Re: elementwise boolean operations

[Przemek Klosowski]

On 05/20/2010 03:51 PM, Judd Storrs wrote:
> function m = graycode(n)
>    m = false(2^n,n) ;
>    m(2,n) = true ;
>    for i=1:n-1
>      j = n-i ;
>      k = 2^i+1:2^(i+1) ;
>      m(k,j) = true ;
>      m(k,j+1:end) = flipdim(m(1:2^i,j+1:end),1) ;
>    endfor
> endfunction

Both Gray and straight binary can be seen as recursively
defined: the top left corner is recursively calculated,
and flipped (for Gray) or copied straight into bottom left
corner, and the leftmost column is filled with 00000...11111

function m=graycode(n)
 if (n==1)  m=[0;1];  return; end
 s=2^(n-1);
 m=false(2*s,n);
 m(1:s,2:end)=graycode(n-1);
 m(s+1:end,2:end)=m(s:-1:1,2:end);
 m(s+1:end,1)=true;
end

function m=binarycode(n)
 if (n==1)  m=[0;1];  return; end
 s=2^(n-1);
 m=false(2*s,n);
 m(1:s,2:end)=binarycode(n-1);
 m(s+1:end,2:end)=m(1:s,2:end);
 m(s+1:end,1)=true;
end

They are competitive with the non-recursive version for time.

Improvement suggestions welcome, e.g. I wonder if there is a nicer way 
of flipping the rows than

m(s+1:end,2:end)=m(s:-1:1,2:end);