Is this a Hadamard matrix?

Question 1

Hadamard matrices is a square matrix whose entries are either +1 or −1 and whose rows are mutually orthogonal.

In other words, it means that each pair of rows has matching entries in exactly half of their columns and mismatched entries in the remaining columns. As a result, the same properties hold for columns as well as rows.

Examples

[ 1 1 1 1
 1 -1 1 -1
 1 1 -1 -1
 1 -1 -1 1 ]

This is a Hadamard matrix, because any pair of rows has matching entries in exactly half of their columns, like:

[ 1 1 1 1
 | / | /
 1 -1 1 -1
 ]

There are 2 |s (matching entries) and 2 /s (mismatched entries). The same logic also applies by columns:

[ 1 ― 1 
 1 ― 1 
 1 ~ -1 
 1 ~ -1 ]

There are two ―s (matching entries) and two ~s (mismatched entries).

This logic can be applied to every pair of rows and columns.

Another example:

[ 1 1
 1 -1 ]

This is a hadamard matrix, because there is only one pair of rows and one pair of columns, but their matching entries is still half of the size of the matrix.

A counterexample like:

[ 1 -1 1 -1
 -1 1 -1 1
 1 -1 1 -1
 -1 1 -1 1 ]

is not a Hadamard matrix, because two pair of rows match in all columns, and four rows don't match in any column. The same logic applies to pair of columns.

Note that

[ 1 ]

is also a Hadamard (1x1) matrix, and your code should handle that (See challenge).

Notes

The matrix inputs are always square
The matrix will only contain -1 or 1. No zeros.

Challenge

The shortest answer that checks if a matrix is a Hadamard matrix wins, because this is code-golf. You may not take input as 1d array, it must be 2d array if your language has a builtin 2d arrays.

Question 2

Is the input matrix guaranteed to contain only 1 and -1?

Question 3

@Arnauld yeah, only 1 and -1. No zeros.

Question 4

You may not take input as 1d array, it must be 2d array - What about languages (like C) that don't have builtin 2d arrays?

Question 5

Please edit the challenge to reflect the input restrictions (only plus/minus 1 values, matrix is square). They are too important to be buried in comments

Question 6

@LuisMendo Okay, I added the notes to the challenge.

Question 7

R, 36 bytes

\(H)any(H%*%t(H)-diag(n<-nrow(H))*n)

Attempt This Online!

Uses the characterization: \$HH^T=nI_n\$.

Outputs swapped TRUE/FALSE.

Question 8

Oh, duh. I'm too rusty!

Question 9

Vyxal, 7 bytes

ÞḊ2$@Π=

Try it Online! -1 thanks to att.

ÞḊ # Is determinant 
 2 # squared
 = # equal to
 $@ # Length of input
 Π # To the power of itself?

Question 10

-1 with det ÞḊ²$@Π=

Question 11

Nekomata + `-e`, 4 bytes

Sđ∙Z

Attempt This Online!

This swaps truthy and falsy values. It outputs False if the input is a Hadamard matrix, and True otherwise.

Sđ∙Z
S Find a subset of the input
 đ whose length is 2
 ∙ and the dot product of the two elements
 Z is nonzero

-e checks if there is a solution.

Nekomata + `-e`, 6 bytes

o∙jZ!!*

Attempt This Online!

o∙jZ!!*
o∙ Outer product with dot product
 This is equivalent to multiplying the matrix by its transpose
 j Join the matrix into a single list
 Z!! Remove zeros
 * Check if it has the same length as the input
 Nekomata does not allow multiplying two lists of different lengths

Question 12

JavaScript (ES6), 60 bytes

Returns false for Hadamard or true for non-Hadamard.

m=>m.some(p=>m.some(q=>p!=q&&p.reduce((t,v,x)=>t+v*q[x],0)))

Try it online!

Commented

m => // m[] = input matrix
m.some(p => // for each row p[] in m[]:
 m.some(q => // for each row q[] in m[]:
 p != q && // test whether p[] is not equal to q[]
 // (comparison of pointers)
 p.reduce( // for each value v at position x in p[],
 (t, v, x) => // using t as the accumulator:
 t + v * q[x], // add v * q[x] to the accumulator,
 // which gives 1 if v = q[x] or -1 otherwise
 0 // start with t = 0
 ) // end of reduce() -> gives 0 if and only if
 // exactly half of the values are matching
 ) // end of inner some()
) // end of outer some()

Question 13

MATL, (削除) 9 (削除ここまで) 7 bytes

Thanks to @alephalpha for a correction (the transpose was originally missing).

t!Y*XR~

Inputs a matrix. Outputs a a matrix of ones (which is truthy) if the input is a Hadamard matrix, or a matrix containing at least a zero (which is falsy) otherwise.

Try it online! Or verify all test cases.

How it works

t % Implicit input: matrix M, of size n, with 1 or −1 entries. Duplicate
! % Transpose
Y* % Matrix product. This is a scalar matrix if and only if M is Hadamard
XR % Set entries on the diagonal and below to 0
~ % Negate each entry. Implicit display

Question 14

Wolfram Language (Mathematica), 22 bytes

#^#&@Tr[1^#]==Det@#^2&

Try it online!

Question 15

Google Sheets, 51 bytes

lambda(a,b,sum(byrow(b,lambda(r,sumproduct(a,r)))))

This is a Google Sheets lambda function that expects two parameters: a should point to a 2D array that contains the first row in the matrix, and b should point to a 2D array that holds the rest of the rows in the matrix.

The function assumes that the matrix in the format specified in the question, and returns 0 if it is a Hadamard matrix and a non-zero value otherwise. In the event of a 1x1 matrix like [[1]], a should point to the matrix and b should point to a null array of zero rows, and the formula will return 0.

screenshot

You can call the function like this:

=lambda(a, b, 
 sum(byrow(b, lambda(r, sumproduct(a, r)))) 
)(A1:D1, A2:D4)

...where a is A1:D1 and bis A2:D4.

Question 16

I don't think the rules allow 2 inputs, unfortunately.

Question 17

Yes, but the challenge is not about parsing the input. It is trivial to get the head and rest from a 2D array. The consensus is to always allow flexible input formats.

Question 18

Charcoal, 12 bytes

×ばつιλ

Attempt This Online! Link is to verbose version of code. Outputs an inverted Charcoal boolean, i.e. - if not Hadamard, nothing if Hadamard, Explanation:

⊙ Input array
 θ Any row satisfies
 ⊙ Input array
 θ Any row satisfies
 κ Outer index
 − Does not equal
 μ Inner index
 ∧ Logical And
 ι Outer row
 ×ばつ Vectorised product with
 λ Inner row
 Σ Has nonzero sum
 Implicitly print

Question 19

PARI/GP, 11 bytes

a->a*a~==#a

Attempt This Online!

Let \$A\$ be the input matrix. Since it is guaranteed to contain only 1 and -1, we only need to check if \$A A^T = n I\$, where \$n\$ is the number of rows in \$A\$. When comparing a matrix with a scalar, PARI/GP will automatically multiply the scalar by the identity matrix of the same size.

Question 20

Python 3, 105 bytes

f=lambda a,x=2:x<1or all(sum(map(int.__eq__,i,j))*2==len(i)for i in a for j in{*a}-{i})*f([*zip(*a)],x-1)

Try it online!

Question 21

05AB1E, (削除) 11 (削除ここまで) 10 bytes

δ*OZ/āDδQQ

Port of @alephalpha's PARI/GP answer, so make sure to upvote that answer as well!

Try it online or verify all test cases.

Explanation:

δ*O # Matrix multiplicate the (implicit) input with itself:
δ # Apply double-vectorized, using two times the (implicit) input-matrix:
 * # Vectorized multiply
 O # Sum each inner row together
 Z # Push the flattened maximum (without popping the matrix)
 / # Divide each inner value by this maximum
 āDδQ # Push an identity-matrix of the same size:
 ā # Push a list in the range [1,length] (without popping)
 D # Duplicate it
 δ # Apply double-vectorized again:
 Q # Check whether the values are equal
 Q # Check if the earlier matrix equals the identity-matrix
 # (after which the result is output implicitly)

Question 22

Haskell, (削除) 47 (削除ここまで), 46 bytes

-1 byte thanks to xnor

f(x:y)=[]==y||all((0==).sum.zipWith(*)x)y&&f y

Try it online!

sum.zipWith(*) computes the scalar product of two vectors
x is the first row, y all the next rows
we check that for all other rows in y the scalar product with the current row x is zero
then we repeat the check discarding the first row (Haskell works well with recursion)

Question 23

null y can be y==[]

Question 24

Ruby, 58 bytes

->m{m.combination(2).all?{|r,s|s.zip(r).sum{|a,b|a*b}==0}}

Try it online!

Question 25

APL+WIN, (削除) 19 (削除ここまで) 20 bytes

+1 to fix problem identified by att

Prompts for matrix. 1 = true, 0 = false

(×ばつ/⍴m)=×ばつ⍉m←⎕

Try it online! Thanks to Dyalog Classic

Question 26

This doesn't work for e.g. ⍉4 4⍴1 1 1 ¯1

Question 27

@att. Thanks. Hopefully fixed.

Question 28

Haskell + hgl, 19 bytes

mF(q0<sm)<xQ(zW ml)

Attempt This Online!

Explanation

This takes all pairs of rows \$(a,b)\$ where \$a\$ comes before \$b\$, computes the dot product and checks that the result is always zero.

xQ computes the pairs
zW ml multiplies the rows pairwise
sm performs the sum to get the dot product
q0 compares it to zero
mF folds using logical and.

Reflection

I'm pretty pleased with this overall. The library is not really built with a lot of linear algebra functions (yet). So this is not a task I should expect this to do very well at. It's not stellar, but it is passable considering everythign. I'm very happy that xQ exists.

There could be a dot product builtin. This would make this much shorter.
xQ should be refactored to take a foldable so this could operate on actual vector types instead of lists.

Question 29

Setanta, 102 bytes

gniomh(m){b:=0le i idir(0,fad@m)le j idir(0,i){s:=0le k idir(0,fad@m)s+=m[i][k]*m[j][k]b=b|s}toradh!b}

try-setanta.ie link

Question 30

Raku (Perl 6) (rakudo), 38 bytes

{!grep {[+] [Z*] $_},.combinations(2)}

Attempt This Online!

Question 31

APL(NARS), 15 chars

{(*/⍴⍵)=×ばつ⍵}

One Hadamard matrix is a M matrix, each elment are +-1 (already proven by input restriction) is quadratic (already proven by input restrictions) that det(M)^2=n^n where nxn is the dimension of the matrix.

{(*/⍴⍵)=×ばつ⍵}
 | ^^ ^------- ×ばつ" make determinant of input ⍵
 | ||--------- ×ばつ⍨" make mult of input on right by itself will be det(⍵)^2 there
 | |------------"=" is equal to
 |----------------"*/⍴⍵" "⍴⍵" get dimensions of input ⍵ and "*/" elevate the first by the second

It should return 1 if the input is a Hadamard matrix, else should return 0; test:

 {((,1)≡∪∣,⍵)∧(*/⍴⍵)=×ばつ⍵}2 2⍴ 1 1 1 ̄1
1
 {(*/⍴⍵)=×ばつ⍵} 4 4⍴ 1 1 1 1 1 ̄1 1 ̄1 1 1 ̄1 ̄1 1 ̄1 ̄1 1
1
 {(*/⍴⍵)=×ばつ⍵} 2 2⍴ 1 1 1 ̄1
1
 {(*/⍴⍵)=×ばつ⍵} 4 4⍴ 1 ̄1 1 ̄1 ̄1 1 ̄1 1 1 ̄1 1 ̄1 ̄1 1 ̄1 1
0
 {(*/⍴⍵)=×ばつ⍵} 1 1⍴1
1

Question 32

Jelly, 7 bytes

Œcḋ/€S¬

Try it online!

Finds all combinations of 2 rows (Œc), calculates the dot product for each (ḋ/€) and sums (S). If this sum is greater than 0, then there exists a pair of rows that is non-orthogonal. Finally, take the logical inverse (¬).

Alternative (8 bytes)

ÆḊ2=L*`$

Try it online!

Tests whether the square determinant (ÆḊ2) is equal to the length of the input, to the power of itself (L*`$). This works as \$\det(H)=\pm n^{n/2} \iff n\$ is a Hadamard matrix.

pajonk 19.4k5 gold badges29 silver badges78 bronze badges · Answer 1 · 2024-07-08 05:43:06Z

8

\$\begingroup\$

R, 36 bytes

\(H)any(H%*%t(H)-diag(n<-nrow(H))*n)

Attempt This Online!

Uses the characterization: \$HH^T=nI_n\$.

Outputs swapped TRUE/FALSE.

Share

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answered Jul 8, 2024 at 5:43

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pajonk

19.4k5 gold badges29 silver badges78 bronze badges

\$\endgroup\$

1

1

\$\begingroup\$ Oh, duh. I'm too rusty! \$\endgroup\$

Giuseppe
– Giuseppe

2024年07月09日 11:39:15 +00:00
Commented Jul 9, 2024 at 11:39

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emanresu A 46.2k5 gold badges111 silver badges257 bronze badges · Answer 2 · 2024-07-07 22:54:05Z

8

\$\begingroup\$

Vyxal, 7 bytes

ÞḊ2$@Π=

Try it Online! -1 thanks to att.

ÞḊ # Is determinant 
 2 # squared
 = # equal to
 $@ # Length of input
 Π # To the power of itself?

Share

Improve this answer

edited Jul 9, 2024 at 7:23

answered Jul 7, 2024 at 22:54

emanresu A's user avatar

emanresu A

46.2k5 gold badges111 silver badges257 bronze badges

\$\endgroup\$

1

2

\$\begingroup\$ -1 with det ÞḊ²$@Π= \$\endgroup\$

att
– att

2024年07月09日 06:36:54 +00:00
Commented Jul 9, 2024 at 6:36

Add a comment |

alephalpha 51.9k7 gold badges75 silver badges196 bronze badges · Answer 3 · 2024-07-08 03:52:32Z

Nekomata + `-e`, 4 bytes

Sđ∙Z

Attempt This Online!

This swaps truthy and falsy values. It outputs False if the input is a Hadamard matrix, and True otherwise.

Sđ∙Z
S Find a subset of the input
 đ whose length is 2
 ∙ and the dot product of the two elements
 Z is nonzero

-e checks if there is a solution.

Nekomata + `-e`, 6 bytes

o∙jZ!!*

Attempt This Online!

o∙jZ!!*
o∙ Outer product with dot product
 This is equivalent to multiplying the matrix by its transpose
 j Join the matrix into a single list
 Z!! Remove zeros
 * Check if it has the same length as the input
 Nekomata does not allow multiplying two lists of different lengths

Arnauld 206k21 gold badges187 silver badges670 bronze badges · Answer 4 · 2024-07-07 21:38:46Z

JavaScript (ES6), 60 bytes

Returns false for Hadamard or true for non-Hadamard.

m=>m.some(p=>m.some(q=>p!=q&&p.reduce((t,v,x)=>t+v*q[x],0)))

Try it online!

Commented

m => // m[] = input matrix
m.some(p => // for each row p[] in m[]:
 m.some(q => // for each row q[] in m[]:
 p != q && // test whether p[] is not equal to q[]
 // (comparison of pointers)
 p.reduce( // for each value v at position x in p[],
 (t, v, x) => // using t as the accumulator:
 t + v * q[x], // add v * q[x] to the accumulator,
 // which gives 1 if v = q[x] or -1 otherwise
 0 // start with t = 0
 ) // end of reduce() -> gives 0 if and only if
 // exactly half of the values are matching
 ) // end of inner some()
) // end of outer some()

Luis Mendo 107k10 gold badges139 silver badges382 bronze badges · Answer 5 · 2024-07-07 22:21:54Z

MATL, (削除) 9 (削除ここまで) 7 bytes

Thanks to @alephalpha for a correction (the transpose was originally missing).

t!Y*XR~

Inputs a matrix. Outputs a a matrix of ones (which is truthy) if the input is a Hadamard matrix, or a matrix containing at least a zero (which is falsy) otherwise.

Try it online! Or verify all test cases.

How it works

t % Implicit input: matrix M, of size n, with 1 or −1 entries. Duplicate
! % Transpose
Y* % Matrix product. This is a scalar matrix if and only if M is Hadamard
XR % Set entries on the diagonal and below to 0
~ % Negate each entry. Implicit display

att 22.8k2 gold badges19 silver badges70 bronze badges · Answer 6 · 2024-07-08 16:34:09Z

4

\$\begingroup\$

Wolfram Language (Mathematica), 22 bytes

#^#&@Tr[1^#]==Det@#^2&

Try it online!

Share

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answered Jul 8, 2024 at 16:34

att's user avatar

att

22.8k2 gold badges19 silver badges70 bronze badges

\$\endgroup\$

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doubleunary 4,6521 gold badge6 silver badges16 bronze badges · Answer 7 · 2024-07-08 19:12:07Z

Google Sheets, 51 bytes

lambda(a,b,sum(byrow(b,lambda(r,sumproduct(a,r)))))

This is a Google Sheets lambda function that expects two parameters: a should point to a 2D array that contains the first row in the matrix, and b should point to a 2D array that holds the rest of the rows in the matrix.

The function assumes that the matrix in the format specified in the question, and returns 0 if it is a Hadamard matrix and a non-zero value otherwise. In the event of a 1x1 matrix like [[1]], a should point to the matrix and b should point to a null array of zero rows, and the formula will return 0.

screenshot

You can call the function like this:

=lambda(a, b, 
 sum(byrow(b, lambda(r, sumproduct(a, r)))) 
)(A1:D1, A2:D4)

...where a is A1:D1 and bis A2:D4.

Yes, but the challenge is not about parsing the input. It is trivial to get the head and rest from a 2D array. The consensus is to always allow flexible input formats.

Neil 184k12 gold badges76 silver badges290 bronze badges · Answer 8 · 2024-07-07 22:13:29Z

Charcoal, 12 bytes

×ばつιλ

Attempt This Online! Link is to verbose version of code. Outputs an inverted Charcoal boolean, i.e. - if not Hadamard, nothing if Hadamard, Explanation:

⊙ Input array
 θ Any row satisfies
 ⊙ Input array
 θ Any row satisfies
 κ Outer index
 − Does not equal
 μ Inner index
 ∧ Logical And
 ι Outer row
 ×ばつ Vectorised product with
 λ Inner row
 Σ Has nonzero sum
 Implicitly print

alephalpha 51.9k7 gold badges75 silver badges196 bronze badges · Answer 9 · 2024-07-08 02:33:21Z

PARI/GP, 11 bytes

a->a*a~==#a

Attempt This Online!

Let \$A\$ be the input matrix. Since it is guaranteed to contain only 1 and -1, we only need to check if \$A A^T = n I\$, where \$n\$ is the number of rows in \$A\$. When comparing a matrix with a scalar, PARI/GP will automatically multiply the scalar by the identity matrix of the same size.

Jitse 8,1243 gold badges21 silver badges45 bronze badges · Answer 10 · 2024-07-08 07:42:15Z

2

\$\begingroup\$

Python 3, 105 bytes

f=lambda a,x=2:x<1or all(sum(map(int.__eq__,i,j))*2==len(i)for i in a for j in{*a}-{i})*f([*zip(*a)],x-1)

Try it online!

Share

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answered Jul 8, 2024 at 7:42

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Jitse

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Kevin Cruijssen 136k14 gold badges154 silver badges394 bronze badges · Answer 11 · 2024-07-08 07:56:29Z

05AB1E, (削除) 11 (削除ここまで) 10 bytes

δ*OZ/āDδQQ

Port of @alephalpha's PARI/GP answer, so make sure to upvote that answer as well!

Try it online or verify all test cases.

Explanation:

δ*O # Matrix multiplicate the (implicit) input with itself:
δ # Apply double-vectorized, using two times the (implicit) input-matrix:
 * # Vectorized multiply
 O # Sum each inner row together
 Z # Push the flattened maximum (without popping the matrix)
 / # Divide each inner value by this maximum
 āDδQ # Push an identity-matrix of the same size:
 ā # Push a list in the range [1,length] (without popping)
 D # Duplicate it
 δ # Apply double-vectorized again:
 Q # Check whether the values are equal
 Q # Check if the earlier matrix equals the identity-matrix
 # (after which the result is output implicitly)

DPD- 4211 silver badge5 bronze badges · Answer 12 · 2024-07-08 22:48:53Z

Haskell, (削除) 47 (削除ここまで), 46 bytes

-1 byte thanks to xnor

f(x:y)=[]==y||all((0==).sum.zipWith(*)x)y&&f y

Try it online!

sum.zipWith(*) computes the scalar product of two vectors
x is the first row, y all the next rows
we check that for all other rows in y the scalar product with the current row x is zero
then we repeat the check discarding the first row (Haskell works well with recursion)

1

\$\begingroup\$ null y can be y==[] \$\endgroup\$

xnor
– xnor

2024年07月09日 03:54:06 +00:00
Commented Jul 9, 2024 at 3:54

G B 23.4k1 gold badge24 silver badges55 bronze badges · Answer 13 · 2024-07-08 10:59:55Z

1

\$\begingroup\$

Ruby, 58 bytes

->m{m.combination(2).all?{|r,s|s.zip(r).sum{|a,b|a*b}==0}}

Try it online!

Share

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answered Jul 8, 2024 at 10:59

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G B

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Graham 7,8061 gold badge15 silver badges13 bronze badges · Answer 14 · 2024-07-08 14:59:32Z

1

\$\begingroup\$

APL+WIN, (削除) 19 (削除ここまで) 20 bytes

+1 to fix problem identified by att

Prompts for matrix. 1 = true, 0 = false

(×ばつ/⍴m)=×ばつ⍉m←⎕

Try it online! Thanks to Dyalog Classic

Share

Improve this answer

edited Jul 9, 2024 at 7:27

answered Jul 8, 2024 at 14:59

Graham's user avatar

Graham

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\$\endgroup\$

2

\$\begingroup\$ This doesn't work for e.g. ⍉4 4⍴1 1 1 ¯1 \$\endgroup\$

att
– att

2024年07月09日 05:07:33 +00:00
Commented Jul 9, 2024 at 5:07
\$\begingroup\$ @att. Thanks. Hopefully fixed. \$\endgroup\$

Graham
– Graham

2024年07月09日 07:28:06 +00:00
Commented Jul 9, 2024 at 7:28

Add a comment |

Wheat Wizard ♦ 103k23 gold badges299 silver badges697 bronze badges · Answer 15 · 2024-07-15 00:59:25Z

Haskell + hgl, 19 bytes

mF(q0<sm)<xQ(zW ml)

Attempt This Online!

Explanation

This takes all pairs of rows \$(a,b)\$ where \$a\$ comes before \$b\$, computes the dot product and checks that the result is always zero.

xQ computes the pairs
zW ml multiplies the rows pairwise
sm performs the sum to get the dot product
q0 compares it to zero
mF folds using logical and.

Reflection

I'm pretty pleased with this overall. The library is not really built with a lot of linear algebra functions (yet). So this is not a task I should expect this to do very well at. It's not stellar, but it is passable considering everythign. I'm very happy that xQ exists.

There could be a dot product builtin. This would make this much shorter.
xQ should be refactored to take a foldable so this could operate on actual vector types instead of lists.

bb94 3,89815 silver badges20 bronze badges · Answer 16 · 2024-08-13 04:30:15Z

1

\$\begingroup\$

Setanta, 102 bytes

gniomh(m){b:=0le i idir(0,fad@m)le j idir(0,i){s:=0le k idir(0,fad@m)s+=m[i][k]*m[j][k]b=b|s}toradh!b}

try-setanta.ie link

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answered Aug 13, 2024 at 4:30

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bb94 3,89815 silver badges20 bronze badges · Answer 17 · 2024-08-13 04:33:33Z

1

\$\begingroup\$

Raku (Perl 6) (rakudo), 38 bytes

{!grep {[+] [Z*] $_},.combinations(2)}

Attempt This Online!

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answered Aug 13, 2024 at 4:33

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Rosario 1,7045 silver badges9 bronze badges · Answer 18 · 2025-01-08 19:49:22Z

APL(NARS), 15 chars

{(*/⍴⍵)=×ばつ⍵}

One Hadamard matrix is a M matrix, each elment are +-1 (already proven by input restriction) is quadratic (already proven by input restrictions) that det(M)^2=n^n where nxn is the dimension of the matrix.

{(*/⍴⍵)=×ばつ⍵}
 | ^^ ^------- ×ばつ" make determinant of input ⍵
 | ||--------- ×ばつ⍨" make mult of input on right by itself will be det(⍵)^2 there
 | |------------"=" is equal to
 |----------------"*/⍴⍵" "⍴⍵" get dimensions of input ⍵ and "*/" elevate the first by the second

It should return 1 if the input is a Hadamard matrix, else should return 0; test:

 {((,1)≡∪∣,⍵)∧(*/⍴⍵)=×ばつ⍵}2 2⍴ 1 1 1 ̄1
1
 {(*/⍴⍵)=×ばつ⍵} 4 4⍴ 1 1 1 1 1 ̄1 1 ̄1 1 1 ̄1 ̄1 1 ̄1 ̄1 1
1
 {(*/⍴⍵)=×ばつ⍵} 2 2⍴ 1 1 1 ̄1
1
 {(*/⍴⍵)=×ばつ⍵} 4 4⍴ 1 ̄1 1 ̄1 ̄1 1 ̄1 1 1 ̄1 1 ̄1 ̄1 1 ̄1 1
0
 {(*/⍴⍵)=×ばつ⍵} 1 1⍴1
1

lolad 9728 silver badges17 bronze badges · Answer 19 · 2025-01-07 23:05:06Z

Jelly, 7 bytes

Œcḋ/€S¬

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Finds all combinations of 2 rows (Œc), calculates the dot product for each (ḋ/€) and sums (S). If this sum is greater than 0, then there exists a pair of rows that is non-orthogonal. Finally, take the logical inverse (¬).

Alternative (8 bytes)

ÆḊ2=L*`$

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Tests whether the square determinant (ÆḊ2) is equal to the length of the input, to the power of itself (L*`$). This works as \$\det(H)=\pm n^{n/2} \iff n\$ is a Hadamard matrix.

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Is this a Hadamard matrix?

Examples

Notes

Challenge

19 Answers 19

R, 36 bytes

Vyxal, 7 bytes

Nekomata + `-e`, 4 bytes

Nekomata + `-e`, 6 bytes

JavaScript (ES6), 60 bytes

Commented

MATL, (削除) 9 (削除ここまで) 7 bytes

How it works

Wolfram Language (Mathematica), 22 bytes

Google Sheets, 51 bytes

Charcoal, 12 bytes

PARI/GP, 11 bytes

Python 3, 105 bytes

05AB1E, (削除) 11 (削除ここまで) 10 bytes

Haskell, (削除) 47 (削除ここまで), 46 bytes

Ruby, 58 bytes

APL+WIN, (削除) 19 (削除ここまで) 20 bytes

Haskell + hgl, 19 bytes

Explanation

Reflection

Setanta, 102 bytes

Raku (Perl 6) (rakudo), 38 bytes

APL(NARS), 15 chars

Jelly, 7 bytes

Alternative (8 bytes)

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Hot Network Questions

Is this a Hadamard matrix?

Examples

Notes

Challenge

19 Answers 19

R, 36 bytes

Vyxal, 7 bytes

Nekomata + -e, 4 bytes

Nekomata + -e, 6 bytes

JavaScript (ES6), 60 bytes

Commented

MATL, (削除) 9 (削除ここまで) 7 bytes

How it works

Wolfram Language (Mathematica), 22 bytes

Google Sheets, 51 bytes

Charcoal, 12 bytes

PARI/GP, 11 bytes

Python 3, 105 bytes

05AB1E, (削除) 11 (削除ここまで) 10 bytes

Haskell, (削除) 47 (削除ここまで), 46 bytes

Ruby, 58 bytes

APL+WIN, (削除) 19 (削除ここまで) 20 bytes

Haskell + hgl, 19 bytes

Explanation

Reflection

Setanta, 102 bytes

Raku (Perl 6) (rakudo), 38 bytes

APL(NARS), 15 chars

Jelly, 7 bytes

Alternative (8 bytes)

Your Answer

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Post as a guest

Post as a guest

Linked

Related

Hot Network Questions

Nekomata + `-e`, 4 bytes

Nekomata + `-e`, 6 bytes