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 −25 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
What is the largest unsigned integer N that fits in 20 bits?
Is (q ↔ r) ∧ (q ↔ ¬r) a tautology, a contradiction or a contingency? Click your answer.
The argument with premises r → q; ¬r and conclusion ¬q is invalid. Click the row of the truth table that is a counterexample.
Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C) = Σm(0, 1, 2, 5, 7). How many product terms N does it have?
A positive-edge-triggered T flip-flop starts with Q = 0 and receives the inputs shown. Click the first clock edge at which Q goes from 1 back to 0.
Find the inverse of 10 modulo 17: the number x with 0 < x < 17 and 10·x ≡ 1 (mod 17).
An RSA key uses p = 19, q = 37 and e = 23. Decrypt the ciphertext c = 216: find the message m.
How many levels does the recursion of merge sort have for an array of n elements?
To show that a new problem X is NP-hard, what must you do?
Why can a binary heap be stored in an array without any pointers?
Kahn's algorithm for topological sorting repeatedly removes vertices with in-degree 0. Which structure holds those vertices?
Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.
Prim’s algorithm builds a minimum spanning tree starting from vertex A. Click the edge it adds as edge number 4.
Third normal form (3NF) removes:
What does EXPLAIN (or EXPLAIN ANALYZE in PostgreSQL) show for a query?
What is the usual order of messages when a client obtains an address from DHCP?
What does TCP fast retransmit do?
The network 192.168.10.0/24 must be divided into at least 3 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
A host has the address 10.20.252.56/29. What is the fourth octet x of its network address?
Answer: U = 231
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: N = 1.049 × 10⁶
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: contradiction
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).
Answer: q = 1, r = 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: N = 3
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.
Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.
Another grouping of the same size exists; it has the same number of terms and literals.
Answer: clock edge 3
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 T flip-flop toggles at each rising edge with T = 1 and holds with T = 0.
Answer: 10−1 mod 17 = 12
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·10 (mod 17).
Answer: m = 524
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: A · About log₂ n
Halving from n down to 1 takes about log₂ n levels, and each level does O(n) merging work, giving O(n log n).
Answer: A · Reduce a known NP-hard problem to X in polynomial time
If a known hard problem can be transformed into X efficiently, a fast algorithm for X would solve the known problem too.
Answer: B · It is a complete binary tree, so positions follow simple index formulas
Completeness means there are no gaps, so the children of index i are always at 2i + 1 and 2i + 2.
Answer: C · A queue (or any collection) of vertices whose in-degree is 0
Ready vertices are kept in a queue; removing one lowers its neighbours' in-degrees, and any that reach 0 are added, giving O(V + E) time.
Answer: neither
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.
Answer: edge DF
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.
Answer: C · Transitive dependencies between non-key attributes
In 3NF no non-key attribute depends on another non-key attribute, such as storing city and postcode where city depends on postcode.
Answer: A · The execution plan chosen by the optimiser, such as scans and join methods
EXPLAIN displays the plan (index or sequential scans, join algorithms, estimated costs); EXPLAIN ANALYZE also runs the query and reports actual times.
Answer: D · 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: D · Resends a segment after three duplicate ACKs, without waiting for the timeout
Three duplicate ACKs strongly suggest a single lost segment, so TCP retransmits it at once. This recovers much faster than waiting for the retransmission timer.
Answer: H = 62
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 = 56
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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