Dear Brendon,

The answer file is in HTML, and works fine at my desk.
I attach the following plain text of the answers.

Regards,
Tad
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Answer to 1998 cosc413 examination

(1) (1.3)

   S={1,2,3,4}, S1={1,3}, S2={1,2,3}, S3={3,4}, S4={2,4}

   B(S)  = 001001001001
   B(S1) = 000001000001
   B(S2) = 000001001001
   B(S3) = 001001000000
   B(S4) = 001000001000

   B(S) = B(S1) + B(S4)  (perform addition as binary numbers)

(1.1) As seen from the above example, the leading 0's preceding
each occurrence of 1 play the role of a buffer zone for carries
for binary addition. If we have an exact cover such as Si, Sj, Sk,
for example, obviously we have

   B(S) = B(Si) + B(Sj) + B(Sk),

since we have no carry and make B(S) by adding these binary numbers.
This is a solution for the corresponding knapsack problem.

(1.2) Suppose we have a solution for the knapsack problem, such as
B(S) = B(Si) + S(Bj) + B(Sk). Then there can be no carry in this 
addition. Otherwise there would be some 1's in some buffer zones.
Since these bynary numbers add up to make B(S), we see Si, Sj, Sk
will form an exact cover of S. Note that we use specific example 
to show a general proof.

(3) Colour the 24 rooms black and white like a chess board, with
the reception room white. Then your Hamiltonian path would alternates
white and black. If there were a Hamiltonian path, the difference 
between the numbers of white and black on your path must be at most
1. In this case, the difference is 2, making a Hamiltonian path
impossible.

(4.3) For a>b>0, we divide a by b, resulting the quation

      a = q * b + r, 0 <= r < b   (q: quotient, r: remainder)

We show that 2*r < a. Suppose b <= a/2. Then r < b <= a/2.
Suppose a/2 < b < a. Then r = a - b < a/2.

(6) This problem belongs to the general framework of Caesar's
cryptosystem by substitution. 

(6.1) There is a plain text BAAC that produces 101111110 under
key1 and key2.

(6.2) If key3 and key4 produce the same cypher text from a plain text,
the plain text must start from A. Then we have the situation as

   key3  A   10
   key4  A   101

The next symbol must be C. To see why, try A and C. Then the situationn
becomes

   key3 AC   10101
   key4 AC   101011

That is, we hit the same situation as before. Thus no identical cypher text
under key3 and key4.

(4.3) Suppose we have two different plain texts that produce an identical
cipher text. Without loss of generality, we can assume one starts with A 
and the other with C. Then we have the situation

   text1  A   10
   text2  C   101

Suppose the next symbol for text1 is A. Then the next symbol for text2 
must be C, resulting in

   text1  AA  1010
   text2  CB  101011

At this stage we can not grow the texts to have the same cypher text.
If the next symbol for text1 is C, that for text2 must be B, resulting in

   text1  AC  10101
   text2  CB  101011

This is the same situation as before.
Thus we conclude key3 is uniquely decipherable.