Permutations of 0s and 1s with Recursion

Question

I am trying to write all the 6-bit permutations of 0s and 1s using recursion (i.e., [0 0 0 0 0 1], [0 0 0 0 1 0], [0 0 0 0 1 1], [0 0 0 1 0 0], ... [1 1 1 1 1 1]). I'm following a tip I got from Yahoo! Answers I've got the following but it just doesn't go all the way to the end and there are duplicate entries.

function [c] = myIEEEBaby_all_decimals(h, n)

% n: number of characters more to add

if n == 0
   converter(h);
   return 
end

in1 = [h 1];
in2 = [h 0];

converter(in1);
converter(in2);

if length(h) < n
    myIEEEBaby_all_decimals([in1], n-1)
    myIEEEBaby_all_decimals([in2], n-1)
end

end

function [d] = converter(IEEE)
% convert custom IEEE representation to decimal

IEEE = [zeros(1,6-length(IEEE)), IEEE];

s = IEEE(1);

characteristic = IEEE([2,3]);
k = find(fliplr(characteristic)) - 1;
c = sum(2.^k);

fraction = IEEE([4:6]);
f = sum(2.^-(find(fliplr(fraction))));

d = (-1)^s*2^(c-1)*(1+f);

disp([num2str(IEEE),' : ', num2str(d)]);

end

The output (MATLAB) is just:

>> myIEEEBaby_all_decimals([],6)
0  0  0  0  0  1 : 0.75
0  0  0  0  0  0 : 0.5
0  0  0  0  1  1 : 0.875
0  0  0  0  1  0 : 0.625
0  0  0  1  1  1 : 0.9375
0  0  0  1  1  0 : 0.6875
0  0  1  1  1  1 : 1.875
0  0  1  1  1  0 : 1.375
0  0  1  1  0  1 : 1.625
0  0  1  1  0  0 : 1.125
0  0  0  1  0  1 : 0.8125
0  0  0  1  0  0 : 0.5625
0  0  1  0  1  1 : 1.75
0  0  1  0  1  0 : 1.25
0  0  1  0  0  1 : 1.5
0  0  1  0  0  0 : 1
0  0  0  0  0  1 : 0.75
0  0  0  0  0  0 : 0.5
0  0  0  0  1  1 : 0.875
0  0  0  0  1  0 : 0.625
0  0  0  1  1  1 : 0.9375
0  0  0  1  1  0 : 0.6875
0  0  0  1  0  1 : 0.8125
0  0  0  1  0  0 : 0.5625
0  0  0  0  0  1 : 0.75
0  0  0  0  0  0 : 0.5
0  0  0  0  1  1 : 0.875
0  0  0  0  1  0 : 0.625
0  0  0  0  0  1 : 0.75
0  0  0  0  0  0 : 0.5

That's not what "permutation" means: http://en.wikipedia.org/wiki/Permutation — Mark Byers, Jan 29 '11 at 19:54

score 1 · Accepted Answer · answered Jan 29 '11 at 19:59

1

Simply iterate from 0 to 2⁶-1 and call Matlab's dec2bin(...) function. You could pad the result with zero's using sprintf(...).

answered Jan 29 '11 at 19:59

Bart Kiers

166,582
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Ah, thanks! That's so much easier than what I thought I needed. – Justin M Jan 31 '11 at 23:27

score 1 · Answer 2 · edited May 23 '17 at 12:04

If you want to compute every possible value of a 6-bit number, there are much more efficient ways to do it than using recursion. Using the function DEC2BIN is one option, but this previous question shows that implementations using the function BITGET or indexing via a look-up table are much faster. For example, here's one possible solution:

allperms = cell2mat(arrayfun(@(b) {bitget((0:2^6-1).',b)},6:-1:1));

Then you can apply your converter function to the result on a row-wise basis:

converter = @(b) (1-2.*b(:,1)).*...             %# The sign contribution
            2.^(b(:,2:3)*[2; 1] - 1).*...       %# The exponent contribution
            (1 + b(:,4:6)*[0.125; 0.25; 0.5]);  %# The fraction contribution
values = converter(allperms);

And you can display the results as follows:

>> disp(sprintf('%d %d %d %d %d %d : %f\n',[allperms values].'))
0 0 0 0 0 0 : 0.500000
0 0 0 0 0 1 : 0.750000
0 0 0 0 1 0 : 0.625000
0 0 0 0 1 1 : 0.875000
0 0 0 1 0 0 : 0.562500
0 0 0 1 0 1 : 0.812500
0 0 0 1 1 0 : 0.687500
0 0 0 1 1 1 : 0.937500
0 0 1 0 0 0 : 1.000000
0 0 1 0 0 1 : 1.500000
0 0 1 0 1 0 : 1.250000
0 0 1 0 1 1 : 1.750000
0 0 1 1 0 0 : 1.125000
0 0 1 1 0 1 : 1.625000
0 0 1 1 1 0 : 1.375000
0 0 1 1 1 1 : 1.875000
0 1 0 0 0 0 : 2.000000
0 1 0 0 0 1 : 3.000000
0 1 0 0 1 0 : 2.500000
0 1 0 0 1 1 : 3.500000
0 1 0 1 0 0 : 2.250000
0 1 0 1 0 1 : 3.250000
0 1 0 1 1 0 : 2.750000
0 1 0 1 1 1 : 3.750000
0 1 1 0 0 0 : 4.000000
0 1 1 0 0 1 : 6.000000
0 1 1 0 1 0 : 5.000000
0 1 1 0 1 1 : 7.000000
0 1 1 1 0 0 : 4.500000
0 1 1 1 0 1 : 6.500000
0 1 1 1 1 0 : 5.500000
0 1 1 1 1 1 : 7.500000
1 0 0 0 0 0 : -0.500000
1 0 0 0 0 1 : -0.750000
1 0 0 0 1 0 : -0.625000
1 0 0 0 1 1 : -0.875000
1 0 0 1 0 0 : -0.562500
1 0 0 1 0 1 : -0.812500
1 0 0 1 1 0 : -0.687500
1 0 0 1 1 1 : -0.937500
1 0 1 0 0 0 : -1.000000
1 0 1 0 0 1 : -1.500000
1 0 1 0 1 0 : -1.250000
1 0 1 0 1 1 : -1.750000
1 0 1 1 0 0 : -1.125000
1 0 1 1 0 1 : -1.625000
1 0 1 1 1 0 : -1.375000
1 0 1 1 1 1 : -1.875000
1 1 0 0 0 0 : -2.000000
1 1 0 0 0 1 : -3.000000
1 1 0 0 1 0 : -2.500000
1 1 0 0 1 1 : -3.500000
1 1 0 1 0 0 : -2.250000
1 1 0 1 0 1 : -3.250000
1 1 0 1 1 0 : -2.750000
1 1 0 1 1 1 : -3.750000
1 1 1 0 0 0 : -4.000000
1 1 1 0 0 1 : -6.000000
1 1 1 0 1 0 : -5.000000
1 1 1 0 1 1 : -7.000000
1 1 1 1 0 0 : -4.500000
1 1 1 1 0 1 : -6.500000
1 1 1 1 1 0 : -5.500000
1 1 1 1 1 1 : -7.500000

Actually the vectorized `dec2bin` seems to be faster than the `bitget` solution as presented here. Try: `x = dec2bin(1:2^6-1); allperms = x-'0';` — Dennis Jaheruddin, Sep 19 '13 at 13:40

score 0 · Answer 3 · answered Sep 19 '13 at 13:35

If you want to write all permutations in a simple way, use dec2bin to get the binary values like @Bart recommended.

However, this kind of operation can often be done significantly more efficient if you use vectorization.

So instead of looping over all values and obtaining the results 1 by one, simply do this:

x = dec2bin(1:2^6-1);

If you want the output to be a matrix, also do this:

x = x - '0';

Permutations of 0s and 1s with Recursion

3 Answers3