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Free A-Level Computer Science guide
For 9618 in 2027–2029, section 6.2 names validation and verification, including byte/block parity, checksum and check digit. State the scheme before calculating a check value: different legitimate schemes give different answers.
Content owner: Michael Print · Written for A-Level learners · Checked against official specifications
Validation checks whether data satisfies allowed rules; verification checks correspondence with its source or a transmitted representation. Neither proves that the underlying real-world claim is true.
Parity, checksums and check digits add redundancy so selected errors can be detected. Detection does not automatically correct the data. A protocol can request/retry transmission after detection, but no simple check guarantees detection of every alteration.
Cambridge International 9618 · 9618 (2027–2029): 6.2; supporting transmission context from 2.1/14.1; ARQ is an explicitly labelled extension
Useful foundations
By the end, you should be able to
This guide uses 2027–2029 syllabus version 2; the checked 2026 version 2 has the same section 6.2 content. Validation methods include range, format, length, presence, existence, limit and check digit.
A range check tests a permitted interval; presence rejects a missing required field. Format checks a pattern; length checks character count; existence looks for an allowed stored value; limit imposes an upper or lower bound.
These checks establish acceptability under rules, rather than factual truth. A plausible but incorrect age can pass its range check.
Entry verification includes visual comparison and double entry. Transfer verification includes byte/block parity and checksum. Double entry detects inconsistent typing, but misses the same wrong value entered twice.
Detection is separate from correction or retransmission. A passed check establishes limited consistency, rather than guaranteeing the data is true or detecting every possible alteration.
Add redundancy
The sender calculates check information from the data using a declared scheme.
Transfer both
The receiver obtains the data and its check information; either may be corrupted.
Check consistency
A failed check reveals an inconsistency. A passed check can still miss errors outside the scheme’s detection ability.
Choose a response
Reject, request another copy or correct only when the error model supports it. Detection alone does not repair the data.
Even parity makes the total number of ones in the checked unit even; odd parity makes it odd. State whether displayed bits already include the parity bit.
Here eight data bits 10110010 contain four ones. Appending even-parity bit 0 gives nine transmitted bits, 101100100. One flipped bit changes the parity and is detected.
Two flips can escape: changing the first two data bits gives 01110010, still four ones. Byte parity detects odd numbers of flips within its unit, but cannot locate the bit or reliably detect even numbers. It is not cryptographic proof.
Two-dimensional even parity checks each row and column. The worked block has four data columns, rather than eight, to keep every bit visible. One corrupted data bit makes its row and column fail.
Under the single-error assumption, their intersection locates the bit. With multiple errors, that conclusion may be wrong: flipping four rectangle corners changes two bits in each affected row/column, leaving all parity checks satisfied.
Additional redundancy provides more information, but the error model determines the conclusions. A failed row/column is not a licence to correct arbitrary multi-bit corruption.
A checksum summarises a sequence through a specified calculation. This teaching scheme sums bytes modulo 256 and chooses a checksum making the final sum zero modulo 256. Actual protocols may use other formulas.
For bytes 100,150,25, sum = 275. Checksum = (−275) MOD 256 = 237; the receiver checks (100+150+25+237) MOD 256 = 0. Do not assume this formula when a task supplies another.
Collisions are possible. Increasing one byte by 1 and decreasing another by 1 preserves this sum, so the check can pass changed data.
Our four-digit scheme applies weights 3,1,3,1 from left to right. Add a digit making the weighted sum divisible by 10. Identifier 4821 gives 27, requiring check digit 3.
Passing establishes consistency with the scheme. It does not prove the item exists or belongs to the user. A check digit and a checksum also do not provide encryption or protection against deliberate tampering.
Automatic Repeat reQuest (ARQ) provides useful context after transmission errors. It is an extension here: the checked 9618 teaching tables do not name ARQ as a requirement.
In stop-and-wait, send a numbered frame and start a timer. The receiver checks it and acknowledges valid reception; the sender retries when no suitable acknowledgement arrives before timeout.
A lost acknowledgement can trigger a duplicate even after correct delivery. Sequence numbers let the receiver recognise it, acknowledge again and avoid delivering the same payload twice.
Some protocols use negative acknowledgements; others use timeout. Start the timer after sending, not after an acknowledgement. Retries address selected failures; permanent disruption needs a failure policy.
Worked example
| Row | Column 1 | Column 2 | Column 3 | Column 4 | Row parity |
|---|---|---|---|---|---|
| 1 | 1 | 0 | 1 | 1 | 1 |
| 2 | 0 | 1 | 1 | 0 | 0 |
| 3 | 1 | 1 | 0 | 0 | 0 |
| Column parity | 0 | 0 | 0 | 1 | 1 |
Worked example
def checksum(values):
if any(type(value) is not int or not 0 <= value <= 255 for value in values):
raise ValueError("Expected byte values")
return (-sum(values)) % 256
def check_digit(identifier):
if len(identifier) != 4 or any(c not in "0123456789" for c in identifier):
raise ValueError("Expected four ASCII digits")
weighted = sum(int(c) * w for c, w in zip(identifier, [3, 1, 3, 1]))
return (-weighted) % 10
data = [100, 150, 25]
check = checksum(data)
print(check)
print((sum(data) + check) % 256)
print(check_digit("4821"))5 original questions total 17 marks. Attempt each before opening the independently written indicative marking guidance.
Distinguish validation and verification using range checking and double entry.
Indicative answer
Eight data bits 11001001 use even parity appended at the end. Give the parity bit, write the transmitted nine-bit unit and explain why two flipped bits can escape detection.
Indicative answer
In this block, row 3 column 4 flips from 0 to 1. Which checks fail and what assumption allows correction?
The last row and column contain even-parity bits. All other cells are data bits; the indicated flip affects one of those data bits.
| Row | Column 1 | Column 2 | Column 3 | Column 4 | Row parity |
|---|---|---|---|---|---|
| 1 | 1 | 0 | 1 | 1 | 1 |
| 2 | 0 | 1 | 1 | 0 | 0 |
| 3 | 1 | 1 | 0 | 0 | 0 |
| Column parity | 0 | 0 | 0 | 1 | 1 |
Indicative answer
Use these schemes to calculate a checksum for bytes 10,20,30 and a check digit for identifier 1234.
Checksum: add the data bytes and choose a checksum from 0–255 so that (sum of bytes + checksum) MOD 256 = 0.
Check digit: multiply the four identifier digits by weights 3,1,3,1 from left to right. Append one digit from 0–9 so that the weighted sum plus that digit is divisible by 10.
Indicative answer
Extension: a valid frame arrives but its acknowledgement is lost. Explain the sender/receiver actions needed to avoid duplicate delivery.
Indicative answer
This guide addresses Cambridge International 9618 9618 (2027–2029): 6.2; supporting transmission context from 2.1/14.1; ARQ is an explicitly labelled extension. Check your examination year and the complete specification for the assessment scope.
These are independently written explanations and practice questions. CompSciTutoring.co.uk is not affiliated with or endorsed by an examination board. The marking guidance is indicative; always check the syllabus for your examination year.
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