# Generate this number table

Given an integer $$\1 < n < 10 \$$ generate a table like below.

For $$\n = 5\$$,

1 2 3 4 5
2 2 3 4 5
3 3 3 4 5
4 4 4 4 5
5 5 5 5 5


For $$\n = 8\$$,

1 2 3 4 5 6 7 8
2 2 3 4 5 6 7 8
3 3 3 4 5 6 7 8
4 4 4 4 5 6 7 8
5 5 5 5 5 6 7 8
6 6 6 6 6 6 7 8
7 7 7 7 7 7 7 8
8 8 8 8 8 8 8 8


Shortest code wins! Trailing whitespace allowed

Output format isn't strict, you can output a 2D list/matrix in any reasonable format.

• Welcome to Code Golf and nice first question! For future reference, we recommend using the Sandbox to get feedback on challenge ideas before posting them to main. Can the output be a list of lists/matrix, or does it have to be formatted like that? If the formatting is required, what should be the output for $n = 11$? Jun 10 at 16:58
• @cairdcoinheringaahing 1 < n < 10 see edit please Jun 10 at 17:01
• What are the permitted output formats? Jun 10 at 17:08
• Related Jun 10 at 18:34
• @cairdcoinheringaahing, I'm actually pretty sure we've had it with numbers, too, but I can't find it. Jun 10 at 21:30

# Jelly, 2 bytes

»þ


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Outputs an n x n matrix. If the output format is required, 4 bytes

This is simply a table of maximums. For each cell $$\A_{i,j}\$$ in the output, the result is equal to $$\\max(i,j)\$$. The 4 byte answer has $$$$ to reuse the argument, and G to format the output as a grid.

# APL (Dyalog Unicode), 5 bytes (SBCS)

Anonymous tacit prefix function.

∘.⌈⍨⍳


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∘.⌈⍨ maximum-table for

⍳integers 1 through the argument

# Wolfram Language (Mathematica), 16 bytes

Max~Array~{#,#}&


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# Factor, 35 bytes

[ [1,b] dup [ max ] cartesian-map ]


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cartesian-map Takes two sequences (in this case, both are [1, n]) and applies a quotation to each pair of elements, resulting in a matrix. In effect, we are mapping max over a coordinate matrix.

## <>^v, 93 89 bytes

i,tTv
vj1<
>I)iIT‹!0jv
v        <
>J)jJI≥v_~ v
>~  v
^  v≤TJ<\23<

^   <

##### Explanation

i pops the top of the stack (by default 0) and stores it into the variable i. , reads an integer from stdin, then t pops it and stores it into variable t. T pushes to the stack the value of the variable t and v redirects the intruction pointer towards the end of the next line. < tells the instruction pointer to go left. 1 pushes 1 onto the stack and j stores it into the variable j, and  prints a newline. Then v and > sends the instruction towards the right on the next line. I pushes the variable i's value onto the stack, ) increments it, i takes it and puts it into i, I pushes the value of i onto the stack and T brings onto the stack the number that was previously read to stdin. ‹ compares them and terminates the program if the grid has finished printing. Then it initializes j to 0 and sends the instruction pointer into a loop that increments j, then prints a space (\23 (pointer is currently going left, that's why it is reversed) pushes 32 (ASCII id of " ") and prints it) (32\ is used instead of " "~ because it is one character shorter), prints j if j is lower than i, or else prints i. When j is greater or equal to t, it exits the loop, prints a new line and goes back into the outer loop.

In the code, the t variable contains the number read from stdin, i is the current line number, and j is the current character number.

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# J, 9 bytes

1+>./~@i.


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## Explanation

1+>./~@i.    input: n
i.    range from 0 to n-1 inclusive
/~       table:
>.            max(n, n)


# R, 30 bytes

function(n)outer(1:n,1:n,pmax)


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# Octave, 19 bytes

@(n)max([1:n]',1:n)


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• For Octave you can actually do this to save 2 bytes: @(n)max(x=1:n,x') Jun 11 at 12:00

# Stax, 3 bytes

╗ìΦ


Run and debug it

# Python 3, 49 bytes

f=lambda x:x*[x]and[a+[x]for a in f(x-1)]+[[x]*x]


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# Vyxal, 6 bytes

ƛ£⁰ƛ¥∴


Try it Online! Returns an array of arrays

ƛ      # (1...input) Map...
£     # Push to register
⁰ƛ   # (1...input) Map...
¥  # Register
∴ # Maximum of the two


Or, for grid output, 10 bytes:

ƛ£⁰ƛ¥∴;Ṅ;⁋


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ƛ       ;  # (1...input) Map...
£         # Push to register
⁰ƛ  ;    # (1...input) Map...
¥      # Push register
∴     # Max
Ṅ   # Joined by spaces
⁋ # Joined by newlines.


# Python 2, 48 bytes

r=range(1,input()+1)
for x in r:print[x]*x+r[x:]


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Prints a list for each row.

# Julia, 19 bytes

!n=max.((1:n)',1:n)


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# K (ngn/k), 10 bytes

{x|\x}1+!:


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Somewhat surprisingly, using a seeded scan x|\x returns the same results as an each-right (x|/:x) or each-left (x|\:x).

• 1+!: generate 1..n from the (implicit) input
• {x|\x} set up a scan, seeded with the input, run over the input

# PowerShell Core, 40 bytes

param($a)1..$a|%{"$(,$_*($_-1)+$_..$a)"}  Try it online! # C# (Visual C# Interactive Compiler), 70 bytes a => new int[a][].Select((x,p)=>new int[a].Select((y,q)=>p>q?p+1:q+1))  I'm sure there is a shorter way to do this, but I can't find it. Try it online! • The first 2 spaces are unnecessary and the first array doesn't need to be multidimentional, saving 4 bytes. Jun 11 at 11:07 # TI-Basic, 38 41 bytes Input N identity(N→[A] For(I,1,N For(J,1,N max(I,J→[A](I,J End End [A]  Output is stored as a matrix in Ans and is displayed. # Rust, 84 bytes |n|(1..=n).fold("".into(),|a,i|(1..=n).fold(a,|b,j|format!("{}{} ",b,i.max(j)))+" ")  Try it online! The output will be a String. The newline is given literally inside quotes, which saves one byte. New contributor Ezhik is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. # C (gcc), 65 bytes i,j;f(n){for(i=1;n/i;)printf(j/n?j=!++i,"%d\n":"%d ",i>++j?i:j);}  Try it online! # Ruby, 35 bytes ->x{(1..x).map{|y|[y]*y+[*y+1..x]}}  Try it online! # Japt, 5 bytes Outputs a 2D-array. õ@õwX  Try it õ@õwX :Implicit input of integer U õ :Range [1,U] @ :Map each X õ : Range [1,U] wX : Max of each with X  Or, if outputting a 1D-array is allowed: # Japt, 4 bytes õ ïw  Try it õ ïw :Implicit input of integer U õ :Range [1,U] ï :Cartesian product with itself w :Reduce each pair by Max  • so close to having õwõ but not quite :( Jun 10 at 17:03 • @hyper-neutrino, wouldn't be a valid Japt programme, anyway. Jun 10 at 17:09 # JavaScript (Node.js), 56 bytes n=>[...Array(n)].map((_,x,a)=>a.map((_,y)=>x>y++?-~x:y))  Try it online! Outputs an array of arrays. Thanks to Shaggy for -5 bytes Thanks to Arnauld for further -2 bytes # JavaScript (Node.js), 71 bytes n=>[...Array(n)].map((_,x,a)=>a.map((_,y)=>x>y++?-~x:y).join).join   Try it online! Outputs string ## History n=>[...Array(n)].map((_,x,a)=>a.map((_,y)=>-~x>++y?-~x:y)) // 58 n=>[...m=Array(n)].map((_,a)=>[...m].map((_,b)=>-~a>++b?-~a:b)) // 63 n=>[...(m=Array(n))].map((_,a)=>[...m].map((_,b)=>-~a>++b?-~a:b)) // 65 n=>[...(m=Array(n)).keys()].map(a=>[...m.keys()].map(b=>-~a>++b?-~a:b)) // 71  • 58 bytes Jun 10 at 21:36 • 56 bytes (based on Shaggy's version) Jun 10 at 21:58 • Damn, @Arnauld, how'd I miss that? I blame being back behind my bar for the first time in 6 months; I'm clearly out of practice at "bar-golfing"! Jun 10 at 23:16 # 05AB1E, 6 bytes Lã€àsä  Try it online! Outputs as a 2D list. Link includes a footer to format the output. Lã€àsä # full program € # list of... à # maximum... € # s... à # of... € # each element of... ã # list of distinct combinations of two elements in... L # [1, 2, 3, ..., # ..., implicit input... L # ]... ã # repeated twice... ä # split into pieces of length... s # implicit input # implicit output  • 4 bytes. (I'm honestly surprised M works here tbh. ;p) 2 days ago # Haskell, 28 bytes f n=(<$>[1..n]).max<$>[1..n]  Try it online! Generates each row by taking the max of each element of [1..n] and a fixed value equal to the row number. map(max 3)[1,2,3,4,5] = [3,3,3,4,5]  We use <$> as an infix synonym for map.

It would be nice to reuse the <$>[1..n], but it runs into type-checking issues 26 bytes, doesn't work f n|q<-(<$>[1..n])=q$q.max  Try it online! If a 1D output is allowed, we can do: 25 bytes f n=max<$>[1..n]<*>[1..n]


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# Python 3, 67 bytes

lambda n:[print(*(max(y,x)+1 for y in range(n))) for x in range(n)]


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# Python 3, 51 49 bytes

lambda n:mgrid[:n,:n].max(0)+1
from numpy import*


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# T-SQL (277 bytes)

Not sure how inputs are handled in SQL but my first declared parameter is the input. I also can't get an online T-SQL compiler that runs this. But this will run in your SSMS just fine.

DECLARE @a int = 7DECLARE @ int=1DECLARE @c int=1DECLARE @b nvarchar(max)='SELECT'WHILE @<=@a BEGIN
SET @c=1WHILE @c<=@a BEGIN
SET @b+=IIF(@c=1,'',',')+STR(IIF(@c<@,@,@c))+'AS['+STR(@c)+']'SET @c+=1
END
SET @b+=IIF(@<@a,'UNION ALL SELECT','')SET @+=1
END
EXEC sp_executesql @b;


It's a naive approach with nothing special, except that it dynamically builds up a query.

The query it generates is:

SELECT         1AS[         1],         2AS[         2],         3AS[         3]UNION ALL SELECT         2AS[         1],         2AS[         2],         3AS[         3]UNION ALL SELECT         3AS[         1],         3AS[         2],         3AS[         3]


# Regenerate, 80 bytes

((${$2+1}|1)( $2){$2-1}( ${$4+1}{1-$4/$~1}| ${$2+1}{1-$2/$~1}){$~1-$2+1}\n){$~1}  Takes the argument from the command-line. Try it here! ### Explanation (Spaces are replaced with underscores to improve visibility.) Each line is generated in three parts. Top-level structure: ( ) Group 1: a line of output (...) Group 2: the first digit (...){...} Group 3: repetitions of the first digit (...){...} Group 4: count up to N \n Add a newline {$~1}  Repeat, generating N lines


Group 2, the first digit on each line:

(         )
${$2+1}     Add 1 to the previous value of group 2
|    Or, if group 2 has not been previously matched...
1   Use 1


Group 3, the rest of the digits on the line that are identical to the first number:

(_$2) The most recent value of group 2 (with a space in front of it) {$2-1}  repeated (group 2) - 1 times


Group 4, the remaining numbers on the line, starting with (group 2) + 1 and counting up to N:

( ${$4+1}{1-$4/$~1}| ${$2+1}{1-$2/$~1}){$~1-$2+1}
(                                     )
_                                                 Space
${$4+1}                                          Previous value of group 4, plus 1
{        }                                Repeat the value this many times:
$4/$~1                                  1 if group 4 == N, 0 otherwise
1-                                        Subtract from 1: 1 -> 0 and 0 -> 1
So if the new value would have been N+1,
|                               If the previous value of group 4 was not
a number, the arithmetic fails, and we
_                              Space
${$2+1}                       Last value of group 2, plus 1
{1-$2/$~1}             Repeat 0 times if group 2 == N
{        }  Repeat the above
$~1-$2+1   N - (group 2) + 1 times


### Worked example

Suppose the input is 3.

Line 1:

• Group 2: not previously matched, so 1
• Group 3:  1 repeated 1-1 = 0 times, so empty string
• Group 4 is repeated 3-1+1 = 3 times:
1. Group 4 not previously matched, so   followed by 2 repeated 1-1/3 = 1 time (the division is integer division)
2. Previous value of group 4 is  2, so   followed by 3 repeated 1-2/3 = 1 time
3. Previous value of group 4 is  3, so   followed by 4 repeated 1-3/3 = 0 times

Result: 1 2 3

Line 2:

• Group 2: previous value is 1, so 2
• Group 3:  2 repeated 2-1 = 1 time
• Group 4 is repeated 3-2+1 = 2 times:
1. Previous value of group 4 is  , which is not a number, so   followed by 3 repeated 1-2/3 = 1 time
2. Previous value of group 4 is  3, so   followed by 4 repeated 1-3/3 = 0 times

Result: 2 2 3

Line 3:

• Group 2: previous value is 2, so 3
• Group 3:  3 repeated 3-1 = 2 times
• Group 4 is repeated 3-3+1 = 1 time:
1. Previous value of group 4 is  , which is not a number, so   followed by 4 repeated 1-3/3 = 0 times

Result: 3 3 3

# Nim, 138 122 bytes

proc x(i:int)=
for x in 1..i:
for j in 1..x:stdout.write($x&" ") for j in x+1..i:stdout.write($j&" ")
echo()


# APL (Dyalog Unicode), 11 bytes

{(⍳⍵)∘.⌈⍳⍵}


Anonymous prefix function

-2 bytes thanks to @Adam in chat

• You can use ∘.⌈⍨⍳⍵ instead of (⍳⍵)∘.⌈⍳⍵, buuut then it's the same as Adam's answer :/
– rues
Jun 10 at 17:12
• @user thanks but Adam already ninja'd the suggestion :P (in chat) Jun 10 at 17:14
• Oh, I see. You don't need the f←, by the way.
– rues
Jun 10 at 17:15
• @user oh sorry (filler) Jun 10 at 17:15
• How to get to a solution as shot as, but different from, mine: Notice that you have two "outer" function applications (⍳ and ⍳), with their results being the arguments to an "inner" dyadic function ∘.⌈. This is perfect for a fork: ⍳∘.⌈⍳