Related Concepts: 01 BCD & Weighted Digital Codes | 03 Gray Code & Code Conversions | 04 Error Control, Parity Generators & Checkers

2.02 Excess-3 Code & Self-Complementing Logic

What is a Self-Complementing Code?

A 4-bit binary code is defined as Self-Complementing if the 1’s complement {bitwise inversion: $0\to 1,1\to 0$} of any decimal digit’s code word yields the exact code word of its 9’s decimal complement ($9-N$).

$\text{Code}(9-N)=\overline{\text{Code}(N)}$

graph LR
    subgraph Self-Complementing Symmetry
        D2[Decimal Digit 2: 0101] -->|Bitwise Inversion| InvD2[1's Comp: 1010]
        InvD2 -->|Equals| D7[Decimal Digit 7 = 9 - 2: 1010]
    end

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1. What is Excess-3 Code?

[concept] Concept: Excess-3 Definition Excess-3 (XS-3) is a non-weighted binary code. Each decimal digit $N$ (0 through 9) is represented by the 4-bit binary value of $N+3$.

• Decimal: $4_{10}$
• Excess-3 representation: $4+3=7_{10}=0111_{\text{XS-3}}$

Why do we use Excess-3?

1. Simplified Subtraction (Self-Complementing): The primary advantage of Excess-3 is that the 1’s complement of an Excess-3 digit is the Excess-3 representation of its 9’s complement (e.g. 1’s complement of 0100 [1] is 1011 [8]). This makes subtraction circuits in early decimal calculators incredibly simple to design.
2. Transmission Reliability: Unlike standard BCD where $0$ is represented by 0000 (which is indistinguishable from a line disconnection or fault), the smallest valid Excess-3 code word is 0011 (decimal 0). This prevents signal transmission line faults from being misread as valid data.

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2. Excess-3 Addition Mechanics

Because Excess-3 values always have an extra offset of $3$ built-in, adding two Excess-3 digits yields: $\text{Raw Sum}=(N_1+3)+(N_2+3)=(N_1+N_2)+6$

To restore the correct Excess-3 bias of $+3$:

• Case A: A carry is generated (Sum $>9$) $⟹$ Add 0011 ($+3$) to correct the sum nibble.
• Case B: No carry is generated (Sum $\le 9$) $⟹$ Subtract 0011 ($-3$) (or add 1101) to correct the sum nibble.

Step-by-Step Addition Examples:

Example 1: Excess-3 Addition with Sum $\le 9$ (No Carry)

Add $2$ and $3$ in Excess-3:

• $2_{\text{XS-3}}=0101$
• $3_{\text{XS-3}}=0110$

0101 & (2 + 3 = 5) \\ +0110 & (3 + 3 = 6) \\ \hline 1011 & (\text{Raw Binary Sum} = 11_{10} \to \text{No Carry}) \\ -0011 & (\text{No Carry} \implies \text{Subtract } 3) \\ \hline \mathbf{1000} & (\text{Corrected Sum} = 8_{10} \implies 5_{\text{decimal}} \text{ in XS-3}) \end{array}$$ **Final Answer:** **$1000_{\text{XS-3}}$** (corresponds to $5_{10}$).

Example 2: Excess-3 Addition with Sum $>9$ (Carry Generated)

Add $6$ and $5$ in Excess-3:

• $6_{\text{XS-3}}=1001$
• $5_{\text{XS-3}}=1000$

1001 & (6 + 3 = 9) \\ +1000 & (5 + 3 = 8) \\ \hline \mathbf{1} \ 0001 & (\text{Raw Binary Sum} \to \text{Carry generated!}) \\ +0011 & (\text{Carry generated} \implies \text{Add } 3 \text{ to sum nibble}) \\ \hline \mathbf{1} \ 0100 & (\text{Carry is pushed to next nibble}) \end{array}$$ Grouped into 4-bit nibbles: $\mathbf{0100\ 0100_{\text{XS-3}}}$ (corresponds to $11_{10}$ in decimal).

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3. Mathematical Proof: Excess-3 is a Self-Complementing Code

Major Exam Proof (2021, 2023, 2025 - 8 to 10 Marks)

Question: Prove mathematically that Excess-3 (XS-3) code is a self-complementing code.

Step-by-Step Proof:

1. Definition of Excess-3 Code: The Excess-3 code word for any decimal digit $N$ ($0\le N\le 9$) is constructed by adding $3_{10}=0011_2$ to its standard 4-bit binary equivalent: $\text{XS-3}(N)=N+3$

2. Evaluating the 9’s Complement of $N$ ($9-N$): The Excess-3 code word for the 9’s complement of $N$ is: $\text{XS-3}(9-N)=(9-N)+3=12-N$

3. Evaluating the 1’s Complement of $\text{XS-3}(N)$: Taking the bitwise 1’s complement of a 4-bit number is mathematically equivalent to subtracting it from $15_{10}$ ($1111_2$): $\overline{\text{XS-3}(N)}=15-\text{XS-3}(N)$ Substituting $\text{XS-3}(N)=N+3$: $\overline{\text{XS-3}(N)}=15-(N+3)=15-N-3=12-N$

4. Comparing Results: $\text{XS-3}(9-N)=12-N$ $\overline{\text{XS-3}(N)}=12-N$ $\therefore \text{XS-3}(9-\mathbf{N})=\overline{\text{XS-3}(\mathbf{N})}■$

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4. Complete 9’s Complement Symmetry Table

Decimal Digit ($N$)	BCD (8421)	Excess-3 Code ($\text{BCD}+3$)	1’s Comp of XS-3	9’s Comp Digit ($9-N$)	XS-3 Code of ($9-N$)
0	0000	0011	1100	9	1100
1	0001	0100	1011	8	1011
2	0010	0101	1010	7	1010
3	0011	0110	1001	6	1001
4	0100	0111	1000	5	1000
5	0101	1000	0111	4	0111
6	0110	1001	0110	3	0110
7	0111	1010	0101	2	0101
8	1000	1011	0100	1	0100
9	1001	1100	0011	0	0011

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5. Design: BCD to Excess-3 Code Converter

Major Exam Design Problem (2018 - 10 Marks)

Question: Design a combinational logic circuit to convert a 4-bit BCD input ($A,B,C,D$) to a 4-bit Excess-3 code output ($W,X,Y,Z$).

Boolean Function Derivation (K-Map Simplification):

• Inputs: $A,B,C,D$ (Don’t care conditions $X=d(10,11,12,13,14,15)$).
• Outputs: $W,X,Y,Z$.

1. Output $Z$ (LSB): $Z={\mathbf{D}}^{\prime }$
2. Output $Y$: $Y=CD+C^{\prime }{D}^{\prime }=(\mathbf{C}\oplus \mathbf{D}{)}^{\prime }$
3. Output $X$: $X=B^{\prime }C+B^{\prime }D+B{C}^{\prime }{D}^{\prime }=\mathbf{B}\oplus (\mathbf{C}+\mathbf{D})$
4. Output $W$ (MSB): $W=A+BC+BD=\mathbf{A}+\mathbf{B}(\mathbf{C}+\mathbf{D})$

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Past Year Questions (PYQs)

• [PYQ 2017, 2021, 2023, 2025]: Define self-complementary code. (04 Marks)
• [PYQ 2021, 2023, 2025]: Prove mathematically that Excess-3 code is a self-complementary code. (08-10 Marks)
• [PYQ 2018]: Design a BCD to Excess-3 code converter circuit. (10 Marks)