ECE 596 C : Handout # 8

Let h(m) = m mod 2. Let m = 5 i.e. 101 binary string, then h(m) = 5 = 25 ≡ 1 mod 2. Now if we change m = 5 to m = 6 i.e. binary string 101 to 110 then h(m) = 6 = 36 ≡ 0 mod 2. So if the message m has changed from even to odd or odd to even then we can detect it. However if m = 5 is modified to m = 7, then h(m) = 7 = 49 ≡ 1 mod 2. This change cannot be detected! Hence we cannot use h(m) = m mod 2 as a good fingerprint of data. We will not always know if it was changed.