CSC2100B Data Structures Tutorials

Lecture I Lecture II Tutorial I Tutorial II
Time M7-8, 2:30 pm - 4:15 pm T2, 9:30 am - 10:15 am T9, 4:30 pm - 5:15 pm H9, 4:30 pm - 5:15 pm
Venue SC L1 ERB LT ERB 404 ERB 404

See announcement at bottom of page (Last updated on 28 Jan 2010)

Week Date Notes Remarks TA Resources
1 12/1 - 14/1 N/A No Tutorial this week N/A N/A
2 19/1 - 21/1 Introduction to C TBD Tom Chao Zhou TBD
3 26/1 - 28/1 Online Judge Server TBD Tom Chao Zhou TBD
4 2/2 -4/2 List, Stack and Queue TBD Tom Chao Zhou TBD
5 9/2 - 11/2 Bianry and AVL Tree TBD Jianye Hao TBD
6 16/2 - 18/2 N/A TBD TBD TBD
7 23/2 - 2/25 Binary Search Trees Implementation TBD Tom Chao Zhou TBD
8 2/3 -4/3 Hashing TBD Tom Chao Zhou TBD
9 9/3 - 11/3 Heap TBD Jianye Hao TBD
10 16/3 - 18/3 Sorting TBD Xin Xin TBD
11 23/3 - 25/3 Code for Homework TBD Xin TBD
12 30/3 - 1/4 TBD TBD TBD TBD
13 6/4 - 8/4 TBD TBD TBD TBD
14 13/4 - 15/4 TBD TBD TBD TBD
15 20/4 - 22/4 Graph TBD Jianye Hao TBD

News

  1. 5 April 2010. The programming mid-term starts from 9:00 am to 1:00 pm on Saturday (10, April), and the venue is in Rm 924, Ho Sin-Hang Engineering Building.
  2. 2 Feb 2010 Document for NON-CSE students to login to Unix using SSH software. Non-CSE SSH
  3. 2 Feb 2010 Written Assignment 1:. Some students asked what is needed to do for problem 1.1(1,3,6) in you 1st written assignment. It is needed to write out the result for the sequence, for example, you could refer to page 42 of lecture notes “http://www.cse.cuhk.edu.hk/~king/csc2100b/PDF/CSC2100BK-Introduction.pdf”.
  4. 29 January 2010. The question id for the 1st programming asgn are 116 and 118.
  5. 29 January 2010. The login id for the Online Judgement System has been sent to you cuhk email(sxxxxxxx@mailserv.cuhk.edu.hk).
  6. 28 January 2010. About the award scheme in the programming asgn: the award scheme is introduced in the TA courses this week, however, it should be based on day, not based on week. This scheme is to be reconfirmed.The score is for reference only, and does not reflect your final grade. Here is an example:

Suppose @award = (0, 4, 3, 2, 1, 0, 0, 0, 0, 0);
the 0th entry is no use, the 1st entry is the number of award times in the 1st day.
pen = # of submissions -1-award[x]; x is the day when you successfully finish the program
if (pen<0){
pen = 0;
}
mark = 30 + (70-$pen*5);
So the rule to calculate the score is:
suppose the student submit the asgn in the 1st day for 10 times, the penalty is pen = 10 - 1 - 4 = 5, then the mark is mark = 30 + （70-5*5）= 75
suppose the student submit the asgn in the 2nd day for 10 times, the penalty is pen = 10 - 1 - 3 = 6, then the mark is mark = 30 + (70-6*5) = 70