Checking a received byte for a parity error

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Quick Questions

Shade in the A or B circle for each question.

1. Even parity is agreed. The receiver gets 10110010. What is the result of the parity check?

A 4 ones, which is even: no error detected

B 4 ones, which is even: error detected

2. Odd parity is agreed. The receiver gets 01101100. What is the result of the parity check?

A 4 ones, which is not odd: error detected

B 4 ones, which is fine: no error detected

3. Even parity is agreed. The receiver gets 11100000. Which statement is correct?

A No error: 3 ones means the check passes

B Error detected: 3 ones is an odd count

4. Which bits does the receiver count when doing a parity check on a byte?

A All 8 bits, including the parity bit

B Only the 7 data bits, leaving out the parity bit

5. During transmission only the parity bit flips, from 0 to 1. The 7 data bits are unchanged. What does the receiver find?

A No error is detected, because the data bits are unchanged

B The count of 1s breaks the rule, so an error is detected

Match the Pairs

Draw a line connecting each item on the left to its match on the right.

Left column		Right column
110110010 (even parity)		AFour is an even count. Odd parity needs an odd count of 1s, so this byte fails the check and an error is detected.
211010100 (odd parity)		B11100000 (even parity)
3A byte arrives under even parity with an odd number of 1s, so the receiver reports an error		CFour 1s is an even count, which breaks the odd rule, so the check fails and an error is detected.
4The parity bit itself can be the bit that changed, and the check still fails		D01100110 -> 11100110 (even)
5Odd parity: 10110100 has four 1s, so it passes		EThe receiver must count the 1s in the whole byte, parity bit included. Ignoring the parity bit gives the wrong result.
6Count the 1s in the 7 data bits only and ignore the parity bit		FFour 1s is an even count, so the even parity check passes and no error is detected.

Definitions

parity check

Fill in the Blanks

Complete each line of code by filling in the missing parts.

received ← 10110010 ones ← 4 IF ones MOD 2 = _ THEN OUTPUT "No error"

received ← 11010110 ones ← 5 IF ones ___ 2 = 1 THEN OUTPUT "No error"

IF ones MOD 2 = 0 THEN OUTPUT "No error" ____ OUTPUT "Error detected"

received ← 10110011 ones ← 5 // parity bit included IF ones MOD 2 __ 0 THEN OUTPUT "Error"

received ← 11010100 ones ← 4 IF ones MOD 2 = 0 THEN ______ "Error"

Spot the Error

Shade in the letter of the line that contains an error.

Question 1

A rule ← "odd"

B ones ← 4

C IF ones MOD 2 = 0 THEN

D OUTPUT "No error detected"

E ENDIF

Question 2

A rule ← "odd"

B ones ← COUNT_ONES(byte)

C IF ones MOD 2 = 1 THEN

D OUTPUT "Error detected"

E ELSE

F OUTPUT "Error detected"

Question 3

A byte ← "11010011"

B ones ← 4

C IF ones MOD 2 = 1 THEN

D OUTPUT "Error detected"

E ENDIF

Question 4

A sent ← "10101010"

B received ← "10101111"

C ones ← 5

D IF ones MOD 2 = 1 THEN

E OUTPUT "Error detected"

Question 5

A rule ← "odd"

B ones ← COUNT_ZEROS(received)

C IF ones MOD 2 = 1 THEN

D OUTPUT "No error detected"

E ELSE

F OUTPUT "Error detected"

Answer Key

Checking a received byte for a parity error

Section A: Quick Questions

A — 4 ones, which is even: no error detected

A — 4 ones, which is not odd: error detected

B — Error detected: 3 ones is an odd count

A — All 8 bits, including the parity bit

B — The count of 1s breaks the rule, so an error is detected

Section B: Match the Pairs

10110010 (even parity) → Four 1s is an even count, so the even parity check passes and no error is detected.

11010100 (odd parity) → Four 1s is an even count, which breaks the odd rule, so the check fails and an error is detected.

A byte arrives under even parity with an odd number of 1s, so the receiver reports an error → 11100000 (even parity)

The parity bit itself can be the bit that changed, and the check still fails → 01100110 -> 11100110 (even)

Odd parity: 10110100 has four 1s, so it passes → Four is an even count. Odd parity needs an odd count of 1s, so this byte fails the check and an error is detected.

Count the 1s in the 7 data bits only and ignore the parity bit → The receiver must count the 1s in the whole byte, parity bit included. Ignoring the parity bit gives the wrong result.

Section C: Definitions

parity check — The receiver counts the 1s in the whole received byte, parity bit included. If the count breaks the agreed even or odd rule, an error has occurred and is detected.

Section D: Fill in the Blanks

received ← 10110010
ones ← 4
IF ones MOD 2 = 0 THEN OUTPUT "No error"

received ← 11010110
ones ← 5
IF ones MOD 2 = 1 THEN OUTPUT "No error"

IF ones MOD 2 = 0 THEN OUTPUT "No error"
ELSE OUTPUT "Error detected"

received ← 10110011
ones ← 5  // parity bit included
IF ones MOD 2 <> 0 THEN OUTPUT "Error"

received ← 11010100
ones ← 4
IF ones MOD 2 = 0 THEN OUTPUT "Error"

Section E: Spot the Error

A — The check IF ones MOD 2 = 0 passes when the count of 1s is even, so the agreed rule is even parity. Line 1 should be rule ← "even".

D — With odd parity, an odd count of 1s means the byte passes the check. Line 4 should output "No error detected".

B — The byte 11010011 contains five 1s, not four. Line 2 should be ones ← 5, so the count matches the byte.

B — Only the last bit (the parity bit) changed, from 0 to 1. The received byte should be 10101011, which has five 1s. The broken line changes a data bit as well and gives six 1s.

B — A parity check counts the 1s in the received byte, not the 0s. Line 2 should be ones ← COUNT_ONES(received).