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 −51 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
What is the most negative integer N that can be stored in 32-bit two's complement?
Is ((r → p) ∧ (r → q)) → (p → q) a tautology, a contradiction or a contingency? Click your answer.
Which statement is equivalent to ¬∃x ∀y R(x, y), with the negation moved all the way inside? Click it.
Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(7, 13, 14, 15). How many literals L does it contain in total (a primed variable counts as one literal)?
Find the minimal SOP form of F(A, B, C, D) = Σm(0, 1, 2, 5, 6, 8, 10, 11, 13) + d(3, 4), where d marks the don't-care conditions. How many literals L does it contain?
Solve 23x ≡ 28 (mod 73). Give the solution x with 0 < x < 73.
An RSA key uses the primes p = 19 and q = 37 and the public exponent e = 13. Find the private exponent d (0 < d < φ(n)).
Which statement about a recursive function without a reachable base case is correct?
Why does any comparison-based search in a sorted array need Ω(log n) comparisons in the worst case?
What is the worst-case height of a binary search tree with n nodes if keys are inserted in sorted order?
Without union by rank or path compression, what is the worst-case cost of a single find in a union-find structure with n elements?
A connected graph has 8 vertices and 11 edges. How many edges k must be removed to leave a spanning tree?
Find the total weight W of a minimum spanning tree of the weighted graph.
What does a common table expression (WITH name AS (...)) provide?
Two-phase locking (2PL) ensures:
Which host is in the same /16 subnet as 10.1.2.3?
In a TCP connection, what does the receive window (rwnd) control?
The network 172.17.0.0/16 is divided into equal subnets with prefix /21. How many subnets N are there?
The network 192.168.77.0/24 is split with VLSM for these LANs: A: 7 hosts, B: 6 hosts, C: 23 hosts, D: 19 hosts, E: 3 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN D?
Answer: U = 205
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 = −2.147 × 10⁹
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: contingency
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: ∀x ∃y ¬R(x, y)
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 = 9
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.
Answer: L = 11
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.
Another grouping of the same size exists; it has the same number of terms and literals.
Answer: x = 52
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: d = 349
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: C · It keeps calling itself until the call stack overflows
Without a base case the recursion never stops; in practice memory for stack frames runs out and the program crashes.
Answer: B · 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: D · n (it becomes a linked list)
Sorted insertions create a degenerate tree with height n, making searches O(n). Balanced trees prevent this.
Answer: C · O(n)
Naive unions can link trees into a long chain, so find may follow n − 1 parent pointers; the two heuristics bring it close to constant.
Answer: k = 4
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: W = 25
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: D · A named temporary result that can be used within one statement
A CTE gives a subquery a name, which makes complex queries easier to read and can be referenced several times in the same statement.
Answer: D · Conflict-serializable schedules
2PL guarantees serializability, but deadlocks remain possible and must be detected or prevented.
Answer: A · 10.1.200.5
With /16 the first 16 bits (10.1) are the network part. Only 10.1.200.5 shares them, so it is on 10.1.0.0/16.
Answer: B · How much unacknowledged data the sender may send, based on the receiver's buffer
The receiver advertises how much buffer space it has; the sender limits unacknowledged data accordingly, preventing receiver overload.
Answer: N = 32
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).
Each /21 subnet has 211 − 2 = 2046 usable hosts.
Answer: x = 32
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
LAN D: 192.168.77.32/27.
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