Jigsaw — 1 Information Representation
Syllabus content for 9618 Topic 1, rebuilt bullet by bullet, with every tested angle mapped onto it.
Legend
Already examined in 9618
Tested in a 9618 paper (2021 onwards). Latest question ID given.
9608 only (not yet in 9618)
Tested under the old 9608 syllabus, still inside the 9618 syllabus wording. Fair game — just untested in the current series. Latest 9608 question ID given.
Not yet tested — inference
In syllabus, not yet asked in either series (or only asked in a much narrower form). Justification given.
From the Save My Exams notes
Content the SME revision notes teach that no past question above covers.
The most examined topic in the paper
Topic 1 carries 128 pages of 9618 Legend — more than Databases. It is also the most predictable: the same half-dozen calculations and conversions recur every series, almost always as Question 1. The marks here are won by speed and accuracy, not by recall, which makes it the one topic where drilling beats reading.
Two habits that protect marks throughout
Always show working. Nearly every calculation question splits its marks as 1 for working, 1 for the answer — so a correct answer with no working caps at 1, and a wrong answer with correct working still scores 1. Read the unit demanded. Questions ask for the answer in kibibytes, mebibytes, megabytes or gigabytes, and the divisor changes with it. Getting the arithmetic right and the unit wrong scores the working mark only.
1.1 Data Representation
1.1.1 Binary magnitudes, and binary versus decimal prefixes
Success
Two acceptable forms, and the second is far quicker to write: By value — a kibibyte = 1024 bytes // 2¹⁰ bytes and a kilobyte = 1000 bytes · a tebibyte = 1024 gibibytes / 1 048 576 kibibytes / 2⁴⁰ bytes whereas a gigabyte = 1000 megabytes / 1 000 000 kilobytes / 10⁹ bytes. By prefix type — “kibi is a binary prefix and kilo is a denary prefix”. One line, full mark.
9618_w24_qp_11_sc_1.a;9618_w23_qp_12_sc_3.a;9618_s23_qp_11_sc_3.d.i
Success
A kibibyte has a binary prefix. Three kibibytes is the same as 3072 bytes. A megabyte has a decimal / denary prefix. Two terabytes is the same as 2000 gigabytes. 1 mark each — note the question mixes a prefix-type answer with an arithmetic answer in the same four marks.
9618_s24_qp_13_sc_1.a
Success
Given 3300 kibibytes, 0.3 megabytes, 3 mebibytes, 3300 kilobytes → the answer is 3300 kibibytes (3300 × 1024 = 3 379 200 bytes, against 3 145 728 for 3 mebibytes and 3 300 000 for 3300 kilobytes). The trap is that mebibytes sound bigger.
9618_s24_qp_12_sc_7.a
Success
8 bits → 1 byte · 8000 bits → 1 kilobyte · 1000 kilobytes → 1 megabyte · 1024 mebibytes → 1 gibibyte · 8192 bits → 1 kibibyte. 1 mark per line. Two of the seven options (1 gigabyte, 1 mebibyte) are distractors with no match.
9618_w21_qp_11_sc_1.a
The pebi / peta row, and the units above tera
The syllabus names only kibi/kilo, mebi/mega, gibi/giga and tebi/tera. SME teaches pebibyte (2⁵⁰) and petabyte (10¹⁵) as the next row. No question in either series has gone above tera — but the four named pairs have all appeared, so the pattern is complete and a fifth row would be a natural extension.
Why the two systems exist
Every question tests the difference; none asks the reason. The explanation — computers address memory in powers of 2, so RAM is quoted in binary prefixes, while storage manufacturers quote capacity in denary prefixes because the numbers look larger — has never been asked, in either series.
The two tables, side by side
Denary (base 10): kilobyte 10³ = 1000 bytes · megabyte 10⁶ · gigabyte 10⁹ · terabyte 10¹² · petabyte 10¹⁵. Binary (base 2): kibibyte 2¹⁰ = 1024 · mebibyte 2²⁰ = 1 048 576 · gibibyte 2³⁰ = 1 073 741 824 · tebibyte 2⁴⁰ · pebibyte 2⁵⁰. SME’s rule of thumb for which to use: be precise about memory (16 GiB RAM really is 16 × 2³⁰ bytes), but a rough estimate is acceptable for storage (a 16 GB memory stick holds 16 × 10⁹ bytes).
1.1.2 Different number systems
Success
3 marks for 6 correct ticks, 2 for 4–5, 1 for 2–3. The hexadecimal value F139 → 16 (four hex digits × 4 bits) · 16 000 000 unique amplitude values → 24 (2²⁴ = 16.7 m) · an IPv4 address → 32 · 256 unique colours → 8 · an IPv6 address → 128 · the denary value 65 000 → 16. A genuinely synoptic question — it needs IP addressing from Topic 2 and colour depth from 1.2.
9618_w25_qp_13_sc_2.a
Success
16 bits → 2¹⁶ // 65 536. The general rule: n bits give 2ⁿ distinct values.
9618_s23_qp_12_sc_4.a
Success
The smallest integer in 8-bit two’s complement → −128 · the largest integer in 8-bit two’s complement → 127 · the largest unsigned integer in 8 bits → 255. 1 mark per line. The distractors (−127, −255, −256, 256, 128) are all near-misses, so the three must be known exactly.
9618_s23_qp_13_sc_7.a
Success
1 mark each. Smallest 1000 0000 (= −128) · largest 0111 1111 (= +127). Note the question asks for the binary patterns, not the denary values — the matching question above asks for the denary. Read which is wanted.
9618_s25_qp_12_sc_2.b.ii;9618_s23_qp_11_sc_3.d.iv
Success
1 mark per benefit, max 2. It is straightforward to convert to and from denary … so it is less complex to encode and decode for programmers · it is easier for digital equipment that displays output information · it can represent monetary values exactly.
9618_w22_qp_13_sc_9.b
Success
Because the denary value in each group of 4 bits is greater than 9 // the value in each nibble is greater than 9. Any nibble from
1010to1111is invalid BCD.9618_w21_qp_12_sc_4.d
One's complement barely exists in 9618
The syllabus names “one’s and two’s complement representation”. Two’s complement appears in almost every paper. One’s complement has been asked exactly once —
9618_s23_qp_12_sc_4.b, giving the 8-bit one’s complement of −120 (working: +120 =0111 1000, answer:1000 0111). Its defining weakness — that it has two representations for zero,0000 0000and1111 1111— has never been examined in either series, though SME teaches it.
Describing what a number base is
Every question uses the bases; none asks what a base means. The definition — a number base is the count of distinct digits the system uses, and each column is a power of the base — plus the fact that one hexadecimal digit maps exactly to one 4-bit nibble, is assumed throughout and asked nowhere.
Why two's complement rather than one's or sign-and-magnitude
Untested in either series. The reason is that two’s complement needs no separate subtraction circuit — the same adder handles any combination of signs — and it has only one representation of zero. SME makes this point explicitly; no mark scheme does.
The four representations, defined
Binary — base 2, digits 0 and 1, column weights 1, 2, 4, 8, 16 … Denary — base 10, digits 0–9. Hexadecimal — base 16, digits 0–9 then A–F (A = 10 … F = 15); each hex digit is exactly one nibble. BCD — each denary digit is separately coded as its own 4-bit pattern, so 2500 becomes
0010 0101 0000 0000. BCD may be stored as one nibble per digit, or two digits packed into a byte. One’s complement: invert every bit. Two’s complement: invert every bit, then add 1 — or, faster, keep every bit up to and including the rightmost 1, then flip the rest. In two’s complement the leftmost column becomes negative (−128 in 8 bits), which is why the range is −128 to +127.
1.1.3 Converting between number bases and representations
The highest-frequency bullet in the entire paper
Fifty-one separate 9618 questions sit under this one bullet, almost all worth 1–2 marks. Nothing here is conceptually hard; the marks are lost to arithmetic slips and to misreading which representation is wanted. Practise all nine conversions below until they are mechanical.
Success
101100111010→ B3A ·110001100111→ C67 ·1110001100111011→ E33B. Split into nibbles from the right, convert each.9618_w25_qp_13_sc_2.d;9618_w25_qp_11_sc_1.a.i;9618_w24_qp_11_sc_1.b.i
Success
1FAB→ 8107 ·C0F→ 3087 (2 marks, with working). Multiply each digit by its power of 16.9618_w24_qp_13_sc_8.a;9618_s24_qp_12_sc_7.c
Success
241 → F1. Divide by 16: quotient is the first digit, remainder the second.
9618_s24_qp_13_sc_1.b
Convert denary to binary and hexadecimal in one question [2]
558 → binary 0010 0010 1110, hexadecimal 22E. 1 mark each. Doing the binary first and grouping into nibbles gives the hex free.
9618_s25_qp_12_sc_2.a
Success
−108 → 1111 1001 0100 · −196 → 1111 0011 1100. Convert the magnitude, pad to 12 bits, then invert and add 1.
9618_w25_qp_13_sc_2.b;9618_w23_qp_12_sc_3.b.i
Success
11100010→ −30 ·10010110→ −106 ·100110010111→ −1641 ·111110111100→ −68 (2 marks, with working).9618_s25_qp_12_sc_2.b.i;9618_w24_qp_11_sc_1.b.ii;9618_w25_qp_11_sc_1.a.iii
Explain how to convert two's complement to denary, and give the value [3]
1 mark per method bullet (max 2) + 1 for the correct conversion. Method 1: flip each bit then add 1 … then convert the new binary number into denary (and negate). Method 2: treat the most significant 1 bit as its corresponding negative denary value … then add the other positive corresponding denary values. For
10011111the answer is −97. Both methods are credited — but the question asks for an explanation, so the method must be written out, not just the answer.9618_w24_qp_13_sc_8.b
Success
108 → 0001 0000 1000 · 964 → 1001 0110 0100. One nibble per digit, including leading zeros.
9618_w25_qp_11_sc_1.a.ii;9618_s23_qp_11_sc_3.d.ii
Success
100001100101→ 865 ·010101110011→ 573. Split into nibbles and read each as a single digit.9618_w23_qp_12_sc_3.b.ii;9618_w24_qp_11_sc_1.b.iii
Success
10010101→ denary 149 → BCD 0001 0100 1001. 1 mark for the denary step, 1 for the BCD — so the intermediate denary value is itself a mark.9618_w22_qp_12_sc_2.a.iii
Success
11001101→ −51. Appears inside Chapter 4 assembly questions as well as Chapter 1.9618_w21_qp_12_sc_4.b
Success
!33 → 0010 0001 / 21 ·L→ 76 /0100 1100/ 4C ·ü252 /1111 1100→ FC. 1 mark per completed space. Three conversions in three marks, each in a different direction.9618_s25_qp_13_sc_1.c.ii
Hexadecimal to binary asked on its own
9618 asks binary→hex repeatedly, and hex→denary repeatedly — but a bare “convert this hexadecimal number into binary” has never been set, despite being the easiest of the nine (each digit becomes one nibble). It appears only inside the character tables.
Converting negative numbers into hexadecimal
All hex conversions in 9618 use positive values. Converting a two’s complement negative into hex (e.g. −68 →
1111 1011 1100→ FBC) has not been asked, though both halves have been separately.
The conversion methods, in brief
Binary → denary: add the column weights where there is a 1. Denary → binary: work left from the highest column that fits, subtracting as you go. Binary ↔ hex: group into nibbles of 4. Denary → hex: divide by 16, quotient then remainder. Hex → denary: first digit × 16, plus second digit. Denary → BCD: code each digit separately into 4 bits. Two’s complement: invert and add 1 — or keep bits up to the rightmost 1, flip the rest.
1.1.4 Binary addition and subtraction
Success
1 mark for the working (the carries), 1 for the answer.
1010 1010 + 0011 0111= 1110 0001, carries1 1 1 1 1. Write the carry row — it is the working mark.9618_w21_qp_11_sc_1.b.i
Success
1 mark per bullet: converting the denary value to binary … the method for the addition … the final answer. For
0010 0011 + 15: 15 =0000 1111, giving 0011 0010. The conversion step is a separate mark from the addition.9618_s21_qp_11_sc_1.c.ii
Success
1 mark per bullet: converting the number being subtracted to two’s complement … adding the values … the final answer. For
0010 0011 − 10: −10 =1111 0110, and0010 0011 + 1111 0110= 0001 1001.9618_s21_qp_11_sc_1.c.iii;9618_w24_qp_11_sc_1.c
Success
1 mark each: converting both numbers to binary · the subtraction method (convert to the negative and add, or direct subtraction) · the correct answer. For 100 − 10:
0110 0100and0000 1010, answer 0101 1010. Both methods are equally credited — direct borrowing is accepted, not only two’s complement.9618_s24_qp_12_sc_7.b
Success
The result is a larger number than can be stored in the given number of bits // the result is greater than 255 (for 8 unsigned bits).
9618_w21_qp_11_sc_1.b.ii
Overflow is named in the syllabus and asked once
The notes say “show understanding of how overflow can occur”. There is exactly one 9618 question on it, worth 1 mark, in the very first series (w21). Nothing has tested: overflow flipping the sign bit in a signed addition, how a programmer would detect it, or why the carry-out of a two’s complement subtraction is discarded rather than being an error. All three are in SME, none in a mark scheme.
Subtraction asked without a method being implied
Every 9618 subtraction question either gives a two’s complement value or says “using binary subtraction”. A question that simply says subtract and leaves the method open — then asks the candidate to justify why two’s complement addition is used instead of borrowing — is the untested conceptual version of this bullet.
Why the extra bit is discarded
SME’s worked example makes the point no mark scheme states: when 48 + (−12) is done in 8-bit two’s complement the sum produces a ninth bit, which is simply ignored, leaving
0010 0100= 36 — the correct answer. In two’s complement arithmetic that carry-out is a by-product of the method, not an overflow. Genuine overflow is when the result itself will not fit in the available bits.
1.1.5 Practical applications of BCD and hexadecimal
Success
1 mark for the application, 1 for a corresponding justification. Financial / banking calculations … because financial transactions use only two decimal places and must be exact, there are no accumulating or rounding errors, and it is difficult to represent decimal values exactly in normal binary. Electronic displays (calculators, digital clocks) … because visual displays only need to show individual digits, and conversion between denary and BCD is straightforward. Storing the date and time in the BIOS … because conversion with denary is easier. Barcode systems … because conversion between denary and BCD can be completed accurately.
9618_s25_qp_12_sc_2.c;9618_w23_qp_12_sc_3.c
Warning
The 9608 framing is broader and adds one credited point 9618 never uses: “when denary numbers need to be electronically coded” · e.g. to operate displays on a calculator where each digit is represented separately · decimal fractions can be accurately represented.
9608_s15_qp_13_sc_1.b.ii;9608_w19_qp_11_sc_5.c.ii;9608_w17_qp_13_sc_1.b.iii
Hexadecimal applications have never been asked — in either series
The bullet is titled “practical applications where Binary Coded Decimal (BCD) and Hexadecimal are used”. Every question in both series asks only about BCD. The hexadecimal half has no question against it anywhere. The available content: hex is a shorthand for binary that is far easier for humans to read and less error-prone to transcribe, because one hex digit replaces four bits. Real uses: MAC addresses, IPv6 addresses, HTML/CSS colour codes (
#FF0000), memory addresses and memory dumps, assembly language operands and error codes, and the ASCII/Unicode code point tables. Several of these are already examined elsewhere in the syllabus (MAC and IPv6 in Topic 2, memory addresses in Topic 4), which makes a synoptic “why is hexadecimal used here” question very easy to set. This is the clearest single gap in Topic 1.
The drawbacks of BCD
Both series ask only for benefits. The costs — BCD wastes storage, since 4 bits represent only 10 of 16 possible patterns, and arithmetic is more complex than in pure binary — have never been asked, though they are the obvious counterweight to the benefits question that is set.
The use-case table
Electronic calculators — keeps numbers in decimal format for easier display and accuracy. Digital clocks and watches — time is naturally decimal (12:45), so display logic is simpler. Banking and financial systems — avoids rounding errors in decimal calculations, especially with money. Older digital and embedded systems — simpler to implement with hardware that drives one digit at a time.
1.1.6 Character sets
Success
1 mark per bullet, max 2. Each character has a unique code · each character in the text is replaced sequentially / in order by its code · the codes are stored in the order they appear in the word.
9618_w25_qp_13_sc_1.b;9618_s24_qp_13_sc_1.d.ii;9618_s21_qp_12_sc_6.b
Success
1 mark per bullet, max 2. ASCII uses 7 / 8 bits; Unicode can use many more — up to 32 · Unicode can represent a wider range of characters, including different languages.
9618_w25_qp_13_sc_1.c
Give one similarity and two differences between ASCII and Unicode [3]
1 mark for the similarity, 2 for differences. Similarity (max 1): both can use 8 bits · both represent each character using a unique code · Unicode contains all the characters ASCII contains // ASCII is a subset of Unicode. Differences (max 2): Unicode goes up to 32 bits per character whereas ASCII is 7 or 8 · Unicode represents a wider range of characters · different languages are represented in Unicode, ASCII is only for one.
9618_w22_qp_11_sc_1.c
Success
A wider range of characters can be represented · so characters from more languages can be represented · and symbols such as emojis can be used.
9618_s25_qp_11_sc_3.b.i
Success
8 / 16 / 32 bits per character · represents 2⁸ / 2¹⁶ etc. characters · represents every language and other characters such as emojis.
9618_s25_qp_13_sc_1.c.i
Success
1 mark for identification, 1 for a matching description. ASCII — 7/8 bits per character // represents 128/256 characters // represents all characters from the Latin alphabet. UNICODE — 8/16/32 bits per character // represents 256/65 536+ characters // represents all characters in all languages.
9618_w24_qp_13_sc_6.a
Success
1 mark for all three correct. ASCII → 7 · extended ASCII → 8 · Unicode → 16/32. All-or-nothing.
9618_s24_qp_13_sc_1.d.i
Success
ASCII = 128 // 2⁷ · Extended ASCII = 256 // 2⁸.
9618_s21_qp_12_sc_6.a
Success
8 // 16 // 32 // 64 — any of these is credited, which is unusually generous.
9618_w25_qp_12_sc_7.b
Success
The Unicode character
ɮ=0010 0111 0110 1110→ denary 10 094 · the Unicode character∑= hex 2140 → denary 8512 · ASCIIh= denary 104 → BCD 0001 0000 0100, hexadecimal 68 · Unicode1= denary 49 → hex 31 · Unicode5→ denary 53. The character-set questions are really conversion questions in disguise.9618_s25_qp_11_sc_3.b.ii;9618_w25_qp_12_sc_7.c.i
Success
Given a word/value lookup table,
0011 1000→ 56 → Science,0011 1100→ 60 → is,0011 1110→ 62 → Amazing!. 1 mark for all three in the correct order.9618_w25_qp_12_sc_6.b
Warning
Only 128 / 256 characters can be represented · it uses values 0 to 127 (or 255 extended) / one byte · many characters used in other languages cannot be represented · in extended ASCII the characters from 128 to 255 may be coded differently on different systems. That last point — extended ASCII is not consistently standardised — appears in no 9618 mark scheme.
9608_w16_qp_13_sc_8.c.i
Warning
Uses 16, 24 or 32 bits / two, three or four bytes · Unicode is designed to be a superset of ASCII · designed so that most characters in other languages can be represented. The word superset is the sharp version of 9618’s “ASCII is a subset of Unicode”.
9608_w16_qp_13_sc_8.c.ii
Warning
The symbols that the computer recognises / uses // a list of characters recognised by the computer hardware and software. 9618 has never asked for the definition itself, only for how one works.
9608_s18_qp_12_sc_4.b.i
Warning
Adds one point 9618 does not credit: Unicode is standardised while ASCII is not. Otherwise the same content as the 9618 difference questions.
9608_s18_qp_12_sc_4.b.ii
Warning
1 mark for working, 1 for the answer. If ASCII
A= 41₁₆, thenZ= A + 25₁₀ = 41₁₆ + 19₁₆ = 5A₁₆. This exploits the fact that character sets are ordered logically — a property 9618 relies on implicitly but has never made the subject of a question.9608_s18_qp_12_sc_4.b.iii;9608_s19_qp_12_sc_3.d.iii
The logical ordering property
SME makes the point that a character set is ordered so that the code for
Bis one more than the code forA, and that in ASCII the sixth bit alone distinguishes upper from lower case (a=0110 0001,A=0100 0001), which makes case conversion a single bit operation. 9608 exploits the ordering in a calculation question; 9618 has never tested either property, even though it is the reason character sets are designed the way they are.
ASCII versus Unicode, feature by feature
Bits: ASCII 7 · extended ASCII 8 · Unicode minimum 16. Characters: 128 · 256 · 65 536+. Covers: standard English keyboard · plus mathematical operators and symbols such as © · all major world languages and emoji. Benefit: ASCII uses far less storage. Drawback: ASCII cannot represent other languages or special characters; Unicode uses considerably more storage. The first 128 Unicode code points are identical to ASCII, which is what makes Unicode backward compatible.
1.2 Multimedia
1.2.1 How data for a bitmapped image is encoded
Success
1 mark each. The image is made of pixels and each pixel has one colour · each colour has a unique binary code · the code for the colour of each pixel is stored in sequence. Note the parallel with character sets and with sound — Cambridge marks all three the same way: unit → unique code → stored in sequence.
9618_s24_qp_12_sc_2.d.i
Success
Pixel — the smallest part of the image // one square or dot of one colour // the smallest addressable element in an image · Colour / bit depth — the number of bits per pixel // the number of bits used to represent each colour // determines the number of colours that can be represented · File header — stores data about the image file / metadata.
9618_s23_qp_11_sc_1.a;9618_w24_qp_12_sc_7.a.i;9618_s21_qp_11_sc_1.a.i
Success
The bit depth of a bitmap image is the number of bits used to store each pixel. Metadata about the image is stored in the header of the file.
9618_w22_qp_13_sc_2.e
Success
The smallest element that makes up an image → pixel · the largest number of different colours representable with a bit depth of 8 bits → 256 // 2⁸ · the term for the dots per inch (dpi) when an image is displayed → screen resolution.
9618_s25_qp_11_sc_3.a
Success
8 bits → 256 // 2⁸. The formula is 2ⁿ, exactly as for unique binary values in 1.1.2.
9618_w24_qp_13_sc_6.b.i
Success
Given that colour depth and image resolution are already there: confirmation that it is a bitmap / the file type · the compression type used · the location / offset of the data within the file · the dimensions, e.g. 100 × 100 pixels.
9618_s23_qp_11_sc_1.b.i
Success
The file will contain metadata · the file will have a header. This is why every file-size question says “calculate an estimate”.
9618_w25_qp_11_sc_7.d.ii
Image resolution versus screen resolution
Both terms are named in the syllabus. Image resolution is examined constantly (every file-size calculation uses it). Screen resolution appears exactly once, as the one-word answer “screen resolution” to a dpi description (
s25_qp_11_sc_3.a) — and once as a distractor inw22_qp_12_sc_8.a, where changing the screen resolution correctly has no effect on the file size. The distinction itself — that image resolution is a property of the file while screen resolution is a property of the display — has never been the subject of a question.
Pixel density, and why it matters
SME adds the layer above screen resolution: pixels per inch (PPI) is the pixel density, which depends on the resolution and the physical size of the display. A 65″ 4K television works out at about 68 PPI, while a modern smartphone exceeds 300 PPI — which is why a phone can be viewed from a few inches without visible pixelation and a television cannot. Not in any mark scheme, but it explains the dpi row that is examined.
1.2.2 Calculating bitmap file size
The formula, and the only thing that changes
File size = image resolution (width × height) × colour depth, giving a size in bits. Then divide: ÷ 8 for bytes · ÷ 1000 per denary step (kB, MB, GB) · ÷ 1024 per binary step (KiB, MiB, GiB). 1 mark for working, 1 mark for the answer, every time. Nine 9618 questions, identical structure.
Success
1000 × 2000 × 16 bits ÷ (8 × 1000 × 1000) = 4 MB · 2 000 000 pixels × 16 ÷ (8 × 1000 × 1000) = 4 MB · 1500 × 3000 × 8 bytes ÷ 1000 ÷ 1000 = 36 MB. Watch the third one: the bit depth is given in bytes, so there is no ÷ 8.
9618_w25_qp_12_sc_6.d;9618_s25_qp_13_sc_1.a.i;9618_s23_qp_11_sc_1.b.ii
Success
4000 × 3000 × 4 bytes = 48 000 000 bytes = 48 MB. The working mark is just
4000 * 3000 * 4— no division by 8, because the depth is already in bytes. The single most common slip in this bullet.9618_s24_qp_13_sc_2.a
Success
2048 × 1024 × 16 ÷ (8 × 1024) = 4096 kibibytes · 512 × 2048 × 8 ÷ (8 × 1024) = 1024 kibibytes // 2¹⁰ kibibytes. Note “a maximum of 256 colours” means a bit depth of 8.
9618_w23_qp_13_sc_2.b;9618_w25_qp_11_sc_7.d.i
Success
2048 × 1024 × 10 ÷ (8 × 1024 × 1024) = 2.5 mebibytes · 1024 × 512 × 8 bits ÷ 8 ÷ 1024 ÷ 1024 = 0.50 mebibytes (3 marks: 1 per working bullet, 1 for the answer).
9618_w23_qp_12_sc_6.c;9618_s21_qp_11_sc_1.a.ii
Success
A video is a sequence of bitmap frames, so the frame rate is simply another multiplier: 4000 × 3000 × 30 frames × 16 ÷ (8 × 1000 × 1000 × 1000) = 0.72 gigabytes.
9618_w25_qp_13_sc_7.b
Working backwards from a file size
Every question gives the resolution and depth and asks for the size. The inverse — given a file size and resolution, find the bit depth; or given a size and a storage limit, find how many images fit — has never been asked in 9618, though it uses exactly the same formula and is a standard way to raise the difficulty.
The file size of a vector graphic
Calculations are only ever asked for bitmaps. The syllabus has no vector-size bullet, so this is correctly untested — but it is worth knowing why: a vector file stores a drawing list of commands, so its size depends on the number of objects, not on the image dimensions.
The worked SME example, including a sound file
Image: 250 000 pixels × 24 bits = 6 000 000 bits = 750 000 bytes = 732 KiB or 750 kB. Sound uses the same discipline: sampling rate × sampling resolution × duration in seconds. For
w22_qp_13_sc_1.b: 50 kHz × (20 × 60 s) × 16 bits = 960 000 000 bits = 120 000 000 bytes = 120 megabytes. That sound calculation is filed under Data Representation in the Legend, not Multimedia — but it is the same formula with time in place of area.
1.2.3 Effects of changing elements of a bitmap image
Success
1 mark each, two for the image and two for the file. Image: there will be fewer shades of colour available … so the image does not match the original, as detail is lost. Image file: fewer bits are used to store each pixel … so less data is stored, therefore the file size is reduced.
9618_s25_qp_13_sc_1.a.ii
Success
1 mark definition + 1 per explanation bullet. Definition: the number of bits used to represent each colour. Explanation: an increase means a greater range of colours (a decrease, a smaller range) · an increase makes the image closer to the original / more realistic (a decrease, less like the original).
9618_w23_qp_11_sc_1.b
Success
1 mark for each explanation. Quality: decreasing the resolution means detail is lost because there are fewer pixels // increasing means more detail because there are more pixels. File size: decreasing the resolution decreases the file size because there are fewer pixels therefore less data // increasing increases it. Each half must carry its because — the bare statement scores nothing.
9618_w24_qp_12_sc_7.a.ii
Success
More pixels can be stored / are available · the image is sharper / less pixelated.
9618_w23_qp_13_sc_2.a
Success
Increased file size.
9618_w24_qp_13_sc_6.b.ii
Success
Colour / bit depth · image resolution. Only these two — a third answer earns nothing.
9618_s24_qp_13_sc_2.c
Success
1 mark for one or two correct rows, 2 marks for all three. Change the colour depth to 16 bits per pixel (from 24) → decreases the file size · change the screen resolution to 1366 × 768 → no change · change the colour of the rectangle from black to red → no change. The two “no change” rows are the question: screen resolution is a property of the display, and changing which colour is used does not change how many bits store it.
9618_w22_qp_12_sc_8.a
The trade-off framed as a decision
Every question asks what happens when a value changes. None asks the candidate to recommend a resolution or depth for a stated purpose and justify it — the format used constantly in Chapter 3 for storage devices. Given that 1.2.5 already asks for a justified choice between bitmap and vector, the same framing applied to resolution and depth is readily available.
1.2.4 How data for a vector graphic is encoded
Success
Drawing list — all the drawing objects in an image // a list that stores the commands / descriptions / mathematical equations required to draw each object · Drawing object — a component created using a formula · Property — an attribute of a drawing object // data about a shape // defines one aspect of the appearance of a drawing object.
9618_w24_qp_12_sc_7.b;9618_w23_qp_11_sc_1.a;9618_s23_qp_11_sc_1.a
Success
1 mark each to max 2. A list of the objects in the drawing · a list that stores the command / description / equation required to draw each object · the properties of each object, e.g. the fill colour, line weight or colour.
9618_s24_qp_12_sc_2.d.ii
Describe each vector term and give an example from a given logo [4]
1 mark for each description, 1 for each valid example. Property — data about the shapes // defines one aspect of the appearance … e.g. black line, white fill, solid line, font of the letter, colour of the triangle. Drawing list — the list of shapes involved in an image … e.g. triangle, capital letter R, rectangle, line. The example must be read off the image given.
9618_w21_qp_12_sc_5.a
Success
2 marks for all 3 correct, 1 mark for 1 correct. Drawing list → data required to create all components in the graphic · Drawing object → a component created using a formula · Property → defines one characteristic of a component.
9618_w23_qp_11_sc_1.a
The relative position of objects
SME lists three things a drawing list holds: the commands to create each object, the attributes/properties of each object, and the relative position of each object. 9618 mark schemes credit the first two repeatedly; position is never mentioned — yet it is what makes a drawing list reproducible, and it is the reason a vector image has no fixed dimensions.
Why a vector graphic scales without loss
Credited as a bare point in the comparison questions (“when vector is enlarged it is recalculated and does not pixelate”), but the mechanism — the file stores equations, not pixels, so on enlargement the shape is recalculated at the new size rather than stretched — has never been the subject of its own question.
What a vector file actually stores
A vector graphic is created from mathematical equations and points; only the mathematics is stored. To draw a circle, the data stored is the centre point (x, y) and the radius — nothing else. The drawing list sits in the file header. Because no dimensions are defined, scaling up causes no loss of quality. Typical examples are logos and clipart; typical file types are
.svg,.ai,.eps.
1.2.5 Justifying the use of a bitmap image or a vector graphic
Success
1 mark per bullet to max 2 for each difference — so two differences, each with an expansion. Composition: bitmap is made of pixels // colours stored for individual pixels · vector stores a set of instructions about how to draw the shape. Scaling: when a bitmap is enlarged the pixels get bigger and it pixelates · when a vector is enlarged it is recalculated and does not pixelate. File size: bitmap files are usually bigger, because data about each pixel must be stored · vector files are smaller, because they contain just the instructions. Compression: bitmaps compress well, with significant reduction · vector graphics do not compress well, because there is little redundant data.
9618_w21_qp_12_sc_5.b.i
Success
1 mark each, max 2. Can be enlarged without pixelation / loss of quality · individual components of the image can be edited · generally a smaller file size.
9618_w22_qp_12_sc_8.b
Warning
1 mark per drawback, 1 for its expansion, max 2 each. A bitmap file is likely to take up more storage space … because the colour of each pixel needs to be stored · a bitmap cannot be enlarged // is difficult to use in different types of document … without the image pixelating · a bitmap would be more difficult to edit … because each pixel would need to be edited separately. That last drawback — editability — appears in no 9618 mark scheme.
9608_s21_qp_11_sc_2.c
Warning
1 mark per bullet, max 2 per reason. Smaller file size … so it can be transferred / downloaded quicker · enlarges without pixelation … because it needs to be used on different screens / devices / resolutions.
9608_w18_qp_12_sc_1.c;9608_s18_qp_11_sc_2.d
Warning
Adds a storage angle 9618 does not use: needs to be large for signs without pixelating · smaller file size means faster transfer rates · smaller file size reduces storage requirements when stored many times on multiple documents.
9608_s20_qp_12_sc_1.b.ii
9618 has asked this bullet once, for 2 marks
There is exactly one 9618 question under “justify the use of a bitmap image or a vector graphic for a given task” —
w22_qp_12_sc_8.b, worth 2 marks. 9608 has five, several at 4 marks with the point-plus-expansion structure. Given that every other Multimedia bullet is examined repeatedly in 9618, this one is overdue, and the 9608 answers above are the model.
Justifying a bitmap over a vector
Every question in both series runs the argument one way: why vector beats bitmap for a logo. The reverse — why a bitmap is right for a photograph (continuous tone, millions of colours, no describable shapes to store as equations, produced directly by a camera sensor) — has never been asked, despite being half of what the bullet says.
The decision questions SME suggests
Three questions settle the choice every time: Does the image need to be resized? · Does it need to be drawn to scale? · Does it need to look real? The first two point to vector, the third to bitmap.
1.2.6 How sound is represented and encoded
Success
1 mark each. The amplitude is recorded a set number of times a second · each instance of an amplitude is given a corresponding binary number · the binary number of each amplitude is saved in sequence. The same three-part shape as the bitmap and character-set answers.
9618_s23_qp_13_sc_3.c.i
Success
1 mark per bullet, max 2. The value / magnitude / size of the analogue sound wave is measured a set number of times each second / at set intervals · each sample / reading / measurement is given a binary number and stored in sequence.
9618_w24_qp_13_sc_6.c.i
Success
The number of times the amplitude is measured per time interval → sampling rate · the number of bits used to store each amplitude measurement → sampling resolution · the type of sound wave before it is recorded by a computer → analogue.
9618_s25_qp_13_sc_1.b
Success
1 mark for 1 correct line, 2 marks for all 3. Sampling → taking measurements at regular intervals and storing the values · Sampling rate → the number of samples taken per second · Sampling resolution → the number of bits used to store each sample.
9618_w23_qp_13_sc_1.b
Success
The number of samples taken per unit time / per second.
9618_w22_qp_11_sc_1.d.i
Success
Data values that are continuously changing // variable // can take any value.
9618_w23_qp_13_sc_1.a
Warning
1 mark per definition. Sampling — measuring the amplitude of the wave at regular / set time intervals · Sampling resolution — the number of bits used to represent each sample · Sampling rate — the number of samples taken per unit of time. 9618 asks for these as a matching exercise; 9608 asks for them to be written out.
9608_w21_qp_11_sc_2.a;9608_s20_qp_12_sc_2.a
Warning
Applied to actual figures: a sampling rate of 88.2 kHz means the sound wave is sampled 88 200 times per second · a sampling resolution of 32 bits means each sample is stored as a 32-bit binary number. Reading the units off a specification is never asked in 9618.
9608_s19_qp_13_sc_5.c
Describe how images and sound are encoded into digital form [4]
Max 3 for image, max 3 for sound. Images: stored as bitmaps · each image is made of pixels · each pixel is a single colour · each colour has a unique binary number · store the sequence of binary numbers for each image / frame. Sound: measure the height / amplitude of the sound wave · a set number of times per second at regular intervals · each amplitude has a unique binary number · store the sequence of binary numbers for each sample. A single question covering both encodings, which 9618 has never set.
9608_w19_qp_12_sc_6.d.i
Warning
1 mark for naming the feature, 1 for describing it, max 2 per feature. Cut / delete — remove part of the sound file · Copy and paste — replicate part of the sound · Amplify — increase the volume of a section · Fading — change the volume of a section so it gets louder or quieter · Change pitch — increase or decrease the frequency of a section · Change sampling resolution — to change the accuracy of the sound or the file size. Sound editing appears nowhere in the 9618 syllabus notes — treat this as background, not as a likely question.
9608_w18_qp_11_sc_1.c;9608_s18_qp_12_sc_5.d
The ADC named as a component
9618 asks repeatedly how analogue becomes digital but never names the device. SME calls the process analogue-to-digital conversion (A2D); Chapter 3 examines the ADC as hardware. The link between the two — that the sampling described here is what the ADC in a microphone does — is never made in a question.
Playing sound back
Every question describes recording. The reverse path — the stored binary values are read in sequence, converted back to voltages by a DAC, and drive a speaker cone — is never asked, though SME describes it and the speaker is in the Chapter 3 syllabus.
1.2.7 Impact of changing the sampling rate and resolution
Success
1 mark each; max 2 for rate and max 2 for resolution. Sampling rate: there are smaller ‘gaps’ in the sound wave // sound is recorded more often · the digital waveform is closer to the analogue waveform · the quantisation errors are smaller. Sampling resolution: there are more bits per sample // a wider range of amplitudes can be stored · each binary amplitude is closer to the analogue amplitude · the digital waveform is closer to the analogue waveform · quantisation errors are smaller. Four points about the rate alone score 2.
9618_s23_qp_13_sc_3.c.ii
Success
1 mark per bullet, max 3. Increase: the number of bits per sample is increased … more values are available to represent each sample // more amplitudes can be represented … each binary amplitude is closer to the analogue amplitude … quantisation errors are reduced … the digital soundwave is closer to the original analogue soundwave. Decrease: the mirror of each.
9618_w23_qp_12_sc_6.b
Success
Improves the accuracy of the sound file · because the digital waveform more closely resembles the analogue waveform · quantisation errors are reduced · it increases the amount of detail stored.
9618_w22_qp_12_sc_6.b.i
Success
Increases the number of bits per sample // a larger range of values · which means the file size increases · makes the sound file more accurate // digital waveform closer to the original · smaller quantisation errors.
9618_w22_qp_11_sc_1.d.ii
Success
Decreases the file size of the sound file · because fewer bits are used to store each sample.
9618_w22_qp_12_sc_6.b.ii
Success
Smaller time gaps between the samples · makes the digital sound wave more accurate · smaller quantisation errors.
9618_w21_qp_11_sc_7.a.i
Success
More samples / data are taken or recorded · so more bits are stored altogether.
9618_w21_qp_11_sc_7.a.ii
Success
1 mark for each correctly connected change box, max 3. Increase the duration → the file size gets bigger · no change to the accuracy. Increase the sampling rate → the file size gets bigger · the accuracy improves. Decrease the sampling resolution → the file size gets smaller · the accuracy worsens. Note each change box needs two lines.
9618_w25_qp_13_sc_1.a
Success
1 mark for 1–2 correct, 2 marks for all 3. Sampling rate 40 kHz → 60 kHz → accuracy increases · duration 20 → 40 minutes → accuracy does not change · sampling resolution 24 → 16 bits → accuracy decreases. The duration row is the trap: longer recordings are bigger, not more accurate.
9618_w22_qp_13_sc_1.a
Success
1 mark each, max 3. Doubling the rate from 44.1 kHz to 88.2 kHz means data transmission will take longer … because there is more data to transmit · the secondary storage device will fill faster … so fewer videos can be stored long-term // videos are overwritten more often. An applied, consequences-based version of the bullet — the only one in 9618.
9618_s24_qp_11_sc_2.d
Warning
1 mark for the definition, 1 for the explanation. Definition: the number of samples taken per unit time // the number of times the amplitude is measured per unit time. Explanation: increasing the sampling rate will increase the accuracy / precision of the digitised sound // will result in smaller quantisation errors. 9618 asks the definition and the effect in separate questions; 9608 combines them.
9608_s17_qp_13_sc_3.a;9608_s17_qp_11_sc_3.a
Warning
Max 2 for the definition, max 2 for the explanation. Definition: the number of distinct values available to encode each sample · specified by the number of bits used to encode one sample · sometimes referred to as bit depth. Explanation: a larger resolution means more values available to store each sample · improves accuracy // decreases the distortion of the sound · means a smaller quantisation error. Two pieces of vocabulary here are 9608-only: “distinct values available” and “distortion”.
9608_s17_qp_12_sc_3.a
Quantisation error named but never defined
The phrase appears in six separate 9618 mark schemes as a credited point. No question in either series asks what a quantisation error is — that each sample must be rounded to the nearest value the resolution can represent, and the difference between the true amplitude and the stored one is the error. It is the single most-credited term in 1.2.7 and the least explained.
Nyquist and choosing a rate
Why a CD samples at 44.1 kHz specifically — roughly twice the upper limit of human hearing — is not in the 9618 syllabus and has never been asked. Worth knowing as context for why “higher is always better” is not how rates are chosen in practice, but do not offer it as a mark-scheme point.
The impact table, in one place
Sampling rate ↑ → more detail, better sound quality · more data, larger file size. Sampling resolution ↑ → bigger range of amplitudes, better quality · more data per sample, larger file size. Duration ↑ → larger file size, no change to accuracy. A typical audio CD: sampling rate 44.1 kHz, sampling resolution 16 bits, recorded in stereo.
1.3 Compression
1.3.1 The need for, and examples of, compression
Success
1 mark each to max 4. Video is data-intensive · the file size needs reducing in order to reduce the amount of bandwidth used · and reduce buffering · this means people are not behind in the conversation · and people with lower bandwidth can still take part. The highest-tariff compression question in the chapter, and the one most tied to Topic 2.
9618_s25_qp_11_sc_2.b.i
Success
1 mark per bullet, max 3. Less storage space is used … so more work can be stored · transmission time is reduced … so the recipient does not have to wait as long for it to arrive · bandwidth usage is reduced … so other transmissions are not adversely affected · less data allowance is used on the email system. Note the structure: each benefit has its own consequence, and the consequence is a separate mark.
9618_w24_qp_11_sc_4.b.i
Success
Reduced bandwidth usage when transmitting · reduced transmission time from email client to email server · reduced storage space on the email · email accounts often have a maximum size for an attachment. That last point is specific to email and is easily missed.
9618_w23_qp_11_sc_1.c
Success
1 mark each. Customers can download in less time · and it takes less of the customer’s bandwidth · the files take up less space on the customer’s storage medium · therefore customers can store more images · and have more space for other files. The answer must be framed from the customer’s side, not the server’s.
9618_s23_qp_11_sc_1.c.i
Success
To reduce the time it takes to download the files from the server // to upload them to the server in the first place · to reduce the storage space used on the web server // on the user’s device.
9618_w23_qp_13_sc_5.b.i
Success
Reduces the file size · faster to transmit / download · the original file is too large for email storage or as an attachment.
9618_w21_qp_11_sc_7.b.i
Give two reasons why a video does not need to be compressed [2]
The inverted question. 1 mark each to max 2. A dedicated connection to the headset // not sharing bandwidth · an already fast connection that can transmit the data without slowing · the video may already be a small file size and does not need further reduction · the video is not saved, so storage is not an issue.
9618_s24_qp_12_sc_2.d.iii
The cost of compressing
Every question asks for benefits. The trade-offs — compression and decompression take processing time, lossy compression permanently destroys data, and an already-compressed file cannot usefully be compressed again — are never asked. The “why not compress” question above (
s24_qp_12_sc_2.d.iii) is the closest, and it is argued from bandwidth rather than from the cost of compressing.
Named compression formats
Not required by the syllabus, but useful as examples: MP3 compresses audio by up to 90% using perceptual music shaping — removing frequencies outside human hearing and quieter sounds masked by louder ones; it is lossy. MP4 stores audio, video, photos and animations and is the standard for streaming. JPEG is the standard lossy bitmap format, and it creates a new file, so the original can no longer be recovered. SVG vector files are XML text, which is why they compress at all.
1.3.2 Lossy and lossless compression, and justifying a method
Success
A sound file compressed by reducing the sampling rate → lossy · because there will be fewer samples per second, so data is permanently lost // so the original sound cannot be re-created. The justification carries the mark; the label alone does not.
9618_w25_qp_12_sc_6.a
Success
No mark for the choice — but lossy is the expected answer. 1 mark each to max 3: reduces file size more than lossless · so significantly less bandwidth / data is needed · so buffering is reduced even more than with lossless · data can be removed which cannot be seen · reducing quality without impacting the experience · for example, the resolution of the video can be reduced // the sample rate of the audio can be reduced.
9618_s25_qp_11_sc_2.b.ii
Success
1 mark per bullet, max 3. Both choices are credited if justified, which is the key feature of this question. Lossy: loss of quality will not be noticed · needs to be viewed in real time, so less bandwidth is needed if the file is smaller · smaller files reduce buffering, so the video plays more smoothly · viewers may watch on different devices, so may not need high resolution. Lossless: the original recording may not have been made in high resolution · could be streaming to high bandwidth devices · the reduction in file size is sufficient for the receiving device · viewers do not want any loss of quality.
9618_w23_qp_12_sc_6.a
Success
1 mark per bullet, max 2. The change may not be noticeable // data removed is usually not noticed by the human eye · for example, changes in shade or detail · it produces a larger decrease in file size compared to lossless.
9618_w24_qp_13_sc_6.b.iii
Success
1 mark each to max 3 — and every point must be comparative. The file takes less storage space on the web server than if lossless was used · is faster to upload/download than if lossless was used · uses less bandwidth to transmit than if lossless was used · consumes less data allowance than if lossless was used.
9618_s24_qp_13_sc_2.b.i
Success
1 mark per bullet. All the data is required // no data can be lost · otherwise the text file will be corrupted / not make sense.
9618_w22_qp_12_sc_8.c.ii
Defining lossy and lossless
Both terms are used in every question; neither has ever been asked for as a definition in 9618. The definitions — lossy permanently discards data and is irreversible; lossless encodes the data so the file can be restored exactly to its original state — are assumed throughout.
Choosing lossless for a sound file
The justification questions cover: lossy for video, lossy for photographs, lossless for text. The fourth combination — lossless for audio, where the recording is a master to be edited further or quality is non-negotiable — is never set, despite sound being examined heavily everywhere else in the topic.
The two methods, defined
Lossy — data is removed to reduce the file size; irreversible; greatly reduces size at the expense of quality; suitable where some quality loss is acceptable, i.e. images, video and sound. In photographs it groups similar colours together, reducing the number of colours without visibly compromising the image. Lossless — data is encoded rather than discarded; reversible, so the file returns to its original state; reduces size less dramatically; usable on all data but essential where loss is unacceptable, i.e. documents and program files. It works by analysing the contents for patterns and repetition. DOCX and PDF apply it automatically.
1.3.3 How a text file, bitmap image, vector graphic and sound file can be compressed
Success
Run-Length Encoding (RLE).
9618_w24_qp_12_sc_7.a.iii
Success
1 mark for naming the method, 1 per description bullet to max 2. Run-length encoding · replaces sequences of the same colour pixel · with the colour code and the number of identical pixels.
9618_s21_qp_11_sc_1.b
Success
1 mark per bullet, max 2. Sequences of consecutive identical colours / pixels · are stored as the colour value and the number of times it occurs consecutively. The two credited words are consecutive and repeating.
9618_w25_qp_11_sc_7.d.iii;9618_s24_qp_13_sc_2.b.ii
Describe one lossless method of compressing a text file [3]
1 mark per bullet, max 3. Run-Length Encoding / RLE · repeated sequences of the same characters are replaced by · a single copy of the character · and a count of the number of characters. The four-bullet structure is worth copying exactly.
9618_w24_qp_11_sc_4.b.ii;9618_w22_qp_11_sc_7.c
Success
1 mark for each correctly completed part. Uncompressed
EA F1 F1 F2 F2 F2 EA→ compressed1EA 2F1 3F2 1EA. Working backwards:2AB 2FF 11D 167→AB AB FF FF 1D 67. Forwards:32 32 80 81 81→232 180 281. The question runs in both directions, so the expansion must be as fluent as the compression.9618_w22_qp_12_sc_8.c.i
Explain why RLE may not reduce the file size, with an example [3]
1 mark each to max 2 for the explanation, 1 mark for an image-related example. RLE stores a colour and the number of times it occurs consecutively · an image may not have many sequences of the same colour · it would need to store each colour and then the count / number 1, which adds data. Example: Red-Green-Blue would become Red 1 Green 1 Blue 1. The sharpest question in the sub-topic, and the only one that tests the limits of the method.
9618_s23_qp_11_sc_1.c.ii
Success
1 mark per bullet to max 2 for each method. Reduce bit depth … reduces the number of bits per colour/pixel, which means each pixel has fewer bits · reduce the colour palette // reduce the number of colours … fewer colours mean fewer bits needed to store each colour · reduce image resolution … fewer pixels per unit measurement means less binary to store.
9618_w21_qp_12_sc_5.b.ii
Success
1 mark for the point, 1 for a matching expansion. Decrease the sample rate … fewer samples / readings stored per second // fewer bits per second stored · decrease the sample resolution … fewer bits per sample // each sample has fewer bits · sound outside the set / human hearing range is removed … fewer measurements are stored // decreases the number of possible binary values so fewer bits are stored.
9618_w24_qp_13_sc_6.c.ii
Success
1 mark per bullet, max 2. Reduce the amplitude to only the range used … limited amplitudes mean fewer bits per sample · run-length encoding … where consecutive sounds are the same, record the binary value of the sound and the number of times it repeats · record the changes instead of the actual sounds. That last method — storing deltas — is credited here and nowhere else.
9618_w21_qp_11_sc_7.b.ii
Compressing a vector graphic — named in the syllabus, never asked
The bullet reads “show understanding of how a text file, bitmap image, vector graphic and sound file can be compressed”. Text, bitmap and sound each have multiple 9618 questions. Vector graphics have none, in either series. The only trace is one line inside a comparison mark scheme: “vector graphic images do not compress well because of little redundant data”. The available content: an SVG is an XML text file, so it can be compressed with the same lossless text methods (producing
.svgz); and a vector image can be simplified by reducing the number of objects or control points, which is lossy. This is the clearest gap in 1.3, and it is a one-mark name away from being examinable.
RLE expressed in binary
9618 asks for RLE in hexadecimal pairs (
w22_qp_12_sc_8.c.i) and in prose. The binary form — the count in a fixed-size field followed by the character’s code, so4Abecomes0000100 1000001— is taught by SME and has never appeared in a question, in either series.
RLE worked three ways
Text:
AAAABBBCCDAA→4A3B2C1D2A. In binary, the count is stored in a fixed-size field (7 or 8 bits) and the character as its 7-bit ASCII value, so4A=0000100 1000001. 1-bit image: frequency/data pairs, e.g.30 11 20 11 20 11 50 11 …— three white then one black, and so on. Pairs can carry across the end of a row onto the start of the next. SME’s worked image drops from 195 bits to 38 bits, a saving of about 80%. Colour image: the RGB value of each colour is stored with the count, e.g.10 0 0 0= ten black pixels,5 255 0 0= five red,3 0 255 0= three green.