array a=[8 6 7] adding 0:1 to the array. possible combination for array should be 2^3=8.
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the output i want is a matrix of values listed below
x(possible combination)= 8 6 7
8 6 8
8 7 7
8 7 8
9 6 7
9 6 8
9 7 7
9 7 8
采纳的回答
a=[8 6 7];
a+dec2bin(0:2^numel(a)-1)-'0'
6 个评论
Nitesh Rana
2022-5-4
编辑：Nitesh Rana
2022-5-4
Use dec2base to convert to base 3 instead of base 2, and substract 1 to make it -1:1 instead of 0:2
a=[8 6 7];
a+dec2base(0:3^numel(a)-1,3)-1-'0'
Nitesh Rana
2022-5-4
Nitesh Rana
2022-5-4
Nitesh Rana
2022-5-4
Well, you already know that the number of rows in the result is b^n where b is the base (in this case 2 or 3) and n is the number of elements you start with. That suggests to me that you also know that counting from 0 to b^n-1 is essentially what the process is. So I start with 0:b^n-1 and convert them to binary or ternary character vectors:
a = [8 6 7]
dec2base(0:3^numel(a)-1,3)
Then subtract the character '0' to get the distance of each character from the character '0', which is equivalent to its numerical value:
dec2base(0:3^numel(a)-1,3)-'0'
Then add that to the original vector a, with adjustment of -1 in base 3 to get -1:1 rather than 0:2.
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