CS Atlas GATE 2027
Digital Logic course
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Digital Logic/Phase 1 · Signed Numbers & Binary Arithmetic

Lesson 1.2.1 · Item 10 of 57

Binary codes: BCD, excess-3, Gray code and parity

Read at your own pace. Try the questions before opening the answers.

On this page

0 · Recall Warm-up

Answer from memory first. Answers are in the Answer Vault at the bottom.

1 · The Hook

🎣

A calculator shows decimal digits, so it often stores each digit separately in 4 bits (BCD). A rotating knob reports its angle with a code where neighbouring positions differ in only one bit (Gray code), so a half-read position is never wildly wrong. And a single extra bit (parity) can catch a flipped bit on a noisy wire. GATE asks conversions in all three.

2 · Predict First

🤔

Gray code for 0 to 3 is 00, 01, 11, 10. What do you notice about each step, including the step from 3 back to 0?

Make a guess in your head. The answer opens Section 3.

3 · Core Concept

💡

Predict First — the answer

Each step changes exactly one bit, even 10 → 00.

🗣️

In plain words

A code is an agreed way of writing information in bits. BCD writes each decimal digit in its own 4 bits. Excess-3 is BCD plus 3. Gray code orders patterns so neighbours differ in one bit. A parity bit makes the number of 1s even (or odd) so a single flipped bit shows up.

1. BCD (binary-coded decimal, 8421 code)

Each decimal digit → 4 bits: 59 → 0101 1001. (Plain binary would be 111011.)

  • Only 10 of the 16 patterns are used; 1010 to 1111 are invalid.
  • BCD addition: add each digit in binary; if a digit sum is more than 9 (or produces a carry), add 6 (0110) to correct it. 8 + 7: 1000 + 0111 = 1111 (> 9) → + 0110 = 1 0101 → BCD 0001 0101 (15).
  • BCD needs more bits than binary: 2026 needs 16 bits in BCD (0010 0000 0010 0110) but 11 bits in binary.

2. Excess-3

Each digit + 3, then 4-bit binary: 59 → 8, 12 → 1000 1100. It is self-complementing: inverting the bits of a digit's code gives the code of 9 − digit (useful for subtraction).

3. Gray code

Consecutive numbers differ in exactly one bit.

  • Binary → Gray: keep the MSB; each next Gray bit = XOR of the two neighbouring binary bits. In one formula: g = b ⊕ (b shifted right by 1). 5 = 101 → 1, 1⊕0, 0⊕1 → 111.
  • Gray → binary: keep the MSB; each next binary bit = previous binary bit ⊕ current Gray bit. 1011 → 1, 1⊕0 = 1, 1⊕1 = 0, 0⊕1 = 1 → 1101.
Decimal01234567
Binary000001010011100101110111
Gray000001011010110111101100

(Gray code is also how K-maps in Phase 4 order their rows and columns.)

4. Parity

  • Even parity: the parity bit makes the total number of 1s even. 1011001 has four 1s → parity bit 0.
  • Odd parity: makes the total odd.
  • A single parity bit detects any odd number of flipped bits but cannot locate or correct them, and misses two flips.

4 · Visual

Binary -> Gray for 1101 (13)

binary :  1   1   0   1
           \ / \ / \ /
XOR    :    0   1   1
Gray   :  1   0   1   1      (MSB copied, then XOR of neighbours)

5 · Worked Examples

Example 1 Easy

Write 59 in BCD and in excess-3.

Solution

BCD 0101 1001; excess-3 1000 1100.

Example 2 Medium

Convert the Gray code 1011 to binary.

Solution

1, 1⊕0 = 1, 1⊕1 = 0, 0⊕1 = 1 → 1101 (13).

Example 3 GATE

In BCD addition, 4 + 5 and 8 + 7 are added digit-wise. Which needs the +6 correction?

Solution

4 + 5 = 9 (1001) is a valid digit: no correction. 8 + 7 = 15 (1111 > 9): add 0110.

6 · GATE Traps

⚠️

  • BCD is digit by digit; don't convert the whole number to binary.
  • Correction +6 is needed when a digit sum exceeds 9 or produces a carry (e.g. 9 + 9 = 1 0010).
  • Gray → binary uses the previous binary bit, not the previous Gray bit.
  • Parity detects odd numbers of errors only.

7 · Memory Hook

🧠

"BCD: 4 bits per digit, add 6 past 9. Gray: one bit changes, g = b ⊕ (b >> 1). Parity: count the 1s."

8 · Connect the Dots

Phase 1 is complete: integers, signed and coded. Real numbers like 3.75 or 0.1 need more: Phase 2 covers fixed-point fractions and the IEEE 754 floating-point format.

9 · Self-Test

GATE format. Try every question before opening the vault.

Q1MCQ1 markBCD

The BCD code for 2026 is:

  • (A) 0010 0000 0010 0011
  • (B) 0110 0010 0000 0010
  • (C) 0111 1110 1010
  • (D) 0010 0000 0010 0110

Q2NAT1 markInvalid BCD

How many of the 16 possible 4-bit patterns are not valid BCD digits?

Numerical answer: ________

Q3MCQ1 markGray code

The 4-bit Gray code for 13 (binary 1101) is:

  • (A) 1101
  • (B) 1011
  • (C) 1110
  • (D) 0111

Q4NAT2 marksGray to binary

The Gray code 1110 is converted to binary. What decimal value results?

Numerical answer: ________

Q5NAT1 markParity

An even-parity bit is added to the 7-bit data 1011001. What is the parity bit?

Numerical answer: ________

10 · Rule Card

  • BCD: 4 bits per decimal digit; 1010–1111 invalid; digit sum > 9 or carry → add 0110.
  • Excess-3: digit + 3; self-complementing.
  • Gray: g = b ⊕ (b >> 1); back: bᵢ = bᵢ₊₁ ⊕ gᵢ (running XOR from the MSB). Neighbours differ in one bit.
  • Even parity makes the count of 1s even; detects odd numbers of flipped bits.

11 · Concept Ledger

Unlocked in this lesson. From here on, these terms may be used.

binary-coded decimalinvalid BCD patternsBCD addition and the +6 correctionexcess-3 codeself-complementing codeGray codebinary to GrayGray to binaryeven and odd parityerror detection

12 · Answer Vault

Answer Vault

Try the questions first. Open when you are ready to check.

Recall Warm-up

  • Subtract the multiplicand.
  • Same-sign operands, opposite-sign result.
  • D.

Self-Test

Q1 Answer: (D)1 mark

One 4-bit group per decimal digit.

  • (A) is wrong: Last digit wrong.
  • (B) is wrong: Reversed.
  • (C) is wrong: That is binary 2026.

Re-read: Lesson 1.2.1 — Binary codes: BCD, excess-3, Gray code and parity

Q2 Answer: 61 mark

1010 to 1111.

Re-read: Lesson 1.2.1 — Binary codes: BCD, excess-3, Gray code and parity

Q3 Answer: (B)1 mark

1, 1⊕1 = 0, 1⊕0 = 1, 0⊕1 = 1.

  • (A) is wrong: Plain binary.
  • (C) is wrong: Wrong.
  • (D) is wrong: Wrong.

Re-read: Lesson 1.2.1 — Binary codes: BCD, excess-3, Gray code and parity

Q4 Answer: 112 marks

1, 1⊕1 = 0, 0⊕1 = 1, 1⊕0 = 1 → 1011.

Re-read: Lesson 1.2.1 — Binary codes: BCD, excess-3, Gray code and parity

Q5 Answer: 01 mark

Four 1s already.

Re-read: Lesson 1.2.1 — Binary codes: BCD, excess-3, Gray code and parity

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