A student in ECSE 222 wants to generate a new fault-tolerant code. The student decided to add two bits {t1t2} at the end of all n-bit binary numbers such that the sum of '1's in the new (n+2) bit number is divisible by three with a remainder of 2. When a single '1' is to be added, t1t2 = 01. how many errors can be detected and corrected using the new code?

Answer rating: 100% (QA)

This code can detect and correct up to two errors The first error can be detected by the sum of 1s in the new number not being divisible by three with ... View the full answer