# Q2(b)(i) · Block-code received words

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/papers/paper-2023-jan-ai-reexam/q2bi

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/papers/paper-2023-jan-ai-reexam--q2bi/

Exact historical variant. Current-syllabus questions solved; excluded topics and source limitations stated.

### Q2(b)(i). For k=2,n=3, the encoding table is 00→000, 01→011, 10→101, 11→110. The sender sends dataword 01 as codeword 011. Explain the receiver outcome if it receives (a)111 and (b)000. [5 marks]

### Answer

### Given

Sent data01, encoded011; valid codewords000,011,101,110.

### To find

Receiver outcome for111 and000.

### Formula and why

A receiver first checks membership in the valid-codeword set, then decodes a valid word.

This even-parity code detects any single-bit error but cannot identify a unique bit to correct.

### Solution

### Step 1. Received111

111 has odd parity and is not a valid word. Relative to011, one bit differs. The receiver detects an error and rejects/requests retransmission; it cannot uniquely infer the original data.

### Step 2. Received000

000 is a valid word and decodes to00. Relative to the actual sent011, two bits changed. Therefore this is an undetected error: the receiver accepts the wrong dataword00.