Compute this fractal matrix

Question 1

The unique-disjointness matrix ( UDISJ(n) ) is a matrix on all pairs of subsets of {1...,n} with entries $$ U_{(A,B)}=\begin{cases} 0, ~ if ~ |A\cap B|=1\\ 1, ~ otherwise \end{cases} $$

Or a bit less mathematical, it is the 2ⁿ times 2ⁿ matrix with a 0 in all entries where both indices have exactly one common 1 bit in their binary representation and a 1 in all other entries.

Example:

 0 1 10 11 100 101 110 111
 _______________________________
0 | 1 1 1 1 1 1 1 1
1 | 1 0 1 0 1 0 1 0
10 | 1 1 0 0 1 1 0 0
11 | 1 0 0 1 1 0 0 1
100 | 1 1 1 1 0 0 0 0
101 | 1 0 1 0 0 1 0 1
110 | 1 1 0 0 0 0 1 1
111 | 1 0 0 1 0 1 1 1

Write a program or function that takes an integer n as input and outputs the corresponding \2ドル^n \times 2^n\$ matrix.

Rules:

The values in the matrix can either be 0 / 1 (as numbers) or truthy/falsey
You are not allowed to permute the rows/columns of the matrix
This is code-golf the shortest solution (per language) wins
Your solution only needs to handle n between 2 and 10 (inclusive)

First 4 elements:

n=1:
1 1
1 
n=2:
1 1 1 1
1 1 
1 1 
1 1
n=3:
1 1 1 1 1 1 1 1
1 1 1 1 
1 1 1 1 
1 1 1 1
1 1 1 1 
1 1 1 1 
1 1 1 1
1 1 1 1 1
n=4:
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 1 1 1 1

Question 2

Python, 72 bytes

lambda m:(q:=range(2**m))and(((i&j).bit_count()!=1for i in q)for j in q)

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Question 3

05AB1E, 10 bytes

o<ÝDδ&b1ö≠

Outputs as a matrix with 1s and 0s.

Port of @mousetail's Python answer, so make sure to upvote that answer as well!

Try it online or verify the first 5 test cases.

Explanation:

o # Push 2 to the power the (implicit) input-list
 <Ý # Pop and push a list in the range [0,2**input)
 Dδ& # Create a bitwise-AND table of it:
 D # Duplicate the list
 δ # Pop both lists and apply double-vectorized:
 & # Bitwise-AND
 b # Then convert each inner value to a binary-string
 1ö # Vectorized-sum its digits by converting from base-1 to a base-10 integer
 ≠ # Check for each that it's NOT equal to 1 (0 if 1; 1 otherwise)
 # (after which the matrix is output implicitly)

Question 4

Not again... Porting now lol

Question 5

@TheThonnu Haha, déjà vu. ;)

Question 6

Jelly, 8 bytes

2ṗ«'`§’n

Try it online!

 (input: n)
2ṗ nth Cartesian power of [1, 2]: for example,
 [[1,1,1,1], [1,1,1,2], [1,1,2,1], ..., [2,2,2,2]] if n=4
 «'` self-table by vectorized minimum
 § sum each vector
 ’ subtract 1
 n not equal to... (implicit argument: n)

Instead of 0b0011 & 0b0110 = 0b0010 we do 1,1,2,2 « 1,2,2,1 = 1,1,2,1.

Instead of checking popcount ≠ 1, we check sum ≠ n+1. §’n is shorter than ’§n1.

Question 7

Vyxal, 66 bits^v2 , 8.25 bytes

Eʁ:v⋏bvṠċ

Try it Online!

-4.75 bytes thanks to TheThonnu

The footer formats the resulting matrix into a nice grid like the test cases. Remove it for just a list of lists. Outputs inverted numbers (0s for 1s and 1s for 0s).

I somehow managed to generate like 3 other fractals while solving this, including an upside down triangle thing with the bottom off-center.

Explained

EDẊ‹sƛƒ⋏bT3;ẇ⁡‎‎⁪⁡⁪⁠⁪⁡⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁪‏‏⁡⁠⁡‌⁢‎‎⁪⁡⁪⁠⁪⁣⁪‏‏⁡⁠⁡‌⁣‎‎⁪⁡⁪⁠⁪⁤⁪‏‏⁡⁠⁡‌⁤‎‎⁪⁡⁪⁠⁪⁢⁡⁪‏‏⁡⁠⁡‌⁢⁡‎‎⁪⁡⁪⁠⁪⁢⁢⁪‏⁠‎⁪⁡⁪⁠⁪⁣⁤⁪‏⁠⁪⁪⁠‎⁪⁡⁪⁠⁪⁤⁣⁪‏‏⁡⁠⁡‌⁢⁢‎‎⁪⁡⁪⁠⁪⁢⁣⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁤⁪‏⁠⁪⁪⁠⁪⁪‏⁡⁠⁡‌⁢⁣‎‎⁪⁡⁪⁠⁪⁣⁡⁪‏⁠‎⁪⁡⁪⁠⁪⁣⁢⁪‏⁠‎⁪⁡⁪⁠⁪⁣⁣⁪‏⁠⁪⁪‏⁡⁠⁡‌⁢⁤‎‎⁪⁡⁪⁠⁪⁤⁡⁪‏⁠‎⁪⁡⁪⁠⁪⁤⁢⁪‏⁠⁪⁪‏⁡⁠⁡‌
ED # ‎⁡Push three copies of n ** 2
 Ẋ # ‎⁢Cartesian product of the range [1, n ** 2] and [1, n ** 2]
 ‹ # ‎⁣with all sublists decremented
 s # ‎⁤and sorted
 ƛ ; # ‎⁢⁡To each pair
 ƒ⋏ # ‎⁢⁢ reduce by bitwise and
 bT3 # ‎⁢⁣ is the length of truthy indices 1?
 ẇ # ‎⁢⁤wrap into 2 ** n chunks
💎

Created with the help of Luminespire.

Question 8

9 bytes (or 8.25 with Vyncode)

Question 9

Thunno 2, 10 bytes

2RẉDȷỌ€ʂ−Q

Try it online!

Outputs a nested list. The footer formats it into the grid. Port of Lynn's Jelly answer.

Explanation

2RẉDȷỌ€ʂ−Q # Implicit input
2Rẉ # [1,2] to the cartesian power of the input
 DȷỌ # Outer product over minimum with itself
 €ʂ # Sum each inner list
 −Q # Decrement, not equal to the input?
 # Implicit output

Old:

OLDȷÆ&ıḃ1ȷcḅ # Implicit input
OL # Push [0..2**n)
 DȷÆ& # Outer product over bitwise AND
 ıḃ1ȷc # Popcount of each
 ḅ # Equals one?
 # Implicit output

Question 10

Python, 66 bytes

lambda m:(q:=range(2**m))and[[(n:=i&j)>n^n-1for i in q]for j in q]

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mousetail's answer, but with bit operations instead of bit_count. We want to check that i&j (which we call n) doesn't have exactly one set bit, which means that it's a not power of two. We do this by checking that n>n^n-1, which is processed as n>n^(n-1). Let's check that this works in every case:

If n is a power of two, then n-1 has disjoint bits set to n and so their XOR is greater than n.
If n is any other positive value, then n and n-1 share the same leading bit, and the XOR cancels this bit making the result smaller than n.
For n==0, the result is 0, the same as n.

Annoyingly Python doesn't let us do the walrus assignment q:=range(2**m) in the iterable of a list comprehension, so the code has it assigned beforehand. loopy walt points out a better way to handle the two levels of iteration is using the classic divmod two-loop trick together with the classic zip/iter chunker.

61 bytes

lambda m:zip(*2**m*[((n:=j&j>>m)>n^n-1for j in range(4**m))])

Attempt This Online!

Question 11

-3 just using the good ol' grouper.

Question 12

Arturo, ^{(削除) 57 54 (削除ここまで)} 50 bytes

$->n[map^2n'x->map^2n=>[0<>^and dec<=<=and&-1x-1]]

Try it!

$->n[ ; a function taking an argument n
 map^2n'x-> ; map over [1..2^n]; assign current elt to x
 map^2n=>[ ; map over [1..2^n]; assign current elt to &
 0<> ; is zero not equal to
 and&-1x-1 ; bitwise and of current elts minus one
 <=<= ; duplicated twice
 dec ; minus one
 and ; bitwise and
 ^ ; NOS to the TOS power
 ] ; end map
] ; end function

Question 13

R, (削除) 55 (削除ここまで) 53 bytes

function(n)(y=outer(x<-1:2^n-1,x,bitwAnd))^0-y%in%2^x

Try it online!

Rather than counting the number of on bits (which is very lengthy in R), tests the equivalent characterization of "equal to a power of 2"; then has to do some reshaping of the vector output of %in% to get back to the right shape matrix.

Question 14

I cannot find anything on meta - is it generally allowed to output a matrix as a flat vector?

Question 15

@pajonk hmm probably not. I didn't notice b/c of the footer, so I'll have to update it.

Question 16

Nim, 119 bytes

import bitops
proc f(n:int)=
 let r=0..<1 shl n;for i in r:
 var s="";for j in r:s&= $int 1!=popcount i and j
 echo s

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Question 17

Jelly, 10 bytes

2*Ḷ&þ`B§n1

Try it online!

How it works

2*Ḷ&þ`B§n1 - Main link. Takes n on the left
2* - 2 ** n
 Ḷ - [0, 1, ..., 2**n)
 þ` - To every pair (a, b) in this range:
 & - Bitwise and
 B - Convert all to binary
 § - Sums
 n1 - Does not equal 1?

Question 18

Pyth, (削除) 15 (削除ここまで) 14 bytes

mmn1s&VdkQ=^U2

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Explanation

mmn1s&VdkQ=^U2Q # implicitly add Q
 # implicitly assign Q = eval(input())
 =^U2Q # Q = repeated cartesian product of [0,1] Q times
m Q # map lambda d over range(Q)
 m Q # map lambda k over range(Q)
 &Vdk # d & k, vectorized
 s # sum bits
 n1 # != 1

Question 19

BQN (CBQN), (削除) 45 bytes (削除ここまで) 22 ‘bytes’

{1≠+ ́ ̈∧⌜ ̃⥊∾⌜⍟(x-1) ̃↕2}

I think the following is slightly more elegant, and it's the same number of characters but, alas, more bytes.

{1≠+ ́∘∧⌜ ̃⥊(↕2)∾⌜⍟x⋈⟨⟩}

We operate only on bit-strings, rather than integers. ⥊⌜⍟(x-1) ̃↕2 gives us the list of binary strings of length x, which is then anded into a table, and summed: + ́∘∧⌜ ̃. We then take only the 1≠ entries.

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Question 20

Though I don't know the specifics of why, every BQN answer I've seen here counts 1 glyph as 1 byte. I think you can call these 22 bytes.

Question 21

@chunes Thanks, that makes quite a difference!

Question 22

Ruby, (削除) 72 56 (削除ここまで) 54 bytes

->n{(w=0...1<<n).map{|i|w.map{|j|a=i&j;(a&a-1)**a>0}}}

Try it online!

This employs one of my favourite bit-twiddling hacks. a&a-1 gives 0 if a has only one bit set. Unfortunately it also gives 0 if a==0 so we use the fact that any number (including 0) raised to the power 0 is 1 to catch this special case: (a&a-1)**a.

So now we have distinction between nonzero and zero numbers. The rules require two distinct values, so we use >0 to convert to true/false and we are done.

The footer in the linked TIO formats the output 2d array line by line, and converts the true/false to 1/0.

Question 23

52 bytes with Ruby 2.7 Numbered Parameters (TIO is stuck on 2.5.5) - ->n{(w=0...1<<n).map{|i|w.map{a=i&_1;(a&a-1)**a>0}}}

Question 24

@ValueInk oh wow thanks, that's new! You really have to check your versions. I still miss when you could write .5 instead of 0.5

Question 25

CG One Handed pointed it out to me earlier this week and it feels like a whole new world. Also, you can still save a byte on 2.5 by using j&=i;(j&j-1)**j>0 for 53 bytes Try it online!

Question 26

Factor + `math.matrices math.unicode`, 57 bytes

[ 2^ dup [ bitand bit-count 1 ≠ ] <matrix-by-indices> ]

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2^ ! two to the input power
dup ! duplicate
[ ... ] <matrix-by-indices> ! create an MxN matrix with indices on top of stack
bitand ! bitwise and
bit-count ! number of on bits
1 ≠ ! not equal to one

Question 27

05AB1E, 15 bytes

oDL<ãε`&b1¢Θ}sô

Try it online!

Outputs a nested list. The footer formats it into the grid. Port of lyxal's Vyxal answer.

-1 thanks to @KevinCruijssen

Explanation

oDL<ãε`&b1¢Θ}sô # Implicit input
oD # Push 2 ** n and duplicate
 L<ã # Cartesian product of [0..2**n) with itself
 ε } # Map over this list:
 `& # Bitwise AND of the pair
 b1¢Θ # Is the popcount equal to 1?
 sô # Split into 2 ** n chunks
 # Implicit output

Question 28

Dâ can be ã (and golfing it some more would result in my 10 bytes answer, which I was already writing before I saw this answer tbh).

Question 29

Nibbles, 11 bytes

.;`,^2$.@!=1+``@`&@$

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Returns a matrix with truthy values (positive numbers) for 1 and falsy values (0) for 0.

Question 30

Raku, 52 bytes

{(^$_ X+&^$_).rotor($_)».&{!$_||$_!=1+<.msb}}o 1+<*

Try it online!

The version of Raku on TIO fails to parse the !$_||$_!=1+<.msb bit, presumably due to a bug, so I've expressed it there as !($_&&$_==1+<.msb), which is equivalent by De Morgan's law, but two bytes longer.

This is an anonymous function consisting of two separate anonymous functions composed together with o. The right-hand function is 1 +< *, which shifts the number 1 leftwards a number of bits given by its argument; that is, two to the power of the argument. That number is passed to the left-hand function, where the variable $_ gets its value.

^$_ is a range of numbers from 0 up to one less than $_. For example, when the number passed to the first function is 3, this is the range 0..7.
X+& gives the cross product of one copy of that range with another using the bitwise-and operator +&.
.rotor($_) takes that flat crossed list and makes it into a square matrix.
».&{ ... } recursively applies the code between the braces to each element of that matrix.
!$_ || $_ != 1 +< .msb tests whether each number is zero (!$_) or consists of only a single binary digit by comparing the number to 1 bit-shifted left a number of places equal to the number's most significant bit (1 +< .msb).

Question 31

JavaScript (ES6), 74 bytes

n=>[...Array(1<<n)].map((_,y,a)=>a.map((_,x)=>(g=k=>k?k%2^g(k/2):1)(x&y)))

Try it online!

Question 32

Charcoal, 18 bytes

≔X2NθEθ⭆θ¬=1Σ↨&ιλ2

Try it online! Link is to verbose version of code. Explanation:

 2 Literal integer `2`
 X Raised to power
 N Input as a number
≔ θ Save in variable
 θ Saved variable
 E Map over implicit range
 θ Saved variable
 ⭆ Map over implicit range and join
 ι Outer value
 & Bitwise And
 λ Inner value
 ↨ 2 Convert to base `2`
 Σ Take the sum
 ¬= Does not equal
 1 Literal integer `1`
 Implicitly print

Question 33

PARI/GP, 51 bytes

n->matrix(2^n,,i,j,sumdigits(bitand(i-1,j-1),2)!=1)

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Question 34

Wolfram Language(Mathematica), 68 bytes

Golfed version. Try it online!

f@n_:=Array[Boole[1!=Tr@IntegerDigits[BitAnd[#-1,#2-1],2]]&,2^{n,n}]

Ungolfed version. Try it onlinne!

f@n_:=Table[If[Total[IntegerDigits[BitAnd[i-1,j-1],2]]!=1,1,0],{i,1,2^n},{j,1,2^n}]

Question 35

JavaScript (Node.js), 65 bytes

n=>[...Array(1<<n)].map((_,y,a)=>a.map((_,x)=>(x&=y)&&+!(x&x-1)))

Try it online!

na

mousetail 14.4k1 gold badge41 silver badges90 bronze badges · Answer 1 · 2023-07-26 13:39:07Z

7

\$\begingroup\$

Python, 72 bytes

lambda m:(q:=range(2**m))and(((i&j).bit_count()!=1for i in q)for j in q)

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Share

Improve this answer

edited Jul 26, 2023 at 13:47

answered Jul 26, 2023 at 13:39

mousetail's user avatar

mousetail

14.4k1 gold badge41 silver badges90 bronze badges

\$\endgroup\$

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Kevin Cruijssen 136k14 gold badges154 silver badges394 bronze badges · Answer 2 · 2023-07-26 14:36:30Z

05AB1E, 10 bytes

o<ÝDδ&b1ö≠

Outputs as a matrix with 1s and 0s.

Port of @mousetail's Python answer, so make sure to upvote that answer as well!

Try it online or verify the first 5 test cases.

Explanation:

o # Push 2 to the power the (implicit) input-list
 <Ý # Pop and push a list in the range [0,2**input)
 Dδ& # Create a bitwise-AND table of it:
 D # Duplicate the list
 δ # Pop both lists and apply double-vectorized:
 & # Bitwise-AND
 b # Then convert each inner value to a binary-string
 1ö # Vectorized-sum its digits by converting from base-1 to a base-10 integer
 ≠ # Check for each that it's NOT equal to 1 (0 if 1; 1 otherwise)
 # (after which the matrix is output implicitly)

1

\$\begingroup\$ Not again... Porting now lol \$\endgroup\$

The Thonnu
– The Thonnu

2023年07月26日 14:38:52 +00:00
Commented Jul 26, 2023 at 14:38
1

\$\begingroup\$ @TheThonnu Haha, déjà vu. ;) \$\endgroup\$

Kevin Cruijssen
– Kevin Cruijssen

2023年07月26日 14:39:49 +00:00
Commented Jul 26, 2023 at 14:39

lynn 69.7k11 gold badges137 silver badges285 bronze badges · Answer 3 · 2023-07-26 18:06:40Z

Jelly, 8 bytes

2ṗ«'`§’n

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 (input: n)
2ṗ nth Cartesian power of [1, 2]: for example,
 [[1,1,1,1], [1,1,1,2], [1,1,2,1], ..., [2,2,2,2]] if n=4
 «'` self-table by vectorized minimum
 § sum each vector
 ’ subtract 1
 n not equal to... (implicit argument: n)

Instead of 0b0011 & 0b0110 = 0b0010 we do 1,1,2,2 « 1,2,2,1 = 1,1,2,1.

Instead of checking popcount ≠ 1, we check sum ≠ n+1. §’n is shorter than ’§n1.

lyxal ♦ 35.6k2 gold badges69 silver badges148 bronze badges · Answer 4 · 2023-07-26 13:30:52Z

Vyxal, 66 bits^v2 , 8.25 bytes

Eʁ:v⋏bvṠċ

Try it Online!

-4.75 bytes thanks to TheThonnu

The footer formats the resulting matrix into a nice grid like the test cases. Remove it for just a list of lists. Outputs inverted numbers (0s for 1s and 1s for 0s).

I somehow managed to generate like 3 other fractals while solving this, including an upside down triangle thing with the bottom off-center.

Explained

EDẊ‹sƛƒ⋏bT3;ẇ⁡‎‎⁪⁡⁪⁠⁪⁡⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁪‏‏⁡⁠⁡‌⁢‎‎⁪⁡⁪⁠⁪⁣⁪‏‏⁡⁠⁡‌⁣‎‎⁪⁡⁪⁠⁪⁤⁪‏‏⁡⁠⁡‌⁤‎‎⁪⁡⁪⁠⁪⁢⁡⁪‏‏⁡⁠⁡‌⁢⁡‎‎⁪⁡⁪⁠⁪⁢⁢⁪‏⁠‎⁪⁡⁪⁠⁪⁣⁤⁪‏⁠⁪⁪⁠‎⁪⁡⁪⁠⁪⁤⁣⁪‏‏⁡⁠⁡‌⁢⁢‎‎⁪⁡⁪⁠⁪⁢⁣⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁤⁪‏⁠⁪⁪⁠⁪⁪‏⁡⁠⁡‌⁢⁣‎‎⁪⁡⁪⁠⁪⁣⁡⁪‏⁠‎⁪⁡⁪⁠⁪⁣⁢⁪‏⁠‎⁪⁡⁪⁠⁪⁣⁣⁪‏⁠⁪⁪‏⁡⁠⁡‌⁢⁤‎‎⁪⁡⁪⁠⁪⁤⁡⁪‏⁠‎⁪⁡⁪⁠⁪⁤⁢⁪‏⁠⁪⁪‏⁡⁠⁡‌
ED # ‎⁡Push three copies of n ** 2
 Ẋ # ‎⁢Cartesian product of the range [1, n ** 2] and [1, n ** 2]
 ‹ # ‎⁣with all sublists decremented
 s # ‎⁤and sorted
 ƛ ; # ‎⁢⁡To each pair
 ƒ⋏ # ‎⁢⁢ reduce by bitwise and
 bT3 # ‎⁢⁣ is the length of truthy indices 1?
 ẇ # ‎⁢⁤wrap into 2 ** n chunks
💎

Created with the help of Luminespire.

\$\begingroup\$ 9 bytes (or 8.25 with Vyncode) \$\endgroup\$

The Thonnu
– The Thonnu

2023年07月26日 14:59:14 +00:00
Commented Jul 26, 2023 at 14:59

The Thonnu 18.7k3 gold badges19 silver badges76 bronze badges · Answer 5 · 2023-07-26 14:16:20Z

Thunno 2, 10 bytes

2RẉDȷỌ€ʂ−Q

Try it online!

Outputs a nested list. The footer formats it into the grid. Port of Lynn's Jelly answer.

Explanation

2RẉDȷỌ€ʂ−Q # Implicit input
2Rẉ # [1,2] to the cartesian power of the input
 DȷỌ # Outer product over minimum with itself
 €ʂ # Sum each inner list
 −Q # Decrement, not equal to the input?
 # Implicit output

Old:

OLDȷÆ&ıḃ1ȷcḅ # Implicit input
OL # Push [0..2**n)
 DȷÆ& # Outer product over bitwise AND
 ıḃ1ȷc # Popcount of each
 ḅ # Equals one?
 # Implicit output

xnor 150k26 gold badges287 silver badges676 bronze badges · Answer 6 · 2023-07-27 05:49:17Z

Python, 66 bytes

lambda m:(q:=range(2**m))and[[(n:=i&j)>n^n-1for i in q]for j in q]

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mousetail's answer, but with bit operations instead of bit_count. We want to check that i&j (which we call n) doesn't have exactly one set bit, which means that it's a not power of two. We do this by checking that n>n^n-1, which is processed as n>n^(n-1). Let's check that this works in every case:

If n is a power of two, then n-1 has disjoint bits set to n and so their XOR is greater than n.
If n is any other positive value, then n and n-1 share the same leading bit, and the XOR cancels this bit making the result smaller than n.
For n==0, the result is 0, the same as n.

Annoyingly Python doesn't let us do the walrus assignment q:=range(2**m) in the iterable of a list comprehension, so the code has it assigned beforehand. loopy walt points out a better way to handle the two levels of iteration is using the classic divmod two-loop trick together with the classic zip/iter chunker.

61 bytes

lambda m:zip(*2**m*[((n:=j&j>>m)>n^n-1for j in range(4**m))])

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\$\begingroup\$ -3 just using the good ol' grouper. \$\endgroup\$

loopy walt
– loopy walt

2023年07月29日 19:09:09 +00:00
Commented Jul 29, 2023 at 19:09

chunes 27.8k3 gold badges32 silver badges55 bronze badges · Answer 7 · 2023-07-27 02:05:54Z

Arturo, ^{(削除) 57 54 (削除ここまで)} 50 bytes

$->n[map^2n'x->map^2n=>[0<>^and dec<=<=and&-1x-1]]

Try it!

$->n[ ; a function taking an argument n
 map^2n'x-> ; map over [1..2^n]; assign current elt to x
 map^2n=>[ ; map over [1..2^n]; assign current elt to &
 0<> ; is zero not equal to
 and&-1x-1 ; bitwise and of current elts minus one
 <=<= ; duplicated twice
 dec ; minus one
 and ; bitwise and
 ^ ; NOS to the TOS power
 ] ; end map
] ; end function

Giuseppe 29.4k3 gold badges33 silver badges106 bronze badges · Answer 8 · 2023-07-26 16:55:22Z

3

\$\begingroup\$

R, (削除) 55 (削除ここまで) 53 bytes

function(n)(y=outer(x<-1:2^n-1,x,bitwAnd))^0-y%in%2^x

Try it online!

Rather than counting the number of on bits (which is very lengthy in R), tests the equivalent characterization of "equal to a power of 2"; then has to do some reshaping of the vector output of %in% to get back to the right shape matrix.

Share

Improve this answer

edited Jul 27, 2023 at 11:10

answered Jul 26, 2023 at 16:55

Giuseppe's user avatar

Giuseppe

29.4k3 gold badges33 silver badges106 bronze badges

\$\endgroup\$

2

\$\begingroup\$ I cannot find anything on meta - is it generally allowed to output a matrix as a flat vector? \$\endgroup\$

pajonk
– pajonk

2023年07月26日 18:14:29 +00:00
Commented Jul 26, 2023 at 18:14
1

\$\begingroup\$ @pajonk hmm probably not. I didn't notice b/c of the footer, so I'll have to update it. \$\endgroup\$

Giuseppe
– Giuseppe

2023年07月26日 18:33:07 +00:00
Commented Jul 26, 2023 at 18:33

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Adamátor 8,2932 gold badges15 silver badges28 bronze badges · Answer 9 · 2023-07-26 15:19:43Z

2

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Nim, 119 bytes

import bitops
proc f(n:int)=
 let r=0..<1 shl n;for i in r:
 var s="";for j in r:s&= $int 1!=popcount i and j
 echo s

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answered Jul 26, 2023 at 15:19

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score 2 · Answer 10 · 2023-07-26 17:45:13Z

Jelly, 10 bytes

2*Ḷ&þ`B§n1

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How it works

2*Ḷ&þ`B§n1 - Main link. Takes n on the left
2* - 2 ** n
 Ḷ - [0, 1, ..., 2**n)
 þ` - To every pair (a, b) in this range:
 & - Bitwise and
 B - Convert all to binary
 § - Sums
 n1 - Does not equal 1?

CursorCoercer 5,56612 silver badges27 bronze badges · Answer 11 · 2023-07-26 14:29:46Z

Pyth, (削除) 15 (削除ここまで) 14 bytes

mmn1s&VdkQ=^U2

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Explanation

mmn1s&VdkQ=^U2Q # implicitly add Q
 # implicitly assign Q = eval(input())
 =^U2Q # Q = repeated cartesian product of [0,1] Q times
m Q # map lambda d over range(Q)
 m Q # map lambda k over range(Q)
 &Vdk # d & k, vectorized
 s # sum bits
 n1 # != 1

LeopardShark 2931 silver badge8 bronze badges · Answer 12 · 2023-07-26 20:59:40Z

BQN (CBQN), (削除) 45 bytes (削除ここまで) 22 ‘bytes’

{1≠+ ́ ̈∧⌜ ̃⥊∾⌜⍟(x-1) ̃↕2}

I think the following is slightly more elegant, and it's the same number of characters but, alas, more bytes.

{1≠+ ́∘∧⌜ ̃⥊(↕2)∾⌜⍟x⋈⟨⟩}

We operate only on bit-strings, rather than integers. ⥊⌜⍟(x-1) ̃↕2 gives us the list of binary strings of length x, which is then anded into a table, and summed: + ́∘∧⌜ ̃. We then take only the 1≠ entries.

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Though I don't know the specifics of why, every BQN answer I've seen here counts 1 glyph as 1 byte. I think you can call these 22 bytes.

Level River St 28.8k4 gold badges40 silver badges112 bronze badges · Answer 13 · 2023-07-26 22:51:10Z

Ruby, (削除) 72 56 (削除ここまで) 54 bytes

->n{(w=0...1<<n).map{|i|w.map{|j|a=i&j;(a&a-1)**a>0}}}

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This employs one of my favourite bit-twiddling hacks. a&a-1 gives 0 if a has only one bit set. Unfortunately it also gives 0 if a==0 so we use the fact that any number (including 0) raised to the power 0 is 1 to catch this special case: (a&a-1)**a.

So now we have distinction between nonzero and zero numbers. The rules require two distinct values, so we use >0 to convert to true/false and we are done.

The footer in the linked TIO formats the output 2d array line by line, and converts the true/false to 1/0.

52 bytes with Ruby 2.7 Numbered Parameters (TIO is stuck on 2.5.5) - ->n{(w=0...1<<n).map{|i|w.map{a=i&_1;(a&a-1)**a>0}}}
@ValueInk oh wow thanks, that's new! You really have to check your versions. I still miss when you could write .5 instead of 0.5
CG One Handed pointed it out to me earlier this week and it feels like a whole new world. Also, you can still save a byte on 2.5 by using j&=i;(j&j-1)**j>0 for 53 bytes Try it online!

chunes 27.8k3 gold badges32 silver badges55 bronze badges · Answer 14 · 2023-07-26 14:13:10Z

Factor + `math.matrices math.unicode`, 57 bytes

[ 2^ dup [ bitand bit-count 1 ≠ ] <matrix-by-indices> ]

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2^ ! two to the input power
dup ! duplicate
[ ... ] <matrix-by-indices> ! create an MxN matrix with indices on top of stack
bitand ! bitwise and
bit-count ! number of on bits
1 ≠ ! not equal to one

The Thonnu 18.7k3 gold badges19 silver badges76 bronze badges · Answer 15 · 2023-07-26 14:24:40Z

05AB1E, 15 bytes

oDL<ãε`&b1¢Θ}sô

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Outputs a nested list. The footer formats it into the grid. Port of lyxal's Vyxal answer.

-1 thanks to @KevinCruijssen

Explanation

oDL<ãε`&b1¢Θ}sô # Implicit input
oD # Push 2 ** n and duplicate
 L<ã # Cartesian product of [0..2**n) with itself
 ε } # Map over this list:
 `& # Bitwise AND of the pair
 b1¢Θ # Is the popcount equal to 1?
 sô # Split into 2 ** n chunks
 # Implicit output

Dâ can be ã (and golfing it some more would result in my 10 bytes answer, which I was already writing before I saw this answer tbh).

Adamátor 8,2932 gold badges15 silver badges28 bronze badges · Answer 16 · 2023-07-26 15:00:19Z

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Nibbles, 11 bytes

.;`,^2$.@!=1+``@`&@$

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Returns a matrix with truthy values (positive numbers) for 1 and falsy values (0) for 0.

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answered Jul 26, 2023 at 15:00

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Sean 8,16612 silver badges16 bronze badges · Answer 17 · 2023-07-26 15:55:32Z

Raku, 52 bytes

{(^$_ X+&^$_).rotor($_)».&{!$_||$_!=1+<.msb}}o 1+<*

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The version of Raku on TIO fails to parse the !$_||$_!=1+<.msb bit, presumably due to a bug, so I've expressed it there as !($_&&$_==1+<.msb), which is equivalent by De Morgan's law, but two bytes longer.

This is an anonymous function consisting of two separate anonymous functions composed together with o. The right-hand function is 1 +< *, which shifts the number 1 leftwards a number of bits given by its argument; that is, two to the power of the argument. That number is passed to the left-hand function, where the variable $_ gets its value.

^$_ is a range of numbers from 0 up to one less than $_. For example, when the number passed to the first function is 3, this is the range 0..7.
X+& gives the cross product of one copy of that range with another using the bitwise-and operator +&.
.rotor($_) takes that flat crossed list and makes it into a square matrix.
».&{ ... } recursively applies the code between the braces to each element of that matrix.
!$_ || $_ != 1 +< .msb tests whether each number is zero (!$_) or consists of only a single binary digit by comparing the number to 1 bit-shifted left a number of places equal to the number's most significant bit (1 +< .msb).

Arnauld 206k21 gold badges187 silver badges670 bronze badges · Answer 18 · 2023-07-26 17:31:16Z

1

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JavaScript (ES6), 74 bytes

n=>[...Array(1<<n)].map((_,y,a)=>a.map((_,x)=>(g=k=>k?k%2^g(k/2):1)(x&y)))

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answered Jul 26, 2023 at 17:31

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Neil 184k12 gold badges76 silver badges290 bronze badges · Answer 19 · 2023-07-26 18:41:46Z

Charcoal, 18 bytes

≔X2NθEθ⭆θ¬=1Σ↨&ιλ2

Try it online! Link is to verbose version of code. Explanation:

 2 Literal integer `2`
 X Raised to power
 N Input as a number
≔ θ Save in variable
 θ Saved variable
 E Map over implicit range
 θ Saved variable
 ⭆ Map over implicit range and join
 ι Outer value
 & Bitwise And
 λ Inner value
 ↨ 2 Convert to base `2`
 Σ Take the sum
 ¬= Does not equal
 1 Literal integer `1`
 Implicitly print

alephalpha 51.9k7 gold badges75 silver badges196 bronze badges · Answer 20 · 2023-07-27 01:38:33Z

1

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PARI/GP, 51 bytes

n->matrix(2^n,,i,j,sumdigits(bitand(i-1,j-1),2)!=1)

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answered Jul 27, 2023 at 1:38

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138 Aspen 7,2671 gold badge5 silver badges18 bronze badges · Answer 21 · 2023-07-30 09:25:12Z

Wolfram Language(Mathematica), 68 bytes

Golfed version. Try it online!

f@n_:=Array[Boole[1!=Tr@IntegerDigits[BitAnd[#-1,#2-1],2]]&,2^{n,n}]

Ungolfed version. Try it onlinne!

f@n_:=Table[If[Total[IntegerDigits[BitAnd[i-1,j-1],2]]!=1,1,0],{i,1,2^n},{j,1,2^n}]

l4m2 32.4k2 gold badges26 silver badges115 bronze badges · Answer 22 · 2023-08-05 16:41:11Z

1

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JavaScript (Node.js), 65 bytes

n=>[...Array(1<<n)].map((_,y,a)=>a.map((_,x)=>(x&=y)&&+!(x&x-1)))

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na

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answered Aug 5, 2023 at 16:41

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Stack Exchange Network

Compute this fractal matrix

Example:

22 Answers 22

Python, 72 bytes

05AB1E, 10 bytes

Jelly, 8 bytes

Vyxal, 66 bits^v2 , 8.25 bytes

Explained

Thunno 2, 10 bytes

Explanation

Python, 66 bytes

61 bytes

Arturo, ^{(削除) 57 54 (削除ここまで)} 50 bytes

R, (削除) 55 (削除ここまで) 53 bytes

Nim, 119 bytes

Jelly, 10 bytes

How it works

Pyth, (削除) 15 (削除ここまで) 14 bytes

Explanation

BQN (CBQN), (削除) 45 bytes (削除ここまで) 22 ‘bytes’

Ruby, (削除) 72 56 (削除ここまで) 54 bytes

Factor + `math.matrices math.unicode`, 57 bytes

05AB1E, 15 bytes

Explanation

Nibbles, 11 bytes

Raku, 52 bytes

JavaScript (ES6), 74 bytes

Charcoal, 18 bytes

PARI/GP, 51 bytes

Wolfram Language(Mathematica), 68 bytes

JavaScript (Node.js), 65 bytes

Your Answer

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Hot Network Questions

Compute this fractal matrix

Example:

22 Answers 22

Python, 72 bytes

05AB1E, 10 bytes

Jelly, 8 bytes

Vyxal, 66 bitsv2 , 8.25 bytes

Explained

Thunno 2, 10 bytes

Explanation

Python, 66 bytes

61 bytes

Arturo, (削除) 57 54 (削除ここまで) 50 bytes

R, (削除) 55 (削除ここまで) 53 bytes

Nim, 119 bytes

Jelly, 10 bytes

How it works

Pyth, (削除) 15 (削除ここまで) 14 bytes

Explanation

BQN (CBQN), (削除) 45 bytes (削除ここまで) 22 ‘bytes’

Ruby, (削除) 72 56 (削除ここまで) 54 bytes

Factor + math.matrices math.unicode, 57 bytes

05AB1E, 15 bytes

Explanation

Nibbles, 11 bytes

Raku, 52 bytes

JavaScript (ES6), 74 bytes

Charcoal, 18 bytes

PARI/GP, 51 bytes

Wolfram Language(Mathematica), 68 bytes

JavaScript (Node.js), 65 bytes

Your Answer

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Vyxal, 66 bits^v2 , 8.25 bytes

Arturo, ^{(削除) 57 54 (削除ここまで)} 50 bytes

Factor + `math.matrices math.unicode`, 57 bytes