如何在代码字中进行所有可能的位更改？ [英] How to make all possible changes of bits in a codeword?

问题描述

我有代码字u，我需要制作一个程序，允许：

为1< = i< = k，对你的k个位中的l进行所有可能的更改。



示例1：

I have codeword u and I need to make a program that allows :
for 1<=i<=k ,make all possible changes of l of the k bits in u.

Example 1:

if u = [1 1 0 1] ; k = 4 and l = 1

我们获得：

we obtain :

u1 = [0 1 0 1] ; u2 = [1 0 0 1] ; u3 = [1 1 1 1] and u4 = [1 1 0 0].

例2：

Example 2 :

if u = [1 0 0 ] ; k = 3  and l = 1

我们获得：

we obtain :

u1 = [0 0 0] ; u2 = [1 1 0] and u3 = [1 0 1].

我试图编程，但我获得：

I tried to program, but i obtain :

if k = 3 and l = 1
u1 = [1 00] ; u2 = [010] and u3 = [001]

我尝试过：



我的代码：

What I have tried:

My code :

l = 1;
ind = nchoosek(1:k, l);
t = size(ind,1);
u1 = zeros(t,k);
u1(bsxfun(@plus, (ind-1)*t, (1:t).')) = 1; 

推荐答案

我不知道MATLAB，你的要求不是很清楚但是下面的C代码实现了示例的输出

I do not know MATLAB and your requirement is not very clear but the following C code achieves the output of the examples

int u = 0x0d;   // 1101
int k = 4;
int l, result;

for(l = 1 << (k - 1); l != 0; l >>= 1)
{
	result = u ^ l; // 0101 = 1101 XOR 1000, 1001 = 1101 XOR 100, 1111 = 1101 XOR 10, 1100 = 1101 XOR 1
}
// it just remains to express 'result' in the form [b1 b2 b3 b4] - where b1 .. b4 are the bit values

上述代码也适用于第二个示例，其中包含3位，当然也有适当的更改u和k的值。

the above code also works for second example with 3 bits with of course appropriate changes in value of u and k.

这篇关于如何在代码字中进行所有可能的位更改？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！