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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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
le 11 Fév 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
le 11 Fév 2016
Modifié(e) : Guillaume
le 11 Fév 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
le 11 Fév 2016
r = VChooseKRO(uint8(0:1), 3);
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Guillaume
le 11 Fév 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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