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Code Golf

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Required fields*

Compute this fractal matrix

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 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

Answer*

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  • \$\begingroup\$ 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}}} \$\endgroup\$ Commented Jul 27, 2023 at 20:10
  • \$\begingroup\$ @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 \$\endgroup\$ Commented Jul 27, 2023 at 21:20
  • \$\begingroup\$ 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! \$\endgroup\$ Commented Jul 27, 2023 at 22:40

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