Problem 1
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(A)

Here we are looking for exactly two of something, so use the general
inclusion exclusion formula:

Sum_{i=0}^{i=floor(k/2)}   (2+i)C(i) * (k-2-i)C(2+i)  * 5^(k-2-2i)


(B)

In this problem count the case AAA seperately...  note the rest of the
string must not contain AA

Then add on the case with AA and AA delimited by non A characters


Problem 2
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The checker should pick a random vector v, then for the supposed inverse, B,
compute A(Bv) and compar this with v.  If they are different, then B is not
A inverse, otherwise there is less than 50% chance it is not the inverse.
This must be repeated at least n times with linearly independent v's to
be sure that B is the inverse.  Each repetition requires two matrix, vector
multiplies, or O(n^2) operations.


Problem 3
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When appending T after S, consider the changes to the first level of nodes
above the actual string.  Only the last above S, and the first above T will
change.  This is a constant number of changes which will be propogated up the
fingerprint tree.  So we have a constant number of changes at each level,
and the number of levels is bounded by log_2 n, and therefore O(log n)
new nodes, and O(log n) time to append.


Problem 4
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(A)  No.  Once one of the good processors has tally >= G then on the next round
all the good processors will have tally >= G, and have all set their votes
permanently.

(B) A processor can halt one round after it has set its vote permanently.