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Encoding and finite representations

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AI-assisted English content review; independent teacher approval pending.

Practice scope and review

Original, AI-authored English practice with worked guidance. Selected topics only; independent teacher review is pending. This is not a complete syllabus, official past-paper collection or predicted exam score. Check your own exam-board specification: these concepts are not certified against a particular board or exam year.

151 original questions across 21 selected topic areas. No account or payment required to practise.

Cambridge International AS & A Level Computer Science (9618), examination year 2027: AI-assisted conceptual alignment to official sources, not teacher or exam-board certification. Syllabus mapping and remaining coverage.

5 questions in this topic

Exact English revisions checked in an AI-assisted academic review; independent teacher approval is still pending.

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Original English exercises targeting selected 9618 2027-29 requirements. Conceptual alignment only; independent teacher approval, practical competence and complete teaching coverage are not established.

Subtopics in this topic (3)
  • Normalised binary floating point
  • BCD versus binary
  • Run-length compression and vector objects

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Question 1A floating-point format has an eight-bit two's-complement mantissa with binary point immediately after its sign bit and a four-bit two's-complement exponent. Value=mantissa*2^exponent. Decode mantissa00101100, exponent0100; normalise without changing its value. With exponent fixed at3, find the nearest representable value to5.6, its mantissa and signed error (stored minus true). Explain precision/range trade-offs if a fixed total bit budget gives more bits to the exponent.6 indicative marks
Construct, trace and explain6 indicative marksNo calculator

A floating-point format has an eight-bit two's-complement mantissa with binary point immediately after its sign bit and a four-bit two's-complement exponent. Value=mantissa*2^exponent. Decode mantissa00101100, exponent0100; normalise without changing its value. With exponent fixed at3, find the nearest representable value to5.6, its mantissa and signed error (stored minus true). Explain precision/range trade-offs if a fixed total bit budget gives more bits to the exponent.

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Worked guidance

  1. The original mantissa weights are0.25+0.0625+0.03125=.34375. Multiplying by16 gives5.5. A positive normalised fractional two's-complement mantissa begins01. Shifting mantissa left once doubles it to.6875 and reducing exponent4 to3 compensates: .6875*8=5.5.
  2. At exponent3, integer mantissa codes89 and90 produce89/128*8=5.5625 and90/128*8=5.625. Distances from5.6 are.0375 and.025, so code90 (01011010) is nearest. Error is5.625-5.6=+.025, not exactly zero.
  3. The finite bit budget cannot represent every real number exactly. Changing exponent allocation alters the range/precision balance; it does not eliminate rounding errors. No claim that every unnormalised bit pattern is an invalid value is needed.

Self-marking points

View marking points
  1. 00101100 means44/128=.34375, exponent0100 means4.
  2. Value is.34375*16=5.5.
  3. Normalise to mantissa01011000 and exponent0011, preserving5.5.
  4. At exponent3 the adjacent-value step is8/128=.0625.
  5. Nearest5.6 is5.625, mantissa01011010; error+.025.
  6. More exponent bits widen exponent range while fewer mantissa bits reduce available precision at a scale.

Examiner tip

Indicative distinct points, not an official Cambridge mark allocation. State bounds, assumptions and intermediate states; practical teacher review remains required.

Question 2Encode decimal59 in eight-bit packed BCD and unsigned binary, and give its hexadecimal integer form. Explain a context where preserving decimal digits is useful. Encode AAAABBCCCCC as (character,count) runs; assume one byte for each character and one for each count and no header. Calculate original/encoded sizes and saving. Finally justify choosing a vector drawing rather than a bitmap for a simple logo repeatedly resized, naming two stored object properties.6 indicative marks
Construct, trace and explain6 indicative marksNo calculator

Encode decimal59 in eight-bit packed BCD and unsigned binary, and give its hexadecimal integer form. Explain a context where preserving decimal digits is useful. Encode AAAABBCCCCC as (character,count) runs; assume one byte for each character and one for each count and no header. Calculate original/encoded sizes and saving. Finally justify choosing a vector drawing rather than a bitmap for a simple logo repeatedly resized, naming two stored object properties.

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Worked guidance

  1. Digit5 is0101 and digit9 is1001, giving packed BCD01011001. Place-value encoding of the number59 is32+16+8+2+1=00111011, hex3B. Interpreting the BCD byte as an ordinary unsigned integer would give89, not59.
  2. The three run records each cost2bytes, for6 total. The uncompressed eleven characters cost11bytes; saving=11-6=5bytes, about45.45% of original size. An alternating string could cost more under this scheme, so no universal compression ratio follows.
  3. For a simple geometric logo, a drawing list can store e.g. a rectangle's coordinates and fill colour. It can be rendered afresh at the chosen resolution; that does not mean the final display uses no pixels or that arbitrary photographs are naturally small vector drawings.

Self-marking points

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  1. BCD is0101 1001, treating each digit separately.
  2. Unsigned binary is00111011, hexadecimal3B; BCD byte is not the binary integer59.
  3. A decimal display/calculator can use digitwise BCD; the choice is contextual, not universal compression.
  4. Runs are(A,4),(B,2),(C,5), original11bytes and encoded6bytes, saving5.
  5. A vector logo stores objects with properties such as line endpoints and fill colour, and can be re-rendered at different sizes without fixed-pixel blockiness.
  6. Detailed photographic shading may favour bitmaps; RLE gains depend on repeated runs and this size comparison excludes headers.

Examiner tip

Indicative distinct points, not an official Cambridge mark allocation. State bounds, assumptions and intermediate states; practical teacher review remains required.

Question 3An 8-bit two's-complement register contains 11101000. Interpret it as signed decimal, then perform one arithmetic right shift (replicate the sign bit) and one logical right shift of the ORIGINAL pattern (insert zero). Give both new patterns and their signed values. Explain the different results.4 indicative marks
Analyse4 indicative marksNo calculator

An 8-bit two's-complement register contains 11101000. Interpret it as signed decimal, then perform one arithmetic right shift (replicate the sign bit) and one logical right shift of the ORIGINAL pattern (insert zero). Give both new patterns and their signed values. Explain the different results.

Cambridge 2027: Direct conceptual match

Checked232-256=-24,244-256=-12 and logical116; no claim that every negative odd shift equals truncation toward zero.

AI-assisted placement. Independent teacher approval pending.

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Worked guidance

  1. The original unsigned sum is 232; subtracting 256 for a leading sign bit gives -24.
  2. Replicating the sign bit gives 11110100: unsigned 244, signed -12. This even negative integer is divided by two exactly in this case.
  3. Zero filling instead gives 01110100, or +116. Logical shifting is a bit-pattern operation and must not be described as signed division for this negative input.

Self-marking points

View marking points
  1. Original value is -24.
  2. Arithmetic result is 11110100, value -12.
  3. Logical result is 01110100, value 116.
  4. Sign extension preserves negative scaling here; zero filling does not.

Examiner tip

Shift each requested result from the original, not from the other shifted value.

Question 4A binary float has an 8-bit two's-complement mantissa with its point immediately after the sign bit, and a 4-bit two's-complement exponent. Value=mantissa*2^exponent. Positive normalised mantissas begin 01. Subnormal numbers are NOT supported. Find the smallest positive normalised value, its bit fields, and explain why 2−102^{-10} is not representable as a positive normalised value in this format.4 indicative marks
Analyse4 indicative marksNo calculator

A binary float has an 8-bit two's-complement mantissa with its point immediately after the sign bit, and a 4-bit two's-complement exponent. Value=mantissa*2^exponent. Positive normalised mantissas begin 01. Subnormal numbers are NOT supported. Find the smallest positive normalised value, its bit fields, and explain why 2−102^{-10} is not representable as a positive normalised value in this format.

Cambridge 2027: Direct conceptual match

Enumerated positive normalised mantissas64..127 divided128 and exponents-8..7: minimum1/512; target1/1024 falls below it.

AI-assisted placement. Independent teacher approval pending.

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Worked guidance

  1. With the specified binary point, 01000000 means 0.1000000 in binary, or 1/2. Smaller positive mantissas would not begin 01 and are not normalised.
  2. The four-bit signed exponent ranges from -8 to +7. Combining its minimum -8 with 1/2 gives (1/2)*2−82^{-8}=2−92^{-9}=1/512.
  3. To retain normalisation at 2^-10 would require exponent -9, which cannot be stored. Because subnormals are excluded, this target is below the smallest positive normalised value; the question does not specify an automatic rounding policy.

Self-marking points

View marking points
  1. Smallest positive normalised mantissa is 01000000, value 1/2.
  2. Minimum exponent is 1000, value -8.
  3. Minimum positive normalised value is 2−92^{-9}=1/512.
  4. 2^-10 needs exponent -9 with mantissa1/2, outside range; this is underflow.

Examiner tip

State whether subnormals are allowed before claiming a smallest positive value.

Question 5A bitmap stores one fixed-length palette index per pixel. It has 120 by 80 pixels and exactly eight available colours. Ignore palette/header bytes and pack indices without row padding. Calculate the minimum index bits and pixel bytes. A ninth colour is then required: recalculate both and the percentage increase in pixel storage.4 indicative marks
Analyse4 indicative marksNo calculator

A bitmap stores one fixed-length palette index per pixel. It has 120 by 80 pixels and exactly eight available colours. Ignore palette/header bytes and pack indices without row padding. Calculate the minimum index bits and pixel bytes. A ninth colour is then required: recalculate both and the percentage increase in pixel storage.

Cambridge 2027: Direct conceptual match

Checked ceil(log2(8))3 versus ceil(log2(9))4, exact packed byte totals3600/4800 and old-base percentage.

AI-assisted placement. Independent teacher approval pending.

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Worked guidance

  1. Three bits supply 232^{3}=8 indices, so the original pixels occupy (120*80*3)/8=3600 bytes under the stated packing rule.
  2. A ninth distinct colour exceeds eight patterns. Four bits provide sixteen indices and require (120*80*4)/8=4800 bytes.
  3. The additional 1200 bytes are one third of the original 3600, approximately 33.33%. Unused index patterns do not remove the extra bit required for every fixed-length pixel index.

Self-marking points

View marking points
  1. Eight colours need 3 bits per index.
  2. 9600 pixels use 28800 bits=3600 bytes.
  3. Nine colours need 4 bits, giving4800 bytes.
  4. Increase is1200/3600*100=33.33% approximately.

Examiner tip

Round the bits-per-index requirement upward before multiplying by pixel count.

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