Inlämningsuppgifter
There will be three/four hand in problems during the course. The deadline is given after each problem, and is about two weeks after publication. (If there is any problems with the deadline, contact Stefan)
- Problem 1: Information Measure: HandIn1.pdf [Deadline: April 3]
- Problem 2: Huffman codes: HandIn2.pdf, LifeOnMars.txt [Deadline: April 16]
- Problem 3: Topic: HandIn3.pdf [Deadline: May 23]
Results
See below table for comment on Hand in 3f
Name |
P1 |
P2 |
P3 |
Passed |
Abdilameer, Omar | 1 | 1 | 1 | 1 |
Alkayyali, Khalil | 1 | 1 | 1 | 1 |
Åberg, Max | 1 | 1 | 1 | 1 |
Bhagavatula, Viswanth | 1 | 1 | 1 | 1 |
Blomqvist, Christopher | 1 | 1 | 1 | 1 |
Bortas, Micaela | 1 | 1 | 1 | 1 |
Desai, Saurabh | 1 | 1 | 1 | 1 |
Enohnyaket, John Ako | 1 | 1 | 1 | 1 |
Gunnarsson, Sara | 1 | 1 | 1 | 1 |
Hedlund, Frida | 1 | 1 | 1 | 1 |
John, Jeena Rachel | 1 | 1 | 1 | 1 |
Jonasson, Ylva | 1 | 1 | 1 | 1 |
Kariminejad, Vesal | 1 | 1 | 1 | 1 |
Karlsson, Linus | 1 | 1 | 1 | 1 |
Krishnamoorthy, Sujitha | 1 | 1 | 1 | 1 |
Lantz, Patrik | 1 | 1 | 1 | 1 |
Lukas, Sabine | 1 | 1 | 1 | 1 |
Moreno Trujillo, Hector Eric | 1 | 1 | 1 | 1 |
Mucha, Jakub | 1 | 1 | 1 | 1 |
Osmani, Ardiana | 1 | 1 | 1 | 1 |
Pettersson, David | 1 | 1 | 1 | 1 |
Rajeshwari Kabbinale | 1 | 1 | 1 | 1 |
Sabea, Sanaa | 1 | 1 | 1 | 1 |
Salman, Mohammad | 1 | 1 | 1 | 1 |
Sekyere, Augustine | 1 | 1 | 1 | 1 |
Sidabras, Tomas | 1 | 1 | 1 | 1 |
Söderberg, Alexander | 1 | 1 | 1 | 1 |
Sundberg, Marcus | 1 | 1 | 1 | 1 |
Teo, Wen Jie | 1 | 1 | 1 | 1 |
Vasileiadis, Athanasios | 1 | 1 | 1 | 1 |
Winquist, Fabian | 1 | 1 | 1 | 1 |
About Problem 3f. There are some confusions about the solution. It should be shown that the syndrom table in d can be used to detect double errors. That is, the decoder should simultaneously be able to correct all single errors and detect all double errors. This is possible if the syndrom for the double errors are not present in the table. Then, the received vector gives a syndrom s. If s=0 there are no errors, but if s!=0 and in the table it is a single error. If s is not in the table it is likely a double error. Since dmin=4 it is not possible to correct a double error but it should be possible to detect.
Then we need to show a double error syndrom is not in the list. One way is to list all syndroms for double errors. Another is to assume e2 is a double error with the same syndrom as the single error e1. Then
e2*H^T = e1*H^T
which gives
0 = e1*H^T+e2*H^T = (e1+e2)*H^T
This means (e1+e2) is a codeword. But since w(e1)=1 and w(e2)=2 we must have w(e1+e2) either 1 or 3. This contradicts the fact that dmin=4, and hence there can be no double error syndroms in the table.