How CSU 370 was graded (Clinger's section)

Assignments 3 and 5 were paired, and assignments 4 and 6 were paired. This means the higher of your two grades for assignments 3 and 5 was used as your grade for each of those two assignments, and the higher of your two grades for assignments 4 and 6 was used as your grade for each of those two assignments.

If your score on the final exam's 21-point programming problem was higher (after the pairing explained above) than your score on some homework assignment, and your score on that assignment was less than perfect, then your score on the final exam's programming problem was used in place of your score on one of the homework assignments (up to the number of points possible for that assignment, usually 20).

The midterm exam was not curved. The final exam was curved by adding 6 points to your uncurved score.

Details of the calculation are written in Scheme.

The final exam's programming problem was interesting enough to deserve some post-exam discussion.

• Here is a corrected statement of the problem.
• Partial credit for this 21-point problem was awarded as follows:
  • 1 point if the body of iterator() returns an instance of a class that implements Iterator
  • 2 points if a class claims to implement Iterator<BLT>, but only 1 point if it claims only to implement Iterator
  • 1 point if that class defines a hasNext() method with the correct type and returns a value of the correct type
  • 1 point if that class defines a Next() method with the correct type and returns a value of the correct type
  • 1 point if that class defines a remove() method with the correct type
  • 1 point if the instance variables of that class are initialized when the iterator() method creates an instance of that class
  • 1 point if the next() method throws a NoSuchElementException if and only if hasNext() is false
  • 1 point if the remove() method throws some kind of exception or actually removes something
  • 6 points if the instance variables of the class that implements Iterator would be adequate to generate all subtrees if their declared (or reasonably obvious) invariants were to hold; less than half the class got this right
  • 2 points if the hasNext() method interprets the instance variables correctly and returns the correct result; this is not possible unless the instance variables are adequate
  • 2 points if the next() method updates the instance variables correctly; this is not possible unless the instance variables are adequate
  • 2 points if the next() method returns the correct results; this is not possible unless the instance variables are adequate
• There are infinitely many correct solutions to this problem. The instructor provides these two examples:
  • solution 0
  • solution 1
• There are infinitely many incorrect non-solutions. The most common mistake was to try to use an instance variable of type BLT to maintain the state of the iterator. (An instance variable of type BLT can be adequate, but only when combined with an invariant that uses the BLT to represent a stack of BLTs or something similar; none of the students who used an instance variable of type BLT were able to think of an adequate invariant. The grading of this problem was made simpler by the Patterson-Hewitt theorem, which says that no finite amount of auxiliary state will help; you have to use recursion somewhere, or else simulate recursion with a stack or similarly unbounded data structure.)