How to generate all 2^n combinations of n choices from a 2-vector?

If I want all combinations of 3 choices of 0 or 1, I do this:

t = combinations([0 1],[0 1],[0 1]);

If I want all combinations of 4 choices, I have to do this:

t = combinations([0 1],[0 1],[0 1],[0,1])

If I want all combinations of 5 choices, I have to do this:

t = combinations([0 1],[0 1],[0 1],[0,1],[0 1])

...and so on. Is there any way to generate all combination of a variable number of choices?

Answer 1

[John D'Errico]

Simple enough. Don't use combinations.

n = 3;
t = dec2bin(0:2^n - 1) - '0'

t = 8×3

0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

Easy enough to make that into a table if you so desire.

Sigh. Do you INSIST on using combinations? This will suffice.

v = repmat({[0 1]},1,n);
t2 = combinations(v{:})

t2 = 8×3 table
    Var1    Var2    Var3
    ____    ____    ____

     0       0       0  
     0       0       1  
     0       1       0  
     0       1       1  
     1       0       0  
     1       0       1  
     1       1       0  
     1       1       1  

I can think of at least a couple other ways if pushed, but they shoiuld work.

[Jerry Guern]

Thank you. Your first solution efficiently solves this specific problem, but I always appreciate it very much when people offer solutions that are easy to generalize.

[Steven Lord]

And of course, if you want a matrix from combinations (assuming all the table variables can be concatenated together) you can do that by indexing into the output from the function call.

n = 3;
t = dec2bin(0:2^n - 1) - '0';
v = repmat({[0 1]},1,n);
t2 = combinations(v{:}).Variables

t2 = 8×3

0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

doTheyMatch = isequal(t, t2)

doTheyMatch = logical
   1

You need to be careful, though, if you have mixed types.

t3 = combinations([0 1], ["apple", "banana"])

t3 = 4×2 table
    Var1      Var2  
    ____    ________

     0      "apple" 
     0      "banana"
     1      "apple" 
     1      "banana"

t4 = t3.Variables % Note that this is a string array, using "0" and "1" not 0 and 1

t4 = 4×2 string array
    "0"    "apple" 
    "0"    "banana"
    "1"    "apple" 
    "1"    "banana"

[Matt J]

The spped performance as a function of n is interesting. combinations() starts to overtake ndgrid() only for n>=20 or so. I would have expected combinations() to benefit from avoiding a cat() operation.

Timing(15)

ans = 0.0017

ans = 8.8375e-04

ans = 6.7575e-04

Timing(20)

ans = 0.1078

ans = 0.0808

ans = 0.1195

function Timing(n)

 timeit(@() methodDEC2BIN(n))

 timeit(@() methodCOMBINATIONS(n)) 

 timeit(@() methodNDGRID(n))
end


function methodDEC2BIN(n)

 t = dec2bin(0:2^n - 1) - '0';

end


function methodCOMBINATIONS(n)

 v = repmat({[0 1]},1,n);
 t = combinations(v{:});

end

function methodNDGRID(n)

 [t{n:-1:1}]=ndgrid([0,1]);
 t=reshape( cat(n+1, t{:}) , [],n);

end

Answer 2

[Matt J]

n=3;

clear t 
[t{n:-1:1}]=ndgrid([0,1]);

t=reshape( cat(n+1, t{:}) , [],n)

t = 8×3

0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>