How do I create a 2^n by n matrix that the rows of the matrix enumerate all the possibilities for a n digit binary number?
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Shawn Miller
el 11 de Feb. de 2016
Comentada: Guillaume
el 11 de Feb. de 2016
For example, for n=3, there are 8 rows, and they are 000,100,010,001,110,101,001,111 (may not in this order). Any fast way to realize it using a function, say f(n)?
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Guillaume
el 11 de Feb. de 2016
Use dec2bin to get the string representation. Subtract '0' from the string to convert characters '0' and '1' to numbers 0 and 1:
n = 3;
dec2bin(0:pow2(n)-1) - '0'
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Guillaume
el 11 de Feb. de 2016
Editada: Guillaume
el 11 de Feb. de 2016
In matlab, when you subtract one character from another, you are in effect subtracting the ASCII values of the character. So, obviously ASCII of '0' - ASCII of '0' is 0. Since '1' follows '0' in the ASCII table. ASCII of '1' - ASCII of '0' is 1. In fact, ASCII of any 'digit' - ASCII of '0' is that digit.
dec2bin will always return strings padded to the minimum number of bits to represent the greatest input. Since you need 3 bits to represent the 4 to 7 of the input array, all strings are returned with 3 bits.
You could have specified the number of bits explicitly if you wanted.
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Jan
el 11 de Feb. de 2016
r = VChooseKRO(uint8(0:1), 3);
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Guillaume
el 11 de Feb. de 2016
You'll have to compile the c code into a mex file. If you don't care for speed (if your n is not huge, speed is not an issue) you may be better off with some of the file exchange submission mentioned by VChooseKRO, or use my solution which has no dependency.
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