Number systems
- Place values
- binary 2k, octal 8k, hex 16k
- To decimal
- Σ dk·bk
- From decimal
- divide by the base, read the remainders bottom-up
- Hex ↔ binary
- one hex digit = 4 bits
- Two’s complement
- invert the bits and add 1
Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
Write −77 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
What is the largest unsigned integer N that fits in 10 bits?
Which formula is logically equivalent to q ∧ (r ∨ p)? Click it.
The argument with premises q → p; p and conclusion q is invalid. Click the row of the truth table that is a counterexample.
Simplify F = A′B + B′C + AC to its minimal sum-of-products form. How many literals L does the simplified expression contain?
For F(A, B, C, D) = Σm(0, 2, 3, 4, 5, 8, 9, 11, 13, 15) + d(6, 10, 14), where d marks the don't-care conditions, find the number N of essential prime implicants.
Find the inverse of 9 modulo 29: the number x with 0 < x < 29 and 9·x ≡ 1 (mod 29).
Use repeated squaring to find 32116 mod 77.
The time complexity of naive recursive Fibonacci (fib(n) = fib(n−1) + fib(n−2)) is:
Why does any comparison-based search in a sorted array need Ω(log n) comparisons in the worst case?
Why is traversing an array often much faster in practice than traversing a linked list of the same length?
A dynamic array doubles when full. Why do many implementations shrink it only when it is one quarter full, not half full?
A tree has 4 vertices of degree 3, 3 vertices of degree 4, 3 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?
Use Dijkstra’s algorithm to find the length d(F) of the shortest path from A to F in the weighted graph.
Why does SELECT price * 2 AS double_price FROM t WHERE double_price > 10 fail in most databases?
For an equality join between two large tables with no useful indexes, which join algorithm does an optimiser often choose?
At which OSI layer does a switch normally operate?
How does a content delivery network (CDN) mainly improve web performance?
The network 192.168.10.0/24 must be divided into at least 12 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
A host has the address 172.20.48.197/19. What is the third octet x of its subnet's broadcast address?
Answer: U = 179
Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.
Answer: N = 1023
Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.
Answer: (q ∧ r) ∨ (q ∧ p)
Logical equivalence: two formulas are equivalent (≡) when they have the same value in every row of the truth table. A tautology is 1 in every row, a contradiction is 0 in every row, and a contingency is neither. Useful laws: De Morgan, ¬(a ∧ b) ≡ ¬a ∨ ¬b and ¬(a ∨ b) ≡ ¬a ∧ ¬b; a → b ≡ ¬a ∨ b ≡ ¬b → ¬a (contrapositive); ¬(a → b) ≡ a ∧ ¬b; and the distributive laws, a ∨ (b ∧ c) ≡ (a ∨ b) ∧ (a ∨ c).
The columns after the bar are the options in the order shown.
Answer: p = 1, q = 0
Quantifiers and arguments: ¬∀x P(x) ≡ ∃x ¬P(x) and ¬∃x P(x) ≡ ∀x ¬P(x). Move a negation inwards, swap every quantifier it passes, and negate the inner formula with the propositional laws, such as ¬(a → b) ≡ a ∧ ¬b. An argument is valid when the conclusion is 1 in every row in which all premises are 1. A single row with true premises and a false conclusion is a counterexample and shows the argument is invalid.
An argument with this form looks like a valid rule, but one row is enough to show that it is not.
Answer: L = 3
Boolean algebra: the rules that simplify expressions are X + X′ = 1 and X·X′ = 0, X + X = X, absorption X + X·Y = X, the combining rule X·Y + X·Y′ = X, X + X′Y = X + Y and the consensus theorem XY + X′Z + YZ = XY + X′Z. De Morgan's laws turn (X·Y)′ into X′ + Y′ and (X + Y)′ into X′·Y′. A Karnaugh map applies these rules graphically: every group of two neighbouring cells is one use of XY + XY′ = X.
Answer: N = 1
Karnaugh maps: the cells are ordered in Gray code (00, 01, 11, 10), so neighbouring cells, including the ones across the edges, differ in exactly one variable. Group the 1s in rectangles of 1, 2, 4, 8 or 16 cells, as large as possible and as few as possible, until every 1 is covered. Each group is one product term: the variables that stay constant inside it. A group of 2ᵏ cells loses k variables. Don't-cares (X) may be used to make groups bigger but never have to be covered. Grouping the 0s instead gives the minimal POS form.
The largest groups that can be formed from 1s and Xs.
Answer: 9−1 mod 29 = 13
Inverses and congruences: a has an inverse modulo n exactly when gcd(a, n) = 1. The extended Euclidean algorithm writes 1 = s·n + t·a, and then a⁻¹ ≡ t (mod n). A congruence a·x ≡ b (mod n) with gcd(a, n) = 1 has the solution x ≡ a⁻¹·b. Two congruences x ≡ r₁ (mod m₁) and x ≡ r₂ (mod m₂) with coprime moduli have one solution modulo m₁·m₂ (Chinese remainder theorem). Euler’s φ(n) counts the numbers from 1 to n that are coprime to n: φ(n) = n·∏(1 − 1/p).
Each remainder r is written as r ≡ t·9 (mod 29).
Answer: 32116 mod 77 = 23
Modular powers and RSA: compute aᵏ mod n by repeated squaring: write k in binary, square, reduce after every step and multiply the powers whose bit is 1. Fermat: aᵖ⁻¹ ≡ 1 (mod p) for a prime p not dividing a, so the exponent can be reduced modulo p − 1. RSA: choose primes p and q, n = p·q and φ(n) = (p − 1)(q − 1); the public exponent e is coprime to φ(n), and the private exponent is d = e⁻¹ mod φ(n). Encrypt with c = mᵉ mod n and decrypt with m = cᵈ mod n.
Answer: D · Exponential
The call tree grows roughly like φⁿ (about 1.618ⁿ). Memoisation reduces it to O(n).
Answer: D · A decision tree with n possible outcomes must have depth at least log₂ n
Each comparison at most halves the set of remaining possibilities, so distinguishing n positions needs at least log₂ n comparisons.
Answer: B · Array elements are contiguous, so they make good use of the CPU cache
Contiguous memory gives good locality of reference, while linked nodes may be scattered and cause many cache misses.
Answer: B · To avoid repeated resizing when pushes and pops alternate near the boundary
Shrinking at one half could trigger a resize on every operation in such a sequence; the gap between 1/4 and full keeps the amortised cost O(1).
Answer: L = 12
Euler, Hamilton and trees: a connected graph has an Euler circuit (every edge once, back to the start) when every vertex has even degree, and an Euler path that is not a circuit when exactly two vertices have odd degree; the path then runs from one odd vertex to the other. A Hamilton circuit visits every vertex once; Kn has (n − 1)!/2 of them. A tree on n vertices has n − 1 edges, and a connected graph with m edges keeps a spanning tree after removing m − n + 1 edges.
Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.
Answer: d(F) = 29
Shortest paths and spanning trees: Dijkstra’s algorithm settles one vertex per round: the unsettled vertex with the smallest tentative distance. Its distance is then final, and every neighbour v gets d(v) = min(d(v), d(u) + w(u, v)). Kruskal’s algorithm builds a minimum spanning tree by taking the edges from light to heavy and skipping any edge that would close a cycle; Prim’s algorithm grows one tree from a start vertex by always adding the lightest edge that leaves it. A spanning tree of n vertices has n − 1 edges.
Each line: the vertex settled, its final distance, and the neighbours whose distance improves.
Answer: D · WHERE is evaluated before the SELECT list, so the alias does not exist yet
Logically FROM and WHERE run before SELECT, so repeat the expression in WHERE or use a subquery; ORDER BY can use the alias because it runs after SELECT.
Answer: B · Hash join
A hash join builds a hash table on one input and probes it with the other, which works well for large equi-joins without indexes.
Answer: A · Data link (layer 2)
A standard switch learns MAC addresses and forwards frames at layer 2. Layer 3 switches also route.
Answer: B · It serves cached content from servers close to the users
CDNs place copies of content at many edge locations and direct users, often via DNS, to a nearby server. This cuts latency and load on the origin.
Answer: H = 14
IPv4 subnetting: an address has 32 bits. The prefix /p says how many leading bits identify the network; the remaining 32 − p bits number the hosts. A subnet has 2³²⁻ᵖ addresses, but the first (network address) and last (broadcast address) cannot be given to hosts, so 2³²⁻ᵖ − 2 hosts fit. Exceptions: a /31 (point-to-point link, RFC 3021) uses both of its 2 addresses for hosts, and a /32 is one single address (a single host or a loopback).
Answer: x = 63
IPv4 subnetting: an address has 32 bits. The prefix /p says how many leading bits identify the network; the remaining 32 − p bits number the hosts. A subnet has 2³²⁻ᵖ addresses, but the first (network address) and last (broadcast address) cannot be given to hosts, so 2³²⁻ᵖ − 2 hosts fit. Exceptions: a /31 (point-to-point link, RFC 3021) uses both of its 2 addresses for hosts, and a /32 is one single address (a single host or a loopback).
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