# GATE | Gate IT 2007 | Question 43

• Difficulty Level : Basic
• Last Updated : 28 Jun, 2021

An error correcting code has the following code words:
00000000, 00001111, 01010101, 10101010, 11110000.
What is the maximum number of bit errors that can be corrected ?
(A) 0
(B) 1
(C) 2
(D) 3

Explanation: While transmitting the data through the channel , noise may be added to the data and thus, may cause errors in the data. Hamming code errors CANNOT be DETECTED if one code converts into another ,so if maximum hamming distance (i.e. the number of 1 s when we XOR 2 hamming codes ) between any two codes is ‘t’ then for detection error should not be more than t-1,otherwise the code could have been converted to another and we may assume that it is a correct hamming code. Similarly,for CORRECTION we should also know which hamming code it was,thus, if the maximum hamming distance is d then d/2 is the partition between two hamming codes from where we can find which code it was.Thus,if we have to correct ‘t’ errors then Max dist = 2*t +1 Max hamming distance is between 01010101 and 10101010 =>8
Thus,
8=2*t+1
t=3.5
We’ll take ceil as taking more than these bits will again make it impossible to correct the error.

This solution is contributed by Shashank Shanker khare

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Another Explanation:
d=(2t+1) t=number of bits could be corrected
maximum hamming distance between any two of the given code = 8 (between 01010101 and 10101010)
t=3.5, t=3

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