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For example, binary has two digit values, 0 and 1, so its number base is 2."},{"text":"The highest digit value available in a number system.","correct":false,"explanation":"This confuses the radix with the largest permitted digit. The radix is the number of available digit values, so base 10 has ten values even though its highest digit is 9."},{"text":"The number of digit positions available in a number system.","correct":false,"explanation":"Digit positions determine how many digits a particular representation contains, not its radix. The radix is the number of distinct digit values available in the number system."},{"text":"The number of bits per digit available in a number system.","correct":false,"explanation":"Bits per digit describes how symbols may be encoded, rather than defining the number base. 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Four bits provide 16 combinations, so one hex digit maps to four bits."},{"text":"One hex digit maps exactly to sixteen bits.","correct":false,"explanation":"This mistakes the number of hexadecimal values for the number of required bits. Hex has 16 digit values, represented by $2^4$ combinations, so each digit maps to four bits."}]},{"id":"a12f7808-692f-40f2-80a4-5c83a99725ac","cardType":"basic","points":1,"front":"What is the binary carry rule for $1 + 1 + carry 1$?","back":"$$11_2$ (write 1 and carry 1).","mcqOptions":[{"text":"$$11_2$ (write 1 and carry 1).","correct":true,"explanation":"The three binary ones total decimal 3, which is $11_2$. The rightmost 1 is written in the current column and the other 1 is carried."},{"text":"$$10_2$ (write 0 and carry 1).","correct":false,"explanation":"This applies the rule for $1+1$ but ignores the incoming carry. Including all three ones gives decimal 3, or $11_2$, so write 1 and carry 1."},{"text":"$$01_2$ (write 1 and carry 0).","correct":false,"explanation":"This treats the three input ones as though their total were only 1. Their total is decimal 3, or $11_2$, so both a written 1 and a carried 1 are required."},{"text":"$$100_2$ (write 0 and carry 10).","correct":false,"explanation":"This overstates the result as decimal 4 rather than decimal 3. The sum $1+1+1$ equals $11_2$, requiring a written 1 and a carry of 1."}]},{"id":"b3da6f79-372b-4e1d-a5e8-d50ab0e3c4f3","cardType":"basic","points":1,"front":"What is the standard technique for performing binary subtraction?","back":"Add the two's complement of the number being subtracted.","mcqOptions":[{"text":"Add the two's complement of the number being subtracted.","correct":true,"explanation":"Taking a number's two's complement produces its negative in a fixed-width binary representation. Adding that negative performs the required subtraction using ordinary binary addition."},{"text":"Subtract the two's complement of the number being subtracted.","correct":false,"explanation":"Subtracting the two's complement effectively subtracts a negative value, producing addition instead. Binary subtraction is performed by adding the two's complement of the number being subtracted."},{"text":"Add the one's complement of the number being subtracted.","correct":false,"explanation":"This omits the final addition of 1 needed to form the two's complement, creating an off-by-one result. The correct method adds the number's two's complement."},{"text":"Add the two's complement of the number being subtracted from.","correct":false,"explanation":"This complements the wrong operand: the minuend rather than the subtrahend. To calculate a subtraction, add the two's complement of the number being subtracted."}]},{"id":"b69cd57b-4df2-4610-ac1a-ed848816449e","cardType":"cloze","points":null,"front":"To find the two's complement of a binary number, [ ] then add 1.","back":"To find the two's complement of a binary number, invert the bits then add 1.","mcqOptions":[{"text":"invert the bits","correct":true,"explanation":"Two's complement is formed by changing every 0 to 1 and every 1 to 0, then adding 1 to the inverted binary value."},{"text":"reverse the bits","correct":false,"explanation":"Reversing changes the order of the bits rather than switching their values. To form the two's complement, invert every bit and then add 1."},{"text":"shift the bits","correct":false,"explanation":"Shifting moves bits between place-value positions and usually multiplies or divides the value by two. Two's complement instead requires inverting every bit, then adding 1."},{"text":"clear the bits","correct":false,"explanation":"Clearing the bits would replace them with zeros and destroy the original pattern. The required first step is to invert each bit, then add 1."}]},{"id":"7d9ce601-6040-4c60-a6bb-f45767aeda0f","cardType":"cloze","points":null,"front":"To find the two's complement of a binary number, invert the bits then [ ].","back":"To find the two's complement of a binary number, invert the bits then add 1.","mcqOptions":[{"text":"add 1","correct":true,"explanation":"After every bit has been inverted, adding 1 completes the two's complement operation. This produces the negative representation of the original value at the chosen bit width."},{"text":"subtract 1","correct":false,"explanation":"Subtracting 1 after inversion does not produce the two's complement and causes the wrong magnitude. The required operation is to invert all bits and then add 1."},{"text":"add 0","correct":false,"explanation":"Adding 0 leaves only the one's complement, so the resulting value is off by one. A two's complement is formed by inverting the bits and adding 1."},{"text":"shift left","correct":false,"explanation":"A left shift changes every bit's place value and generally doubles an unsigned value. It is not part of complementation; after inversion, the correct step is to add 1."}]},{"id":"27c196fb-f855-45ad-875c-6ca5a78c4742","cardType":"basic","points":2,"front":"How does the leftmost bit differ between sign-and-magnitude and two's complement?","back":"

Sign-and-magnitude: a sign marker only
Two's complement: a negative place value

","mcqOptions":[{"text":"

Sign-and-magnitude: a sign marker only
Two's complement: a negative place value

","correct":true,"explanation":"In sign-and-magnitude, the leftmost bit separately identifies the sign. In two's complement, that bit contributes a negative place value, such as $-128$ in an 8-bit number."},{"text":"

Sign-and-magnitude: a negative place value only
Two's complement: a sign marker

","correct":false,"explanation":"This reverses the roles of the leftmost bit. Sign-and-magnitude uses it only as a sign marker, whereas two's complement assigns it a negative place value."},{"text":"

Sign-and-magnitude: a sign marker only
Two's complement: a positive place value

","correct":false,"explanation":"The sign-and-magnitude description is correct, but the two's complement bit is not positive. Its negative place value allows negative numbers to be represented and processed using binary addition."},{"text":"

Sign-and-magnitude: a positive place value only
Two's complement: a sign marker

","correct":false,"explanation":"Neither description is correct. Sign-and-magnitude reserves the leftmost bit as a sign marker, while two's complement makes it a negative place value within the number itself."}]},{"id":"19ae48eb-d7fc-416e-9124-fdd0f921f3db","cardType":"basic","points":2,"front":"What do the mantissa and exponent represent in floating point?","back":"

Mantissa: the significant digits
Exponent: the position of the binary point

","mcqOptions":[{"text":"

Mantissa: the significant digits
Exponent: the position of the binary point

","correct":true,"explanation":"The mantissa stores the meaningful digits and therefore controls precision. The exponent scales the number by specifying how far the binary point is effectively moved."},{"text":"

Mantissa: the position of the binary point
Exponent: the significant digits

","correct":false,"explanation":"This swaps the two components. The mantissa contains the significant digits and determines precision, while the exponent determines the binary point's effective position and the number's scale."},{"text":"

Mantissa: the integer part
Exponent: the fractional part

","correct":false,"explanation":"This incorrectly treats floating-point storage as a fixed split between integer and fractional parts. The mantissa stores significant digits, while the exponent controls the binary point's position."},{"text":"

Mantissa: the range of values
Exponent: the precision of the value

","correct":false,"explanation":"This reverses their practical effects and does not state what they represent. The mantissa holds significant digits, affecting precision, while the exponent positions the binary point, affecting range."}]},{"id":"8290ba7e-be59-43f4-bbd4-556e1d996023","cardType":"basic","points":1,"front":"How does increasing the number of exponent bits affect the representation?","back":"Increases the range of values that can be stored.","mcqOptions":[{"text":"Increases the range of values that can be stored.","correct":true,"explanation":"Exponent bits determine the scale of floating-point numbers, so adding them allows much smaller and larger values to be represented. The mantissa primarily controls precision."},{"text":"Increases the precision of the stored values.","correct":false,"explanation":"This confuses exponent bits with mantissa bits. Mantissa bits increase precision, whereas increasing exponent bits increases the range of values that can be stored."},{"text":"Increases the accuracy of the stored values.","correct":false,"explanation":"Accuracy is not the main property controlled by exponent length. Exponent bits increase the range of representable values; mantissa bits determine how precisely values are stored."},{"text":"Decreases the range of values that can be stored.","correct":false,"explanation":"Adding exponent bits provides more possible exponent values, extending rather than reducing the representable range. The correct effect is an increased range of stored values."}]},{"id":"6a889c66-cae6-4c20-9581-08f8df1f130c","cardType":"basic","points":1,"front":"How does increasing the number of mantissa bits affect the representation?","back":"Increases the precision of the stored values.","mcqOptions":[{"text":"Increases the precision of the stored values.","correct":true,"explanation":"Mantissa bits store the significant digits of a floating-point number. More mantissa bits allow more digits to be represented, increasing the precision of stored values."},{"text":"Increases the range of values that can be stored.","correct":false,"explanation":"This describes the role of exponent bits, which control scale and range. Increasing mantissa bits instead increases the precision of the stored values."},{"text":"Decreases the precision of the stored values.","correct":false,"explanation":"Adding mantissa bits gives the number more significant digits, so precision increases rather than decreases. The correct effect is increased precision of the stored values."},{"text":"Increases the number of signs that can be stored.","correct":false,"explanation":"The sign bit determines whether a value is positive or negative, not the mantissa. More mantissa bits increase the precision of the stored values."}]},{"id":"7980a29f-4d20-4413-ad53-cbfd06ecd47c","cardType":"cloze","points":null,"front":"A normalised positive mantissa begins with the bits [ ].","back":"A normalised positive mantissa begins with the bits 0.1.","mcqOptions":[{"text":"0.1","correct":true,"explanation":"A normalised positive binary mantissa begins with 0.1. This ensures the first significant bit is 1 and gives the representation a standard, non-redundant form."},{"text":"1.0","correct":false,"explanation":"The pattern 1.0 is used for a normalised negative mantissa in this representation. A normalised positive mantissa begins with 0.1."},{"text":"0.0","correct":false,"explanation":"Starting with 0.0 would indicate an additional leading zero rather than a normalised positive mantissa. The required beginning is 0.1, with the first significant bit immediately after the binary point."},{"text":"1.1","correct":false,"explanation":"The pattern 1.1 does not represent the required sign convention for a normalised positive mantissa. The standard positive beginning is 0.1."}]},{"id":"6ff6fd8b-5461-4612-bd73-3999c973519c","cardType":"cloze","points":null,"front":"A normalised negative mantissa begins with the bits [ ].","back":"A normalised negative mantissa begins with the bits 1.0.","mcqOptions":[{"text":"1.0","correct":true,"explanation":"A normalised negative binary mantissa begins with 1.0 in this sign-and-magnitude-style representation. The leading 1 indicates the negative sign, followed by the binary point and zero."},{"text":"0.1","correct":false,"explanation":"The pattern 0.1 is the beginning of a normalised positive mantissa, not a negative one. A normalised negative mantissa begins with 1.0."},{"text":"1.1","correct":false,"explanation":"Although the leading 1 suggests a negative value, the normalised negative form begins specifically with 1.0. The second bit must therefore be zero, not one."},{"text":"0.0","correct":false,"explanation":"A beginning of 0.0 represents neither the required normalised negative pattern nor a significant leading bit. The correct beginning for a negative mantissa is 1.0."}]},{"id":"f0c496c6-4186-406b-ae4f-f56ffceaa9fa","cardType":"basic","points":1,"front":"Identify the OCR symbols for AND, OR, and XOR.","back":"

AND: $∧$
OR: $∨$
XOR: $⊻$

","mcqOptions":[{"text":"

AND: $∧$
OR: $∨$
XOR: $⊻$

","correct":true,"explanation":"The OCR symbols are ∧ for AND, ∨ for OR, and ⊻ for XOR. These symbols correspond to the three logical operations in the stated order."},{"text":"

AND: $∨$
OR: $∧$
XOR: $⊻$

","correct":false,"explanation":"The AND and OR symbols have been swapped. OCR uses ∧ for AND and ∨ for OR, while ⊻ correctly represents XOR."},{"text":"

AND: $∧$
OR: $⊻$
XOR: $∨$

","correct":false,"explanation":"The symbols for OR and XOR have been exchanged. The correct mapping is ∧ for AND, ∨ for OR, and ⊻ for XOR."},{"text":"

AND: $⊻$
OR: $∨$
XOR: $∧$

","correct":false,"explanation":"This assigns the XOR symbol to AND and the AND symbol to XOR. The correct mapping is ∧ for AND, ∨ for OR, and ⊻ for XOR."}]},{"id":"52dca29b-da02-4cf8-be19-a0ac7c74d3e6","cardType":"basic","points":1,"front":"State De Morgan’s Law for the expression $¬(A ∧ B)$.","back":"$$¬A ∨ ¬B$","mcqOptions":[{"text":"$$¬A ∨ ¬B$","correct":true,"explanation":"De Morgan’s Law changes AND to OR and negates each individual term. Therefore, negating the whole expression ¬(A ∧ B) gives ¬A ∨ ¬B."},{"text":"$$¬A ∧ ¬B$","correct":false,"explanation":"This negates both terms but leaves AND unchanged. De Morgan’s Law also changes AND to OR, so the correct result is ¬A ∨ ¬B."},{"text":"$$¬A ∨ B$","correct":false,"explanation":"Only A has been negated, so the expression is not equivalent to the original negated bracket. Both terms must be negated and AND must become OR: ¬A ∨ ¬B."},{"text":"$$A ∨ B$","correct":false,"explanation":"The individual terms have not been negated, even though the whole bracket is negated. De Morgan’s Law requires both negations and changes AND to OR, giving ¬A ∨ ¬B."}]},{"id":"aad227ff-0edc-4208-b5c2-56e3d2f11289","cardType":"basic","points":1,"front":"Apply the Distribution Law to $A ∧ (B ∨ C)$.","back":"$$(A ∧ B) ∨ (A ∧ C)$","mcqOptions":[{"text":"$$(A ∧ B) ∨ (A ∧ C)$","correct":true,"explanation":"The Distribution Law distributes A across the bracket: A is combined separately with B and C, and the resulting terms are joined with OR, giving (A ∧ B) ∨ (A ∧ C)."},{"text":"$$(A ∨ B) ∧ (A ∨ C)$","correct":false,"explanation":"This uses the other distribution pattern, distributing OR rather than the given AND. Applying the law to A ∧ (B ∨ C) gives (A ∧ B) ∨ (A ∧ C)."},{"text":"$$(A ∧ B) ∧ (A ∧ C)$","correct":false,"explanation":"The two distributed terms must be joined by OR because the original bracket contains B ∨ C. Replacing that connective with AND changes the expression; the correct result uses ∨."},{"text":"$$A ∧ (B ∧ C)$","correct":false,"explanation":"This changes the OR inside the bracket to AND instead of distributing A across both terms. The correct expansion is (A ∧ B) ∨ (A ∧ C)."}]},{"id":"b4bba1d0-94a9-427a-ac03-bd2285061801","cardType":"basic","points":1,"front":"State the Double Negation law.","back":"$$¬¬A ≡ A$","mcqOptions":[{"text":"$$¬¬A ≡ A$","correct":true,"explanation":"Double Negation states that applying NOT twice returns the original Boolean value. Therefore, ¬¬A is logically equivalent to A, written ¬¬A ≡ A."},{"text":"$$¬¬A ≡ ¬A$","correct":false,"explanation":"This removes only one of the two negations. Double Negation removes both, leaving the original expression, so the correct law is ¬¬A ≡ A."},{"text":"$$¬¬A ≡ ¬¬¬A$","correct":false,"explanation":"The expression on the right still contains three negations and is not generally equivalent to the original double negation. Applying Double Negation gives ¬¬A ≡ A."},{"text":"$$¬¬A ≡ ¬(A ∧ A)$","correct":false,"explanation":"This introduces an AND operation that was not present and does not express the Double Negation law. Two NOT operations cancel, giving ¬¬A ≡ A."}]},{"id":"f41c1648-60f4-4e1f-b3df-70b5419481f0","cardType":"basic","points":2,"front":"What are the rules for valid Karnaugh map groups?","back":"

Rectangles of size powers of 2 ($1, 2, 4, 8...$)
Groups can wrap around edges

","mcqOptions":[{"text":"

Rectangles of size powers of 2 ($1, 2, 4, 8...$)
Groups can wrap around edges

","correct":true,"explanation":"Valid groups contain a power-of-two number of adjacent cells and form rectangles. Opposite edges are considered adjacent, so groups may wrap around the map."},{"text":"

Rectangles of size multiples of 2 ($2, 4, 6, 8...$)
Groups cannot wrap around edges

","correct":false,"explanation":"Confusing powers with multiples incorrectly permits sizes such as 6 and excludes single cells. Groups must have sizes $1, 2, 4, 8...$, and they may wrap around edges."},{"text":"

Squares only of size powers of 2 ($1, 4, 16, 64...$)
Groups can connect diagonally

","correct":false,"explanation":"Karnaugh map groups may be rectangular, not only square, and diagonal cells are not adjacent. Valid rectangles contain $1, 2, 4, 8...$ cells and may wrap around edges."},{"text":"

Rectangles of size powers of 3 ($1, 3, 9, 27...$)
Groups must be separate from each other

","correct":false,"explanation":"Groups use powers of 2 rather than powers of 3, and overlapping groups are permitted. Valid rectangular groups may also wrap around the map’s edges."}]},{"id":"5c01e687-c2b6-49cd-9b52-01118d8b2f6a","cardType":"basic","points":1,"front":"What does Big-O notation measure?","back":"The growth rate of resource use (time/space) as input size $n$ increases.","mcqOptions":[{"text":"The growth rate of resource use (time/space) as input size $n$ increases.","correct":true,"explanation":"Big-O describes how an algorithm’s time or space requirements scale as the input grows. It focuses on the dominant growth pattern rather than exact resource usage."},{"text":"The exact amount of resource use (time/space) for input size $n$.","correct":false,"explanation":"Big-O does not give an exact resource count because constants and lower-order terms are omitted. It describes the growth rate of time or space as $n$ increases."},{"text":"The input limit of a program (data/storage) before size $n$ is reached.","correct":false,"explanation":"This confuses complexity with a system’s capacity limit. Big-O measures how time or space resource use grows as the input size $n$ increases."},{"text":"The processor speed needed (cycles/seconds) as input size $n$ increases.","correct":false,"explanation":"Big-O is independent of a particular processor’s clock speed. It measures the growth rate of an algorithm’s time or space requirements as $n$ increases."}]},{"id":"f56fca52-64e4-4010-a6ad-28a71ee71d34","cardType":"cloze","points":null,"front":"Big-O notation simplifies comparisons by ignoring [ ] and smaller terms.","back":"Big-O notation simplifies comparisons by ignoring constants and smaller terms.","mcqOptions":[{"text":"constants","correct":true,"explanation":"Constant factors do not affect the long-term growth class, so Big-O omits them. For example, both $n$ and $5n$ have linear complexity $O(n)$."},{"text":"variables","correct":false,"explanation":"Variables such as $n$ determine how resource use changes as input grows, so they cannot be discarded. Big-O ignores constants and lower-order terms instead."},{"text":"exponents","correct":false,"explanation":"Exponents strongly affect growth and must not be ignored: $O(n)$ and $O(n^2)$ represent different complexity classes. Big-O ignores constants and smaller terms."},{"text":"operations","correct":false,"explanation":"The number of operations is used to derive an algorithm’s growth rate rather than being automatically ignored. Big-O specifically removes constant factors and smaller terms."}]},{"id":"374df412-1063-4f86-b0cb-d1866496dd6b","cardType":"cloze","points":null,"front":"Big-O notation simplifies comparisons by ignoring constants and [ ].","back":"Big-O notation simplifies comparisons by ignoring constants and smaller terms.","mcqOptions":[{"text":"smaller terms","correct":true,"explanation":"As $n$ becomes large, the highest-order term dominates growth. Big-O therefore omits lower-order, smaller terms as well as constant factors."},{"text":"largest terms","correct":false,"explanation":"The largest, dominant term determines the complexity class and must be retained. Big-O ignores constants and smaller terms because their relative effect diminishes as $n$ grows."},{"text":"dominant terms","correct":false,"explanation":"Discarding the dominant term would remove the part that determines the growth rate. Big-O retains that term while ignoring constants and smaller terms."},{"text":"input variables","correct":false,"explanation":"The input variable $n$ is essential because complexity describes scaling as $n$ increases. Big-O simplifies expressions by ignoring constants and smaller terms, not input variables."}]},{"id":"a281f2be-d98a-4bc0-8807-3b15d1455416","cardType":"basic","points":1,"front":"What is the typical time complexity of an algorithm with one loop from $0$ to $n-1$?","back":"$$O(n)$ (linear time).","mcqOptions":[{"text":"$$O(n)$ (linear time).","correct":true,"explanation":"The loop performs one iteration for each value from $0$ to $n-1$, giving $n$ iterations. Its running time therefore grows linearly with $n$."},{"text":"$$O(1)$ (constant time).","correct":false,"explanation":"Treating the loop as one instruction ignores that its body runs $n$ times. The number of operations grows with the input, so the complexity is $O(n)$."},{"text":"$$O(\\log n)$ (logarithmic time).","correct":false,"explanation":"Logarithmic complexity occurs when the remaining problem is repeatedly divided, as in binary search. This loop advances one item at a time, so it is $O(n)$."},{"text":"$$O(n^2)$ (polynomial time).","correct":false,"explanation":"Quadratic time usually results from two nested loops that each run $n$ times. A single loop performs only $n$ iterations, giving $O(n)$."}]},{"id":"be7060a5-ac56-443f-8d60-194b3302a41b","cardType":"basic","points":1,"front":"What is the time complexity of two nested loops, both iterating $n$ times?","back":"$$O(n^2)$ (polynomial time).","mcqOptions":[{"text":"$$O(n^2)$ (polynomial time).","correct":true,"explanation":"For each of the outer loop’s $n$ iterations, the inner loop runs $n$ times. Multiplying them gives $n \\times n=n^2$ operations."},{"text":"$$O(n)$ (linear time).","correct":false,"explanation":"Adding the loop lengths would apply to two sequential loops, not nested loops. Nesting multiplies their iteration counts, producing $n \\times n=n^2$ and therefore $O(n^2)$."},{"text":"$$O(\\log n)$ (logarithmic time).","correct":false,"explanation":"Neither loop repeatedly halves the problem, so logarithmic growth does not apply. Two loops nested over $n$ items execute the body $n^2$ times."},{"text":"$$O(n^3)$ (polynomial time).","correct":false,"explanation":"Cubic complexity would normally require three nested loops that each run $n$ times. Two such loops multiply to $n^2$, giving $O(n^2)$."}]},{"id":"1d339bbc-5e7b-4eb5-9461-3d2eeeef5e65","cardType":"basic","points":2,"front":"Under what conditions is binary search $O(\\log n)$?","back":"

The data is sorted
The middle item can be accessed efficiently

","mcqOptions":[{"text":"

The data is sorted
The middle item can be accessed efficiently

","correct":true,"explanation":"Sorting allows comparison with the middle item to eliminate half the remaining search area. Efficient middle access ensures each halving step does not add linear overhead."},{"text":"

The data is unsorted
The middle item can be accessed efficiently

","correct":false,"explanation":"Efficient middle access is insufficient if the data is unsorted, because a comparison cannot identify which half to discard. Binary search requires sorted data and efficient middle access."},{"text":"

The data is sorted
The items can only be accessed sequentially

","correct":false,"explanation":"Although sorting is required, sequential access makes repeatedly reaching the middle expensive and can remove the logarithmic advantage. The middle item must be accessed efficiently."},{"text":"

The data is hashed
The first item can be accessed efficiently

","correct":false,"explanation":"Hashing supports a different lookup method and does not provide the ordered halves needed for binary search. Binary search requires sorted data and efficient access to the middle item."}]}]}],"examCountdown":{"examDate":"2027-05-11","paperLabel":"Paper 1"},"heading":"Mathematical skills","title":"Mathematical skills Flashcards","topbar":{"backHref":"/as-level/computer-science/ocr/flashcards"},"topicSidebar":"$L36"}]}],"topicId":"a0f7c825-892f-5923-b584-f64707b2ca74","walled":"$L37"}],"$L38","$L39"]