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.
Set the bits so they store −76 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
A system must give every one of 2566 items its own binary code of the same length. What is the smallest number of bits b needed?
Which formula is logically equivalent to (q ∨ r) → p? Click it.
Which statement is equivalent to ¬∃x (F(x) → G(x)), with the negation moved all the way inside? Click it.
Simplify F = B′C + A′B′C + A′BC + AB′C to its minimal sum-of-products form. How many literals L does the simplified expression contain?
A positive-edge-triggered JK flip-flop starts with Q = 0 and receives the inputs shown at 7 rising clock edges. How many times N does Q change state?
Solve 36x ≡ 33 (mod 89). Give the solution x with 0 < x < 89.
How many solutions N does 36x ≡ 24 (mod 40) have in {0, 1, …, 39}?
The time complexity of naive recursive Fibonacci (fib(n) = fib(n−1) + fib(n−2)) is:
Which of these satisfiability problems is known to be solvable in polynomial time?
How much memory does an adjacency list use for a graph with V vertices and E edges?
How does a radix tree (compressed trie) improve on a plain trie?
A tree has 4 vertices of degree 3, 4 vertices of degree 4, 3 vertices of degree 5, 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(D) of the shortest path from A to D in the weighted graph.
Second normal form (2NF) removes:
After customers LEFT JOIN orders, why use COUNT(o.id) rather than COUNT(*) to count orders per customer?
What is the usual order of messages when a client obtains an address from DHCP?
DNS queries normally use UDP. When does DNS typically use TCP instead?
A department needs 1990 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
A host has the address 10.20.110.254/28. What is the fourth octet x of its network address?
Answer: U = 180
Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.
Answer: b = 12 bit
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 → p) ∧ (r → 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: ∀x (F(x) ∧ ¬G(x))
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.
The other options either keep a quantifier unchanged, change the order of the quantifiers, or negate the inner formula wrongly.
Answer: L = 4
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 = 4
Flip-flops and counters: a flip-flop changes only at the active clock edge. D copies its input (Q⁺ = D); T toggles when T = 1; JK holds for 00, resets for 01, sets for 10 and toggles for 11. An n-bit binary counter has 2ⁿ states and counts modulo 2ⁿ; a mod-N counter goes 0, 1, …, N − 1 and back to 0, and needs the smallest n with 2ⁿ ≥ N flip-flops. Each stage of a ripple counter halves the frequency, so the last output of a mod-N counter runs at fclk/N. A Johnson counter of n flip-flops has 2n states, a ring counter n.
A JK flip-flop: 00 holds, 01 resets (Q = 0), 10 sets (Q = 1), 11 toggles.
Answer: x = 38
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).
Answer: N = 4
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).
Answer: B · Exponential
The call tree grows roughly like φⁿ (about 1.618ⁿ). Memoisation reduces it to O(n).
Answer: B · 2-SAT, where each clause has at most two literals
2-SAT is in P (linear time via strongly connected components), while 3-SAT is NP-complete.
Answer: B · O(V + E)
Each vertex has a list and each edge appears once (directed) or twice (undirected), giving O(V + E).
Answer: D · It merges chains of single-child nodes into one edge labelled with a string
Collapsing paths with no branching reduces the number of nodes and the memory used while keeping prefix search.
Answer: L = 23
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(D) = 13
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: C · Partial dependencies of non-key columns on part of a composite key
2NF requires 1NF and that every non-key attribute depends on the whole key, not just part of it.
Answer: C · COUNT(*) counts the NULL-filled row as 1 for customers without orders
The left join gives one row with NULL order columns for such customers; COUNT(o.id) skips the NULL and correctly reports 0.
Answer: B · Discover, Offer, Request, Acknowledge
The client broadcasts Discover, servers reply with Offers, the client Requests one offer and the chosen server confirms with an Acknowledge.
Answer: C · For responses too large for UDP and for zone transfers
If a UDP response is truncated, the resolver retries over TCP. Zone transfers between name servers also use TCP because they carry lots of data.
Answer: p = 21
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 = 240
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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