# Q2A · Block coding and all pairwise distances

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/papers/tee-paper-final/q2a

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/papers/tee-paper-final--q2a/

TEE · Actual TEE paper · all questions · current syllabus 702AI0C023

### Actual PYQ · Q2A. (i) Explain the block coding method for error detection and correction. (2 marks) (ii) Two binary codewords are used: C1=1001101 and C2=1110001. Compute their Hamming distance. (2 marks) (iii) C3=1010101 is added. Compute its pairwise distances from the other codewords. (2 marks) (iv) Based on the minimum Hamming distance, determine how many errors the code can detect and correct; justify your answer. (4 marks) [10 marks]

#### Answer

**(i) Block coding.** Map each k-bit dataword to a distinct n-bit codeword by adding n−k redundant bits. Only the valid codeword subset represents data. A received invalid pattern detects an error; sufficiently separated codewords permit unique correction within the guaranteed radius. For a chosen code, the minimum Hamming distance measures that separation.

## Given

| Codeword | Bits |
| --- | --- |
| C1 | 1001101 |
| C2 | 1110001 |
| C3 | 1010101 |

## To find

All pair distances, dmin and the maximum guaranteed numbers of errors.

## Formula and why



$$
d(a,b)=\mathrm{weight}(a\oplus b)
$$





$$
s=d_{\min}-1,\quad t=\lfloor(d_{\min}-1)/2\rfloor
$$



XOR exposes differing positions. Use the closest codeword pair, because it is the weakest separation in the code.

## Step-by-step solution

#### Step 1. XOR every distinct pair

| Pair | First | Second | XOR | Distance |
| --- | --- | --- | --- | --- |
| C1 vs C2 | 1001101 | 1110001 | 0111100 | 4 |
| C1 vs C3 | 1001101 | 1010101 | 0011000 | 2 |
| C2 vs C3 | 1110001 | 1010101 | 0100100 | 2 |

#### Step 2. Take the minimum

dmin = min(4, 2, 2) = 2.

#### Step 3. Calculate the guarantees

Detect = 2−1 = 1; correct = floor((2−1)/2) = 0.

#### Step 4. Justify geometrically

2 bit changes can reach another valid codeword, so only 1 are guaranteed detectable. Correction regions require 2t < 2.

**Final answer:** dmin=2; detect up to 1 errors; correct up to 0 errors.

#### What this result means

A distance of one particular pair does not set the whole-code capability unless it is the minimum.

Source: Final · 5 December 2025, original question paper page 1. Original marks retained; solution independently authored.