← IGCSE Computer Science Paper 1

Working out the parity bit for a byte

Count the 1s in the 7 data bits, then set the leftmost parity bit to 0 or 1 so the total number of 1s in the whole 8-bit byte matches the agreed rule (even or odd).

Part of: IGCSE Computer Science 0478 — 2.2 Methods of Error Detection — Module 14, ACARA Digital Technologies Years 7–8 — Wired and Wireless Networks — Module 64

Understanding Working out the parity bit for a byte in IGCSE Computer Science Paper 1

What it does: Two 1s is already even, so a leftmost 0 is added to the 7 data bits 1000001, giving 01000001. Three 1s is odd, so a leftmost 1 makes four 1s in total. The byte is 11000011. 1110000 has three 1s, which is already odd. The parity bit must be 0, so the byte is 01110000. With odd parity the total number of 1s must be odd. Four 1s needs one more, so the parity bit is 1.

Common mistakes: The data bits 1011001 contain four 1s, not five. The count should be 4, so even parity needs a parity bit of 0. There are already three 1s, which is odd. Odd parity needs no extra 1, so the parity bit must be 0. A parity bit of 1 would make four 1s, which is even.

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This module's key terms are also in a printable crossword (Modules 155–164) and word search (Modules 142–164).

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Learning
Duo
Even parity. The 7 data bits are 1011001. What is the parity bit?
0, because there are four 1s, which is already even1, because there are four 1s and it must be made even
Matches
Even parity, data 1100100 -> 01100100
↓
The data has three 1s, which is odd. The parity bit must be 1 to make four 1s, so the byte should be 11100100.
Applying
Meanings
Term
count the 1s
Meaning
Count the 1s in the 7 data bits, then set the leftmost parity bit to 0 or 1 so the total number of 1s in the whole 8-bit byte matches the agreed rule (even or odd).
Practice
Data bits: 1000001
Testing
Gaps
Even parity bit:
Check Answer
Spots
2Data bits: 1011001
3Count of 1s: 5

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Quick Questions

1. Even parity. The 7 data bits are 1011001. What is the parity bit?
A. 0, because there are four 1s, which is already even ✓
B. 1, because there are four 1s and it must be made even
Count the 1s in the 7 data bits: 1011001 has four 1s. Four is already even, so the parity bit is 0 and the byte is 01011001.
2. Odd parity. The 7 data bits are 1011001. What is the parity bit?
A. 0, because four 1s is already odd
B. 1, to make the total number of 1s five (odd) ✓
The data bits contain four 1s (even). Odd parity needs an odd total, so the parity bit is 1 and the byte is 11011001.
3. Even parity. The data bits are 0110100. Which full 8-bit byte is correct?
A. 00110100
B. 10110100 ✓
0110100 has three 1s (odd). To make the total even, the parity bit is 1, written in the leftmost place: 10110100.
4. Odd parity. The data bits are 1110000. Which full 8-bit byte is correct?
A. 01110000 ✓
B. 11110000
1110000 has three 1s, which is already odd. The parity bit must be 0, so the byte is 01110000.
5. Where is the parity bit written in an 8-bit byte with 7 data bits?
A. In the leftmost place, in front of the 7 data bits ✓
B. In the rightmost place, after the 7 data bits
In this course the parity bit takes the leftmost place of the byte, and the 7 data bits follow it.

Match the Pairs

Even parity, data 1011001 -> 01011001 → The data has four 1s, which is already even, so the parity bit is 0 and the whole byte keeps an even number of 1s.
Odd parity, data 1011001 -> 11011001 → The data has four 1s, which is even, so the parity bit is 1 to make five 1s, an odd total.
ASCII 'A' is 1000001 in 7 bits. It has two 1s, so with even parity the parity bit is 0. → 01000001
ASCII 'C' is 1000011 in 7 bits. It has three 1s, so with even parity the parity bit is 1. → 11000011
Even parity, data 1100100 -> 01100100 → The data has three 1s, which is odd. The parity bit must be 1 to make four 1s, so the byte should be 11100100.
Odd parity, data 1111000 -> 11110001 → The parity bit was put at the right-hand end. It belongs in the leftmost place, so the byte should be 11111000.

Key Term

count the 1s
Count the 1s in the 7 data bits, then set the leftmost parity bit to 0 or 1 so the total number of 1s in the whole 8-bit byte matches the agreed rule (even or odd).

Fill in the Blanks

1. Data: 1011001 Count of 1s: 4 Even parity bit: 0
Hint: Four is already an even count, so decide which bit value keeps the total even.
2. Data: 1100110 (four 1s) Odd parity bit: 1
Hint: Type the bit value that changes an even count of 1s into an odd count.
3. Char A = 1000001 (ASCII, 7 bits) Even parity byte: 01000001
Hint: Type the leftmost parity bit for the ASCII character, which already has an even number of 1s.
4. Char C = 1000011 (three 1s) Odd parity byte: 01000011
Hint: Type the leftmost parity bit when the 7 data bits already have an odd number of 1s and the rule is odd parity.
5. Data: 0101010 (three 1s) Bit for even parity: 1 Total 1s in byte: 4
Hint: Type the total number of 1s in the full byte after the parity bit is added.

Spot the Error

Question 1
A Rule: even parity
B Data bits: 1011001
C Count of 1s: 5
D Parity bit: 0
E Byte: 01011001
Line C — The data bits 1011001 contain four 1s, not five. The count should be 4, so even parity needs a parity bit of 0.
Question 2
A Rule: odd parity
B Data bits: 0110100
C Count of 1s: 3
D Parity bit: 1
E Byte: 00110100
Line D — There are already three 1s, which is odd. Odd parity needs no extra 1, so the parity bit must be 0. A parity bit of 1 would make four 1s, which is even.
Question 3
A Rule: odd parity
B Data bits: 1110000
C Count of 1s: 3
D Parity bit: 1
E Byte: 11110000
Line A — The parity bit of 1 makes the total four 1s, which is even. So the rule must be even parity. With odd parity, the parity bit would have been 0.
Question 4
A Character: A (ASCII 65)
B 7 data bits: 1000011
C Count of 1s: 2
D Even parity bit: 0
E Byte: 01000001
Line B — ASCII 65 is 1000001 in 7 bits, which has two 1s. The broken line shows 1000011, which is ASCII 67 (the letter C).
Question 5
A Character: C (ASCII 68)
B 7 data bits: 1000011
C Count of 1s: 3
D Even parity bit: 1
E Byte: 11000011
Line A — The data bits 1000011 are ASCII 67, which is the letter C. ASCII 68 is D (1000100), so the code number on line 1 does not match the rest of the program.

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