Computer Networks · TEE · practice

Single-error correction · example 1

Unit 3 · Errors, block coding and CRC · optional practice

Optional extension · no matching calculation in the two supplied papers

Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.

Practice question. Encode data 1011 using even-parity Hamming coding. Number positions left-to-right starting at 1 and insert data in that order. If bit position 5 flips, show how the receiver locates and corrects it. [7 practice marks]

Answer

Given

Data=1011; even parity; error position=5; only one bit flips.

To find

Redundant bits, complete codeword and syndrome correction.

Formula and why

2^r\ge m+r+1

\mathrm{syndrome}=s_1+2s_2+4s_4+\cdots

Parity positions correspond to binary index bits, so a single flipped position changes precisely its index checks.

Step-by-step solution

Step 1. Choose redundancy

m=4; r=3: 2³=8 ≥ 4+3+1=8; one fewer parity bit fails.

Step 2. Place data and compute checks

Check	Covered positions (including check)	Parity bit
p1	1, 3, 5, 7	0
p2	2, 3, 6, 7	1
p4	4, 5, 6, 7	0

Position	1	2	3	4	5	6	7
Encoded bit	0	1	1	0	0	1	1

Step 3. Introduce and check the error

Received = 0110111

Syndrome check	Received-bit XOR	Result
s1	0 XOR 1 XOR 1 XOR 1	1
s2	1 XOR 1 XOR 1 XOR 1	0
s4	0 XOR 1 XOR 1 XOR 1	1

Step 4. Locate and flip

Syndrome integer=5. Flip position 5; recovered codeword=0110011; extracted data=1011.

Final answer: Codeword 0110011; syndrome 5; corrected data 1011.