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 −112 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
An uncompressed image is 1024 × 768 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
You want to prove (p ∧ q) → ¬r by contrapositive. Which implication do you prove instead? Click it.
The premises r ∨ q; ¬r are given. Which conclusion follows from them? Click it.
In this Karnaugh map exactly one 1 has no neighbouring 1, so it stays a group of one cell (a full minterm of four literals) in the minimal SOP form. Neighbours across the map edges count. Click that cell.
Find the minimal SOP form of F(A, B, C, D) = Σm(2, 3, 5, 10, 11, 12) + d(4, 6), where d marks the don't-care conditions. How many product terms N does it have?
Solve 45x ≡ 1 (mod 49). Give the solution x with 0 < x < 49.
An RSA key uses p = 41, q = 43 and e = 17. Decrypt the ciphertext c = 479: find the message m.
What is the worst-case time complexity of quicksort?
A problem is NP-complete if it is:
How long does it take to check whether the edge (u, v) exists using an adjacency matrix?
Which structure lets you compute the maximum of every window of size k in an array in O(n) total?
A connected graph has 7 vertices and 14 edges. How many edges k must be removed to leave a spanning tree?
Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 4; A itself is round 1.
An employees table has a column manager_id that refers to another employee. How do you list each employee next to their manager's name?
Boyce–Codd normal form (BCNF) requires that for every non-trivial functional dependency X → Y:
HTTP is stateless. How do websites usually keep a user logged in across requests?
How does fragmentation differ in IPv6 compared with IPv4?
A subnet uses the mask 255.255.240.0. How many usable host addresses H does it have?
The network 192.168.108.0/24 is split with VLSM for these LANs: A: 10 hosts, B: 21 hosts, C: 47 hosts, D: 8 hosts, E: 4 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 = 144
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: M = 1536 KiB
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: ¬¬r → ¬(p ∧ q)
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 contrapositive is equivalent to the original statement, so proving it proves the statement.
Answer: q
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.
Answer: AB = 01, CD = 10
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.
A 1 with a neighbouring 1 can join a group of two or more.
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.
The largest groups that can be formed from 1s and Xs.
Answer: x = 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).
Answer: m = 380
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 · O(n²)
With consistently bad pivots (e.g. already sorted input and first-element pivot) partitions are unbalanced, giving O(n²). The average case is O(n log n).
Answer: C · In NP and every NP problem reduces to it in polynomial time
NP-complete problems are the hardest in NP; a polynomial algorithm for one would give one for all of NP.
Answer: C · O(1)
The matrix entry at row u, column v tells you immediately whether the edge exists; an adjacency list may need to scan u's neighbours.
Answer: B · A deque holding indices in decreasing order of value
The front of the deque is always the window's maximum; each index enters and leaves the deque once, so the work is linear.
Answer: k = 8
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: G
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 · Join the employees table with itself using two aliases
A self-join such as FROM employees e JOIN employees m ON e.manager_id = m.id lets one alias be the employee and the other the manager.
Answer: D · X is a superkey
BCNF demands that every determinant is a candidate key (superkey); 3NF allows an exception when Y is part of a candidate key.
Answer: C · With cookies holding a session identifier or token
The server sets a cookie containing a session ID or token, and the browser returns it with each request so the server can recognise the user.
Answer: D · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
An IPv6 router drops an oversized packet and returns ICMPv6 Packet Too Big. The source then sends smaller packets or uses a fragment extension header.
Answer: H = 4094
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 = 112
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: 192.168.108.112/28.
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