CSCI 6610		  Automata and Formal Languages		     Spring 2004

		     	   News about the midterm test


--------------------------- SCORES AND GRADE SCALE -----------------------------

	Class average raw score was 91 out of 120 points possible.

	Class averages on questions 1, 2, 3, 4, and 5 as % of the possible were:
	  97%, 67%, 62%, 94%, and 76%, respectively.

	Highest score in class was 111 points.

	Grade scale: A >= 85%, B >= 72.5%, C >= 60%, D >= 50%. 
	  This will be the percentage-to-grade conversion mapping for all
	  scaled scores in the course.

	Scaling: the raw scores will be treated as being out of 100 (instead of
	  out of 120), so the class average is 91% scaled.


---------------------------------- ANSWERS -------------------------------------

Question 1  --  2 points per part.

1(a)  Y
1(b)  Y
1(c)  Y
1(d)  N
1(e)  Y
1(f)  Y
1(g)  N
1(h)  N
1(i)  Y
1(j)  N
1(k)  Y
1(l)  N
1(m)  Y
1(n)  Y
1(o)  N


Question 2  --  7, 10, and 13 points for (a), (b), and (c) respectively.

2(a)  There is a recursive function f such that 
      (for all w in Sigma*)(w in A iff f(w) in B).

2(b)  Disprove, by counterexample.  Let A be empty (recursive) and B be A_{TM}
      (not recursive), for instance.  Then A is mapping reducible to B by the
      function which always returns the value <M_0,1>, where M_0 is some fixed
      TM with input alphabet {0,1} which does not accept 1.

2(c)  Disprove, by contradiction.  
	First, C is RE; we can recognize C by a NTM N which rejects any input
      x which is not of the form <M> where M is a TM with input alphabet {0,1}.
      For any input of this form, N guesses a string w over {0,1} 
      nondeterministically.  N rejects w if it has an even number of 1's.
      Otherwize N simulates M on the input w and accepts <M> iff M accepts w.
	Now from class we know that E_{TM} is not RE and that any set mapping
      reducible to an RE set is RE, so E_{TM} can't be mapping reducible to C.
      
      
Question 3  --  5, 5, and 20 points for (a), (b), and (c) respectively.

3(a)  A DTM with special oracle input and output tapes, and a special query
      state.  Operation depends on an oracle set in the query state, as follows;
      1 or 0 is written on the oracle output tape in one step, depending
      on whether or not the string written on the oracle input tape belongs to
      the oracle set.

3(b)  A is decided by some oracle TM M when M is operated according to B as the
      oracle set.
      
3(c)  We can show directly that E_{TM} is mapping reducible to complement(C);
      Turing reducibility to C then follows from the class facts that mapping
      reducibility implies Turing reducibility, C is Turing equivalent to its
      complement, and Turing reducibility is transitive.
	For the mapping reduction, let f(x) = <M_1> whenever x which is not of
      the form <M> where M is a TM with input alphabet {0,1}, where M_1 is a 
      fixed TM with L(M_1) = {1} (so <M_1> is not in C).  In case x = <M> where
      M is a TM with input alphabet {0,1}, let f(x) = <D(N(M))> where D(N) is a 
      DTM equivalent to N and N(M) is an NTM with input alphabet {0,1} which 
      rejects every input except 1, and on input 1 it guesses a string over M's
      input alphabet and accepts 1 iff M accepts the guessed string (by 
      simulating M on it).  
	A complete answer goes on to verify that x is in E_{TM} iff f(x) is not
      in C, and that f is recursive.


Question 4  --  5 points.  L is in NP iff there is some NTM decider for L which
      has polynomial time complexity.  Or equivalently, iff there is a 
      polynomial time verifier for L.


Question 5  --  5, 10, and 10 points for (a), (b), and (c) respectively.

5(a)  k + Sum (i from 1 to k) log_2(x_i).

5(b)  A certificate is a string ptn in {0,1}^k, which gives a partition of the 
      input sequence into X = { x_i : ptn[i] = 1} and Y = { x_i : ptn[i] = 0}.
      If n is the input length as given in part (a) then evaluating SUM(X) and
      SUM(Y) and checking that they are equal can be done in time O(n^3), O(n^2)
      or even O(n) with due care.  In any case this gives a polynomial time
      verifier for PARTITION, so it is in the class P.
      
5(c)  PARTITION is very like SUBSET-SUM.  It is not hard to see that 
      SUBSET-SUM is polynomial time reducible to PARTITION, which is therefore
      NP-complete.  So it is unlikely to be in the class P, as that would imply
      that P = NP.
	Note that for part (a), Sum (i from 1 to k) x_i is NOT reasonable as a
      measure of the length of the input.  That would correspond to 
      UNARY-PARTITION, which is in P for the same reason as UNARY-SUBSET_SUM
      (just let the target for the subset sum be one half of the sum of all the
      input numbers).