大门|大门-CS-2017（第二组）|问题55

考虑一个二进制代码，它只包含四个有效代码字，如下所示。 00000, 01011, 10101, 11110

设码的最小汉明距离为p，可由码纠正的最大错误位数为q。p和q的值为：

（A） p=3，q=1 （B） p=3，q=2 （C） p=4，q=1 （D） p=4，q=2 答复: （A） 说明： 我们需要找到最小值 汉明距离 （相应位位置的差异）

00000, 01011, 10101, 11110

For two binary strings, hamming distance is number
of ones in XOR of the two strings.

Hamming distance of first and second is 3, so is 
for first and third. Hamming distance of first and
fourth is 4.

Hamming distance of second and third is 4, and 
second and fourth is 3.

Hamming distance of third and fourth is 3.

Thus a code with minimum Hamming distance d 
between its codewords can detect at most d-1
errors and can correct ⌊(d-1)/2⌋ errors.

Here d = 3. So number of errors that can be
corrected is 1.

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