General Comments
================

A few people did not quite understand the meaning of <=_m.
They would do a proof by contradiction: Assume B is
decidable and construct a decider for A.  Then, they would
say they showed A <=_m B.   But this is not enough.
One consequence of A <=_m B is that if B is decidable then
A is decidable, but A <=_m B has consequences even if B
is not decidable.  (For example, if B is semi-decidable, so
is A, and many similar consequences hold.)



Question 1
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This question was well done.  Not much to say.

Question 2
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Many people tried to prove this problem was not decidable.
There was one group, who I assume descussed the problem,
said to simulate a TM on every other square skipping the nth
square.  This does not work because the head can move only
one square at a time.

Question 3
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Most people had the right intuition, neither problem is
semi-decidable.  Unfortunatly, many people did not get the
reduction right.  There were no common wrong solutions.

Question 4
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A significant number of people did not understand the facts
that were given.  Some people took the three languages and
combined the two that are not regular, and took the
complement.  Others did not understand that to take the
complement of the languages they had to be solvable on a
deterministic PDA.  They ignored the deterministic part and
took the complement of a general CFG.