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 1111 0000 stores a signed integer in two's complement. What is its value N in decimal?
Set the bits to the 8-bit result of 1011 0111 AND 1110 1101. What is the result's value R in decimal?
In how many of the 8 rows of its truth table is (r → q) ∧ ((p → q) → r) true? Call this number N.
The premises b ∨ c; ¬b ∨ a 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.
A synchronous decade (mod-10) counter counts 0, 1, …, 9 and then returns to 0. It starts at 0. Which count Q (in decimal) does it show after 19 clock pulses?
Find the inverse of 20 modulo 31: the number x with 0 < x < 31 and 20·x ≡ 1 (mod 31).
An RSA key uses p = 5, q = 19 and e = 11. Decrypt the ciphertext c = 26: find the message m.
What is the amortised time complexity of appending to a dynamic array that doubles its capacity when full?
Which statement about selecting the k-th smallest element of an unsorted array is correct?
What is true of the adjacency matrix of an undirected graph?
A segment tree is typically used to:
A connected graph has 20 vertices and 31 edges. How many edges k must be removed to leave a spanning tree?
How many walks of length 3 go from A to D? A walk may repeat vertices and edges. Call the number N.
First normal form (1NF) requires that:
Why might a query with WHERE YEAR(order_date) = 2025 not use an index on order_date?
At which OSI layer does a switch normally operate?
In a TCP connection, what does the receive window (rwnd) control?
A department needs 221 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
A router has routes to these networks: 10.158.16.0/24, 10.158.17.0/24, 10.158.18.0/24, 10.158.19.0/24, 10.158.20.0/24, 10.158.21.0/24, 10.158.22.0/24, 10.158.23.0/24. What is the longest prefix length p of a single summary route that covers all of them?
Answer: N = −16
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: R = 165
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.
AND gives 1 only where both bits are 1.
Answer: N = 3
Truth tables: a formula with n different variables has 2ⁿ rows, one for every combination of truth values (1 = true, 0 = false). Work from the inside out and give every subformula its own column. ¬a flips the value; a ∧ b is 1 only when both are 1; a ∨ b is 0 only when both are 0; a → b is 0 only when a = 1 and b = 0; a ↔ b is 1 when both sides are equal.
Answer: c ∨ a
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 = 00, CD = 00
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: Q = 9
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 10 pulses bring the counter back to the same state.
Answer: 20−1 mod 31 = 14
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·20 (mod 31).
Answer: m = 11
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 · O(1)
Copying all elements happens only at powers of two; spread over all appends, the average cost per append is constant.
Answer: D · Quickselect runs in expected O(n), and median-of-medians guarantees O(n) worst case
Quickselect's expected work is n + n/2 + n/4 + ... = O(n) with random pivots; choosing the pivot as a median of medians makes the worst case linear.
Answer: B · It is symmetric about the main diagonal
In an undirected graph edge {u, v} means both M[u][v] and M[v][u] are set, so the matrix equals its transpose.
Answer: B · Answer range queries (such as range sums) and updates in O(log n)
A segment tree stores aggregates for intervals, supporting range queries and point updates in O(log n) time.
Answer: k = 12
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 = 2
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: C · Each column contains atomic values and there are no repeating groups
1NF requires atomic values per cell and a unique row identification, without repeating groups.
Answer: B · Applying a function to the column prevents a simple index range lookup
The expression is not "sargable"; using order_date >= '2025-01-01' AND order_date < '2026-01-01' allows an index range scan.
Answer: A · Data link (layer 2)
A standard switch learns MAC addresses and forwards frames at layer 2. Layer 3 switches also route.
Answer: D · 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: p = 24
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: p = 21
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Summary route: 10.158.16.0/21
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