Output diagonal positions of me squared

Question

Given a number n, Output an ordered list of 1-based indices falling on either of the diagonals of an n*n square matrix.

Example:

For an input of 3:

The square shall be:

1 2 3
4 5 6
7 8 9

Now we select all the indices represented by \, / or X (# or non-diagonal positions are rejected)

\ # /
# X #
/ # \

The output shall be:

[1,3,5,7,9]

Test cases:

1=>[1]
2=>[1,2,3,4]
3=>[1,3,5,7,9]
4=>[1,4,6,7,10,11,13,16]
5=>[1,5,7,9,13,17,19,21,25]

There will be no accepted answer. I want to know the shortest code for each language.

Answer 1

Octave, 28 bytes

@(n)find((I=eye(n))+flip(I))

Anonymous function that inputs a number and outputs a column vector of numbers.

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Answer 2

JavaScript (ES6), 48 bytes

Outputs a dash-separated list of integers as a string.

f=(n,k=n*n)=>--k?f(n,k)+(k%~-n&&k%-~n?'':~k):'1'

Formatted and commented

f = (n, k = n * n) => // given n and starting with k = n²
  --k ?               // decrement k; if it does not equal zero:
    f(n, k) + (       //   return the result of a recursive call followed by:
      k % ~-n &&      //     if both k % (n - 1) and
      k % -~n ?       //             k % (n + 1) are non-zero:
        ''            //       an empty string
      :               //     else:
        ~k            //       -(k + 1) (instantly coerced to a string)
    )                 //   end of iteration
  :                   // else:
    '1'               //   return '1' and stop recursion

Test cases

f=(n,k=n*n)=>--k?f(n,k)+(k%~-n&&k%-~n?'':~k):'1'

;[1,2,3,4,5].forEach(n => console.log(n, '->', f(n)))

Answer 3

R, 38 35 34 38 bytes

3 bytes saved when I remembered about the existence of the which function... , 1 byte saved thanks to @Rift

d=diag(n<-scan());which(d|d[n:1,])

+4 bytes for the argument ec=T when called as a full program by source()

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Explanation:

n<-scan()            # take input
d=diag(n);           # create an identity matrix (ones on diagonal, zeros elsewhere)
d|d[n:1,]            # coerce d to logical and combine (OR) with a flipped version
which([d|d[n:1,]])   # Find indices for T values in the logical expression above

Answer 4

Jelly, 8 bytes

⁼þ`+Ṛ$ẎT

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Uses Luis Mendo's algorithm on his MATL answer.

Answer 5

Octave, 41 37 bytes

This works in MATLAB too by the way. No sneaky Octave specific functionality :)

@(x)unique([x:x-1:x^2-1;1:x+1:x*x+1])

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Explanation:

Instead of creating a square matrix, and find the two diagonals, I figured I rather calculate the diagonals directly instead. This was 17 bytes shorter! =)

@(x)                                   % Anonymous function that takes 'x' as input
    unique(...                   ...)  % unique gives the unique elements, sorted
           [x:x-1:x^2-1                % The anti-diagonal (is that the correct word?)
                       ;               % New row
                        1:x+1:x*x+1])  % The regular diagonal

This is what it looks like, without unique:

ans =    
    6   11   16   21   26   31
    1    8   15   22   29   36

_{Yes, I should probably have flipped the order of the diagonals to make it more human-friendly.}

Answer 6

MATL, 6 bytes

XytP+f

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Explanation

Same approach as my Octave answer.

Consider input 3 as an example.

Xy   % Implicit input. Identity matrix of that size
     % STACK: [1 0 0;
               0 1 0;
               0 0 1]
t    % Duplicate
     % STACK: [1 0 0
               0 1 0
               0 0 1],
              [1 0 0
               0 1 0
               0 0 1]
P    % Flip vertically
     % STACK: [1 0 0
               0 1 0
               0 0 1],
              [0 0 1
               0 1 0
               1 0 0]
+    % Add
     % STACK: [1 0 1
               0 2 0
               1 0 1]
f    % Linear indices of nonzero entries. Implicit display  
     % STACK:[1; 3; 5; 7; 9]

Linear indexing is column-major, 1-based. For more information see length-12 snippet here.

Answer 7

Python 2, 54 53 bytes

lambda n:[i+1for i in range(n*n)if i%-~n<1or i%~-n<1]

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Answer 8

Mathematica, 42 bytes

Union@Flatten@Table[{i,#+1-i}+i#-#,{i,#}]&

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@KellyLowder golfed it down to..

Mathematica, 37 bytes

##&@@@Table[{i-#,1-i}+i#,{i,#}]⋃{}&

and @alephalpha threw away the table!

Mathematica, 34 bytes

Union@@Range[{1,#},#^2,{#+1,#-1}]&

Answer 9

Octave, 68 54 bytes

Thanks to @Stewie Griffin for saving 14 bytes!

@(x)unique([diag(m=reshape(1:x^2,x,x)),diag(flip(m))])

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MATLAB, 68 bytes

x=input('');m=reshape([1:x*x],x,x);unique([diag(m) diag(flipud(m))])

Explanation:

@(x)                               % Anonymous function
m=reshape([1:x*x],x,x);            % Create a vector from 1 to x^2 and
                                   % reshape it into an x*x matrix.
diag(m)                            % Find the values on the diagonal.
diag(flip(m))                      % Flip the matrix upside down and
                                   % find the values on the diagonal.
unique([])                         % Place the values from both diagonals
                                   % into a vector and remove duplicates.

Answer 10

Proton, 41 bytes

n=>[i+1for i:0..n*n if!(i%-~n)or!(i%~-n)]

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Answer 11

MATL, 14 bytes

U:GeGXytPY|*Xz

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Answer 12

C (gcc), 65 58 bytes

-7 bytes thanks to Titus!

f(n,i){for(i=0;i<n*n;i++)i%-~n&&i%~-n||printf("%d ",i+1);}

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Answer 13

Javascript, 73 63 bytes

old version

n=>[...Array(y=n*n).keys(),y].filter(x=>(--x/n|0)==x%n||(x/n|0)==n-x%n-1)

Saved 10 bytes thanks to @Shaggy

n=>[...Array(n*n)].map((_,y)=>y+1).filter(x=>!(--x%-~n&&x%~-n))

First time golfing! here's hoping I didn't mess up too badly.

let f=n=>[...Array(n*n)].map((_,y)=>y+1).filter(x=>!(--x%-~n&&x%~-n));
[1,2,3,4,5].forEach(n=>console.log(f(n)))

Answer 14

C# (.NET Core), 97 83 bytes

f=>{var s="[";for(int i=0;i<n*n-1;)s+=i%-~n<1|i++%~-n<1?i+",":"";return s+n*n+"]";}

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The change here is based on the shift between numbers to find. The two shifts starting at 0 are n-1 and n+1, so if n=5, the numbers for n-1 would be 0,4,8,12,16,20 and for n+1 would be 0,6,12,18,24. Combining these and giving 1-indexing (instead of 0-indexing) gives 1,5,7,9,13,17,19,21,25. The offset from n is achieved using bitwise negation (bitwise complement operation), where ~-n==n-1 and -~n==n+1.

Old Version

f=>{var s="[";for(int i=0;i<n*n-1;i++)s+=(i/n!=i%n&&n-1-i/n!=i%n?"":i+1+",");return s+$"{n*n}]";}

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This approach uses the column and row indices for determining if the numbers are on the diagonals. i/n gives the row index, and i%n gives the column index.

Returning Only The Number Array

If constructing only the number array is deemed to count towards the byte cost, then the following could be done, based on Dennis.Verweij's suggestion (using System.Linq; adds an extra 18 bytes):

C# (.NET Core), 66+18=84 bytes

x=>Enumerable.Range(1,x*x).Where(v=>~-v%~-x<1|~-v%-~x<1).ToArray()

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Answer 15

Pyth, 20 18 bytes

^{(Here is the initial version.)}

hMf|<%ThQ1<%TtQ1U*

Test Suite.

Pyth, 18 bytes

hMf>1*%ThQ%T|tQ1U*

Test Suite.

Answer 16

Perl 5, 56 + 1 (-n) = 57 bytes

!(($_+1+$_/$,)%$,&&$_%($,+1))&&say++$_ for 0..($,=$_)**2

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Answer 17

Java (OpenJDK 8), 71 bytes

n->{for(int i=0;i<n*n;i++)if(i%-~n<1||i%~-n<1)System.out.println(i+1);}

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Port of scottinet's answer.

Answer 18

Japt, 16 bytes

Can't seem to do better than this but I'm sure it's possible. Had to sacrifice 2 bytes for the unnecessary requirement that we use 1-indexing.

²õ f@´XvUÉ ªXvUÄ

Test it

Answer 19

Octave, 32 bytes

@(n)find((m=abs(--n:-2:-n))==m')

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Answer 20

PHP, 56 54+1 bytes

+1 byte for -R flag

for(;$z**.5<$n=$argn;$z++)$z%-~$n&&$z%~-$n||print~+$z;

prints numbers prepended by dashes. Run as pipe with -nR or try it online.

requires PHP 5.6 or later for ** operator.
Add one byte for older PHP: Replace ;$z**.5<$n=$argn with $z=$argn;$z<$n*$n.

Answer 21

Ruby, 45 bytes

->n{(n*n).times{|i|i%-~n>0&&i%~-n>0||p(i+1)}}

Works internally as zero indexed. checks if i modulo n+1 or n-1 is 0, if so prints i+1.