# Double trace of a square matrix

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]

• Loosely related: Sum the diagonals, Transposes and diagonals, Generalized matrix trace Oct 24, 2021 at 10:39
• I suggest adding a test case for [[4,4,4,4],[4,4,4,4],[4,4,4,4],[4,4,4,4]]. In J, (+.|.)@= works to create a mask of the traces for most arrays, except for arrays with repeated rows. Oct 26, 2021 at 0:08
• @ConorO'Brien Added Oct 26, 2021 at 11:29
• is there a challenge for just the anti-diagonal? and maybe a problem to check if a matrix is a band matrix would be interesting.
– qwr
Oct 27, 2021 at 2:07
• @qwr The second one you propose sounds interesting! Oct 27, 2021 at 9:13

# APL (Dyalog 18.0), 17 bytes

Anonymous lambda.

{≢⍸⍵×∨∘⌽⍨∘.=⍨⍳≢⍵}


Try it online!

{}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)

# K (ngn/k), 15 bytes

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


Try it online!

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


# 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 those two indices
ḋ⁸   -- for each resulting list, take the dot product with the corresponding row vector of the input matrix
S  -- sum the results


# R, 47 bytes

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


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# 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
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)


• Interestingly this alternative using a set also ties at 62
– xnor
Oct 25, 2021 at 6:48
• Wouldn't the 2nd program work for python 3.8+. I am not sure which feature requires only python 3.10. Oct 26, 2021 at 13:10
• @Mohammad, The use of assignment expressions in sequence indexes was only introduced in python 3.10 Oct 26, 2021 at 13:17

# 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))


Try it online!

• @LuisMendo It does! Thanks:) Oct 24, 2021 at 11:33

# 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!

# Pari/GP, 42 bytes

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


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# Vyxald, 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


# 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

• ovs is right, which is why all those other answers are awkwardly long. (Otherwise Jelly would be 4 bytes) Oct 25, 2021 at 7:45
• (To clarify: the one that doesn't work is the 7-byter at the top. The 10-byter works fine with odd sizes.) Oct 25, 2021 at 7:56
• @ovs fixed by now Oct 25, 2021 at 8:43
• @Bubbler yeah check now it is the same long looking answer now Oct 25, 2021 at 8:44
• @KevinCruijssen thanks for the 1 byte save! new builtin learned Oct 25, 2021 at 8:44

# 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


# 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;}


Try it online!

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.

• i;s;f(m,l)int*m;{for(s=i=-l;i;)s+=m[i-i*l]-~m[++i*~l];s-=l%2*m[l*l/2];} for 71
– att
Oct 25, 2021 at 18:08
• @att Oh, that's sweet - thanks! :D Oct 25, 2021 at 18:17
• 61 bytes by using my Ruby approach Oct 25, 2021 at 20:44
• 59 Bytes by using *m+=i%~l&&i%~-l?.. Oct 26, 2021 at 7:55
• @AZTECCO Fantastic - thanks! :D Oct 26, 2021 at 10:39

# tinylisp, 132 109 bytes

(load library
(d L length
(d f(q((M)(s(a(trace M)(trace(reverse M)))(*(odd?(L M))(nth(nth M(/(L M)2))(/(L M)2


Try it online!

-23 bytes thanks to DLosc.

• @DLosc Thanks! I must have been thinking about a different problem when I did this. Feb 3 at 19:54

# Pip-x, 2421 20 bytes

YEY#a$+$+a*HV:y+1+Ry


Attempt This Online! The matrix should be given as a command-line argument in the following form:

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


### Explanation

YEY#a$+$+a*HV:y+1+Ry
#a                 Size of the matrix
EY                   Identity matrix of that size
Y                     Yank into the y variable
Ry  y reversed (backward identity matrix)
1+    Add 1 to each value
y+      Add 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
$+ Fold on addition (sum each column)$+               Fold on addition (sum that list of sums)


# 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


# 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.

# Pyth, 18 bytes

s.e+@bk&-hyklb@_bk


Try it online!

## 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))

# 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
)            // end of reduce()


# JavaScript ES2022, 46 bytes

Suggested by @tsh:

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


(Doesn't work on TIO.)

• You can give Array.prototype.at a try which is introduced by ES2022: m=>m.reduce((t,r,i)=>t+r[i]+r.at(~i,r[i]=0),0)
– tsh
Oct 26, 2021 at 2:01