# Q2A · CRC polynomial division

Subject: Computer Networks

Exam: TEE

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

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

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

### Actual PYQ · Q2A. Consider a message represented by the polynomial M(x)=x5+x4+x. Consider a generating polynomial G(x)=x3+x2+1. Generate a 3-bit CRC and show the transmitted frame. How is error detected by CRC? [10 marks]

#### Answer

## Given

Message=110010; generator=1101; generator degree r=3. This is textbook systematic CRC with zero initial remainder and no reflection/final XOR.

## To find

Exactly r CRC bits, the transmitted frame and receiver syndromes.

## Formula and why



$$
R=(M\,2^r)\bmod_2 G,\qquad T=(M\,2^r)\oplus R
$$



Modulo-2 division uses aligned XOR, not ordinary decimal division.

## Step-by-step solution

#### Step 1. Preserve all polynomial coefficients

Message bits=110010; generator bits=1101. Leading exponent is 3, so CRC length is 3.

#### Step 2. Append zeros

110010 followed by 3 zeros = 110010000.

#### Step 3. Perform each nonzero XOR subtraction

| Before | Aligned generator | After XOR |
| --- | --- | --- |
| 110010000 | 110100000 | 000110000 |
| 000110000 | 000110100 | 000000100 |

Remainder padded to 3 bits = **100**. Leading zero columns need no subtraction.

#### Step 4. Construct the transmitted frame

Message + CRC = 110010 + 100 = **110010100**.

#### Step 5. Check the full frame

| Before | Aligned generator | After XOR |
| --- | --- | --- |
| 110010100 | 110100000 | 000110100 |
| 000110100 | 000110100 | 000000000 |

Receiver remainder = 000; the uncorrupted frame passes.

#### Step 6. Demonstrate detection

Flip the last bit: 110010101. Dividing by 1101 leaves 001, which is nonzero, so this error is detected. A zero syndrome would mean no detected error, not proof of no corruption.

**Final answer:** CRC=100; transmitted frame=110010100; uncorrupted receiver remainder=000.

#### What this result means

Some software CRCs use different initial/reflection/final conventions; this answer follows the polynomial-division question’s stated textbook model.

Scheme correction: the remainder 100 is correct, but one scheme line prints 110010111. The correctly appended frame is 110010100; the receiver division shown above verifies it.

Source: Re-exam · 4 February 2026, original question paper page 1. Original marks retained; solution independently authored.