all subsets
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[code:1xko7hvu]
data aa;
input n1 $ 1;
call symputx('nobs', _N_);
datalines;
a
b
c
d
e
f
g
;
run;
proc transpose data=aa out=bb;
var n1;
run;
proc means data=bb noprint;
class col1-col&nobs;
output out=pattern(drop=_type_ _freq_);
run;
[/code:1xko7hvu]
data ahuige(keep=chars);
do i=0 to 2**&len-1;
chars=put("&chars",&len..);
do j=1 to &len;
if substr(put(i,binary&len..),j,1)^=1 then substr(chars,j,1)=' ';
end;
output;
end;
run;[/code:3k1d7ljr]
I forgot binary representation. Here's another one. Is it possible to reduce the time complexity from O(n*2^n) to O(2^n) ?
[code:3b0s05xp]
proc iml;
chars = {'a' 'b' 'c' 'd' 'e' 'f' 'g'};
x = repeat(0,1,ncol(chars));
do xi=1 to 2**ncol(chars)-1;
if x[1]=0 then x[1]=1;
else do;
n0 = x[>:<]; *the position of first 0;
x[1:(n0-1)] = 0;
x[n0] = 1;
end;
y=rowcatc((chars[loc(x=1)])`);
print x / y;
end;
From the appearance of your source code it seems the inner loop is gone away. I have no idea about iml. But it seems the function rowcatc does this for you(correct me if I am wrong).
however, the rowcatc itself or its counterpart should have a mechanism to scan the whole vector and concatenate them conditionally.
It might not reduce the complexity substantially. At least on my computer it takes more time to run. Anyway, it is a nice try buddy.
I was just mimicking your code, so the inner loop is still there. I think the 'loc' function needs to visit every entry of vector 'x' to identify all the 1's, also 'x[>:<]' may have to search the whole vector to identify the first 0.
Anyway, my question is simply STUPID, because 2^n is already so bad that adding a factor of n doesn't make it worse.
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