Repetitions of
                                                            # Literal `#`
                                                            ⪫ Join with spaces
                                                            ≔ θ Assign to variable
                                                            θ Retrieve variable
                                                            E Map over characters
                                                            θ Retrieve variable
                                                            ⭆ Replace characters with
                                                            ⌊ Minimum of
                                                            ⟦ List of
                                                            ι Row character
                                                            λ Column character
                                                            Implicitly print each row on its own line





                                                            share|improve this answer









                                                            $endgroup$
















                                                              0












                                                              0








                                                              0





                                                              $begingroup$


                                                              Charcoal, 18 bytes



                                                              ≔⪫EN×#⊕ι θEθ⭆θ⌊⟦ιλ


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



                                                                 N                Input number
                                                              E Map over implicit range
                                                              ι Current value
                                                              ⊕ Incremented
                                                              × Repetitions of
                                                              # Literal `#`
                                                              ⪫ Join with spaces
                                                              ≔ θ Assign to variable
                                                              θ Retrieve variable
                                                              E Map over characters
                                                              θ Retrieve variable
                                                              ⭆ Replace characters with
                                                              ⌊ Minimum of
                                                              ⟦ List of
                                                              ι Row character
                                                              λ Column character
                                                              Implicitly print each row on its own line





                                                              share|improve this answer









                                                              $endgroup$




                                                              Charcoal, 18 bytes



                                                              ≔⪫EN×#⊕ι θEθ⭆θ⌊⟦ιλ


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



                                                                 N                Input number
                                                              E Map over implicit range
                                                              ι Current value
                                                              ⊕ Incremented
                                                              × Repetitions of
                                                              # Literal `#`
                                                              ⪫ Join with spaces
                                                              ≔ θ Assign to variable
                                                              θ Retrieve variable
                                                              E Map over characters
                                                              θ Retrieve variable
                                                              ⭆ Replace characters with
                                                              ⌊ Minimum of
                                                              ⟦ List of
                                                              ι Row character
                                                              λ Column character
                                                              Implicitly print each row on its own line






                                                              share|improve this answer












                                                              share|improve this answer



                                                              share|improve this answer










                                                              answered 4 hours ago









                                                              NeilNeil

                                                              81.7k745178




                                                              81.7k745178























                                                                  0












                                                                  $begingroup$


                                                                  Perl 6, 35 bytes





                                                                  {[~](("
                                                                  "~[~] *xx$_)xx$_)>>.say}


                                                                  Try it online!



                                                                  Anonymous code block that takes a number and prints the multiplication table with *s and a leading newline.






                                                                  share|improve this answer









                                                                  $endgroup$


















                                                                    0












                                                                    $begingroup$


                                                                    Perl 6, 35 bytes





                                                                    {[~](("
                                                                    "~[~] *xx$_)xx$_)>>.say}


                                                                    Try it online!



                                                                    Anonymous code block that takes a number and prints the multiplication table with *s and a leading newline.






                                                                    share|improve this answer









                                                                    $endgroup$
















                                                                      0












                                                                      0








                                                                      0





                                                                      $begingroup$


                                                                      Perl 6, 35 bytes





                                                                      {[~](("
                                                                      "~[~] *xx$_)xx$_)>>.say}


                                                                      Try it online!



                                                                      Anonymous code block that takes a number and prints the multiplication table with *s and a leading newline.






                                                                      share|improve this answer









                                                                      $endgroup$




                                                                      Perl 6, 35 bytes





                                                                      {[~](("
                                                                      "~[~] *xx$_)xx$_)>>.say}


                                                                      Try it online!



                                                                      Anonymous code block that takes a number and prints the multiplication table with *s and a leading newline.







                                                                      share|improve this answer












                                                                      share|improve this answer



                                                                      share|improve this answer










                                                                      answered 3 hours ago









                                                                      Jo KingJo King

                                                                      24.9k359128




                                                                      24.9k359128























                                                                          0












                                                                          $begingroup$


                                                                          Mouse-2002, 79 bytes



                                                                          Abusing Mouse's macros to repeat functionality.



                                                                          ?1+n:#P,n,j,k,b#P,j,y,k,#P,n,i,' ,#P,i,x,35;;;;$P0 2%:(1%.2%.-^4%3%!'2%.1+2%:)@


                                                                          Try it online!






                                                                          share|improve this answer









                                                                          $endgroup$


















                                                                            0












                                                                            $begingroup$


                                                                            Mouse-2002, 79 bytes



                                                                            Abusing Mouse's macros to repeat functionality.



                                                                            ?1+n:#P,n,j,k,b#P,j,y,k,#P,n,i,' ,#P,i,x,35;;;;$P0 2%:(1%.2%.-^4%3%!'2%.1+2%:)@


                                                                            Try it online!






                                                                            share|improve this answer









                                                                            $endgroup$
















                                                                              0












                                                                              0








                                                                              0





                                                                              $begingroup$


                                                                              Mouse-2002, 79 bytes



                                                                              Abusing Mouse's macros to repeat functionality.



                                                                              ?1+n:#P,n,j,k,b#P,j,y,k,#P,n,i,' ,#P,i,x,35;;;;$P0 2%:(1%.2%.-^4%3%!'2%.1+2%:)@


                                                                              Try it online!






                                                                              share|improve this answer









                                                                              $endgroup$




                                                                              Mouse-2002, 79 bytes



                                                                              Abusing Mouse's macros to repeat functionality.



                                                                              ?1+n:#P,n,j,k,b#P,j,y,k,#P,n,i,' ,#P,i,x,35;;;;$P0 2%:(1%.2%.-^4%3%!'2%.1+2%:)@


                                                                              Try it online!







                                                                              share|improve this answer












                                                                              share|improve this answer



                                                                              share|improve this answer










                                                                              answered 3 hours ago









                                                                              MooseOnTheRocksMooseOnTheRocks

                                                                              1714




                                                                              1714






























                                                                                  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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