Grading Scheme for Quiz 2

Part I: Out of 3 points (fifth rule inapplicable, not math!)
	-1: Forgot to circle the tutorial.
	-2: Forgot either the name or the student number
	-3: If neither name or number on the quiz

Part II:
   MTMS = multitape TM with stay
   TM = ordinary TM
   
   Emulating TM with MTMS: Out of 2 points, 0.5 if "I don't know"
	May use any model known to be equivalent; list is below
	-2 if not included
	-1 if massively complex
	-1 if not indicating encoding of TM configuration
	-1 if not indicating emulation of TM transition/move

   Emulating MTMS with TM: Out of 5 points
	May use any model known to be equivalent; allowed list is
		TM, semi-infinite TM, multitrack TM: w/ or w/o stay
		multitape TM, NTM, 2-D TM, multihead TM: w/o stay only
	-2 if assuming unproven equivalence in proof
	-1 if using different model than used above to show equivalence
	    and not arguing equivalence between these models
	-1 if model being used not described clearly enough
	-2 if not indicating encoding of MTMS configuration
	-3 if not indicating how a MTMS transition/move is emulated

	A correct proof will score between 3 and 5 points.
	An incorrect proof will score between 0 and 2 points.

	5: The proof is correct (possibly missing a minor detail or 2)
	4: The proof is mostly correct but is missing some important points
	      examples: Using a different model than for reverse
			 direction without explicitly arguing equivalence
			Assumes model equivalences without stating them
	3: The proof is very incomplete but is on the right track
	      example: Not indicating how the tape contents and head
			positions are emulated
	2: The proof fails to prove the equivalence but is a good effort
	1: Not a proof, but shows understanding of how to show TM equivalence
	      example: Argues on similarities in structure of machines
	0: The proof does not demonstrate understanding of the material.


Grader's comments:

1) 1 out of 7 for only stating the idea that "stay = move-right +
move-left" only if it shows understanding of a TM equivalence proof.

2) 2 out of 7 for case 1 if it is accompanied with the (not-going-to
work) idea of adding some extra states to handle the "stay" move.

3) Many of the solutions were based on the same idea as it was in the
proof for showing the equivalence of 2-D TM and ordinary TM. In this
case, the mark will vary from 3 to 5 depending on the missing parts of
the
proof (the remaining 2 points is for emulating TM with MTMS).

Note: If you have used this approach and again you mentioned that
you can simulate a stay with a right-move + a left-move, then it
indicates that you have not understood the proof properly. In this
case, you will get at most 3 out of 5.