Double trace of a square matrix

Question

Inspired by a question (now closed) at Stack Overflow.

Given a square matrix, let its double trace be defined as the sum of the entries from its main diagonal and its anti-diagonal. These are marked with X in the following examples:

X · · X
· X X · 
· X X ·
X · · X

X · · · X
· X · X ·
· · X · ·
· X · X ·
X · · · X

Note that for odd n, the central entry, which belongs to both diagonals, is counted only once.

Rules

• The matrix size can be any positive integer.
• The matrix will only contain non-negative integers.
• Any reasonable input format can be used. If the matrix is taken as an array (even a flat one) its size cannot be taken as a separate input.
• Input and output means are flexible as usual. Programs or functions are allowed. Standard loopholes are forbidden.
• Shortest wins.

Test cases

5
   ->  5

3 5
4 0
     ->  12

 7  6 10
20 13 44
 5  0  1
          ->  36

4 4 4 4
4 4 4 4
4 4 4 4
4 4 4 4
          ->  32

23  4 21  5
24  7  0  7
14 22 24 16
 4  7  9 12
             ->  97

22 12 10 11  1
 8  9  0  5 17
 5  7 15  4  3
 5  3  7  0 25
 9 15 19  3 21
                -> 85

Inputs in other formats:

[[5]]
[[3,5],[4,0]]
[[7,6,10],[20,13,44],[5,0,1]]
[[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4]]
[[23,4,21,5],[24,7,0,7],[14,22,24,16],[4,7,9,12]]
[[22,12,10,11,1],[8,9,0,5,17],[5,7,15,4,3],[5,3,7,0,25],[9,15,19,3,21]]

[5]
[3 5; 4 0]
[7 6 10; 20 13 44; 5 0 1]
[4 4 4 4; 4 4 4 4; 4 4 4 4; 4 4 4 4]
[23 4 21 5; 24 7 0 7; 14 22 24 16; 4 7 9 12]
[22 12 10 11 1; 8 9 0 5 17; 5 7 15 4 3; 5 3 7 0 25; 9 15 19 3 21]

[5]
[3,5,4,0]
[7,6,10,20,13,44,5,0,1]
[4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4]
[23,4,21,5,24,7,0,7,14,22,24,16,4,7,9,12]
[22,12,10,11,1,8,9,0,5,17,5,7,15,4,3,5,3,7,0,25,9,15,19,3,21]

Answer 1

APL (Dyalog 18.0), 17 bytes

Anonymous lambda.

{≢⍸⍵×∨∘⌽⍨∘.=⍨⍳≢⍵}

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{…} dfn; argument is ⍵

 ≢⍵ number of rows (and columns) in the argument

 ⍳ indices 1 through that

 ∘.=⍨ equality table (gives identity matrix)

 ∨∘⌽⍨ OR with mirrored self (indicates both diagonals)

 ⍵× multiply with argument

 ≢⍸ sum (lit. length of list of indices where each index is repeated as many times as its corresponding number in that)

Answer 2

Jelly, 8 bytes

TżU$Ṭḋ⁸S

Try it online!

T         -- truthy indices of the argument, since every row is non-empty, this is [1 .. len(z)]
 żU$      -- zip with its reverse
    Ṭ     -- for each pair of integers, create a list with 1's at the indices given
     ḋ⁸   -- for each resulting list, take the dor product with the correspondin row vector of the input matrix
       S  -- sum the results

Answer 3

Python 3, 62 bytes

lambda m:sum(r[i]+r[~i]*(i!=len(m)+~i)for i,r in enumerate(m))

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Explanation

lambda m:
                                      for i,r in enumerate(m)  # iterate m with i = row index and r = row
                 +r[~i]*(i!=len(m)+~i)                         # add r[len(m) - i - 1] if the index does not equal i
             r[i]                                              # add r[i]
         sum(                                                ) # sum numbers generated by the loop

---

Python 3.10, ⁵⁹ 58 bytes

lambda m,i=0:sum(r[i:=i-1]+r[~i]*(~i!=len(m)+i)for r in m)

^{If you know a online python 3.10 executor which allows sharing code snippets please leave a comment.}

Answer 4

R, 47 bytes

function(m)sum(m[(z=diag(y<-nrow(m)))|z[y:1,]])

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Answer 5

Octave, 36 34 bytes

This uses logical indexing. As an index we use an identity matrix of the size of the input and or it with it's flipped version. Thanks @LuisMendo for -2 bytes!

@(x)sum(x(flip(e=eye(size(x)))|e))

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Answer 6

K (ngn/k), 15 bytes

{+//x*|/|:\=#x}

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Useful to have an identity matrix primitive.

Explanation:

{+//x*|/|:\=#x}
           =#x    / identity matrix with size of x
        |:\       / append reverse                          
      |/          / OR both matrices
    x*            / multiply by original matrix
 +//              / sum all elements

Answer 7

MATL, 13 11 bytes

This uses logical indexing. As an index we use an identity matrix of the size of the input and or it with it's flipped version, just like in my Octave answer. Thanks @LuisMendo for -2 bytes:)

tZyXytPY|)s

Explanation

t            push input twice to stack
 Zy          get size of matrix
   Xy        get an identity matrix of that size
     t       duplicate the identity matrix
      P      flip one of the identty matrices
       Y|    logical OR the two identity matrices
         )   perform (logical) indexing to the original matrix
          s  compute the sum

Try it online!

Answer 8

Pari/GP, 42 bytes

m->sum(i=1,#m,m[i,i]+m[i,j=#m+1-i]*(i!=j))

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Answer 9

Vyxal d, 12 bytes

ż:Ẋv≈?Lẇ:R⋎*

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Some of Vyxal's behaviour is cursed.

ż            # 1...length
 :Ẋ          # Cartesian product with self
   v≈        # Vectorised all_same 
             # Only values along the diagonal will be truthy, the rest will be falsy, creating an identity matriix
     ?Lẇ     # Make it a square (molding doesn't work, for some reason)
        :R⋎  # Or with reversed copy
           * # Multiply (vectorised)
             # (d flag) deep sum of this

Answer 10

05AB1E, 9 bytes

āDδQÂ~*˜O

Try it online!

-1 thanks to @KevinCruijssen

Wish to have the identity matrix builtin

g               Length
 L              Range
  D             Dup
   δ            Outer product with
    Q           Equals?
     Â          Bifurcate
      ~         Bitwise OR
       *        Multiply with input
        ˜       Flatten
         O      Sum

Answer 11

Ruby, 64 bytes

->m{(r=0...l=m.size).sum{|y|r.sum{|x|x==y||x-~y==l ?m[y][x]:0}}}

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->m{              # lambda taking a 2d array and returning diagonal sum
(r=0...l=m.size)  # range 0..length 
.sum{|y|r.sum{|x| # sum the result of passing each coordinates

x==y||x-~y==l ?   # if on diagonals 
m[y][x]:0         # take value else 0

Answer 12

C (gcc), ^{81 \$\cdots\$ 71} 59 bytes

i;f(m,l)int*m;{for(i=l*l;--i;)*m+=i%~l&&i%~-l?0:m[i];l=*m;}

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Saved a bytes thanks to att!!!
Saved 12 bytes thanks to AZTECCO!!!

Inputs a pointer to a flattened square array and the number of rows (because pointers in C carry no length info).
Returns its double trace.

Answer 13

Charcoal, 18 bytes

ＩΣＥθΣΦι∨⁼κμ⁼⁺κμ⊖Ｌθ

Try it online! Link is to verbose version of code. Explanation: Filters out elements not on either main diagonal, then takes the sum.

   θ                Input matrix
  Ｅ                 Map over rows
      ι             Current row
     Φ              Filtered where
         κ          Row index
        ⁼           Equals
          μ         Column index
       ∨            Logical Or
             κ      Row index
            ⁺       Plus
              μ     Column index
           ⁼        Equals
                 θ  Input matrix
                Ｌ   Length
               ⊖    Decremented
    Σ               Take the sum
 Σ                  Take the sum
Ｉ                   Cast to string
                    Implicitly print

Answer 14

JavaScript (ES6), 51 bytes

m=>m.reduce((t,r,i)=>t+r[i]+r.reverse(r[i]=0)[i],0)

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Commented

m =>         // m[] = matrix
m.reduce(    // reduce():
  ( t,       //   t   = sum
    r,       //   r[] = current row
    i        //   i   = row index, used as a column index
  ) =>       //
  t + r[i] + //   add r[i] to t
  r.reverse( //   reverse r[] ...
    r[i] = 0 //     ... but only once r[i] has been cleared
             //     so that it's not counted twice
  )[i],      //   add r[i] again
  0          //   start with t = 0
)            // end of reduce()

Answer 15

Raku, 43 bytes

{sum .kv.flatmap:{@^v[unique $^k,@v-$k-1]}}

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A function which takes its input in $_ as a list-of-lists.

• .kv returns an interleaved sequence of each row's index and the row, ie: 0, the first row, 1, the second row, etc.
• .flatmap: { ... } passes the key and value to the brace-delimited anonymous function and flattens the return values. That function takes the index of each row in $k and the row itself in @v, each argument declared with a "twigil" $^/@^ on its first lexical appearance.
• $k and @v - $k - 1 are the indices of the elements of each row that go into the double trace. @v, the row, evaluates to its number of elements in a numerical context (the subtraction operator).
• unique selects only the unique indices. This eliminates the double index at the center of the matrix, if it has an odd dimension.
• @v[...] is a slice that returns the row elements at the unique indices.

Answer 16

Pyth, 18 bytes

s.e+@bk&-hyklb@_bk

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Explanation

s.e+@bk&-hyklb@_bk
 .e                 # Iterate through implicit input with (b = row, k = index)
    @bk             # b[k]
       &-hyklb      # if ((2 * k) + 1) - len(b) != 0
   +          @_bk  #   plus b[::-1][k]
s                   # sum results from the iteration

The ((2 * k) + 1) - len(b) is derived from rearranging k - (len(b) - (k + 1))

Answer 17

Pip -x, 24 21 bytes

YEY#a$+$ALa*HV:y+URVy

Run it at Replit! The matrix should be given as a command-line argument in the following form:

[[7;6;10];[20;13;44];[5;0;1]]

which may need to be wrapped in ' single quotes.

Or, verify all test cases with an equivalent version in Pip Classic: Try it online!

Explanation

YEY#a$+$ALa*HV:y+URVy
   #a                  Size of the matrix
 EY                    Identity matrix of that size
Y                      Yank into the y variable
                  RVy  y reversed (backward identity matrix)
                 U     With each value incremented
               y+      Added to y (identity matrix)
            HV:        Each value halved (rounded down)
                       This creates a matrix where both diagonals are 1's and
                       everything else is 0's
          a*           Multiply the input matrix itemwise by the 1/0 matrix
       $AL             Fold on append-list (flatten)
     $+                Fold on addition (sum)