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.
The 8-bit pattern 1001 0110 stores a signed integer in two's complement. What is its value N in decimal?
An uncompressed image is 512 × 512 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
You want to prove ¬r → ¬q by contrapositive. Which implication do you prove instead? Click it.
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(1, 2, 6, 7, 9, 12). How many literals L does it contain in total (a primed variable counts as one literal)?
For F(A, B, C, D) = Σm(5, 9, 12, 13) + d(0, 2, 3, 4, 11), where d marks the don't-care conditions, find the number N of essential prime implicants.
Find the inverse of 14 modulo 23: the number x with 0 < x < 23 and 14·x ≡ 1 (mod 23).
37 is prime. Use Fermat’s little theorem to find 1856515 mod 37.
A recurrence T(n) = 2T(n/2) + O(n) solves to:
What are the time and space costs of computing the n-th Fibonacci number with a loop that keeps only the last two values?
What balance condition does an AVL tree maintain at every node?
Which problem of linear probing does double hashing mainly reduce?
Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.
Find the total weight W of a minimum spanning tree of the weighted graph.
Why can a table have only one clustered index?
What is a covering index for a query?
A router has routes for 10.0.0.0/8 and 10.1.0.0/16. Which route does it use for a packet to 10.1.5.5?
What is the main benefit of persistent connections (keep-alive) in HTTP/1.1?
A department needs 1282 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
The network 192.168.184.0/24 is split with VLSM for these LANs: A: 23 hosts, B: 27 hosts, C: 5 hosts, D: 126 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 B?
Answer: N = −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.
The other bits keep their normal positive values.
Answer: M = 256 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: ¬¬q → ¬¬r
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: ∀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 = 13
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: 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: 14−1 mod 23 = 5
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·14 (mod 23).
Answer: 1856515 mod 37 = 5
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 · O(n log n)
By the master theorem (a = 2, b = 2, f(n) = n), T(n) = Θ(n log n).
Answer: B · O(n) time and O(1) space
Each step needs only the two previous values, so the table of the bottom-up method can be replaced by two variables.
Answer: C · The heights of the left and right subtrees differ by at most 1
The balance factor (height left − height right) must be −1, 0 or 1; rotations restore it after inserts and deletes.
Answer: A · Primary clustering, where long runs of occupied slots form
Linear probing steps by 1, so occupied runs grow and merge; double hashing uses a second hash for the step size so different keys probe different sequences.
Answer: Euler path, no circuit
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.
The Euler path starts at one odd vertex and ends at the other.
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: C · The rows can be stored physically in only one order
A clustered index orders the table's data by its key, and data can only be kept in one order; other indexes are non-clustered.
Answer: D · An index that contains all columns the query needs, so the table itself need not be read
If all referenced columns are in the index, the DBMS can do an index-only scan and skip fetching table rows, which saves I/O.
Answer: D · 10.1.0.0/16, because it is the longest matching prefix
IP forwarding uses longest prefix match: among all routes that match the destination, the one with the longest mask is chosen.
Answer: D · Several requests reuse one TCP connection, avoiding repeated handshakes
Reusing a connection avoids a new handshake and slow start for each object, so pages with many resources load faster.
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 = 128
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 B: 192.168.184.128/27.
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