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ECE-C490 Security in Computing
Winter term 2005
Homework #6 Solutions
Due in class on Tuesday, March 1st
1. Consider the following code:
V = {0000000, 0001111, 0010011, 0011100, 0100101, 0101010, 0110110,
0111001}:
a. Find d(V) = smallest distance between any two vectors in the code. In this
case, 3 is the smallest number of bits that are different between any two
vectors.
b. What is the multiplicity of errors that can be corrected? If two errors are
present, we could accidentally “correct” the vector to the wrong value,
since the minimum distance between any two vectors is 3. Thus only 1
error can be corrected.
c. What is the multiplicity of errors that can be detected? We can still detect
up to two errors. If three were present, in some cases we would have
simply changed the original vector to another valid one.
d. Correct the message w = 0111010. Closest to 0101010 (differs by only
one bit). Therefore we correct to 0101010.
e. List five of the messages that would be corrected to 0000000 if received.
Any vector that differs by only one bit would be corrected to 0000000. For
example, 1000000, 0100000, etc.
f. List five of the messages that would be corrected to 0011100 if received.
Same idea as part (e). Possible answers: 1011100, 0111100, 0001100, etc.
2. Consider the group < J+7, +7 >. Find 3-1 and 5-1. The identity for this group is 0. (3
+7 4) = 0, so 3-1 = 4. Similarly, (5 +7 2) = 0, so 5-1 = 2.
3. Consider the group <J+4, +4>. Find the identity for this group, as well as 2-1. The
identity for this group is also 0, thus (2 +4 2) = 0, so 2-1 = 2.
4. Is <J+8, +10> a group? Why or why not? Closure does not hold for this set; (8 +10
1) = 9, which is not an element of the set. Thus it is not a group.
5. Is the set of numbers {-3, -2, -1, 0, 1, 2, 3} a group with respect to addition mod
4? Why or why not? Closure holds for this set, but there is no unique inverse. The
identity is 0, but for each element of the set except 0, there are multiple inverses
we can use to obtain 0. For example, (1 +4 (-1)) = 0, but (1 +4 3) = 0 as well. Thus
this set is not a group.