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.
Write −22 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
What is the most negative integer N that can be stored in 24-bit two's complement?
You want to prove (p ∨ q) → r by contrapositive. Which implication do you prove instead? Click it.
The premises p ∨ q; ¬p ∨ r are given. Which conclusion follows from them? Click it.
Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C, D) = Σm(1, 2, 3, 5, 7, 12, 13, 14, 15). How many literals L does it contain in total (a primed variable counts as one literal)?
A 4-bit binary up counter starts at 8 (decimal, Q = 1000). Which count Q (in decimal) does it show after 19 clock pulses?
Solve 18x ≡ 15 (mod 23). Give the solution x with 0 < x < 23.
An RSA key uses p = 17, q = 23 and e = 17. Decrypt the ciphertext c = 199: find the message m.
What is the time complexity of the standard dynamic programming algorithm for the longest common subsequence of strings of lengths m and n?
Insertion sort runs in O(n) on sorted input. Does this contradict the Ω(n log n) lower bound for comparison sorting?
What is the cost of inserting an element in the middle of an array of n elements?
Inserting a key into an AVL tree makes a node's left subtree too tall, and the new key went into the left subtree of its left child (left-left case). Which fix is applied?
A connected graph has 16 vertices and 16 edges. How many edges k must be removed to leave a spanning tree?
How many walks of length 3 go from A to B? A walk may repeat vertices and edges. Call the number N.
What does SELECT name, CASE WHEN grade >= 5.5 THEN 'pass' ELSE 'fail' END FROM results return?
Which SQL feature is best suited to finding every employee below a manager in an organisation tree of unknown depth?
NAT (network address translation) is mainly used to:
Why does IPv6 not need NAT for address shortage?
The network 10.165.0.0/22 is divided into equal subnets with prefix /25. How many subnets N are there?
The network 192.168.79.0/24 is split with VLSM for these LANs: A: 51 hosts, B: 3 hosts, C: 10 hosts, D: 27 hosts, E: 15 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 E?
Answer: U = 234
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: N = −8.389 × 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: ¬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 ∨ r
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: L = 7
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: Q = 11
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.
Every 16 pulses bring the counter back to the same state.
Answer: x = 20
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 = 97
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: B · O(mn)
The table has (m + 1)(n + 1) cells and each is computed in O(1) from its neighbours.
Answer: A · No, the lower bound applies to the worst case over all inputs, not to every input
The bound says every comparison sort needs about n log n comparisons on some input of size n; special inputs can be much faster.
Answer: B · O(n), because later elements must shift
All elements after the insertion point move one position to the right, which takes time proportional to n.
Answer: A · A single right rotation at the unbalanced node
In the left-left case one right rotation lifts the left child up and restores balance; zig-zag cases need a double rotation.
Answer: k = 1
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: N = 6
Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.
Answer: D · Each name with the label 'pass' or 'fail' depending on the grade
A CASE expression evaluates its conditions per row and returns the matching value, so every row gets a label.
Answer: B · A recursive common table expression (WITH RECURSIVE)
A recursive CTE starts with an anchor query and repeatedly joins the previous result to the table, so it handles hierarchies of any depth.
Answer: D · Let many devices with private addresses share one public IP address
NAT rewrites addresses (and ports) so private networks can access the internet through a smaller number of public addresses.
Answer: B · Its 128-bit address space is large enough to give every device a global address
IPv6 has about 3.4 × 10³⁸ addresses, so the scarcity that drove NAT in IPv4 disappears; firewalls still provide security.
Answer: N = 8
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 /25 subnet has 27 − 2 = 126 usable hosts.
Answer: x = 96
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 E: 192.168.79.96/27.
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