N-dimensional N^N array filled with N

Question

In: Enough memory and a positive integer N

Out: N-dimensional N^N array filled with N, where N^N means N terms of N-by-N-by-N-by...

Examples:

1: [1] which is a 1D array (a list) of length 1, containing a single 1

2: [[2,2],[2,2]] which is a 2D array (a table) with 2 rows and 2 columns, filled with 2s

3: [[[3,3,3],[3,3,3],[3,3,3]],[[3,3,3],[3,3,3],[3,3,3]],[[3,3,3],[3,3,3],[3,3,3]]] which is a 3D array (a cube) with 3 layers, 3 rows, and 3 columns, filled with 3s

4: [[[[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4]],[[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4]],[[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4]],[[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4]]],[[[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4]],[[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4]],[[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4]],[[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4]]],[[[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4]],[[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4]],[[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4]],[[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4]]],[[[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4]],[[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4]],[[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4]],[[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4]]]]

5 and 6: Please see one of the answers.

Answer 1

Python, 32 bytes

lambda n:eval('['*n+'n'+']*n'*n)

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Makes a string like "[[[n]*n]*n]*n" with n multiplcations, and evaluates it as Python code. Since the evaluation happens within the function scope, the variable name n evaluates to the function input.

Answer 2

J, 4 bytes

$~#~

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Explanation

$~#~  Input: integer n
  #~  Create n copies of n
$~    Shape n into an array with dimensions n copies of n

Answer 3

APL (Dyalog APL), 4 bytes

⍴⍨⍴⊢

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Answer 4

Mathematica, 22 20 bytes

(t=Table)@@t[#,#+1]&

(* or *)

Table@@Table[#,#+1]&

Answer 5

R, 26

This is the obvious answer but perhaps there is something cleverer?

n=scan();array(n,rep(n,n))

Answer 6

JavaScript (ES6),  44  40 bytes

f=(n,k=i=n)=>i--?f(n,Array(n).fill(k)):k

Demo

f=(n,k=i=n)=>i--?f(n,Array(n).fill(k)):k

console.log(JSON.stringify(f(1)))
console.log(JSON.stringify(f(2)))
console.log(JSON.stringify(f(3)))
console.log(JSON.stringify(f(4)))

Answer 7

05AB1E (legacy), 6 5 bytes

-1 thanks to Kevin Cruijssen

F¹.D)

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F     # For 0 .. input
 ¹.D) # Push <input> copies of the result of the last step as an array

Answer 8

Haskell, 52 bytes

f n=iterate(filter(>'"').show.(<$[1..n]))(show n)!!n

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Inspired by @nimi's answer, but using more predefined functions.

• Uses iterate and !! instead of a recursive help function.
• Instead of constructing list delimiters "by hand", uses filter(>'"').show to format a list of strings, then stripping away the extra " characters.

Answer 9

Octave, 35 33 25 23 20 bytes

@(N)ones(N+!(1:N))*N

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@(N)ones(N*ones(1,N))*N

@(N)repmat(N,N*ones(1,N))

Thanks to @LuisMendo saved 8 bytes

@(N)ones(num2cell(!(1:N)+N){:})*N

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Previous answer:

@(N)repmat(N,num2cell(!(1:N)+N){:})

Try it online!

Answer 10

Haskell, 62 bytes

n#0=show n
n#l='[':tail((',':)=<<n#(l-1)<$[1..n])++"]"
f n=n#n

Usage example: f 2 -> "[[2,2],[2,2]]". Try it online!.

Haskell's strict type system prevents a function that returns nested lists of different depths, so I construct the result as a string.

How it works:

n#l=                         n with the current level l is
    '[':                     a literal [ followed by
           n#(l-1)<$[1..n]   n copies of   n # (l-1)
        (',':)=<<            each prepended by a , and flattened into a single list
      tail                   and the first , removed
                  ++"]"      followed by a literal ]

n#0=show n                   the base case is n as a string

f n=n#n                      main function, start with level n         

Answer 11

Python 2, 36 bytes

-2 bytes thanks to @CalculatorFeline

a=n=input()
exec"a=[a]*n;"*n
print a

Try it online!

Answer 12

MATL, 8 bytes

ttY"l$l*

Try it at MATL Online (I have added some code showing the actual size of the output since all n-dimensional outputs in MATL are shown as 2D matrices where all dimensions > 2 are flattened into the second dimension).

Explanation

        % Implicitly grab the input (N)
tt      % Make two copies of N
Y"      % Perform run-length decoding to create N copies of N
l$1     % Create a matrix of ones that is this size  
*       % Multiply this matrix of ones by N
        % Implicitly display the result  

Answer 13

CJam, 12 bytes

ri:X{aX*}X*p

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Explanation

ri:X          Read an integer from input, store it in X (leaves it on the stack)
    {   }X*   Execute this block X times:
     a          Wrap the top of stack in an array
      X*        Repeat the array X times
           p  Print nicely

Answer 14

Jelly, 5 bytes

⁾Wẋẋv

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How?

⁾Wẋẋv - Main link: n                            e.g.       3
⁾Wẋ   - character pair literal ['W','ẋ']                  "Wẋ"
   ẋ  - repeat list n times                               "WẋWẋWẋ"
    v - evaluate as Jelly code with input n          eval("WẋWẋWẋ", 3)
      - ...
        WẋWẋ... - toEval: n                e.g. 3
        W        - wrap                        [3]
         ẋ       - repeat list n times         [3,3,3]
          Wẋ     - wrap and repeat            [[3,3,3],[3,3,3],[3,3,3]]
            ...  - n times total             [[[3,3,3],[3,3,3],[3,3,3]],[[3,3,3],[3,3,3],[3,3,3]],[[3,3,3],[3,3,3],[3,3,3]]]

Answer 15

Java 97 96 95 bytes

Object c(int n,int i){Object[]a=new Object[n];for(int j=0;j<n;)a[j++]=i<2?n:c(n,i-1);return a;}

---

Ungolfed:

public class N_Dim {

    public static Object create(int n) {
        return create(n, n);
    }

    public static Object create(int n, int i) {
        Object[] array = new Object[n];
        for(int j=0;j<n;j++) {
            array[j] = i<2?n:create(n, i - 1);
        }
        return array;
    }

    public static void main(String[] args) {
        System.out.println(Arrays.deepToString((Object[]) create(3)));
    }

}

Answer 16

JavaScript (ES6), 38 bytes

f=(n,m=n)=>m?Array(n).fill(f(n,m-1)):n

The memory-hungry version of this is 45 bytes:

f=(n,m=n)=>m?[...Array(n)].map(_=>f(n,m-1)):n

Answer 17

Bash + GNU utilities, 117 bytes

n=$[$1**$1]
seq -f$1o%.fd$n+1-p $n|dc|rev|sed -r "s/(0+|$[$1-1]*).*$/\1/;s/^(0*)/\1$1/;s/^1/[1]/"|tr \\n0$[$1-1] \ []

Try it online!

---

The program essentially counts from 0 to (n^n)-1 in base n, where n is the input. For each base-n number k in the count, it does the following:

1. If k ends with at least one digit 0, print a '[' for each digit 0 at the end of k.
2. Print n.
3. If k ends with at least one digit n-1, print a ']' for each digit n-1 at the end of k.

(The value n=1 needs to have brackets added as a special case. This input value also generates some output to stderr, which can be ignored under standard PPCG rules.)

Maybe there's a shorter way to implement this idea.

---

Sample run:

./array 3
[[[3 3 3] [3 3 3] [3 3 3]] [[3 3 3] [3 3 3] [3 3 3]] [[3 3 3] [3 3 3] [3 3 3]]]

Answer 18

Jelly, 4 bytes

R»µ¡

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R»µ¡
R     Range. 2 -> [1, 2]
 »    Max between left arg and right arg. Vectorizes. -> [2, 2]
  µ   Separates into a new chain.
   ¡  Repeat 2 times. After another iteration this yields [[2, 2], [2, 2]].

Same thing but with a single monad and no need for the chain separator:

4 bytes

»€`¡

Answer 19

Jelly, 5 bytes

Wẋ³µ¡

Try it online!

Explanation

         Implicit input: N
   µ¡    Apply N times to input:
Wẋ³        “N copies of”

Answer 20

Python 3, 57 53 50 38 bytes

f=lambda n,c=0:n-c and[f(n,c+1)*n]or 1

Try it online!

---

-4 bytes thanks to @CalculatorFeline

---

34 bytes:

f=lambda c,n:c and[f(c-1,n)*n]or 1

Needs to be called as f(4,4)

Answer 21

Ruby, 28 26 bytes

Thanks to Cyoce for saving 2 bytes!

->n{eval'['*n+'n'+']*n'*n}

Stolen shamelessly from xnor's excellent answer.

Answer 22

Ruby, 27 bytes

->a{(z=a).times{z=[z]*a};z}

Only 1 byte more but using a different approach instead of the 'eval' trick from xnor's wonderful Python answer.

Answer 23

PHP, 70 62 bytes

This is the simplest I can come up with.

for(;$i++<$n=$argv[1];)$F=array_fill(0,$n,$F?:$n);print_r($F);

Takes the input as the first argument and prints the resulting array on the screen.

---

Thanks to @user59178 for saving me 8 bytes!

Answer 24

Perl 6, 25 bytes

{($^n,{$_ xx$n}...*)[$n]}

Starts with n, and iteratively applies the "repeat n times" transformation n times, each time creating an additional level of List nesting.

Try it online!

Answer 25

Clojure, 36 bytes

#(nth(iterate(fn[a](repeat % a))%)%)

Iterates function which repeats its argument n times, it produces infinite sequence of such elements and then takes its nth element.

See it online

Answer 26

Rebol, 45 bytes

func[n][array/initial append/dup copy[]n n n]

Answer 27

Batch, 141 bytes

@set t=.
@for /l %%i in (2,1,%1)do @call set t=%%t%%,.
@set s=%1
@for /l %%i in (1,1,%1)do @call call set s=[%%%%t:.=%%s%%%%%%]
@echo %s%

Batch doesn't actually have arrays so this just prints the string representation of an array. Explanation: The first two lines build up a repeated pattern of N .s separated by N-1 ,s in the variable t. The fourth line then uses this as a substitution pattern N times to create the N-dimensional array. The double call is necessary because of how the for and set statements work. First, the for command substitutes variables. As it happens, all of my % signs are doubled, so this does nothing except to unquote them all, resulting in call call set s=[%%t:.=%s%%%]. It then repeats the resulting statement N times. Each time, the call command substitutes variables. At this point, the s variable only has a single set of %s, so it gets substituted, resulting in (e.g.) call set s=[%t:.=[2,2]%]. The inner call then substitutes the t variable, resulting in (e.g.) set s=[[2,2],[2,2]], performing the desired assignment. The final value of s is then printed.

Answer 28

Clojure, 49 bytes

(defmacro r[n]`(->> ~n ~@(repeat n`(repeat ~n))))

Not the shortest Clojure example, but I amused myself with the quoting and unquoting.

Answer 29

I, 7 bytes

I got this from my colleague, the creator of I.

#Bbhph~

#Bb     the copy # function Bound to binding
   hp   hook the argument to (the right of) the power function (repeat)
     h~ hook the argument to the left ~ (of the entire resulting function)

Try it online!

Answer 30

Common Lisp, 128 102 95 79 bytes

(defun f(x &optional y)(if(if y(< y 2))x(fill(make-list x)(f x(if y(1- y)x)))))

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Answer 31

MathGolf, 5 4 bytes

_{a*

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Explanation using n = 2 (note that the outermost [] are not arrays, but are just there to show the stack)

_      duplicate TOS (stack is [2, 2])
 {     loop 2 times (stack is [2])
  a    wrap in array ([2] => [[2]], [[2, 2]] => [[[2, 2]]])
   *   pop a, b : push(a*b) ([[2]] => [[2, 2]], [[[2, 2]]] => [[[2, 2], [2, 2]]])

Answer 32

Perl 6, 61 bytes

my $x=prompt(0);my @a=$x xx$x;"@a=[@a] xx $x;".EVAL xx$x-1;

Big rippof from the Python 2 answer, but converted :P

Answer 33

Clojure, 63 bytes

#(loop[a(repeat % %)d 1](if(= d %)a(recur(repeat % a)(inc d))))

This is a lambda function, usage is like so:

(#(...) {input_no})

...where {input_no} is replaced with the number.

Output for 3 is like this:

(((3 3 3) (3 3 3) (3 3 3)) ((3 3 3) (3 3 3) (3 3 3)) ((3 3 3) (3 3 3) (3 3 3)))

This uses Clojure's definition of lists, which are denoted as ().

Ungolfed code and explanation:

; Defines the function
(defn layered [n]
  ; Begins a loop with a variable depth of 1,
  ; and a list of n elements which are all n
  (loop [depth 1
         array (repeat n n)]
    ; If "depth" is equal to n, return the list
    (if (= depth n) array
      ; Else, continue on with the loop, with
      ; an incremented "depth"...
      (recur (inc depth)
        ; ...and a list which contains the
        ; list repeated n times
        (repeat n array)))))

Answer 34

SNOBOL4 (CSNOBOL4), 59 bytes

	DEFINE('F(N)')
F	F =ARRAY(DUPL(N ',',N - 1) N,N)	:(RETURN)

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Defines a function that returns an ARRAY with the appropriate dimensions and filled with value N.

Answer 35

Haskell, 51 bytes

Other than the existing Haskell solutions this constructs a usable data type not just a string representation thereof (and is shorter):

data L=N[L]|E Int
f n=iterate(N.(<$[1..n]))(E n)!!n

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Explanation / Ungolfed

The reason why this is an interesting challenge in Haskell is that a solution to this challenge needs to return different deeply nested lists and Haskell (without importing external modules) does not support this. One workaround which turned out to be the golfiest, is to define our own data type:

data NestedList = Nest [NestedList] | Entry Int

Now we can just define a function f :: Int -> NestedList which is able to represent all the required data types:

pow n = iterate (Nest . replicate n) (Entry n) !! n

Answer 36

Japt, 14 bytes

_òU}g[UpU ÆUÃ]

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Output is wrapped in an extra singleton array.

Explanation:

          Æ Ã     #Create an array
           U      #Where every element is U
      UpU         #And the length is U^U
_  }g[       ]    #Repeat this function U times:
 òU               # Cut into slices of length U

Answer 37

Python 3, 89 Bytes

def f(x):
 r=[x for _ in range(x)]
 for _ in range(x-1):r=[r for _ in range(x)]
 print(r)

Answer 38

Lithp, 82 bytes

(def f #N::((def i #N,I,K::((if(> I 0)((i N(- I 1)(list-fill N K)))K)))(i N N N)))

Try it online!

I tried implementing this as a list comprehension, but couldn't comprehend how to do it correctly. Instead I went for an implementation based on the JavaScript answer. Unfortunately, my language is fairly long-winded, and this is complicated by the fact that I lack shorthand and other useful features.

A more readable version is available at the Try it Online link.

Answer 39

Hoon, 56 bytes

|=
n/@
=+
i=1
|-
?:
=(n i)
(reap n n)
(reap n $(i +(i)))

Create a new function that takes an atom n. Make a variable i starting at 1, and start a loop: if i==n return a list with n elements of n, else return a list with n elements of the value returned by recursing to the start of the loop with i = i + 1.

I'm a little bit upset that there's not really anything you can do to golf this in Hoon :/ The standard trick of using unnamed variables doesn't apply because it's longer to use lark syntax for once, due to the loop shifting the location of i.

> =f |=
  n/@
  =+
  i=1
  |-
  ?:
  =(n i)
  (reap n n)
  (reap n $(i +(i)))
> (f 1)
~[1]
> (f 2)
~[~[2 2] ~[2 2]]
> (f 3)
~[~[~[3 3 3] ~[3 3 3] ~[3 3 3]] ~[~[3 3 3] ~[3 3 3] ~[3 3 3]] ~[~[3 3 3] ~[3 3 3] ~[3 3 3]]]
> (f 4)
~[
  ~[
    ~[~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4]]
    ~[~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4]]
    ~[~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4]]
    ~[~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4]]
  ]
  ~[
    ~[~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4]]
    ~[~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4]]
    ~[~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4]]
    ~[~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4]]
  ]
  ~[
    ~[~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4]]
    ~[~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4]]
    ~[~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4]]
    ~[~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4]]
  ]
  ~[
    ~[~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4]]
    ~[~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4]]
    ~[~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4]]
    ~[~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4] ~[4 4 4 4]]
  ]
]

Answer 40

WendyScript, 34 bytes

#:(x){<<j=[x]*x#i:1->x j=[j]*x/>j}
#:(x){<<j=[x]*x#i:1->x j=[j]*x/>j}(2) // [[2, 2], [2, 2]]

Try it online!

Answer 41

12-basic, 42 bytes

A=[N=INPUT()]*N
FOR I=2TO N
A=[A]*N
NEXT?A

Answer 42

V, 14 bytes

é,"aPÀñDÀpys0]

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input

é,               # write comma
  "aP            # paste register A (input)
     Àñ          # loop À (number in register A) times
       D         # cut to end of line
        Àp       # paste À times
          ysB]   # surround everything before comma in square brackets

Answer 43

C (GCC), 53 bytes

Since C "has" multidimensional arrays (i.e. chunks of memory plus automatic offset calculation), this is almost a reasonable solution. Still, it's mostly a joke and borders on non-competing.

f(n){int s=pow(n,n),*p=malloc(s*8);wmemset(p,n,s);}

Try It Online

In the TIO I go through the ceremony of casting the result (which is an int) to the appropriate array type despite the fact that my array print function (appropriately) takes a generic pointer.

While I can't demonstrate that the structure of the array is determined by f (it's not), I do include a flat printout of the elements of the output for an input of 3 using the multidimensional array element access syntax, to show that the layout of the returned memory matches that of the corresponding array type.

Byte count

• function: 51 bytes
• compiler flags (-lm): 2 bytes

Unportability

• The result pointer is returned through an int, so this may not work for all heap addresses.
• By using wmemset I require that the sizes of int and wchar_t are equal, and the scaling in the argument to malloc requires the sizes to be at most 8 bytes.
• Implicit declaration madness. I don't even know why passing ints to an unprototyped pow that actually takes doubles seems to work correctly.
• The function "returns" by (I think) just leaving the return value of wmemset in the proper register. This behavior may be particular to the x86 architecture.

Answer 44

Julia 0.6, 23 bytes

n->fill(n,fill(n,n)...)

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The first argument to fill is the value to fill the new array with. The following arguments to that are the dimensions of the new array. Here we create those dimensions themselves using another fill call: we create an array of n ns, then splat that with ... so that the first fill call gets n number of dimension arguments, each with value n - so it creates an n-dimensional array where each dimension length is n, and filled with the value n.

Answer 45

Pip -p, 12 bytes

YaLaY[y]RLay

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Explanation

YaLaY[y]RLay
Y             Yank into the y variable
 a            the first command-line argument, a
  La          Fixed-iterations loop, do the following a times:
     [y]       y wrapped in a list
        RLa    repeated a times (results in a list containing a copies of y)
    Y          Yank that list back into y
           y  After the loop, print y

Answer 46

Perl 5, 34 bytes

sub{eval'@_=[(@_)x$x];'x($x="@_")}

Try it online!

Answer 47

Lua, 85 bytes

function f(n,d,x)d=d or 2 x={}for i=1,n do x[i]=d>n and n or f(n,d+1)end return x end

Try it online!

(The TIO link contains a bit of extra code to print the table as Lua has no built-in way of printing a table's contents)

Explanation

function f(n, d, x)
  d = d or 2
  x = {}
  for i = 1, n do
    x[i] = d > n and n or f(n, d + 1)
  end
  return x
end

As you can probably see, f is a recursive function which takes the number N as n, and two other arguments, d and x. d is the depth, and is not used in the initial call, in which case it defaults to 2. x is not actually used as a parameter, but is there because the function requires a temporary local variable, and it is shorter to declare it as a parameter than to use the local keyword.