CSCI 2720			Data Structures			     Spring 2004

		     	       News about test 2


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

	Class average raw score was 69 out of 100 points possible.

	Class averages on questions 1-5, 6, 7, 8, 9, 10, 11, and 12 as raw score 
	  % of the possible were: 82%, 74%, 40%, 77%, 89%, 50%, 67%, and 46%,
	  respectively.

	Highest score in class was 90 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: calculate the raw score as % out of 90 to obtain the scaled
	  score for test 2.  So a raw score of 70 scales to 77.777...%, which
	  lies just below the middle of the B range.


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

1(a)  T
1(b)  F
1(c)  F
1(d)  T
2(a)  F  
2(b)  F
2(c)  T
3     F
4(a)  F
4(b)  F
4(c)  F
4(d)  T
5(a)  T
5(b)  F
5(c)  T

6     At the top level, an array 0..9 of column pointers, all null except at 0,
      2 and 9;  each of those points to the start of a singly linked list 
      containing the non-default values for the corresponding column, e.g.,
	0[*]-->[ 5,  0,*]-->[57,500,/]
	2[*]-->[11, 55,*]-->[22,101,*]-->[90,427,/]
	9[*]-->[11, 13,/]

7(a)  1 + 2 + 3 + ... + n = n*(n+1)/2

7(b)  { if ( i < j ) return 0.0;
	return T[ n*(i-1) + (j-1) - i*(i-1)/2 ]; }

8(a)  The pairings are ((((NT)S)(IO))(AE)); order doesn't matter, e.g. (TN)
      and (NT) are equally correct and indicate that T and N are sibling leaves.

8(b)  198 bits.


9     A#LIONS#LITTLE#LITTER (using # for the space symbol)

10       0 1 2 3 4 5 6 
        +-+-+-+-+-+-+-+
      A |1|0|3|2|0|6|0|
        +-+-+-+-+-+-+-+
      B |0|1|0|0|4|0|3|
        +-+-+-+-+-+-+-+
      C |1|2|3|4|5|6|7|
        +-+-+-+-+-+-+-+
      
11(a) First add LC(30)-->(20), then perform double rotation on the path
      LC(50)-->(6) and RC(6)-->(15).
      
11(b) First replace (70) by (80) and remove the node LC(90), where (80) was
      originally.  Next, rebalance at (50), which can be done by a single
      rotation along the edge LC(50)-->(6) or else by a double rotation on
      the path LC(50)-->(6) and RC(6)-->(15).
      
12(a) Three splits occur, increasing the height from 2 to 3.  The parts which
      changed are now (root_pointer)-->(35), RC(35)-->(70), LC(35)-->(18),
      RC(18)-->(25), LC(18)-->(10), RC(25)-->(30), LC(25)-->(20), RC(10)-->(15),
      and LC(10)-->(5).  Note that rotations are never used in a 2-3 tree 
      insertion.

12(b) First replace 70 by 80 in the root, leaving the leaf which contained 80
      to now be empty.  So the parent and sibling keys are combined into a 
      single node (90,95) with an empty parent node.  That is combined with the
      sibling node (45) and the parent key 80 to form a new 3-node (45,80) with
      LC(45,80)-->(40), MC(45,80)-->(50,60) and RC(45,80)-->(90,95).  The root
      has become a 2-node, so no more adjustment is needed.  In the final tree
      we have (root_pointer)-->(35), RC(35)-->(45,80), and LC(35)-->(10,18).
      The subtree at (10,18) is unchanged.