Print a physical multiplication table

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

edited yesterday

asked yesterday

dzaima

15.6k21959

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

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

$begingroup$
Nitpick: ∙ (U+2219: BULLET OPERATOR) isn't present in the ASCII character set. Nor is • (U+2022: BULLET) or ⋅ (U+22C5: DOT OPERATOR) or · (U+00B7: MIDDLE DOT). :)
$endgroup$
– Andreas Rejbrand
13 hours ago

add a comment |

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

edited yesterday

asked yesterday

dzaima

15.6k21959

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

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

$begingroup$
Nitpick: ∙ (U+2219: BULLET OPERATOR) isn't present in the ASCII character set. Nor is • (U+2022: BULLET) or ⋅ (U+22C5: DOT OPERATOR) or · (U+00B7: MIDDLE DOT). :)
$endgroup$
– Andreas Rejbrand
13 hours ago

add a comment |

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

edited yesterday

asked yesterday

dzaima

15.6k21959

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

edited yesterday

asked yesterday

dzaima

15.6k21959

edited yesterday

asked yesterday

dzaima

15.6k21959

edited yesterday

asked yesterday

dzaima

15.6k21959

asked yesterday

dzaima

15.6k21959

asked yesterday

dzaima

15.6k21959

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

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

$begingroup$
Nitpick: ∙ (U+2219: BULLET OPERATOR) isn't present in the ASCII character set. Nor is • (U+2022: BULLET) or ⋅ (U+22C5: DOT OPERATOR) or · (U+00B7: MIDDLE DOT). :)
$endgroup$
– Andreas Rejbrand
13 hours ago

add a comment |

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

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

$begingroup$
Nitpick: ∙ (U+2219: BULLET OPERATOR) isn't present in the ASCII character set. Nor is • (U+2022: BULLET) or ⋅ (U+22C5: DOT OPERATOR) or · (U+00B7: MIDDLE DOT). :)
$endgroup$
– Andreas Rejbrand
13 hours ago

Can we have extra row/columns of background characters in front, rather than at the end of the table?

– Kirill L.
yesterday

@KirillL. sure, as long as the rows line up

– dzaima
yesterday

Nitpick: ∙ (U+2219: BULLET OPERATOR) isn't present in the ASCII character set. Nor is • (U+2022: BULLET) or ⋅ (U+22C5: DOT OPERATOR) or · (U+00B7: MIDDLE DOT). :)

– Andreas Rejbrand
13 hours ago

add a comment |

21 Answers
21

active

oldest

votes

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.

edited yesterday

answered yesterday

Kirill L.

5,5031525

$begingroup$
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.
yesterday

$begingroup$
Oh, can they? I missed that !
$endgroup$
– digEmAll
yesterday

add a comment |

Haskell, 43 bytes

f n=map=<<flip(map.max)$show.(10^)=<<[1..n]

Try it online!

A clever approach by Ørjan Johansen outputting with 0's and 1's, generating each 10...00 part as the string representation of a power of 10.

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!

edited 2 hours ago

answered yesterday

xnor

92.4k18188446

3

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

1

$begingroup$
Just realized, flip(map.max)=mapM max.
$endgroup$
– Ørjan Johansen
2 hours ago

$begingroup$
@ØrjanJohansen Whoa, how does that work? I think you could do with posting an answer of your own :-)
$endgroup$
– xnor
2 hours ago

$begingroup$
Set the monad to (->) b, then mapM :: (a -> b -> c) -> [a] -> b -> [c].
$endgroup$
– Ørjan Johansen
1 hour ago

add a comment |

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

edited yesterday

answered yesterday

Suever

9,8021446

add a comment |

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:

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

edited yesterday

answered yesterday

Arnauld

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
yesterday

add a comment |

Jelly, 8 bytes

1ẋⱮj0ȧþ`

Try it online!

answered yesterday

Erik the Outgolfer

32.5k429105

add a comment |

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!

answered yesterday

xnor

92.4k18188446

add a comment |

APL (Dyalog Unicode), 12 10 12 bytes^SBCS

∘.×⍨∊,,⎕⍴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

edited yesterday

answered yesterday

Sherlock9

8,06411860

1

$begingroup$
you could take input as ⎕ to avoid the { }
$endgroup$
– ngn
yesterday

add a comment |

Wolfram Language (Mathematica), 41 bytes

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

Try it online!

edited yesterday

answered yesterday

shrap

1013

add a comment |

PowerShell, 42 bytes

($r=1..$args[0])|%{"$($r|%{'#'*$_})`n"*$_}

Try it online!

edited 5 hours ago

answered 8 hours ago

Andrei Odegov

46926

$begingroup$
40 bytes :)
$endgroup$
– mazzy
7 hours ago

$begingroup$
@mazzy, cleverly :)
$endgroup$
– Andrei Odegov
7 hours ago

add a comment |

J, 17 bytes

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

Try it online!

answered yesterday

Galen Ivanov

7,16211034

$begingroup$
did not know about this I. trick!
$endgroup$
– Jonah
yesterday

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

add a comment |

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 Ṭ 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.

answered yesterday

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
yesterday

1

$begingroup$
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
yesterday

$begingroup$
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
yesterday

$begingroup$
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
yesterday

1

$begingroup$
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
yesterday

add a comment |

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!

answered yesterday

MooseOnTheRocks

1914

add a comment |

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

edited yesterday

answered yesterday

nimi

32.4k32389

$begingroup$
Do the 0 row and column help?
$endgroup$
– dfeuer
yesterday

1

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

add a comment |

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

edited yesterday

answered yesterday

Graham

2,52678

add a comment |

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

answered yesterday

Neil

81.7k745178

add a comment |

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.

answered yesterday

Jo King

24.9k359128

add a comment |

Brain-Flak, 168 bytes

(({})<>){(({})<{({}[()]<(<>({})<>){(({})<{({}[(())])}>[()])}{}{}(([(()()()()())(){}]){})>)}{}(({}))>[()])}{}{}{}(){{}({}<>((((()()){}){}){}){})<>()}<>{({}<>)<>}<>{}

Try it online!

answered 21 hours ago

MegaTom

3,5421424

add a comment |

05AB1E, 9 bytes

Defines a program $ftexttt{: }color{purple}{texttt{Nat}} →color{purple}{texttt{List}}texttt{[}color{purple}{texttt{List}}texttt{[}color{purple}{texttt{Nat}}texttt{]]}$.

Code:

$L×0ýSDδ*

Uses the 05AB1E-encoding. Try it online! or use the pretty-printed version.

Explanation:



$              # Push the number 1 and the input n

 L             # Create the list [1, 2, 3, ..., n]

  ×            # Vectorized string multiplication: 1 × [1, 2, 3, ..., n]

                 This would result in ["1", "11", "111", ..., "1" × n]

   0ý          # Join the resulting list with '0', resulting in "10110111011110111110..."

     S         # Split into single digits: [1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, ...]

      Dδ*      # Multiplication table with itself

answered 18 hours ago

Adnan

35.7k562225

add a comment |

Python 3, 88 bytes

n=range(int(input()))

for j in n:

 l=''

 for i in n:l+='#'*-~i+' '

 print((l+'n')*-~j)

Try it online!

answered 16 hours ago

Kerosenic

New contributor

$begingroup$
You can save bytes by declaring l outside the loop. Try it online!
$endgroup$
– Jo King
14 hours ago

$begingroup$
82 bytes.
$endgroup$
– Jonathan Frech
11 hours ago

add a comment |

C# (.NET Core), 208 bytes, N+1 indexed.

using C=System.Console;class M{static void Main(){int i=int.Parse(C.ReadLine()),j=1,k=1;var l="";var t=l;for(;j<=i;j++,k=0)while(k++<j)l+=k<j?"#":" ";for(i=0;i<l.Length;t=l[i++]=='#'?l+"n":"n")C.Write(t);}}

Try it online!

I feel quite like this is not the best approach, but it's what I have at the moment, so here we are.

The program (ab)uses nested loops to build the initial string, and then reads through that string (since it's rectangular!) to determine whether to output the string or just a newline.

answered 4 hours ago

Stackstuck

1277

$begingroup$
202?
$endgroup$
– ASCII-only
1 hour ago

$begingroup$
166, indexing fixed
$endgroup$
– ASCII-only
55 mins ago

$begingroup$
also *square not rectangular
$endgroup$
– ASCII-only
37 mins ago

$begingroup$
Squares are rectangles!
$endgroup$
– Stackstuck
27 mins ago

$begingroup$
also you should post this as your own answer, you did most of the work on it @ASCII-only.
$endgroup$
– Stackstuck
22 mins ago

add a comment |

SmileBASIC, 83 77 bytes

Graphical output.
Input is N-1

INPUT N

FOR J=0TO N

X=0FOR I=0TO N

GFILL X,Y,X+I,Y+J

X=X+I+2NEXT

Y=Y+J+2NEXT

edited 4 hours ago

answered 4 hours ago

12Me21

5,61711336

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "200"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

If this is an answer to a challenge…

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

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

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

More generally…

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

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

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodegolf.stackexchange.com%2fquestions%2f181582%2fprint-a-physical-multiplication-table%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

21 Answers
21

active

oldest

votes

21 Answers
21

active

oldest

votes

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.

edited yesterday

answered yesterday

Kirill L.

5,5031525

$begingroup$
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.
yesterday

$begingroup$
Oh, can they? I missed that !
$endgroup$
– digEmAll
yesterday

add a comment |

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.

edited yesterday

answered yesterday

Kirill L.

5,5031525

$begingroup$
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.
yesterday

$begingroup$
Oh, can they? I missed that !
$endgroup$
– digEmAll
yesterday

add a comment |

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.

edited yesterday

answered yesterday

Kirill L.

5,5031525

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.

edited yesterday

answered yesterday

Kirill L.

5,5031525

edited yesterday

answered yesterday

Kirill L.

5,5031525

answered yesterday

Kirill L.

5,5031525

answered yesterday

Kirill L.

5,5031525

$begingroup$
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.
yesterday

$begingroup$
Oh, can they? I missed that !
$endgroup$
– digEmAll
yesterday

add a comment |

$begingroup$
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.
yesterday

$begingroup$
Oh, can they? I missed that !
$endgroup$
– digEmAll
yesterday

Thanks, BTW it can probably even be 30, if the extra blank row can be in front, rather than at the end.

– Kirill L.
yesterday

Oh, can they? I missed that !

– digEmAll
yesterday

add a comment |

Haskell, 43 bytes

f n=map=<<flip(map.max)$show.(10^)=<<[1..n]

Try it online!

A clever approach by Ørjan Johansen outputting with 0's and 1's, generating each 10...00 part as the string representation of a power of 10.

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!

edited 2 hours ago

answered yesterday

xnor

92.4k18188446

3

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

1

$begingroup$
Just realized, flip(map.max)=mapM max.
$endgroup$
– Ørjan Johansen
2 hours ago

$begingroup$
@ØrjanJohansen Whoa, how does that work? I think you could do with posting an answer of your own :-)
$endgroup$
– xnor
2 hours ago

$begingroup$
Set the monad to (->) b, then mapM :: (a -> b -> c) -> [a] -> b -> [c].
$endgroup$
– Ørjan Johansen
1 hour ago

add a comment |

Haskell, 43 bytes

f n=map=<<flip(map.max)$show.(10^)=<<[1..n]

Try it online!

A clever approach by Ørjan Johansen outputting with 0's and 1's, generating each 10...00 part as the string representation of a power of 10.

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!

edited 2 hours ago

answered yesterday

xnor

92.4k18188446

3

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

1

$begingroup$
Just realized, flip(map.max)=mapM max.
$endgroup$
– Ørjan Johansen
2 hours ago

$begingroup$
@ØrjanJohansen Whoa, how does that work? I think you could do with posting an answer of your own :-)
$endgroup$
– xnor
2 hours ago

$begingroup$
Set the monad to (->) b, then mapM :: (a -> b -> c) -> [a] -> b -> [c].
$endgroup$
– Ørjan Johansen
1 hour ago

add a comment |

Haskell, 43 bytes

f n=map=<<flip(map.max)$show.(10^)=<<[1..n]

Try it online!

A clever approach by Ørjan Johansen outputting with 0's and 1's, generating each 10...00 part as the string representation of a power of 10.

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!

edited 2 hours ago

answered yesterday

xnor

92.4k18188446

Haskell, 43 bytes

f n=map=<<flip(map.max)$show.(10^)=<<[1..n]

Try it online!

A clever approach by Ørjan Johansen outputting with 0's and 1's, generating each 10...00 part as the string representation of a power of 10.

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!

edited 2 hours ago

answered yesterday

xnor

92.4k18188446

edited 2 hours ago

answered yesterday

xnor

92.4k18188446

answered yesterday

xnor

92.4k18188446

answered yesterday

xnor

92.4k18188446

3

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

1

$begingroup$
Just realized, flip(map.max)=mapM max.
$endgroup$
– Ørjan Johansen
2 hours ago

$begingroup$
@ØrjanJohansen Whoa, how does that work? I think you could do with posting an answer of your own :-)
$endgroup$
– xnor
2 hours ago

$begingroup$
Set the monad to (->) b, then mapM :: (a -> b -> c) -> [a] -> b -> [c].
$endgroup$
– Ørjan Johansen
1 hour ago

add a comment |

3

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

1

$begingroup$
Just realized, flip(map.max)=mapM max.
$endgroup$
– Ørjan Johansen
2 hours ago

$begingroup$
@ØrjanJohansen Whoa, how does that work? I think you could do with posting an answer of your own :-)
$endgroup$
– xnor
2 hours ago

$begingroup$
Set the monad to (->) b, then mapM :: (a -> b -> c) -> [a] -> b -> [c].
$endgroup$
– Ørjan Johansen
1 hour ago

This can be shortened with my old triangular number trick: 43 bytes

– Ørjan Johansen
yesterday

Just realized, flip(map.max)=mapM max.

– Ørjan Johansen
2 hours ago

@ØrjanJohansen Whoa, how does that work? I think you could do with posting an answer of your own :-)

– xnor
2 hours ago

Set the monad to (->) b, then mapM :: (a -> b -> c) -> [a] -> b -> [c].

– Ørjan Johansen
1 hour ago

add a comment |

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

edited yesterday

answered yesterday

Suever

9,8021446

add a comment |

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

edited yesterday

answered yesterday

Suever

9,8021446

add a comment |

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

edited yesterday

answered yesterday

Suever

9,8021446

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

edited yesterday

answered yesterday

Suever

9,8021446

edited yesterday

answered yesterday

Suever

9,8021446

answered yesterday

Suever

9,8021446

answered yesterday

Suever

9,8021446

add a comment |

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:

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

edited yesterday

answered yesterday

Arnauld

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
yesterday

add a comment |

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:

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

edited yesterday

answered yesterday

Arnauld

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
yesterday

add a comment |

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:

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

edited yesterday

answered yesterday

Arnauld

78.8k795327

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:

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

edited yesterday

answered yesterday

Arnauld

78.8k795327

edited yesterday

answered yesterday

Arnauld

78.8k795327

answered yesterday

Arnauld

78.8k795327

answered yesterday

Arnauld

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
yesterday

add a comment |

$begingroup$
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
yesterday

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.

– Erik the Outgolfer
yesterday

add a comment |

Jelly, 8 bytes

1ẋⱮj0ȧþ`

Try it online!

answered yesterday

Erik the Outgolfer

32.5k429105

add a comment |

Jelly, 8 bytes

1ẋⱮj0ȧþ`

Try it online!

answered yesterday

Erik the Outgolfer

32.5k429105

add a comment |

Jelly, 8 bytes

1ẋⱮj0ȧþ`

Try it online!

answered yesterday

Erik the Outgolfer

32.5k429105

Jelly, 8 bytes

1ẋⱮj0ȧþ`

Try it online!

answered yesterday

Erik the Outgolfer

32.5k429105

answered yesterday

Erik the Outgolfer

32.5k429105

answered yesterday

Erik the Outgolfer

32.5k429105

answered yesterday

Erik the Outgolfer

32.5k429105

add a comment |

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!

answered yesterday

xnor

92.4k18188446

add a comment |

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!

answered yesterday

xnor

92.4k18188446

add a comment |

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!

answered yesterday

xnor

92.4k18188446

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!

answered yesterday

xnor

92.4k18188446

answered yesterday

xnor

92.4k18188446

answered yesterday

xnor

92.4k18188446

answered yesterday

xnor

92.4k18188446

add a comment |

APL (Dyalog Unicode), 12 10 12 bytes^SBCS

∘.×⍨∊,,⎕⍴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

edited yesterday

answered yesterday

Sherlock9

8,06411860

1

$begingroup$
you could take input as ⎕ to avoid the { }
$endgroup$
– ngn
yesterday

add a comment |

APL (Dyalog Unicode), 12 10 12 bytes^SBCS

∘.×⍨∊,,⎕⍴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

edited yesterday

answered yesterday

Sherlock9

8,06411860

1

$begingroup$
you could take input as ⎕ to avoid the { }
$endgroup$
– ngn
yesterday

add a comment |

APL (Dyalog Unicode), 12 10 12 bytes^SBCS

∘.×⍨∊,,⎕⍴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

edited yesterday

answered yesterday

Sherlock9

8,06411860

APL (Dyalog Unicode), 12 10 12 bytes^SBCS

∘.×⍨∊,,⎕⍴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

edited yesterday

answered yesterday

Sherlock9

8,06411860

edited yesterday

answered yesterday

Sherlock9

8,06411860

answered yesterday

Sherlock9

8,06411860

answered yesterday

Sherlock9

8,06411860

1

$begingroup$
you could take input as ⎕ to avoid the { }
$endgroup$
– ngn
yesterday

add a comment |

1

$begingroup$
you could take input as ⎕ to avoid the { }
$endgroup$
– ngn
yesterday

you could take input as ⎕ to avoid the { }

– ngn
yesterday

add a comment |

Wolfram Language (Mathematica), 41 bytes

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

Try it online!

edited yesterday

answered yesterday

shrap

1013

add a comment |

Wolfram Language (Mathematica), 41 bytes

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

Try it online!

edited yesterday

answered yesterday

shrap

1013

add a comment |

Wolfram Language (Mathematica), 41 bytes

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

Try it online!

edited yesterday

answered yesterday

shrap

1013

Wolfram Language (Mathematica), 41 bytes

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

Try it online!

edited yesterday

answered yesterday

shrap

1013

edited yesterday

answered yesterday

shrap

1013

answered yesterday

shrap

1013

answered yesterday

shrap

1013

add a comment |

PowerShell, 42 bytes

($r=1..$args[0])|%{"$($r|%{'#'*$_})`n"*$_}

Try it online!

edited 5 hours ago

answered 8 hours ago

Andrei Odegov

46926

$begingroup$
40 bytes :)
$endgroup$
– mazzy
7 hours ago

$begingroup$
@mazzy, cleverly :)
$endgroup$
– Andrei Odegov
7 hours ago

add a comment |

PowerShell, 42 bytes

($r=1..$args[0])|%{"$($r|%{'#'*$_})`n"*$_}

Try it online!

edited 5 hours ago

answered 8 hours ago

Andrei Odegov

46926

$begingroup$
40 bytes :)
$endgroup$
– mazzy
7 hours ago

$begingroup$
@mazzy, cleverly :)
$endgroup$
– Andrei Odegov
7 hours ago

add a comment |

PowerShell, 42 bytes

($r=1..$args[0])|%{"$($r|%{'#'*$_})`n"*$_}

Try it online!

edited 5 hours ago

answered 8 hours ago

Andrei Odegov

46926

PowerShell, 42 bytes

($r=1..$args[0])|%{"$($r|%{'#'*$_})`n"*$_}

Try it online!

edited 5 hours ago

answered 8 hours ago

Andrei Odegov

46926

edited 5 hours ago

answered 8 hours ago

Andrei Odegov

46926

answered 8 hours ago

Andrei Odegov

46926

answered 8 hours ago

Andrei Odegov

46926

$begingroup$
40 bytes :)
$endgroup$
– mazzy
7 hours ago

$begingroup$
@mazzy, cleverly :)
$endgroup$
– Andrei Odegov
7 hours ago

add a comment |

$begingroup$
40 bytes :)
$endgroup$
– mazzy
7 hours ago

$begingroup$
@mazzy, cleverly :)
$endgroup$
– Andrei Odegov
7 hours ago

40 bytes :)

– mazzy
7 hours ago

@mazzy, cleverly :)

– Andrei Odegov
7 hours ago

add a comment |

J, 17 bytes

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

Try it online!

answered yesterday

Galen Ivanov

7,16211034

$begingroup$
did not know about this I. trick!
$endgroup$
– Jonah
yesterday

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

add a comment |

J, 17 bytes

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

Try it online!

answered yesterday

Galen Ivanov

7,16211034

$begingroup$
did not know about this I. trick!
$endgroup$
– Jonah
yesterday

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

add a comment |

J, 17 bytes

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

Try it online!

answered yesterday

Galen Ivanov

7,16211034

J, 17 bytes

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

Try it online!

answered yesterday

Galen Ivanov

7,16211034

answered yesterday

Galen Ivanov

7,16211034

answered yesterday

Galen Ivanov

7,16211034

answered yesterday

Galen Ivanov

7,16211034

$begingroup$
did not know about this I. trick!
$endgroup$
– Jonah
yesterday

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

add a comment |

$begingroup$
did not know about this I. trick!
$endgroup$
– Jonah
yesterday

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

did not know about this I. trick!

– Jonah
yesterday

@Jonah I was waiting for an opportunity to use it :)

– Galen Ivanov
22 hours ago

add a comment |

Jelly, 7 bytes

‘RÄṬ|þ`

Try it online!

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

answered yesterday

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
yesterday

1

$begingroup$
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
yesterday

$begingroup$
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
yesterday

$begingroup$
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
yesterday

1

$begingroup$
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
yesterday

add a comment |

Jelly, 7 bytes

‘RÄṬ|þ`

Try it online!

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

answered yesterday

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
yesterday

1

$begingroup$
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
yesterday

$begingroup$
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
yesterday

$begingroup$
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
yesterday

1

$begingroup$
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
yesterday

add a comment |

Jelly, 7 bytes

‘RÄṬ|þ`

Try it online!

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

answered yesterday

community wiki

ais523

Jelly, 7 bytes

‘RÄṬ|þ`

Try it online!

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

answered yesterday

community wiki

ais523

answered yesterday

community wiki

ais523

community wiki

ais523

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
yesterday

1

$begingroup$
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
yesterday

$begingroup$
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
yesterday

$begingroup$
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
yesterday

1

$begingroup$
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
yesterday

add a comment |

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

1

$begingroup$
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
yesterday

$begingroup$
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
yesterday

$begingroup$
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
yesterday

1

$begingroup$
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
yesterday

Why is this a community wiki?! If you want to waive rep you could just give it to Erik the Outgolfer

– Jonathan Allan
yesterday

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.

– ais523
yesterday

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.

– ais523
yesterday

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.

– Jonathan Allan
yesterday

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.

– ais523
yesterday

add a comment |

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!

answered yesterday

MooseOnTheRocks

1914

add a comment |

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!

answered yesterday

MooseOnTheRocks

1914

add a comment |

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!

answered yesterday

MooseOnTheRocks

1914

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!

answered yesterday

MooseOnTheRocks

1914

answered yesterday

MooseOnTheRocks

1914

answered yesterday

MooseOnTheRocks

1914

answered yesterday

MooseOnTheRocks

1914

add a comment |

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

edited yesterday

answered yesterday

nimi

32.4k32389

$begingroup$
Do the 0 row and column help?
$endgroup$
– dfeuer
yesterday

1

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

add a comment |

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

edited yesterday

answered yesterday

nimi

32.4k32389

$begingroup$
Do the 0 row and column help?
$endgroup$
– dfeuer
yesterday

1

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

add a comment |

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

edited yesterday

answered yesterday

nimi

32.4k32389

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

edited yesterday

answered yesterday

nimi

32.4k32389

edited yesterday

answered yesterday

nimi

32.4k32389

answered yesterday

nimi

32.4k32389

answered yesterday

nimi

32.4k32389

$begingroup$
Do the 0 row and column help?
$endgroup$
– dfeuer
yesterday

1

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

add a comment |

$begingroup$
Do the 0 row and column help?
$endgroup$
– dfeuer
yesterday

1

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

Do the 0 row and column help?

– dfeuer
yesterday

@dfeuer: yes, they save a byte. See the first version of my answer.

– nimi
yesterday

add a comment |

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

edited yesterday

answered yesterday

Graham

2,52678

add a comment |

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

edited yesterday

answered yesterday

Graham

2,52678

add a comment |

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

edited yesterday

answered yesterday

Graham

2,52678

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

edited yesterday

answered yesterday

Graham

2,52678

edited yesterday

answered yesterday

Graham

2,52678

answered yesterday

Graham

2,52678

answered yesterday

Graham

2,52678

add a comment |

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

answered yesterday

Neil

81.7k745178

add a comment |

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

answered yesterday

Neil

81.7k745178

add a comment |

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

answered yesterday

Neil

81.7k745178

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

answered yesterday

Neil

81.7k745178

answered yesterday

Neil

81.7k745178

answered yesterday

Neil

81.7k745178

answered yesterday

Neil

81.7k745178

add a comment |

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.

answered yesterday

Jo King

24.9k359128

add a comment |

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.

answered yesterday

Jo King

24.9k359128

add a comment |

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.

answered yesterday

Jo King

24.9k359128

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.

answered yesterday

Jo King

24.9k359128

answered yesterday

Jo King

24.9k359128

answered yesterday

Jo King

24.9k359128

answered yesterday

Jo King

24.9k359128

add a comment |

Brain-Flak, 168 bytes

(({})<>){(({})<{({}[()]<(<>({})<>){(({})<{({}[(())])}>[()])}{}{}(([(()()()()())(){}]){})>)}{}(({}))>[()])}{}{}{}(){{}({}<>((((()()){}){}){}){})<>()}<>{({}<>)<>}<>{}

Try it online!

answered 21 hours ago

MegaTom

3,5421424

add a comment |

Brain-Flak, 168 bytes

(({})<>){(({})<{({}[()]<(<>({})<>){(({})<{({}[(())])}>[()])}{}{}(([(()()()()())(){}]){})>)}{}(({}))>[()])}{}{}{}(){{}({}<>((((()()){}){}){}){})<>()}<>{({}<>)<>}<>{}

Try it online!

answered 21 hours ago

MegaTom

3,5421424

add a comment |

Brain-Flak, 168 bytes

(({})<>){(({})<{({}[()]<(<>({})<>){(({})<{({}[(())])}>[()])}{}{}(([(()()()()())(){}]){})>)}{}(({}))>[()])}{}{}{}(){{}({}<>((((()()){}){}){}){})<>()}<>{({}<>)<>}<>{}

Try it online!

answered 21 hours ago

MegaTom

3,5421424

Brain-Flak, 168 bytes

(({})<>){(({})<{({}[()]<(<>({})<>){(({})<{({}[(())])}>[()])}{}{}(([(()()()()())(){}]){})>)}{}(({}))>[()])}{}{}{}(){{}({}<>((((()()){}){}){}){})<>()}<>{({}<>)<>}<>{}

Try it online!

answered 21 hours ago

MegaTom

3,5421424

answered 21 hours ago

MegaTom

3,5421424

answered 21 hours ago

MegaTom

3,5421424

answered 21 hours ago

MegaTom

3,5421424

add a comment |

05AB1E, 9 bytes

Defines a program $ftexttt{: }color{purple}{texttt{Nat}} →color{purple}{texttt{List}}texttt{[}color{purple}{texttt{List}}texttt{[}color{purple}{texttt{Nat}}texttt{]]}$.

Code:

$L×0ýSDδ*

Uses the 05AB1E-encoding. Try it online! or use the pretty-printed version.

Explanation:



$              # Push the number 1 and the input n

 L             # Create the list [1, 2, 3, ..., n]

  ×            # Vectorized string multiplication: 1 × [1, 2, 3, ..., n]

                 This would result in ["1", "11", "111", ..., "1" × n]

   0ý          # Join the resulting list with '0', resulting in "10110111011110111110..."

     S         # Split into single digits: [1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, ...]

      Dδ*      # Multiplication table with itself

answered 18 hours ago

Adnan

35.7k562225

add a comment |

05AB1E, 9 bytes

Defines a program $ftexttt{: }color{purple}{texttt{Nat}} →color{purple}{texttt{List}}texttt{[}color{purple}{texttt{List}}texttt{[}color{purple}{texttt{Nat}}texttt{]]}$.

Code:

$L×0ýSDδ*

Uses the 05AB1E-encoding. Try it online! or use the pretty-printed version.

Explanation:



$              # Push the number 1 and the input n

 L             # Create the list [1, 2, 3, ..., n]

  ×            # Vectorized string multiplication: 1 × [1, 2, 3, ..., n]

                 This would result in ["1", "11", "111", ..., "1" × n]

   0ý          # Join the resulting list with '0', resulting in "10110111011110111110..."

     S         # Split into single digits: [1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, ...]

      Dδ*      # Multiplication table with itself

answered 18 hours ago

Adnan

35.7k562225

add a comment |

05AB1E, 9 bytes

Defines a program $ftexttt{: }color{purple}{texttt{Nat}} →color{purple}{texttt{List}}texttt{[}color{purple}{texttt{List}}texttt{[}color{purple}{texttt{Nat}}texttt{]]}$.

Code:

$L×0ýSDδ*

Uses the 05AB1E-encoding. Try it online! or use the pretty-printed version.

Explanation:



$              # Push the number 1 and the input n

 L             # Create the list [1, 2, 3, ..., n]

  ×            # Vectorized string multiplication: 1 × [1, 2, 3, ..., n]

                 This would result in ["1", "11", "111", ..., "1" × n]

   0ý          # Join the resulting list with '0', resulting in "10110111011110111110..."

     S         # Split into single digits: [1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, ...]

      Dδ*      # Multiplication table with itself

answered 18 hours ago

Adnan

35.7k562225

05AB1E, 9 bytes

Defines a program $ftexttt{: }color{purple}{texttt{Nat}} →color{purple}{texttt{List}}texttt{[}color{purple}{texttt{List}}texttt{[}color{purple}{texttt{Nat}}texttt{]]}$.

Code:

$L×0ýSDδ*

Uses the 05AB1E-encoding. Try it online! or use the pretty-printed version.

Explanation:



$              # Push the number 1 and the input n

 L             # Create the list [1, 2, 3, ..., n]

  ×            # Vectorized string multiplication: 1 × [1, 2, 3, ..., n]

                 This would result in ["1", "11", "111", ..., "1" × n]

   0ý          # Join the resulting list with '0', resulting in "10110111011110111110..."

     S         # Split into single digits: [1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, ...]

      Dδ*      # Multiplication table with itself

answered 18 hours ago

Adnan

35.7k562225

answered 18 hours ago

Adnan

35.7k562225

answered 18 hours ago

Adnan

35.7k562225

answered 18 hours ago

Adnan

35.7k562225

add a comment |

Python 3, 88 bytes

n=range(int(input()))

for j in n:

 l=''

 for i in n:l+='#'*-~i+' '

 print((l+'n')*-~j)

Try it online!

answered 16 hours ago

Kerosenic

New contributor

$begingroup$
You can save bytes by declaring l outside the loop. Try it online!
$endgroup$
– Jo King
14 hours ago

$begingroup$
82 bytes.
$endgroup$
– Jonathan Frech
11 hours ago

add a comment |

Python 3, 88 bytes

n=range(int(input()))

for j in n:

 l=''

 for i in n:l+='#'*-~i+' '

 print((l+'n')*-~j)

Try it online!

answered 16 hours ago

Kerosenic

New contributor

$begingroup$
You can save bytes by declaring l outside the loop. Try it online!
$endgroup$
– Jo King
14 hours ago

$begingroup$
82 bytes.
$endgroup$
– Jonathan Frech
11 hours ago

add a comment |

Python 3, 88 bytes

n=range(int(input()))

for j in n:

 l=''

 for i in n:l+='#'*-~i+' '

 print((l+'n')*-~j)

Try it online!

answered 16 hours ago

Kerosenic

New contributor

Python 3, 88 bytes

n=range(int(input()))

for j in n:

 l=''

 for i in n:l+='#'*-~i+' '

 print((l+'n')*-~j)

Try it online!

answered 16 hours ago

Kerosenic

New contributor

answered 16 hours ago

Kerosenic

New contributor

answered 16 hours ago

Kerosenic

answered 16 hours ago

Kerosenic

New contributor

Kerosenic is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$begingroup$
You can save bytes by declaring l outside the loop. Try it online!
$endgroup$
– Jo King
14 hours ago

$begingroup$
82 bytes.
$endgroup$
– Jonathan Frech
11 hours ago

add a comment |

$begingroup$
You can save bytes by declaring l outside the loop. Try it online!
$endgroup$
– Jo King
14 hours ago

$begingroup$
82 bytes.
$endgroup$
– Jonathan Frech
11 hours ago

You can save bytes by declaring l outside the loop. Try it online!

– Jo King
14 hours ago

82 bytes.

– Jonathan Frech
11 hours ago

add a comment |

C# (.NET Core), 208 bytes, N+1 indexed.

using C=System.Console;class M{static void Main(){int i=int.Parse(C.ReadLine()),j=1,k=1;var l="";var t=l;for(;j<=i;j++,k=0)while(k++<j)l+=k<j?"#":" ";for(i=0;i<l.Length;t=l[i++]=='#'?l+"n":"n")C.Write(t);}}

Try it online!

I feel quite like this is not the best approach, but it's what I have at the moment, so here we are.

The program (ab)uses nested loops to build the initial string, and then reads through that string (since it's rectangular!) to determine whether to output the string or just a newline.

answered 4 hours ago

Stackstuck

1277

$begingroup$
202?
$endgroup$
– ASCII-only
1 hour ago

$begingroup$
166, indexing fixed
$endgroup$
– ASCII-only
55 mins ago

$begingroup$
also *square not rectangular
$endgroup$
– ASCII-only
37 mins ago

$begingroup$
Squares are rectangles!
$endgroup$
– Stackstuck
27 mins ago

$begingroup$
also you should post this as your own answer, you did most of the work on it @ASCII-only.
$endgroup$
– Stackstuck
22 mins ago

add a comment |

C# (.NET Core), 208 bytes, N+1 indexed.

using C=System.Console;class M{static void Main(){int i=int.Parse(C.ReadLine()),j=1,k=1;var l="";var t=l;for(;j<=i;j++,k=0)while(k++<j)l+=k<j?"#":" ";for(i=0;i<l.Length;t=l[i++]=='#'?l+"n":"n")C.Write(t);}}

Try it online!

I feel quite like this is not the best approach, but it's what I have at the moment, so here we are.

The program (ab)uses nested loops to build the initial string, and then reads through that string (since it's rectangular!) to determine whether to output the string or just a newline.

answered 4 hours ago

Stackstuck

1277

$begingroup$
202?
$endgroup$
– ASCII-only
1 hour ago

$begingroup$
166, indexing fixed
$endgroup$
– ASCII-only
55 mins ago

$begingroup$
also *square not rectangular
$endgroup$
– ASCII-only
37 mins ago

$begingroup$
Squares are rectangles!
$endgroup$
– Stackstuck
27 mins ago

$begingroup$
also you should post this as your own answer, you did most of the work on it @ASCII-only.
$endgroup$
– Stackstuck
22 mins ago

add a comment |

C# (.NET Core), 208 bytes, N+1 indexed.

using C=System.Console;class M{static void Main(){int i=int.Parse(C.ReadLine()),j=1,k=1;var l="";var t=l;for(;j<=i;j++,k=0)while(k++<j)l+=k<j?"#":" ";for(i=0;i<l.Length;t=l[i++]=='#'?l+"n":"n")C.Write(t);}}

Try it online!

I feel quite like this is not the best approach, but it's what I have at the moment, so here we are.

The program (ab)uses nested loops to build the initial string, and then reads through that string (since it's rectangular!) to determine whether to output the string or just a newline.

answered 4 hours ago

Stackstuck

1277

C# (.NET Core), 208 bytes, N+1 indexed.

using C=System.Console;class M{static void Main(){int i=int.Parse(C.ReadLine()),j=1,k=1;var l="";var t=l;for(;j<=i;j++,k=0)while(k++<j)l+=k<j?"#":" ";for(i=0;i<l.Length;t=l[i++]=='#'?l+"n":"n")C.Write(t);}}

Try it online!

I feel quite like this is not the best approach, but it's what I have at the moment, so here we are.

The program (ab)uses nested loops to build the initial string, and then reads through that string (since it's rectangular!) to determine whether to output the string or just a newline.

answered 4 hours ago

Stackstuck

1277

answered 4 hours ago

Stackstuck

1277

answered 4 hours ago

Stackstuck

1277

answered 4 hours ago

Stackstuck

1277

$begingroup$
202?
$endgroup$
– ASCII-only
1 hour ago

$begingroup$
166, indexing fixed
$endgroup$
– ASCII-only
55 mins ago

$begingroup$
also *square not rectangular
$endgroup$
– ASCII-only
37 mins ago

$begingroup$
Squares are rectangles!
$endgroup$
– Stackstuck
27 mins ago

$begingroup$
also you should post this as your own answer, you did most of the work on it @ASCII-only.
$endgroup$
– Stackstuck
22 mins ago

add a comment |

$begingroup$
202?
$endgroup$
– ASCII-only
1 hour ago

$begingroup$
166, indexing fixed
$endgroup$
– ASCII-only
55 mins ago

$begingroup$
also *square not rectangular
$endgroup$
– ASCII-only
37 mins ago

$begingroup$
Squares are rectangles!
$endgroup$
– Stackstuck
27 mins ago

$begingroup$
also you should post this as your own answer, you did most of the work on it @ASCII-only.
$endgroup$
– Stackstuck
22 mins ago

202?

– ASCII-only
1 hour ago

166, indexing fixed

– ASCII-only
55 mins ago

also *square not rectangular

– ASCII-only
37 mins ago

Squares are rectangles!

– Stackstuck
27 mins ago

also you should post this as your own answer, you did most of the work on it @ASCII-only.

– Stackstuck
22 mins ago

add a comment |

SmileBASIC, 83 77 bytes

Graphical output.
Input is N-1

INPUT N

FOR J=0TO N

X=0FOR I=0TO N

GFILL X,Y,X+I,Y+J

X=X+I+2NEXT

Y=Y+J+2NEXT

edited 4 hours ago

answered 4 hours ago

12Me21

5,61711336

add a comment |

SmileBASIC, 83 77 bytes

Graphical output.
Input is N-1

INPUT N

FOR J=0TO N

X=0FOR I=0TO N

GFILL X,Y,X+I,Y+J

X=X+I+2NEXT

Y=Y+J+2NEXT

edited 4 hours ago

answered 4 hours ago

12Me21

5,61711336

add a comment |

SmileBASIC, 83 77 bytes

Graphical output.
Input is N-1

INPUT N

FOR J=0TO N

X=0FOR I=0TO N

GFILL X,Y,X+I,Y+J

X=X+I+2NEXT

Y=Y+J+2NEXT

edited 4 hours ago

answered 4 hours ago

12Me21

5,61711336

SmileBASIC, 83 77 bytes

Graphical output.
Input is N-1

INPUT N

FOR J=0TO N

X=0FOR I=0TO N

GFILL X,Y,X+I,Y+J

X=X+I+2NEXT

Y=Y+J+2NEXT

edited 4 hours ago

answered 4 hours ago

12Me21

5,61711336

edited 4 hours ago

answered 4 hours ago

12Me21

5,61711336

answered 4 hours ago

12Me21

5,61711336

answered 4 hours ago

12Me21

5,61711336

add a comment |

If this is an answer to a challenge…

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

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

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

More generally…

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

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

draft saved

draft discarded

If this is an answer to a challenge…

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

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

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

More generally…

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

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

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

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

Print a physical multiplication table

Rules

Examples

Rules

Examples

Rules

Examples

Rules

Examples

21 Answers 21

R, 56 54 43 36 30 bytes

Haskell, 43 bytes

MATL, 14 10 bytes

JavaScript (ES6), 73 72 69 bytes

JavaScript (ES7), 87 83 82 bytes

How?

Jelly, 8 bytes

Python 2, 67 bytes

APL (Dyalog Unicode), 12 10 12 bytesSBCS

Explanation

Wolfram Language (Mathematica), 41 bytes

PowerShell, 42 bytes

J, 17 bytes

Jelly, 7 bytes

Explanation

Mouse-2002, 79 bytes

Haskell, 69 68 bytes

APL+WIN, 29 bytes

Charcoal, 18 bytes

Perl 6, 35 bytes

Brain-Flak, 168 bytes

05AB1E, 9 bytes

Code:

Explanation:

Python 3, 88 bytes

C# (.NET Core), 208 bytes, N+1 indexed.

SmileBASIC, 83 77 bytes

Your Answer

Sign up or log in

Post as a guest

Post as a guest

21 Answers 21

21 Answers 21

R, 56 54 43 36 30 bytes

R, 56 54 43 36 30 bytes

R, 56 54 43 36 30 bytes

R, 56 54 43 36 30 bytes

Haskell, 43 bytes

Haskell, 43 bytes

Haskell, 43 bytes

Haskell, 43 bytes

MATL, 14 10 bytes

MATL, 14 10 bytes

MATL, 14 10 bytes

MATL, 14 10 bytes

JavaScript (ES6), 73 72 69 bytes

JavaScript (ES7), 87 83 82 bytes

How?

JavaScript (ES6), 73 72 69 bytes

JavaScript (ES7), 87 83 82 bytes

How?

JavaScript (ES6), 73 72 69 bytes

JavaScript (ES7), 87 83 82 bytes

How?

JavaScript (ES6), 73 72 69 bytes

JavaScript (ES7), 87 83 82 bytes

How?

Jelly, 8 bytes

Jelly, 8 bytes

Jelly, 8 bytes

Jelly, 8 bytes

Python 2, 67 bytes

Python 2, 67 bytes

Python 2, 67 bytes

Python 2, 67 bytes

APL (Dyalog Unicode), 12 10 12 bytesSBCS

Explanation

APL (Dyalog Unicode), 12 10 12 bytesSBCS

Explanation

APL (Dyalog Unicode), 12 10 12 bytesSBCS

21 Answers
21

APL (Dyalog Unicode), 12 10 12 bytes^SBCS

21 Answers
21

21 Answers
21

APL (Dyalog Unicode), 12 10 12 bytes^SBCS

APL (Dyalog Unicode), 12 10 12 bytes^SBCS

APL (Dyalog Unicode), 12 10 12 bytes^SBCS

APL (Dyalog Unicode), 12 10 12 bytes^SBCS