Print a NxN integer involute

Question

Given an integer \$N\$, you must print a \$N\times N\$ integer involute with the numbers increasing in a clockwise rotation. You can start with either 0 or 1 at the top left, increasing as you move towards the centre.

Examples

Input => 1
Output => 
0

Input => 2
Output => 
0 1
3 2

Input => 5
Output => 
 0  1  2  3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10  9 8 

Input => 10
Output => 
 0  1  2  3  4  5  6  7  8  9
35 36 37 38 39 40 41 42 43 10
34 63 64 65 66 67 68 69 44 11
33 62 83 84 85 86 87 70 45 12
32 61 82 95 96 97 88 71 46 13
31 60 81 94 99 98 89 72 47 14
30 59 80 93 92 91 90 73 48 15
29 58 79 78 77 76 75 74 49 16
28 57 56 55 54 53 52 51 50 17
27 26 25 24 23 22 21 20 19 18

You may output a 2 dimensional array, or a grid of numbers.

Challenge inspired by Article by Eugene McDonnell

This is code-golf so the goal is to minimize your source code with answers being scored in bytes.

Answer 1

Jelly, 13 bytes

⁸JW;ṚZ+LƲƲ2¡¡

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A nilad function or full program that takes a number from stdin and returns the 1-based involute of size n × n.

How it works

⁸JW;ṚZ+LƲƲ2¡¡
⁸                Empty list ([])
          2¡¡    Take n from stdin and repeat 2n times starting from the above:
                   Given previous matrix m,
    ṚZ+LƲ          Rotate m and add the number of rows of m
 JW;     Ʋ         Prepend a row of 1..(number of rows of m)

The involute can be constructed iteratively as follows:

Start with m = [] (think of it as 0-row, 0-col matrix)
Repeat 2n times:
    If m has r rows, add r to every cell of m
    Rotate m 90deg clockwise
    Attach 0..r-1 (or 1..r) as a new row at the top of m

After every step, the matrix m is always an involute for some rectangular size (square at even steps). For example, using 1-based indexing, doing a single step on

1 2 3
8 9 4
7 6 5

gives

 1  2  3
10 11  4
 9 12  5
 8  7  6

and then

 1  2  3  4
12 13 14  5
11 16 15  6
10  9  8  7

Answer 2

MATL, 12 11 bytes

UG1YL-GoQ&P

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MATL (like MATLAB) has a spiral matrix which spirals clockwise from the center, so this performs the necessary flips and subtracts the matrix from N^2.

Probably Luis Mendo knows how to golf this... Thanks to Luis Mendo for −1 byte.

	% Implicit input N
U	% Square
G1YL	% Push the N×N spiral matrix (S)
	% clockwise from the center, starting at 1
-	% N^2 - S, element-wise
GoQ	% Push mod(N,2)+1
&P	% Flip along that dimension:
	% for odd N, flips the rows; for even N, flips the columns
	% Implicit output

Answer 3

APL (Dyalog Unicode), ^{18 17} 16 bytes

(⍉⍳⍨,⊖+≢)⍣2⍣⎕⊤⍨⍬

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Full program. TIO uses ⍵ instead of ⎕ for easier test case demonstration. For the core algorithm, refer to my Jelly answer.

The only additional trick here is the ⊤⍨⍬, which is one of the shortest expressions that give a 0-by-0 matrix (0 0⍴0). This one in particular is handy because it doesn't mess up by stranding with the input value.

-1 byte using ⍳∘≢ → ⍳⍨; the rows are guaranteed to be distinct, so finding indices of items in itself gives the vector of indices for the leading axis.

Also -1 byte by moving ⍉ to the end of the train to save a ∘. Add first row to a transposed matrix == Add first column and then transpose it.

Answer 4

Python, 73 bytes (@ovs)

f=lambda n,m=0,p=-1:n*[n]and((*range(m,n),),*zip(*f(2*n-m+p,n,~p)[::-1]))

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Old Python, 74 bytes (@DialFrost)

f=lambda n,m=0,p=-1:n and((*range(m,n),),*zip(*f(2*n-m+p,n,~p)[::-1]))or()

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Old Python, 76 bytes

f=lambda n,m=0,p=-1:n>m and((*range(m,n),),*zip(*f(2*n-m+p,n,~p)[::-1]))or()

-15 by reversing order of recursion.

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Old Python, 91 bytes

lambda n:g((n*n-1,))
g=lambda*s:(l:=s[0][0])and g((*range(l-len(s),l),),*zip(*s[::-1]))or s

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Builds the spiral from inside to out by recursively rotating 90° and adding a row on the top.

Answer 5

BQN, 29 bytes^SBCS

{⍉⌽∘⍉∘∾´≍¨⌽(/«⥊2↕↕𝕩+1)⊔⌽↕𝕩⋆2}

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A port of Bubbler's Jelly answer comes in at 20 bytes:

{(⊒˜∾⍉∘⌽+≠)⍟2⍟𝕩↕0‿0}

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Answer 6

R, 84 81 67 bytes

Or R>=4.1, 60 bytes by replacing the word function with a \.

Edit: -14 bytes thanks to @Giuseppe inspired by @Bubbler's Jelly answer.

function(n){m=t(1)
while(T<n)m=rbind(1:n,(T=nrow(m))+t(m[T:1,]))
m}

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Uses the common "rotate and add a row on top" approach.

See also @Giuseppe's direct construction of the matrix.

Answer 7

Charcoal, 48 41 40 bytes

Ｆ⊗Ｎ≔⁺⟦Ｅυλ⟧Ｅ∧υ⌊υ⁺ＬυＥ⮌υ§μλυＥυ⪫Ｅι◧ＩλＬ⌈Ｅυ⌈ν 

Try it online! Link is to verbose version of code. Edit: Saved 4 bytes by porting @Bubbler's observation that you can vectorised add the current length of the array to all of its elements on each iteration, 3 bytes by special-casing an empty array to avoid having to perform the first step manually, and 1 byte by golfing my rotation code. Explanation:

Ｆ⊗Ｎ

Repeat 2n times.

≔⁺⟦Ｅυλ⟧Ｅ∧υ⌊υ⁺ＬυＥ⮌υ§μλυ

Rotate the spiral while adding its length to it values and prefix an additional row of numbers from 0 up to its length. (Unfortunately Charcoal won't let me vectorised add the length to the whole matrix at once but fortunately I can at least do it a row at a time as part of the rotation.) Also handle the edge case of the initially empty spiral which produces a zero-length array.

Ｅυ⪫Ｅι◧ＩλＬ⌈Ｅυ⌈ν 

Display the final spiral as a grid.

Answer 8

R, 78 bytes

function(n)matrix(order(cumsum(rep(rep(c(n,1,-n,-1),,x<-2*n-1),n-1:x%/%2))),n)

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Direct construction of the matrix, no rotations required.

Answer 9

APL (Dyalog Unicode), 35 bytes

{(⍵*2)-⌽{⍉⌽⍵,(⌈/,⍵)+⍳≢⍵}⍣(2×⍵-1)⍪0}

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Answer 10

JavaScript (Node.js), 91 bytes

f=(n,s=[[n*n-1]],[l]=t=s[0])=>l?f(n,[0,...t].map((_,i)=>s.map(r=>r[i-1]||--l).reverse())):s

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A port of loopy walt's Python answer. Although I don't really understand what happening.

Answer 11

JavaScript (ES7),  111  101 bytes

Saved 9 bytes thanks to @tsh

Returns a matrix. The results are 1-indexed.

n=>[...Array(n)].map((x=-n/2,y,a)=>a.map(_=>n*n-(i=4*(++x*x>y*y?x:y)**2)+(x>y||-1)*(i**.5+x+y),y+=x))

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Answer 12

R, 187 183 151 bytes

function(n){w=which
X=diag(0,n)
X[n,]=m=n:1
while(!min(X)){i=w(!X)[1]
j=w(!!X[-1:-i])
X[i-1+1:j]=max(X)+j:1
X=t(X)[m,]}
`if`(n%%2,t(t(X)[m,])[m,],X)-1}

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A clumsier version of rotate and fill!

Answer 13

Julia, 46 bytes

m\n=[1:n n.+rotl90(m')]'
!n=[n<2||!~-n\~-n\n;]

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the simple rotate and add row, with some trickery with ' (transpose)
starts with 1

Answer 14

JavaScript (Node.js), 108 bytes

n=>[...Array(n)].map((_,Y,a)=>a.map(g=(y=Y,x,_,s=n)=>y?--s-x?s-y?x?s*4+g(y-1,x-1,_,s-1):4*s-y:3*s-x:s+y:x));

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Answer 15

Jelly,  34 33  32 bytes

^{I would have thought Jelly would be better at this, maybe a more mathematical approach will triumph?}

ŒMḢḢ
+þ`¬ZṚ$ṛ¬Ç$¦€Ç¦‘ɼ$ḣÇȦƊ?ƬȦƇṂ

A monadic Link accepting a positive integer that yields a list of lists of positive integers (top-left 1 option).

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How?

Start with an \$N\times N\$ table of zeros, then repeatedly either fill in the next number in the first zero of the first row that contains a maximal value, unless that row is filled in which case rotate. This process terminates with a full table but rotated further than we want, so we collect all the states along the way then pick the one we need out (using ȦƇṂ).

ŒMḢḢ - Helper Link: list of list of integers (current state)
ŒM   - multidimensional indices of maximal elements
  Ḣ  - head
   Ḣ - head -> row index containing the maximum value

+þ`¬ZṚ$ṛ¬Ç$¦€Ç¦‘ɼ$ḣÇȦƊ?ƬȦƇṂ - Link: positive integer, N
+þ`                         - [1..N] addition table with [1..N]
   ¬                        - logical NOT -> N×N table of zeros
                       Ƭ    - collect inputs while distinct applying:
                      ?     -   if...
                     Ɗ      -   ...condition: last three links as a monad:
                   Ç        -     call our helper link
                  ḣ         -     head (the current state) to that index
                    Ȧ       -     all truthy when flattened?
      $                     -   ...then: last two links as a monad:
    Z                       -     transpose
     Ṛ                      -     reverse
                                  - this rotates when we've finished a row
                 $          -   ...else: last two links as a monad:
                ɼ           -     apply to the register (initially 0) & yield:
               ‘            -       increment
              ¦             -     sparse application...
             Ç              -     ...to indices: call our helper link
            €               -     ...apply: for each:
           ¦                -       sparse application...
          $                 -       ...to indices: last two links as a monad:
        ¬                   -         logical NOT (vectorises)
         Ç                  -         call our helper link
       ṛ                    -       ...apply: the incremented register value
                         Ƈ  - filter keep those for which:
                        Ȧ   -   all truthy when flattened?
                          Ṃ - minimum (of the rotate tables we have left)

---

Another approach, but even more lengthy at 39:

RµṚĖm€0F;Ɱ"ḊṚ’$ṖṚ4ƭ³’Ḥ¤Ð¡UÐeẎIƇŒṬ€×"J$S

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This one works by building a list of the two-dimensional indices ordered by their final value then composing the table by adding up tables that are all zeros except the bottom-rightmost entry which is the number we want there in the end.

Answer 16

05AB1E, 13 bytes

¯I·FāUøí¬g+Xš

1-based.

Port of @Bubbler's Jelly answer, so make sure to upvote him as well!

Try it online or verify all test cases.

Explanation:

¯             # Start with an empty list []
 I·           # Push the input and double it
   F          # Pop and loop that many times:
    ā         #  Push a list in the range [1,length] (without popping the matrix)
     U        #  Pop and store this list in variable `X`
      øí      #  Rotate the matrix 90 degrees clockwise:
      ø       #   Zip/transpose; swapping rows/columns
       í      #   Reverse each row
        ¬     #  Push the first row (without popping the matrix)
         g    #  Pop and push its length
          +   #  Add that to each value in the matrix
           Xš #  Prepend `X` as first row
              # (after the loop, the resulting matrix is output implicitly)

Answer 17

Factor + math.matrices, 82 79 bytes

[ 2 * { } [ dup length [ m+n flip [ reverse ] map ] keep iota prefix ] repeat ]

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Port of @Bubbler's excellent Jelly answer.

Answer 18

Wolfram Language (Mathematica), 47 bytes

-13 bytes thanks to @att.

Nest[Reverse[{--i,##}&@@@#]&,{{}},i=2#-1]-i&

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